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11. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$,
$$
B=\left\{a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, a_{4}^{2}, a_{5}^{2}\right\},
$$
where $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are 5 different positive integers, and
$$
\begin{array}{l}
a_{1}<a_{2}<a_{3}<a_{4}<a_{5}, \\
A \cap B=\left\{a_{1}, a_{4}\right\}, a_{1}+a_{4}=10 .
\end{array}
$$
If the sum of all elements in $A \cup B$ is 256, then the number of sets $A$ that satisfy the conditions is $\qquad$
|
11.2.
Since $a_{1}^{2}=a_{1}$, therefore, $a_{1}=1, a_{4}=9$.
Since $B$ contains 9, $A$ contains 3.
If $a_{3}=3$, then $a_{2}=2$.
Thus, $a_{5}+a_{5}^{2}=146$. No positive integer solution.
If $a_{2}=3$, since $10 \leqslant a_{5} \leqslant 11$, then $a_{2}^{2} \neq a_{5}$.
Thus, $a_{3}+a_{3}^{2}+a_{5}+a_{5}^{2}=152$.
Also, $a_{3} \geqslant 4$, when $a_{5}=10$, $a_{3}=6$; when $a_{5}=$ 11, $a_{3}=4$. Therefore, there are 2 sets $A$ that satisfy the conditions, which are $\{1,3,4,9,11\},\{1,3,6,9,10\}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{0}=0$, $a_{1}=1$, for all positive integers $n$, we have
$$
a_{n+1}=2 a_{n}+2007 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2008 \mid a_{n}$.
|
14. Solution 1: Let $m=2008, a_{n+1}=2 a_{n}+2007 a_{n-1}$, the characteristic equation is $\lambda^{2}-2 \lambda-2007=0$, with characteristic roots $1 \pm \sqrt{m}$. Combining $a_{0}=0, a_{1}=1$, we get
$$
a_{n}=\frac{1}{2 \sqrt{m}}\left[(1+\sqrt{m})^{n}-(1-\sqrt{m})^{n}\right] \text {. }
$$
By the binomial theorem,
$$
a_{n}=\frac{1}{2 \sqrt{m}}\left[\sum_{k=0}^{n} \mathrm{C}_{n}^{k} m^{\frac{k}{2}}-\sum_{k=0}^{n}(-1)^{k} \mathrm{C}_{n}^{k} m^{\frac{k}{2}}\right] \text {. }
$$
When $n$ is odd,
$$
a_{n}=\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3} m+\cdots+\mathrm{C}_{n}^{n-2} m^{\frac{n-3}{2}}+\mathrm{C}_{n}^{n} m^{\frac{n-1}{2}} \text {; }
$$
When $n$ is even,
$$
a_{n}=\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3} m+\cdots+\mathrm{C}_{n}^{n-3} m^{\frac{n-4}{2}}+\mathrm{C}_{n}^{n-1} m^{\frac{n-2}{2}} \text {. }
$$
Thus, $m\left|a_{n} \Leftrightarrow m\right| \mathrm{C}_{n}^{1}$, i.e., $2008 \mid n$.
Therefore, the smallest positive integer that satisfies the condition is 2008.
Solution 2: The following are all $a_{n}$ modulo 2008, then $a_{n+1} \equiv 2 a_{n}-a_{n-1}$, i.e.,
$$
a_{n+1}-a_{n} \equiv a_{n}-a_{n-1} .
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2008.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, $2008 \mid n$.
Therefore, the smallest positive integer that satisfies the condition is 2008.
|
2008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. There are 10 students standing in a row, and their birthdays are in different months. There are $n$ teachers who will select these students to join $n$ interest groups. Each student is selected by exactly one teacher, and the order of the students is maintained. Each teacher must select students whose birthdays are in months that are either strictly increasing or strictly decreasing (selecting one or two students is also considered strictly increasing or decreasing). Each teacher should select as many students as possible. For all possible orderings of the students, find the minimum value of $n$.
|
15. If $n \leqslant 3$, let's assume the birth months of these 10 students are $1,2, \cdots, 10$.
When the students are sorted by their birthdays as $4,3,2,1,7,6,5,9, 8,10$, there exists at least one teacher who must select two students from the first four. Since the birth months of these two students are decreasing, and the birth months of the last six students are all greater than those of the first four, this teacher cannot select any of the last six students; among the remaining no more than two teachers, there must be one who has to select two students from the fifth to the seventh. Similarly, this teacher cannot select any of the last three students; the remaining no more than one teacher also cannot select any of the last three students, leading to a contradiction.
Below is the proof: For any distinct ordered sequence of real numbers $a_{1}, a_{2}, \cdots, a_{m}$, when $m \geqslant 5$, there must exist three numbers $a_{i}, a_{j}, a_{k}(i<a_{j}>a_{k}$.
Let the maximum and minimum numbers be $a_{s}$ and $a_{t}$. Without loss of generality, assume $s<a_{s+1}>a_{t} ; s+1=t$.
Since $m \geqslant 5$, there must be at least two numbers either before $a_{s}, a_{s+1}$ or after $a_{s}, a_{s+1}$.
Without loss of generality, assume there are two numbers $a_{s+2}$ and $a_{s+3}$ after $a_{s}, a_{s+1}$. Thus,
$$
a_{s}>a_{s+2}>a_{s+3} \text { or } a_{s+1}<a_{s+2}<a_{s+3}
$$
must hold.
Using the above conclusion, when $n=4$, the first teacher can select at least 3 students; if the remaining students are greater than or equal to 5, the second teacher can also select at least 3 students; at this point, the number of remaining students does not exceed 4, and can be selected by the two teachers.
Therefore, the minimum value of $n$ is 4.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{2}{3}, a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)} \text {. }
$$
Then $a_{2007}=$
|
8.1343352 .
Given $a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}$, squaring both sides yields $3\left(a_{n+1}-a_{n}\right)^{2}=2\left(a_{n+1}+a_{n}\right)$.
Also, $3\left(a_{n}-a_{n-1}\right)^{2}=2\left(a_{n}+a_{n-1}\right)$, subtracting the two equations gives
$$
\begin{array}{l}
3\left(a_{n+1}-a_{n-1}\right)\left(a_{n+1}-2 a_{n}+a_{n-1}\right) \\
=2\left(a_{n+1}-a_{n-1}\right) .
\end{array}
$$
Given $a_{1}=\frac{2}{3}, a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}$, we find $a_{2}=2$.
From the recursive relation, it is easy to see that the sequence $\left\{a_{n}\right\}$ is a monotonically increasing sequence, so $a_{n+1}-a_{n-1} \neq 0$.
Hence $3\left(a_{n+1}-2 a_{n}+a_{n-1}\right)=2$
$$
\Rightarrow\left(a_{n+1}-a_{n}\right)-\left(a_{n}-a_{n-1}\right)=\frac{2}{3} \text {. }
$$
Therefore, the sequence $\left\{a_{n+1}-a_{n}\right\}$ is an arithmetic sequence with the first term $a_{2}-a_{1}=\frac{4}{3}$ and common difference $\frac{2}{3}$.
Thus, $a_{n+1}-a_{n}=\frac{4}{3}+\frac{2}{3}(n-1)=\frac{2}{3}(n+1)$.
Hence, $a_{n}=a_{1}+\frac{2}{3}(2+3+\cdots+n)=\frac{1}{3} n(n+1)$.
Therefore, $a_{2007}=1343352$.
|
1343352
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the complex number $z_{1}=(6-a)+(4-b) \mathrm{i}$,
$$
\begin{array}{l}
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value, $3 a+4 b=$ $\qquad$
|
9.12.
It is easy to find that $z_{1}+z_{2}+z_{3}=12+9 \mathrm{i}$. Therefore,
$$
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=15 \text {. }
$$
The equality holds if and only if $\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9}$. Solving this, we get $a=\frac{7}{3}, b=\frac{5}{4}$.
Thus, $3 a+4 b=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the function
$$
y=\frac{225}{4 \sin ^{2} x}+\frac{2}{\cos x}
$$
has a minimum value of
|
10.68
Since $x \in\left(0, \frac{\pi}{2}\right)$, we have $\sin x>0, \cos x>0$. Let $k>0$, then
$$
\begin{array}{l}
y=\frac{225}{4 \sin ^{2} x}+k \sin ^{2} x+\frac{1}{\cos x}+\frac{1}{\cos x}+ \\
\quad k \cos ^{2} x-k \\
\geqslant 15 \sqrt{k}+3 \sqrt[3]{k}-k .
\end{array}
$$
Equality in (1) holds if and only if
$$
\left\{\begin{array} { l }
{ \frac { 2 2 5 } { 4 \operatorname { s i n } ^ { 2 } x } = k \operatorname { s i n } ^ { 2 } x , } \\
{ \frac { 1 } { \operatorname { c o s } x } = k \operatorname { c o s } ^ { 2 } x }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin ^{2} x=\frac{15}{2 \sqrt{k}}, \\
\cos ^{2} x=\frac{1}{\sqrt[3]{k^{2}}}
\end{array}\right.\right.
$$
At this point, $\frac{15}{2 \sqrt{k}}+\frac{1}{\sqrt[3]{k^{2}}}=1$.
Let $\frac{1}{k}=t^{6}$, then $2 t^{4}+15 t^{3}-2=0$. And
$$
\begin{array}{l}
2 t^{4}+15 t^{3}-2=2 t^{4}-t^{3}+16 t^{3}-2 \\
=t^{3}(2 t-1)+2(2 t-1)\left(4 t^{2}+2 t+1\right) \\
=(2 t-1)\left(t^{3}+8 t^{2}+4 t+2\right),
\end{array}
$$
Thus, $(2 t-1)\left(t^{3}+8 t^{2}+4 t+2\right)=0$.
Noting that $0<t^{6}=\frac{1}{k} \leqslant 1$, it is easy to see that the only root satisfying the condition is $t=\frac{1}{2}$.
When $t=\frac{1}{2}$, $k=\frac{1}{t^{6}}=64$, and equality in (1) is achieved.
Therefore, the minimum value of the function $y=\frac{225}{4 \sin ^{2} x}+\frac{2}{\cos x}$ is $15 \sqrt{64}+3 \sqrt[3]{64}-64=68$.
|
68
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. For the function $f(x)=\sqrt{a x^{2}+b x}$, there exists a positive number $b$, such that the domain and range of $f(x)$ are the same. Then the value of the non-zero real number $a$ is $\qquad$.
|
11. -4 .
If $a>0$, for the positive number $b$, the domain of $f(x)$ is
$$
D=\left(-\infty,-\frac{b}{a}\right] \cup[0,+\infty) .
$$
However, the range of $f(x)$, $A \subseteq[0,+\infty)$, so $D \neq A$, which does not meet the requirement.
If $a>0$, so, $a=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Real numbers $x, y$ satisfy $\tan x=x, \tan y=y$, and $|x| \neq|y|$. Then the value of $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}$ is
|
II, 11.0.
From the given, we have
$$
\begin{array}{l}
\frac{\sin (x+y)}{x+y}=\frac{\sin (x+y)}{\tan x+\tan y} \\
=\frac{\sin (x+y)}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}=\cos x \cdot \cos y .
\end{array}
$$
Similarly, $\frac{\sin (x-y)}{x-y}=\cos x \cdot \cos y$. Therefore, $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Definition: The length of the interval $\left[x_{1}, x_{2}\right]\left(x_{1}<x_{2}\right)$ is $x_{2}-x_{1}$. Given that the domain of the function $y=\left|\log _{\frac{1}{2}} x\right|$ is $[a, b]$, and the range is $[0,2]$. Then the difference between the maximum and minimum values of the length of the interval $[a, b]$ is $\qquad$.
|
12.3.
The graphs of the functions $y=$ $\left|\log _{\frac{1}{2}} x\right|$ and $y=2$ are shown in Figure 7.
From $y=0$, we get $x=1$;
From $y=2$, we get $x=\frac{1}{4}$ or $x=4$.
When $a=\frac{1}{4}, b=1$, $(b-a)_{\text {min }}=\frac{3}{4}$;
When $a=\frac{1}{4}, b=4$, $(b-a)_{\max }=\frac{15}{4}$.
Therefore, $(b-a)_{\max }-(b-a)_{\min }=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Given the parabola $y=x^{2}+3 x+c$ passes through two points $(m, 0)$ and $(n, 0)$, and
$$
m^{3}+3 m^{2}+(c-2) m-2 n-c=8,
$$
the parabola intersects the hyperbola $y=\frac{k}{x}$ at the point $(1, d)$.
(1) Find the equations of the parabola and the hyperbola;
(2) Given points $P_{1}, P_{2}, \cdots, P_{2008}$ are all on the hyperbola $y=\frac{k}{x}(x>0)$, with their x-coordinates being $a$, $2a, \cdots, 2008a$, and $O$ being the origin, let $S_{1}=$ $S_{\triangle P_{1} P_{2} O}, S_{2}=S_{\triangle P_{1} P_{3} O}, \cdots \cdots$ Point $Q$ is on the hyperbola $y=\frac{k}{x}(x<0)$, and a perpendicular line $Q M \perp y$-axis is drawn from $Q$ to $M$, let $S=S_{\triangle Q M O}$. Find
$$
S_{1}+S_{2}+\cdots+S_{2007}+\frac{S}{2}+\frac{S}{3}+\cdots+\frac{S}{2008}
$$
the value.
|
(1) According to the problem, we have
$$
\left\{\begin{array}{l}
m\left(m^{2}+3 m+c\right)-2(m+n)-c=8 \\
m^{2}+3 m+c=0, \\
m+n=-3 .
\end{array}\right.
$$
Substituting equations (2) and (3) into equation (1) gives $c=-2$.
Therefore, the equation of the parabola is $y=x^{2}+3 x-2$.
Since the intersection point of the parabola and the hyperbola is $(1, d)$, we have
$$
\left\{\begin{array}{l}
d = 1^{2} + 3 \times 1 - 2, \\
d = \frac{k}{1}
\end{array} \Rightarrow \left\{\begin{array}{l}
d=2, \\
k=2 .
\end{array}\right.\right.
$$
Therefore, the equation of the hyperbola is $y=\frac{2}{x}$.
(2) Since points $P_{1}$ and $P_{n+1} (n=1,2, \cdots, 2007)$ are on the hyperbola $y=\frac{2}{x} (x>0)$, and their x-coordinates are $a$ and $(n+1)a$ respectively, the y-coordinates of points $P_{1}$ and $P_{n+1}$ are $\frac{2}{a}$ and $\frac{2}{(n+1)a}$ respectively.
As shown in Figure 6, draw a line $a_{1} \parallel x$-axis through point $P_{1}$ intersecting the y-axis at $A_{1}$, and draw a line $b_{n+1} \parallel y$-axis through $P_{n+1}$ intersecting the x-axis at $B_{n+1}$ and intersecting $a_{1}$ at $C_{n+1}$. Then
$$
\begin{aligned}
& S_{n}=S_{P_{1} P_{n+1} o} \\
= & (n+1) a \cdot \frac{2}{a}-\frac{1}{2} a \cdot \frac{2}{a}- \\
& \frac{1}{2}(n+1) a \cdot \frac{2}{(n+1) a}- \\
& \frac{1}{2}[(n+1) a-a]\left[\frac{2}{a}-\frac{2}{(n+1) a}\right] \\
= & n+\frac{n}{n+1} .
\end{aligned}
$$
Let $Q(e, f) (e<0)$. Then $f=\frac{2}{e}$.
Thus, $S=S_{\triangle Q M O}=\frac{1}{2}|e| \cdot|f|=\frac{1}{2} e f=1$.
$$
\begin{array}{l}
\text { Therefore, } S_{1}+S_{2}+\cdots+S_{2007}+\frac{S}{2}+\frac{S}{3}+\cdots+\frac{S}{2008} \\
=\left(1+\frac{1}{2}\right)+\left(2+\frac{2}{3}\right)+\cdots+ \\
\left(2007+\frac{2007}{2008}\right)+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2008} \\
= 1+2+\cdots+2007+1 \times 2007 \\
= \frac{(1+2007) \times 2007}{2}+2007 \\
= 2017035 .
\end{array}
$$
|
2017035
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The largest solution and the smallest solution of the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$ sum to $(\quad)$ (where, $[x]$ denotes the greatest integer not exceeding $x$, the same applies below).
(A) 85
(B) -85
(C) 42
(D) -42
|
4.B.
Let $x=42 p+q(p, q$ be integers, $0 \leqslant q \leqslant 41)$.
Substituting $x$ into the original equation, we get
$$
p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q \text {. }
$$
For each different $q$, a unique ordered pair $(p, q)$ is determined, and thus, $x$ is also unique. To compare the sizes of $x$, we should first compare the sizes of $p$, and if $p$ is equal, then compare the sizes of $q$.
$$
\begin{array}{l}
\text { Since } p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q \\
\leqslant \frac{q}{2}+\frac{q}{3}+\frac{q}{7}-q=-\frac{q}{42} \leqslant 0,
\end{array}
$$
Therefore, the maximum value of $p$ is only achieved when $q=0$, and the maximum value is 0.
Thus, the largest solution for $x$ is 0.
$$
\begin{array}{l}
\text { Also, } p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q \\
\geqslant \frac{q-1}{2}+\frac{q-2}{3}+\frac{q-6}{7}-q=-\frac{q+85}{42} \\
\geqslant-3,
\end{array}
$$
Therefore, $p \geqslant-3$.
When and only when $q=41$, $p=-3$.
Thus, the smallest solution for $x$ is -85.
In summary, the sum of the largest and smallest solutions is
-85.
|
-85
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the circumcenter, incenter, and orthocenter of non-isosceles $\triangle ABC$ be $O$, $I$, and $H$, respectively, with the circumradius being $1$ and $\angle A=60^{\circ}$. Then the circumradius of $\triangle OIH$ is $\qquad$.
|
If $\triangle A B C$ is an acute triangle, since
$$
\angle B O C=\angle B I C=\angle B H C=120^{\circ} \text {, }
$$
then, $O$, $I$, $H$, $B$, and $C$ are concyclic;
if $\triangle A B C$ is an obtuse triangle, since
$$
\angle B O C=\angle B I C=120^{\circ}, \angle B H C=60^{\circ} \text {, }
$$
and $H$ is on the opposite side of $B C$ from $O$, $I$, and $A$, then, $O$, $I$, $H$, $B$, and $C$ are concyclic.
Therefore, the circumradius of $\triangle O I H$ is the same as the circumradius of $\triangle B H C$, which equals the circumradius of $\triangle A B C$, which is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}(n \geqslant 1)$ satisfy
$$
a_{n+1}=2 b_{n}-a_{n}, b_{n+1}=2 a_{n}-b_{n}
$$
$(n=1,2, \cdots)$. If $a_{1}=2007, a_{n}>0(n=2,3$, $\cdots$ ), then $b_{1}$ equals $\qquad$ .
|
2.2007.
Since $b_{n}=\frac{1}{2}\left(a_{n}+a_{n+1}\right)$, we have
$$
a_{n+2}+2 a_{n+1}-3 a_{n}=0 \text{. }
$$
By the method of characteristic roots, we get
$$
a_{n}=\frac{2007+b_{1}}{2}+\frac{b_{1}-2007}{6}(-3)^{n} \text{. }
$$
If for any positive integer $n$, we have $a_{n}>0$, then it must be that $b_{1}=2007$.
|
2007
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $\sum_{k=1}^{207}\left[\sqrt[4]{\frac{2007}{k}}\right]$ The value is $\qquad$
|
3.2167 .
The value sought is the number of lattice points within the region bounded by the curve $y=\sqrt[4]{\frac{2007}{x}}$ and $x>0$, $y>0$.
Changing the column-wise counting to row-wise counting, the value sought is
$$
\begin{array}{l}
\sum_{i=1}^{6}\left[\frac{2007}{i^{4}}\right]=2007+125+24+7+3+1 \\
=2167
\end{array}
$$
|
2167
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given an even function $f: \mathbf{Z} \rightarrow \mathbf{Z}$ that satisfies $f(1)=1$, $f(2007) \neq 1$, and for any integers $a, b$,
$$
f(a+b) \leqslant \max \{f(a), f(b)\} \text {. }
$$
Then the possible value of $f(2008)$ is $\qquad$
|
5.1.
Since $f(2) \leqslant \max \{f(1), f(1)\}=1$, assume for a positive integer $k$ greater than or equal to 2, we have $f(k) \leqslant 1$, then
$$
f(k+1) \leqslant \max \{f(k), f(1)\}=1 \text {. }
$$
Therefore, for all positive integers $n$, we have $f(n) \leqslant 1$.
Since $f$ is an even function, we can conclude that for all non-zero integers $n$, we have $f(n) \leqslant 1$.
Since $f(2007) \neq 1$, it follows that $f(2007)<1$.
Thus, $1=f(1)=f(2008+(-2007))$
$\leqslant \max \{f(2008), f(-2007)\} \leqslant 1$.
Since $f(-2007)=f(2007)<1$, it must be that $f(2008)=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For a positive integer $n \geqslant 2007$, a complex number $z$ satisfies
$$
(a+1) z^{n+1}+a \text { i } z^{n}+a \text { i } z-(a+1)=0 \text {, }
$$
where the real number $a>-\frac{1}{2}$. Then the value of $|z|$ is $\qquad$ .
|
6.1.
Given $z^{n}[(a+1) z+a \mathrm{i}]=a+1-a \mathrm{i} z$, then $|z|^{n}|(a+1) z+a \mathrm{i}|=|a+1-a \mathrm{i} z|$. Let $z=x+y \mathrm{i}$, where $x, y$ are real numbers. Then
$$
\begin{array}{l}
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2} \\
=(2 a+1)\left(|z|^{2}-1\right) .
\end{array}
$$
If $|z|>1$, since $a>-\frac{1}{2}$, then
$$
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2}>0 \text {. }
$$
Thus, $|z|<1$, which is a contradiction.
Similarly, if $|z|<1$, it also leads to a contradiction.
Therefore, $|z|=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(a+b+2 c)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8 .
\end{array}
$$
(2003, USA Mathematical Olympiad)
|
Analysis: Let's assume $a+b+c=1$. Then the original inequality transforms into
$$
\begin{array}{l}
\frac{(1+a)^{2}}{2 a^{2}+(1-a)^{2}}+\frac{(1+b)^{2}}{2 b^{2}+(1-b)^{2}}+ \\
\frac{(1+c)^{2}}{2 c^{2}+(1-c)^{2}} \leqslant 8 .
\end{array}
$$
Let $f(x)=\frac{(1+x)^{2}}{2 x^{2}+(1-x)^{2}}$
$$
=\frac{1}{3}\left[1+\frac{8 x+2}{3\left(x-\frac{1}{3}\right)^{2}+\frac{2}{3}}\right] \leqslant \frac{12 x+4}{3},
$$
Then we have
$$
\begin{array}{l}
f(a)+f(b)+f(c) \\
\leqslant \frac{12(a+b+c)+12}{3}=8
\end{array}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find the smallest positive integer $n$ such that: If each vertex of a regular $n$-gon is arbitrarily colored with one of the three colors red, yellow, or blue, then there must exist four vertices of the same color that form the vertices of an isosceles trapezoid (a convex quadrilateral with one pair of parallel sides and the other two sides equal in length but not parallel is called an isosceles trapezoid).
(Cold Gangsong, Contributed)
|
5. The smallest value of $n$ sought is 17.
First, prove: When $n=17$, the conclusion holds.
Proof by contradiction.
Assume there exists a method to three-color the vertices of a regular 17-gon such that no four vertices of the same color form the vertices of an isosceles trapezoid.
Since $\left[\frac{17-1}{3}\right]+1=6$, there must exist a set of 6 vertices of the same color. Without loss of generality, assume these are yellow. Connecting these 6 points pairwise, we get $\mathrm{C}_{6}^{2}=15$ line segments. Since these line segments can have only $\left[\frac{17}{2}\right]=8$ different lengths, one of the following two scenarios must occur:
(1) There are three line segments of the same length.
Note that $3 \times 17$ does not allow these three line segments to share a common vertex. Therefore, there exist two line segments with no common vertices. The four vertices of these two line segments satisfy the problem's requirements, leading to a contradiction.
(2) There are 7 pairs of line segments of the same length.
By assumption, each pair of line segments of the same length must share a common yellow vertex; otherwise, we could find four yellow vertices that satisfy the problem's requirements. According to the pigeonhole principle, there must be two pairs of line segments that share the same yellow vertex. The other four vertices of these four line segments must form the vertices of an isosceles trapezoid, leading to a contradiction.
Therefore, when $n=17$, the conclusion holds.
Next, construct a coloring method for $n \leqslant 16$ that does not satisfy the problem's requirements. Let $A_{1}, A_{2}, \cdots, A_{n}$ represent the vertices of the regular $n$-gon (in clockwise order), and $M_{1}, M_{2}, M_{3}$ represent the sets of vertices of the three colors.
$$
\begin{array}{l}
\text { When } n=16 \text {, let } \\
M_{1}=\left\{A_{5}, A_{8}, A_{13}, A_{14}, A_{16}\right\}, \\
M_{2}=\left\{A_{3}, A_{6}, A_{7}, A_{11}, A_{15}\right\}, \\
M_{3}=\left\{A_{1}, A_{2}, A_{4}, A_{9}, A_{10}, A_{12}\right\} .
\end{array}
$$
For $M_{1}$, the distances from $A_{14}$ to the other 4 vertices are all different, and these 4 vertices form the vertices of a rectangle, not an isosceles trapezoid. For $M_{3}$, the 6 vertices are the endpoints of 3 diameters, so any 4 vertices either form the vertices of a rectangle or a non-isosceles quadrilateral.
$$
\begin{array}{l}
\text { When } n=15 \text {, let } \\
M_{1}=\left\{A_{1}, A_{2}, A_{3}, A_{5}, A_{8}\right\}, \\
M_{2}=\left\{A_{6}, A_{9}, A_{13}, A_{14}, A_{15}\right\}, \\
M_{3}=\left\{A_{4}, A_{7}, A_{10}, A_{11}, A_{12}\right\},
\end{array}
$$
Each $M_{i}$ contains no 4 points that form the vertices of an isosceles trapezoid.
$$
\begin{array}{l}
\text { When } n=14 \text {, let } \\
M_{1}=\left\{A_{1}, A_{3}, A_{8}, A_{10}, A_{14}\right\}, \\
M_{2}=\left\{A_{4}, A_{5}, A_{7}, A_{11}, A_{12}\right\}, \\
M_{3}=\left\{A_{2}, A_{6}, A_{9}, A_{13}\right\},
\end{array}
$$
Each $M_{i}$ contains no 4 points that form the vertices of an isosceles trapezoid.
When $n=13$, let
$$
\begin{array}{l}
M_{1}=\left\{A_{5}, A_{6}, A_{7}, A_{10}\right\}, \\
M_{2}=\left\{A_{1}, A_{8}, A_{11}, A_{12}\right\}, \\
M_{3}=\left\{A_{2}, A_{3}, A_{4}, A_{9}, A_{13}\right\},
\end{array}
$$
Each $M_{i}$ contains no 4 points that form the vertices of an isosceles trapezoid.
In the above cases, removing vertex $A_{13}$ and keeping the coloring method unchanged gives a coloring method for $n=12$; then, removing vertex $A_{12}$ gives a coloring method for $n=11$; and removing vertex $A_{11}$ gives a coloring method for $n=10$.
When $n \leqslant 9$, we can ensure that the number of vertices of each color is less than 4, so there are no 4 vertices of the same color that form the vertices of an isosceles trapezoid.
The examples constructed above show that $n \leqslant 16$ does not satisfy the problem's requirements.
In conclusion, the smallest value of $n$ sought is 17.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. As shown in Figure 4, in $\triangle A B C$, $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=3, C D=2$, then $S_{\triangle A B C}$ $=$ $\qquad$ .
|
10.15 .
Solution 1: Let $A B=c, A C=b$. Then
$$
\begin{aligned}
A D^{2} & =c^{2}-9=b^{2}-4, b=\sqrt{c^{2}-5}, \\
A D & =\sqrt{c^{2}-9} .
\end{aligned}
$$
As shown in Figure 11, draw $C E \perp A B$ at point $E$, $\triangle A E C$ is an isosceles right triangle. Thus,
$$
\begin{array}{l}
E C=\frac{\sqrt{2}}{2} A C=\frac{\sqrt{2}}{2} b \\
=\frac{\sqrt{2}}{2} \sqrt{c^{2}-5}, \\
2 S_{\triangle A B C}=\frac{\sqrt{2}}{2} c \cdot \sqrt{c^{2}-5}=5 \sqrt{c^{2}-9} .
\end{array}
$$
Rearranging gives $\left(c^{2}-10\right)\left(c^{2}-45\right)=0$.
Solving yields $c_{1}=\sqrt{10}$, so $h=1$ or $h=6$.
When $h=1$, $\angle B A D>\angle A B C, \angle C A D>$ $\angle A C B$, so, $\angle B A C>\angle A B C+\angle A C B$.
Thus, $\angle B A C>\frac{1}{2} \times 180^{\circ}=90^{\circ}$, which contradicts $\angle B A C$ $=45^{\circ}$.
Therefore, $A D=6$.
So, $S_{\triangle A B C}=\frac{1}{2} \times 5 \times 6=15$.
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In a dormitory of a school, there are several students, one of whom serves as the dorm leader. During New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator. Each dormitory administrator also gives a greeting card back to the dorm leader. In this way, a total of 51 greeting cards were used. How many students live in this dormitory?
|
14. Let there be $x$ students in this dormitory, and $y$ administrators in the dormitory building $\left(x, y \in \mathbf{N}_{+}\right)$. According to the problem, we have
$$
x(x-1)+x y+y=51 \text{. }
$$
Simplifying, we get $x^{2}+(y-1) x+y-51=0$.
Thus, $\Delta=(y-1)^{2}-4(y-51)$
$$
=y^{2}-6 y+205=(y-3)^{2}+196 \text{. }
$$
Since $x \in \mathbf{N}_{+}$, $\Delta$ must be a perfect square.
Let $(y-3)^{2}+196=k^{2}(k \in \mathbf{N})$. Then
$$
(y-3+k)(y-3-k)=-196 \text{, }
$$
where $y-3+k$ and $y-3-k$ have the same parity, and $y-3+k \geqslant y-3-k$.
So, $\left\{\begin{array}{l}y-3+k=2, \\ y-3-k=-98\end{array}\right.$
or $\left\{\begin{array}{l}y-3+k=98, \\ y-3-k=-2\end{array}\right.$
or $\left\{\begin{array}{l}y-3+k=14, \\ y-3-k=-14 .\end{array}\right.$
From equation set (1), we get $y=-45$, which is not valid, so we discard it;
From equation set (2), we get $y=51$, in which case the original equation becomes $x^{2}+50 x=0$, solving for $x_{1}=-50, x_{2}=0$, both of which are not valid, so we discard them;
From equation set (3), we get $y=3$, in which case the original equation becomes $x^{2}+2 x-48=0$, solving for $x_{1}=-8$ (not valid, discard), $x_{2}=6$.
Answer: There are 6 students in this dormitory.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive integers, and $a_{1}<a_{2}<\cdots<a_{n} \leqslant 2007$, to ensure that there always exist four distinct numbers $a_{i} γ a_{j} γ a_{k} γ a_{l}$, such that $a_{i}+a_{j}=a_{k}+a_{l}=a_{n}$, what is the minimum value of $n$? And explain the reason.
|
15. Let $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005$, $a_{1004}=2007$, a total of 1004 odd numbers. Clearly, the sum of any two numbers is not equal to 2007.
If we add the number 2006 to the above 1004 numbers, i.e., $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005, a_{1004}=$ $2006, a_{1000}=2007$, at this point, there exists only one pair of numbers $a_{1}=1, a_{1004}=2006$, whose sum is
$$
a_{1006}=2007 \text {. }
$$
Therefore, the minimum value of $n$ is no less than 1006.
Next, we prove that when $n \geqslant 1006$, there must exist four numbers that satisfy the condition.
When $a_{n}=2007$, since
$$
2007=1+2006=2+2005=\cdots=1007+1004 \text {, }
$$
this indicates that 2007 can only be decomposed into 1003 different sums of two positive integers. Therefore, when $n \geqslant 1006$, i.e., $n-1$ $\geqslant 1005$, among the at least 1005 numbers $a_{1}, a_{2}, \cdots, a_{n-1}$ excluding $a_{n}=2007$, at least two of the 1003 different sums $a_{i}+a_{j} γ a_{k}+a_{l}$ and their four addends $a_{i}$ γ $a_{j} γ a_{k} γ a_{l}$ are included, which are the four positive integers required by the problem.
When $a_{n}<2007$, if $a_{n}=2 m-1$, then $a_{n}$ can be expressed as $m$ different sums of two different positive integers; if $a_{n}=2 m$, then $a_{n}$ can be expressed as $m-1$ different sums of two different positive integers. Since $a_{n}<2007$, we have $m \leqslant 1003$. Excluding $a_{n}$, among the at least 1005 numbers $a_{1}, a_{2}, \cdots, a_{n-1}$, at least two of the no more than 1003 different sums and their four addends are included, which are the four numbers required by the problem.
In summary, the minimum value of $n$ is 1006.
|
1006
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) Each point on the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist 4 points of different colors that are concyclic.
(Zhang Limin, problem contributor)
|
Obviously, $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arcs $\overparen{A B}$ (including $A$ but not $B$), $\overparen{B C}$ (including $B$ but not $C$), and $\overparen{C A}$ (including $C$ but not $A$) with colors $1, 2,$ and $3$ respectively, and color all other points in the plane with color $4$. This satisfies the conditions and there do not exist four points of different colors on the same circle.
Therefore, $n \neq 4, n \geqslant 5$.
When $n=5$, by condition (2), there exists a line $l$ on which there are exactly two colors of points. Without loss of generality, assume that the points on line $l$ are only of colors $1$ and $2$. By condition (1), there exist points $A, B, C$ of colors $3, 4, 5$ respectively that are not collinear. Let the circle passing through points $A, B, C$ be $\odot O$.
If $\odot O$ intersects line $l$, then there exist four points of different colors on the same circle;
If $\odot O$ is disjoint from line $l$ and $\odot O$ has points of colors $1$ and $2$, then there exist four points of different colors on the same circle;
If $\odot O$ is disjoint from line $l$ and $\odot O$ has no points of colors $1$ and $2$, as shown in Figure 2, draw a perpendicular from $O$ to line $l$ intersecting $l$ at point $D$. Assume the color of $D$ is $1$, and the perpendicular intersects $\odot O$ at points $E$ and $S$. Assume the color of $E$ is $3$, and consider a point $F$ on line $l$ with color $2$. The line $FS$ intersects $\odot O$ at point $G$. Since $EG \perp GF$, points $D, E, F, G$ are concyclic.
If $G$ is not a point of color $3$, then there exist four points of different colors on the same circle;
If $G$ is a point of color $3$, then one of $B$ or $C$ must be different from $S$ (assume it is $B$). The line $SB$ intersects line $l$ at point $H$. Since $EB \perp BH$, points $B, E, D, H$ are concyclic.
If $H$ is a point of color $2$, then $B, H, D, E$ are four points of different colors on the same circle;
If $H$ is a point of color $1$, since
$$
SB \cdot SH = SE \cdot SD = SG \cdot SF,
$$
points $B, H, F, G$ are concyclic.
Thus, $B, H, F, G$ are four points of different colors on the same circle.
In summary, when $n=5$, there exist four points of different colors on the same circle.
Therefore, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (25 points) Given
$$
f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x .
$$
(1) Solve the equation: $f(x)=0$;
(2) Find the number of subsets of the set
$$
M=\left\{n \mid f\left(n^{2}-214 n-1998\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
(Li Tiehan, problem contributor)
|
(1) For any $0 < x_1 < x_2$, we have $\frac{x_1 + 1}{x_2 + 1} > \frac{x_1}{x_2}$, thus $\lg \frac{x_1 + 1}{x_2 + 1} > \lg \frac{x_1}{x_2}$. Therefore,
$$
\begin{array}{l}
f(x_1) - f(x_2) > \lg \frac{x_1}{x_2} - \log \frac{x_1}{x_2} \\
= \lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9}.
\end{array}
$$
Since $0 < \frac{x_1}{x_2} < 1$, we have $\lg \frac{x_1}{x_2} < 0$. Therefore,
$$
\lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9} > \lg \frac{x_1}{x_2} - \lg \frac{x_1}{x_2} = 0.
$$
Hence, $f(x)$ is a decreasing function on $(0, +\infty)$.
Noting that $f(9) = 0$, we have:
When $x > 9$, $f(x) < f(9) = 0$.
When $x < 9$, $f(x) > f(9) = 0$.
Therefore, $f(x) = 0$ has and only has one root $x = 9$.
$$
\begin{array}{l}
\text{(2) From } f(n^2 - 214n - 1998) \geq 0 \\
\Rightarrow f(n^2 - 214n - 1998) \geq f(9). \\
\text{Then }\left\{\begin{array}{l}
n^2 - 214n - 1998 \leq 9, \\
n^2 - 214n - 1998 > 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
n^2 - 214n - 2007 \leq 0, \\
n^2 - 214n - 1998 > 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
(n - 223)(n + 9) \leq 0, \\
(n - 107)^2 > 1998 + 107^2 = 13447 > 115^2
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leq n \leq 223, \\
n > 222 \text{ or } n < -8
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leq n \leq 223, \\
n \geq 223 \text{ or } n \leq -9.
\end{array}\right. \\
\end{array}
$$
Thus, $n = 223$ or $n = -9$.
Hence, $M = \{-9, 223\}$.
Therefore, the number of subsets of $M$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Seven, (25 points) Let $n$ be a positive integer, $a=[\sqrt{n}]$ (where $[x]$ denotes the greatest integer not exceeding $x$). Find the maximum value of $n$ that satisfies the following conditions:
(1) $n$ is not a perfect square;
(2) $a^{3} \mid n^{2}$.
(Zhang Tongjun
Zhu Yachun, problem contributor)
|
Seven, from (1) we get $a<\sqrt{n}<a+1$, then $a^{2}<n<a^{2}+2 a+1$,
that is
$a^{2}+1 \leqslant n \leqslant a^{2}+2 a$.
Let $n=a^{2}+t(t \in\{1,2, \cdots, 2 a\})$.
From (2) we have
$a^{3}\left|\left(a^{4}+2 a^{2} t+t^{2}\right) \Rightarrow a^{2}\right| t^{2} \Rightarrow a \mid t$.
Furthermore, $a^{3} \mid \left(a^{4}+2 a^{2} t+t^{2}\right) \Rightarrow a^{3} \mid t^{2}$.
Let $t^{2}=k a^{3}$, then $t=a \sqrt{k a}$.
Since $t, a, k \in \mathbf{N}_{+}$, we have $\sqrt{k a} \in \mathbf{N}_{+}$.
From $t \in\{1,2 ; \cdots, 2 a\}$, we have $t=a \sqrt{k a} \leqslant 2 a$, that is $\sqrt{k a} \leqslant 2$.
Therefore, $\sqrt{k a}=1$ or $2, k a \leqslant 4, a \leqslant 4$.
Since $n=a^{2}+t$, and $a \leqslant 4, t \leqslant 2 a$, we can let $a=4, t=2 a=8$. Then $n=a^{2}+t=16+8=24$ is the maximum.
Upon verification, $n=24$ satisfies conditions (1) and (2).
Therefore, the maximum value of $n$ is 24.
|
24
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in Figure 2, in the polyhedron ABCDEF, it is known that quadrilateral $ABCD$ is a square with side length 3, $EF$ $/ / AB, EF=\frac{3}{2}$. If the volume of the polyhedron is $\frac{15}{2}$, then the distance between $EF$ and $AC$ is
|
9.2 .
Take the midpoints of $A B$ and $C D$ as $M$ and $N$, and connect $F M$, $F N$, and $M N$.
Since $E F / / A M$ and $E F = \frac{3}{2} = A M$, we know that the polyhedron $A D E - M N F$ is a triangular prism.
Let the distance between $E F$ and $A C$ be $h$. From
$$
V_{A D E-M N F} + V_{F-B C M M} = \frac{15}{2} \text {, }
$$
we get $\frac{1}{2} \cdot h \cdot \frac{9}{2} + \frac{1}{3} \cdot h \cdot \frac{9}{2} = \frac{15}{2}$.
Solving for $h$, we get $h = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given five points $A, B, C, D, E$ on a plane, no three of which are collinear. By connecting these points with 4 lines, each point is an endpoint of at least one line segment. The number of different connection methods is $\qquad$ kinds.
|
10.135.
In Figure 5, 4 connection methods meet the requirements of the problem (the diagram only represents the connection form between points and lines, without considering the position of the points).
Case (1), based on the choice of the central point, there are 5 connection methods;
Case (2), it can be considered as the permutation of 5 points $A, B, C, D, E$, but one permutation and its reverse permutation are the same, and they are in a one-to-one correspondence, so there are $\frac{5!}{2}=60$ connection methods;
Case (3), first, the choice of the branching point has 5 options, secondly, the selection of the two points for the fork has $\mathrm{C}_{4}^{2}=6$ options, and finally, the order of the remaining two connected points is different, with 2! options. In total, there are $5 \times 6 \times 2=60$ methods;
Case (4), choosing three points to construct a triangle, there are $\mathrm{C}_{5}^{3}=$ 10 options.
In total, there are $5+60+60+10=135$ connection methods.
|
135
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given $\tan \alpha+\log _{2}(2 \tan \alpha-6)=8$, $\tan \beta+2^{\tan \beta-1}=5$. Then $\tan \alpha+\tan \beta$ equals $\qquad$ .
|
11.8 .
Let $t=\log _{2}(2 \tan \alpha-6)$. Then $t+2^{t-1}=5$. Also, $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, and $\tan \beta+2^{\tan \beta-1}=5$, so $\tan \beta=t$.
Therefore, $\tan \alpha+\tan \beta=\tan \alpha+t=8$.
Hence, $\tan \alpha+\tan \beta=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with the left vertex $A$ and the right focus $F$. Let $P$ be any point on the hyperbola in the first quadrant. If $\angle P F A=2 \angle F A P$ always holds, then the eccentricity $e$ of the hyperbola is
|
12.2.
From the problem, we can take a point $P$ on the hyperbola such that $P F$ is perpendicular to the $x$-axis, giving $P(c, y)$.
Then $\frac{c^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Hence, $y^{2}=\frac{b^{4}}{a^{2}}$.
Since $y>0$, we have $y=\frac{b^{2}}{a}=\frac{c^{2}-a^{2}}{a}$.
Given that $\angle P F A=2 \angle F A P=\frac{\pi}{2}$, we know that $\triangle A F P$ is an isosceles triangle, so $A F=P F$, which means $a+c=\frac{c^{2}-a^{2}}{a}$.
Rearranging gives $c^{2}-a c-2 a^{2}=0$, or
$e^{2}-e-2=0$.
Since $e>1$, solving gives $e=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1} a_{n}+a_{n+1}-2 a_{n}=0 \text {. }
$$
For any positive integer $n$, we have
$\sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)<M$ (where $M$ is a constant and an integer).
Find the minimum value of $M$.
|
Three, 13. From the problem, for $n \in \mathbf{N}_{+}, a_{n} \neq 0$, and
\[
\frac{1}{a_{n+1}}=\frac{1}{2}+\frac{1}{2 a_{n}},
\]
which means
\[
\frac{1}{a_{n+1}}-1=\frac{1}{2}\left(\frac{1}{a_{n}}-1\right).
\]
Given $a_{1}=2$, we have $\frac{1}{a_{1}}-1=-\frac{1}{2}$.
Therefore, the sequence $\left\{\frac{1}{a_{n}}-1\right\}$ is a geometric sequence with the first term $-\frac{1}{2}$ and common ratio $\frac{1}{2}$. Thus,
\[
\frac{1}{a_{n}}-1=-\frac{1}{2} \times\left(\frac{1}{2}\right)^{n-1}=-\left(\frac{1}{2}\right)^{n},
\]
which implies
\[
a_{n}=\frac{2^{n}}{2^{n}-1}.
\]
Hence,
\[
a_{i}\left(a_{i}-1\right)=\frac{2^{i}}{\left(2^{i}-1\right)^{2}}(i=1,2, \cdots, n).
\]
For $i \geqslant 2$, we have
\[
\begin{array}{l}
a_{i}\left(a_{i}-1\right)=\frac{2^{i}}{\left(2^{i}-1\right)^{2}}<\frac{2^{i}}{\left(2^{i}-1\right)\left(2^{i}-2\right)} \\
=\frac{2^{i-1}}{\left(2^{i}-1\right)\left(2^{i-1}-1\right)}=\frac{1}{2^{i-1}-1}-\frac{1}{2^{i}-1} . \\
\text { Therefore, } \sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)=\sum_{i=1}^{n} \frac{2^{i}}{\left(2^{i}-1\right)^{2}} \\
<\frac{2^{1}}{\left(2^{1}-1\right)^{2}}+\sum_{i=2}^{n}\left(\frac{1}{2^{i-1}-1}-\frac{1}{2^{i}-1}\right) \\
=3-\frac{1}{2^{n}-1}<3 . \\
\text { Also, } \sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)=\sum_{i=1}^{n} \frac{2^{i}}{\left(2^{i}-1\right)^{2}} . \\
\geqslant \frac{2^{1}}{\left(2^{1}-1\right)^{2}}=2 .
\end{array}
\]
Therefore, the minimum value of $M$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the sequence
$$
b_{n}=\frac{1}{3 \sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right](n=0,1, \cdots) \text {. }
$$
(1) For what values of $n$ is $b_{n}$ an integer?
(2) If $n$ is odd and $2^{-\frac{2 n}{3}} b_{n}$ is an integer, what is $n$?
|
15. (1) From
$$
\begin{array}{l}
b_{n}=\frac{2}{3 \sqrt{3}}\left[n \sqrt{3}+\mathrm{C}_{n}^{3}(\sqrt{3})^{3}+\mathrm{C}_{n}^{6}(\sqrt{3})^{5}+\cdots\right] \\
=2\left(\frac{n}{3}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5} \cdot 3+\cdots\right),
\end{array}
$$
$b_{n}$ is an integer if and only if $31 n$.
(2) First, $b_{n}$ is an integer, $31 n$. Second,
$$
\begin{array}{l}
\frac{1}{\sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right] \\
= \frac{1}{\sqrt{3}}[(1+\sqrt{3})-(1-\sqrt{3})]\left[(1+\sqrt{3})^{n-1}+\right. \\
\left.(1+\sqrt{3})^{n-2}(1-\sqrt{3})+\cdots+(1-\sqrt{3})^{n-1}\right] \\
= 2\left\{\left[(1+\sqrt{3})^{n-1}+(1-\sqrt{3})^{n-1}\right]-\right. \\
2\left[(1+\sqrt{3})^{n-3}+(1-\sqrt{3})^{n-3}\right]+\cdots+ \\
\left.(-2)^{\frac{n-1}{2}}\right\} \\
= 2\left\{2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-1}{2}}+(2-\sqrt{3})^{\frac{n-1}{2}}\right]-\right. \\
2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-3}{2}}+(2-\sqrt{3})^{\frac{n-3}{2}}\right]+\cdots+ \\
\left.(-2)^{\frac{n-1}{2}}\right\} .
\end{array}
$$
Since $(2+\sqrt{3})^{k}+(2-\sqrt{3})^{k}$
$$
=2\left(2^{k}+\mathrm{C}_{k}^{2} 2^{k-2} \cdot 3+\cdots\right)
$$
is even, therefore,
$$
\begin{array}{l}
2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-1}{2}}+(2-\sqrt{3})^{\frac{n-1}{2}}\right]- \\
2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-3}{2}}+\cdots+(2-\sqrt{3})^{\frac{n-3}{2}}\right]+\cdots+ \\
(-2)^{\frac{n-1}{2}}
\end{array}
$$
is an integer, and the power of 2 is the lowest in the last term $(-2)^{\frac{n-1}{2}}$, i.e., the power of 2 in $\frac{1}{\sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right]$ is $\frac{n-1}{2}+1=\frac{n+1}{2}$.
Since $\frac{n+1}{2} \leqslant \frac{2 n}{3}$, equality holds only when $n \geqslant 3$, so, $2^{-\frac{2 n}{3}} b_{n}$ is an integer if and only if $n=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 2, $C D$ is the altitude on the hypotenuse $A B$ of Rt $\triangle A B C$, $I_{1} γ I_{2} γ I_{3} γ$ are the incenter of $\triangle A B C γ \triangle A C D$ γ $\triangle B C D$ respectively, $A C = 20, B C = 15$. Then the area of $\triangle I_{1} I_{2} I_{3}$ is ( ).
(A) 4
(B) 4.5
(C) 5
(D) 5.5
|
6.C.
As shown in Figure 6, connect $A I_{1}$,
$B I_{1}$, $I_{2} D$, $I_{3} D$. Draw $I_{1} G \perp A B$ at point $G$,
$I_{2} E \perp A B$ at point $E$, and $I_{3} F \perp A B$
at point $F$. In the right triangle $\triangle A B C$, it is easy to see that
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{20^{2}+15^{2}}=25 \text {. }
$$
Since $C D \perp A B$, by the projection theorem, we have
$$
A D=\frac{A C^{2}}{A B}=\frac{20^{2}}{25}=16 \text {. }
$$
Thus, $B D=A B-A D=9$,
$$
C D=\sqrt{A C^{2}-A D^{2}}=\sqrt{20^{2}-16^{2}}=12 \text {. }
$$
Since $I_{1}$, $I_{2}$, and $I_{3}$ are the incenter of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ respectively, $I_{2}$ lies on $A I_{1}$, $I_{3}$ lies on $B I_{1}$, and $I_{1} G$, $I_{2} E$, $I_{3} F$ are the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ respectively. $D I_{2}$ and $D I_{3}$ are the angle bisectors of $\angle A D C$ and $\angle B D C$ respectively. Therefore,
$$
\begin{array}{l}
\angle A D I_{2}=\angle I_{2} D C=45^{\circ}, \\
\angle B D I_{3}=\angle C D I_{3}=45^{\circ} \\
\Rightarrow \angle I_{2} D I_{3}=\angle I_{2} D C+\angle C D I_{3}=90^{\circ} \\
\Rightarrow I_{2} D \perp I_{3} D .
\end{array}
$$
Since $I_{1} G$ is the inradius of the right triangle $\triangle A B C$, we have
$$
I_{1} G=\frac{1}{2}(A C+B C-A B)=5 \text {. }
$$
Similarly, $I_{2} E=4$, $I_{3} F=3$.
Thus, $I_{2} D=\frac{I_{2} E}{\sin \angle A D I_{2}}=\frac{4}{\sin 45^{\circ}}=4 \sqrt{2}$.
Similarly, $I_{3} D=3 \sqrt{2}$.
Therefore, $S_{\triangle I_{1} I_{2} I_{3}}$
$$
\begin{aligned}
= & S_{\triangle A B B_{1}}-S_{\triangle \triangle I_{2}}-S_{\triangle B D I_{3}}-S_{\triangle D I_{2} I_{3}} \\
= & \frac{1}{2} A B \cdot I_{1} G-\frac{1}{2} A D \cdot I_{2} E- \\
& \frac{1}{2} B D \cdot I_{3} F-\frac{1}{2} I_{2} D \cdot I_{3} D \\
= & 5 .
\end{aligned}
$$
|
5
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Given that $n$ and $k$ are positive integers, and satisfy the inequality
$$
\frac{1}{7}<\frac{n-k}{n+k}<\frac{63}{439} .
$$
If for a given positive integer $n$, there is only one positive integer $k$ that makes the inequality true. Find the maximum and minimum values of all positive integers $n$ that meet the requirement.
|
From the known inequality, we have
$$
\frac{188}{251} n < k < \frac{189}{251} n.
$$
When $n=252, 253, 254$, substituting into equation (1) respectively, we get
$$
\begin{array}{l}
188 \frac{188}{251} < k < 189, \\
189 \frac{125}{251} < k < 189 \frac{3}{4}, \\
190 \frac{62}{251} < k < 190 \frac{1}{2},
\end{array}
$$
none of which meet the requirements.
When $n=255$, $190 \frac{250}{251} < k < 191 \frac{1}{4}$.
Thus, there is a unique positive integer $k=191$.
Therefore, $n=255$ is the smallest value that meets the condition.
Hence, the maximum value of all positive integers $n$ that meet the requirement is 2008, and the minimum value is 191.
|
2008
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $a, b$ are constants. If the parabola $C$:
$$
y=\left(t^{2}+t+1\right) x^{2}-2(a+t)^{2} x+t^{2}+3 a t+b
$$
passes through the fixed point $P(1,0)$ for any real number $t$, find the value of $t$ when the chord intercepted by the parabola $C$ on the $x$-axis is the longest.
|
$=1.2$.
Substituting $P(1,0)$ into the equation of the parabola $C$ yields
$$
\left(t^{2}+t+1\right)-2(a+t)^{2}+\left(t^{2}+3 a t+b\right)=0 \text {, }
$$
which simplifies to $t(1-a)+\left(1-2 a^{2}+b\right)=0$,
holding for all $t$.
Thus, $1-a=0$, and $1-2 a^{2}+b=0$.
Solving these, we get $a=1, b=1$.
Substituting into the equation of the parabola $C$ gives
$$
y=\left(t^{2}+t+1\right) x^{2}-2(1+t)^{2} x+\left(t^{2}+3 t+1\right) \text {. }
$$
Setting $y=0$, we have
$$
\left(t^{2}+t+1\right) x^{2}-2(1+t)^{2} x+\left(t^{2}+3 t+1\right)=0 \text {. }
$$
Let the roots of this equation be $x_{1}$ and $x_{2}$, then
$$
\begin{array}{l}
|A B|=\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\left|\frac{2 t}{t^{2}+t+1}\right| .
\end{array}
$$
Noting that the discriminant of this quadratic equation $\Delta=4 t^{2}>0$, to find the maximum value of $|A B|$, we categorize by $t$.
(1) When $t>0$,
$$
|A B|=\frac{2 t}{t^{2}+t+1}=\frac{2}{t+\frac{1}{t}+1} \leqslant \frac{2}{3} \text {; }
$$
(2) When $t<0$,
$$
\begin{array}{l}
|A B|=-\frac{2 t}{t^{2}+t+1}=\frac{2}{-t-\frac{1}{t}-1} \\
\leqslant \frac{2}{-1+2 \sqrt{(-t) \frac{1}{-t}}}=2 .
\end{array}
$$
When $t=-1$, $|A B|_{\text {max }}=2$.
From (1) and (2), we know that when $t=-1$, the chord length $|A B|$ reaches its maximum value of 2.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. To make $p=x^{4}+6 x^{3}+11 x^{2}+3 x+32$ a perfect square of an integer, then the integer $x$ has $\qquad$ solutions.
|
2.0 .
From the problem, we know that $p \equiv x^{4}-x^{2}+2(\bmod 3)$.
Since $x^{2} \equiv 0,1(\bmod 3)$, thus,
$$
p \equiv x^{2}\left(x^{2}-1\right)+2 \equiv 2(\bmod 3) \text {. }
$$
Therefore, $p$ is not a perfect square.
|
2.0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the expansion of $(\sqrt[5]{3}+\sqrt[3]{5})^{100}$, there are $\qquad$ terms that are rational numbers.
|
4.7.
The general term of the binomial theorem expansion is
$$
\mathrm{C}_{100}^{r} 3^{\frac{1}{5}(100-r)} 5^{\frac{r}{3}}(0 \leqslant r \leqslant 100),
$$
it is a rational number if and only if
$$
3|r, 5|(100-r) \Rightarrow 15 \mid r \text {. }
$$
It is easy to see that there are 7 such $r$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $x, y \in \mathbf{R}$. Then
$$
\cos (x+y)+2 \cos x+2 \cos y
$$
the minimum value is $\qquad$ $\therefore$.
|
$$
\begin{array}{l}
\text { 5. }-3 \text {. } \\
\cos (x+y)+2 \cos x+2 \cos y+3 \\
=\left(1+\cos \frac{x+y}{2}\right)^{2}\left(1+\cos \frac{x-y}{2}\right)+ \\
\quad\left(1-\cos \frac{x+y}{2}\right)^{2}\left(1-\cos \frac{x-y}{2}\right) \\
\geqslant 0 .
\end{array}
$$
It is clear that equality can be achieved.
Therefore, the minimum value of the given algebraic expression is -3.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, let $P$ be a point inside the convex quadrilateral $A B C D$. Draw perpendiculars from $P$ to $A B$, $B C$, $C D$, and $D A$, with the feet of the perpendiculars being $E$, $F$, $G$, and $H$, respectively. Given that $A H=3$, $H D=4$, $D G=1$, $G C=5$, $C F=6$, $F B=4$, and $B E-A E=1$. Then the perimeter of quadrilateral $A B C D$ is $\qquad$
|
6.34.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
P A^{2}-A H^{2}=P D^{2}-D H^{2}, \\
P D^{2}-D G^{2}=P C^{2}-C G^{2}, \\
P C^{2}-C F^{2}=P B^{2}-B F^{2}, \\
P B^{2}-B E^{2}=P A^{2}-A E^{2} .
\end{array}
$$
Adding the above four equations, we get
$$
\begin{array}{l}
A H^{2}+D G^{2}+C F^{2}+B E^{2} \\
=A E^{2}+B F^{2}+C G^{2}+D H^{2} \\
\Rightarrow 9+1+36+B E^{2}=A E^{2}+16+25+16 \\
\Rightarrow(B E-A E)(B E+A E)=11 .
\end{array}
$$
Since $B E-A E=1$, we have $B E+A E=11$, which means $A B=11$.
Therefore, the perimeter of quadrilateral $A B C D$ is 34.
|
34
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Positive integers $x$ not exceeding 1000, such that the sum of the digits of both $x$ and $x+1$ are odd. Then the number of positive integers $x$ that satisfy the condition is $\qquad$.
|
8.46.
Obviously, $x \neq 1000$.
Let $x=\overline{a b c}$, where $a, b, c \in\{0,1, \cdots, 9\}$, and not all are zero. $S(x)=a+b+c$ is the sum of the digits of $x$.
(1) If $c \neq 9$, then
$$
\begin{array}{l}
S(x)=a+b+c, \\
S(x+1)=a+b+c+1 ;
\end{array}
$$
(2) If $c=9, b \neq 9$, then
$$
S(x)=a+b+9, S(x+1)=a+b+1 \text {; }
$$
(3) If $b=c=9, a \neq 9$, then
$$
S(x)=a+18, S(x+1)=a+1 \text {; }
$$
(4) If $a=b=c=9$, then
$$
S(x)=27, S(x+1)=1 \text {. }
$$
From this, we can see that $S(x)$ and $S(x+1)$ are both odd only in cases (2) and (4).
In case (2), $a+b$ is even, so $a$ and $b$ are of the same parity. This results in 45 $x$ values that satisfy the condition.
In case (4), only one $x=999$ satisfies the condition.
In total, there are 46 $x$ values that satisfy the condition.
|
46
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $k$ is a positive integer not exceeding 50, such that for any positive integer $n, 2 \times 3^{6 n}+k \times 2^{3 n+1}-1$ is always divisible by 7. Then the number of such positive integers $k$ is $\qquad$.
|
9.7.
$$
\begin{array}{l}
2 \times 3^{6 n}+k \times 2^{3 n+1}-1 \\
=2 \times 27^{2 n}+2 k \times 8^{n}-1 \\
\equiv 2 \times(-1)^{2 n}+2 k-1 \\
\equiv 2 k+1(\bmod 7) .
\end{array}
$$
But $2 \times 3^{6 n}+k \times 2^{3 n+1}-1 \equiv 0(\bmod 7)$, then
$$
2 k+1 \equiv 0(\bmod 7) \text {, }
$$
which means $2 k+1=7 m$ (where $m$ is an odd number).
Since $1 \leqslant k \leqslant 50$, we have $3 \leqslant 7 m \leqslant 101$.
Thus, $m=1,3, \cdots, 13$, and the corresponding $k=3,10, \cdots, 45$, a total of 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (20 points) In a box, there are a total of 88 small balls in three colors: red, yellow, and black. It is known that by randomly taking out 24 balls, it can be guaranteed that at least 10 of the balls are of the same color. Under the condition that this is true, regardless of how the balls of each color are distributed, what is the minimum number of balls that must be randomly taken out to ensure that at least 20 balls are of the same color?
|
First, prove that taking out only 43 balls is not enough.
In fact, when there are 42 red balls, 41 yellow balls, and 5 black balls in the box, if you take any 24 balls, then the number of red and yellow balls is at least $24-5=19$, so there must be at least 10 red or yellow balls; but taking 19 red balls, 19 yellow balls, and 5 black balls, a total of 43 balls, there are no 20 balls of the same color.
Next, prove that taking 44 balls from the box, there must be 20 balls of the same color. Let the number of red, yellow, and black balls in the box be $x$, $y$, and $z$ respectively, and assume $x \geqslant y \geqslant z$.
(1) If $z \leqslant 5$, then among the 44 balls taken out, the number of red and yellow balls is at least $44-5=39$, so there must be at least 20 red or yellow balls.
(2) If $z=6$, when $y \leqslant 8$, the number of red balls among the 44 balls is greater than or equal to $44-8-6>20$; when $y \geqslant 9$, taking 9 red balls, 9 yellow balls, and 6 black balls, there are no 10 balls of the same color among the 24 balls, which does not meet the conditions of the problem.
(3) If $z=7$, when $y \geqslant 8$, taking 9 red balls, 8 yellow balls, and 7 black balls, there are no 10 balls of the same color among the 24 balls, which does not meet the conditions of the problem; when $y=7$, the number of red balls among the 44 balls is greater than or equal to $44-7-7>20$.
(4) If $z \geqslant 8$, then taking 8 red balls, 8 yellow balls, and 8 black balls, it is known that the conditions of the problem are not met.
In summary, at least 44 balls must be taken out to ensure that there are 20 balls of the same color.
(Xiong Bin, Li Deyuan, Gu Hongda, Liu Hongkun, Ye Shengyang)
|
44
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Try to find all positive integers $k$, such that for any positive integers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a$, $b$, and $c$.
(2002, Girls' Mathematical Olympiad)
|
Explanation: First, from the inequality relationship between $a b+b c+c a$ and $a^{2}+b^{2}+c^{2}$, we derive $k>5$. Then, by constructing an example, we find that $k \leqslant 6$. Therefore, since $k \in \mathbf{Z}_{+}$, we conclude that $k=6$. Finally, we provide the proof.
Notice that for any positive real numbers $a, b, c$, we have
$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a.
$$
On one hand, from
$$
\begin{array}{l}
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) \\
\geqslant 5(a b+b c+c a),
\end{array}
$$
we get $k>5$.
Since $k \in \mathbf{Z}_{+}$, we have $k \geqslant 6$.
On the other hand, because a triangle with side lengths $1, 1, 2$ does not exist, by the problem statement, we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right),
$$
which implies $k \leqslant 6$.
Thus, we have $k=6$.
Next, we prove that $k=6$ satisfies the condition.
Assume without loss of generality that $a \leqslant b \leqslant c$.
From $6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, we get
$5 c^{2}-6(a+b) c+\left(5 a^{2}+5 b^{2}-6 a b\right)<0$.
By $\Delta=64\left[a b-(a-b)^{2}\right]$
$$
\leqslant 64 a b \leqslant 16(a+b)^{2},
$$
we have $c<\frac{6(a+b)+\sqrt{\Delta}}{10}$
$$
\leqslant \frac{6(a+b)+4(a+b)}{10}=a+b.
$$
Therefore, a triangle with side lengths $a, b, c$ exists.
Hence, the positive integer $k$ that satisfies the condition is $k=6$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On a plane, 2007 non-coincident lines $l_{1}, l_{2}, \cdots, l_{2007}$ are drawn, always following the rule of alternating perpendicular and parallel $\left(l_{2} \perp l_{1}, l_{3} / / l_{2}, l_{4} \perp l_{3}, l_{5} / /\right.$ $\left.l_{4}, \cdots\right)$. These 2007 non-coincident lines have a total of intersection points.
|
5.
$$
1007012
$$
|
1007012
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (15 points) Given $T=\{1,2, \cdots, 8\}$. For $A \subseteq T, A \neq \varnothing$, define $S(A)$ as the sum of all elements in $A$. Question: How many non-empty subsets $A$ of $T$ are there such that $S(A)$ is a multiple of 3 but not a multiple of 5?
(Chen Yonggao provided)
|
For $\varnothing$, define $S(\varnothing)=0$. Let
$$
T_{0}=\{3,6\}, T_{1}=\{1,4,7\}, T_{2}=\{2,5,8\} \text {. }
$$
For $A \subseteq \dot{T}$, let
$$
A_{0}=A \cap T_{0}, A_{1}=A \cap T_{1}, A_{2}=A \cap T_{2} .
$$
Then $S(A)=S\left(A_{0}\right)+S\left(A_{1}\right)+S\left(A_{2}\right)$
$$
\equiv\left|A_{1}\right|-\left|A_{2}\right|(\bmod 3) \text {. }
$$
Therefore, $3|S(A) \Leftrightarrow| A_{1}|\equiv| A_{2} \mid(\bmod 3)$. There are the following cases:
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ | A _ { 1 } | = 0 , } \\
{ | A _ { 2 } | = 0 ; }
\end{array} \quad \left\{\begin{array} { l }
{ | A _ { 1 } | = 0 , } \\
{ | A _ { 2 } | = 3 ; }
\end{array} \left\{\begin{array}{l}
\left|A_{1}\right|=3, \\
\left|A_{2}\right|=0 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l }
{ | A _ { 1 } | = 3 , } \\
{ | A _ { 2 } | = 3 ; }
\end{array} \left\{\begin{array} { l }
{ | A _ { 1 } | = 1 , } \\
{ | A _ { 2 } | = 1 ; }
\end{array} \left\{\begin{array}{l}
\left|A_{1}\right|=2, \\
\left|A_{2}\right|=2 .
\end{array}\right.\right.\right.
\end{array}
$$
Thus, the number of non-empty subsets $A$ that satisfy $3 \mid S(A)$ is
$$
2^{2}\left(\mathrm{C}_{3}^{0} \mathrm{C}_{3}^{0}+\mathrm{C}_{3}^{0} \mathrm{C}_{3}^{3}+\mathrm{C}_{3}^{3} \mathrm{C}_{3}^{0}+\mathrm{C}_{3}^{3} \mathrm{C}_{3}^{3}+\mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}+\right.
$$
$\left.C_{3}^{2} C_{3}^{2}\right)-1=87$.
If $3|S(A), 5| S(A)$, then $15 \mid S(A)$.
By $S(T)=36$, we know that the possible values of $S(A)$ that satisfy $3|S(A) γ 5| S(A)$ are $15 γ 30$. And
$$
\begin{array}{l}
15=8+7=8+6+1=8+5+2 \\
=8+4+3=8+4+2+1=7+6+2 \\
=7+5+3=7+5+2+1=7+4+3+1 \\
=6+5+4=6+5+3+1=6+4+3+2 \\
=5+4+3+2+1, \\
36-30=6=5+1=4+2=3+2+1 .
\end{array}
$$
Therefore, the number of $A$ that satisfy $3|S(A) γ 5| S(A) γ A \neq \varnothing$ is 17.
Thus, the number of $A$ that satisfies the conditions is $87-17=70$.
|
70
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the smallest real number $m$, such that for any positive numbers $a, b, c$ satisfying $a+b+c=1$, we have
$$
m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
(3rd China Southeast Mathematical Olympiad)
|
Explanation: First find the minimum value of $m$, then provide a proof.
When $a=b=c=\frac{1}{2}$, from the given inequality we have
$$
m \geqslant 27.
$$
Next, we prove that the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1,
$$
holds for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
It is easy to see that for any positive real numbers $a, b, c$, we have
$$
\begin{array}{l}
a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, \\
b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, \\
c^{3}+a^{3} \geqslant c^{2} a+c a^{2}.
\end{array}
$$
Adding these three inequalities, we get
$$
\begin{array}{l}
2\left(a^{3}+b^{3}+c^{3}\right) \\
\geqslant a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2}.
\end{array}
$$
Thus,
$$
3\left(a^{3}+b^{3}+c^{3}\right)
\begin{aligned}
\geqslant & a^{3}+b^{3}+c^{3}+a^{2} b+b^{2} c+c^{2} a+ \\
& a b^{2}+b c^{2}+c a^{2} \\
= & (a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \\
= & a^{2}+b^{2}+c^{2}.
\end{aligned}
$$
Therefore,
$$
6\left(a^{2}+b^{2}+c^{2}\right)+1
\begin{array}{l}
=6\left(a^{2}+b^{2}+c^{2}\right)+(a+b+c)^{2} \\
\leqslant 6\left(a^{2}+b^{2}+c^{2}\right)+3\left(a^{2}+b^{2}+c^{2}\right) \\
=9\left(a^{2}+b^{2}+c^{2}\right) \\
\leqslant 27\left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$
Thus, the minimum value of $m$ is 27.
Note: First, by setting $a=b=c=\frac{1}{3}$, we get $m \geqslant 27$, then we prove that 27 is the minimum value of $m$. The proof starts from the inequalities $a^{3}+b^{3} \geqslant a^{2} b+a b^{2}$, etc., to derive
$$
3\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2}+b^{2}+c^{2},
$$
and thus prove the conclusion.
|
27
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y$ satisfy the equation
$$
x^{2}-3 x y+3 y^{2}+4 x-18 y+52=0 \text {. }
$$
then the units digit of $y^{x}$ is $\qquad$ .
|
$=1.4$.
From the given equation, we have
$$
x^{2}-(3 y-4) x+\left(3 y^{2}-18 y+52\right)=0 \text {. }
$$
Since $x$ is a real number, then
$$
\Delta=(3 y-4)^{2}-4\left(3 y^{2}-18 y+52\right) \geqslant 0 \text {, }
$$
which simplifies to $-3(y-8)^{2} \geqslant 0$.
Thus, $y=8$.
Substituting $y=8$ into the original equation, we get $x^{2}-20 x+100=0$.
Solving this, we find $x=10$.
Therefore, $y^{z}=8^{10}=2^{30}=\left(2^{4}\right)^{7} \times 2^{2}=4 \times 16^{7}$.
Since the units digit of $16^{7}$ is 6, the units digit of $4 \times 16^{7}$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 3, select a point $P$ inside $\triangle A B C$, and draw line segments $D E, F G$, and $H M$ through $P$ parallel to the three sides of $\triangle A B C$. The areas of the resulting $\square A F P M, \square B D P G$, and $\square C E P H$ are $28, 12$, and $42$ respectively. Then the area of $\triangle A B C$ is $\qquad$
|
2.144.
Let the areas of $\triangle D P M$, $\triangle P G H$, and $\triangle P E F$ be $S_{1}$, $S_{2}$, and $S_{3}$, respectively. According to the problem,
$\triangle D P M \backsim \triangle P E F \backsim \triangle D E A$.
Therefore, $\frac{\sqrt{S_{1}}}{\sqrt{S_{\triangle A D E}}}=\frac{D P}{D E}$, $\frac{\sqrt{S_{3}}}{\sqrt{S_{\triangle D E E}}}=\frac{P E}{D E}$.
Thus, $\frac{\sqrt{S_{1}}+\sqrt{S_{3}}}{\sqrt{S_{\triangle A D E}}}=\frac{D P+P E}{D E}=\frac{D E}{D E}=1$, which means $\sqrt{S_{1}}+\sqrt{S_{3}}=\sqrt{S_{\triangle A D E}}$.
Since $S_{\square A F P M}=S_{\triangle A D E}-\left(S_{1}+S_{3}\right)$, we have $S_{1} S_{3}=\frac{1}{4} S_{\square A F P M}^{2}=\frac{1}{4} \times 28^{2}$.
Similarly, $S_{3} S_{2}=\frac{1}{4} S_{\square C E P H}^{2}=\frac{1}{4} \times 42^{2}$,
$S_{2} S_{1}=\frac{1}{4} S_{\square B D P G}^{2}=\frac{1}{4} \times 12^{2}$.
Thus, $S_{1}=4$, $S_{2}=9$, $S_{3}=49$.
Therefore, $S_{\triangle A B C}$
$$
\begin{array}{l}
=S_{\text {DAFPM }}+S_{\text {DBSPG }}+S_{\text {DCXPH }}+S_{1}+S_{2}+S_{3} \\
=28+12+42+4+9+49=144 .
\end{array}
$$
|
144
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $M=\frac{8}{\sqrt{2008}-44}$, $a$ is the integer part of $M$, and $b$ is the fractional part of $M$. Then
$$
a^{2}+3(\sqrt{2008}+37) a b+10=
$$
|
3.2008 .
Since $M=\frac{8}{\sqrt{2008}-44}=\frac{\sqrt{2008}+44}{9}$
$$
\begin{array}{l}
=9+\frac{\sqrt{2008}-37}{9}, \\
0<\frac{\sqrt{2008}-37}{9}<1,
\end{array}
$$
Therefore, $a=9, b=\frac{\sqrt{2008}-37}{9}$.
Then $a b=\sqrt{2008}-37$.
Hence $a^{2}+3(\sqrt{2008}+37) a b+10$
$$
\begin{array}{l}
=9^{2}+3(\sqrt{2008}+37)(\sqrt{2008}-37)+10 \\
=2008
\end{array}
$$
|
2008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 4, given that the equilateral $\triangle ABC$ is inscribed in $\odot O, AB$ $=86$. If point $E$ is on side $AB$, and through $E$ a line $DG \parallel BC$ intersects $\odot O$ at points $D, G$, and intersects $AC$ at point $F$, and let $AE=x, DE$ $=y$. If $x, y$ are both positive integers, then $y=$ $\qquad$
|
4.12.
From the problem, we know that $E F=A E=x$.
By the symmetry of the circle, $F G=D E=y$.
By the intersecting chords theorem, we have $A E \cdot E B=D E \cdot E G$, which means $x(86-x)=y(x+y)$.
If $x$ is odd, then $x(86-x)$ is also odd, and in this case, $y(x+y)$ is even. Therefore, $x$ must be even, and $y$ must also be even.
From $x(86-x)=y(x+y)$, we get
$x^{2}+(y-86) x+y^{2}=0$.
Thus, $x=\frac{86-y \pm \sqrt{(y-86)^{2}-4 y^{2}}}{2}$.
By $\Delta=(y-86)^{2}-4 y^{2} \geqslant 0$, we get
$0<y \leqslant 28$.
Since $x$ is a positive integer, $\Delta$ must be a perfect square.
Since $\Delta=(86-3 y)(86+y)$, and $y=2,4$, $\cdots, 28$, it is verified that only when $y=12$, $\Delta$ is a perfect square. At this time, $x=2$ or 72.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Given prime numbers $p$ and $q$ such that the algebraic expressions $\frac{2 p+1}{q}$ and $\frac{2 q-3}{p}$ are both natural numbers. Try to find the value of $p^{2} q$.
---
The translation is provided as requested, maintaining the original format and line breaks.
|
Let $p \geqslant q$, then we have $1 \leqslant \frac{2 q-3}{p}<2$.
Thus, it can only be $\frac{2 q-3}{p}=1$, which means $p=2 q-3$.
At this point, $\frac{2 p+1}{q}=\frac{4 q-5}{q}=4-\frac{5}{q}$.
To make $\frac{2 p+1}{q}$ a natural number, it can only be $q=5$, hence $p=7$.
Now, let $p<q$, in this case, $1 \leqslant \frac{2 p+1}{q}<3$, which can be divided into the following two scenarios:
(1) $\frac{2 p+1}{q}=1, q=2 p+1$.
At this point, $\frac{2 q-3}{p}=\frac{4 p-1}{p}$, which gives $p=1$, which does not meet the requirements.
(2) $\frac{2 p+1}{q}=2$, i.e., $2 p+1=2 q$, the left side is odd, the right side is even, a contradiction.
Therefore, the only $p$ and $q$ that satisfy the conditions are $p=7$ and $q=5$.
Thus, $p^{2} q=49 \times 5=245$.
|
245
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 5, in $\odot O$, $AB$ and $CD$ are two perpendicular diameters. Point $E$ is on radius $OA$, and point $F$ is on the extension of radius $OB$, such that $OE = BF$. Lines $CE$ and $CF$ intersect $\odot O$ at points $G$ and $H$, respectively. Lines $AG$ and $AH$ intersect line $CD$ at points $N$ and $M$, respectively. Prove that:
$$
\frac{DM}{MC} - \frac{DN}{NC} = 1.
$$
|
As shown in Figure 7, connect $D G$ and $D H$, and extend $C G$ to point $P$. Since $A B$ and $C D$ are perpendicular diameters,
we have
$$
\overparen{A C}=\overparen{A D}=\overparen{B C}=\overparen{B D}.
$$
Thus,
$$
\begin{array}{l}
\angle A H C=\angle A H D \\
=\angle A G C \\
=\angle P G N=45^{\circ}.
\end{array}
$$
Since $A, H, D, G$ are concyclic,
$$
\angle D G N=\angle A H D=45^{\circ}.
$$
Therefore, $\angle C H M=\angle D H M$, and $\angle P G N=\angle D G N$.
Thus, $H M$ is the angle bisector of $\triangle C H D$, and $G N$ is the external angle bisector of $\triangle C G D$.
So, $\frac{D M}{M C}=\frac{H D}{H C}$, and $\frac{D N}{N C}=\frac{G D}{G C}$.
Hence, $\frac{D M}{M C}-\frac{D N}{N C}=\frac{H D}{H C}-\frac{G D}{G C}$.
By the similarity of Rt $\triangle C D H \sim$ Rt $\triangle C F O$ and Rt $\triangle C D G \sim$ Rt $\triangle C E O$, we get
$$
\frac{H D}{H C}=\frac{O F}{O C}, \frac{G D}{G C}=\frac{O E}{O C}.
$$
Therefore, $\frac{H D}{H C}-\frac{G D}{G C}=\frac{O F}{O C}-\frac{O E}{O C}=\frac{O F-O E}{O C}$.
Since $O E=B F$ and $O F=O B+B F$, we have $O F-O E=O B=O C$.
Thus, $\frac{D M}{M C}-\frac{D N}{N C}=1$.
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the geometric sequence $z_{1}, z_{2}, \cdots, z_{n}, \cdots$ be such that $z_{1}=$ $1, z_{2}=a+b \mathrm{i}, z_{3}=b \mathrm{i}(a, b \in \mathbf{R}, ab>0)$. Then the smallest natural number $n$ for which $z_{1} z_{2} \cdots z_{n}<0$ is $\qquad$ .
|
$Ni, 1.8$.
$$
\begin{array}{l}
\text { Given } z_{2}^{2}=z_{1} z_{3} \Rightarrow(a+b \mathrm{i})^{2}=b \mathrm{i} \\
\Rightarrow a^{2}-b^{2}+2 a b \mathrm{i}=b \mathrm{i} \\
\Rightarrow\left\{\begin{array} { l }
{ a ^ { 2 } = b ^ { 2 } , } \\
{ 2 a b = b }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\frac{1}{2}, \\
b=\frac{1}{2}
\end{array}\right.\right. \\
\Rightarrow \text { common ratio } q=\frac{z_{2}}{z_{1}}=\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+\mathrm{i} \sin \frac{\pi}{4}\right) \\
\Rightarrow z_{1} z_{2} \cdots z_{n}=\left(z_{1}\right)^{n} q^{\frac{n(n-1)}{2}} \\
=\left(\frac{\sqrt{2}}{2}\right)^{\frac{n(n-1)}{2}}\left[\cos \frac{n(n-1)}{8} \pi+i \sin \frac{n(n-1)}{8} \pi\right] \\
<0 \\
\Rightarrow\left\{\begin{array}{l}
\cos \frac{n(n-1)}{8} \pi=-1, \\
\sin \frac{n(n-1)}{8} \pi=0
\end{array}\right. \\
\Rightarrow \frac{n(n-1)}{8}=k \text { (odd number). } \\
\end{array}
$$
The smallest value of $n$ that satisfies the condition is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A has a box, inside there are 4 balls in total, red and white; Person B has a box, inside there are 2 red balls, 1 white ball, and 1 yellow ball. Now, A randomly takes 2 balls from his box, B randomly takes 1 ball from his box. If the 3 balls drawn are all of different colors, then A wins. To ensure A has the highest probability of winning, the number of red balls in A's box should be
|
3.2.
Suppose box A contains $n(n \geqslant 1)$ red balls, then it has $4-n$ white balls.
Therefore, the probability of A winning is
$$
P=\frac{\mathrm{C}_{n}^{1} \mathrm{C}_{4-n}^{1}}{\mathrm{C}_{4}^{2} \mathrm{C}_{4}^{1}}=\frac{1}{24} n(4-n) \text {. }
$$
Since $\sqrt{n(4-n)} \leqslant \frac{n+4-n}{2}=2$, that is,
$$
n(4-n) \leqslant 4 \text {, }
$$
the equality holds if and only if $n=2$, at which $P$ is maximized.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the ellipse $\frac{y^{2}}{a^{2}}+$ $\frac{x^{2}}{b^{2}}=1(a>b>0)$, $m=\left(\frac{x_{1}}{b}, \frac{y_{1}}{a}\right)$, $n$ $=\left(\frac{x_{2}}{b}, \frac{y_{2}}{a}\right)$, and $\boldsymbol{m} \cdot \boldsymbol{n}=0$. The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$, the length of the minor axis is $2$, and $O$ is the origin. Then the area of $\triangle A O B$ is - $\qquad$
|
4.1.
Given $e^{2}=\frac{c^{2}}{a^{2}}=\frac{3}{4}, b=1$, then $a^{2}=b^{2}+c^{2}=4$.
Also, $\boldsymbol{m} \cdot \boldsymbol{n}=0 \Rightarrow y_{1} y_{2}=-4 x_{1} x_{2}$.
Substituting the coordinates of points $A$ and $B$ into the ellipse equation, we get
$$
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{4}+x_{1}^{2}=1 \\
\frac{y_{2}^{2}}{4}+x_{2}^{2}=1 .
\end{array}\right.
$$
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2}|O A| \cdot|O B| \sin \angle A O B \\
=\frac{1}{2}|O A| \cdot|O B| \sqrt{1-\cos ^{2} \angle A O B} \\
=\frac{1}{2}|O A| \cdot|O B| \sqrt{1-\left(\frac{O A \cdot O B}{|O A| \cdot|O B|}\right)^{2}} \\
=\frac{1}{2} \sqrt{|O A|^{2} \cdot|O B|^{2}-(O A \cdot O B)^{2}} \\
=\frac{1}{2} \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)-\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}} .
\end{array}
$$
Substituting equations (1), (2), and (3) into the above equation, we get
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2} \sqrt{\left(4-3 x_{1}^{2}\right)\left(4-3 x_{2}^{2}\right)-\left(-3 x_{1} x_{2}\right)^{2}} \\
=\sqrt{4-3\left(x_{1}^{2}+x_{2}^{2}\right)} .
\end{array}
$$
From (1) $)^{2}$, we easily get $y_{1}^{2} y_{2}^{2}=16 x_{1}^{2} x_{2}^{2}$, i.e.,
$$
16\left(1-x_{1}^{2}\right)\left(1-x_{2}^{2}\right)=16 x_{1}^{2} x_{2}^{2} \text {. }
$$
Thus, $x_{1}^{2}+x_{2}^{2}=1$.
Therefore, $S_{\triangle A O B}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the Cartesian coordinate system, there is a parabola $y=$ $x^{2}-(5 c-3) x-c$ and three points $A\left(-\frac{1}{2} c, \frac{5}{2} c\right)$, $B\left(\frac{1}{2} c, \frac{9}{2} c\right)$, $C(2 c, 0)$, where $c>0$. There exists a point $P$ on the parabola such that the quadrilateral with vertices $A$, $B$, $C$, and $P$ is a parallelogram. Then the number of such points $P$ is $\qquad$.
|
4.3.
(1) If $A B$ is the diagonal, then $P_{1}(-2 c, 7 c)$. For $P_{1}$ to be on the parabola, it must satisfy
$$
7 c=(-2 c)^{2}-(5 c-3)(-2 c)-c \text {. }
$$
Solving this, we get $c_{1}=0$ (discard), $c_{2}=1$.
Thus, $c=1$, and at this point, $P_{1}(-2,7)$.
(2) If $B C$ is the diagonal, then $P_{2}(3 c, 2 c)$. Similarly, we get $c=1$, and at this point, $P_{2}(3,2)$.
(3) If $A C$ is the diagonal, then $P_{3}(c,-2 c)$, we get $c=1$, and at this point, $P_{3}(1,-2)$.
In summary, there exist three points $P_{1}(-2,7)$, $P_{2}(3,2)$, and $P_{3}(1,-2)$, such that the quadrilateral with vertices $A$, $B$, $C$, and $P$ is a parallelogram.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 2, $EF$ intersects the diagonal $AC$ of $\square ABCD$ at point $G$,
intersects $AB$, $AD$ at points
$E$, $F$, and connects $CE$,
$CF$, $BG$. If $\frac{1}{S_{\triangle ACE}}+$
$\frac{1}{S_{\triangle ACF}}=\frac{\lambda}{S_{\triangle ABG}}$, find
the value of $\lambda$.
|
II. As shown in Figure 8, draw $EM \parallel BC$ intersecting $AC$ at $M$, then $\frac{AB}{AE}=\frac{AC}{AM}$.
From $\frac{AD}{AF}=\frac{BC}{AF}$
$=\frac{BC}{EM} \cdot \frac{EM}{AF}$
$=\frac{AC}{AM} \cdot \frac{GM}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AM}+\frac{AC}{AM} \cdot \frac{GM}{AG}$
$=\frac{AC}{AM}\left(1+\frac{GM}{AG}\right)=\frac{AC}{AM} \cdot \frac{AM}{AG}=\frac{AC}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AG}$
$\Rightarrow \frac{AB \cdot AG}{AC \cdot AE}+\frac{AD \cdot AG}{AC \cdot AF}=1$.
Since $\frac{AB \cdot AG}{AC \cdot AE}=\frac{S_{\triangle ABG}}{S_{\triangle ACE}}, \frac{AD \cdot AG}{AC \cdot AF}=\frac{S_{\triangle ADG}}{S_{\triangle ACF}}$, thus
$\frac{S_{\triangle ABG}}{S_{\triangle ACE}}+\frac{S_{\triangle ADG}}{S_{\triangle ACF}}=1$.
It is easy to prove that $\triangle ABC \cong \triangle CDA$, then
$S_{\triangle ABC}=S_{\triangle ADC}$.
Draw $BP \perp AC$, and $DQ \perp AC$, with feet of the perpendiculars at $P$ and $Q$ respectively.
From $S_{\triangle ABC}=\frac{1}{2} AC \cdot BP, S_{\triangle ADC}=\frac{1}{2} AC \cdot DQ$, we get
$BP=DQ$.
Also, $S_{\triangle ABG}=\frac{1}{2} AG \cdot BP, S_{\triangle ADG}=\frac{1}{2} AG \cdot DQ$,
thus $S_{\triangle ABG}=S_{\triangle ADG}$
$\Rightarrow \frac{S_{\triangle ABG}}{S_{\triangle ACE}}+\frac{S_{\triangle ABG}}{S_{\triangle ACF}}=1 \Rightarrow \lambda=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
---
In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
|
Three, as shown in Figure 9, let the sides of $\triangle ABC$ be $BC = a$, $CA = b$, and $AB = c$. Let $CD = h$, $AD = x$, and $BD = y$. The radii of the two incircles are $r_1$ and $r_2$. Then,
$$
\begin{array}{l}
RS = |RD - SD| \\
= |r_1 - r_2|. \\
\text{Since } b = AC = AP + CP = AE + CR \\
= (x - r_1) + (h - r_1),
\end{array}
$$
Thus, $r_1 = \frac{x + h - b}{2}$.
Similarly, $r_2 = \frac{y + h - a}{2}$.
Therefore, $|RS| = |r_1 - r_2|$
$$
\begin{array}{l}
= \left| \frac{x + h - b}{2} - \frac{y + h - a}{2} \right| \\
= \frac{1}{2} |(x - y) + (a - b)|.
\end{array}
$$
Also, $a^2 - y^2 = h^2 = b^2 - x^2$, so $x^2 - y^2 = b^2 - a^2$.
Since $x + y = c$, we have
$$
x - y = \frac{(b + a)(b - a)}{c}.
$$
Substituting equation (2) into equation (1) gives
$$
\begin{array}{l}
RS = \frac{1}{2} \left| \frac{(b + a)(b - a)}{c} + (a - b) \right| \\
= \frac{|b - a|}{2c} |a + b - c|.
\end{array}
$$
Since $a$, $b$, and $c$ are three consecutive natural numbers, let these three consecutive natural numbers be $n$, $n+1$, and $n+2$.
Since equation (3) is symmetric with respect to $a$ and $b$, we can assume without loss of generality that $a > b$. There are three cases:
(1) $a = n+1$, $b = n$, $c = n+2$, in this case, $RS = \frac{1 \times (n-1)}{2(n+2)}$ is not an integer.
(2) $a = n+2$, $b = n$, $c = n+1$, in this case, $RS = \frac{2(n+1)}{2(n+1)} = 1$, there are infinitely many such $n$, so there are infinitely many triangles that satisfy the condition.
(3) $a = n+2$, $b = n+1$, $c = n$, in this case, $RS = \frac{n+3}{2n}$, for $RS$ to be a positive integer, then $n+3 \geq 2n$.
Thus, $n \leq 3$.
Since $n$ is a positive integer, then $n = 1, 2, 3$.
When $n = 1$, $a = 3$, $b = 2$, $c = 1$, in this case, $b + c = a$ cannot form a triangle, so $n \neq 1$.
When $n = 2$, $RS = \frac{5}{4}$ is not an integer.
When $n = 3$, $RS = 1$.
In summary, $RS = 1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of ways to choose 5 different and non-adjacent numbers from the set $\{1,2, \cdots, 25\}$ is $\qquad$ kinds.
|
4.20349 .
Let $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ be a subset of 5 non-adjacent numbers from $\{1,2, \cdots, 25\}$, and
$a_{i+1}-a_{i} \geqslant 2(i=1,2,3,4)$.
Let $a_{i+1}^{\prime}-a_{i}^{\prime}=a_{i+1}-a_{i}-1$, i.e.,
$\left(a_{i+1}-a_{i+1}^{\prime}\right)-\left(a_{i}-a_{i}^{\prime}\right)=1$.
Take $a_{1}^{\prime}=a_{1}$, then $a_{i}^{\prime}=a_{i}-i+1$.
Clearly, $a_{i+1}^{\prime}-a_{i}^{\prime}=a_{i+1}-a_{i}-1 \geqslant 1$, and
$$
1 \leqslant a_{1}^{\prime}<a_{2}^{\prime}<a_{3}^{\prime}<a_{4}^{\prime}<a_{5}^{\prime} \leqslant 21 \text {. }
$$
Thus, the solution is $\mathrm{C}_{21}^{5}=20349$.
|
20349
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the increasing sequence $1,3,4,9,10,12,13, \cdots$ where each term is either a power of 3 or the sum of several different powers of 3. Then the 100th term of this sequence is $\qquad$ .
|
5.981.
The terms of this sequence are
$$
a_{0}+a_{1} 3+a_{2} 3^{2}+\cdots+a_{n} 3^{n},
$$
where $a_{i} \in\{0,1\}, i=1,2, \cdots, n$.
When $n=5$, there can be $2^{6}-1=63$ numbers, the 64th term is $3^{6}=729$. Starting from the 65th term, there are $2^{5}-1=31$ terms that do not contain $3^{5}=$ 243, the 96th term is $729+$ $243=972$, followed by $729+243+1,729+243+3$, $729+243+1+3$, the 100th term is $729+243+9=$ 981.
|
981
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $f(x)$ be defined on $\mathbf{N}_{+}$, with its range $B \subseteq$ $\mathbf{N}_{+}$, and for any $n \in \mathbf{N}_{+}$, we have
$$
f(n+1)>f(n) \text {, and } f(f(n))=3 n \text {. }
$$
Then $f(10)+f(11)=$ $\qquad$
|
6.39.
Given $f(f(1))=3$, we know $f(f(f(1)))=f(3)$.
If $f(1)=1$, then $3=f(f(1))=f(1)=1$, which is a contradiction.
Therefore, $2 \leqslant f(1)<f(2) \leqslant f(f(1))=3$.
Thus, $f(2)=3, f(1)=2$,
$f(3)=f(f(2))=6$,
$f(6)=f(f(3))=9$.
Also, $6=f(3)<f(4)<f(5)<f(6)=9$, so
$f(4)=7, f(5)=8$,
$f(7)=f(f(4))=12$,
$f(12)=f(f(7))=21$.
Since $f(9)=f(f(6))=18$,
$18=f(9)<f(10)<f(11)<f(12)=21$,
Therefore, $f(10)=19, f(11)=20$.
Thus, $f(10)+f(11)=39$.
|
39
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Let $A=\{1,2, \cdots, 30\}$. Find the smallest positive integer $n$, such that for any 11 subsets of $A$, if the union of any 5 of them has at least $n$ elements, then there must exist 3 of these 11 subsets whose intersection is non-empty.
|
Three, the minimum value of $n$ is 22.
First, prove: $n \geqslant 22$.
$$
\begin{array}{l}
\quad \text { Let } A_{i}=\{i, i+5, i+10, i+15, i+20, i+25\}, \\
i=1,2, \cdots, 5, \\
\quad B_{j}=\{j, j+6, j+12, j+18, j+24\}, j= \\
1,2, \cdots, 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Clearly, }\left|A_{i}\right|=6(i=1,2, \cdots, 5), \\
\left|A_{i} \cap A_{j}\right|=0(1 \leqslant i<j \leqslant 5), \\
\left|B_{j}\right|=5(j=1,2, \cdots, 6), \\
\left|B_{i} \cap B_{j}\right|=0(1 \leqslant i<j \leqslant 6), \\
\left|A_{i} \cap B_{j}\right|=1(1 \leqslant i \leqslant 5,1 \leqslant j \leqslant 6) .
\end{array}
$$
Thus, for any three subsets, there must be two that are both $A_{i}$ or both $B_{j}$, and their intersection is the empty set. For any five subsets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{s}}, B_{j_{1}}, B_{j_{2}}, \cdots$, $B_{j_{t}}(s+t=5)$, we have
$$
\begin{aligned}
& \left|A_{i_{1}} \cup A_{i_{2}} \cup \cdots \cup A_{i_{s}} \cup B_{j_{1}} \cup B_{j_{2}} \cup \cdots \cup B_{j_{t}}\right| \\
= & \left|A_{i_{1}}\right|+\left|A_{i_{2}}\right|+\cdots+\left|A_{i_{s}}\right|+\left|B_{j_{1}}\right|+ \\
& \left|B_{j_{2}}\right|+\cdots+\left|B_{j_{t}}\right|-s t \\
= & 6 s+5 t-s t=6 s+5(5-s)-s(5-s) \\
= & s^{2}-4 s+25 \geqslant 21,
\end{aligned}
$$
and the intersection of any three subsets is the empty set, hence $n \geqslant 22$.
Next, prove: $n=22$ satisfies the conditions.
Assume there exist 11 subsets of $A$, such that the union of any five of them contains at least 22 elements, and the intersection of any three is the empty set. Since each element belongs to at most two subsets, we can assume each element belongs to exactly two subsets (otherwise, add some elements to some subsets, and the conditions still hold). By the pigeonhole principle, among these 11 subsets, there must be one subset (let it be $B$) that contains at least $\left[\frac{2 \times 30}{11}\right]+1=6$ elements. Let the other 10 subsets be $B_{1}, B_{2}, \cdots, B_{10}$, then, any five subsets not containing $B$ correspond to at least 22 elements in $A$, and all 5-subset groups not containing $B$ together correspond to at least $22 \mathrm{C}_{10}^{5}$ elements.
On the other hand, for an element $a$, if $a \notin B$, then two of $B_{1}, B_{2}, \cdots, B_{10}$ contain $a$, so $a$ is counted $\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}$ times; if $a \in B$, then one of $B_{1}, B_{2}, \cdots, B_{10}$ contains $a$, so $a$ is counted $\mathrm{C}_{10}^{5}-\mathrm{C}_{9}^{5}$ times. Thus,
$$
\begin{array}{l}
22 \mathrm{C}_{10}^{5} \leqslant(30-|B|)\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}\right)+|B|\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{9}^{5}\right) \\
=30\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}\right)-|B|\left(\mathrm{C}_{9}^{5}-\mathrm{C}_{8}^{5}\right) \\
\leqslant 30\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}\right)-6\left(\mathrm{C}_{9}^{5}-\mathrm{C}_{8}^{5}\right) .
\end{array}
$$
This leads to $25 \leqslant 24$, a contradiction.
In conclusion, the minimum value of $n$ is 22.
|
22
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 In $\triangle A B C$, $A^{\prime} γ B^{\prime} γ C^{\prime}$ are on $B C γ C A γ A B$ respectively. Given that $A A^{\prime} γ B B^{\prime} γ C C^{\prime}$ concur at $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} \cdot \frac{C O}{O C^{\prime}}$.
(10th American Invitational Mathematics Examination)
|
Let $\angle A O B^{\prime}=\theta_{1}, \angle A O C^{\prime}=\theta_{2}$.
As shown in Figure 3, from $S_{\triangle A O B}+$
$S_{\triangle C O B^{\prime}}=S_{\triangle M O C}$, we get
$$
\begin{array}{l}
\frac{1}{2} \rho_{1} \rho_{2} \sin \theta_{1}+ \\
\frac{1}{2} \rho_{1} \rho_{6} \sin \left[180^{\circ}-\right. \\
\left.\left(\theta_{1}+\theta_{2}\right)\right] \\
=\frac{1}{2} \rho_{2} \rho_{6} \sin \left(180^{\circ}-\theta_{2}\right) .
\end{array}
$$
Then $\rho_{1} \rho_{2} \sin \theta_{1}-\rho_{2} \rho_{6} \sin \theta_{2}+$
$$
\rho_{1} \rho_{6} \sin \left(\theta_{1}+\theta_{2}\right)=0 \text {. }
$$
Similarly, we get
$$
\begin{array}{l}
-\rho_{2} \rho_{4} \sin \theta_{1}+\rho_{2} \rho_{3} \sin \theta_{2}+\rho_{3} \rho_{4} \sin \left(\theta_{1}+\theta_{2}\right)=0 . \\
\rho_{4} \rho_{5} \sin \theta_{1}+\rho_{5} \rho_{6} \sin \theta_{2}-\rho_{4} \rho_{6} \sin \left(\theta_{1}+\theta_{2}\right)=0 .
\end{array}
$$
Considering equations (1), (2), and (3) as a homogeneous linear system of equations in $\sin \theta_{1}, \sin \theta_{2}$, and $\sin \left(\theta_{1}+\theta_{2}\right)$, and knowing that there must be a non-zero solution, we have
$$
\left|\begin{array}{ccc}
\rho_{1} \rho_{2} & -\rho_{2} \rho_{6} & \rho_{1} \rho_{6} \\
-\rho_{2} \rho_{4} & \rho_{2} \rho_{3} & \rho_{3} \rho_{4} \\
\rho_{4} \rho_{5} & \rho_{5} \rho_{6} & -\rho_{4} \rho_{6}
\end{array}\right|=0 .
$$
Expanding and dividing both sides by $\rho_{2} \rho_{4} \rho_{6}$, we get
$$
\rho_{2} \rho_{4} \rho_{6}=\rho_{1} \rho_{2} \rho_{3}+2 \rho_{1} \rho_{3} \rho_{5}+\rho_{3} \rho_{4} \rho_{5}+\rho_{1} \rho_{5} \rho_{6} \text {. }
$$
Let $\rho_{2}=k_{1} \rho_{5}, \rho_{4}=k_{2} \rho_{1}, \rho_{6}=k_{3} \rho_{3}$.
Substituting into equation (4) and dividing by $\rho_{1} \rho_{3} \rho_{5}$, we get
$$
k_{1} k_{2} k_{3}=k_{1}+k_{2}+k_{3}+2 \text {. }
$$
Given $k_{1}+k_{2}+k_{3}=92$, we have $k_{1} k_{2} k_{3}=94$. Therefore, $\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} \cdot \frac{C O}{O C^{\prime}}=94$.
|
94
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given a sequence of numbers $a_{1}, a_{2}, \cdots, a_{100}$, where $a_{3} = 9, a_{7} = -7, a_{98} = -1$, and the sum of any three consecutive numbers is a constant. Then the value of $a_{1} + a_{2} + \cdots + a_{100}$ is ( ).
(A) 0
(B) 40
(C) 32
(D) 26
|
5.D.
According to the problem, we have
$$
\begin{array}{l}
a_{1}=a_{4}=a_{7}=\cdots=a_{100}=-7, \\
a_{2}=a_{5}=a_{8}=\cdots=a_{98}=-1, \\
a_{3}=a_{6}=a_{9}=\cdots=a_{99}=9 .
\end{array}
$$
Then $\bar{a}_{1}+a_{2}+\cdots+a_{100}$
$$
=33(9-7-1)-7=26 \text {. }
$$
|
26
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
13. Ordered pairs of positive integers $(a, b)(a<b)$ satisfy $a+b=2008$, and $a, b$ are coprime. Then the number of pairs $(a, b)$ that satisfy the condition is $\qquad$.
|
13.500.
Since $2008=2^{3} \times 251$, therefore, $a$ cannot be an even number, and it cannot be a multiple of 251. And $a \leqslant 1003$, so the number of $a$ is $1003-501-2=500$.
|
500
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 As shown in Figure 5, in Rt $\triangle C A B$, $\angle A=$ $90^{\circ}, \angle B γ \angle C$ are bisected and intersect at $F$, and intersect the opposite sides at points $D γ E$. Find $S_{\text {quadrilateral } B C D E}: S_{\triangle B F C}$.
|
Solution: Let $A B=c, A C=b, B C=a$. By the property of the internal angle bisector, we have
$$
\frac{A E}{E B}=\frac{b}{a} \text {. }
$$
Then $\frac{A E}{c}=\frac{b}{a+b}$, i.e., $A E=\frac{b c}{a+b}$.
Thus, $B E=c-A E=\frac{a c}{a+b}$.
Therefore, $S_{\triangle C B E}=\frac{1}{2} B E \cdot b=\frac{a b c}{2(a+b)}$.
And $\frac{C F}{F E}=\frac{a}{B E}=\frac{a+b}{c}$, so
$$
\frac{C F}{C E}=\frac{a+b}{a+b+c}, \frac{S_{\triangle C B F}}{S_{\triangle C B E}}=\frac{a+b}{a+b+c},
$$
$$
S_{\triangle B P C}=\frac{a b c}{2(a+b+c)} \text {. }
$$
Similarly, we can get $A D=\frac{b c}{a+c}$.
Then $S_{\text {quadrilateral } B C D E}=\frac{1}{2} b c-\frac{1}{2} \cdot \frac{b c}{a+c} \cdot \frac{b c}{a+b}$
$$
=\frac{a b c(a+b+c)}{2(a+b)(a+c)} \text {. }
$$
Noting that $a^{2}=b^{2}+c^{2}$, then
$$
\begin{array}{l}
\frac{S_{\text {quadrilateralBCDE }}}{S_{\triangle B F C}}=\frac{(a+b+c)^{2}}{(a+b)(a+c)} \\
=\frac{a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a}{a^{2}+a b+a c+b c} \\
=\frac{2 a^{2}+2 a b+2 b c+2 c a}{a^{2}+a b+a c+b c}=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (15 points) Let $\left\{a_{n}\right\}$ be an integer sequence, and it satisfies: for any $n\left(n \in \mathbf{N}_{+}\right)$, we have
$$
(n-1) a_{n+1}=(n+1) a_{n}-2(n-1) \text {, }
$$
and $2008 \mid a_{2000}$. Find the smallest positive integer $n(n \geqslant 2)$, such that $2008 \mid a_{n}$.
|
When $n=1$, we have $a_{1}=0$.
When $n \geqslant 2$, the original equation transforms to
$$
a_{n+1}=\frac{n+1}{n-1} a_{n}-2 \text{. }
$$
Let $b_{n}=\frac{a_{n}}{n-1}$, then $n b_{n+1}=(n+1) b_{n}-2$.
Thus, for $n \geqslant 2$, we have
$$
b_{n+1}-2=\frac{n+1}{n}\left(b_{n}-2\right) \text{. }
$$
From equation (2), we know
$$
\begin{array}{l}
b_{n}-2=\frac{n}{n-1} \cdot \frac{n-1}{n-2} \cdots \cdot \frac{3}{2}\left(b_{2}-2\right)=\left(\frac{b_{2}}{2}-1\right) n \\
\Rightarrow a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] .
\end{array}
$$
Given $2008 \mid a_{2000}$, we have $\frac{a_{2}}{2} \in \mathbf{Z}$.
Thus, let $a_{2}=2 k(k \in \mathbf{Z})$.
From equation (3), we get $a_{n}=(n-1)[(k-1) n+2]$.
Given $2008 \mid a_{2007}$
$$
\begin{array}{l}
\Rightarrow 2006(2007 k-2005) \equiv 0(\bmod 2008) \\
\Rightarrow k \equiv 3(\bmod 1004) \\
\Rightarrow a_{n}=(n-1)[(1004 m+2) n+2](m \in \mathbf{Z}) .
\end{array}
$$
Then $20081 a_{n}$
$$
\begin{array}{l}
\Leftrightarrow 20081(n-1)[(1004 m+2) n+2] \\
\Rightarrow 1004 \mid(n-1)(n+1) .
\end{array}
$$
Thus, $2 \mid n$.
Let $n=2 l+1(l \in \mathbf{Z}) \Rightarrow 251 \mid l(l+1)$.
Since 251 is a prime number, and $(l, l+1)=1$, we have
$$
l+1 \geqslant 251 \Rightarrow l_{\text {min }}=250 \Rightarrow n_{\text {min }}=501 .
$$
|
501
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $f$ is a mapping from the set $M=\{a, b, c, d\}$ to $N=\{0, 1, 2\}$, and
$$
f(a)+f(b)+f(c)+f(d)=4.
$$
Then the number of different mappings is $($ .
(A) 13
(B) 19
(C) 21
(D) 23
|
-1.B.
(1) When the image set is $\{1\}$, we have $f(a)=f(b)=$ $f(c)=f(d)=1$, there is 1 such mapping;
(2) When the image set is $\{0,2\}$, $f(a) γ f(b)$ γ $f(c) γ f(d)$ have 2 zeros and 2 twos. Enumerating, we find there are 6 such mappings;
(3) When the image set is $\{0,1,2\}$, $f(a) γ f(b)$ γ $f(c) γ f(d)$ have 1 zero, 2 ones, and 1 two. Enumerating, we find there are 12 such mappings.
In summary, there are $1+6+12=19$ different mappings.
|
19
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Define an operation of sets $A$ and $B$: $A * B=$ $\left\{x \mid x=x_{1}+x_{2}\right.$, where, $\left.x_{1} \in A, x_{2} \in B\right\}$. If $A$ $=\{1,2,3\}, B=\{1,2\}$, then the sum of all elements in $A * B$ is $\qquad$
|
δΊγ7.14.
$A * B$ δΈε
η΄ δΈΊ $2,3,4,5$, ζ
ε
Άζζε
η΄ ζ°εδΉεδΈΊ 14 .
---
Second, 7.14.
The elements of $A * B$ are $2,3,4,5$, so the sum of all the digits of its elements is 14.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. The sequence $1,1,2,1,1,3,1,1,1,4,1,1,1,1$, $5, \cdots, \underbrace{1,1, \cdots, 1}_{n-1 \uparrow}, n, \cdots$ has the sum of its first 2007 terms as
|
9.3898 .
In the sequence, from number 1 to $n$ there are $n+\frac{n(n-1)}{2}$ terms, thus, from number 1 to 62 there are $62+\frac{62 \times(62-1)}{2}$ $=1953$ terms, followed by 54 ones.
Therefore, the sum of the first 2007 terms is
$$
\begin{array}{l}
(1+2+\cdots+62)+(1+2+\cdots+61)+54 \\
=3898
\end{array}
$$
|
3898
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given that the equation $x^{3}+3 x^{2}-x+a$ $=0$ has three real roots that form an arithmetic sequence. Then the real number $a=$ $\qquad$
|
12. -3 .
Let these three roots be $b-d$, $b$, and $b+d$. Then
$$
\begin{array}{l}
x^{3}+3 x^{2}-x+a \\
=(x-b+d)(x-b)(x-b-d),
\end{array}
$$
i.e., $3 x^{2}-x+a$
$$
=-3 b x^{2}+\left(3 b^{2}-d^{2}\right) x-b^{3}+b d^{2} \text {. }
$$
Comparing coefficients, we get
$$
-3 b=3,3 b^{2}-d^{2}=-1,-b^{3}+b d^{2}=a \text {. }
$$
Thus, $a=-3$.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (i) (Grade 10) $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=$ $\qquad$ .
(ii) (Grade 11) Using the digits $1,2,3,4,5,6$ to number the six faces of a cube. If a numbered cube can be rotated to match the numbering of another cube, they are considered the same numbering method. Therefore, the number of different numbering methods is $\qquad$ kinds.
|
9. (i) $\sqrt{2}$.
Since $\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}$
$=\cos 40^{\circ} \sqrt{1-\cos 80^{\circ}}$
$=\cos 40^{\circ} \cdot \sqrt{2} \sin 40^{\circ}=\frac{\sqrt{2}}{2} \sin 80^{\circ}=\frac{\sqrt{2}}{2} \cos 10^{\circ}$,
therefore, $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=\sqrt{2}$.
(ii) 30.
Place the face marked with 6 on the table, the face opposite to 6 can be marked in 5 different ways, and the four middle faces can be marked with the remaining four numbers, considering the rotation of the cube around the vertical axis through the center of the base, there are 3! ways to mark the numbers. Therefore, the total number of different marking methods is $5 \times 3!=30$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If $\frac{1^{2}+3^{2}+\cdots+(2 n-1)^{2}}{2^{2}+4^{2}+\cdots+(2 n)^{2}}=\frac{13}{14}$, then
$$
n=
$$
|
10.20.
From $\sum_{k=1}^{2 n} k^{2}=\frac{27}{14} \sum_{k=1}^{n}(2 k)^{2}$, we get
$$
\begin{array}{l}
\frac{2 n(2 n+1)(4 n+1)}{6} \\
=\frac{27}{14} \times 4 \times \frac{n(n+1)(2 n+1)}{6},
\end{array}
$$
which means $4 n+1=\frac{27}{7}(n+1)$.
Therefore, $n=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The incenter of $\triangle A B C$ is $I$, and the angle bisector of $\angle B$ intersects $A C$ at point $P$. If $A P+A B=B C$, and $A B=3, B C=$ 5, then the value of $A I$ is $\qquad$ .
|
2.2.
As shown in Figure 7, on segment $B C$, take $B A^{\prime}=B A$, extend $A I$ to intersect $B C$ at $Q$, and connect $P A^{\prime}$.
$$
\begin{array}{l}
\text { Given } B A^{\prime}=B A, \\
\angle A B P=\angle A^{\prime} B P, \\
B P=B P,
\end{array}
$$
we have $\triangle A B P \cong \triangle A^{\prime} B P$.
Thus, $A^{\prime} P=A P=B C-B A=B C-B A^{\prime}=A^{\prime} C$.
Therefore, $\angle A^{\prime} P C=\angle C$.
Since $\angle B A C=\angle B A^{\prime} P=\angle C+\angle A^{\prime} P C=$ $2 \angle C$, we have
$$
\begin{array}{l}
\angle A P B=\angle P B C+\angle C \\
=\frac{1}{2} \angle B+\angle C=\frac{1}{2}(\angle B+\angle A). \\
\text { Also, } \angle A I P=\angle I A B+\angle I B A \\
=\frac{1}{2}(\angle A+\angle B),
\end{array}
$$
Thus, $\angle A P B=\angle A I P$, and $A I=A P$.
Since $A P=B C-A B=5-3=2$, we have $A I=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) Find the smallest positive integer $n$ such that: for any $n$ points $A_{1}, A_{2}$, $\cdots, A_{n}$ taken on the circumference of $\odot O$, among the $\mathrm{C}_{n}^{2}$ angles $\angle A_{i} O A_{j}(1 \leqslant i<j \leqslant n)$, at least 2007 are not greater than $120^{\circ}$.
|
12. First, when $n=90$, as shown in Figure 7, let $AB$ be the diameter of $\odot O$. Take 45 points near points $A$ and $B$, respectively. At this time, there are only $2 \mathrm{C}_{45}^{2}=45 \times 44=1980$ angles that do not exceed $120^{\circ}$. Therefore, $n=90$ does not satisfy the problem's requirements.
Second, when $n=91$, we will prove that there are at least 2007 angles that do not exceed $120^{\circ}$.
For 91 points $A_{1}, A_{2}, \cdots, A_{91}$ on the circumference, if $\angle A_{i} O A_{j}>120^{\circ}$, then connect $A_{i} A_{j}$, thus forming a graph $G$. Suppose graph $G$ has $e$ edges.
When $\angle A_{i} O A_{j}>120^{\circ}, \angle A_{j} O A_{k}>120^{\circ}$, then $\angle A_{i} O A_{k} \leq 120^{\circ}$. Therefore, if $e \geqslant 1$, without loss of generality, assume $A_{1}$ and $A_{2}$ are connected by an edge. Since the graph has no triangles, for any point $A_{i} (3 \leqslant i \leqslant 91)$, it can be connected to at most one of $A_{1}$ or $A_{2}$. Thus,
$$
d\left(A_{1}\right)+d\left(A_{2}\right) \leqslant 89+2=91,
$$
where $d(A)$ denotes the number of edges from $A$.
$$
\text{Also, } d\left(A_{1}\right)+d\left(A_{2}\right)+\cdots+d\left(A_{91}\right)=2 e \text{, }
$$
and for every pair of vertices $A_{i}, A_{j}$ connected by an edge in graph $G$, we have $d\left(A_{i}\right)+d\left(A_{j}\right) \leqslant 91$.
Thus, summing over all edges, we get
$$
\left(d\left(A_{1}\right)\right)^{2}+\left(d\left(A_{2}\right)\right)^{2}+\cdots+\left(d\left(A_{91}\right)\right)^{2}
$$
$\leqslant 91 e$.
By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
91\left[\left(d\left(A_{1}\right)\right)^{2}+\left(d\left(A_{2}\right)\right)^{2}+\cdots+\left(d\left(A_{91}\right)\right)^{2}\right] \\
\geqslant\left[d\left(A_{1}\right)+d\left(A_{2}\right) \cdots+d\left(A_{91}\right)\right]^{2}=4 e^{2} . \\
\text{Thus, } \frac{4 e^{2}}{91} \leqslant\left(d\left(A_{1}\right)\right)^{2}+\left(d\left(A_{2}\right)\right)^{2}+\cdots+\left(d\left(A_{91}\right)\right)^{2} \\
\leqslant 91 e, \\
e \leqslant \frac{91^{2}}{4}2007
$$
pairs of points, which are not connected by an edge. Therefore, the corresponding angles do not exceed $120^{\circ}$.
In conclusion, the minimum value of $n$ is 91.
(Xiong Bin, Gu Hongda, Li Dayuan, Liu Hongkun, Ye Shengyang)
|
91
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Calculate: $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=$
|
9.4 .
$$
\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=\frac{2 \sin \left(30^{\circ}-10^{\circ}\right)}{\frac{1}{2} \sin 20^{\circ}}=4
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. To cut a rectangular prism into $k$ tetrahedra, the minimum value of $k$ is
|
11.5 .
According to the equivalence, we only need to consider the cutting situation of a unit cube.
On the one hand, first, we need to show that 4 is not enough. If there were 4, since all faces of a tetrahedron are triangles and are not parallel to each other, the top face of the cube would have to be cut into at least two triangles, and the bottom face would also have to be cut into at least two triangles. The area of each triangle is less than or equal to $\frac{1}{2}$, and these four triangles must belong to four different tetrahedrons. The height of a tetrahedron with such a triangle as its base is less than or equal to 1.
Therefore, the sum of the volumes of the four different tetrahedrons is less than or equal to $4\left(\frac{1}{3} \times \frac{1}{2} \times 1\right)=\frac{2}{3}<1$, which does not meet the requirement.
Thus, $k \geqslant 5$.
On the other hand, as shown in Figure 5, the unit cube can be cut into 5 tetrahedrons, for example, by removing a tetrahedron $A_{1} B C_{1} D$ from the center of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, leaving four tetrahedrons at the corners.
In total, there are 5 tetrahedrons.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$ .
|
12.133113.
We call this kind of number a "good number". There are 3 one-digit good numbers; 12 two-digit good numbers, which is $3 \times 4$; 48 three-digit good numbers, which is $3 \times 4^2$; $\cdots \cdots \cdot k$-digit good numbers have $3 \times 4^{k-1}$ numbers $(k=1$, $2, \cdots)$. Let $S_{n}=3 \sum_{k=1}^{n} 4^{k-1}$.
Since $S_{5}<2007<S_{6}, 2007-S_{5}=984$, the 2007th good number is the 984th six-digit good number; and among the six-digit good numbers, those starting with 1 total $4^{5}=1024$ numbers, and those starting with $10, 11, 12, 13$ each have $4^{4}=256$ numbers, so the first two digits of the 2007th good number are 13, and it is the $984-3 \times 256=216$th number among those starting with 13; and those starting with $130, 131, 132, 133$ each have 64 numbers, so the first three digits of $a_{2007}$ are 133, and it is the $216-3 \times 64=24$th number among those starting with 133; and those starting with 1330, $1331, 1332, 1333$ each have 16 numbers, so the first four digits of $a_{2007}$ are 1331, and it is the 24 $16=8$th number among those starting with 1331; then the first five digits of $a_{200}$ are 13311, and it is the $8-4=4$th number among those starting with 13311. Therefore, $a_{2007}=133113$.
|
133113
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. If a four-digit number $n=\overline{a b c d}$ has digits $a, b, c, d$ such that any three of these digits can form the lengths of the sides of a triangle, then $n$ is called a "four-digit triangular number". Find the total number of four-digit triangular numbers.
|
15. Call $(a, b, c, d)$ a digital group of $n$, then
$$
a, b, c, d \in M=\{1,2, \cdots, 9\} \text {. }
$$
(1) When the digital group contains only one value, i.e., $(a, a, a, a)(a=1,2, \cdots, 9)$, there are 9 values of $n$.
(2) When the digital group contains exactly two values $a, b(a>b)$.
(i) The digital group is of type $(a, a, a, b)$, then any three digits can form a triangle. For each $a \in\{2,3, \cdots, 9\}, b$ can take $a-1$ values, so the number of digital groups is $\sum_{a=2}^{9}(a-1)=36$. For each group $(a, a, a, b), b$ has 4 placement methods, thus, there are $36 \times 4=144$ such $n$.
(ii) The digital group is of type $(a, b, b, b)$, according to the triangle formation condition, $b < a < 2b$.
(i) The digital group is of type $(a, b, c, c)$, according to the triangle formation condition, then $c < b < a < 2c$, there are 14 such $(a, b, c, c)$, each group has $A_4^2 = 12$ placement methods, thus, there are $14 \times 12 = 168$ such $n$.
(ii) The digital group is of type $(a, b, b, c)$, $c < b < a < b+c$, this condition is equivalent to the number of ways to form a triangle by taking 3 different numbers from $M=\{1,2, \cdots, 9\}$, there are 34 groups, each group has $A_4^2 = 12$ placement methods, thus, there are $34 \times 12 = 408$ such $n$.
(iii) The digital group is of type $(a, a, b, c)$, $c < b < a < b+c$, same as case (ii), there are $34 A_4^2 = 408$ such $n$ values.
In total, there are $168 + 408 + 408 = 984$ such $n$.
(4) $a, b, c, d$ are all different, then $d < c < b < a < c+d$, there are 16 such $a, b, c, d$, each group has $4!$ arrangements, thus, there are $16 \times 4! = 384$ such $n$ values.
In summary, the total number of four-digit triangular numbers $n$ is
$$
9 + 304 + 984 + 384 = 1681 \text{ (values). }
$$
(Provided by Tao Pingsheng)
|
1681
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. There are 2008 viewing stations set up along a circular marathon track, labeled as $c_{1}, c_{2}, \cdots, c_{2008}$ in a clockwise direction. These stations divide the track into 2008 segments. An athlete places mascots numbered $1, 2, \cdots, 2008$ at these stations in the following manner: He first places the 1st mascot at $c_{1}$, then runs clockwise over 29 segments and places the 2nd mascot at the station $c_{30}$; he then runs clockwise over 29 segments again and places the 3rd mascot at the station $c_{59}$, and so on. What is the number of the mascot placed at station $c_{2008}$?
(A) 2008 (B) 1896 (C) 1732 (D) 1731
|
- 1. C.
Let the mascot number placed at site $c_{k}$ be $x$, $x \in\{1,2, \cdots, 2008\}$. Then $29(x-1)+1 \equiv k(\bmod 2008)$.
When $k=2008$, we get
$29(x-1)+1 \equiv 0(\bmod 2008)$.
Let $29(x-1)+1=2008 y$, i.e., $x=\frac{2008 y-1}{29}+1=69 y+1+\frac{7 y-1}{29}$.
It is easy to see that when $y=-4+29 t, t \in \mathbf{Z}$, $x \in \mathbf{Z}$.
Taking $t=1$, we get $y=25$, at this time,
$$
x=69 y+1+\frac{7 y-1}{29}=1732 .
$$
|
1732
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
II. Fill-in-the-blank Questions (9 points each, total 54 points)
1. Remove all perfect squares and cubes from the natural numbers, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2008}=$ $\qquad$
|
Two, 1.2062.
First, remove numbers of the form $n^{6}$. In the intervals $\left(1^{6}, 2^{6}\right),\left(2^{6}, 3^{6}\right),\left(3^{6}, 4^{6}\right), \cdots$, there are no more overlaps between square numbers and cube numbers.
Since $\sqrt{2^{6}}=2^{3}=8, \sqrt[3]{2^{6}}=4$, removing the interval endpoints, we know that $\left(1^{6}, 2^{6}\right)$ contains 6 square numbers and 2 cube numbers. Similarly, $\left(2^{6}, 3^{6}\right)$ contains 18 square numbers and 4 cube numbers.
Therefore, in the interval $\left[1,3^{6}\right]$, there are a total of 33 square and cube numbers.
Since $2008+33>45^{2}$, and $45^{2}=2025,46^{2}=$ $2116,12^{3}=1728,13^{3}=2197$, the interval $(2025$, $2116)$ contains no square or cube numbers. In $\left(3^{6}, 45^{2}\right]$, there are 18 square numbers $\left(28^{2}, 29^{2}, \cdots, 45^{2}\right)$ and 3 cube numbers $\left(10^{3}, 11^{3}, 12^{3}\right)$. Thus, $2008+33+18+3=2062$, so $a_{2008}=2062$.
|
2062
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then, in the plane, the area of the figure formed by all points satisfying $[x]^{2}+[y]^{2}=50$ is
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ
|
5.12.
First, consider the first quadrant.
From $50=1^{2}+7^{2}=7^{2}+1^{2}=5^{2}+5^{2}$, we get $([x],[y])=(1,7),(7,1),(5,5)$.
And from $[x]=1,[y]=7$, we get the unit square $1 \leqslant x<2,7 \leqslant y<8$, whose area is 1.
Similarly, from $([x],[y])=(7,1)$ and $(5,5)$, we also get one unit square each, and these three squares do not overlap.
If we consider all four quadrants, we get a total of 12 non-overlapping unit squares, with a total area of 12.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits be denoted as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". The sum of all coincidence numbers is $\qquad$ .
|
6.531.
Let $n=\overline{a_{1} a_{2} \cdots a_{k}}\left(a_{1} \neq 0\right)$.
From $n-s(n)=p(n)$, we get
$$
\sum_{i=1}^{k-1} a_{i}\left(10^{k-i}-1\right)=a_{1} a_{2} \cdots a_{k} \text {, }
$$
which means $a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right)+m=0$,
where $m=a_{2}\left(10^{k-2}-1\right)+\cdots+a_{k-1}(10-1) \geqslant 0$.
If $k \geqslant 3$, since
$$
10^{k-1}-1-a_{2} a_{3} \cdots a_{k} \geqslant 10^{k-1}-1-9^{k-1}>0 \text {, }
$$
it contradicts equation (1). Therefore, $k \leqslant 2$.
When $k=1$, $n=s(n)$, which does not meet the condition.
Thus, $k=2$.
Hence, we can set $n=\overline{a_{1} a_{2}}=10 a_{1}+a_{2}$. From
$$
a_{1}+a_{2}+a_{1} a_{2}=10 a_{1}+a_{2} \text {, }
$$
we get $9 a_{1}=a_{1} a_{2}$.
Therefore, $a_{2}=9, a_{1} \in\{1,2, \cdots, 9\}$, which means all coincidental numbers are $19,29, \cdots, 99$, and their sum is 531.
|
531
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (20 points) The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=\left\{\begin{array}{l}
a_{n}-n, a_{n}>n ; \\
a_{n}+n, a_{n} \leqslant n .
\end{array}\right.
$$
(1) Find $a_{2008}$;
(2) Find the smallest positive integer $n$, such that $a_{n}=2008$.
|
Five, the initial values of the easy sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & $2 \mid 3$ & & & 6 & 7 & 8 & & 10 & 11 & 12 & & \\
\hline$a$ & 1 & 2 & 41 & & 10 & & 11 & & $12 \mid$ & 2 & 13 & 1 & 14 \\
\hline
\end{tabular}
Next, focus on the subscripts $n_{k}$ such that $a_{n_{k}}=1$: $n_{1}=1$, $n_{2}=4, n_{3}=13, \cdots \cdots$ They satisfy the following recursive relation:
$$
n_{k+1}=3 n_{k}+1(k=1,2, \cdots) \text {. }
$$
We will use induction on $k$.
When $k=1,2$, equation (1) holds.
Assume $a_{n_{k}}=1$, then by the condition
$$
\begin{array}{l}
a_{n_{k}+1}=n_{k}+1, a_{n_{k}+2}=2 n_{k}+2, \\
a_{n_{k}+3}=n_{k}, a_{n_{k}+4}=2 n_{k}+3, \cdots \cdots
\end{array}
$$
By induction, we get
$$
\begin{array}{c}
a_{n_{k}+2 m-1}=n_{k}+2-m\left(m=1,2, \cdots, n_{k}+1\right), \\
a_{n_{k}+2 m}=2 n_{k}+1+m\left(m=1,2, \cdots, n_{k}\right) .
\end{array}
$$
Thus, when $m=n_{k}+1$,
$$
a_{3 n_{k}+1}=n_{k}+2-\left(n_{k}+1\right)=1 \text {. }
$$
Therefore, $n_{k+1}=3 n_{k}+1(k=1,2, \cdots)$, i.e., equation (1) holds.
From equation (1), we get $2 n_{k+1}+1=3\left(2 n_{k}+1\right)$.
Let $2 n_{k}+1=x_{k}$. Then $x_{k+1}=3 x_{k}, x_{1}=3$.
So, $x_{k}=3^{k}$.
Therefore, $n_{k}=\frac{3^{k}-1}{2}(k=1,2, \cdots)$.
And $n_{7}=\frac{3^{7}-1}{2}=1093, n_{8}=\frac{3^{8}-1}{2}=3280$, thus $n_{7}<2008<n_{8}$.
Also, $2008=n_{7}+2 \times 458-1$, so by equation (2), $a_{2008}=n_{7}+2-458=637$.
From equation (2), we know that when $n \leqslant 3 n_{k}=n_{k+1}-1$, $a_{n} \leqslant 3 n_{k}+1=n_{k+1}$.
Therefore, when $n<n_{7}$, $a_{n} \leqslant n_{7}=1093$.
And when $n_{7} \leqslant n<n_{8}$, either $a_{n} \leqslant 1094$, or $a_{n} \geqslant 2 \times 1094$, i.e., $a_{n}$ cannot take the value 2008.
Furthermore, consider the case $n_{8} \leqslant n<n_{9}$.
From $n_{8}+2-m=2008$, we get $m=1274$.
By equation (2),
$a_{5827}=a_{n_{8}+2 m-1}=n_{8}+2-m=2008$.
Thus, the smallest $n$ such that $a_{n}=2008$ is 5827.
|
5827
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $x, y, z$ be non-negative real numbers, and $x+y+z=$ 2. Then the sum of the maximum and minimum values of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ is $\qquad$ .
|
2.1.
Since $x, y, z$ are non-negative real numbers, we have
$$
A=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant 0.
$$
When $x=y=0, z=2$, $A=0$, thus the minimum value of $A$ is 0.
Assume $A$ reaches its maximum value at $(x, y, z)$, without loss of generality, let $x \leqslant y \leqslant z$, prove: $x=0$.
In fact, if $x>0$, then let $x^{\prime}=0, y^{\prime}=y+x, z^{\prime}=z$, at this time,
$$
\begin{array}{l}
A^{\prime}=x^{\prime 2} y^{\prime 2}+y^{\prime 2} z^{\prime 2}+z^{\prime 2} x^{\prime 2}=(y+x)^{2} z^{2} \\
=y^{2} z^{2}+z^{2} x^{2}+2 x y z^{2} \\
\geqslant y^{2} z^{2}+z^{2} x^{2}+2 x y(x y) \\
>y^{2} z^{2}+z^{2} x^{2}+x y(x y)=A,
\end{array}
$$
which contradicts the maximality of $A$.
When $x=0$,
$$
y+z=2, A=y^{2} z^{2} \leqslant\left(\frac{y+z}{2}\right)^{4}=1,
$$
the equality holds when $y=z=1, x=0$.
Therefore, the maximum value of $A$ is 1.
Thus, the sum of the maximum and minimum values is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a, b$ be any two distinct positive integers. Then the minimum value of $\left|a b\left(a^{4}-b^{4}\right)\right|$ is $\qquad$ .
|
3.30.
$$
\begin{array}{l}
\text { Let } A=a b\left(a^{4}-b^{4}\right) \\
=a b(a-b)(a+b)\left(a^{2}+b^{2}\right) .
\end{array}
$$
First, prove: $2 \mid A$.
If $a$ and $b$ are both odd or both even, then $2 \mid (a-b)$, so $2 \mid A$;
If $a$ and $b$ are of different parity, then one of $a$ or $b$ is even, so $2 \mid ab$, hence $2 \mid A$.
Next, prove: $3 \mid A$.
If one of $a$ or $b$ is a multiple of 3, then $3 \mid ab$, so $3 \mid A$;
If neither $a$ nor $b$ is a multiple of 3, then
$a \equiv b \pmod{3}$ or $a \equiv -b \pmod{3}$.
So, $3 \mid (a-b)(a+b)$, hence $3 \mid A$.
Finally, prove: $5 \mid A$.
By $a \equiv 0, \pm 1, \pm 2 \pmod{5}$, we get
$$
a^{2} \equiv 0, \pm 1 \pmod{5} \text {. }
$$
Similarly, $b^{2} \equiv 0, \pm 1 \pmod{5}$.
Thus, $ab \equiv 0 \pmod{5}$ or $a^{2}-b^{2} \equiv 0 \pmod{5}$ or $a^{2}+b^{2} \equiv 0 \pmod{5}$.
Therefore, $5 \mid A$.
Since $2, 3, 5$ are pairwise coprime, we have
$30 \mid a b\left(a^{4}-b^{4}\right)$.
Thus, $\left|a b\left(a^{4}-b^{4}\right)\right| \geqslant 30$.
When $a=2, b=1$, $a b\left(a^{4}-b^{4}\right)=30$, so the minimum value of $\left|a b\left(a^{4}-b^{4}\right)\right|$ is 30.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Find the smallest positive integer $t$, such that for any convex $n$-gon $A_{1} A_{2} \cdots A_{n}$, as long as $n \geqslant t$, there must exist three points $A_{i} γ A_{j} γ A_{k}(1 \leqslant i<j<k \leqslant n)$, such that the area of $\triangle A_{i} A_{j} A_{k}$ is no more than $\frac{1}{n}$ of the area of the convex $n$-gon $A_{1} A_{2} \cdots A_{n}$.
|
Three, first prove a lemma.
Lemma For any convex hexagon $A_{1} A_{2} \cdots A_{6}$, there exists $1 \leqslant i\frac{S}{k+1}$, then
$S_{\text {pentagon } A_{1} A_{2} \cdots A_{k}}<S-\frac{S}{k+1}=\frac{k S}{k+1}$.
By the induction hypothesis, there must be $1 \leqslant i<j<r \leqslant n$, such that $S_{\triangle A_{i} A_{j} A_{1}} \leqslant \frac{1}{k} \cdot \frac{k S}{k+1}=\frac{S}{k+1}$.
The conclusion holds.
In summary, the minimum value of $t$ is 6.
(Feng Yuefeng, Shenzhen Senior High School, 518040)
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x^{2}+y^{2} \leqslant 1$. Then the maximum value of the function $z=$ $\frac{\cos x+\cos y}{1+\cos x y}$ is $\qquad$ .
|
2 1.1.
Assume $x \geqslant 0, y \geqslant 0$. When $0 \leqslant x \leqslant 1$, $0 \leqslant x y \leqslant y<1<\pi$, so $\cos x y \geqslant \cos y$.
Therefore, $\cos x+\cos y \leqslant 1+\cos x y$, where the equality holds when $x=y=0$.
Hence $z_{\max }=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (20 points) As shown in Figure 1, $EF$ is a chord of the parabola $\Gamma: y^{2}=2px$. Tangents to $\Gamma$ at points $E$ and $F$ intersect at point $C$. Points $A$ and $B$ are on the rays $EC$ and $CF$ respectively, and $\frac{EC}{CA}=\frac{CF}{FB}=\lambda$.
(1) Prove that the line $AB$ is tangent to $\Gamma$;
(2) Let $AB$ be tangent to $\Gamma$ at point $G$. Find $\frac{S_{\triangle EFG}}{S_{\triangle ABC}}$.
|
(1) Let $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then the equations of the tangents $E C$ and $C F$ are
$$
y_{1} y=p\left(x+x_{1}\right), y_{2} y=p\left(x+x_{2}\right).
$$
By solving the system of equations, we get
$$
C\left(\frac{y_{1} y_{2}}{2 p}, \frac{y_{1}+y_{2}}{2}\right).
$$
From $\frac{E C}{C A}=\frac{C F}{F B}=\lambda$, we have
$$
\frac{\boldsymbol{E} \boldsymbol{A}}{\boldsymbol{A C}}=\frac{\boldsymbol{C B}}{\boldsymbol{B F}}=-(1+\lambda).
$$
Let $A\left(m_{1}, n_{1}\right), B\left(m_{2}, n_{2}\right)$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
m_{1}=\frac{x_{1}-(1+\lambda) \frac{y_{1} y_{2}}{2 p}}{1-(1+\lambda)}=\frac{(1+\lambda) y_{1} y_{2}-y_{1}^{2}}{2 p \lambda}, \\
n_{1}=\frac{y_{1}-(1+\lambda) \frac{y_{1}+y_{2}}{2}}{1-(1+\lambda)}=\frac{(1+\lambda) y_{2}-(1-\lambda) y_{1}}{2 \lambda} ;
\end{array}\right. \\
\left\{\begin{array}{l}
m_{2}=\frac{\frac{y_{1} y_{2}}{2 p}-(1+\lambda) x_{2}}{1-(1+\lambda)}=\frac{(1+\lambda) y_{2}^{2}-y_{1} y_{2}}{2 p \lambda}, \\
n_{2}=\frac{\frac{y_{1}+y_{2}}{2}-(1+\lambda) y_{2}}{1-(1+\lambda)}=\frac{(1+2 \lambda) y_{2}-y_{1}}{2 \lambda} .
\end{array}\right. \\
\end{array}
$$
Thus, $n_{1}-n_{2}=\frac{y_{1}-y_{2}}{2}$,
$m_{1}-m_{2}=\frac{\left(y_{1}-y_{2}\right)\left[(1+\lambda) y_{2}-y_{1}\right]}{2 p \lambda}$.
Hence $\frac{n_{1}-n_{2}}{m_{1}-m_{2}}=\frac{p \lambda}{(1+\lambda) y_{2}-y_{1}}$.
Then $n_{1}-\frac{n_{1}-n_{2}}{m_{1}-m_{2}} \cdot m_{1}$
$$
\begin{aligned}
& =\frac{(1+\lambda) y_{2}-(1-\lambda) y_{1}}{2 \lambda}- \\
& \frac{p \lambda}{(1+\lambda) y_{2}-y_{1}} \cdot \frac{(1+\lambda) y_{1} y_{2}-y_{1}^{2}}{2 p \lambda} \\
= & \frac{(1+\lambda) y_{2}-y_{1}}{2 \lambda} .
\end{aligned}
$$
Therefore, the equation of the line $A B$ is
$$
y=\frac{n_{1}-n_{2}}{m_{1}-m_{2}} x+n_{1}-\frac{n_{1}-n_{2}}{m_{1}-m_{2}} \cdot m_{1}
$$
which can be written as
$$
y=\frac{p \lambda}{(1+\lambda) y_{2}-y_{1}} x+\frac{(1+\lambda) y_{2}-y_{1}}{2 \lambda} .
$$
By combining this with the equation $y^{2}=2 p x$ and eliminating $x$, we get
$$
y^{2}-\frac{2\left[(1+\lambda) y_{2}-y_{1}\right]}{\lambda} y+\frac{\left[(1+\lambda) y_{2}-y_{1}\right]^{2}}{\lambda^{2}}
$$
$=0$.
Since its discriminant
$$
\begin{array}{l}
\Delta=4\left[\frac{(1+\lambda) y_{2}-y_{1}}{\lambda}\right]^{2}-4 \cdot \frac{\left[(1+\lambda) y_{2}-y_{1}\right]^{2}}{\lambda^{2}} \\
=0,
\end{array}
$$
the line $A B$ is tangent to $\Gamma$.
(2) Let the y-coordinate of the tangent point $G$ be $n$. From (1), we have
$$
\begin{array}{l}
n=\frac{(1+\lambda) y_{2}-y_{1}}{\lambda} . \\
\text { Hence } \frac{A B}{B G}=\frac{n_{2}-n_{1}}{n-n_{2}}
\end{array}
$$
$$
\begin{array}{l}
=\frac{\frac{y_{2}-y_{1}}{2}}{\frac{(1+\lambda) y_{2}-y_{1}}{\lambda}-\frac{(1+2 \lambda) y_{2}-y_{1}}{2 \lambda}} \\
=\frac{\frac{y_{2}-y_{1}}{2}}{\frac{y_{2}-y_{1}}{2 \lambda}}=\lambda .
\end{array}
$$
Thus, $\frac{S_{\triangle A B G}}{S_{\triangle A B C}}=\frac{|A E| \cdot|A G|}{|A C| \cdot|A B|}$
$$
=(1+\lambda) \frac{1+\lambda}{\lambda}=\frac{(1+\lambda)^{2}}{\lambda}.
$$
Therefore, $\frac{S_{\text {quadrilateral } B C B C}}{S_{\triangle A B C}}=\frac{(1+\lambda)^{2}-\lambda}{\lambda}=\frac{\lambda^{2}+\lambda+1}{\lambda}$.
$$
\begin{array}{l}
\text { And } S_{\triangle C E F}=\frac{|E C| \cdot|C F|}{|C A| \cdot|C B|} S_{\triangle A B C} \\
=\lambda \cdot \frac{\lambda}{1+\lambda} S_{\triangle A B C}=\frac{\lambda^{2}}{1+\lambda} S_{\triangle A B C}, \\
S_{\triangle B G F}=\frac{|F B| \cdot|B G|}{|C B| \cdot|A B|} S_{\triangle A B C} \\
=\frac{1}{1+\lambda} \cdot \frac{1}{\lambda} S_{\triangle A B C}=\frac{1}{\lambda \cdot(1+\lambda)} S_{\triangle A B C}, \\
\text { Hence } S_{\triangle E F G}=S_{\text {quadrilateral } C C B C}-S_{\triangle C E F}-S_{\triangle B G F} \\
=2 S_{\triangle A B C}, \\
\text { i.e., } \frac{S_{\triangle E F G}}{S_{\triangle A B C}}=2 .
\end{array}
$$
|
2
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) The most recent mathematics competition consisted of 6 problems, with each correct answer scoring 7 points and each incorrect (or unanswered) question scoring 0 points. After the competition, a certain team scored a total of 161 points, and it was found during the score tallying that: any two participants in the team had at most two problems answered correctly in common, and no three participants answered the same two problems correctly. How many participants are there in the team at least?
Please retain the original text's line breaks and format, and output the translation result directly.
|
We find that by moving a point in the first column to another column in the same row, we can reduce the number of columns in Figure 6. For example, by making the move $(6,1) \rightarrow(6,2)$, we can simultaneously make the moves $(4,10) \rightarrow(6,3)$, $(3,9) \rightarrow(6,4)$, and $(5,9) \rightarrow(6,7)$, thus obtaining Figure 7 with 23 red points. Similarly, we can obtain Figure 8. This indicates that the minimum value of $n$ is no more than 7.
Three, suppose the team has $n$ players, denoted as $a_{1}$, $a_{2}$, ..., $a_{n}$, and the 6 problems are numbered as $1, 2, \cdots, 6$. We create a $6 \times n$ grid table, where if player $a_{i} (i=1,2, \cdots, n)$ answers the $j$-th problem $(j=1,2, \cdots, 6)$ correctly, we color the center of the small grid $(j, i)$ in the $j$-th row and $i$-th column red. Our problem is to find the minimum value of $n$ such that in a $6 \times n$ grid table, there are at least 23 red points without any "horizontal" 6 points.
If the first column has 6 red points, then the subsequent columns can have at most 2 red points each. Since $C_{6}^{2}=15>9$, we can take columns 2 to 10, where columns 2 to 9 each have 2 red points, and column 10 has 1 red point (as shown in Figure 6), which satisfies the problem's conditions. This indicates that the minimum value of $n$ is no more than 10.
Next, we prove that the minimum value of $n$ is greater than 6.
For a grid table with exactly 6 columns, by the pigeonhole principle, at least one column must have no fewer than 4 red points. Without loss of generality, assume the first column, and the first 4 rows of the first column have red points. If a column has 2 red points, we call it a "row pair" in that column. Thus, in the first 4 rows, each of the 5 columns other than the first column can have only one row pair. Therefore, the first 4 rows have a total of $C_{4}^{2}+5=11$ row pairs. Consider the last two rows: if the first column has additional red points, then the row with red points cannot have any other red points. If the first column has 2 additional red points, this can add 9 row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+9=20$; if the first column has 1 additional red point, assume the 5th row of the first column has a red point, then even if the 6th row has red points in all other columns except the first column, this can add $C_{4}^{1}+5 \times 2=14$ row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+14=25$; if the first column has no additional red points, then in the last two rows, there can be at most 2 row pairs, which occupy two columns, and in the remaining three columns, each column can have at most 1 red point, thus adding $2 \times 5+3 \times 2=16$ row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+16=27$. This indicates that 27 is the maximum possible total number of row pairs.
Let the number of red points in the $i$-th column be $x_{i} (i=1,2, \cdots, 6)$, and $\sum_{i=1}^{6} x_{i}=k$. Then the total number of row pairs is $\sum_{i=1}^{6} C_{x_{i}}^{2} \leqslant 27$, i.e.,
$$
\sum_{i=1}^{6} x_{i}^{2}-\sum_{i=1}^{6} x_{i} \leqslant 54.
$$
By the Cauchy-Schwarz inequality, we have
$$
\sum_{i=1}^{6} x_{i}^{2} \geqslant \frac{1}{6}\left(\sum_{i=1}^{6} x_{i}\right)^{2}=\frac{1}{6} k^{2}.
$$
Therefore, $\frac{k^{2}}{6} \leqslant k+54$.
Solving this, we get $3-3 \sqrt{37} \leqslant k \leqslant 3+3 \sqrt{37}$.
Since $k$ is a positive integer, $k \leqslant 21$. This indicates that the maximum number of red points in a $6 \times 6$ grid table is 21.
Furthermore, when $n \leqslant 5$, the total number of red points in the grid table is no more than $4 \times 5=20$. This indicates that the minimum value of $n$ is at least 7. In summary, the team must have at least 7 players.
(Liu Shixiong, Zhongshan Affiliated High School of South China Normal University, 528447)
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $f(x)$ is an odd function on the interval $\left[t, t^{2}-2 t-2\right]$, then the value of $t$ is $\qquad$
|
3. -1 .
The domain of an odd function is symmetric about the origin, and the right endpoint of the interval is not less than the left endpoint. Therefore,
$$
\left\{\begin{array} { l }
{ - t = t ^ { 2 } - 2 t - 2 > 0 , } \\
{ t \leqslant t ^ { 2 } - 2 t - 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t^{2}-t-2=0, \\
t<0, \\
t^{2}-3 t-2 \geqslant 0 .
\end{array}\right.\right.
$$
Solving this, we get $t \doteq-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The number of all distinct real solutions of the equation $2 x^{3}+6 x^{2}+12 x+8+x^{3} \sqrt{x^{2}+1}$ $+(x+2)^{3} \sqrt{(x+2)^{2}+1}=0$ is $\qquad$.
|
6.1.
Rewrite the equation as
$$
\begin{array}{l}
x^{3}+x^{3} \sqrt{x^{2}+1}+(x+2)^{3}+ \\
(x+2)^{3} \sqrt{(x+2)^{2}+1}=0 . \\
\text { Let } f(x)=x^{3}+x^{3} \sqrt{x^{2}+1} \\
=x^{3}\left(1+\sqrt{x^{2}+1}\right) .
\end{array}
$$
Then equation (1) can be rewritten as
$$
f(x+2)=-f(x) \text {. }
$$
Clearly, $f(-x)=-f(x)$.
From equations (2) and (3), we get
$$
f(x+2)=f(-x) \text {. }
$$
Clearly, $g(x)=x^{3}$ is an increasing function, and $h(x)=1+\sqrt{x^{2}+1}$ is an increasing function for $x \geqslant 0$ and a decreasing function for $x \leqslant 0$, and $h(x) \geqslant 2(x \in \mathbf{R})$. Therefore, when $x \geqslant 0$, $f(x)=g(x) h(x)$ is an increasing function; when $x \leqslant 0$, $f(x)=g(x) h(x)$ is also an increasing function.
If $x_{2}<0<x_{1}$, then $f\left(x_{2}\right)<f(0)<f\left(x_{1}\right)$. Therefore, $f(x)$ is an increasing function on $\mathbf{R}$.
Thus, from equation (4) we get $x+2=-x$, i.e., $x=-1$.
Clearly, $x=-1$ satisfies equation (1). Hence, the given equation has a unique real root $x=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(50 points)(1) Prove that there exist infinitely many pairs of positive integers $x, y$ such that $x^{2}-101 y^{2}=-1$.
(2) Given $0.30102<\lg 2<0.30103$,
$$
0.32220<\lg 2.1<0.32221 \text{. }
$$
Determine the first 6 digits before the decimal point and the first 131 digits after the decimal point of the number $(10+\sqrt{101})^{101}$ (without using calculators or similar tools).
|
(1) Let $A_{2 n-1}=(\sqrt{101}-10)^{2 n-1}$, $B_{2 n-1}=(\sqrt{101}+10)^{2 n-1}(n=1,2, \cdots)$.
Then $A_{2 n-1} B_{2 n-1}=1$.
Also, let $B_{2 n-1}=x_{2 n-1}+y_{2 n-1} \sqrt{101}$, where
$x_{2 n-1}, y_{2 n-1} \in \mathbf{N}_{+}$. By the binomial theorem,
$A_{2 n-1}=-x_{2 n-1}+y_{2 n-1} \sqrt{101}$.
Substituting $A_{2 n-1}$ and $B_{2 n-1}$ into equation (1) gives
$$
x_{2 n-1}^{2}-101 y_{2 n-1}^{2}=-1 \text {. }
$$
Therefore, there are infinitely many pairs of positive integers $X=x_{2 n-1}, Y=$ $y_{2 n-1}(n=1,2, \cdots)$, such that
$$
X^{2}-101 Y^{2}=-1 \text {. }
$$
(2) From (1) we get $B_{2 n-1}-A_{2 n-1}=2 x_{2 n-1}$, i.e.,
$$
B_{2 n-1}=2 x_{n}+A_{2 n-1} \text {. }
$$
Since $2 x_{n} \in \mathbf{Z}$, $A_{2 n-1}$ is the fractional part of $B_{2 n-1}$, and $2 x_{2 n-1}$ is the integer part of $B_{2 n-1}$. Hence,
$$
\begin{array}{l}
\frac{1}{21}<\sqrt{101}-10=\frac{1}{\sqrt{101}+10}<\frac{1}{10+10}=\frac{1}{20} \\
\Rightarrow 21^{-101}=\left(\frac{1}{21}\right)^{101}<A_{101}=(\sqrt{101}-10)^{101} \\
<\left(\frac{1}{20}\right)^{101}=20^{-101} \\
\Rightarrow-101(1+\lg 2.1)=-101 \lg 21 \\
<\lg A_{101}<101 \lg 20=-101(1+\lg 2) \\
\Rightarrow-133.54321=-101 \times 1.32221 \\
<\lg A_{101}<-101 \times 1.30102 \\
=-131.40302 \\
\Rightarrow 10^{-134} \times 10^{0.45679}<A_{101}<10^{-132} \times 10^{0.99698}
\end{array}
$$
Therefore, the 131 digits after the decimal point of $A_{101}$
are all 0. Thus, the 131 digits after the decimal point of $B_{101}=(\sqrt{101}+10)^{101}$ are all 0.
Next, we find the remainder when $2 x_{101}$ is divided by $10^{6}$.
By the binomial theorem,
$$
\begin{aligned}
& 2 x_{101} \\
= & 2\left(10^{101}+\mathrm{C}_{101}^{2} \times 101 \times 10^{99}+\cdots+\mathrm{C}_{101}^{96} \times 101^{48} \times 10^{5}+\right. \\
& \left.\mathrm{C}_{101}^{98} \times 101^{49} \times 10^{3}+\mathrm{C}_{101}^{100} \times 101^{50} \times 10\right) \\
\equiv & 2\left[\frac{101 \times 100 \times 99 \times 98 \times 97}{5 \times 4 \times 3 \times 2} \times 101^{48} \times 10^{5}+\right. \\
& \left.\frac{101 \times 100 \times 99}{3 \times 2}(100+1)^{49} \times 10^{3}+101^{11} \times 10\right]\left(\bmod 10^{6}\right) \\
\equiv & 101 \times 33 \times 49 \times 97 \times 101^{48} \times 10^{6}+ \\
& 33 \times(100+1)^{50} \times 10^{5}+2 \times 101^{51} \times 10\left(\bmod 10^{6}\right) \\
\equiv & 33 \times 10^{5}+2 \times(100+1)^{51} \times 10\left(\bmod 10^{6}\right) \\
\equiv & 3 \times 10^{5}+2\left(1+100 \times 51+100^{2} C_{51}^{2}\right) \times 10\left(\bmod 10^{6}\right) \\
\equiv & 3 \times 10^{5}+20+2000 \times 51+2 \times 10^{5} \times \frac{51 \times 50}{2}\left(\bmod 10^{6}\right) \\
\equiv & 3 \times 10^{5}+20+102000\left(\bmod 10^{6}\right) \\
\equiv & 300000+20+102000\left(\bmod 10^{6}\right) \\
\equiv & 402020\left(\bmod 10^{6}\right) .
\end{aligned}
$$
Therefore, the first 6 digits before the decimal point of $B_{101}=(\sqrt{101}+10)^{101}$ are 402020.
In summary, the first 6 digits before the decimal point of the number $(\sqrt{101}+10)^{101}$ are 402020, and the 131 digits after the decimal point are all 0.
|
402020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Xiao Wang walks along the street at a uniform speed and finds that a No. 18 bus passes him from behind every 6 min, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals, then, the interval between dispatches is $\qquad$ $\min$.
|
7.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, and the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, with the distance between two consecutive buses traveling in the same direction being $s \mathrm{~m}$.
From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s
\end{array}\right.
$$
Solving these equations, we get $s=4 x$, i.e., $\frac{s}{x}=4$.
Therefore, the interval time for bus No. 18 to depart from the main station is $4 \mathrm{~min}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 1, in $\triangle A B C$, $A B=7, A C$ $=11, M$ is the midpoint of $B C$, $A D$ is the angle bisector of $\angle B A C$, $M F / / A D$. Then the length of $F C$ is $\qquad$
|
8.9.
As shown in Figure 5, let $N$ be the midpoint of $AC$, and connect $MN$. Then $MN \parallel AB$.
Also, $MF \parallel AD$, so,
$\angle FMN = \angle BAD = \angle DAC = \angle MFN$.
Therefore, $FN = MN = \frac{1}{2} AB$.
Thus, $FC = FN + NC = \frac{1}{2} AB + \frac{1}{2} AC = 9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12.B. Given that $a$ and $b$ are positive integers, the quadratic equation $x^{2}-2 a x+b=0$ has two real roots $x_{1}$ and $x_{2}$, and the quadratic equation $y^{2}+2 a y+b=0$ has two real roots $y_{1}$ and $y_{2}$. It is also given that $x_{1} y_{1}-x_{2} y_{2}=2008$. Find the minimum value of $b$.
|
12.B. For the equation in $x$, $x^{2}-2 a x+b=0$, the roots are $a \pm \sqrt{a^{2}-b}$. For the equation in $y$, $y^{2}+2 a y+b=0$, the roots are $-a \pm \sqrt{a^{2}-b}$.
Let $\sqrt{a^{2}-b}=t$.
Then when $x_{1}=a+t, x_{2}=a-t, y_{1}=-a+t$, $y_{2}=-a-t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the condition;
When $x_{1}=a-t, x_{2}=a+t, y_{1}=-a-t$, $y_{2}=-a+t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the condition;
When $x_{1}=a-t, x_{2}=a+t, y_{1}=-a+t$, $y_{2}=-a-t$, we get $x_{1} y_{1}-x_{2} y_{2}=4 a t$;
When $x_{1}=a+t, x_{2}=a-t, y_{1}=-a-t$, $y_{2}=-a+t$, we get $x_{1} y_{1}-x_{2} y_{2}=-4 a t$.
Since $t=\sqrt{a^{2}-b}>0$, we have $a t=502$.
Given that $a$ is a positive integer, we know that $t$ is a rational number, and thus, $t$ is an integer.
From $a t=502$, we get $a=251, t=2$, so the minimum value of $b$ is $b=a^{2}-t^{2}=251^{2}-2^{2}=62997$.
Therefore, the minimum value of $b$ is 62997.
|
62997
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. B. As shown in Figure 2, in $\triangle ABC$, the lengths of the three sides are $BC = a$, $CA = b$, $AB = c$, where $a$, $b$, and $c$ are all integers, and the greatest common divisor of $a$ and $b$ is 2. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the perimeter of $\triangle ABC$.
|
13. B. As shown in Figure 8, extend $G I$, intersecting sides $B C$ and $C A$ at points $P$ and $Q$ respectively. Let the projections of the centroid $G$ on sides $B C$ and $C A$ be $E$ and $F$, and the inradius of $\triangle A B C$ be $r$. The lengths of the altitudes from $B C$ and $C A$ are $h_{a}$ and $h_{b}$ respectively. It is easy to see that $C P = C Q$.
From $S_{\triangle P Q C} = S_{\triangle G P C} + S_{\triangle C Q C}$, we get
$$
2 r = G E + G F = \frac{1}{3}\left(h_{a} + h_{b}\right),
$$
which means $2 \cdot \frac{2 S_{\triangle A B C}}{a + b + c} = \frac{1}{3}\left(\frac{2 S_{\triangle B B C}}{a} + \frac{2 S_{\triangle A B C}}{b}\right)$.
Thus, $a + b + c = \frac{6 a b}{a + b}$.
Since the centroid $G$ and the incenter $I$ of $\triangle A B C$ do not coincide, $\triangle A B C$ is not an equilateral triangle, and $b \neq a$. Otherwise, if $a = b = 2$, then $c = 2$, which is a contradiction.
Assume without loss of generality that $a > b$.
Since $(a, b) = 2$, let $a = 2 a_{1}$ and $b = 2 b_{1}$, with $\left(a_{1}, b_{1}\right) = 1$. Therefore, $\frac{6 a b}{a + b} = \frac{12 a_{1} b_{1}}{a_{1} + b_{1}}$ is an integer. Hence, $\left(a_{1} + b_{1}\right) \mid 12$, which means $(a + b) \mid 24$.
Therefore, only when $a = 14$ and $b = 10$, we get $c = 11$, satisfying the conditions.
Thus, the perimeter of $\triangle A B C$ is 35.
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given that the area of quadrilateral $ABCD$ is 32, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is 16.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(2003, National Junior High School Mathematics League)
|
Analysis: Note that the number of quadrilaterals $ABCD$ is determined by the number of triples of the lengths of $AB$, $CD$, and $AC$. Therefore, we can start by determining the shape of the quadrilateral $ABCD$.
Solution: (1) As shown in Figure 2, let
$$
AB = a, CD = b, AC = l
$$
$(a, b, l$ are all positive integers, and $a \leq b$). Let the height from $AB$ in $\triangle ABC$ be $h_a$, and the height from $DC$ in $\triangle ACD$ be $h_b$. Then
$$
\begin{array}{l}
S_{\text{quadrilateral } ABCD} = S_{\triangle ABC} + S_{\triangle ACD} \\
= \frac{1}{2} a h_a + \frac{1}{2} b h_b \leq \frac{1}{2} (a + b) l.
\end{array}
$$
Equality holds if and only if $h_a = h_b = l$. At this point, $AB \parallel CD$, meaning that quadrilateral $ABCD$ is a trapezoid or a parallelogram, and $AC$ is their height.
From the given information,
$$
64 \leq (a + b) l.
$$
Also, from the problem statement,
$$
a + b = 16 - l.
$$
Substituting equation (2) into equation (1) gives $(16 - l) l \geq 64$, which simplifies to
$$
(l - 8)^2 \leq 0.
$$
Thus, $l = 8$. At this point, $a + b = 8$.
Therefore, there are four such quadrilaterals:
$$
\begin{array}{l}
a = 1, b = 7, l = 8; a = 2, b = 6, l = 8; \\
a = 3, b = 5, l = 8; a = 4, b = 4, l = 8.
\end{array}
$$
(2) In quadrilateral $ABCD$, $AC \perp AB$, $AC \perp CD$. If $AB = a$, from the conclusion in (1),
$$
\begin{array}{l}
CD = 8 - a, BC^2 = AB^2 + AC^2 = a^2 + 8^2, \\
AD^2 = CD^2 + AC^2 = (8 - a)^2 + 8^2,
\end{array}
$$
where $a$ is a positive integer, and $1 \leq a \leq 4$.
Therefore, the sum of the squares of the side lengths of such a quadrilateral is
$$
\begin{array}{l}
AB^2 + BC^2 + CD^2 + DA^2 \\
= a^2 + (a^2 + 8^2) + (8 - a)^2 + [(8 - a)^2 + 8^2] \\
= 4a^2 - 32a + 256 = 4(a - 4)^2 + 192 \geq 192.
\end{array}
$$
Thus, when $a = 4 (= b)$, the above sum of squares is minimized, and the minimum value is 192.
|
192
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14.A. Choose $n$ numbers from $1,2, \cdots, 9$. Among them, there must be some numbers (at least one, or possibly all) whose sum is divisible by 10. Find the minimum value of $n$.
|
14. A. When $n=4$, the numbers $1,3,5,8$ do not have any subset of numbers whose sum is divisible by 10.
When $n=5$, let $a_{1}, a_{2}, \cdots, a_{5}$ be five different numbers from $1,2, \cdots, 9$. If the sum of any subset of these numbers cannot be divisible by 10, then $a_{1}, a_{2}, \cdots, a_{5}$ cannot simultaneously contain 1 and 9, 2 and 8, 3 and 7, 4 and 6. Therefore, $a_{1}, a_{2}, \cdots, a_{5}$ must contain the number 5.
If $a_{1}, a_{2}, \cdots, a_{5}$ contains 1, then it does not contain 9. Thus, it does not contain $4 (4+1+5=10)$, so it contains 6; thus, it does not contain $3 (3+6+1=10)$, so it contains 7; thus, it does not contain $2 (2+1+7=10)$, so it contains 8. However, $5+7+8=20$ is a multiple of 10, which is a contradiction.
If $a_{1}, a_{2}, \cdots, a_{5}$ contains 9, then it does not contain 1. Thus, it does not contain $6 (6+9+5=20)$, so it contains 4; thus, it does not contain $7 (7+4+9=20)$, so it contains 3; thus, it does not contain $8 (8+9+3=10)$, so it contains 2. However, $5+3+2=10$ is a multiple of 10, which is a contradiction.
In summary, the minimum value of $n$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 As shown in Figure $3, \odot O$ has a diameter whose length is the largest integer root of the quadratic equation in $x$
$$
x^{2}+2(k-2) x+k
$$
$=0$ (where $k$ is an integer). $P$ is a point outside $\odot O$. A tangent $PA$ and a secant $PBC$ are drawn from point $P$ to $\odot O$, with $A$ being the point of tangency, and $B, C$ being the points where the secant $PBC$ intersects $\odot O$. If the lengths of $PA, PB, PC$ are all positive integers, and the length of $PB$ is not a composite number, find the value of $PA^{2}+PB^{2}+PC^{2}$.
|
Analysis: First, find the value of the diameter, then discuss the cases based on $P B$ not being a composite number.
Solution: Let the two integer roots of the equation $x^{2}+2(k-2) x+k=0$ be $x_{1}$ and $x_{2}$, with $x_{1}>x_{2}$. Then
$$
x_{1}+x_{2}=-2(k-2), x_{1} x_{2}=k \text {. }
$$
From the above two equations, we know $2 x_{1} x_{2}+x_{1}+x_{2}=4$, which means
$$
\left(2 x_{1}+1\right)\left(2 x_{2}+1\right)=9=9 \times 1 \text {. }
$$
Since $x_{1}>x_{2}$, then $2 x_{1}+1=9,2 x_{2}+1=1$, which gives $x_{1}=4, x_{2}=0$.
Thus, the diameter of $\odot O$ is 4.
Note that $0<B C \leqslant 4$, and $P A, P B, P C$ are positive integers, so $B C=P C-P B$ is a positive integer.
Therefore, $B C=1,2,3,4$.
By the Power of a Point theorem, we have
$$
P A^{2}=P B \cdot P C=P B(P B+B C) \text {. }
$$
(1) When $B C=1$, from equation (1) we get
$$
P A^{2}=P B^{2}+P B \text {. }
$$
Thus, $P B^{2}<P A^{2}<(P B+1)^{2}$, which is a contradiction.
(2) When $B C=2$, from equation (1) we get
$$
P A^{2}=P B^{2}+2 P B \text {. }
$$
Thus, $P B^{2}<P A^{2}<(P B+1)^{2}$, which is a contradiction.
(3) When $B C=3$, from equation (1) we get
$$
P A^{2}=P B^{2}+3 P B \text {, }
$$
which means $(P A-P B)(P A+P B)=3 P B$.
Since $P B$ is not a composite number, and $P A-P B<P A+P B$, the only possibilities are
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ P A - P B = 1 , } \\
{ P A + P B = 3 P B ; }
\end{array} \quad \left\{\begin{array}{l}
P A-P B=3, \\
P A+P B=P B ;
\end{array}\right.\right. \\
\left\{\begin{array}{l}
P A-P B=P B, \\
P A+P B=3 .
\end{array}\right.
\end{array}
$$
Solving these, we get $P A=2, P B=1$.
In this case, $P C=P B+B C=1+3=4$.
Therefore, $P A^{2}+P B^{2}+P C^{2}=21$.
(4) When $B C=4$, from equation (1) we get
$$
P A^{2}=P B^{2}+4 P B \text {. }
$$
Thus, $(P B+1)^{2}<P B^{2}+4 P B=P A^{2}$
$$
<(P B+2)^{2},
$$
which is a contradiction.
In summary, $P A^{2}+P B^{2}+P C^{2}=21$.
|
21
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a, b$ are positive numbers, and
$$
a^{2009}+b^{2009}=a^{2007}+b^{2007} \text {, }
$$
then the maximum value of $a^{2}+b^{2}$ is $\qquad$
|
Given $a, b$, without loss of generality, assume $a \geqslant b>0$.
Then $a^{2} \geqslant b^{2}, a^{2007} \geqslant b^{2007}$.
$$
\begin{array}{l}
\text { At this point, }\left(a^{2007}-b^{2007}\right)\left(a^{2}-b^{2}\right) \geqslant 0 \\
\Rightarrow a^{2009}+b^{2009} \geqslant a^{2} b^{2007}+a^{2007} b^{2} \\
\Rightarrow\left(a^{2007}+b^{2007}\right)\left(a^{2}+b^{2}\right) \\
\quad=a^{2009}+b^{2009}+a^{2} b^{2007}+a^{2007} b^{2} \\
\quad \leqslant 2\left(a^{2009}+b^{2009}\right) .
\end{array}
$$
Considering $a^{2009}+b^{2009}=a^{2007}+b^{2007}$, and both sides are positive, so $a^{2}+b^{2} \leqslant 2$.
When $a=b=1$, $a^{2}+b^{2}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Given $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. Find the integer part of $x^{6}+x^{5}+2 x^{4}-4 x^{3}+3 x^{2}+4 x-4$.
|
Given $x>0$.
If $\sqrt{1+x}>x$, then
$$
\begin{array}{l}
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}>\sqrt{1+\sqrt{1+x}} \\
>\sqrt{1+x},
\end{array}
$$
which contradicts the assumption;
If $\sqrt{1+x}<x$, then
$$
\begin{array}{l}
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}<\sqrt{1+\sqrt{1+x}} \\
<\sqrt{1+x},
\end{array}
$$
which also contradicts the assumption.
Therefore, $\sqrt{1+x}=x$.
Solving gives $x=\frac{1+\sqrt{5}}{2}$ (negative value is discarded).
$$
\begin{aligned}
\text { Hence the original expression }= & \left(x^{2}-x-1\right)\left(x^{4}+2 x^{3}+5 x^{2}+\right. \\
& 3 x+11)+18 x+7 \\
= & 18 \times \frac{1+\sqrt{5}}{2}+7=16+9 \sqrt{5}=16+\sqrt{405} .
\end{aligned}
$$
Since $20<\sqrt{405}<21$, the integer part of the original expression is 36.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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