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4. Among the following four propositions:
(1) A quadrilateral with one pair of opposite sides equal and one pair of opposite angles equal is a parallelogram;
(2) A quadrilateral with one pair of opposite sides equal and one diagonal bisecting the other diagonal is a parallelogram;
(3) A quadrilateral with one pair of o... | 4. (4).
Propositions (1), (2), and (3) can be refuted with the following counterexamples:
Proposition (1): The quadrilateral $ABCD$ in Figure 5(a), where $\triangle ABD \cong \triangle CDE$.
Proposition (2): As shown in Figure 5(b), construct an isosceles $\triangle ADE$, extend the base $ED$ to any point $O$, and us... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If 6 pieces of $1 \times 2$ paper are used to cover a $3 \times 4$ grid, the number of different ways to cover it is. | 4.11.
As shown in Figure 8, the cells of the grid are numbered. Let $M(a, b)$ denote the number of ways to cover the entire grid when the cells numbered $a$ and $b$ (which are adjacent) are covered by the same piece of paper. We focus on the covering of cell 8. It is given that $M(8,5) = 2, M(8,11) = 3, M(8,7) = M(8,9... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Second question As shown in Figure 8, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center point of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed ... | Solution: The minimum number of chess pieces to be removed is 11.
Define the square at the $x$-th column and $y$-th row as $(x, y)$ $(1 \leqslant x \leqslant 8,1 \leqslant y \leqslant 7, x, y \in \mathbf{Z})$.
A group of 5 consecutive chess pieces is called a "good group". The problem requires that no chess pieces for... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. Equation
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+ \\
\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}+\frac{x^{3}+4^{3}}{(x+4)^{3}}+\frac{x^{3}+9^{3}}{(x+9)^{3}}\right]=1
\end{array}
$$
The number of distinct non-zero integer solutions is | 10.4 .
Using $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, the original equation is
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+1+\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}-\right. \\
\left.1+\frac{x^{3}+4^{3}}{(x+4)^{3}}-1+\frac{x^{3}+9^{3}}{(x+9)^{3}}-1\right]=0 \\
\Leftrightarrow \frac{x^{3}+49 x... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$,
$$
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 |
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 |
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 |
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 |
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 |
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 $f(x)=\frac{(1+x)^{2}}{2 x^{2}+(1-x)^{2}}$
$$
=\frac{1}{3}\left[1+\frac{8 x... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
14. In a dormitory of a school, there are several students, one of whom serves as the dorm leader. During New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator. Each dormitory administrator also gives a gre... | 14. Let there be $x$ students in this dormitory, and $y$ administrators in the dormitory building $\left(x, y \in \mathbf{N}_{+}\right)$. According to the problem, we have
$$
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 |
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 |
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
$$
V_{A D E-M N F} + V_{F-B C M M} = \frac{15}{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. Given $\tan \alpha+\log _{2}(2 \tan \alpha-6)=8$, $\tan \beta+2^{\tan \beta-1}=5$. Then $\tan \alpha+\tan \beta$ equals $\qquad$ . | 11.8 .
Let $t=\log _{2}(2 \tan \alpha-6)$. Then $t+2^{t-1}=5$. Also, $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, and $\tan \beta+2^{\tan \beta-1}=5$, so $\tan \beta=t$.
Therefore, $\tan \alpha+\tan \beta=\tan \alpha+t=8$.
Hence, $\tan \alpha+\tan \beta=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with the left vertex $A$ and the right focus $F$. Let $P$ be any point on the hyperbola in the first quadrant. If $\angle P F A=2 \angle F A P$ always holds, then the eccentricity $e$ of the hyperbola is | 12.2.
From the problem, we can take a point $P$ on the hyperbola such that $P F$ is perpendicular to the $x$-axis, giving $P(c, y)$.
Then $\frac{c^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Hence, $y^{2}=\frac{b^{4}}{a^{2}}$.
Since $y>0$, we have $y=\frac{b^{2}}{a}=\frac{c^{2}-a^{2}}{a}$.
Given that $\angle P F A=2 \angle F ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
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 $\triangle A B C$, $I_{1} γ I_{2} γ I_{3} γ$ 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
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{20^{2}+15^{2}}=25 \text {.... | 5 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. $a, b$ are constants. If the parabola $C$:
$$
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}
$$
It is clear that equality can be achieved.
Therefore... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $k$ is a positive integer not exceeding 50, such that for any positive integer $n, 2 \times 3^{6 n}+k \times 2^{3 n+1}-1$ is always divisible by 7. Then the number of such positive integers $k$ is $\qquad$. | 9.7.
$$
\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 {, }
$$
which means $2 k+1=7 m$ (whe... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Try to find all positive integers $k$, such that for any positive integers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a$, $b$, and $c$.
(2002, Girls' Mathematical Olympiad) | Explanation: First, from the inequality relationship between $a b+b c+c a$ and $a^{2}+b^{2}+c^{2}$, we derive $k>5$. Then, by constructing an example, we find that $k \leqslant 6$. Therefore, since $k \in \mathbf{Z}_{+}$, we conclude that $k=6$. Finally, we provide the proof.
Notice that for any positive real numbers ... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y$ satisfy the equation
$$
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 {, }
$$
which simplifies to $-3(y-8)^{2} \geqslant 0$.
Thus, $y=8$.
Substituting $y=8$ into the original equat... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 4, given that the equilateral $\triangle ABC$ is inscribed in $\odot O, AB$ $=86$. If point $E$ is on side $AB$, and through $E$ a line $DG \parallel BC$ intersects $\odot O$ at points $D, G$, and intersects $AC$ at point $F$, and let $AE=x, DE$ $=y$. If $x, y$ are both positive integers, then $y=... | 4.12.
From the problem, we know that $E F=A E=x$.
By the symmetry of the circle, $F G=D E=y$.
By the intersecting chords theorem, we have $A E \cdot E B=D E \cdot E G$, which means $x(86-x)=y(x+y)$.
If $x$ is odd, then $x(86-x)$ is also odd, and in this case, $y(x+y)$ is even. Therefore, $x$ must be even, and $y$ must... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 5, in $\odot O$, $AB$ and $CD$ are two perpendicular diameters. Point $E$ is on radius $OA$, and point $F$ is on the extension of radius $OB$, such that $OE = BF$. Lines $CE$ and $CF$ intersect $\odot O$ at points $G$ and $H$, respectively. Lines $AG$ and $AH$ intersect line $CD$ at p... | As shown in Figure 7, connect $D G$ and $D H$, and extend $C G$ to point $P$. Since $A B$ and $C D$ are perpendicular diameters,
we have
$$
\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 |
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, $S_{\triangle C B E}=\frac{1}{2} B E \cdot b=\frac{a b c}{2(a+b)}$.
An... | 2 | Geometry | 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
$$
-3 b=3,3 b^{2}-d^{2}=-1,-b^{3}+b d^{2}=a \text {. }... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The incenter of $\triangle A B C$ is $I$, and the angle bisector of $\angle B$ intersects $A C$ at point $P$. If $A P+A B=B C$, and $A B=3, B C=$ 5, then the value of $A I$ is $\qquad$ . | 2.2.
As shown in Figure 7, on segment $B C$, take $B A^{\prime}=B A$, extend $A I$ to intersect $B C$ at $Q$, and connect $P A^{\prime}$.
$$
\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}
$$
we have $\triangle A B P \cong \triangle A^{\prime} B P$.
Th... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Calculate: $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=$ | 9.4 .
$$
\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
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. To cut a rectangular prism into $k$ tetrahedra, the minimum value of $k$ is | 11.5 .
According to the equivalence, we only need to consider the cutting situation of a unit cube.
On the one hand, first, we need to show that 4 is not enough. If there were 4, since all faces of a tetrahedron are triangles and are not parallel to each other, the top face of the cube would have to be cut into at le... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then, in the plane, the area of the figure formed by all points satisfying $[x]^{2}+[y]^{2}=50$ is
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ | 5.12.
First, consider the first quadrant.
From $50=1^{2}+7^{2}=7^{2}+1^{2}=5^{2}+5^{2}$, we get $([x],[y])=(1,7),(7,1),(5,5)$.
And from $[x]=1,[y]=7$, we get the unit square $1 \leqslant x<2,7 \leqslant y<8$, whose area is 1.
Similarly, from $([x],[y])=(7,1)$ and $(5,5)$, we also get one unit square each, and these t... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $x, y, z$ be non-negative real numbers, and $x+y+z=$ 2. Then the sum of the maximum and minimum values of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ is $\qquad$ . | 2.1.
Since $x, y, z$ are non-negative real numbers, we have
$$
A=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant 0.
$$
When $x=y=0, z=2$, $A=0$, thus the minimum value of $A$ is 0.
Assume $A$ reaches its maximum value at $(x, y, z)$, without loss of generality, let $x \leqslant y \leqslant z$, prove: $x=0$.
In fact, ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $t$, such that for any convex $n$-gon $A_{1} A_{2} \cdots A_{n}$, as long as $n \geqslant t$, there must exist three points $A_{i} γ A_{j} γ A_{k}(1 \leqslant i<j<k \leqslant n)$, such that the area of $\triangle A_{i} A_{j} A_{k}$ is no more than $\frac{1}{n}$ of t... | Three, first prove a lemma.
Lemma For any convex hexagon $A_{1} A_{2} \cdots A_{6}$, there exists $1 \leqslant i\frac{S}{k+1}$, then
$S_{\text {pentagon } A_{1} A_{2} \cdots A_{k}}<S-\frac{S}{k+1}=\frac{k S}{k+1}$.
By the induction hypothesis, there must be $1 \leqslant i<j<r \leqslant n$, such that $S_{\triangle A_{i}... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x^{2}+y^{2} \leqslant 1$. Then the maximum value of the function $z=$ $\frac{\cos x+\cos y}{1+\cos x y}$ is $\qquad$ . | 2 1.1.
Assume $x \geqslant 0, y \geqslant 0$. When $0 \leqslant x \leqslant 1$, $0 \leqslant x y \leqslant y<1<\pi$, so $\cos x y \geqslant \cos y$.
Therefore, $\cos x+\cos y \leqslant 1+\cos x y$, where the equality holds when $x=y=0$.
Hence $z_{\max }=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) As shown in Figure 1, $EF$ is a chord of the parabola $\Gamma: y^{2}=2px$. Tangents to $\Gamma$ at points $E$ and $F$ intersect at point $C$. Points $A$ and $B$ are on the rays $EC$ and $CF$ respectively, and $\frac{EC}{CA}=\frac{CF}{FB}=\lambda$.
(1) Prove that the line $AB$ is tangent to $\Gamma$;
(... | (1) Let $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then the equations of the tangents $E C$ and $C F$ are
$$
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 |
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
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 1, in $\triangle A B C$, $A B=7, A C$ $=11, M$ is the midpoint of $B C$, $A D$ is the angle bisector of $\angle B A C$, $M F / / A D$. Then the length of $F C$ is $\qquad$ | 8.9.
As shown in Figure 5, let $N$ be the midpoint of $AC$, and connect $MN$. Then $MN \parallel AB$.
Also, $MF \parallel AD$, so,
$\angle FMN = \angle BAD = \angle DAC = \angle MFN$.
Therefore, $FN = MN = \frac{1}{2} AB$.
Thus, $FC = FN + NC = \frac{1}{2} AB + \frac{1}{2} AC = 9$. | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14.A. Choose $n$ numbers from $1,2, \cdots, 9$. Among them, there must be some numbers (at least one, or possibly all) whose sum is divisible by 10. Find the minimum value of $n$. | 14. A. When $n=4$, the numbers $1,3,5,8$ do not have any subset of numbers whose sum is divisible by 10.
When $n=5$, let $a_{1}, a_{2}, \cdots, a_{5}$ be five different numbers from $1,2, \cdots, 9$. If the sum of any subset of these numbers cannot be divisible by 10, then $a_{1}, a_{2}, \cdots, a_{5}$ cannot simultan... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a, b$ are positive numbers, and
$$
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 |
2. As shown in Figure 1, given that $G$ is the centroid of $\triangle A B O$. If $P Q$ passes through point $G$, and
$$
\begin{array}{l}
O A=a, O B=b, \\
O P=m a, O Q=n b,
\end{array}
$$
then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ | 2.3.
From $O M=\frac{1}{2}(a+b)$, we know
$$
O G=\frac{2}{3} O M=\frac{1}{3}(a+b) .
$$
From the collinearity of points $P, G, Q$, we have $\boldsymbol{P G}=\lambda \boldsymbol{G} \boldsymbol{Q}$.
And $P G=O G-O P=\frac{1}{3}(a+b)-\dot{m} a$
$$
\begin{array}{l}
=\left(\frac{1}{3}-m\right) a+\frac{1}{3} b, \\
G Q=O Q-O... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. If $p$, $q$, $\frac{2 p-1}{q}$, $\frac{2 q-1}{p}$ are all integers, and $p>1$, $q>1$. Then $p+q=$ $\qquad$ . | 3.8 .
If $p=q$, then
$$
\frac{2 p-1}{q}=\frac{2 p-1}{p}=2-\frac{1}{p} \text {. }
$$
Given $p>1$, then $\frac{2 p-1}{q}$ is not an integer, which contradicts the problem statement. Therefore, $p \neq q$.
By symmetry, without loss of generality, assume $p>q$. Let
$$
\begin{array}{l}
\frac{2 p-1}{q}=m, \\
\frac{2 q-1}{p... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) Find the smallest real number $A$, such that for each quadratic trinomial $f(x)$ satisfying the condition $|f(x)| \leqslant 1(0 \leqslant x \leqslant 1)$, the inequality $f^{\prime}(0) \leqslant A$ holds. | Three, let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {. }
$$
From the problem, we know
$$
|f(0)| \leqslant 1,\left|f\left(\frac{1}{2}\right)\right| \leqslant 1,|f(1)| \leqslant 1 .
$$
Notice that $f(0)=c, f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$,
$$
\begin{array}{l}
f(1)=a+b+c, \\
f... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $\alpha^{2008}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
(2005, China Western Mathematical Olympiad) | Explanation: Let $S_{n}=\alpha^{n}+\beta^{n}\left(n \in \mathbf{N}_{+}\right)$,
$$
\left(\sigma_{1}, \sigma_{2}\right)=(\alpha+\beta, \alpha \beta) \text {. }
$$
Then $S_{1}=\sigma_{1}$,
$$
S_{2}=\sigma_{1} S_{1}+2(-1)^{3} \sigma_{2}=\sigma_{1}^{2}-2 \sigma_{2}, \cdots \cdots
$$
By Newton's formula, we have
$$
\alpha... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given a positive integer $n(n>3)$, let real numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n} \geqslant n, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2} .
\end{array}
$$
Find the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
(28th Uni... | The above equation holds for $a \geqslant 2$.
Try taking $a=2$, at this point, $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ $=2$. From this, we can conjecture that the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is 2.
Solution: First, we prove that for any real numbers $a_{1}, a_{2}, \cdots... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given real numbers $x, y$ satisfy $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$. Then the value of $3 x^{2}-2 y^{2}+3 x-3 y-2007$ is ( ).
(A) -2008
(B) 2008
(C) -1
(D) 1 | 6.D.
From $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$,
we get
$$
\begin{array}{l}
x-\sqrt{x^{2}-2008} \\
=\frac{2008}{y-\sqrt{y^{2}-2008}}=y+\sqrt{y^{2}-2008}, \\
y-\sqrt{y^{2}-2008}=\frac{2008}{x-\sqrt{x^{2}-2008}} \\
=x+\sqrt{x^{2}-2008} .
\end{array}
$$
From the above two equations, we ... | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ | $$
\begin{array}{l}
\text { Given } a^{2}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\frac{3-\sqrt{5}}{2}=1-a \\
\Rightarrow a^{2}+a=1 . \\
\text { Therefore, } \frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a} \\
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a \cdot a^{2}-a} \\
=\frac{a^{3}-2 a^{3}-1+2}{a(1-... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the squares of positive integers $1,2, \cdots$ in a sequence: $149162536496481100121144 \cdots$, the digit at the 1st position is 1, the digit at the 5th position is 6, the digit at the 10th position is 4, the digit at the 2008th position is $\qquad$. | 4.1.
$1^{2}$ to $3^{2}$, each result occupies 1 digit, totaling $1 \times 3$ $=3$ digits;
$4^{2}$ to $9^{2}$, each result occupies 2 digits, totaling $2 \times 6$ $=12$ digits;
$10^{2}$ to $31^{2}$, each result occupies 3 digits, totaling $3 \times$ $22=66$ digits;
$32^{2}$ to $99^{2}$, each result occupies 4 digits, t... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. There is a 6-row $n$-column matrix composed of 0s and 1s, where each row contains exactly 5 ones, and the number of columns in which any two rows both have a 1 is at most 2. Find the minimum value of $n$. | 15. First, the total number of 1s in the matrix is $5 \times 6=$ 30.
Let the number of 1s in the $k$-th column be $l_{k}$. Then
$$
\sum_{k=1}^{n} l_{k}=30 \text {. }
$$
For any $1 \leqslant i<j \leqslant 6$ and $1 \leqslant k \leqslant n$, consider the triplet $\{i, j, k\}$: such that the $i$-th row and the $j$-th row... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 15, in trapezoid $A B C D$, $A D / / B C (B C > A D), \angle D=90^{\circ}, B C=$ $C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$ .
(2004, National Junior High School Mathematics Competition) | (Hint: First, complete the trapezoid $ABCD$ into a square $CBFD$, then rotate $\triangle ABF$ $90^{\circ}$ around point $B$ to the position of $\triangle BCG$. It is easy to see that $\triangle ABE \cong \triangle GBE$. Therefore, $AE = EG = EC + AF = 10$. Let $EC = x$. Then $AF = 10 - x$, $DE = 12 - x$, $AD = 2 + x$. ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the quadratic function $f(x)=a x^{2}+b x+c, a$ $\in \mathbf{N}_{+}, c \geqslant 1, a+b+c \geqslant 1$, the equation $a x^{2}+b x+c$ $=0$ has two distinct positive roots less than 1. Then the minimum value of $a$ is | 6.5.
Let the two roots of the equation $a x^{2}+b x+c=0$ be $x_{1}$ and $x_{2}$, and $0<x_{1}<x_{2}<1, a>4$.
Therefore, $a \geqslant 5$.
Take $f(x)=5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)$, which meets the conditions. Hence, the minimum value of $a$ is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Figure 4 is a "sheep's head" pattern, the method of which is: starting from square (1), using one of its sides as the hypotenuse, an isosceles right triangle is constructed outward, and then using its legs as sides, squares (2) and (2)' are constructed outward, ... and so on. If the side length of square (1) ... | Solution: By the Pythagorean theorem, the side length of square (2) is $64 \times \frac{\sqrt{2}}{2}=32 \sqrt{2}(\mathrm{~cm})$,
the side length of square (3) is
$$
32 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{2}=32(\mathrm{~cm}) \text {, }
$$
the side length of square (4) is
$$
32 \... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Let $a$ be an integer such that the equation $a x^{2}-(a+5) x+a+7=0$ has at least one rational root. Find all possible rational roots of the equation.
| When $a=0$, the rational root of the equation is $x=\frac{7}{5}$.
The following considers the case where $a \neq 0$. In this case, the original equation is a quadratic equation, and by the discriminant
$$
(a+5)^{2}-4 a(a+7) \geqslant 0,
$$
which simplifies to $3 a^{2}+18 a-25 \leqslant 0$.
Solving this, we get $-\frac... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+(1+x)^{3}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+b_{2} x^{2}+\cdots+b_{n} x^{n},
\end{array}
$$
and it satisfies $b_{1}+b_{2}+\cdots+b_{n}=26$. Then a possible value of the positive integer $n$ is $\qquad$ | Take $x=0$, we get $b_{0}=n$.
Take $x=1$, we get
$$
2+2^{2}+2^{3}+\cdots+2^{n}=b_{0}+b_{1}+b_{2}+\cdots+b_{n} \text {, }
$$
i.e., $\frac{2\left(1-2^{n}\right)}{1-2}=n+26$.
Thus, $2\left(2^{n}-1\right)=n+26$.
Therefore, $n=4$ is a possible value. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that the function $f(x)$ is defined on $\mathbf{R}$, and satisfies:
(1) $f(x)$ is an even function;
(2) For any $x \in \mathbf{R}$, $f(x+4) = f(x)$, and when $x \in [0,2]$, $f(x) = x + 2$.
Then the distance between the two closest points of intersection between the line $y=4$ and the graph of the function $f(... | 9.4.
Draw the graph of the function $f(x)$ (as shown in Figure 3).
From Figure 3, it is easy to see that the distance between the two closest intersection points of the line $y=4$ and the graph of the function $f(x)$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let the function $f(x)=x^{2}+a x+b(a, b$ be real constants). It is known that the inequality $|f(x)| \leqslant 12 x^{2}+4 x-30|$ holds for any real number $x$. Define the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ as:
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, 2 a_{n}=f\left(a_{n-1}\right)+15(n=2,3, \c... | 15. (1) Let the two real roots of the equation $2 x^{2}+4 x-30=0$ be $\alpha, \beta$. Then $\alpha+\beta=-2, \alpha \beta=-15$.
Taking $x=\alpha$ in $|f(x)| \leqslant\left|2 x^{2}+4 x-30\right|$, we get $|f(x)| \leqslant 0$, hence $f(\alpha)=0$.
Similarly, $f(\beta)=0$.
Therefore, $f(x)=(x-\alpha)(x-\beta)$
$$
=x^{2}-... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
16. In four-dimensional space, the distance between point $A\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ and point $B\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ is defined as
$$
A B=\sqrt{\sum_{i=1}^{4}\left(a_{i}-b_{i}\right)^{2}} .
$$
Consider the set of points
$I=\left\{P\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \mid c_{i}=... | 16. Construct the following 8 points:
$$
\begin{array}{l}
P_{1}(0,0,0,0), P_{2}(0,1,0,0), P_{3}(0,0,0,1), \\
P_{4}(0,0,1,1), P_{5}(1,1,0,0), P_{6}(1,1,1,0), \\
P_{7}(1,1,1,1), P_{8}(1,0,1,1).
\end{array}
$$
By calculation, it is known that no three of these points can form an equilateral triangle, so, $n_{\min } \geqs... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $a$ is an integer, $14 a^{2}-12 a-27 \mid$ is a prime number. Then the sum of all possible values of $a$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. D.
From the problem, we know
$$
\left|4 a^{2}-12 a-27\right|=|(2 a+3)(2 a-9)|
$$
is a prime number.
Therefore, $2 a+3= \pm 1$ or $2 a-9= \pm 1$, which means $a=-1,-2$ or $a=5,4$.
Thus, the sum of all possible values of $a$ is 6. | 6 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given $a+\frac{1}{a+1}=b+\frac{1}{b-1}-2$, and $a-$ $b+2 \neq 0$. Then the value of $a b-a+b$ is $\qquad$ . | $=γ 1.2$.
Let $a+1=x$,
$$
b-1=y \text {. }
$$
Then $x-y$
$$
\begin{array}{l}
=a-b+2 \\
\neq 0 .
\end{array}
$$
Substitute into the given equation,
we get $x+\frac{1}{x}=y+\frac{1}{y}$,
which means $(x-y)(x y-1)=0$.
Since $x-y \neq 0$, it follows that $x y=1$.
Therefore, $(a+1)(b-1)=1$, which means $a b-a+b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let the perfect square $y^{2}$ be the sum of the squares of 11 consecutive integers. Then the minimum value of $|y|$ is $\qquad$ . | Solution: Let the middle number of 11 consecutive integers be $x$. Then
$$
\begin{aligned}
y^{2}= & (x-5)^{2}+(x-4)^{2}+\cdots+x^{2}+\cdots+ \\
& (x+4)^{2}+(x+5)^{2} \\
= & x^{2}+2\left(x^{2}+1^{2}\right)+2\left(x^{2}+2^{2}\right)+\cdots+2\left(x^{2}+5^{2}\right) \\
= & 11 x^{2}+2\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\rig... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 3. Given real numbers } a, b, c. \text { If } \\ \frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 c a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1, \\ \text { then }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2008}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2008}+ \\ \left(\frac{a^{2}+b^{... | 3.3.
Notice
$$
\begin{array}{l}
\frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 a a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+1+\frac{(c-a)^{2}-b^{2}}{2 c a}+1+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}-1=1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+\frac{(c-a)^{2}-b^{2}}{2 ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given that $n$ is a positive integer greater than 10, and set $A$ contains $2n$ elements. If the family of sets
$$
\left\{A_{i} \subseteq A \mid i=1,2, \cdots, m\right\}
$$
satisfies the following two conditions, it is called "suitable":
(1) For any $i=1,2, \cdots, m, \operatorname{Card}\left(A_{i}\... | The maximum positive integer $m=4$.
Take two non-complementary $n$-element subsets $A_{1}, A_{2}$ of $A$, and let $A_{3}, A_{4}$ be the complements of $A_{1}, A_{2}$, respectively. From these four sets, any three sets will always have two sets that are complementary, hence the intersection of these three sets is the em... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 There is a four-digit number
$$
N=\overline{(a+1) a(a+2)(a+3)} \text {, }
$$
which is a perfect square. Find $a$. | Solution: Note that the last digit of a perfect square can only be $0, 1, 4, 5, 6, 9$, thus
$$
a+3 \equiv 0,1,4,5,6,9(\bmod 10) .
$$
Also, $a \geqslant 0, a+3 \leqslant 9$, so $a=1,2,3,6$.
Therefore, the last two digits of $N$ are
$$
\overline{(a+2)(a+3)}=34,45,56,89 \text {. }
$$
Since the last two digits of a perfe... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given real numbers $a, b, c, d$, and $a \neq b, c \neq d$. If the equations: $a^{2}+a c=2, b^{2}+b c=2, c^{2}+a c=$ $4, d^{2}+a d=4$ all hold, then the value of $6 a+2 b+3 c+2 d$ is $\qquad$. | 9.0 .
$$
\begin{array}{l}
\text { Given }\left(a^{2}+a c\right)-\left(b^{2}+b c\right)=2-2=0, \\
\left(c^{2}+a c\right)-\left(d^{2}+a d\right)=4-4=0,
\end{array}
$$
we get
$$
\begin{array}{l}
(a-b)(a+b+c)=0, \\
(c-d)(a+c+d)=0 .
\end{array}
$$
Since $a \neq b, c \neq d$, we have
$$
a+b+c=0, a+c+d=0 .
$$
Thus, $b=d=-(... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that $a$ and $b$ are real numbers, and $a^{2}+a b+b^{2}=3$. If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. | Three, 11. Let $a^{2}-a b+b^{2}=k$. From
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=3, \\
a^{2}-a b+b^{2}=k
\end{array} \Rightarrow a b=\frac{3-k}{2} .\right.
$$
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$$
=3+\frac{3-k}{2}=\frac{9-k}{2} \text {. }
$$
Since $(a+b)^{2} \geqslant 0$, i.e., $\frac{9-k}{2} \geqs... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$. Then the value of $\frac{\tan \alpha}{\tan \beta}$ is | $$
\begin{array}{l}
\sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\
= 3(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta), \\
\sin \alpha \cdot \cos \beta = 2 \cos \alpha \cdot \sin \beta. \\
\text{Therefore, } \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \c... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let the four-digit number $\overline{a b c d}$ be a perfect square, and its arithmetic square root can be expressed as $\sqrt{\overline{a b c d}}=\overline{a b}+\sqrt{\overline{c d}}$. How many such four-digit numbers are there? | Given $\sqrt{a b c d}=\overline{a b}+\sqrt{\overline{c d}}$, we know
$$
\begin{array}{l}
(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}+\overline{c d} \\
=\overline{a b c d}=100 \overline{a b}+\overline{c d} .
\end{array}
$$
Thus, $(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}=100 \overline{a b}$, whi... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The sum of all non-negative integer solutions to the inequality $|2 x-1|<6$ with respect to $x$ is $\qquad$ . | 2.1.6.
The original inequality is equivalent to $\left\{\begin{array}{l}2 x-1-6 .\end{array}\right.$
Solving it, we get $-\frac{5}{2}<x<\frac{7}{2}$.
Therefore, all non-negative integer solutions that satisfy the condition are $x=0,1,2,3$.
Thus, the sum of all non-negative integer solutions is 6. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) Let the real number $x$ satisfy
$$
\frac{3 x-1}{2}-\frac{4 x-2}{3} \geqslant \frac{6 x-3}{5}-\frac{13}{10} \text {. }
$$
Find the minimum value of $2|x-1|+|x+4|$. | Four, multiplying both sides of the original inequality by 30 gives
$$
15(3 x-1)-10(4 x-2) \geqslant 6(6 x-3)-39 \text {. }
$$
Solving this yields $x \leqslant 2$.
Let $y=2|x-1|+|x+4|$.
(1) When $x \leqslant-4$,
$$
y=-2(x-1)-(x+4)=-3 x-2 \text {. }
$$
Therefore, the minimum value of $y$ is $(-3) \times(-4)-2=$ 10, at... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A tangent line is drawn from the left focus $F$ of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ to the circle $x^{2}+y^{2}=9$, with the point of tangency being $T$. Extend $F T$ to intersect the right branch of the hyperbola at point $P$. If $M$ is the midpoint of segment $F P$, and $O$ is the origin, find th... | 14. Without loss of generality, place point $P$ in the first quadrant. As shown in Figure 2, let $F^{\prime}$ be the right focus of the hyperbola, and connect $P F^{\prime}$. Since $M$ and $O$ are the midpoints of $F P$ and $F F^{\prime}$ respectively, we have $|M O|=\frac{1}{2}\left|P F^{\prime}\right|$. By the defini... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9.5. On an infinitely large chessboard, the distance between any two squares is defined as the minimum number of steps a King needs to move from one square to another. Given three squares that are pairwise 100 steps apart, find the number of squares that are 50 steps away from each of these three squares. | 9.5. Let the side length of a small square be 1.
Represent a small square by the coordinates of its center. Let the absolute difference of the x-coordinates of two small squares be $x$, and the absolute difference of the y-coordinates be $y$.
It is not hard to see that a king can reach from one square to another in $... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $x$, $m$, and $n$ are positive integers, $m+n=5$, and $x^{2}+m$ and $\left|x^{2}-n\right|$ are both prime numbers. Then the number of possible values of $x$ is $\qquad$ . | 3.2.
From the problem, $m$ can take the values $1,2,3,4$, and accordingly, $n$ can be $4,3,2,1$, with $m$ and $n$ being one odd and one even.
Thus, $x^{2}+m$ and $\left|x^{2}-n\right|$ are one odd and one even.
Since $x^{2}+m$ and $\left|x^{2}-n\right|$ are both prime numbers, it follows that $x^{2}+m=2$ or $\left|x^{... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $f(x)=a x+b(a, b$ be real numbers),
$$
\begin{array}{l}
f_{1}(x)=f(x), \\
f_{n+1}(x)=f\left(f_{n}(x)\right)(n=1,2, \cdots) .
\end{array}
$$
If $f_{7}(x)=128 x+381$, then $a+b=$ | Ni.7.5.
From the problem, we know
$$
\begin{array}{l}
f_{n}(x)=a^{n} x+\left(a^{n-1}+a^{n-2}+\cdots+a+1\right) b \\
=a^{n} x+\frac{a^{n}-1}{a-1} \cdot b .
\end{array}
$$
From $f_{7}(x)=128 x+381$, we get
$$
a^{7}=128, \frac{a^{7}-1}{a-1} \cdot b=381 \text {. }
$$
Therefore, $a=2, b=3, a+b=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$. | 15. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), and \( C(0, c) \), and assume \( b > c \).
The equation of line \( PB \) is \( y - b = \frac{y_{0} - b}{x_{0}} x \).
Simplifying, we get \( \left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0 \).
The distance from the circle center \((1,0)\) to \( PB \) is 1, so
\[ ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Table 1, in the use of computers, codes are arranged according to certain rules, and they are infinite from left to right and from top to bottom. In Table 1, the code 100 appears $\qquad$ times.
\begin{tabular}{|c|c|c|c|c|c|}
\hline 1 & 1 & 1 & 1 & 1 & $\cdots$ \\
\hline 1 & 2 & 3 & 4 & 5 & $\cdots$ \\
\... | From Table 1, we know that the $m$-th row is an arithmetic sequence with the first term 1 and common difference $m-1$. Therefore, the $n$-th number in the $m$-th row is
$$
\begin{array}{l}
a_{m n}=1+(n-1)(m-1) . \\
\text { Let } a_{m n}=100 \text {. Then } \\
(m-1)(n-1)=99=3^{2} \times 11 \text {. }
\end{array}
$$
The... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (40 points) Among nine visually identical gold coins, one weighs $a$, seven weigh $b$, and the last one weighs $c$, and $a < b < c$. Using a balance scale, find the minimum number of weighings required to identify the coin weighing $a$ and the coin weighing $c$.
To solve this problem, we need to devise a strategy t... | 9. Solution 1: First, label the nine coins as
$A, B, C, D, E, F, G, H, I$.
Weigh $(A, B, C, D)$ against $(E, F, G, H)$. If they balance, then $a+c=2b$. If they do not balance, assume without loss of generality that $(A, B, C, D) > (E, F, G, H)$.
For the second weighing, weigh $(A, B)$ against $(C, D)$ and $(E, F)$ aga... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of values of $n$ $\left(n \in \mathbf{N}_{+}\right)$ for which the equation $x^{2}-6 x-2^{n}=0$ has integer solutions is $\qquad$ | 2.1.
$x=3 \pm \sqrt{3^{2}+2^{n}}$, where $3^{2}+2^{n}$ is a perfect square. Clearly, $n \geqslant 2$.
When $n \geqslant 2$, we can assume
$$
2^{n}+3^{2}=(2 k+1)^{2}\left(k \in \mathbf{N}_{+}, k \geqslant 2\right),
$$
i.e., $2^{n-2}=(k+2)(k-1)$.
It is evident that $k-1=1, k=2, n=4$.
The number of values of $n$ that mak... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Divide $1,2, \cdots, 9$ into three groups, each containing three numbers, such that the sum of the numbers in each group is a prime number.
(1) Prove that there must be two groups with equal sums;
(2) Find the number of all different ways to divide them.
| Three, (1) Since the sum of three different numbers in $1,2, \cdots, 9$ is between 6 and 24, the prime numbers among them are only 7, 11, 13, 17, 19, 23, these six. Now, these six numbers are divided into two categories based on their remainders when divided by 3:
$A=\{7,13,19\}$, where each number leaves a remainder o... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a$, $b$, $c$ satisfy $(a+b)(b+c)(c+a)=0$ and $abc<0$. Then the value of the algebraic expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}$ is | $=.1 .1$.
From $(a+b)(b+c)(c+a)=0$, we know that at least two of $a, b, c$ are opposite numbers. Also, since $abc<0$, it follows that among $a, b, c$, there must be two positive and one negative.
Thus, the value of $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In Rt $\triangle A B C$, $F$ is the midpoint of the hypotenuse $A B$, and $D, E$ are points on sides $C A, C B$ respectively, such that $\angle D F E=$ $90^{\circ}$. If $A D=3, B E=4$, then the length of segment $D E$ is $\qquad$ | 3.5 .
As shown in Figure 6, extend $D F$ to point $G$ such that $D F = F G$, and connect $G B$ and $G E$.
Given $A F = F B$, we have
$\triangle A D F \cong \triangle B G F$
$\Rightarrow B G = A D = 3$
$\Rightarrow \angle A D F = \angle B G F$
$\Rightarrow A D \parallel G B$
$\Rightarrow \angle G B E + \angle A C B = 1... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (25 points) In the expression β $\square 1 \square 2 \square 3 \square 4 \square 5 \square 6 \square 7$ $\square 8 \square 9$ β, fill in the small squares with β+β or β-β signs. If the algebraic sum can be $n$, then the number $n$ is called a "representable number"; otherwise, it is called an "unrepresentable num... | (1) Since $+1-2-3+4+5-6+7-8$ $+9=7$, therefore, 7 is a number that can be represented.
$$
\text { Also, }+1+2+3+4+5+6+7+8+9=45
$$
is an odd number, and for any two integers $a$ and $b$, $a+b$ and $a-b$ have the same parity. Therefore, no matter how the β $+\cdots \times$ - β signs are filled, the algebraic sum must be ... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Given $a-b=1, a^{2}-b^{2}=-1$. Then $a^{2008}-b^{2008}=$ | 12. -1 .
Given $a^{2}-b^{2}=(a+b)(a-b)=-1$, and $a-b=1$, then $a+b=-1$.
Therefore, $\left\{\begin{array}{l}a+b=-1 \\ a-b=1 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}a=0, \\ b=-1 .\end{array}\right.$ Hence, $a^{2 \alpha R}-b^{20 R}=0^{2008}-(-1)^{2008}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given that $a$ and $b$ are real numbers, and $ab=1, a \neq 1$, let $M=\frac{a}{a+1}+\frac{b}{b+1}, N=\frac{1}{a+1}+\frac{1}{b+1}$. Then the value of $M-N$ is $\qquad$. | 15.0.
Given $a b=1, a \neq 1$, so,
$$
\begin{array}{l}
M=\frac{a}{a+1}+\frac{b}{b+1}=\frac{a}{a+a b}+\frac{b}{b+a b} \\
=\frac{a}{a(1+b)}+\frac{b}{b(1+a)}=\frac{1}{a+1}+\frac{1}{b+1}=N .
\end{array}
$$
Thus, $M-N=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$ and $b$ be integers, and the equation
$$
a x^{2}+b x+1=0
$$
has two distinct positive roots both less than 1. Then the minimum value of $a$ is | 2. 5 .
Let the two roots of the equation be $x_{1}, x_{2}$.
From $x_{1} x_{2}=\frac{1}{a}>0$, we know $a>0$.
Also, $f(0)=1>0$, so according to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a>0, \\
00 .
\end{array}\right.
$$
Since $a$ is a positive integer, from equations (2) and (3), we get
$-(a+1)<b... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a, b, c, d, e$ be distinct positive odd numbers. If the equation
$$
(x-a)(x-b)(x-c)(x-d)(x-e)=2009
$$
has an integer root $x$, then the last digit of $a+b+c+d+e$ is ( ).
(A) 1
(B) 3
(C) 7
(D) 9 | 5.I).
For any integer $x, x-a, x-b, x-c, x-$ $d, x-e$ are $I$ distinct integers. And expressing 2009 as the product of 7 $I$ distinct integers has only one unique form:
$$
2009=1 \times(-1) \times 7 \times(-7) \times 41.
$$
From this, $x$ is even.
Let $x=2 m$. Then
$$
\begin{array}{l}
\{2 m-a, 2 m-b, 2 m-c, 2 m-d, 2 ... | 9 | Algebra | MCQ | Yes | Yes | cn_contest | false |
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