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1. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies
$$
f(f(x)+f(y))=f(x)+y(\forall x, y \in \mathbf{R}) \text {. }
$$
then $f(2011)=$ $\qquad$ .
|
$-1.2011$.
Let $x=0$ in equation (1), we get
$$
f(f(0)+f(y))=f(0)+y,
$$
which shows that $f$ is surjective.
Let $x$ be the zero root $a$ of $f(x)$ in equation (1). Then $f(f(y))=y$.
Replace $x$ with $f(x)$ in equation (1) to get
$$
f(x+f(y))=x+y \text {. }
$$
Let $y=a$ in equation (2) to get $f(x)=x+a$.
Substitute into equation (1) to get
$$
f(x+a+y+a)=x+a+y \text {, }
$$
which simplifies to $x+y+3 a=x+y+a \Rightarrow a=0$.
Thus, $f(x)=x$.
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}, a_{2}, \cdots, a_{2011}$ be positive real numbers, $S=\sum_{i=1}^{2011} a_{i}$, and
$$
\begin{array}{l}
(S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
=4\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{2011} a_{1}}\right)^{2} .
\end{array}
$$
Then $S=$ . $\qquad$
|
3. 2011.
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& (S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
= & \left(a_{1}+a_{2}+\cdots+a_{2011}+\frac{1+\cdots+1}{2011}\right) . \\
& \left(a_{2}+a_{3}+\cdots+a_{1}+a_{1} a_{2}+\cdots+a_{2011} a_{1}\right) \\
\geqslant & {\left[2\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{2011} a_{1}}\right)\right]^{2} . }
\end{aligned}
$$
Since equality holds, we have
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\cdots=\frac{a_{2011}}{a_{1}}=\frac{1}{a_{1} a_{2}}=\cdots=\frac{1}{a_{2011} a_{1}} \text {. }
$$
Thus, $a_{1}=a_{2}=\cdots=a_{2011}=1$.
Therefore, $S=2011$.
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. From the set $\{1,2, \cdots, 2011\}$, randomly select 1005 different numbers such that their sum is 1021035. Then, there are at least some odd numbers among them.
|
6.5.
From 1 to 2011, taking 1005 even numbers, their sum is $\frac{1005(2+2010)}{2}=1011030$.
Replacing $2,4,6,8,10$ with 2003, 2005, $2007,2009,2011$, increases the sum by
$$
2001 \times 5=10005,
$$
making the total sum 1021035.
Thus, there are at least 5 odd numbers.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Inside a large sphere with a radius of 4, 24 cubes with edge lengths of 1 have been placed arbitrarily. Prove: At least 4 small spheres with a radius of $\frac{1}{2}$ can still be placed inside the large sphere, such that these small spheres and the cubes are all within the large sphere and do not overlap with each other.
|
11. To place a small ball with a radius of $\frac{1}{2}$ completely inside the large ball $D(O, 4)$, the distance from its center to the surface of the large ball should be no less than $\frac{1}{2}$, meaning the center of the small ball should be within the ball $D\left(0,4-\frac{1}{2}\right)$. The volume of this ball is
$$
\frac{4 \pi}{3}\left(4-\frac{1}{2}\right)^{3}=\frac{343 \pi}{6}.
$$
Additionally, the center of the small ball should be no less than $\frac{1}{2}$ away from any surface of the cube. Therefore, the volume controlled by each cube within the large ball is
$$
3 \pi\left(\frac{1}{2}\right)^{2}+\frac{4 \pi}{3}\left(\frac{1}{2}\right)^{3}+4=\frac{11 \pi}{12}+4,
$$
which means the center of the small ball should not be in the $\frac{1}{4}$ cylinder (12 of them) with the vertex of the cube as the central axis and $\frac{1}{2}$ as the radius, the $\frac{1}{8}$ sphere (8 of them) with the vertex as the center and $\frac{1}{2}$ as the radius, the rectangular prism (6 of them) with the face as the base and $\frac{1}{2}$ as the height, and within the cube itself.
Similarly, the volume controlled by each small ball within the large ball is $\frac{4 \pi}{3}$.
If $l$ small balls have already been placed, the sufficient condition for placing the $(l+1)$-th small ball is
$$
\frac{343 \pi}{6}-24\left(\frac{11 \pi}{12}+4\right)-l \cdot \frac{4 \pi}{3} \geqslant 0,
$$
which simplifies to $l \leqslant \frac{3}{4}\left(\frac{343}{6}-22\right)-96 \cdot \frac{3}{4 \pi}$
$$
=\frac{211}{8}-\frac{72}{\pi} \approx 3.457.
$$
Therefore, at least 4 small balls can be placed.
|
4
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Try to find the last two non-zero digits of 2011!.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let the last two non-zero digits of a positive integer $k$ be denoted as $f(k)$. Below, we discuss under modulo 100.
Clearly, $5 \times f(k!)$, and if $5 \times f(k), 5 \times f(l)$, then $f(k l) \equiv f(k) f(l)$. Therefore,
$$
\begin{array}{l}
f(2011!) \equiv f\left(\frac{2011!}{2000!}\right) f(2000!) \\
\equiv 68 f(2000!), \\
f(2000!) \\
\equiv f\left(\prod_{i \in A_{1}} g_{220}(i)\right) f\left(\prod_{i \in \Lambda_{2}} g_{200}(i)\right) f\left(g_{2000}(10)\right),
\end{array}
$$
where, $A_{1}=\{1,2,3,7,8,9\}, A_{2}=\{4,5,6\}$,
$$
g_{k}(i)=\prod_{j=0}^{k-1}(10 j+i) .
$$
It is easy to see that, $g_{k}(i) g_{k}(10-i) \equiv i^{k}(10-i)^{k}(i=$ $1,2,3,4)$, and when $20 \mid k$,
$$
\begin{array}{l}
g_{k}(i) \equiv i^{k} \equiv\left\{\begin{array}{ll}
1, & i=1,3,7,9 ; \\
76, & i=2,4,6,8 .
\end{array}\right. \\
\text { Then } f\left(\prod_{i \in \Lambda_{1}} g_{200}(i)\right) \equiv 76^{2} \equiv 76, \\
f\left(\prod_{i \in \lambda_{2}} g_{200}(i)\right) \\
\equiv\left(\prod_{j=0}^{10} \frac{10 j+4}{2}\right)\left(\prod_{j=0}^{100} \frac{10 j+6}{2}\right) f\left(2^{200} \prod_{j=0}^{109}(10 j+5)\right) \\
\equiv\left(\prod_{i \in A_{1} \backslash 1,91} g_{100}(i)\right) f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right) \\
\equiv 76 f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right), \\
f\left(g_{200}(10)\right)=f(200!) .
\end{array}
$$
$$
\begin{array}{l}
\quad \text { Hence } f(2000!) \\
\equiv 76 f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right) f(200!) \\
\equiv 76 f(400!) \\
\equiv 76 f\left(\prod_{i \in \Lambda_{1}} g_{40}(i)\right) f\left(\prod_{i \in \lambda_{2}} g_{40}(i)\right) f\left(g_{40}(10)\right) . \\
\text { Also } f\left(\prod_{i \in \Lambda_{1}} g_{40}(i)\right) \equiv 76, \\
f\left(\prod_{i \in \Lambda_{2}} g_{40}(i)\right) \\
\equiv\left(\prod_{i \in A_{1} \backslash 1,91} g_{20}(i)\right) f\left(2^{40} \prod_{j=1}^{40}(2 j-1)\right) \\
\equiv 76 f\left(2^{40} \prod_{j=1}^{40}(2 j-1)\right), \\
f\left(g_{40}(10)\right)=f(40!), \\
\text { Hence } f(2000!) \equiv 76 f(80!) \\
\equiv 76 f\left(\prod_{i \in \Lambda_{1}} g_{8}(i)\right) f\left(\prod_{i \in \Lambda_{2}} g_{8}(i)\right) f\left(g_{8}(10)\right) . \\
\text { And } f\left(\prod_{i \in A_{1}} g_{8}(i)\right) \equiv \prod_{i=1}^{3} g_{8}(i) g_{8}(10-i) \\
\equiv \prod_{i=1}^{3} i^{8}(10-i)^{8} \equiv 76,
\end{array}
$$
$$
\begin{array}{l}
f\left(\prod_{i \in A_{2}} g_{8}(i)\right)=\left(\prod_{\left.i \in A_{1} \backslash 1,9\right\}} g_{4}(i)\right) f\left(2^{8} \prod_{j=1}^{8}(2 j-1)\right) \\
=2^{4} \times 7^{4} \times 3^{4} \times 8^{4} f\left(2^{8} \times 15!!\right) \equiv 44, \\
f\left(g_{8}(10)\right)=f(8!)=32,
\end{array}
$$
Therefore, $f(2000!) \equiv 76^{2} \times 44 \times 32 \equiv 8$.
Thus, $f(2011!) \equiv 68 \times 8 \equiv 44$.
|
44
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For any positive integer $n(n \geqslant 2)$, try to find:
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right]
\end{array}
$$
the value.
|
Notice
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\sum_{k=2}^{n} \log _{\frac{3}{2}} \frac{k^{3}+1}{k^{3}-1}=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{k^{3}+1}{k^{3}-1} \\
=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{(k+1)\left(k^{2}-k+1\right)}{(k-1)\left(k^{2}+k+1\right)} \\
=\log _{\frac{3}{2}}\left(\prod_{k=2}^{n} \frac{k+1}{k-1} \cdot \prod_{k=2}^{n} \frac{k^{2}-k+1}{k^{2}+k+1}\right), \\
\prod_{k=2}^{n} \frac{k+1}{k-1}=\frac{n(n+1)}{2}, \\
\prod_{k=2}^{n} \frac{k^{2}-k+1}{k^{2}+k+1}=\prod_{k=2}^{n} \frac{(k-1)^{2}+(k-1)+1}{k^{2}+k+1} \\
=\frac{3}{n^{2}+n+1} .
\end{array}
$$
Also, $1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}=\frac{n^{4}+2 n^{3}+3 n^{2}+2 n+1}{n^{2}(n+1)^{2}}$
$$
=\left[\frac{n^{2}+n+1}{n(n+1)}\right]^{2} \text {, }
$$
then $\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(1+n)^{2}}\right]$
$$
=\log _{\frac{3}{2}} \frac{n^{2}+n+1}{n(n+1)} \text {. }
$$
Substituting equations (2) and (3) into equation (1) gives
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\log _{\frac{3}{2}}\left[\frac{n(n+1)}{2} \cdot \frac{3}{n^{2}+n+1}\right] .
\end{array}
$$
From equations (4) and (5), we get
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\quad \frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right] \\
=\log _{\frac{3}{2}}\left[\frac{n(n+1)}{2} \cdot \frac{3}{n^{2}+n+1}\right]+\log _{\frac{3}{2}} \frac{n^{2}+n+1}{n(n+1)} \\
=\log _{\frac{3}{2}} \frac{3}{2}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In $\triangle A B C$, it is known that $A B=A C, \angle C$'s bisector $C D$ intersects $A B$ at point $D, B D, B C, C D$ are three consecutive integers. Find the perimeter of $\triangle A B C$.
|
Let $A B=A C=b, B C=a$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
B D=\frac{a b}{a+b}, \\
B C-B D=a-\frac{a b}{a+b}=\frac{a^{2}}{a+b} \\
=k \in\{1,2\} .
\end{array}
$$
Following Example 1, we get
$$
C D^{2}=B D(B D+B C)=\frac{a^{2} b(a+2 b)}{(a+b)^{2}} \text {. }
$$
If $k=1$, then $b=a^{2}-a$.
Thus, $C D^{2}=(a-1)(2 a-1)>(a-2)^{2}$.
Therefore, $(a-1)(2 a-1)=(a+1)^{2}$.
Solving this, we get $a=5, b=20, a+2 b=45$.
If $k=2$, similarly, we get $a=3, b=\frac{3}{2}$, which is discarded.
Therefore, the perimeter of $\triangle A B C$ is 45.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Assuming the Earth rotates once around the axis connecting the North Pole and the South Pole in 23 hours 56 minutes 4 seconds, and the Earth's equatorial radius is $6378.1 \mathrm{~km}$. Then, when you stand on the equator and rotate with the Earth, the linear velocity around the axis is $\qquad$ meters/second (rounded to meters, $\pi$ taken as 3.1416$)$.
|
1. 465
|
465
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) The numbers $1,2, \cdots, 666$ are written on a blackboard. In the first step, the first eight numbers: $1,2, \cdots, 8$, are erased, and the sum of these numbers, 36, is written after 666; in the second step, the next eight numbers: $9,10, \cdots, 16$, are erased, and the sum of these numbers, 100, is written at the end; this process continues (i.e., in each step, the first eight numbers are erased, and the sum of these eight numbers is written at the end).
(1) How many steps are required until only one number remains on the blackboard?
(2) When only one number remains on the blackboard, find the sum of all numbers that have appeared on the blackboard (if a number appears multiple times, it should be counted repeatedly).
|
(1) Since each step reduces the numbers by seven, after $\frac{666-1}{7}=95$ steps, only one number remains.
(2) From $666-512=154$, it follows that after $\frac{154}{7}=22$ steps, there are 512 numbers left.
In 22 steps, a total of $22 \times 8=176$ numbers are crossed out, and their sum is
$1+2+\cdots+176=88 \times 177$.
Let $S=1+2+\cdots+666=333 \times 667$.
Then after 22 steps, the sum of the remaining 512 numbers is still $S$.
Assume there were originally $8^{k}$ numbers, with their sum being $x$. Then after $8^{k-1}$ steps, all the original $8^{k}$ numbers are crossed out, and the sum of the remaining $8^{k-1}$ numbers on the blackboard is still $x$.
Therefore, when the blackboard is left with only one number, the sum of all the numbers is $(k+1) x$.
Thus, when only one number remains on the blackboard, the sum of all the numbers that have appeared on the blackboard is
$$
\begin{array}{l}
88 \times 177+4 S=88 \times 177+4 \times 333 \times 667 \\
=904020
\end{array}
$$
|
904020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the inverse function of $y=f(x+1)$ is
$$
\begin{array}{c}
y=f^{-1}(x+1) \text {, and } f(1)=4007 \text {. Then } \\
f(1998)=
\end{array}
$$
|
II, 7.2010.
From $y=f^{-1}(x+1)$, we get $x+1=f(y)$, which means $x=f(y)-1$.
Thus, the inverse function of $y=f^{-1}(x+1)$ is $y=f(x)-1$. Therefore, $f(x+1)-f(x)=-1$.
Let $x=1,2, \cdots, 1997$, add up all the equations and simplify to get $f(1998)-f(1)=-1997$.
Hence, $f(1998)=2010$.
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given the quadratic function
$$
y=a x^{2}+b x+c \geqslant 0(a<b) \text {. }
$$
Then the minimum value of $M=\frac{a+2 b+4 c}{b-a}$ is $\qquad$
|
11. 8.
From the conditions, it is easy to see that $a>0, b^{2}-4 a c \leqslant 0$.
Notice that
$$
\begin{array}{l}
M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\
\geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} .
\end{array}
$$
Let $t=\frac{b}{a}$. Then $t>1$. Thus,
$$
\begin{array}{l}
M \geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)}=\frac{t^{2}+2 t+1}{t-1} \\
=(t-1)+\frac{4}{t-1}+4 \\
\geqslant 2 \sqrt{4}+4=8 .
\end{array}
$$
The equality holds if and only if $t=3, b^{2}=$ $4 a c$, i.e., $b=3 a, c=\frac{9}{4} a$.
Therefore, the minimum value of $M$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. (15 points) As shown in Figure 2, it is known that the ellipse $C$ passes through the point $M(2,1)$, with the two foci at $(-\sqrt{6}, 0)$ and $(\sqrt{6}, 0)$. $O$ is the origin, and a line $l$ parallel to $OM$ intersects the ellipse $C$ at two different points $A$ and $B$.
(1) Find the maximum value of the area of $\triangle OAB$;
(2) Prove that the lines $MA$, $MB$, and the $x$-axis form an isosceles triangle.
|
16. (1) Let the equation of the ellipse $C$ be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
From the problem, we have
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - b ^ { 2 } = 6 , } \\
{ \frac { 4 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2}=8, \\
b^{2}=2 .
\end{array}\right.\right.
$$
Therefore, the equation of the ellipse is
$$
\frac{x^{2}}{8}+\frac{y^{2}}{2}=1 \text {. }
$$
Since the line $l \parallel O M$, we can set
$$
l: y=\frac{1}{2} x+m \text {. }
$$
Substituting the above equation into (1) gives
$$
x^{2}+2 m x+2 m^{2}-4=0 \text {. }
$$
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=-2 m, x_{1} x_{2}=2 m^{2}-4 \text {. }
$$
Since the line $l$ intersects the ellipse $C$ at two distinct points $A$ and $B$, we have
$$
\Delta=(2 m)^{2}-4\left(2 m^{2}-4\right)>0 \text {. }
$$
Thus, $m \in(-2,2)$, and $m \neq 0$.
Therefore, $S_{\triangle O A B}=\frac{1}{2}|m|\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=\frac{1}{2}|m| \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=|m| \sqrt{4-m^{2}}=\sqrt{m^{2}\left(4-m^{2}\right)}
\end{array}
$$
$\leqslant 4$.
The equality holds if and only if $m^{2}=4-m^{2}$, i.e., $m= \pm \sqrt{2}$.
Therefore, the maximum area of $\triangle O A B$ is 4.
(2) Let the slopes of the lines $M A$ and $M B$ be $k_{1}$ and $k_{2}$, respectively. Then
$$
k_{1}=\frac{y_{1}-1}{x_{1}-2}, k_{2}=\frac{y_{2}-1}{x_{2}-2} .
$$
We only need to prove: $k_{1}+k_{2}=0$.
In fact,
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{\frac{1}{2} x_{1}+m-1}{x_{1}-2}+\frac{\frac{1}{2} x_{2}+m-1}{x_{2}-2} \\
=1+m\left(\frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}\right) \\
=1+m \cdot \frac{\left(x_{1}+x_{2}\right)-4}{x_{1} x_{2}-2\left(x_{1}+x_{2}\right)+4} \\
=1+m \cdot \frac{-2 m-4}{2 m^{2}-4-2(-2 m)+4}=0 .
\end{array}
$$
Therefore, the lines $M A$ and $M B$ form an isosceles triangle with the $x$-axis.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. (15 points) Given the function
$$
f(x)=\frac{1}{2} m x^{2}-2 x+1+\ln (x+1)(m \geqslant 1) \text {. }
$$
(1) If the curve $C: y=f(x)$ has a tangent line $l$ at point $P(0,1)$ that intersects $C$ at only one point, find the value of $m$;
(2) Prove that the function $f(x)$ has a decreasing interval $[a, b]$, and find the range of the length $t=b-a$ of the decreasing interval.
|
17. (1) Note that the domain of the function $f(x)$ is $(-1,+\infty)$,
$$
f^{\prime}(x)=m x-2+\frac{1}{x+1}, f^{\prime}(0)=-1 .
$$
Therefore, the slope of the tangent line $l$ at the point of tangency $P(0,1)$ is -1.
Thus, the equation of the tangent line is $y=-x+1$.
Since the tangent line $l$ intersects the curve $C$ at only one point, the equation
$$
\frac{1}{2} m x^{2}-x+\ln (x+1)=0
$$
has exactly one real solution.
Clearly, $x=0$ is a solution to the equation.
Let $g(x)=\frac{1}{2} m x^{2}-x+\ln (x+1)$. Then
$$
g^{\prime}(x)=m x-1+\frac{1}{x+1}=\frac{m x\left[x-\left(\frac{1}{m}-1\right)\right]}{x+1} \text {. }
$$
When $m=1$, $g^{\prime}(x)=\frac{x^{2}}{x+1} \geqslant 0$ (with equality only when $x=0$), so $g(x)$ is monotonically increasing on $(-1,+\infty)$, meaning $x=0$ is the only real solution to the equation.
When $m>1$, from
$$
g^{\prime}(x)=\frac{m x\left[x-\left(\frac{1}{m}-1\right)\right]}{x+1}=0 \text {, }
$$
we get $x_{1}=0, x_{2}=\frac{1}{m}-1 \in(-1,0)$.
On the interval $\left(-1, x_{2}\right)$, $g^{\prime}(x)>0$, and on the interval $\left(x_{2}, 0\right)$, $g^{\prime}(x)g(0)=0$.
As $x \rightarrow-1$, $g(x) \rightarrow-\infty$, so $g(x)=0$ has another solution in $\left(-1, x_{2}\right)$, which is a contradiction.
In summary, we have $m=1$.
(2) Note that
$$
\begin{array}{l}
f^{\prime}(x)=\frac{m x^{2}+(m-2) x-1}{x+1}(x>-1) . \\
\text { Hence } f^{\prime}(x)0$, and the axis of symmetry is
$$
\begin{array}{l}
x=-\frac{1}{2}+\frac{1}{m}>-1, \\
h(-1)=m-(m-2)-1=1>0,
\end{array}
$$
Therefore, the equation $h(x)=0$ has two distinct real roots $x_{1} 、 x_{2}$ in $(-1,+\infty)$, meaning the solution set of equation (1) is $\left(x_{1} 、 x_{2}\right)$.
Thus, the interval of monotonic decrease for the function $f(x)$ is $\left[x_{1}, x_{2}\right]$.
Then $t=x_{2}-x_{1}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$
$$
=\sqrt{\frac{\Delta}{m^{2}}}=\sqrt{1+\frac{4}{m^{2}}} \text {. }
$$
Since $m \geqslant 1$, we have $1<\sqrt{1+\frac{4}{m^{2}}} \leqslant \sqrt{5}$.
Therefore, the range of the length $t$ of the interval of monotonic decrease for the function $y=f(x)$ is $(1, \sqrt{5}]$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $a, b (a \neq b)$, and they satisfy
$$
\begin{array}{l}
(a+1)^{2}=3-3(a+1), \\
3(b+1)=3-(b+1)^{2} .
\end{array}
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13
|
【Analysis】Transform the known two equations into
$$
\begin{array}{l}
(a+1)^{2}+3(a+1)-3=0, \\
(b+1)^{2}+3(b+1)-3=0 .
\end{array}
$$
It can be seen that $a$ and $b$ are the two roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$
Solution: From equation (1) in the analysis, simplifying and rearranging yields
$$
x^{2}+5 x+1=0 \text {. }
$$
Since $\Delta=25-4>0, a+b=-5, a b=1$, therefore, both $a$ and $b$ are negative.
Then $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=-\frac{a^{2}+b^{2}}{\sqrt{a b}}$
$$
=-\frac{(a+b)^{2}-2 a b}{\sqrt{a b}}=-23 \text {. }
$$
Therefore, the answer is B.
|
-23
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let real numbers $s, t$ satisfy
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, \\
t^{2}+99 t+19=0(s t \neq 1) . \\
\text { Find the value of } \frac{s t+4 s+1}{t} \text { . }
\end{array}
$$
(1999, National Junior High School Mathematics Competition)
|
【Analysis】Transform the first equation of the known equations, and combine it with the second equation to find that $\frac{1}{s} 、 t(s t \neq 1)$ are the two roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Since $s \neq 0$, the first equation can be transformed into
$$
\left(\frac{1}{s}\right)^{2}+99\left(\frac{1}{s}\right)+19=0 \text {. }
$$
Also, $t^{2}+99 t+19=0$, and $s t \neq 1$, so $\frac{1}{s} 、 t$ are the two distinct real roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Thus, $\frac{1}{s}+t=-99, \frac{1}{s} \cdot t=19$, which means
$$
s t+1=-99 s, t=19 s . \\
\text { Therefore, } \frac{s t+4 s+1}{t}=\frac{-99 s+4 s}{19 s}=-5 .
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, for a $7 \times 7$ grid of small squares, how many of the small squares can be shaded such that no two shaded squares are adjacent? Adjacent means they share a side.
How many of the small squares can be shaded?
|
Four, color at most 26 small squares.
The coloring in Figure 2 satisfies the conditions.
The following proves: At most 26 small squares can be colored.
First, according to the problem, for any $2 \times 2$ square grid, at most two of the small squares can be colored; for a $3 \times 3$ square grid, at most 5 of the small squares can be colored to meet the conditions.
Second, for a $5 \times 5$ square grid, it is based on a $3 \times 3$ square grid with a border of width 2 added (as shown in Figure 3). This border can be divided into 4 $2 \times 2$ square grids, with the bottom-right two squares, one being reused and the other not used. Therefore, at most 9 additional squares can be colored, and when adding 9 colored squares, the bottom-right square must be colored, while the reused square is not colored. Thus, a $5 \times 5$ square grid can have at most 14 squares colored, and when 14 squares are colored, the bottom-right square is colored.
Finally, a $7 \times 7$ square grid is based on a $5 \times 5$ square grid with a border of width 2 added (as shown in Figure 4).
Similarly, at most 13 additional squares can be colored, and when adding 13 colored squares, the bottom-right square must be colored, while the reused square is not colored. Thus, at most 27 squares can be colored. At this point, assume the square in the last row and the second-to-last column is not colored (otherwise, the square in the second-to-last row and the last column is not colored). According to the previous analysis, the two squares in the last two rows and the third-to-last column are colored, which is a contradiction.
Therefore, at most 26 squares can be colored.
In summary, at most 26 small squares can be colored.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value?
|
Prompt: Example 6. From the problem, we know that $x$ and $y$ are the two real roots of the equation
$$
t^{2}-(z-1) t+z^{2}-7 z+14=0
$$
By the discriminant $\Delta \geqslant 0$, we get
$$
3 z^{2}-26 z+55 \leqslant 0 \text {. }
$$
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\text { Also, } x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(z-1)^{2}-2\left(z^{2}-7 z+14\right) \\
=(z-6)^{2}+9,
\end{array}
$$
Therefore, when $z=5$, the maximum value of $x^{2}+y^{2}$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
the integer part.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let
then $2<\sqrt[3]{23}=a_{1}<3, a_{n+1}=\sqrt[3]{23+a_{n}}$.
By mathematical induction, it is easy to prove
$$
2<a_{n}<3\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
In fact,
the integer part is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given real numbers $x, y$ satisfy
$$
x^{2}+3 y^{2}-12 y+12=0 \text {. }
$$
then the value of $y^{x}$ is $\qquad$
|
Hint: Treat the known equation as a quadratic equation in $y$ (the main variable)
$$
3 y^{2}-12 y+\left(12+x^{2}\right)=0 \text {. }
$$
From $\Delta=-12 x^{2} \geqslant 0$, and since $x^{2} \geqslant 0$, then $x=0$.
Thus, $y=2$. Therefore, $y^{x}=2^{0}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given real numbers $a, b, c$ satisfy
$$
a=2 b+\sqrt{2}, a b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 \text {. }
$$
Then $\frac{b c}{a}=$ $\qquad$ .
|
Hint: Eliminate $a$ from the two known equations, then treat $c$ as a constant, to obtain a quadratic equation in $b$
$$
2 b^{2}+\sqrt{2} b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 .
$$
From the discriminant $\Delta \geqslant 0$, we get $c=0$.
Therefore, $\frac{b c}{a}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \\
\frac{x}{5^{3}+4^{3}}+\frac{y}{5^{3}+6^{3}}=1,
\end{array}
$$
then $x+y=$ $\qquad$
(2005, National Junior High School Mathematics Competition)
|
【Analysis】It is easy to notice that the denominators of the two given equations contain $3^{3}$ and $5^{3}$ respectively. Therefore, we can consider $3^{3}$ and $5^{3}$ as the roots of a certain equation.
Solution From the given conditions, it is easy to see that $3^{3}$ and $5^{3}$ are the roots of the equation in terms of $t$:
$$
\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1 \text {, }
$$
which is equivalent to the quadratic equation
$$
t^{2}+\left(4^{3}+6^{3}-x-y\right) t+\left(4^{3} \times 6^{3}-4^{3} y-6^{3} x\right)=0.
$$
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
3^{3}+5^{3}=-\left(4^{3}+6^{3}-x-y\right) . \\
\text { Therefore, } x+y=3^{3}+4^{3}+5^{3}+6^{3}=432 .
\end{array}
$$
|
432
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given
$\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$.
Then $\frac{b+c}{a}=$ $\qquad$
|
From the given equation, we have
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$
When $a \neq b$, by the discriminant of a quadratic equation, we know that the quadratic equation in $x$
$$
(a-b) x^{2}+(b-c) x+(c-a)=0
$$
has two equal real roots.
$$
\text { Also, }(a-b)+(b-c)+(c-a)=0 \text {, thus, }
$$
$x=1$ is a root of equation (1).
Therefore, both roots of equation (1) are $x=1$.
So, $1 \times 1=\frac{c-a}{a-b} \Rightarrow \frac{b+c}{a}=2$.
When $a=b$, we have
$$
(b-c)^{2}=0 \Rightarrow b=c \text {. }
$$
In this case, $\frac{b+c}{a}=\frac{b+b}{b}=2$.
In summary, $\frac{b+c}{a}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason.
|
If 2010 can be written as the sum of squares of $k$ prime numbers, taking the sum of the squares of the smallest 10 distinct primes, then
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010 .
\end{array}
$$
Therefore, $k \leqslant 9$.
It is easy to see that there is only one even prime number 2, and the rest are odd. The square of an odd number is congruent to 1 modulo 8.
Since 2010 is congruent to 2 modulo 8, but the sum of the squares of nine different primes is congruent to 1 or 4 modulo 8, and the sum of the squares of eight different primes is congruent to 0 or 3 modulo 8, hence $k \leqslant 7$.
When $k=7$, after trial calculation, we get
$$
2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}=2010 \text {. }
$$
In summary, 2010 can be written as the sum of squares of $k$ distinct prime numbers, and the maximum value of $k$ is 7.
In fact, it can also be proven that $k \neq 1,2, \cdots, 6$.
Therefore, 2010 can only be expressed as the sum of squares of seven distinct prime numbers.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If the function $f(x)=\ln \frac{\mathrm{e} x}{\mathrm{e}-x}$, then $\sum_{k=1}^{2010} f\left(\frac{k e}{2011}\right)=$ $\qquad$ .
|
13.2010.
Notice
$$
\begin{array}{l}
f(x)+f(\mathrm{e}-x) \\
=\ln \left[\frac{\mathrm{e} x}{\mathrm{e}-x} \cdot \frac{\mathrm{e}(\mathrm{e}-x)}{\mathrm{e}-(\mathrm{e}-x)}\right]=2 . \\
\text { Therefore } \sum_{k=1}^{2010} f\left(\frac{k \mathrm{e}}{2011}\right) \\
=\sum_{k=1}^{1005}\left(f\left(\frac{k \mathrm{e}}{2011}\right)+f\left(\mathrm{e}-\frac{k \mathrm{e}}{2011}\right)\right) \\
=2 \times 1005=2010 .
\end{array}
$$
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. (15 points) In a certain grade, $n$ students participate in Chinese and
Mathematics exams, with scores ranging from 0 to 100 for each subject. Suppose no two students have exactly the same scores (i.e., at least one subject score is different). Additionally, “A is better than B” means that student A’s scores in both Chinese and Mathematics are higher than student B’s scores in both subjects. Question: What is the smallest value of $n$ such that there must exist three students (let’s call them A, B, and C) where A is better than B, and B is better than C.
|
19. Establish a Cartesian coordinate system $x O y$.
If a student's Chinese score is $i$ points, and mathematics score is $j$ points, let it correspond to the integer point $(i, j)$ on the plane, called a "score point". Thus, the examination results of $n$ students are mapped to $n$ score points within the range
$$
0 \leqslant x \leqslant 100,0 \leqslant y \leqslant 100
$$
on the plane.
Consider 201 lines on the plane:
$$
y=x \pm b(b=0,1, \cdots, 100) \text {. }
$$
If a line has three score points, it indicates that there are three students, A, B, and C, such that A is better than B, and B is better than C.
Clearly, the lines $y=x+100$ and $y=x-100$ can each have at most one score point; the lines $y=x+99$ and $y=x-99$ can each have at most two score points.
Since $2 \times(201-2)+1 \times 2=400$, when $n>400$, there must be a line with three score points.
Thus, the minimum value of $n$, $n_{0} \leqslant 401$.
Let the set
$S=\{(i, j) \mid i=0,1 ; j=0,1, \cdots, 100\}$;
$T=\{(i, j) \mid i=0,1, \cdots, 100 ; j=0,1\}$.
Clearly, $|S \cup T|=400$, and there do not exist three score points on the same line in $S \cup T$.
Therefore, $n_{0} \geqslant 401$.
Thus, $n_{0}=401$.
|
401
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given points $M(0,2)$ and $N(-3,6)$, the distances from these points to line $l$ are $1$ and $4$, respectively. The number of lines $l$ that satisfy these conditions is . $\qquad$
|
2.3.
It is easy to get $M N=5$.
Then the circle $\odot M$ with radius 1 is externally tangent to the circle $\odot N$ with radius 4. Therefore, there are 3 lines $l$ that satisfy the condition.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n+1}=\frac{1+a_{n}}{1-a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Let $T_{n}=a_{1} a_{2} \cdots a_{n}$. Then $T_{2010}=$ $\qquad$
|
5. -6 .
It is easy to get $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}$, $a_{1} a_{2} a_{3} a_{4}=1$.
Also, $a_{5}=2=a_{1}$, by induction it is easy to know $a_{n+4}=a_{n}\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $T_{2010}=T_{4 \times 502+2}=a_{1} a_{2}=-6$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$
|
Given $a=\frac{-1+\sqrt{5}}{2}$, we can obtain the corresponding quadratic equation $a^{2}+a-1=0$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{\left(a^{2}+a-1\right)\left(a^{3}-a\right)-2(a-1)}{a\left(a^{2}-1\right)} \\
=\frac{-2}{a^{2}+a}=-2 .
\end{array}
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given that $a$ and $b$ are integers, and satisfy $a-b$ is a prime number, $ab$ is a perfect square. If $a \geqslant 2011$, find the minimum value of $a$.
|
Three, let $a-b=m$ (where $m$ is a prime number) and $ab=n^2$ (where $n$ is a positive integer).
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} \times 1 .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive integers, and $2a-m+2n > 2a-m-2n$ (since $m$ is a prime number), we have:
$$
2a-m+2n=m^{2}, 2a-m-2n=1 .
$$
Solving these, we get $a=\frac{(m+1)^{2}}{4}, n=\frac{m^{2}-1}{4}$.
Thus, $b=a-m=\frac{(m-1)^{2}}{4}$.
Given $a \geqslant 2011$, i.e., $\frac{(m+1)^{2}}{4} \geqslant 2011$.
Considering $m$ is a prime number, we get $m \geqslant 89$.
In this case, $a \geqslant \frac{(89+1)^{2}}{4}=2025$.
When $a=2025$,
$$
m=89, b=1936, n=1980 \text {. }
$$
Therefore, the minimum value of $a$ is 2025.
|
2025
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the sequence $\left\{x_{n}\right\}$ :
$$
1,3,3,3,5,5,5,5,5, \cdots
$$
formed by all positive odd numbers arranged from smallest to largest, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c+d=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
4.3.
For $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, $x_{n}=2 k+1, k=[\sqrt{n-1}]$,
therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Thus, $(a, b, c, d)=(2,1,-1,1)$.
So $a+b+c+d=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. $\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is
$\qquad$ .
|
6.1.
Original expression $=\sum_{i=0}^{50} C_{50}^{i} \cdot \sum_{j=0}^{50} C_{50}^{j}=\left(2^{50}\right)^{2}=2^{100}$.
And $2^{5} \equiv 1(\bmod 31)$, so
Original expression $=2^{100} \equiv 1(\bmod 31)$.
Therefore, the remainder is 1.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. There are 20 teams participating in the national league. Question: What is the minimum number of matches that must be played so that in any group of three teams, at least two teams have played against each other?
|
Let team $A$ have the minimum number of matches (which is $k$ matches). Then
(1) There are $k$ teams that have played against team $A$, and each of these teams has played at least $k$ matches;
(2) There are $19-k$ teams that have not played against team $A$, and these teams must all have played against each other, otherwise, the two teams that have not played against each other and team $A$ would not meet the requirements. Therefore, the minimum number of matches is
$$
\begin{array}{l}
N=\frac{1}{2}\left[k+k^{2}+(19-k)(18-k)\right] \\
=(k-9)^{2}+90 \geqslant 90 .
\end{array}
$$
Next, we prove that 90 matches can meet the requirements of the problem.
Divide the 20 teams into two groups, each with 10 teams, and each group conducts a single round-robin tournament.
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the integers $1,2, \cdots, 2011$, the number of integers that can be expressed in the form $[x[x]]$ is $\qquad$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
|
2. 1.990.
Let $x=k+a, k=[x], 0 \leqslant a<1$. Then $[x[x]]=[k(k+a)]=k^{2}+[k a]$.
Therefore, $k^{2}, k^{2}+1, \cdots, k^{2}+k-1(k=1,2$, $\cdots, 44)$ can all be expressed in the form of $[x[x]]$. Hence, the number of integers that meet the requirement is
$$
1+2+\cdots+44=990 \text { (numbers). }
$$
|
990
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a certain kingdom, there are 32 knights, some of whom are servants to other knights. Each servant can have at most one master, and each master must be richer than any of his servants. If a knight has at least four servants, he is ennobled as a noble. If it is stipulated that a servant of $A$'s servant is not a servant of $A$, then the maximum possible number of nobles is $\qquad$
|
2.7.
According to the problem, the richest knight is not a servant of any other knight, so at most 31 knights can be servants of other knights.
Since each noble has at least four servants, there can be at most 7 nobles.
Number the 32 knights as $1,2, \cdots, 32$, with wealth decreasing as the number increases, then
$$
\begin{array}{l}
1(2,3,4,5) ; \quad 5(6,7,8,9) ; \\
9(10,11,12,13) ; 13(14,15,16,17) ; \\
17(18,19,20,21) ; 21(22,23,24,25) ; \\
25(26,27,28,29),
\end{array}
$$
Each knight outside the parentheses is a noble, and the knights inside the parentheses are servants.
This shows that 7 nobles are possible.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x_{1}=x_{2011}=1$, $\left|x_{n+1}\right|=\left|x_{n}+1\right|(n=1,2, \cdots, 2010)$.
Then $x_{1}+x_{2}+\cdots+x_{2010}=$ $\qquad$
|
3. -1005 .
From the given, it is easy to obtain
$$
x_{n+1}^{2}=x_{n}^{2}+2 x_{n}+1(n=1,2, \cdots, 2010) \text {. }
$$
By summing up and organizing these 2010 equations, we get
$$
\begin{array}{l}
x_{2011}^{2}=2\left(x_{1}+x_{2}+\cdots+x_{2010}\right)+2011 . \\
\text { Therefore, } x_{1}+x_{2}+\cdots+x_{2010}=-1005 .
\end{array}
$$
|
-1005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Among the positive integers not greater than 100, the ordered integer pairs $(m, n)$ that satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n}
$$
are $\qquad$ pairs.
|
4. 170 .
Notice that $\sqrt{2} n-1<m<\sqrt{2}(n+1)$.
For each $n$, the number of $m$ is given by
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1
\end{array}
$$
Since $100<71 \sqrt{2}<101<72 \sqrt{2}$, we have $n \leqslant 71$.
But when $n=71$, $m=100$.
Therefore, the number of $m$ is
$$
\begin{array}{l}
\sum_{n=1}^{70}([\sqrt{2}(n+1)]-[\sqrt{2} n]+1)+1 \\
=[71 \sqrt{2}]-[\sqrt{2}]+71 \\
=100-1+71=170 .
\end{array}
$$
|
170
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) When $1<x<5$, find the minimum value of the algebraic expression
$$
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}}
$$
|
$$
\begin{array}{l}
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}} \\
=\sqrt{\frac{\left(7 x^{2}-2 x-5\right)\left(-7 x^{2}+34 x+5\right)}{x^{4}}} \\
=\sqrt{\left(7 x-\frac{5}{x}-2\right)\left(-7 x+\frac{5}{x}+34\right)} \text {. } \\
\end{array}
$$
Let $t=7 x-\frac{5}{x}$. Then
$$
\begin{array}{l}
\text { Equation (1) }=\sqrt{(t-2)(34-t)} \\
=\sqrt{-t^{2}+36 t-68} \\
=\sqrt{-(t-18)^{2}+256} \leqslant 16 .
\end{array}
$$
When $t=7 x-\frac{5}{x}=18$, i.e., $x=\frac{9+2 \sqrt{29}}{7}$,
the expression reaches its maximum value of 16.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Given 15 quadratic equations $x^{2}-p_{i} x+q_{i}=0(i=1,2, \cdots, 15)$ with coefficients $p_{i} 、 q_{i}$ taking values from $1,2, \cdots, 30$, and these coefficients are all distinct. If an equation has a root greater than 20, it is called a "good equation." Find the maximum number of good equations.
|
Second, if there exists an equation $x^{2}-p_{i} x+q_{i}=0$ with two roots $x_{1}, x_{2}$, then from $x_{1}+x_{2}=p_{i}>0, x_{1} x_{2}=q_{i}>0$, we know that both roots are positive.
Next, if a certain equation has a root greater than 20, then $p_{i}=x_{1}+x_{2}>20$. However, among the numbers $1,2, \cdots, 30$, there are only 10 numbers greater than 20, so the number of good equations is at most 10.
Finally, consider
$$
x^{2}-(20+k) x+k=0(k=1,2, \cdots, 10) \text {, }
$$
these ten equations. Their larger root
$$
\begin{array}{l}
x=\frac{20+k+\sqrt{(20+k)^{2}-4 k}}{2} \\
=\frac{20+k+\sqrt{k^{2}+36 k+400}}{2} \\
>\frac{20+k+k+18}{2}=k+19 \geqslant 20 .
\end{array}
$$
Therefore, the maximum number of good equations is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 2, given points $A$ and $B$ are two distinct points outside circle $\odot O$, point $P$ is on $\odot O$, and $PA$, $PB$ intersect $\odot O$ at points $D$ and $C$ respectively, different from point $P$, and $AD \cdot AP = BC \cdot BP$.
(1) Prove: $\triangle OAB$ is an isosceles triangle;
(2) Let $p$ be a prime number, and $m$ be a positive integer. If $AD \cdot AP = p(2p + 1)$, $OA = m - 1$, and the radius of $\odot O$ is 3, find the length of $OA$.
|
(1) Draw $A T$ tangent to $\odot O$ at point $T$, and connect $O T$.
Let the radius of $\odot O$ be $R$. Then
$$
A D \cdot A P=A T^{2}=O A^{2}-R^{2} \text {. }
$$
Similarly, $B C \cdot B P=O B^{2}-R^{2}$.
From the given, it is easy to see that $O A=O B$.
(2) From the proof in (1), we know
$$
p(2 p+1)=(m-1)^{2}-9 \text {, }
$$
which means $p(2 p+1)=(m-4)(m+2)>0$.
Since $p$ is a prime and $(p, 2 p+1)=1$, we have
$$
p \mid(m-4) \text { or } p \mid(m+2)
$$
exactly one of them holds.
If $p \mid(m-4)$, then $m-4=k p\left(k \in \mathbf{N}_{+}\right)$.
If $k=1$, then
$$
2 p+1=p+6 \Rightarrow p=5, m=9, O A=8 \text {. }
$$
If $k \geqslant 2$, then $m+2=k p+6>2 p+1$, a contradiction.
If $p \mid(m+2)$, then $m+2=k p\left(k \in \mathbf{N}_{+}\right)$.
If $k \leqslant 2$, then $m-4=k p-6<2 p+1$, a contradiction.
Thus, $k \geqslant 3$. In this case,
$$
\begin{array}{l}
2 p+1=k(k p-6) \geqslant 3(3 p-6) \\
\Rightarrow 7 p \leqslant 19 \Rightarrow p=2 .
\end{array}
$$
However, when $p=2$, $(m-4)(m+2)=10$ has no positive integer solutions, a contradiction.
In summary, $O A=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the inequality
$$
3 x+4 \sqrt{x y} \leqslant a(x+y)
$$
holds for all positive numbers $x$ and $y$. Then the minimum value of the real number $a$ is . $\qquad$
|
-1.4 .
From the problem, we know that $a \geqslant\left(\frac{3 x+4 \sqrt{x y}}{x+y}\right)_{\text {max }}$.
$$
\text { Also, } \frac{3 x+4 \sqrt{x y}}{x+y} \leqslant \frac{3 x+(x+4 y)}{x+y}=4 \text {, }
$$
with equality holding if and only if $x=4 y>0$.
Therefore, the minimum value of $a$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the equation in $x$
$$
x^{2}+(a-2010) x+a=0 \quad (a \neq 0)
$$
has two integer roots. Then the value of the real number $a$ is $\qquad$.
|
3.4024.
Let the roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=-(a-2010), x_{1} x_{2}=a \text {. }
$$
Then $x_{1} x_{2}+x_{1}+x_{2}=2010$, which means
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=2011 \text {. }
$$
Since 2011 is a prime number, we have
$$
\left\{\begin{array} { l }
{ x _ { 1 } = 0 , } \\
{ x _ { 2 } = 2 0 1 0 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=-2012, \\
x_{2}=-2 .
\end{array}\right.\right.
$$
Thus, $a=0$ (discard) or $a=4024$.
|
4024
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $[x]$ is the greatest integer not exceeding the real number $x$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{3}{2}, a_{n+1}=a_{n}^{2}-a_{n}+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then $m=\left[\sum_{k=1}^{2011} \frac{1}{a_{k}}\right]$ is . $\qquad$
|
4.1.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}^{2}-a_{n}+1 \\
\Rightarrow a_{n+1}-1=a_{n}\left(a_{n}-1\right) \\
\Rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\text { Then } \frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1} . \\
\text { Therefore } m=\left[\frac{1}{a_{1}-1}-\frac{1}{a_{2012}-1}\right]=\left[2-\frac{1}{a_{2012}-1}\right]=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) In a certain drill activity, the drill leader arranged $n$ students, numbered $1 \sim n (n>3)$, in a circular formation, and they performed a $1-2-3$ cyclic counting. The drill leader recorded the numbers of the students who reported, and required students who reported 1 and 2 to leave the formation, while students who reported 3 remained in the formation, and their numbers were changed to the sum of the numbers of the three students in this cycle of counting. The counting continued in this manner. When the number of students remaining on the playground was no more than two, the counting activity ended. The drill leader recorded the numbers of the students who remained on the playground (for example, if nine students numbered $1 \sim 9$ were arranged in a circular formation, after the counting ended, only the student originally numbered 9 remained on the playground, at which point his number was 45, and the data recorded by the drill leader were $1,2,3,4,5,6,7,8,9,6,15,24,45$). It is known that 2011 students participated in the drill.
(1) What was the initial number of the student who remained on the playground?
(2) Find the sum of the numbers recorded by the drill leader.
|
Three, the numbers recorded by the leader are $a_{1}$, $a_{2}, \cdots, a_{n}, a_{n+1}, \cdots, a_{m}$, with their sum being $S_{n}$. Thus,
(1) After each 1-2-3 counting cycle, the total number of students decreases by 2, but the sum of the numbers remains unchanged.
(2) If $n=3^{k}$, after $3^{k-1}$ 1-2-3 counting cycles, $3^{k-1}$ new numbers are generated, and the sum of these numbers is $\frac{(1+n) n}{2}$; after another $3^{k-2}$ counting cycles, $3^{k-2}$ new numbers are generated, and the sum of these numbers is also $\frac{(1+n) n}{2}; \cdots \cdots$ After a total of
$$
3^{k-1}+3^{k-2}+\cdots+3^{1}+1=\frac{3^{k}-1}{2}
$$
counting cycles, only one student remains, and his number is $\frac{(1+n) n}{2}$, and his original number is $n$ (i.e., when $n=3^{k}$, the last student remains on the field).
From (1), we know $S_{n}=(k+1) \frac{(1+n) n}{2}$.
(3) If the number of students is $3 k$, and the number of the last student is $a_{p}$, then after $k$ counting cycles, $k$ new numbers are generated, and the number of students becomes $k$, with the first student's number being $a_{p+1}$.
For a general odd number $n(n>3)$, let $k$ and $r$ satisfy $n=3^{k}+2 r\left(0 \leqslant r<3^{k}\right)$.
Then $3^{k}<n<3^{k+1}, 3 r<n$. One group, the sum of their numbers is
$$
M=1+2+\cdots+3 r=\frac{3 r(3 r+1)}{2} .
$$
Let the initial sum of student numbers be $N=\frac{n(n+1)}{2}$.
From (1) and (3), we know that after $r$ 1-2-3 counting cycles, the number of students becomes $3^{k}$, with the first student's number being $a_{3 r+1}$, called the second group; after another $3^{k-1}$ counting cycles, the number of students becomes $3^{k-1}$, called the third group; repeat the counting cycles until the number of students is 1, which is the $k$ +2 group.
The original number of the last student remaining on the field is $3 r$, and
$$
\begin{array}{l}
S_{n}=M+(k+1) N \\
=\frac{3 r(3 r+1)}{2}+(k+1) \frac{(1+n) n}{2} .
\end{array}
$$
When $n=2011$, since $n=2011=3^{6}+2 \times 641$, we have $k=6, r=641$.
The original number of the last student remaining on the field is 1923. Therefore, $S_{2011}=16011388$.
|
16011388
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given $a, b, c > 0, a^{2}+b^{2}+c^{2}=14$. Prove: $a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27} \geqslant 14$.
|
Prove the construction of the $3 \times 5$ matrix
$$
\left(\begin{array}{ccccc}
a^{5} & a^{5} & 1 & 1 & 1 \\
\frac{b^{5}}{8} & \frac{b^{5}}{8} & 4 & 4 & 4 \\
\frac{c^{5}}{27} & \frac{c^{5}}{27} & 9 & 9 & 9
\end{array}\right) .
$$
Using Carleman's inequality, we get
$$
\begin{array}{l}
{\left[\left(a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27}\right)^{2}(1+4+9)^{3}\right]^{\frac{1}{5}}} \\
\geqslant\left(a^{5} \cdot a^{5} \cdot 1^{3}\right)^{\frac{1}{5}}+\left(\frac{b^{5}}{8} \cdot \frac{b^{5}}{8} \cdot 4^{3}\right)^{\frac{1}{5}}+\left(\frac{c^{5}}{27} \cdot \frac{c^{5}}{27} \cdot 9^{3}\right)^{\frac{1}{5}} .
\end{array}
$$
Since $a^{2}+b^{2}+c^{2}=14$, we have
$$
a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27} \geqslant\left[\frac{\left(a^{2}+b^{2}+c^{2}\right)^{5}}{14^{3}}\right]^{\frac{1}{2}}=14 .
$$
|
14
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the product of the first 2011 positive integers
$$
1 \times 2 \times \cdots \times 2011
$$
can be divided by $2010^{k}$, then the maximum value of the positive integer $k$ is
|
4.30.
$$
2010=2 \times 3 \times 5 \times 67 \text {. }
$$
In $1 \times 2 \times \cdots \times 2011$, the exponent of 67 is
$$
\left[\frac{2011}{67}\right]+\left[\frac{2011}{67^{2}}\right]+\cdots=30 \text {. }
$$
Obviously, the exponents of $2,3,5$ in $1 \times 2 \times \cdots \times 2011$ are all greater than 30.
Therefore, the maximum value of $k$ is 30.
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given integers $p$ and $q$ satisfy $p+q=2010$, and the quadratic equation $67 x^{2}+p x+q=0$ has two positive integer roots. Then $p=$ $\qquad$ .
|
7. -2278 .
Let the two positive integer roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$.
Then $x_{1}+x_{2}=-\frac{p}{67}, x_{1} x_{2}=\frac{q}{67}$.
Thus, $x_{1} x_{2}-x_{1}-x_{2}=\frac{p+q}{67}=\frac{2010}{67}=30$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=31 \\
\Rightarrow\left\{\begin{array}{l}
x_{1}-1=1, \\
x_{2}-1=31 .
\end{array}\right.
\end{array}
$$
Therefore, $x_{1}+x_{2}=34, p=-34 \times 67=-2278$.
|
-2278
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Place the nine digits $1,2, \cdots, 9$ into the nine small squares in Figure 4, so that the seven three-digit numbers $\overline{a b c} 、 \overline{d e f} 、 \overline{g h i} 、 \overline{a d g} 、 \overline{b e h} 、 \overline{c f i}$ and $\overline{a e i}$ are all divisible by 11. Find the maximum value of the three-digit number $\overline{c e g}$.
|
12. According to the problem, for modulo 11 we have
$$
\begin{array}{l}
a+c \equiv b, d+f \equiv e, g+i \equiv h, a+g \equiv d, \\
b+h \equiv e, c+i \equiv f, a+i \equiv e . \\
\text { Then }(a+c)+(d+f)+(g+i)+(b+h)+e \\
\equiv b+h+3 e \equiv 4 e(\bmod 11) .
\end{array}
$$
The left side of the above equation is
$$
1+2+\cdots+9=45 \equiv 1(\bmod 11) \text {. }
$$
So $4 e \equiv 1(\bmod 11), e=3$.
Thus, $d+f \equiv b+h \equiv a+i \equiv 3(\bmod 11)$.
Therefore, $\{d, f\} 、\{b, h\} 、\{a, i\}$ are permutations of $\{1,2\} 、\{6,8\}$ 、 $\{5,9\}$.
Hence, $\{c, g\}$ is $\{4,7\}$.
To maximize $\overline{c e g}$, we can take
$$
c=7, g=4 \text {. }
$$
Figure 8 is one valid arrangement that satisfies the problem.
Therefore, the maximum value of $\overline{c e g}$ is 734.
|
734
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A natural number is called a "good number" if it is exactly 2007 more than the sum of its digits. Then the sum of all good numbers is $\qquad$ .
|
5.20145.
Let $f(n)=n-S(n)$ ($S(n)$ is the sum of the digits of the natural number $n$). Then the function $f(n)$ is a non-strictly increasing function, and
$$
\begin{array}{l}
f(2009)<f(2010) \\
=f(2011)=\cdots=f(2019)=2007 \\
<f(2020) .
\end{array}
$$
Therefore, there are only 10 natural numbers that satisfy the condition, and their sum is
$$
2010+2011+\cdots+2019=20145 .
$$
|
20145
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sum of 12 distinct positive integers is 2010. Then the maximum value of the greatest common divisor of these positive integers is . $\qquad$
|
$-1.15$.
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d$, where $\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then $2010=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$S \geqslant 1+2+\cdots+12=78$.
Also, $S \mid 2010$, and $2010=2 \times 3 \times 5 \times 67$, so the smallest positive divisor of 2010 greater than 77 is $2 \times 67=134$.
Therefore, $d \leqslant 15$, and $d=15$ can be achieved, by setting
$$
\left(a_{1}, a_{2}, \cdots, a_{11}, a_{12}\right)=(1,2, \cdots, 11,68)
$$
This completes the translation.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n}+a_{n+1}=-n^{2} \text {. }
$$
then $a_{15}=$ $\qquad$
|
7. -104 .
Rewrite $a_{n}+a_{n+1}=-n^{2}$ as
$$
\left(a_{n}+\frac{n^{2}}{2}-\frac{n}{2}\right)+\left[a_{n+1}+\frac{(n+1)^{2}}{2}-\frac{n+1}{2}\right]=0 \text {. }
$$
Let $b_{n}=a_{n}+\frac{n^{2}}{2}-\frac{n}{2}$. Then
$$
b_{1}=1 \text {, and } b_{n+1}=-b_{n} \text {. }
$$
Therefore, $b_{2 k-1}=1, b_{2 k}=-1(k=1,2, \cdots)$.
Thus, $b_{15}=1$, which means
$$
a_{15}=1+\frac{15}{2}-\frac{15^{2}}{2}=-104 .
$$
|
-104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If a four-digit number $n$ contains at most two different digits among its four digits, then $n$ is called a "simple four-digit number" (such as 5555 and 3313). Then, the number of simple four-digit numbers is
|
8. 576.
If the four digits of a four-digit number are all the same, then there are nine such four-digit numbers.
If the four digits of a four-digit number have two different values, the first digit \( a \in \{1,2, \cdots, 9\} \) has 9 possible choices. After choosing \( a \), select \( b \in \{0,1, \cdots, 9\} \) and \( b \neq a \), which gives 9 ways to choose \( b \); for the remaining three positions, each can be filled with \( a \) or \( b \), but they cannot all be filled with \( a \), giving \( 2^3 - 1 = 7 \) ways to fill them. Therefore, the number of four-digit numbers with exactly two different digits is
$$
9 \times 9 \times 7 = 567 \text{ (numbers). }
$$
Thus, the total number of simple four-digit numbers is \( 9 + 567 = 576 \).
|
576
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and
$$
f(x)=f(1-x) \text {. }
$$
Then $f(2010)=$ $\qquad$ .
|
$$
\begin{array}{l}
f(0)=0, \\
f(x+1)=f(-x)=-f(x) .
\end{array}
$$
From the conditions, we have $f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2010)=f(0)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ .
|
10.4.
The figure determined by $3|x+1|+2|y-1| \leqslant 6$ is the quadrilateral $ABCD$ and its interior, where,
$$
A(-1,4) 、 B(1,1) 、 C(-1,-2) 、 D(-3,1) \text {. }
$$
By the knowledge of linear programming, the maximum value of $2x-3y$ is 4.
The maximum value can be achieved when $x=-1, y=-2$.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. The capacity of a set refers to the sum of its elements. Then the total capacity of all non-empty sets $A$ that satisfy the condition “ $A \subseteq\{1,2, \cdots, 7\}$, and if $a \in A$ then $8-a \in A$ ” is
(Answer with a specific number).
|
12.224.
First, find the single-element and two-element sets that satisfy the conditions:
$$
A_{1}=\{4\}, A_{2}=\{1,7\}, A_{3}=\{2,6\}, A_{4}=\{3,5\} .
$$
Then, any combination of elements from these four sets also meets the requirements.
Therefore, the total sum of elements in all sets $A$ that satisfy the conditions is
$$
(4+8+8+8) \times 2^{3}=224 \text {. }
$$
|
224
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3: There is an $8 \times 8$ chessboard, and at the start, each of the 64 small squares contains a "castle" chess piece. If a castle chess piece can attack an odd number of other castle chess pieces still on the board, it is removed. Question: What is the maximum number of castle chess pieces that can be removed (a castle chess piece can attack other pieces in the same row or column without any obstruction)?
|
First, prove that the pieces on the four corners of the chessboard will not be taken away. Since a castle on a corner can attack at most two other castles, if this castle is taken away, it means that the castle can only attack one. Without loss of generality, we can assume that the castle in the upper left corner is the first to be taken away among these four, which means that either the first column or the first row has been completely cleared. In other words, the castle in the upper right corner or the lower left corner has been taken away, which is impossible.
When there are five castles left, the castles outside these four corners can only attack two or zero castles, so at least five castles will remain. Taking the castles in the order of the numbers in Figure 5 is one way to end up with five castles left. Therefore, the maximum number of castles that can be taken away is $64-5=59$.
|
59
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the sequence $\left\{8 \times\left(-\frac{1}{3}\right)^{n-1}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the smallest integer $n$ that satisfies the inequality
$$
\left|S_{n}-6\right|<\frac{1}{125}
$$
is
|
2. 7
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For $0<x<1$, if the complex number
$$
z=\sqrt{x}+\mathrm{i} \sqrt{\sin x}
$$
corresponds to a point, then the number of such points inside the unit circle is $n=$
|
6. 1 .
From the point on the unit circle, we have
$$
x+\sin x=1(00(x \in(0,1))$, which means $\varphi(x)$ is a strictly increasing function.
Also, $\varphi(0)=-10$, so the equation $x+\sin x=1$ has exactly one real root in $(0,1)$.
Therefore, $n=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010}$.
Then $m=$ $\qquad$ $(\bmod 2011)$.
|
8. 0 .
Notice
$$
\begin{array}{l}
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010} \\
=\left(\frac{1}{1}+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+\cdots+ \\
\left(\frac{1}{1005}+\frac{1}{1006}\right) \\
= \frac{2011}{1 \times 2010}+\frac{2011}{2 \times 2009}+\cdots+\frac{2011}{1005 \times 1006} \\
= 2011 \times \frac{q}{2010!},
\end{array}
$$
where $q$ is a positive integer.
Thus, $2011 n q=m \times 2010!$.
This shows that 2011 divides $m \times 2010!$, but 2011 is a prime number and does not divide 2010!, so 2011 must divide $m$, giving $m \equiv 0(\bmod 2011)$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Take the subset $A_{i}=\left\{a_{i}, a_{i+1}, \cdots, a_{i+59}\right\}(i=1,2, \cdots$, 70 ) of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}$, where $a_{70+i}=a_{i}$. If there exist $k$ sets among $A_{1}, A_{2}, \cdots, A_{70}$ such that the intersection of any seven of them is non-empty, find the maximum value of $k$.
---
The problem is asking to find the maximum number of sets $k$ from the collection $\{A_1, A_2, \ldots, A_{70}\}$ such that the intersection of any seven of these sets is non-empty. Each set $A_i$ is defined as a subset of $S$ containing 60 consecutive elements, with the indices wrapping around modulo 70.
|
Given:
$$
A_{1} \cap A_{2} \cap \cdots \cap A_{60}=\left\{a_{60}\right\} \text {, }
$$
Furthermore, the intersection of any seven sets from $A_{1}, A_{2}, \cdots, A_{\infty}$ is non-empty, hence $k \geqslant 60$.
We will now prove that if $k>60$, it cannot be guaranteed that the intersection of any seven sets from the $k$ selected sets is non-empty. Therefore, the maximum value of $k$ is 60.
Divide the known sets $A_{1}, A_{2}, \cdots, A_{70}$ into 10 groups (constructing pigeonholes):
$$
B_{i}=\left\{A_{i}, A_{10+i}, \cdots, A_{60+i}\right\}(i=1,2, \cdots, 10) .
$$
Since the union of the complements $\overline{A_{j}}$ of any seven sets $A_{j}$ within each $B_{i}$ satisfies
$$
\begin{array}{l}
\overline{A_{i}} \cup \overline{A_{10+i}} \cup \cdots \cup \overline{A_{60+i}} \\
=\left\{a_{i+60}, a_{i+61}, \cdots, a_{i+69}\right\} \cup \\
\left\{a_{i+70}, a_{i+11}, \cdots, a_{i+79}\right\} \cup \cdots \cup \\
\left\{a_{i+120}, a_{i+121}, \cdots, a_{i+129}\right\} \\
=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}=S\left(a_{70+i}=a_{i}\right), \\
\end{array}
$$
it follows that the intersection of any seven sets within each $B_{i}$ is the empty set:
$$
A_{i} \cap A_{10+i} \cap \cdots \cap A_{60+i}=\varnothing \text {. }
$$
For $k>60$, we have $k=6 \times 10+r(1 \leqslant r \leqslant 9)$.
By the pigeonhole principle, any selection of $k$ sets from $A_{1}, A_{2}, \cdots, A_{70}$ must include seven sets from the same $B_{i}(1 \leqslant i \leqslant 10)$.
This contradicts the condition that the intersection of any seven sets is non-empty.
Therefore, the maximum value of $k$ is 60.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Consider the matrix
$$
\left(a_{i j}\right)_{n \times n}\left(a_{i j} \in\{1,2,3\}\right) \text {. }
$$
If $a_{i j}$ is such that its row $i$ and column $j$ both contain at least three elements (including $a_{i j}$) that are equal to $a_{i j}$, then the element $a_{i j}$ is called "good". If the matrix $\left(a_{i j}\right)_{n \times n}$ contains at least one good element, find the minimum value of $n$.
|
The minimum value of $n$ is 7.
When $n=6$, the matrix
$\left(\begin{array}{llllll}1 & 1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 3 & 2 & 2 \\ 2 & 2 & 1 & 1 & 3 & 3 \\ 2 & 2 & 3 & 3 & 1 & 1 \\ 3 & 3 & 2 & 2 & 1 & 1 \\ 3 & 3 & 1 & 1 & 2 & 2\end{array}\right)$
has no good elements, so $n \geqslant 7$.
Below, we use proof by contradiction to show that when $n=7$, the matrix $\left(a_{i j}\right)$ must have at least one good element.
Assume that no element in the matrix $\left(a_{i j}\right)$ is good.
Define: If a number $m$ appears at least three times in a row (or column), then we say that the number $m$ controls that row (or column).
It is easy to see that the number of rows (or columns) controlled by the number $m$ is at most 4.
First, we prove a lemma:
Lemma: If the number 1 controls the first three rows, then the 1s in these three rows must appear in at least five columns.
Proof: Since the number 1 appears at least 9 times in the first three rows, by the pigeonhole principle, it must appear in at least five columns. Now, we show that five columns are impossible.
If the number 1 appears in the first five columns, consider the submatrix $T$ formed by the first five columns and the last four rows of the original matrix. Then, $T$ contains at most one 1.
If there is a row in $T$ where 2 appears at least three times and another row where 3 appears at least three times, then there must be a column where 1, 2, and 3 each appear three times. Therefore, in this column, 1, 2, and 3 each appear at most twice, which is a contradiction.
Assume that there are three rows in $T$ where 2 appears at least three times, then in the first five columns, the numbers 1 and 2 each appear at most twice in each column. Thus, in the first five columns, the number 3 appears at least three times in each column, which is a contradiction.
Therefore, this scenario is impossible.
We prove in two steps:
(1) If the number $m$ controls at least 3 rows (or columns), then the number of elements in the matrix $\left(a_{i j}\right)$ that are equal to $m$ is at most 16.
We discuss two cases (assume $m=1$ without loss of generality).
(i) The number 1 controls four rows.
Assume these are the first four rows, then the number 1 appears at least 12 times in the first four rows. By the pigeonhole principle, it must appear in at least six columns.
(1) If the number 1 appears in seven columns, then the number of 1s is at most 14.
(2) If the number 1 appears in six columns, then the number of 1s is at most $6 \times 2 + 3 = 15$.
(ii) The number 1 controls three rows.
Assume these are the first three rows. By the lemma, the number 1 appears in at least six columns.
(1) If the number 1 appears in seven columns, then the number of 1s is at most 14.
(2) If the number 1 appears in six columns, then the number of 1s is at most $6 \times 2 + 4 = 16$.
(2) For $m=1,2,3$, the number of rows or columns controlled by $m$ is at least 3.
By the pigeonhole principle, assume 1 controls at least 3 rows.
We show that the number of rows or columns controlled by 2 and 3 is at least 3.
We discuss two cases:
(i) The number 1 controls four rows.
Assume these are the first four rows, then the number 1 appears at least 12 times in the first four rows. By the pigeonhole principle, it must appear in at least six columns.
(1) If the number 1 appears in seven columns, then 2 and 3 each control three columns and four columns, respectively.
(2) If the number 1 appears in the first six columns, consider the submatrix $T$ formed by the first six columns and the last three rows of the original matrix. $T$ contains only 2 and 3.
If the first row of $T$ contains 3 twos and 3 threes, then 2 and 3 each control three columns; if the first row of $T$ contains 4 twos, then 3 controls four columns, and 2 controls three rows (since in the last two rows of $T$, it is impossible for one row to contain more than 3 threes).
(ii) The number 1 controls three rows.
Assume these are the first three rows. By the lemma, the number 1 appears in at least six columns.
(1) If the number 1 appears in seven columns, then 2 and 3 each control three columns and four columns, respectively.
(2) If the number 1 appears in the first six columns, consider the submatrix $T$ formed by the first six columns and the last four rows of the original matrix. $T$ contains at most 3 ones, so there must be a row in $T$ that does not contain 1.
If this row contains 3 twos and 3 threes, then 2 and 3 each control three columns; if this row contains 4 twos, then 3 controls four columns.
Since 1 appears at most 3 times, 2 must control at least three rows in the last four rows.
By (1) and (2), in the matrix $\left(a_{i j}\right)$, the number of rows or columns controlled by the numbers $m=1,2,3$ is at least 3.
Thus, the numbers 1, 2, and 3 each appear at most 16 times in the matrix $\left(a_{i j}\right)$.
But when $n=7$, the matrix $\left(a_{i j}\right)$ has 49 elements, which is a contradiction.
Therefore, when $n=7$, the matrix $\left(a_{i j}\right)$ must have at least one good element.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
410 translators are invited to an international mathematics conference. Each translator is proficient in exactly two of the five languages: Greek, Slovenian, Vietnamese, Spanish, and German, and no two translators are proficient in the same pair of languages. The translators are to be assigned to five rooms, with two translators in each room, and these two translators must be proficient in the same language. How many different allocation schemes are there (considering all possible allocation schemes of 5 pairs of translators arranged in five rooms as the same scheme)?
|
Construct a graph $G$, where points $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ represent five languages, and any two points (such as $x_{1}, x_{2}$) are connected by an edge $(x_{1} x_{2})$ representing a translator proficient in these two languages.
By the problem statement, graph $G$ is a simple complete graph.
Next, orient all edges of graph $G$ according to the following rule: if the common language of the room where the translator $x_{1} x_{2}$ is located is $x_{1}$, then the edge $x_{1} x_{2}$ is oriented as $x_{2} \rightarrow x_{1}$, which can also be denoted as $\overrightarrow{x_{2} x_{1}}$.
By the problem statement, graph $G$ is a simple directed graph, and the in-degrees of all points in graph $G$ are even (0, 2, or 4).
(1) If there are two points $x_{1}, x_{2}$ with in-degree 4, then the edge $x_{1} x_{2}$ is both $x_{1} \rightarrow x_{2}$ and $x_{2} \rightarrow x_{1}$, which is a contradiction.
(2) If the in-degree of each point is 2, then the out-degree is also 2.
Take any point $x_{1}$, and draw two edges $\overrightarrow{x_{1} x_{2}}$ and $\overrightarrow{x_{4} x_{5}}$.
As shown in Figure 2, place $x_{i} (i=1,2, \cdots, 5)$ in a circle.
$x_{3} \rightarrow x_{4}, x_{3} \rightarrow x_{5};$
$x_{2} \rightarrow x_{4}, x_{3} \rightarrow x_{4};$
From $\overrightarrow{x_{3} x_{5}}$ and $\overrightarrow{x_{4} x_{5}}$, we have
$x_{5} \rightarrow x_{1}, x_{5} \rightarrow x_{2}$.
Thus, we obtain graph $G^{\prime}$, which corresponds one-to-one with a 5-cycle arrangement.
Therefore, there are $4! = 24$ allocation schemes.
(3) If there is a point $x_{1}$ with in-degree 4, then there must be another point $x_{2}$ with out-degree 4, and the remaining three points have in-degree 2.
Clearly, there are edges
$\overrightarrow{x_{i} x_{1}} (i=2,3,4,5), \overrightarrow{x_{2} x_{j}} (j=3,4,5)$.
Thus, $x_{3}, x_{4}, x_{5}$ form a unidirectional cycle, which corresponds one-to-one with a 3-cycle arrangement, and $\overrightarrow{x_{i} x_{1}}$ can be divided into two groups in three ways.
Therefore, there are $5 \times 4 \times 2! \times 3 = 120$ allocation schemes.
In summary, there are 144 allocation schemes.
|
144
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A. Let $a=\sqrt{7}-1$. Then the value of the algebraic expression $3 a^{3}+12 a^{2}-$ $6 a-12$ is ( ).
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$
|
,- 1. A. A.
Notice
$$
a=\sqrt{7}-1 \Rightarrow a^{2}+2 a-6=0 \text {. }
$$
Then $3 a^{3}+12 a^{2}-6 a-12$
$$
=\left(a^{2}+2 a-6\right)(3 a+6)+24=24 .
$$
|
24
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. B. Given that $\angle A, \angle B$ are two acute angles, and satisfy
$$
\begin{array}{l}
\sin ^{2} A+\cos ^{2} B=\frac{5}{4} t, \\
\cos ^{2} A+\sin ^{2} B=\frac{3}{4} t^{2} .
\end{array}
$$
Then the sum of all possible real values of $t$ is ().
(A) $-\frac{8}{3}$
(B) $-\frac{5}{3}$
(C) 1
(D) $\frac{11}{3}$
|
3. B. C.
Adding the two equations yields $3 t^{2}+5 t=8$.
Solving gives $t=1, t=-\frac{8}{3}$ (discard).
When $t=1$, $\angle A=45^{\circ}, \angle B=30^{\circ}$ satisfies the given equations.
Therefore, the sum of all possible values of the real number $t$ is 1.
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. A. Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{99^{3}}$. Then the integer part of $4 S$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
|
5. A. A.
When $k=2,3, \cdots, 99$, we have
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {, }
$$
Thus, $1<S=1+\sum_{k=2}^{29} \frac{1}{k^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99 \times 100}\right)<\frac{5}{4} \text {. }
$$
Therefore, $4<4 S<5$.
Hence, the integer part of $4 S$ is 4.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. B. Given that the lengths of the two legs are integers $a$ and $b$ $(b<2011)$. Then the number of right triangles with the hypotenuse length $b+1$ is
|
6. B. 31.
By the Pythagorean theorem, we have
$$
a^{2}=(b+1)^{2}-b^{2}=2 b+1 \text{. }
$$
Given $b<2011$, we know that $a$ is an odd number in the interval $(1, \sqrt{4023})$, so $a$ must be $3, 5, \cdots, 63$.
Therefore, there are 31 right-angled triangles that satisfy the conditions.
|
31
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A. As shown in Figure 1, points $A$ and $B$ lie on the line $y=x$. Parallel lines to the $y$-axis through $A$ and $B$ intersect the hyperbola $y=\frac{1}{x} (x>0)$ at points $C$ and $D$. If $BD=2AC$, then the value of $4OC^2-OD^2$ is $\qquad$
|
8. A. 6 .
Let points $C(a, b), D(c, d)$.
Then points $A(a, a), B(c, c)$.
Since points $C, D$ are on the hyperbola $y=\frac{1}{x}$, we have $a b=1, c d=1$.
Given $B D=2 A C$
$$
\begin{array}{l}
\Rightarrow|c-d|=2|a-b| \\
\Rightarrow c^{2}-2 c d+d^{2}=4\left(a^{2}-2 a b+b^{2}\right) \\
\Rightarrow 4\left(a^{2}+b^{2}\right)-\left(c^{2}+d^{2}\right)=8 a b-2 c d=6 \\
\Rightarrow 4 O C^{2}-O D^{2}=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A. As shown in Figure 3, in the right triangle $\triangle ABC$, the hypotenuse $AB$ is 35 units long, and the square $CDEF$ is inscribed in $\triangle ABC$ with a side length of 12. Then the perimeter of $\triangle ABC$ is $\qquad$
|
10. A. 84.
Let $BC = a, AC = b$. Then,
$$
a^{2} + b^{2} = 35^{2} = 1225.
$$
Since Rt $\triangle AFE \sim \text{Rt} \triangle ACB$, we have,
$$
\frac{FE}{CB} = \frac{AF}{AC} \Rightarrow \frac{12}{a} = \frac{b-12}{b}.
$$
Thus, $12(a + b) = ab$.
From equations (1) and (2), we get
$$
\begin{array}{l}
(a + b)^{2} = a^{2} + b^{2} + 2ab \\
= 1225 + 24(a + b).
\end{array}
$$
Solving this, we get $a + b = -25$ (discard), 49.
Therefore, $a + b + c = 49 + 35 = 84$.
|
84
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. B. Let the four-digit number
$\overline{a b c d}$ satisfy
$$
a^{3}+b^{3}+c^{3}+d^{3}+1=10 c+d .
$$
Then the number of such four-digit numbers is $\qquad$
|
10. B. 5 .
From $d^{3} \geqslant d$ we know $c^{3}+1 \leqslant 10 c \Rightarrow 1 \leqslant c \leqslant 3$.
If $c=3$, then $a^{3}+b^{3}+d^{3}=2+d$.
Thus, $d=1$ or 0. Therefore, $a=b=1$, which gives us $1131, 1130$ as solutions.
If $c=2$, then $a^{3}+b^{3}+d^{3}=11+d$. Thus, $d \leqslant 2$.
When $d=2$, $a^{3}+b^{3}=5$, no solution;
When $d=1$ or 0, $a^{3}+b^{3}=11$, no solution.
If $c=1$, then $a^{3}+b^{3}+d^{3}=8+d$. Thus, $d \leqslant 2$.
When $d=2$, $a=b=1$, which gives us $1112$ as a solution;
When $d=1$ or 0, $a=2, b=0$, which gives us $2011, 2010$ as solutions.
In summary, there are 5 four-digit numbers that satisfy the conditions.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. B. If five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are randomly selected from $1,2, \cdots, n$, and one of these integers is always a prime number, find the maximum value of $n$.
|
13. B. When $n \geqslant 49$, take the integers $1, 2^{2}, 3^{2}, 5^{2}, 7^{2}$. These five integers are five pairwise coprime distinct integers, but none of them are prime.
When $n=48$, in the integers $1, 2, \cdots, 48$, take any five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are all not prime, then at least four of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are composite (let these be $a_{1}, a_{2}, a_{3}, a_{4}$).
Let the smallest prime factors of $a_{1}, a_{2}, a_{3}, a_{4}$ be $p_{1}, p_{2}, p_{3}, p_{4}$, respectively.
Since $a_{1}, a_{2}, a_{3}, a_{4}$ are pairwise coprime, $p_{1}, p_{2}, p_{3}, p_{4}$ are pairwise distinct.
Let $p$ be the largest of $p_{1}, p_{2}, p_{3}, p_{4}$. Then $p \geqslant 7$.
Since $a_{1}, a_{2}, a_{3}, a_{4}$ are composite, there must exist an $a_{i} \geqslant p^{2} \geqslant 7^{2}=49$, which is a contradiction.
Therefore, at least one of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ must be prime.
In conclusion, the maximum value of the positive integer $n$ is 48.
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $f(x)$ be a polynomial with integer coefficients, $f(0)=11$, and 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 .
$$
Then the maximum value of $n$ is
|
7.3.
Let $g(x)=f(x)-2010$. Then $x_{1}, x_{2}, \cdots, x_{n}$ are all roots of $g(x)=0$.
Thus, $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) \cdot q(x)$,
where $q(x)$ is a polynomial with integer coefficients. Therefore,
$$
\begin{array}{l}
g(0)=11-2010=-1999 \\
=\prod_{i=1}^{n}\left(-x_{i}\right) q(0) .
\end{array}
$$
Since 1999 is a prime number, -1999 can be at most the product of three different integers, i.e., $-1 \times 1 \times 1999$, hence $n \leqslant 3$.
When $n=3$,
$$
\begin{array}{l}
g(x)=(x+1)(x-1)(x+1999), \\
f(x)=(x+1)(x-1)(x+1999)+2010 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (14 points) As shown in Figure 2, the corridor is 3 m wide, the angle between the corridors is 120°, the ground is level, and the ends of the corridor are sufficiently long. Question: What is the maximum length of a horizontal rod (neglecting its thickness) that can pass through the corridor?
|
II. 9. As shown in Figure 4, draw any horizontal line through the inner vertex $P$ of the corridor corner, intersecting the outer sides of the corridor at points $A$ and $B$. The length of a wooden rod that can pass through the corridor in a horizontal position is less than or equal to $AB$.
Let $\angle B A Q=\alpha$. Then
$$
\begin{array}{l}
\angle A B Q=60^{\circ}-\alpha, \\
A B=A P+P B=\frac{3}{\sin \alpha}+\frac{3}{\sin \left(60^{\circ}-\alpha\right)} .
\end{array}
$$
As $\alpha$ varies, the minimum value of the above expression is the maximum length of the wooden rod that can pass through the corridor in a horizontal position.
By the arithmetic mean inequality and product-to-sum formulas, we get
$$
\begin{array}{l}
A B \geqslant 6 \sqrt{\frac{1}{\sin \alpha \cdot \sin \left(60^{\circ}-\alpha\right)}} \\
=6 \sqrt{\frac{1}{\frac{1}{2}\left[\cos \left(2 \alpha-60^{\circ}\right)-\cos 60^{\circ}\right]}} \\
\geqslant 6 \sqrt{\frac{2}{1-\frac{1}{2}}}=12,
\end{array}
$$
and when $\alpha=30^{\circ}$, $A B=12$.
Therefore, the maximum length of a wooden rod that can pass through the corridor in a horizontal position is $12 \mathrm{~m}$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) Given a positive integer $n$ that satisfies the following condition: for each positive integer $m$ in the open interval $(0,2009)$, there always exists a positive integer $k$, such that
$$
\frac{m}{2009}<\frac{k}{n}<\frac{m+1}{2010} \text {. }
$$
Find the minimum value of such $n$.
|
12. Notice
$$
\begin{array}{l}
\frac{m}{2009}2010 k
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
m n+1 \leqslant 2009 k, \\
m n+n-1 \geqslant 2010 k
\end{array}\right. \\
\Rightarrow 2009(m n+n-1) \geqslant 2009 \times 2010 k \\
\geqslant 2010(m n+1) \\
\Rightarrow 2009 m n+2009 n-2009 \\
\geqslant 2010 m n+2010 \\
\Rightarrow \geqslant \frac{4019}{2009-m} .
\end{array}
$$
Since the above inequality holds for every positive integer $m$ in the interval $(0,2009)$, we have
$$
n \geqslant \frac{4019}{2009-2008}=4019 \text {. }
$$
On the other hand, according to the inequality “$a, b, c, d \in \mathbf{R}_{+}$, $\frac{a}{b}<\frac{c}{d} \Rightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$”, we know
$$
\frac{m}{2009}<\frac{m+(m+1)}{2009+2010}<\frac{m+1}{2010},
$$
which means $\frac{m}{2009}<\frac{2 m+1}{4019}<\frac{m+1}{2010}$
holds for every positive integer $m$ in the interval $(0,2009)$.
Therefore, the minimum value of $n$ is 4019.
|
4019
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $k_{1}<k_{2}<\cdots<k_{n}$ be non-negative integers, satisfying $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=227$.
Then $k_{1}+k_{2}+\cdots+k_{n}=$ $\qquad$
|
- 1. 19.
Notice that
$$
\begin{array}{l}
227=1+2+32+64+128 \\
=2^{0}+2^{1}+2^{5}+2^{6}+2^{7} .
\end{array}
$$
Therefore, $k_{1}+k_{2}+\cdots+k_{n}=0+1+5+6+7=19$.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a>0$, the graphs of the functions $f(x)=|x+2a|$ and $g(x)=|x-a|$ intersect at point $C$, and they intersect the $y$-axis at points $A$ and $B$ respectively. If the area of $\triangle ABC$ is 1, then $a=$ $\qquad$ .
|
2. 2 .
From the graphs of $f(x)$ and $g(x)$, we know that $\triangle ABC$ is an isosceles right triangle with base $a$, so its area is $\frac{a^{2}}{4}=1$. Therefore, $a=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the function $y=x^{3}$, the tangent line at $x=a_{k}$ intersects the $x$-axis at point $a_{k+1}$. If $a_{1}=1, S_{n}=\sum_{i=1}^{n} a_{i}$, then $\lim _{n \rightarrow \infty} S_{n}$ $=$ . $\qquad$
|
4.3.
It is known that $y^{\prime}=3 x^{2}$.
Therefore, the equation of the tangent line to $y=x^{3}$ at $x=a_{k}$ is $y-a_{k}^{3}=3 a_{k}^{2}\left(x-a_{k}\right)$.
Thus, the above equation intersects the $x$-axis at the point $\left(a_{k+1}, 0\right)$. Hence, $-a_{k}^{3}=3 a_{k}^{2}\left(a_{k+1}-a_{k}\right)$.
From this, we get $a_{k+1}=\frac{2}{3} a_{k}$.
Given $a_{1}=1$, thus $\lim _{n \rightarrow \infty} S_{n}=\frac{1}{1-\frac{2}{3}}=3$.
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all $x, y, z \in \mathbf{R}$
$$
f(x+y)+f(y+z)+f(z+x) \geqslant 3 f(x+2 y+z) .
$$
Then $f(1)-f(0)=$ $\qquad$
|
5.0.
Let $x=-y=z$, we get
$$
f(2 x) \geqslant f(0) \Rightarrow f(1) \geqslant f(0) \text {. }
$$
Let $x=y=-z$, we get
$$
f(0) \geqslant f(2 x) \Rightarrow f(0) \geqslant f(1) \text {. }
$$
Thus, $f(1)-f(0)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (18 points) Let $S$ be a set of distinct quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, where $a_{i}=0$ or 1 $(i=1,2,3,4)$. It is known that the number of elements in $S$ does not exceed 15, and satisfies:
if $\left(a_{1}, a_{2}, a_{3}, a_{4}\right) 、\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$, then $\left(c_{1}, c_{2}, c_{3}, c_{4}\right) 、\left(d_{1}, d_{2}, d_{3}, d_{4}\right) \in S$,
where, $c_{i}=\max \left\{a_{i}, b_{i}\right\}, d_{i}=\min \left\{a_{i}, b_{i}\right\}(i=1,2,3,4)$. Find the maximum number of elements in the set $S$.
|
12. Clearly, there are 16 possible quadruples. Since at least one quadruple is not in $S$, it follows that $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$ must have at least one that is not in $S$. Otherwise, by the given conditions, all quadruples would be in $S$.
Assume $(1,0,0,0) \notin S$.
In this case, by the given conditions,
$(1,1,0,0)$, $(1,0,1,0)$, $(1,0,0,1)$
at least two of these cannot be in $S$ (assume $(1,1,0,0)$ and $(1,0,1,0)$ are not in $S$). Then $(1,1,1,0)$ and $(1,1,0,1)$ cannot both be in $S$ (assume $(1,1,1,0)$ is not in $S$). Therefore, the number of elements in $S$ is at most $16-4=12$.
Let $S$ be the set of all 16 possible quadruples after removing the above 4 quadruples.
Next, we prove by contradiction that $S$ satisfies the conditions of the problem.
Take any $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, $\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$.
(1) If $\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \notin S$, then $c_{1}=1$, $c_{4}=0$.
Thus, $1 \in \{a_{1}, b_{1}\}$, $a_{4}=b_{4}=0$.
Assume $a_{1}=1$, then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is among the 4 quadruples removed, which is a contradiction.
(2) If $\left(d_{1}, d_{2}, d_{3}, d_{4}\right) \notin S$, then $d_{1}=1$, $d_{4}=0$.
Thus, $a_{1}=b_{1}=1$, $0 \in \{a_{4}, b_{4}\}$.
Assume $a_{4}=0$, then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is among the 4 quadruples removed, which is a contradiction.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let the sequence of rational numbers $\left\{a_{n}\right\}$ be defined as follows:
$a_{k}=\frac{x_{k}}{y_{k}}$, where $x_{1}=y_{1}=1$, and
if $y_{k}=1$, then $x_{k+1}=1, y_{k+1}=x_{k}+1$;
if $y_{k} \neq 1$, then $x_{k+1}=x_{k}+1, y_{k+1}=y_{k}-1$.
How many terms in the first 2011 terms of this sequence are positive integers?
|
11. The sequence is
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots \text {. }
$$
Group it as follows:
$$
\begin{array}{l}
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right),\left(\frac{1}{3}, \frac{2}{2}, \frac{3}{1}\right), \cdots, \\
\left(\frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}\right), \cdots
\end{array}
$$
Let any number in the $k$-th group be $\frac{b}{a}(a, b \in \mathbf{N}_{+})$. Then $a+b=k+1$.
$$
\begin{array}{l}
\text { Hence } \frac{b}{a} \in \mathbf{N}_{+} \Leftrightarrow b=a t\left(t \in \mathbf{N}_{+}\right) \\
\Leftrightarrow k+1=(t+1) a .
\end{array}
$$
Thus, each positive integer term $\frac{b}{a}$ in the $k$-th group corresponds one-to-one with a divisor of $k+1$ greater than 1.
Therefore, the number of positive integer terms in the $k$-th group is the number of distinct divisors of $k+1$ greater than 1.
By calculation, the first 63 groups contain
$$
\begin{array}{l}
1 \times 18+2 \times 4+3 \times 20+4 \times 1+5 \times 10+ \\
6 \times 1+7 \times 6+8 \times 1+9 \times 1+11 \times 1 \\
=216
\end{array}
$$
positive integer terms.
Notice that
$$
1+2+\cdots+63=2016 \text {. }
$$
Thus, the 2011th term of the sequence belongs to the 63rd group, and is the 6th number from the end.
Therefore, the terms in the 63rd group that do not belong to the first 2011 terms are
$$
\left(\frac{59}{5}, \frac{60}{4}, \frac{61}{3}, \frac{62}{2}, \frac{63}{1}\right) \text {, }
$$
among which, there are three integers.
In summary, the first 2011 terms of the sequence contain 213 terms that are positive integers.
|
213
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) A dance troupe has $n(n \geqslant 5)$ actors, and they have arranged some performances, each of which is performed by four actors on stage. In one performance, they found that: it is possible to appropriately arrange several performances so that every two actors in the troupe perform on stage together exactly once during this performance. Find the minimum value of $n$.
---
The translation maintains the original formatting and structure of the source text.
|
Three, use $n$ points to represent $n$ actors.
If two actors have performed on the same stage once, then connect the corresponding points with an edge. Thus, the condition of this problem is equivalent to:
Being able to partition the complete graph $K_{n}$ of order $n$ into several complete graphs $K_{4}$ of order 4, such that each edge belongs to exactly one $K_{4}$.
First, from the problem statement, we know $\mathrm{C}_{4}^{2} I \mathrm{C}_{n}^{2}$, i.e.,
$12 \ln (n-1)$.
Therefore, $n \neq 5, 6, 7, 8, 10, 11$.
Second, consider the edges containing point $A$ (as an endpoint), there are $n-1$ such edges, and each edge belongs to exactly one $K_{4}$, thus, there are $n-1$ $K_{4}$s containing point $A$ (as an endpoint). However, each $K_{4}$ containing point $A$ has three edges containing point $A$, thus, each $K_{4}$ is counted 3 times.
Thus, $31(n-1)$.
Therefore, $n \neq 9, 12$.
Finally, represent 13 points using $0,1, \cdots, 12$. For $m=0,1, \cdots, 12$, let $m, m+1, m+5, m+11$ form a $K_{4}$ (points are considered modulo 13). Then the 13 $K_{4}$s form a valid partition.
In summary, the minimum value of $n$ is 13.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. From the numbers $1,2, \cdots, 2014$, what is the maximum number of numbers that can be selected such that none of the selected numbers is 19 times another?
|
According to the problem, if $k$ and $19 k$ cannot both appear in $1,2, \cdots, 2014$, and since $2014=19 \times 106$, and $106=5 \times 19+11$, then choose
$$
1,2,3,4,5,106,107, \cdots, 2013 \text {, }
$$
These 1913 numbers satisfy the requirement.
The numbers not chosen are $6,7, \cdots, 105,2014$, a total of 101 numbers.
Since $2014=1913+101$, if 1914 numbers are selected, then in the following 101 pairs of numbers
$$
(6,6 \times 19),(7,7 \times 19), \cdots,(106,106 \times 19)
$$
there must be a pair where both numbers are selected.
Thus, there must be one number that is 19 times the other.
Therefore, from the 2014 numbers $1,2, \cdots, 2014$, the maximum number of numbers that can be selected is 1913.
|
1913
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $T \subseteq\{1,2, \cdots, 25\}$. If for any two distinct elements $a, b (a \neq b)$ in $T$, their product $ab$ is not a perfect square, find the maximum number of elements in $T$, and the number of subsets $T$ that satisfy this condition.
|
Prompt: Divide the set $\{1,2, \cdots, 25\}$ into several subsets such that the product of any two elements in the same subset is a perfect square, and the product of any two elements in different subsets is not a perfect square.
To maximize the number of elements in $T$, let
$$
\begin{array}{l}
A_{1}=\{1,4,9,16,25\}, A_{2}=\{2,8,18\}, \\
A_{3}=\{3,12\}, A_{4}=\{5,20\}, A_{5}=\{6,24\}, \\
A_{6}=\{7\}, A_{7}=\{10\}, A_{8}=\{11\}, A_{9}=\{13\}, \\
A_{10}=\{14\}, A_{11}=\{15\}, A_{12}=\{17\}, A_{13}=\{19\}, \\
A_{14}=\{21\}, A_{15}=\{22\}, A_{16}=\{23\} .
\end{array}
$$
Taking one number from each of these 16 sets to form the set $T$ meets the requirement.
Therefore, the set $T$ can have at most 16 elements.
If the set $T$ has more than 16 elements, then there must be two elements from the same set $A_{k}(k=1,2, \cdots, 16)$, thus, the product of these two elements will be a perfect square.
Hence, the number of sets $T$ that satisfy the requirement is
$$
5 \times 3 \times 2 \times 2 \times 2=120 \text { (sets). }
$$
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the smallest positive integer $n$ such that for any $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
(1991, Australian Mathematical Olympiad)
|
Let $M=\left\{a_{i} \mid a_{i}=0,1, \cdots, 995\right\}$.
Since $a_{i}+a_{j} \leqslant 995+994=1989<1991$, $0<\left|a_{i}-a_{j}\right| \leqslant 995$,
then the sum and difference of any two numbers in $M$ are not multiples of 1991.
Therefore, $n \geqslant 997$.
Let $a_{1}, a_{2}, \cdots, a_{997}$ be any 997 integers.
If there exist $a_{i} 、 a_{j}$ such that $a_{i} \equiv a_{j}(\bmod 1991)$, then the difference $a_{i}-a_{j}$ is divisible by 1991.
If any two numbers $a_{i} 、 a_{j}$ are not congruent modulo 1991, consider the smallest absolute residues of 1991
$$
-995,-994, \cdots,-1,0,1, \cdots, 994,995 \text {. }
$$
Divide them into 996 groups:
$$
\{-995,995\},\{-994,994\}, \cdots,\{-1,1\},\{0\} \text {. }
$$
Thus, there are at least two different numbers $a_{i} 、 a_{j}$ such that
$$
a_{i}=-a_{j}(\bmod 1991) \text {. }
$$
In this case, 1991 ! $\left(a_{i}+a_{j}\right)$.
|
997
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the units digit of $\left(7^{2004}+36\right)^{818}$.
$(2004$, Shanghai Jiao Tong University Independent Recruitment Examination)
|
$$
\begin{array}{l}
\text { Solution: Since } 7^{4} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004}+36 \equiv 7(\bmod 10) .
\end{array}
$$
$$
\text { Therefore, the original expression } \equiv 7^{818} \equiv 7^{2} \equiv 9(\bmod 10) \text {. }
$$
Thus, the unit digit is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $u$ be a root of the equation
$$
x^{3}-3 x+10=0
$$
Let $f(x)$ be a quadratic polynomial with rational coefficients, and
$$
\alpha=\frac{1}{2}\left(u^{2}+u-2\right), f(\alpha)=u .
$$
Find $f(0)$.
(2010, Five Schools Joint Examination for Independent Enrollment)
|
From the given, we have $u^{3}-3 u+10=0$. Then
$$
\begin{aligned}
\alpha^{2} &=\frac{1}{4}\left(u^{2}+u-2\right)^{2} \\
&=\frac{1}{4}\left(u^{4}+2 u^{3}-3 u^{2}-4 u+4\right) \\
&=\frac{1}{4}\left[u(3 u-10)+2(3 u-10)-3 u^{2}-4 u+4\right] \\
&=-4-2 u .
\end{aligned}
$$
Let $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{Q}, a \neq 0)$. Then
$$
\begin{aligned}
u &=f(\alpha)=a \alpha^{2}+b \alpha+c \\
&=a(-4-2 u)+\frac{b}{2}\left(u^{2}+u-2\right)+c \\
&\Rightarrow b u^{2}+(b-4 a-2) u-(8 a+2 b-2 c)=0 . \\
\text { Let } k &=(b-4 a-2), \\
l & =-(8 a+2 b-2 c) .
\end{aligned}
$$
$$
\begin{aligned}
\text { Then } b u^{2}+k u+l &=0(k, l \in \mathbf{Q}) . \\
\text { Hence } (b u-k)\left(b u^{2}+k u+l\right) &=
\end{aligned}
$$
$$
\begin{aligned}
&=b^{2} u^{3}+\left(b l-k^{2}\right) u-k l \\
&=b^{2}(3 u-10)+\left(b l-k^{2}\right) u-k l \\
&=\left(3 b^{2}+b l-k^{2}\right) u-\left(10 b^{2}+k l\right)=0 .
\end{aligned}
$$
Since $\pm 1, \pm 2, \pm 5, \pm 10$ are not roots of equation (1), then $u$ cannot be a rational number. Therefore,
$$
\left\{\begin{array}{l}
3 b^{2}+b l-k^{2}=0 \\
10 b^{2}+k l=0 .
\end{array}\right.
$$
Assume $b \neq 0$. Then,
$$
\begin{aligned}
k^{3} &=3 b^{2} k+b k l=3 b^{2} k-10 b^{3} \\
&\Rightarrow\left(\frac{k}{b}\right)^{3}-3\left(\frac{k}{b}\right)+10=0 .
\end{aligned}
$$
Thus, $\frac{k}{b}$ is a rational root of equation (1), which is impossible, so $b=0$.
$$
\begin{aligned}
\text { By } b u^{2}+k u+l &=0 \Rightarrow k u+l=0 \\
&\Rightarrow\left\{\begin{array}{l}
k=0, \\
l=0
\end{array} \Rightarrow \left\{\begin{array}{l}
-4 a-2=0, \\
8 a-2 c=0
\end{array}\right.\right. \\
&\Rightarrow\left\{\begin{array}{l}
a=-\frac{1}{2}, \\
c=-2 .
\end{array}\right.
\end{aligned}
$$
Therefore, $f(x)=-\frac{1}{2} x^{2}-2, f(0)=-2$.
In advanced algebra, a polynomial that cannot be expressed as the product of polynomials with rational coefficients is called an irreducible polynomial over the field of rational numbers, such as $x^{3}-3 x+10$ in equation (1), which is an irreducible polynomial over the field of rational numbers. By the degree theory in advanced algebra, we know that $x^{3}-3 x+10$ is the minimal polynomial of $u$. Hence, $u^{2}, u, 1$ are linearly independent over the field of rational numbers. Therefore, from $b u^{2}+k u+l=0$, we can deduce that $b=k=l=0$, which leads to the conclusion.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given real numbers $x, y, z$ satisfy
$$
x y z=32, x+y+z=4 \text {. }
$$
Then the minimum value of $|x|+|y|+|z|$ is $\qquad$ .
(2010, Hubei Province High School Mathematics Competition)
|
Answer: 12.
The text above has been translated into English, maintaining the original text's line breaks and format. Here is the direct output of the translation result.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given positive integers $a, b$ satisfy
$$
|b-2|+b-2=0,|a-b|+a-b=0,
$$
and $a \neq b$. Then the value of $a b$ is $\qquad$ .
|
2.1.2.
From the given conditions, we know that $a>0, b-2 \leqslant 0, a-b \leqslant 0$. Therefore, $a<b \leqslant 2$. Hence, $a=1, b=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the side lengths of a trapezoid are $3,4,5,6$. Then the area of this trapezoid is $\qquad$ .
|
3. 18.
First, determine the lengths of the two bases. As shown in Figure 7, let trapezoid $ABCD$ have $AD$ and $BC$ as the upper and lower bases, respectively. Draw $AE \parallel CD$. Then $BE$ is the difference between the upper and lower bases.
In $\triangle ABE$, $AB - AE = AB - CD < BE = BC - AD$. Therefore, $AB = 5, CD = 4, AD = 3, BC = 6$. In $\triangle ABE$, $AB = 5, BE = 3, AE = CD = 4$. Thus, $AE$ is the height of the trapezoid.
Therefore, $S_{\text{trapezoid}} = \frac{1}{2} \times (3 + 6) \times 4 = 18$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 3, in $\triangle ABC$, it is given that $D$ is a point on side $BC$ such that $AD = AC$, and $E$ is the midpoint of side $AD$ such that $\angle BAD = \angle ACE$. If $S_{\triangle BDE} = 1$, then $S_{\triangle ABC}$ is $\qquad$.
|
4.4.
Notice that
$$
\begin{array}{l}
\angle E C D=\angle A C D-\angle A C E \\
=\angle A D C-\angle B A D=\angle A B C .
\end{array}
$$
Therefore, $\triangle E C D \backsim \triangle A B C$.
$$
\text { Then } \frac{C D}{B C}=\frac{E D}{A C}=\frac{E D}{A D}=\frac{1}{2} \text {. }
$$
Thus, $D$ is the midpoint of side $B C$.
Hence $S_{\triangle A B C}=2 S_{\triangle A C D}=4 S_{\triangle E C D}=4 S_{\triangle D B E}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) For a certain project, Team A alone needs 12 days to complete, and Team B alone needs 9 days to complete. If the two teams are scheduled to work in whole days, how many schemes are there to ensure that the project is completed within 8 days?
|
Three, let teams A and B work for $x, y$ days respectively to just meet the requirements. Then we have
$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{9}=1, \\
0 \leqslant x \leqslant 8, \\
0 \leqslant y \leqslant 8,
\end{array}\right.
$$
and $x, y$ are both integers.
From $\frac{x}{12}+\frac{y}{9}=1$, we get $x=12-y-\frac{y}{3}$.
Since $x$ and $y$ are both integers, and $0 \leqslant y \leqslant 8$, so, $y=0$ or 3 or 6.
When $y=0$, $x=12$ (discard);
When $y=3$, $x=8$;
When $y=6$, $x=4$.
Therefore, there are two solutions, one is A works for 8 days, B works for 3 days; the other is A works for 4 days, B works for 6 days.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the polynomial $f(x)$ satisfy: for any $x \in \mathbf{R}$, we have
$$
f(x+1)+f(x-1)=2 x^{2}-4 x .
$$
Then the minimum value of $f(x)$ is $\qquad$
|
- 1. - -2 .
Since $f(x)$ is a polynomial, we know that $f(x+1)$ and $f(x-1)$ have the same degree as $f(x)$. Therefore, $f(x)$ is a quadratic polynomial (let's assume $f(x) = ax^2 + bx + c$). Then,
$$
\begin{array}{l}
f(x+1) + f(x-1) \\
= 2ax^2 + 2bx + 2(a + c) = 2x^2 - 4x.
\end{array}
$$
Thus, $a = 1, b = -2, c = -1$.
Hence, $f(x) = x^2 - 2x - 1 = (x-1)^2 - 2 \geqslant -2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (25 points) For a set $M=\left\{p_{1}, p_{2}, \cdots, p_{2_{n}}\right\}$ consisting of $2 n$ prime numbers, its elements can be paired to form $n$ products, resulting in an $n$-element set. If
$$
\begin{aligned}
A & =\left\{a_{1} a_{2}, a_{3} a_{4}, \cdots, a_{2 n-1} a_{2 n}\right\} \\
\text { and } \quad B & =\left\{b_{1} b_{2}, b_{3} b_{4}, \cdots, b_{2 n-1} b_{2 n}\right\}
\end{aligned}
$$
are two $n$-element sets obtained in this way, where
$$
\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}=\left\{b_{1}, b_{2}, \cdots, b_{2 n}\right\}=M,
$$
and $A \cap B=\varnothing$, then the set pair $\{A, B\}$ is called a "couplet" formed by $M$ (for example, from the four-element set $\{a, b, c, d\}$, three couplets can be formed:
$$
\begin{array}{l}
\{a b, c d\} \sim\{a c, b d\}, \\
\{a b, c d\} \sim\{a d, b c\}, \\
\{a c, b d\} \sim\{a d, b c\} .
\end{array}
$$
Find the number of couplets that can be formed from the six-element prime set $M=\{a, b, c, d, e, f\}$.
|
11. Six elements can form fifteen different "slips of paper," listed as follows:
$$
\begin{array}{l}
\{a b, c d, e f\},\{a b, c e, d f\},\{a b, c f, d e\}, \\
\{a c, b d, e f\},\{a c, b e, d f\},\{a c, b f, d e\}, \\
\{a d, b c, e f\},\{a d, b e, c f\},\{a d, b f, c e\}, \\
\{a e, b c, d f\},\{a e, b d, c f\},\{a e, b f, c d\}, \\
\{a f, b c, d e\},\{a f, b d, c e\},\{a f, b e, c d\} .
\end{array}
$$
Consider the slip of paper at the intersection of the $i$-th row and the $j$-th column as a coordinate point, denoted as $(i, j)$.
For the point $(1,1)$ in the first row, it pairs with two points in each of the following rows:
$$
\begin{array}{l}
(1,1) \sim(2,2),(1,1) \sim(2,3) ; \\
(1,1) \sim(3,2),(1,1) \sim(3,3) ; \\
(1,1) \sim(4,1),(1,1) \sim(4,2) ; \\
(1,1) \sim(5,1),(1,1) \sim(5,2) .
\end{array}
$$
This results in $4 \times 2=8$ pairs.
Similarly, the points $(1,2)$ and $(1,3)$ each pair with 8 points in the following four rows.
Thus, the three points in the first row form $3 \times 4 \times 2=24$ pairs with the points in the following four rows;
Each point in the second row pairs with two points in each of the following rows, resulting in $3 \times 3 \times 2=18$ pairs;
Each point in the third row pairs with two points in each of the following rows, resulting in $3 \times 2 \times 2=12$ pairs;
Each point in the fourth row pairs with two points in the following row, resulting in $3 \times 1 \times 2=6$ pairs.
Therefore, the total number of pairs is $6 \times(1+2+3+4)=60$, meaning that 60 couplets can be created from the set $M$.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010 x=4020, \\
(y-1)^{2011}+(y-1)^{2009}+2010 y=0,
\end{array}\right.
$$
then $x+y=$ $\qquad$ .
|
2. 2 .
Transform the original system of equations into
$$
\begin{array}{l}
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010(x-1)=2010, \\
(y-1)^{2011}+(y-1)^{2009}+2010(y-1)=-2010 .
\end{array}\right. \\
\text { Let } f(t)=t^{2011}+t^{2009}+2010 t(t \in \mathbf{R}) .
\end{array}
$$
Then the function $f(t)$ is a monotonically increasing odd function.
From $f(x-1)=-f(y-1)$, we get
$$
x-1=-(y-1) \Rightarrow x+y=2 .
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {, }
$$
then there exists an integer $k$, such that the following equations hold for
$\qquad$ number of them.
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$;
(2) $\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}=\frac{1}{a^{2 k+1}+b^{2 k+1}+c^{2 k+1}}$;
(3) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k}=\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}$;
(4) $\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}=\frac{1}{a^{2 k}+b^{2 k}+c^{2 k}}$.
|
3. 2 .
From the given equation, we have
$$
\begin{array}{l}
\frac{(b c+a c+a b)(a+b+c)-a b c}{a b c(a+b+c)}=0 . \\
\text { Let } P(a, b, c) \\
=(b c+a c+a b)(a+b+c)-a b c \\
=(a+b)(b+c)(c+a) .
\end{array}
$$
From $P(a, b, c)=0$, we get
$$
a=-b \text { or } b=-c \text { or } c=-a \text {. }
$$
Verification shows that equations (1) and (2) hold, while equations (3) and (4) do not hold.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$, and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$. Draw $PC \perp x$-axis and $PD \perp y$-axis, with the feet of the perpendiculars being $C$ and $D$, respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is
|
4. 24.
Notice that
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\
=2\left(x+\frac{9}{x}\right)+12 \geqslant 24 .
\end{array}
$$
The equality holds if and only if $x=3$. Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $k_{1}, k_{2}, \cdots, k_{n}$ are $n$ distinct positive integers, and satisfy $\sum_{i=1}^{n} k_{i}^{3}=2024$. Then the maximum value of the positive integer $n$ is $\qquad$ .
|
6. 8 .
From the problem, we know that when $n \geqslant 9$, we have
$$
\sum_{i=1}^{n} k_{i}^{3} \geqslant \sum_{i=1}^{n} i^{3} \geqslant \sum_{i=1}^{9} i^{3}=2025>2024,
$$
which is a contradiction. Therefore, $n \leqslant 8$.
Also, $2^{3}+3^{3}+\cdots+9^{3}=2024$, so the maximum value of the positive integer $n$ is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $m, n$ be given positive integers. In each square of an $m \times n$ chessboard, fill a number according to the following rule: first, fill the numbers in the 1st row and the 1st column arbitrarily, then, for any other square, let the number filled in it be $x$, and the number in the 1st row in the same column as $x$ be $y$, the number in the 1st column in the same row as $x$ be $z$, and the number in the 1st row and 1st column be $a$, then $a + x = y + z$. Thus, a unique number can be filled in each square. After all squares are filled with numbers, take any rectangle on the chessboard, then the probability that the sum of the numbers in the two diagonal corners of the rectangle is equal is $\qquad$
|
6.1.
Let the cell at the $i$-th row and $j$-th column of the chessboard be denoted as $a_{i j}$, and the number filled in cell $a_{i j}$ is also represented by $a_{i j}$.
Consider any rectangle on the chessboard, and let the four corner cells of this rectangle be $a_{i j}, a_{i t}, a_{s j}, a_{s t} (i<s, j<t)$.
By the rule of filling numbers, we have
$$
\begin{array}{l}
a_{i j}=a_{1 j}+a_{i 1}-a_{11}, a_{i t}=a_{1 t}+a_{i 1}-a_{11}, \\
a_{s j}=a_{1 j}+a_{s 1}-a_{11}, a_{s t}=a_{1 t}+a_{s 1}-a_{11} . \\
\text { Then } a_{i j}+a_{s t}=\left(a_{1 j}+a_{i 1}-a_{11}\right)+\left(a_{1 t}+a_{s 1}-a_{11}\right) \\
=a_{1 j}+a_{i 1}+a_{1 t}+a_{s 1}-2 a_{11}, \\
a_{i t}+a_{s j}=\left(a_{1 t}+a_{i 1}-a_{11}\right)+\left(a_{1 j}+a_{s 1}-a_{11}\right) \\
=a_{1 j}+a_{i 1}+a_{1 t}+a_{s 1}-2 a_{11} .
\end{array}
$$
Therefore, $a_{i j}+a_{s t}=a_{i t}+a_{s j}$
Thus, the required probability is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 There are three types of goods, A, B, and C. If you buy 3 pieces of A, 7 pieces of B, and 1 piece of C, it costs a total of 315 yuan; if you buy 4 pieces of A, 10 pieces of B, and 1 piece of C, it costs a total of 420 yuan. Question: How much would it cost to buy one piece each of A, B, and C?
|
Let the unit prices of A, B, and C be $x$, $y$, and $z$ yuan, respectively. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
3 x+7 y+z=315, \\
4 x+10 y+z=420 .
\end{array}\right.
$$
The problem actually only requires finding the value of $x+y+z$, without necessarily solving for $x$, $y$, and $z$ individually. Therefore, we should try to isolate the value of $x+y+z$.
The system of equations (1) can be equivalently transformed into
$$
\left\{\begin{array}{l}
2(x+3 y)+(x+y+z)=315, \\
3(x+3 y)+(x+y+z)=420 .
\end{array}\right.
$$
It is easy to see that, $x+y+z=105$.
Therefore, purchasing one each of A, B, and C would cost 105 yuan.
|
105
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $p$ is a prime number greater than 5, and $m$ is the smallest non-negative remainder when $\left(p^{2}+5 p+5\right)^{2}$ is divided by 120. Then the units digit of $2009^{m}$ is $\qquad$
|
Notice that
$$
\begin{array}{l}
\left(p^{2}+5 p+5\right)^{2}=\left[\left(p^{2}+5 p+5\right)^{2}-1\right]+1 \\
=\left(p^{2}+5 p+6\right)\left(p^{2}+5 p+4\right)+1 \\
=(p+1)(p+2)(p+3)(p+4)+1 .
\end{array}
$$
Let $M=(p+1)(p+2)(p+3)(p+4)$.
Clearly, $5! \mid p M$.
Since $p$ is a prime number greater than 5, 120 divides $M$.
Therefore, $\left(p^{2}+5 p+5\right)^{2}$ always has a remainder of 1 when divided by 120, i.e., $m=1$.
Thus, $2009^{m}=2009$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For $\triangle A B C$, squares are constructed outward on its three sides $a, b, c$, with their areas denoted as $S_{a}, S_{b}, S_{c}$ respectively. If $a+b+c=18$, then the minimum value of $S_{a}+S_{b}+S_{c}$ is $\qquad$
|
2. 108.
$$
\begin{array}{l}
\text { Given }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0 \\
\Rightarrow 2\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2(a b+b c+c a) \\
\Rightarrow 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2} \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant \frac{18^{2}}{3}=108 . \\
\text { Therefore, }\left(S_{a}+S_{b}+S_{c}\right)_{\text {min }}=108 .
\end{array}
$$
|
108
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the real numbers $x, y, z, w$ satisfy
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}=1, \\
\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$
then $x^{2}+y^{2}+z^{2}+w^{2}=$
|
3. 36 .
It is known that $2^{2}$ and $4^{2}$ are the two roots of the equation with respect to $t$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}=1
$$
which means they are the two roots of the equation:
$$
t^{2}-\left(1^{2}+3^{2}+x^{2}+y^{2}\right) t+1^{2} \times 3^{2}+3^{2} x^{2}+1^{2} \times y^{2}=0
$$
Therefore,
$$
\begin{array}{l}
1^{2}+3^{2}+x^{2}+y^{2}=2^{2}+4^{2} \\
\Rightarrow x^{2}+y^{2}=2^{2}-1^{2}+4^{2}-3^{2}=10 .
\end{array}
$$
Similarly, $z^{2}+w^{2}=6^{2}-5^{2}+8^{2}-7^{2}=26$.
Thus, $x^{2}+y^{2}+z^{2}+w^{2}=36$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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