problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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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
$$
\begin{array}{l}
a_{1}<a_{2}<a_{3}<a_{4}<a_{5}, \\
A \cap B=\left\{a_{1}, a_{4}\right\}... | 11.2.
Since $a_{1}^{2}=a_{1}$, therefore, $a_{1}=1, a_{4}=9$.
Since $B$ contains 9, $A$ contains 3.
If $a_{3}=3$, then $a_{2}=2$.
Thus, $a_{5}+a_{5}^{2}=146$. No positive integer solution.
If $a_{2}=3$, since $10 \leqslant a_{5} \leqslant 11$, then $a_{2}^{2} \neq a_{5}$.
Thus, $a_{3}+a_{3}^{2}+a_{5}+a_{5}^{2}=152$.
Al... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{0}=0$, $a_{1}=1$, for all positive integers $n$, we have
$$
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 {. }
$$
By the binomial theorem,
... | 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. There are 10 students standing in a row, and their birthdays are in different months. There are $n$ teachers who will select these students to join $n$ interest groups. Each student is selected by exactly one teacher, and the order of the students is maintained. Each teacher must select students whose birthdays are... | 15. If $n \leqslant 3$, let's assume the birth months of these 10 students are $1,2, \cdots, 10$.
When the students are sorted by their birthdays as $4,3,2,1,7,6,5,9, 8,10$, there exists at least one teacher who must select two students from the first four. Since the birth months of these two students are decreasing, ... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
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}\... | 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 ge... | 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
$$
\le... | 68 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. For the function $f(x)=\sqrt{a x^{2}+b x}$, there exists a positive number $b$, such that the domain and range of $f(x)$ are the same. Then the value of the non-zero real number $a$ is $\qquad$. | 11. -4 .
If $a>0$, for the positive number $b$, the domain of $f(x)$ is
$$
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}
$$
Similarly, $\frac{\sin (x-y)}{x-y}=\cos x \cdot \cos y$. Therefore, $\frac{\sin (x+y)}{x+y}-\frac{\sin (x... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Definition: The length of the interval $\left[x_{1}, x_{2}\right]\left(x_{1}<x_{2}\right)$ is $x_{2}-x_{1}$. Given that the domain of the function $y=\left|\log _{\frac{1}{2}} x\right|$ is $[a, b]$, and the range is $[0,2]$. Then the difference between the maximum and minimum values of the length of the interval $[... | 12.3.
The graphs of the functions $y=$ $\left|\log _{\frac{1}{2}} x\right|$ and $y=2$ are shown in Figure 7.
From $y=0$, we get $x=1$;
From $y=2$, we get $x=\frac{1}{4}$ or $x=4$.
When $a=\frac{1}{4}, b=1$, $(b-a)_{\text {min }}=\frac{3}{4}$;
When $a=\frac{1}{4}, b=4$, $(b-a)_{\max }=\frac{15}{4}$.
Therefore, $(b-a)_... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Given the parabola $y=x^{2}+3 x+c$ passes through two points $(m, 0)$ and $(n, 0)$, and
$$
m^{3}+3 m^{2}+(c-2) m-2 n-c=8,
$$
the parabola intersects the hyperbola $y=\frac{k}{x}$ at the point $(1, d)$.
(1) Find the equations of the parabola and the hyperbola;
(2) Given points $P_{1}, P_{2}, \cdots, P_... | (1) According to the problem, we have
$$
\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.
$$
Substituting equations (2) and (3) into equation (1) gives $c=-2$.
Therefore, the equation of the parabola is $y=x^{2}+3 x-2$.
Since the intersection point of the pa... | 2017035 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The largest solution and the smallest solution of the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$ sum to $(\quad)$ (where, $[x]$ denotes the greatest integer not exceeding $x$, the same applies below).
(A) 85
(B) -85
(C) 42
(D) -42 | 4.B.
Let $x=42 p+q(p, q$ be integers, $0 \leqslant q \leqslant 41)$.
Substituting $x$ into the original equation, we get
$$
p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q \text {. }
$$
For each different $q$, a unique ordered pair $(p, q)$ is determined, and thus, $x$ is also unique. To... | -85 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Let the circumcenter, incenter, and orthocenter of non-isosceles $\triangle ABC$ be $O$, $I$, and $H$, respectively, with the circumradius being $1$ and $\angle A=60^{\circ}$. Then the circumradius of $\triangle OIH$ is $\qquad$. | If $\triangle A B C$ is an acute triangle, since
$$
\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 {, }
$$
and $H$ is on the oppo... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}(n \geqslant 1)$ satisfy
$$
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{a... | 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... | 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 $|z|>1$, sin... | 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 $f(x)=\frac{(1+x)^{2}}{2 x^{2}+(1-x)^{2}}$
$$
=\frac{1}{3}\left[1+\frac{8 x... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
5. Find the smallest positive integer $n$ such that: If each vertex of a regular $n$-gon is arbitrarily colored with one of the three colors red, yellow, or blue, then there must exist four vertices of the same color that form the vertices of an isosceles trapezoid (a convex quadrilateral with one pair of parallel side... | 5. The smallest value of $n$ sought is 17.
First, prove: When $n=17$, the conclusion holds.
Proof by contradiction.
Assume there exists a method to three-color the vertices of a regular 17-gon such that no four vertices of the same color form the vertices of an isosceles trapezoid.
Since $\left[\frac{17-1}{3}\right]+... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. As shown in Figure 4, in $\triangle A B C$, $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=3, C D=2$, then $S_{\triangle A B C}$ $=$ $\qquad$ . | 10.15 .
Solution 1: Let $A B=c, A C=b$. Then
$$
\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=\fr... | 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 gre... | 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, $\Delta=(y-1)^{2}-4(y-51)$
$$
=y^{2}-6 y+205=(y-3)^{2}+196 \text{... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive integers, and $a_{1}<a_{2}<\cdots<a_{n} \leqslant 2007$, to ensure that there always exist four distinct numbers $a_{i} 、 a_{j} 、 a_{k} 、 a_{l}$, such that $a_{i}+a_{j}=a_{k}+a_{l}=a_{n}$, what is the minimum value of $n$? And explain the reason. | 15. Let $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005$, $a_{1004}=2007$, a total of 1004 odd numbers. Clearly, the sum of any two numbers is not equal to 2007.
If we add the number 2006 to the above 1004 numbers, i.e., $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005, a_{1004}=$ $2006, a_{1000}=2007$, at this point, there exists o... | 1006 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) Each point on the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$... | Obviously, $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arcs $\overparen{A B}$ (including $A$ but not $B$), $\overparen{B C}$ (including $B$ but not $C$), and $\overparen{C A}$ (including $C$ but not $A$) with colors $1, 2,$ and $3$ respectively, and color all oth... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Six. (25 points) Given
$$
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}
$$
Since $0 <... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Seven, (25 points) Let $n$ be a positive integer, $a=[\sqrt{n}]$ (where $[x]$ denotes the greatest integer not exceeding $x$). Find the maximum value of $n$ that satisfies the following conditions:
(1) $n$ is not a perfect square;
(2) $a^{3} \mid n^{2}$.
(Zhang Tongjun
Zhu Yachun, problem contributor) | Seven, from (1) we get $a<\sqrt{n}<a+1$, then $a^{2}<n<a^{2}+2 a+1$,
that is
$a^{2}+1 \leqslant n \leqslant a^{2}+2 a$.
Let $n=a^{2}+t(t \in\{1,2, \cdots, 2 a\})$.
From (2) we have
$a^{3}\left|\left(a^{4}+2 a^{2} t+t^{2}\right) \Rightarrow a^{2}\right| t^{2} \Rightarrow a \mid t$.
Furthermore, $a^{3} \mid \left(a^{4}+2... | 24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. As shown in Figure 2, in the polyhedron ABCDEF, it is known that quadrilateral $ABCD$ is a square with side length 3, $EF$ $/ / AB, EF=\frac{3}{2}$. If the volume of the polyhedron is $\frac{15}{2}$, then the distance between $EF$ and $AC$ is | 9.2 .
Take the midpoints of $A B$ and $C D$ as $M$ and $N$, and connect $F M$, $F N$, and $M N$.
Since $E F / / A M$ and $E F = \frac{3}{2} = A M$, we know that the polyhedron $A D E - M N F$ is a triangular prism.
Let the distance between $E F$ and $A C$ be $h$. From
$$
V_{A D E-M N F} + V_{F-B C M M} = \frac{15}{2}... | 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... | 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 ... | 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\... | 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, $b_{n}$ ... | 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 $\triangle A B C$, $I_{1} 、 I_{2} 、 I_{3} 、$ are the incenter of $\triangle A B C 、 \triangle A C D$ 、 $\triangle B C D$ 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
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{20^{2}+15^{2}}=25 \text {.... | 5 | Geometry | MCQ | Yes | Yes | cn_contest | false |
One, (20 points) Given that $n$ and $k$ are positive integers, and satisfy the inequality
$$
\frac{1}{7}<\frac{n-k}{n+k}<\frac{63}{439} .
$$
If for a given positive integer $n$, there is only one positive integer $k$ that makes the inequality true. Find the maximum and minimum values of all positive integers $n$ that ... | From the known inequality, we have
$$
\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}
$$
no... | 2008 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. $a, b$ are constants. If the parabola $C$:
$$
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 {, }
$$
which simplifies to $t(1-a)+\left(1-2 a^{2}+b\right)=0$,
holding for all $t$.
Thus, $1-a=0$, and $1-2 a^{2}+b=0$.
Solving these, we get $a=1, b=1$.
Substituting int... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. To make $p=x^{4}+6 x^{3}+11 x^{2}+3 x+32$ a perfect square of an integer, then the integer $x$ has $\qquad$ solutions. | 2.0 .
From the problem, we know that $p \equiv x^{4}-x^{2}+2(\bmod 3)$.
Since $x^{2} \equiv 0,1(\bmod 3)$, thus,
$$
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}
$$
It is clear that equality can be achieved.
Therefore... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, let $P$ be a point inside the convex quadrilateral $A B C D$. Draw perpendiculars from $P$ to $A B$, $B C$, $C D$, and $D A$, with the feet of the perpendiculars being $E$, $F$, $G$, and $H$, respectively. Given that $A H=3$, $H D=4$, $D G=1$, $G C=5$, $C F=6$, $F B=4$, and $B E-A E=1$. Then th... | 6.34.
By the Pythagorean theorem, we have
$$
\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... | 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
$$
S(x)=a+b+9, S(x+1)=a+b+1 \text ... | 46 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $k$ is a positive integer not exceeding 50, such that for any positive integer $n, 2 \times 3^{6 n}+k \times 2^{3 n+1}-1$ is always divisible by 7. Then the number of such positive integers $k$ is $\qquad$. | 9.7.
$$
\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 {, }
$$
which means $2 k+1=7 m$ (whe... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (20 points) In a box, there are a total of 88 small balls in three colors: red, yellow, and black. It is known that by randomly taking out 24 balls, it can be guaranteed that at least 10 of the balls are of the same color. Under the condition that this is true, regardless of how the balls of each color are distri... | First, prove that taking out only 43 balls is not enough.
In fact, when there are 42 red balls, 41 yellow balls, and 5 black balls in the box, if you take any 24 balls, then the number of red and yellow balls is at least $24-5=19$, so there must be at least 10 red or yellow balls; but taking 19 red balls, 19 yellow bal... | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Try to find all positive integers $k$, such that for any positive integers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a$, $b$, and $c$.
(2002, Girls' Mathematical Olympiad) | Explanation: First, from the inequality relationship between $a b+b c+c a$ and $a^{2}+b^{2}+c^{2}$, we derive $k>5$. Then, by constructing an example, we find that $k \leqslant 6$. Therefore, since $k \in \mathbf{Z}_{+}$, we conclude that $k=6$. Finally, we provide the proof.
Notice that for any positive real numbers ... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. On a plane, 2007 non-coincident lines $l_{1}, l_{2}, \cdots, l_{2007}$ are drawn, always following the rule of alternating perpendicular and parallel $\left(l_{2} \perp l_{1}, l_{3} / / l_{2}, l_{4} \perp l_{3}, l_{5} / /\right.$ $\left.l_{4}, \cdots\right)$. These 2007 non-coincident lines have a total of intersect... | 5.
$$
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... | 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 $a, b, c$... | 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 equat... | 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 $\tr... | 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... | 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 $a^{2}+3(\sqrt{2008}+37) a b+10$
$$
\begin{array}{l}
=9^{2}+3(\sqrt{2008}... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 4, given that the equilateral $\triangle ABC$ is inscribed in $\odot O, AB$ $=86$. If point $E$ is on side $AB$, and through $E$ a line $DG \parallel BC$ intersects $\odot O$ at points $D, G$, and intersects $AC$ at point $F$, and let $AE=x, DE$ $=y$. If $x, y$ are both positive integers, then $y=... | 4.12.
From the problem, we know that $E F=A E=x$.
By the symmetry of the circle, $F G=D E=y$.
By the intersecting chords theorem, we have $A E \cdot E B=D E \cdot E G$, which means $x(86-x)=y(x+y)$.
If $x$ is odd, then $x(86-x)$ is also odd, and in this case, $y(x+y)$ is even. Therefore, $x$ must be even, and $y$ must... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given prime numbers $p$ and $q$ such that the algebraic expressions $\frac{2 p+1}{q}$ and $\frac{2 q-3}{p}$ are both natural numbers. Try to find the value of $p^{2} q$.
---
The translation is provided as requested, maintaining the original format and line breaks. | Let $p \geqslant q$, then we have $1 \leqslant \frac{2 q-3}{p}<2$.
Thus, it can only be $\frac{2 q-3}{p}=1$, which means $p=2 q-3$.
At this point, $\frac{2 p+1}{q}=\frac{4 q-5}{q}=4-\frac{5}{q}$.
To make $\frac{2 p+1}{q}$ a natural number, it can only be $q=5$, hence $p=7$.
Now, let $p<q$, in this case, $1 \leqslant \... | 245 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 5, in $\odot O$, $AB$ and $CD$ are two perpendicular diameters. Point $E$ is on radius $OA$, and point $F$ is on the extension of radius $OB$, such that $OE = BF$. Lines $CE$ and $CF$ intersect $\odot O$ at points $G$ and $H$, respectively. Lines $AG$ and $AH$ intersect line $CD$ at p... | As shown in Figure 7, connect $D G$ and $D H$, and extend $C G$ to point $P$. Since $A B$ and $C D$ are perpendicular diameters,
we have
$$
\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}
$$... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
1. Let the geometric sequence $z_{1}, z_{2}, \cdots, z_{n}, \cdots$ be such that $z_{1}=$ $1, z_{2}=a+b \mathrm{i}, z_{3}=b \mathrm{i}(a, b \in \mathbf{R}, ab>0)$. Then the smallest natural number $n$ for which $z_{1} z_{2} \cdots z_{n}<0$ is $\qquad$ . | $Ni, 1.8$.
$$
\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},... | 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... | 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,
$... | 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 ... | 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
$$
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{4}+x_{1}^{2}=1 \\
\... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In the Cartesian coordinate system, there is a parabola $y=$ $x^{2}-(5 c-3) x-c$ and three points $A\left(-\frac{1}{2} c, \frac{5}{2} c\right)$, $B\left(\frac{1}{2} c, \frac{9}{2} c\right)$, $C(2 c, 0)$, where $c>0$. There exists a point $P$ on the parabola such that the quadrilateral with vertices $A$, $B$, $C$, an... | 4.3.
(1) If $A B$ is the diagonal, then $P_{1}(-2 c, 7 c)$. For $P_{1}$ to be on the parabola, it must satisfy
$$
7 c=(-2 c)^{2}-(5 c-3)(-2 c)-c \text {. }
$$
Solving this, we get $c_{1}=0$ (discard), $c_{2}=1$.
Thus, $c=1$, and at this point, $P_{1}(-2,7)$.
(2) If $B C$ is the diagonal, then $P_{2}(3 c, 2 c)$. Simila... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 2, $EF$ intersects the diagonal $AC$ of $\square ABCD$ at point $G$,
intersects $AB$, $AD$ at points
$E$, $F$, and connects $CE$,
$CF$, $BG$. If $\frac{1}{S_{\triangle ACE}}+$
$\frac{1}{S_{\triangle ACF}}=\frac{\lambda}{S_{\triangle ABG}}$, find
the value of $\lambda$. | II. As shown in Figure 8, draw $EM \parallel BC$ intersecting $AC$ at $M$, then $\frac{AB}{AE}=\frac{AC}{AM}$.
From $\frac{AD}{AF}=\frac{BC}{AF}$
$=\frac{BC}{EM} \cdot \frac{EM}{AF}$
$=\frac{AC}{AM} \cdot \frac{GM}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AM}+\frac{AC}{AM} \cdot \frac{GM}{AG}$
$=\frac{AC... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
---
In $\tr... | Three, as shown in Figure 9, let the sides of $\triangle ABC$ be $BC = a$, $CA = b$, and $AB = c$. Let $CD = h$, $AD = x$, and $BD = y$. The radii of the two incircles are $r_1$ and $r_2$. Then,
$$
\begin{array}{l}
RS = |RD - SD| \\
= |r_1 - r_2|. \\
\text{Since } b = AC = AP + CP = AE + CR \\
= (x - r_1) + (h - r_1),
... | 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 $... | 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 i... | 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))... | 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: $n \geqslant 22$.
$$
\begin{array}{l}
\quad \text { Let } A_{i}=\{i, i+5, i+10, i+15, i+20, i+25\}, \\
i=1,2, \cdots, 5, \\
\quad B_{j}=\{j, j+6, j+12, j+18, j+24\}, j= \\
1,2, \cdots, 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Clearly, }\left|A_{i}\right|=6(i=1,... | 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 In $\triangle A B C$, $A^{\prime} 、 B^{\prime} 、 C^{\prime}$ are on $B C 、 C A 、 A B$ respectively. Given that $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$ concur at $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \fra... | Let $\angle A O B^{\prime}=\theta_{1}, \angle A O C^{\prime}=\theta_{2}$.
As shown in Figure 3, from $S_{\triangle A O B}+$
$S_{\triangle C O B^{\prime}}=S_{\triangle M O C}$, we get
$$
\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. \\
\le... | 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 $\triangle C A B$, $\angle A=$ $90^{\circ}, \angle B 、 \angle C$ are bisected and intersect at $F$, and intersect the opposite sides at points $D 、 E$. 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, $S_{\triangle C B E}=\frac{1}{2} B E \cdot b=\frac{a b c}{2(a+b)}$.
An... | 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 k... | 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 $\{0,2\}$, $f(a) 、 f(b)$ 、 $f(c) 、 f(d)$ have 2 zeros and 2 twos. Enumerating, we find there are 6 such mappings;
(3) When the image set is $\{0,1,2\}$, $f(a) 、 f(b)$ 、 $f(c) 、 f(d)$ have 1 ze... | 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 $\qquad$ | 二、7.14.
$A * B$ 中元素为 $2,3,4,5$, 故其所有元素数字之和为 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
$$
-3 b=3,3 b^{2}-d^{2}=-1,-b^{3}+b d^{2}=a \text {. }... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (i) (Grade 10) $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=$ $\qquad$ .
(ii) (Grade 11) Using the digits $1,2,3,4,5,6$ to number the six faces of a cube. If a numbered cube can be rotated to match the numbering of another cube, they are considered the same numbering method. Therefore, the numbe... | 9. (i) $\sqrt{2}$.
Since $\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}$
$=\cos 40^{\circ} \sqrt{1-\cos 80^{\circ}}$
$=\cos 40^{\circ} \cdot \sqrt{2} \sin 40^{\circ}=\frac{\sqrt{2}}{2} \sin 80^{\circ}=\frac{\sqrt{2}}{2} \cos 10^{\circ}$,
therefore, $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=\sqrt{2}$... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. If $\frac{1^{2}+3^{2}+\cdots+(2 n-1)^{2}}{2^{2}+4^{2}+\cdots+(2 n)^{2}}=\frac{13}{14}$, then
$$
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}
$$
we have $\triangle A B P \cong \triangle A^{\prime} B P$.
Th... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) Find the smallest positive integer $n$ such that: for any $n$ points $A_{1}, A_{2}$, $\cdots, A_{n}$ taken on the circumference of $\odot O$, among the $\mathrm{C}_{n}^{2}$ angles $\angle A_{i} O A_{j}(1 \leqslant i<j \leqslant n)$, at least 2007 are not greater than $120^{\circ}$. | 12. First, when $n=90$, as shown in Figure 7, let $AB$ be the diameter of $\odot O$. Take 45 points near points $A$ and $B$, respectively. At this time, there are only $2 \mathrm{C}_{45}^{2}=45 \times 44=1980$ angles that do not exceed $120^{\circ}$. Therefore, $n=90$ does not satisfy the problem's requirements.
Second... | 91 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Calculate: $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=$ | 9.4 .
$$
\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
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. To cut a rectangular prism into $k$ tetrahedra, the minimum value of $k$ is | 11.5 .
According to the equivalence, we only need to consider the cutting situation of a unit cube.
On the one hand, first, we need to show that 4 is not enough. If there were 4, since all faces of a tetrahedron are triangles and are not parallel to each other, the top face of the cube would have to be cut into at le... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$ . | 12.133113.
We call this kind of number a "good number". There are 3 one-digit good numbers; 12 two-digit good numbers, which is $3 \times 4$; 48 three-digit good numbers, which is $3 \times 4^2$; $\cdots \cdots \cdot k$-digit good numbers have $3 \times 4^{k-1}$ numbers $(k=1$, $2, \cdots)$. Let $S_{n}=3 \sum_{k=1}^{n... | 133113 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. If a four-digit number $n=\overline{a b c d}$ has digits $a, b, c, d$ such that any three of these digits can form the lengths of the sides of a triangle, then $n$ is called a "four-digit triangular number". Find the total number of four-digit triangular numbers. | 15. Call $(a, b, c, d)$ a digital group of $n$, then
$$
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... | 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 ... | - 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 \mathb... | 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... | 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
保留源文本的换行和格式，直接输出翻译结果。 | 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 t... | 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 {, }
$$
which means $a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right)+m=0$,
where $m=a_{2}\left(10^{k-2}-1\right)+\cdots+a_{k-1}(... | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$
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.
$$
(1) Find $a_{2008}$;
(2) Find the smallest positive integer $n$, such that $a_{n}=2008$. | Five, the initial values of the easy sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & $2 \mid 3$ & & & 6 & 7 & 8 & & 10 & 11 & 12 & & \\
\hline$a$ & 1 & 2 & 41 & & 10 & & 11 & & $12 \mid$ & 2 & 13 & 1 & 14 \\
\hline
\end{tabular}
Next, focus on the subscripts $n_{k}$ such ... | 5827 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $x, y, z$ be non-negative real numbers, and $x+y+z=$ 2. Then the sum of the maximum and minimum values of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ is $\qquad$ . | 2.1.
Since $x, y, z$ are non-negative real numbers, we have
$$
A=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant 0.
$$
When $x=y=0, z=2$, $A=0$, thus the minimum value of $A$ is 0.
Assume $A$ reaches its maximum value at $(x, y, z)$, without loss of generality, let $x \leqslant y \leqslant z$, prove: $x=0$.
In fact, ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a, b$ be any two distinct positive integers. Then the minimum value of $\left|a b\left(a^{4}-b^{4}\right)\right|$ is $\qquad$ .
| 3.30.
$$
\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}
$$
First, prove: $2 \mid A$.
If $a$ and $b$ are both odd or both even, then $2 \mid (a-b)$, so $2 \mid A$;
If $a$ and $b$ are of different parity, then one of $a$ or $b$ is even, so $2 \mid ab$... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $t$, such that for any convex $n$-gon $A_{1} A_{2} \cdots A_{n}$, as long as $n \geqslant t$, there must exist three points $A_{i} 、 A_{j} 、 A_{k}(1 \leqslant i<j<k \leqslant n)$, such that the area of $\triangle A_{i} A_{j} A_{k}$ is no more than $\frac{1}{n}$ of t... | Three, first prove a lemma.
Lemma For any convex hexagon $A_{1} A_{2} \cdots A_{6}$, there exists $1 \leqslant i\frac{S}{k+1}$, then
$S_{\text {pentagon } A_{1} A_{2} \cdots A_{k}}<S-\frac{S}{k+1}=\frac{k S}{k+1}$.
By the induction hypothesis, there must be $1 \leqslant i<j<r \leqslant n$, such that $S_{\triangle A_{i}... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x^{2}+y^{2} \leqslant 1$. Then the maximum value of the function $z=$ $\frac{\cos x+\cos y}{1+\cos x y}$ is $\qquad$ . | 2 1.1.
Assume $x \geqslant 0, y \geqslant 0$. When $0 \leqslant x \leqslant 1$, $0 \leqslant x y \leqslant y<1<\pi$, so $\cos x y \geqslant \cos y$.
Therefore, $\cos x+\cos y \leqslant 1+\cos x y$, where the equality holds when $x=y=0$.
Hence $z_{\max }=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) As shown in Figure 1, $EF$ is a chord of the parabola $\Gamma: y^{2}=2px$. Tangents to $\Gamma$ at points $E$ and $F$ intersect at point $C$. Points $A$ and $B$ are on the rays $EC$ and $CF$ respectively, and $\frac{EC}{CA}=\frac{CF}{FB}=\lambda$.
(1) Prove that the line $AB$ is tangent to $\Gamma$;
(... | (1) Let $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then the equations of the tangents $E C$ and $C F$ are
$$
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).
$$
From $\frac{E C}{C... | 2 | Geometry | proof | Yes | Yes | cn_contest | false |
Three. (50 points) The most recent mathematics competition consisted of 6 problems, with each correct answer scoring 7 points and each incorrect (or unanswered) question scoring 0 points. After the competition, a certain team scored a total of 161 points, and it was found during the score tallying that: any two partici... | We find that by moving a point in the first column to another column in the same row, we can reduce the number of columns in Figure 6. For example, by making the move $(6,1) \rightarrow(6,2)$, we can simultaneously make the moves $(4,10) \rightarrow(6,3)$, $(3,9) \rightarrow(6,4)$, and $(5,9) \rightarrow(6,7)$, thus ob... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If $f(x)$ is an odd function on the interval $\left[t, t^{2}-2 t-2\right]$, then the value of $t$ is $\qquad$ | 3. -1 .
The domain of an odd function is symmetric about the origin, and the right endpoint of the interval is not less than the left endpoint. Therefore,
$$
\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}-... | -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)... | -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+\s... | (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... | 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 ... | 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
... | 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 c... | 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... | 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}$ respe... | 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 Mathemat... | 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
$$
$(a, b, l$ are all positive integers,... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14.A. Choose $n$ numbers from $1,2, \cdots, 9$. Among them, there must be some numbers (at least one, or possibly all) whose sum is divisible by 10. Find the minimum value of $n$. | 14. A. When $n=4$, the numbers $1,3,5,8$ do not have any subset of numbers whose sum is divisible by 10.
When $n=5$, let $a_{1}, a_{2}, \cdots, a_{5}$ be five different numbers from $1,2, \cdots, 9$. If the sum of any subset of these numbers cannot be divisible by 10, then $a_{1}, a_{2}, \cdots, a_{5}$ cannot simultan... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 As shown in Figure $3, \odot O$ has a diameter whose length is the largest integer root of the quadratic equation in $x$
$$
x^{2}+2(k-2) x+k
$$
$=0$ (where $k$ is an integer). $P$ is a point outside $\odot O$. A tangent $PA$ and a secant $PBC$ are drawn from point $P$ to $\odot O$, with $A$ being the point of... | Analysis: First, find the value of the diameter, then discuss the cases based on $P B$ not being a composite number.
Solution: Let the two integer roots of the equation $x^{2}+2(k-2) x+k=0$ be $x_{1}$ and $x_{2}$, with $x_{1}>x_{2}$. Then
$$
x_{1}+x_{2}=-2(k-2), x_{1} x_{2}=k \text {. }
$$
From the above two equation... | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a, b$ are positive numbers, and
$$
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} \\
\Right... | 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 contradi... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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