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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 opposite angles equal and the diagonal connecting the vertices of this pair of opposite angles being bisected by the other diagonal is a parallelogram;
(4) A quadrilateral with one pair of opposite angles equal and the diagonal connecting the vertices of this pair of opposite angles bisecting the other diagonal is a parallelogram.
Among them, the correct proposition numbers are $\qquad$
|
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 use $O$ as the intersection of the diagonals to construct $\square ABCE$. In this case, the quadrilateral $ABCD$ satisfies the conditions $AD = (AE =) BC$ and $AO = CO$, but it is not a parallelogram.
Proposition (3): The quadrilateral $ABCD$ in Figure 5(c), where $A$ and $C$ are any two points on the perpendicular bisector of $BD$.
The following is a proof that Proposition (4) is correct.
As shown in Figure 5(d), given $\angle BAD = \angle DCB$ and $OB = OD$.
With point $O$ as the center, rotate $\triangle ABD$ counterclockwise by $180^\circ$. Since $OB = OD$, point $D$ coincides with $B$, and point $B$ coincides with $D$, while point $A$ coincides with some point $A_1$ on ray $OC$. If $A_1$ is not $C$, then $\angle BA_1D > \angle BCD$ (if $A_1$ is inside segment $OC$) or $\angle BA_1D < \angle BCD$ (if $A_1$ is on the extension of $OC$), both of which contradict $\angle BA_1D = \angle BAD = \angle BCD$. Therefore, $A_1$ must be $C$, i.e., $OA = OA_1 = OC$. Thus, quadrilateral $ABCD$ is a parallelogram.
|
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 $M(8,5) = 2, M(8,11) = 3, M(8,7) = M(8,9) = 3$.
Therefore, the total number of different ways to cover the entire grid is 11.
|
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 so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally) in sequence. How many chess pieces must be removed at minimum to meet this requirement? And explain the reasoning.
|
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 form a "good group".
Let the number of removed chess pieces be $n$.
Lemma 1: There must be a removed chess piece in each row and each column. If no chess piece is removed in a certain row (column), then there will be more than one good group.
Lemma 2: If only 1 chess piece is removed in a certain column, then this chess piece must be in the 3rd, 4th, or 5th row of that column.
Lemma 3: If only 1 chess piece is removed in a certain row, then this chess piece must be in the 4th or 5th column of that row.
Proof of lemmas is omitted.
When $n \leqslant 7$, by the pigeonhole principle, there must be 1 column without any removed chess pieces. By Lemma 1, there must be a good group, so $n \leqslant 7$ does not meet the requirement.
When $n=8$, by Lemma 1, exactly 1 chess piece is removed from each column. By Lemma 2, these 8 chess pieces are all in the 3rd, 4th, or 5th rows. At this point, the 1st, 2nd, 6th, and 7th rows have no removed chess pieces, so there exists a good group. Therefore, $n=8$ does not meet the requirement.
When $n=9$, if 3 or more chess pieces are removed in a certain column, then in the remaining 7 columns, fewer than or equal to 6 chess pieces are removed, which means there must be a column without any removed chess pieces, and thus a good group exists. Therefore, at most 2 chess pieces can be removed in one column. By Lemma 1, a chess piece must be removed from each column. Since $9=1 \times 7+2$, it is only possible that 7 columns each have 1 chess piece removed, and the remaining 1 column has 2 chess pieces removed.
By Lemma 2, the 7 chess pieces in these 7 columns are all in the 3rd, 4th, or 5th rows, and there are 2 remaining chess pieces in the 1st, 2nd, 6th, and 7th rows. Therefore, there must be a row without any removed chess pieces, and thus a good group exists. Therefore, $n=9$ does not meet the requirement.
When $n=10$, if 4 or more chess pieces are removed in a certain column, then in the remaining 7 columns, fewer than or equal to 6 chess pieces are removed, which means there must be a column without any removed chess pieces, and thus a good group exists; if 3 chess pieces are removed in a certain column, then in the remaining 7 columns, 7 chess pieces are removed. By Lemma 1 and Lemma 2, exactly 1 chess piece is removed from each of these 7 columns, and these 7 chess pieces are distributed in the 3rd, 4th, and 5th rows. At this point, the 1st, 2nd, 6th, and 7th rows have no removed chess pieces, and there are 3 remaining chess pieces, which means there must be a row without any removed chess pieces, and thus a good group exists; by Lemma 1 and $10=1 \times 6+2 \times 2$, it is only possible that 6 columns each have 1 chess piece removed, and 2 columns each have 2 chess pieces removed. By Lemma 2, the 6 chess pieces in these 6 columns are distributed in the 3rd, 4th, and 5th rows, and the remaining 4 chess pieces are distributed in the 1st, 2nd, 6th, and 7th rows. By Lemma 1, these 4 chess pieces correspond to the 1st, 2nd, 6th, and 7th rows, one-to-one, without overlap. By Lemma 3, the two columns with 2 chess pieces removed are the 4th and 5th columns. Since the two chess pieces in the same column are distributed in the 1st, 2nd, 6th, and 7th rows, the possible combinations of removed chess pieces are:
$$
\begin{array}{l}
(4,1) \text { and }(4,6),(4,2) \text { and }(4,7), \\
(4,1) \text { and }(4,7),(4,2) \text { and }(4,6),
\end{array}
$$
Among them, $(4,1)$ and $(4,7)$ do not meet the requirement (because $(4,2)$, $(4,3), \cdots,(4,6)$ form a good group).
Similarly, $(5,1)$ and $(5,7)$ do not meet the requirement.
Also, because only 1 chess piece is removed in each row, $(4,2)$ and $(4,6)$ do not meet the requirement.
By symmetry, assume the removed chess pieces are
$(4,1),(4,6),(5,2),(5,7)$.
At this point, as shown in Figure 9 ("β" represents a chess piece, "Γ" represents a removed chess piece, and blank squares represent undetermined positions).
Since only 1 chess piece is removed in the 3rd column, at most 1 "Γ" is in $(3,3),(3,5)$, so in $(1,1),(2,2),(3,3),(4,4)$, $(5,5)$ and $(1,7),(2,6),(3,5),(4,4),(5,3)$, there must be 1 good group.
Therefore, $n=10$ does not meet the requirement.
When $n=11$, an example can be constructed as shown in Figure 10.
In summary, the minimum number of chess pieces to be removed is 11.
|
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
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+1+\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}-\right. \\
\left.1+\frac{x^{3}+4^{3}}{(x+4)^{3}}-1+\frac{x^{3}+9^{3}}{(x+9)^{3}}-1\right]=0 \\
\Leftrightarrow \frac{x^{3}+49 x}{(x+1)(x+4)(x+9)}- \\
\quad\left[\frac{x}{(x+1)^{2}}+\frac{4 x}{(x+4)^{2}}+\frac{9 x}{(x+9)^{2}}\right]=0 .
\end{array}
$$
Dividing both sides of the equation by $x$, and rearranging, we get
$$
x\left(x^{4}-98 x^{2}-288 x+385\right)=0 \text {. }
$$
Dividing by $x$ again, we have
$$
\left(x^{2}-31\right)^{2}-(6 x+24)^{2}=0 \text {, }
$$
which is $\left(x^{2}+6 x-7\right)\left(x^{2}-6 x-55\right)=0$.
Thus, $(x+7)(x-1)(x+5)(x-11)=0$.
Upon verification, $x_{1}=-7, x_{2}=1, x_{3}=-5, x_{4}=11$ are all roots of the original equation. Therefore, the original equation has 4 integer roots.
|
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
$$
\begin{array}{l}
a_{1}<a_{2}<a_{3}<a_{4}<a_{5}, \\
A \cap B=\left\{a_{1}, a_{4}\right\}, a_{1}+a_{4}=10 .
\end{array}
$$
If the sum of all elements in $A \cup B$ is 256, then the number of sets $A$ that satisfy the conditions is $\qquad$
|
11.2.
Since $a_{1}^{2}=a_{1}$, therefore, $a_{1}=1, a_{4}=9$.
Since $B$ contains 9, $A$ contains 3.
If $a_{3}=3$, then $a_{2}=2$.
Thus, $a_{5}+a_{5}^{2}=146$. No positive integer solution.
If $a_{2}=3$, since $10 \leqslant a_{5} \leqslant 11$, then $a_{2}^{2} \neq a_{5}$.
Thus, $a_{3}+a_{3}^{2}+a_{5}+a_{5}^{2}=152$.
Also, $a_{3} \geqslant 4$, when $a_{5}=10$, $a_{3}=6$; when $a_{5}=$ 11, $a_{3}=4$. Therefore, there are 2 sets $A$ that satisfy the conditions, which are $\{1,3,4,9,11\},\{1,3,6,9,10\}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. There are 10 students standing in a row, and their birthdays are in different months. There are $n$ teachers who will select these students to join $n$ interest groups. Each student is selected by exactly one teacher, and the order of the students is maintained. Each teacher must select students whose birthdays are in months that are either strictly increasing or strictly decreasing (selecting one or two students is also considered strictly increasing or decreasing). Each teacher should select as many students as possible. For all possible orderings of the students, find the minimum value of $n$.
|
15. If $n \leqslant 3$, let's assume the birth months of these 10 students are $1,2, \cdots, 10$.
When the students are sorted by their birthdays as $4,3,2,1,7,6,5,9, 8,10$, there exists at least one teacher who must select two students from the first four. Since the birth months of these two students are decreasing, and the birth months of the last six students are all greater than those of the first four, this teacher cannot select any of the last six students; among the remaining no more than two teachers, there must be one who has to select two students from the fifth to the seventh. Similarly, this teacher cannot select any of the last three students; the remaining no more than one teacher also cannot select any of the last three students, leading to a contradiction.
Below is the proof: For any distinct ordered sequence of real numbers $a_{1}, a_{2}, \cdots, a_{m}$, when $m \geqslant 5$, there must exist three numbers $a_{i}, a_{j}, a_{k}(i<a_{j}>a_{k}$.
Let the maximum and minimum numbers be $a_{s}$ and $a_{t}$. Without loss of generality, assume $s<a_{s+1}>a_{t} ; s+1=t$.
Since $m \geqslant 5$, there must be at least two numbers either before $a_{s}, a_{s+1}$ or after $a_{s}, a_{s+1}$.
Without loss of generality, assume there are two numbers $a_{s+2}$ and $a_{s+3}$ after $a_{s}, a_{s+1}$. Thus,
$$
a_{s}>a_{s+2}>a_{s+3} \text { or } a_{s+1}<a_{s+2}<a_{s+3}
$$
must hold.
Using the above conclusion, when $n=4$, the first teacher can select at least 3 students; if the remaining students are greater than or equal to 5, the second teacher can also select at least 3 students; at this point, the number of remaining students does not exceed 4, and can be selected by the two teachers.
Therefore, the minimum value of $n$ is 4.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the complex number $z_{1}=(6-a)+(4-b) \mathrm{i}$,
$$
\begin{array}{l}
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value, $3 a+4 b=$ $\qquad$
|
9.12.
It is easy to find that $z_{1}+z_{2}+z_{3}=12+9 \mathrm{i}$. Therefore,
$$
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=15 \text {. }
$$
The equality holds if and only if $\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9}$. Solving this, we get $a=\frac{7}{3}, b=\frac{5}{4}$.
Thus, $3 a+4 b=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. For the function $f(x)=\sqrt{a x^{2}+b x}$, there exists a positive number $b$, such that the domain and range of $f(x)$ are the same. Then the value of the non-zero real number $a$ is $\qquad$.
|
11. -4 .
If $a>0$, for the positive number $b$, the domain of $f(x)$ is
$$
D=\left(-\infty,-\frac{b}{a}\right] \cup[0,+\infty) .
$$
However, the range of $f(x)$, $A \subseteq[0,+\infty)$, so $D \neq A$, which does not meet the requirement.
If $a>0$, so, $a=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Real numbers $x, y$ satisfy $\tan x=x, \tan y=y$, and $|x| \neq|y|$. Then the value of $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}$ is
|
II, 11.0.
From the given, we have
$$
\begin{array}{l}
\frac{\sin (x+y)}{x+y}=\frac{\sin (x+y)}{\tan x+\tan y} \\
=\frac{\sin (x+y)}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}=\cos x \cdot \cos y .
\end{array}
$$
Similarly, $\frac{\sin (x-y)}{x-y}=\cos x \cdot \cos y$. Therefore, $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Definition: The length of the interval $\left[x_{1}, x_{2}\right]\left(x_{1}<x_{2}\right)$ is $x_{2}-x_{1}$. Given that the domain of the function $y=\left|\log _{\frac{1}{2}} x\right|$ is $[a, b]$, and the range is $[0,2]$. Then the difference between the maximum and minimum values of the length of the interval $[a, b]$ is $\qquad$.
|
12.3.
The graphs of the functions $y=$ $\left|\log _{\frac{1}{2}} x\right|$ and $y=2$ are shown in Figure 7.
From $y=0$, we get $x=1$;
From $y=2$, we get $x=\frac{1}{4}$ or $x=4$.
When $a=\frac{1}{4}, b=1$, $(b-a)_{\text {min }}=\frac{3}{4}$;
When $a=\frac{1}{4}, b=4$, $(b-a)_{\max }=\frac{15}{4}$.
Therefore, $(b-a)_{\max }-(b-a)_{\min }=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the circumcenter, incenter, and orthocenter of non-isosceles $\triangle ABC$ be $O$, $I$, and $H$, respectively, with the circumradius being $1$ and $\angle A=60^{\circ}$. Then the circumradius of $\triangle OIH$ is $\qquad$.
|
If $\triangle A B C$ is an acute triangle, since
$$
\angle B O C=\angle B I C=\angle B H C=120^{\circ} \text {, }
$$
then, $O$, $I$, $H$, $B$, and $C$ are concyclic;
if $\triangle A B C$ is an obtuse triangle, since
$$
\angle B O C=\angle B I C=120^{\circ}, \angle B H C=60^{\circ} \text {, }
$$
and $H$ is on the opposite side of $B C$ from $O$, $I$, and $A$, then, $O$, $I$, $H$, $B$, and $C$ are concyclic.
Therefore, the circumradius of $\triangle O I H$ is the same as the circumradius of $\triangle B H C$, which equals the circumradius of $\triangle A B C$, which is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given an even function $f: \mathbf{Z} \rightarrow \mathbf{Z}$ that satisfies $f(1)=1$, $f(2007) \neq 1$, and for any integers $a, b$,
$$
f(a+b) \leqslant \max \{f(a), f(b)\} \text {. }
$$
Then the possible value of $f(2008)$ is $\qquad$
|
5.1.
Since $f(2) \leqslant \max \{f(1), f(1)\}=1$, assume for a positive integer $k$ greater than or equal to 2, we have $f(k) \leqslant 1$, then
$$
f(k+1) \leqslant \max \{f(k), f(1)\}=1 \text {. }
$$
Therefore, for all positive integers $n$, we have $f(n) \leqslant 1$.
Since $f$ is an even function, we can conclude that for all non-zero integers $n$, we have $f(n) \leqslant 1$.
Since $f(2007) \neq 1$, it follows that $f(2007)<1$.
Thus, $1=f(1)=f(2008+(-2007))$
$\leqslant \max \{f(2008), f(-2007)\} \leqslant 1$.
Since $f(-2007)=f(2007)<1$, it must be that $f(2008)=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For a positive integer $n \geqslant 2007$, a complex number $z$ satisfies
$$
(a+1) z^{n+1}+a \text { i } z^{n}+a \text { i } z-(a+1)=0 \text {, }
$$
where the real number $a>-\frac{1}{2}$. Then the value of $|z|$ is $\qquad$ .
|
6.1.
Given $z^{n}[(a+1) z+a \mathrm{i}]=a+1-a \mathrm{i} z$, then $|z|^{n}|(a+1) z+a \mathrm{i}|=|a+1-a \mathrm{i} z|$. Let $z=x+y \mathrm{i}$, where $x, y$ are real numbers. Then
$$
\begin{array}{l}
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2} \\
=(2 a+1)\left(|z|^{2}-1\right) .
\end{array}
$$
If $|z|>1$, since $a>-\frac{1}{2}$, then
$$
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2}>0 \text {. }
$$
Thus, $|z|<1$, which is a contradiction.
Similarly, if $|z|<1$, it also leads to a contradiction.
Therefore, $|z|=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(a+b+2 c)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8 .
\end{array}
$$
(2003, USA Mathematical Olympiad)
|
Analysis: Let's assume $a+b+c=1$. Then the original inequality transforms into
$$
\begin{array}{l}
\frac{(1+a)^{2}}{2 a^{2}+(1-a)^{2}}+\frac{(1+b)^{2}}{2 b^{2}+(1-b)^{2}}+ \\
\frac{(1+c)^{2}}{2 c^{2}+(1-c)^{2}} \leqslant 8 .
\end{array}
$$
Let $f(x)=\frac{(1+x)^{2}}{2 x^{2}+(1-x)^{2}}$
$$
=\frac{1}{3}\left[1+\frac{8 x+2}{3\left(x-\frac{1}{3}\right)^{2}+\frac{2}{3}}\right] \leqslant \frac{12 x+4}{3},
$$
Then we have
$$
\begin{array}{l}
f(a)+f(b)+f(c) \\
\leqslant \frac{12(a+b+c)+12}{3}=8
\end{array}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
14. In a dormitory of a school, there are several students, one of whom serves as the dorm leader. During New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator. Each dormitory administrator also gives a greeting card back to the dorm leader. In this way, a total of 51 greeting cards were used. How many students live in this dormitory?
|
14. Let there be $x$ students in this dormitory, and $y$ administrators in the dormitory building $\left(x, y \in \mathbf{N}_{+}\right)$. According to the problem, we have
$$
x(x-1)+x y+y=51 \text{. }
$$
Simplifying, we get $x^{2}+(y-1) x+y-51=0$.
Thus, $\Delta=(y-1)^{2}-4(y-51)$
$$
=y^{2}-6 y+205=(y-3)^{2}+196 \text{. }
$$
Since $x \in \mathbf{N}_{+}$, $\Delta$ must be a perfect square.
Let $(y-3)^{2}+196=k^{2}(k \in \mathbf{N})$. Then
$$
(y-3+k)(y-3-k)=-196 \text{, }
$$
where $y-3+k$ and $y-3-k$ have the same parity, and $y-3+k \geqslant y-3-k$.
So, $\left\{\begin{array}{l}y-3+k=2, \\ y-3-k=-98\end{array}\right.$
or $\left\{\begin{array}{l}y-3+k=98, \\ y-3-k=-2\end{array}\right.$
or $\left\{\begin{array}{l}y-3+k=14, \\ y-3-k=-14 .\end{array}\right.$
From equation set (1), we get $y=-45$, which is not valid, so we discard it;
From equation set (2), we get $y=51$, in which case the original equation becomes $x^{2}+50 x=0$, solving for $x_{1}=-50, x_{2}=0$, both of which are not valid, so we discard them;
From equation set (3), we get $y=3$, in which case the original equation becomes $x^{2}+2 x-48=0$, solving for $x_{1}=-8$ (not valid, discard), $x_{2}=6$.
Answer: There are 6 students in this dormitory.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) Each point on the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist 4 points of different colors that are concyclic.
(Zhang Limin, problem contributor)
|
Obviously, $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arcs $\overparen{A B}$ (including $A$ but not $B$), $\overparen{B C}$ (including $B$ but not $C$), and $\overparen{C A}$ (including $C$ but not $A$) with colors $1, 2,$ and $3$ respectively, and color all other points in the plane with color $4$. This satisfies the conditions and there do not exist four points of different colors on the same circle.
Therefore, $n \neq 4, n \geqslant 5$.
When $n=5$, by condition (2), there exists a line $l$ on which there are exactly two colors of points. Without loss of generality, assume that the points on line $l$ are only of colors $1$ and $2$. By condition (1), there exist points $A, B, C$ of colors $3, 4, 5$ respectively that are not collinear. Let the circle passing through points $A, B, C$ be $\odot O$.
If $\odot O$ intersects line $l$, then there exist four points of different colors on the same circle;
If $\odot O$ is disjoint from line $l$ and $\odot O$ has points of colors $1$ and $2$, then there exist four points of different colors on the same circle;
If $\odot O$ is disjoint from line $l$ and $\odot O$ has no points of colors $1$ and $2$, as shown in Figure 2, draw a perpendicular from $O$ to line $l$ intersecting $l$ at point $D$. Assume the color of $D$ is $1$, and the perpendicular intersects $\odot O$ at points $E$ and $S$. Assume the color of $E$ is $3$, and consider a point $F$ on line $l$ with color $2$. The line $FS$ intersects $\odot O$ at point $G$. Since $EG \perp GF$, points $D, E, F, G$ are concyclic.
If $G$ is not a point of color $3$, then there exist four points of different colors on the same circle;
If $G$ is a point of color $3$, then one of $B$ or $C$ must be different from $S$ (assume it is $B$). The line $SB$ intersects line $l$ at point $H$. Since $EB \perp BH$, points $B, E, D, H$ are concyclic.
If $H$ is a point of color $2$, then $B, H, D, E$ are four points of different colors on the same circle;
If $H$ is a point of color $1$, since
$$
SB \cdot SH = SE \cdot SD = SG \cdot SF,
$$
points $B, H, F, G$ are concyclic.
Thus, $B, H, F, G$ are four points of different colors on the same circle.
In summary, when $n=5$, there exist four points of different colors on the same circle.
Therefore, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (25 points) Given
$$
f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x .
$$
(1) Solve the equation: $f(x)=0$;
(2) Find the number of subsets of the set
$$
M=\left\{n \mid f\left(n^{2}-214 n-1998\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
(Li Tiehan, problem contributor)
|
(1) For any $0 < x_1 < x_2$, we have $\frac{x_1 + 1}{x_2 + 1} > \frac{x_1}{x_2}$, thus $\lg \frac{x_1 + 1}{x_2 + 1} > \lg \frac{x_1}{x_2}$. Therefore,
$$
\begin{array}{l}
f(x_1) - f(x_2) > \lg \frac{x_1}{x_2} - \log \frac{x_1}{x_2} \\
= \lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9}.
\end{array}
$$
Since $0 < \frac{x_1}{x_2} < 1$, we have $\lg \frac{x_1}{x_2} < 0$. Therefore,
$$
\lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9} > \lg \frac{x_1}{x_2} - \lg \frac{x_1}{x_2} = 0.
$$
Hence, $f(x)$ is a decreasing function on $(0, +\infty)$.
Noting that $f(9) = 0$, we have:
When $x > 9$, $f(x) < f(9) = 0$.
When $x < 9$, $f(x) > f(9) = 0$.
Therefore, $f(x) = 0$ has and only has one root $x = 9$.
$$
\begin{array}{l}
\text{(2) From } f(n^2 - 214n - 1998) \geq 0 \\
\Rightarrow f(n^2 - 214n - 1998) \geq f(9). \\
\text{Then }\left\{\begin{array}{l}
n^2 - 214n - 1998 \leq 9, \\
n^2 - 214n - 1998 > 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
n^2 - 214n - 2007 \leq 0, \\
n^2 - 214n - 1998 > 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
(n - 223)(n + 9) \leq 0, \\
(n - 107)^2 > 1998 + 107^2 = 13447 > 115^2
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leq n \leq 223, \\
n > 222 \text{ or } n < -8
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leq n \leq 223, \\
n \geq 223 \text{ or } n \leq -9.
\end{array}\right. \\
\end{array}
$$
Thus, $n = 223$ or $n = -9$.
Hence, $M = \{-9, 223\}$.
Therefore, the number of subsets of $M$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in Figure 2, in the polyhedron ABCDEF, it is known that quadrilateral $ABCD$ is a square with side length 3, $EF$ $/ / AB, EF=\frac{3}{2}$. If the volume of the polyhedron is $\frac{15}{2}$, then the distance between $EF$ and $AC$ is
|
9.2 .
Take the midpoints of $A B$ and $C D$ as $M$ and $N$, and connect $F M$, $F N$, and $M N$.
Since $E F / / A M$ and $E F = \frac{3}{2} = A M$, we know that the polyhedron $A D E - M N F$ is a triangular prism.
Let the distance between $E F$ and $A C$ be $h$. From
$$
V_{A D E-M N F} + V_{F-B C M M} = \frac{15}{2} \text {, }
$$
we get $\frac{1}{2} \cdot h \cdot \frac{9}{2} + \frac{1}{3} \cdot h \cdot \frac{9}{2} = \frac{15}{2}$.
Solving for $h$, we get $h = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given $\tan \alpha+\log _{2}(2 \tan \alpha-6)=8$, $\tan \beta+2^{\tan \beta-1}=5$. Then $\tan \alpha+\tan \beta$ equals $\qquad$ .
|
11.8 .
Let $t=\log _{2}(2 \tan \alpha-6)$. Then $t+2^{t-1}=5$. Also, $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, and $\tan \beta+2^{\tan \beta-1}=5$, so $\tan \beta=t$.
Therefore, $\tan \alpha+\tan \beta=\tan \alpha+t=8$.
Hence, $\tan \alpha+\tan \beta=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with the left vertex $A$ and the right focus $F$. Let $P$ be any point on the hyperbola in the first quadrant. If $\angle P F A=2 \angle F A P$ always holds, then the eccentricity $e$ of the hyperbola is
|
12.2.
From the problem, we can take a point $P$ on the hyperbola such that $P F$ is perpendicular to the $x$-axis, giving $P(c, y)$.
Then $\frac{c^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Hence, $y^{2}=\frac{b^{4}}{a^{2}}$.
Since $y>0$, we have $y=\frac{b^{2}}{a}=\frac{c^{2}-a^{2}}{a}$.
Given that $\angle P F A=2 \angle F A P=\frac{\pi}{2}$, we know that $\triangle A F P$ is an isosceles triangle, so $A F=P F$, which means $a+c=\frac{c^{2}-a^{2}}{a}$.
Rearranging gives $c^{2}-a c-2 a^{2}=0$, or
$e^{2}-e-2=0$.
Since $e>1$, solving gives $e=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1} a_{n}+a_{n+1}-2 a_{n}=0 \text {. }
$$
For any positive integer $n$, we have
$\sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)<M$ (where $M$ is a constant and an integer).
Find the minimum value of $M$.
|
Three, 13. From the problem, for $n \in \mathbf{N}_{+}, a_{n} \neq 0$, and
\[
\frac{1}{a_{n+1}}=\frac{1}{2}+\frac{1}{2 a_{n}},
\]
which means
\[
\frac{1}{a_{n+1}}-1=\frac{1}{2}\left(\frac{1}{a_{n}}-1\right).
\]
Given $a_{1}=2$, we have $\frac{1}{a_{1}}-1=-\frac{1}{2}$.
Therefore, the sequence $\left\{\frac{1}{a_{n}}-1\right\}$ is a geometric sequence with the first term $-\frac{1}{2}$ and common ratio $\frac{1}{2}$. Thus,
\[
\frac{1}{a_{n}}-1=-\frac{1}{2} \times\left(\frac{1}{2}\right)^{n-1}=-\left(\frac{1}{2}\right)^{n},
\]
which implies
\[
a_{n}=\frac{2^{n}}{2^{n}-1}.
\]
Hence,
\[
a_{i}\left(a_{i}-1\right)=\frac{2^{i}}{\left(2^{i}-1\right)^{2}}(i=1,2, \cdots, n).
\]
For $i \geqslant 2$, we have
\[
\begin{array}{l}
a_{i}\left(a_{i}-1\right)=\frac{2^{i}}{\left(2^{i}-1\right)^{2}}<\frac{2^{i}}{\left(2^{i}-1\right)\left(2^{i}-2\right)} \\
=\frac{2^{i-1}}{\left(2^{i}-1\right)\left(2^{i-1}-1\right)}=\frac{1}{2^{i-1}-1}-\frac{1}{2^{i}-1} . \\
\text { Therefore, } \sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)=\sum_{i=1}^{n} \frac{2^{i}}{\left(2^{i}-1\right)^{2}} \\
<\frac{2^{1}}{\left(2^{1}-1\right)^{2}}+\sum_{i=2}^{n}\left(\frac{1}{2^{i-1}-1}-\frac{1}{2^{i}-1}\right) \\
=3-\frac{1}{2^{n}-1}<3 . \\
\text { Also, } \sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)=\sum_{i=1}^{n} \frac{2^{i}}{\left(2^{i}-1\right)^{2}} . \\
\geqslant \frac{2^{1}}{\left(2^{1}-1\right)^{2}}=2 .
\end{array}
\]
Therefore, the minimum value of $M$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the sequence
$$
b_{n}=\frac{1}{3 \sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right](n=0,1, \cdots) \text {. }
$$
(1) For what values of $n$ is $b_{n}$ an integer?
(2) If $n$ is odd and $2^{-\frac{2 n}{3}} b_{n}$ is an integer, what is $n$?
|
15. (1) From
$$
\begin{array}{l}
b_{n}=\frac{2}{3 \sqrt{3}}\left[n \sqrt{3}+\mathrm{C}_{n}^{3}(\sqrt{3})^{3}+\mathrm{C}_{n}^{6}(\sqrt{3})^{5}+\cdots\right] \\
=2\left(\frac{n}{3}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5} \cdot 3+\cdots\right),
\end{array}
$$
$b_{n}$ is an integer if and only if $31 n$.
(2) First, $b_{n}$ is an integer, $31 n$. Second,
$$
\begin{array}{l}
\frac{1}{\sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right] \\
= \frac{1}{\sqrt{3}}[(1+\sqrt{3})-(1-\sqrt{3})]\left[(1+\sqrt{3})^{n-1}+\right. \\
\left.(1+\sqrt{3})^{n-2}(1-\sqrt{3})+\cdots+(1-\sqrt{3})^{n-1}\right] \\
= 2\left\{\left[(1+\sqrt{3})^{n-1}+(1-\sqrt{3})^{n-1}\right]-\right. \\
2\left[(1+\sqrt{3})^{n-3}+(1-\sqrt{3})^{n-3}\right]+\cdots+ \\
\left.(-2)^{\frac{n-1}{2}}\right\} \\
= 2\left\{2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-1}{2}}+(2-\sqrt{3})^{\frac{n-1}{2}}\right]-\right. \\
2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-3}{2}}+(2-\sqrt{3})^{\frac{n-3}{2}}\right]+\cdots+ \\
\left.(-2)^{\frac{n-1}{2}}\right\} .
\end{array}
$$
Since $(2+\sqrt{3})^{k}+(2-\sqrt{3})^{k}$
$$
=2\left(2^{k}+\mathrm{C}_{k}^{2} 2^{k-2} \cdot 3+\cdots\right)
$$
is even, therefore,
$$
\begin{array}{l}
2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-1}{2}}+(2-\sqrt{3})^{\frac{n-1}{2}}\right]- \\
2^{\frac{n-1}{2}}\left[(2+\sqrt{3})^{\frac{n-3}{2}}+\cdots+(2-\sqrt{3})^{\frac{n-3}{2}}\right]+\cdots+ \\
(-2)^{\frac{n-1}{2}}
\end{array}
$$
is an integer, and the power of 2 is the lowest in the last term $(-2)^{\frac{n-1}{2}}$, i.e., the power of 2 in $\frac{1}{\sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right]$ is $\frac{n-1}{2}+1=\frac{n+1}{2}$.
Since $\frac{n+1}{2} \leqslant \frac{2 n}{3}$, equality holds only when $n \geqslant 3$, so, $2^{-\frac{2 n}{3}} b_{n}$ is an integer if and only if $n=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 2, $C D$ is the altitude on the hypotenuse $A B$ of Rt $\triangle A B C$, $I_{1} γ I_{2} γ I_{3} γ$ are the incenter of $\triangle A B C γ \triangle A C D$ γ $\triangle B C D$ respectively, $A C = 20, B C = 15$. Then the area of $\triangle I_{1} I_{2} I_{3}$ is ( ).
(A) 4
(B) 4.5
(C) 5
(D) 5.5
|
6.C.
As shown in Figure 6, connect $A I_{1}$,
$B I_{1}$, $I_{2} D$, $I_{3} D$. Draw $I_{1} G \perp A B$ at point $G$,
$I_{2} E \perp A B$ at point $E$, and $I_{3} F \perp A B$
at point $F$. In the right triangle $\triangle A B C$, it is easy to see that
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{20^{2}+15^{2}}=25 \text {. }
$$
Since $C D \perp A B$, by the projection theorem, we have
$$
A D=\frac{A C^{2}}{A B}=\frac{20^{2}}{25}=16 \text {. }
$$
Thus, $B D=A B-A D=9$,
$$
C D=\sqrt{A C^{2}-A D^{2}}=\sqrt{20^{2}-16^{2}}=12 \text {. }
$$
Since $I_{1}$, $I_{2}$, and $I_{3}$ are the incenter of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ respectively, $I_{2}$ lies on $A I_{1}$, $I_{3}$ lies on $B I_{1}$, and $I_{1} G$, $I_{2} E$, $I_{3} F$ are the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ respectively. $D I_{2}$ and $D I_{3}$ are the angle bisectors of $\angle A D C$ and $\angle B D C$ respectively. Therefore,
$$
\begin{array}{l}
\angle A D I_{2}=\angle I_{2} D C=45^{\circ}, \\
\angle B D I_{3}=\angle C D I_{3}=45^{\circ} \\
\Rightarrow \angle I_{2} D I_{3}=\angle I_{2} D C+\angle C D I_{3}=90^{\circ} \\
\Rightarrow I_{2} D \perp I_{3} D .
\end{array}
$$
Since $I_{1} G$ is the inradius of the right triangle $\triangle A B C$, we have
$$
I_{1} G=\frac{1}{2}(A C+B C-A B)=5 \text {. }
$$
Similarly, $I_{2} E=4$, $I_{3} F=3$.
Thus, $I_{2} D=\frac{I_{2} E}{\sin \angle A D I_{2}}=\frac{4}{\sin 45^{\circ}}=4 \sqrt{2}$.
Similarly, $I_{3} D=3 \sqrt{2}$.
Therefore, $S_{\triangle I_{1} I_{2} I_{3}}$
$$
\begin{aligned}
= & S_{\triangle A B B_{1}}-S_{\triangle \triangle I_{2}}-S_{\triangle B D I_{3}}-S_{\triangle D I_{2} I_{3}} \\
= & \frac{1}{2} A B \cdot I_{1} G-\frac{1}{2} A D \cdot I_{2} E- \\
& \frac{1}{2} B D \cdot I_{3} F-\frac{1}{2} I_{2} D \cdot I_{3} D \\
= & 5 .
\end{aligned}
$$
|
5
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. $a, b$ are constants. If the parabola $C$:
$$
y=\left(t^{2}+t+1\right) x^{2}-2(a+t)^{2} x+t^{2}+3 a t+b
$$
passes through the fixed point $P(1,0)$ for any real number $t$, find the value of $t$ when the chord intercepted by the parabola $C$ on the $x$-axis is the longest.
|
$=1.2$.
Substituting $P(1,0)$ into the equation of the parabola $C$ yields
$$
\left(t^{2}+t+1\right)-2(a+t)^{2}+\left(t^{2}+3 a t+b\right)=0 \text {, }
$$
which simplifies to $t(1-a)+\left(1-2 a^{2}+b\right)=0$,
holding for all $t$.
Thus, $1-a=0$, and $1-2 a^{2}+b=0$.
Solving these, we get $a=1, b=1$.
Substituting into the equation of the parabola $C$ gives
$$
y=\left(t^{2}+t+1\right) x^{2}-2(1+t)^{2} x+\left(t^{2}+3 t+1\right) \text {. }
$$
Setting $y=0$, we have
$$
\left(t^{2}+t+1\right) x^{2}-2(1+t)^{2} x+\left(t^{2}+3 t+1\right)=0 \text {. }
$$
Let the roots of this equation be $x_{1}$ and $x_{2}$, then
$$
\begin{array}{l}
|A B|=\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\left|\frac{2 t}{t^{2}+t+1}\right| .
\end{array}
$$
Noting that the discriminant of this quadratic equation $\Delta=4 t^{2}>0$, to find the maximum value of $|A B|$, we categorize by $t$.
(1) When $t>0$,
$$
|A B|=\frac{2 t}{t^{2}+t+1}=\frac{2}{t+\frac{1}{t}+1} \leqslant \frac{2}{3} \text {; }
$$
(2) When $t<0$,
$$
\begin{array}{l}
|A B|=-\frac{2 t}{t^{2}+t+1}=\frac{2}{-t-\frac{1}{t}-1} \\
\leqslant \frac{2}{-1+2 \sqrt{(-t) \frac{1}{-t}}}=2 .
\end{array}
$$
When $t=-1$, $|A B|_{\text {max }}=2$.
From (1) and (2), we know that when $t=-1$, the chord length $|A B|$ reaches its maximum value of 2.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. To make $p=x^{4}+6 x^{3}+11 x^{2}+3 x+32$ a perfect square of an integer, then the integer $x$ has $\qquad$ solutions.
|
2.0 .
From the problem, we know that $p \equiv x^{4}-x^{2}+2(\bmod 3)$.
Since $x^{2} \equiv 0,1(\bmod 3)$, thus,
$$
p \equiv x^{2}\left(x^{2}-1\right)+2 \equiv 2(\bmod 3) \text {. }
$$
Therefore, $p$ is not a perfect square.
|
2.0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the expansion of $(\sqrt[5]{3}+\sqrt[3]{5})^{100}$, there are $\qquad$ terms that are rational numbers.
|
4.7.
The general term of the binomial theorem expansion is
$$
\mathrm{C}_{100}^{r} 3^{\frac{1}{5}(100-r)} 5^{\frac{r}{3}}(0 \leqslant r \leqslant 100),
$$
it is a rational number if and only if
$$
3|r, 5|(100-r) \Rightarrow 15 \mid r \text {. }
$$
It is easy to see that there are 7 such $r$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $x, y \in \mathbf{R}$. Then
$$
\cos (x+y)+2 \cos x+2 \cos y
$$
the minimum value is $\qquad$ $\therefore$.
|
$$
\begin{array}{l}
\text { 5. }-3 \text {. } \\
\cos (x+y)+2 \cos x+2 \cos y+3 \\
=\left(1+\cos \frac{x+y}{2}\right)^{2}\left(1+\cos \frac{x-y}{2}\right)+ \\
\quad\left(1-\cos \frac{x+y}{2}\right)^{2}\left(1-\cos \frac{x-y}{2}\right) \\
\geqslant 0 .
\end{array}
$$
It is clear that equality can be achieved.
Therefore, the minimum value of the given algebraic expression is -3.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $k$ is a positive integer not exceeding 50, such that for any positive integer $n, 2 \times 3^{6 n}+k \times 2^{3 n+1}-1$ is always divisible by 7. Then the number of such positive integers $k$ is $\qquad$.
|
9.7.
$$
\begin{array}{l}
2 \times 3^{6 n}+k \times 2^{3 n+1}-1 \\
=2 \times 27^{2 n}+2 k \times 8^{n}-1 \\
\equiv 2 \times(-1)^{2 n}+2 k-1 \\
\equiv 2 k+1(\bmod 7) .
\end{array}
$$
But $2 \times 3^{6 n}+k \times 2^{3 n+1}-1 \equiv 0(\bmod 7)$, then
$$
2 k+1 \equiv 0(\bmod 7) \text {, }
$$
which means $2 k+1=7 m$ (where $m$ is an odd number).
Since $1 \leqslant k \leqslant 50$, we have $3 \leqslant 7 m \leqslant 101$.
Thus, $m=1,3, \cdots, 13$, and the corresponding $k=3,10, \cdots, 45$, a total of 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Try to find all positive integers $k$, such that for any positive integers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a$, $b$, and $c$.
(2002, Girls' Mathematical Olympiad)
|
Explanation: First, from the inequality relationship between $a b+b c+c a$ and $a^{2}+b^{2}+c^{2}$, we derive $k>5$. Then, by constructing an example, we find that $k \leqslant 6$. Therefore, since $k \in \mathbf{Z}_{+}$, we conclude that $k=6$. Finally, we provide the proof.
Notice that for any positive real numbers $a, b, c$, we have
$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a.
$$
On one hand, from
$$
\begin{array}{l}
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) \\
\geqslant 5(a b+b c+c a),
\end{array}
$$
we get $k>5$.
Since $k \in \mathbf{Z}_{+}$, we have $k \geqslant 6$.
On the other hand, because a triangle with side lengths $1, 1, 2$ does not exist, by the problem statement, we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right),
$$
which implies $k \leqslant 6$.
Thus, we have $k=6$.
Next, we prove that $k=6$ satisfies the condition.
Assume without loss of generality that $a \leqslant b \leqslant c$.
From $6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, we get
$5 c^{2}-6(a+b) c+\left(5 a^{2}+5 b^{2}-6 a b\right)<0$.
By $\Delta=64\left[a b-(a-b)^{2}\right]$
$$
\leqslant 64 a b \leqslant 16(a+b)^{2},
$$
we have $c<\frac{6(a+b)+\sqrt{\Delta}}{10}$
$$
\leqslant \frac{6(a+b)+4(a+b)}{10}=a+b.
$$
Therefore, a triangle with side lengths $a, b, c$ exists.
Hence, the positive integer $k$ that satisfies the condition is $k=6$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y$ satisfy the equation
$$
x^{2}-3 x y+3 y^{2}+4 x-18 y+52=0 \text {. }
$$
then the units digit of $y^{x}$ is $\qquad$ .
|
$=1.4$.
From the given equation, we have
$$
x^{2}-(3 y-4) x+\left(3 y^{2}-18 y+52\right)=0 \text {. }
$$
Since $x$ is a real number, then
$$
\Delta=(3 y-4)^{2}-4\left(3 y^{2}-18 y+52\right) \geqslant 0 \text {, }
$$
which simplifies to $-3(y-8)^{2} \geqslant 0$.
Thus, $y=8$.
Substituting $y=8$ into the original equation, we get $x^{2}-20 x+100=0$.
Solving this, we find $x=10$.
Therefore, $y^{z}=8^{10}=2^{30}=\left(2^{4}\right)^{7} \times 2^{2}=4 \times 16^{7}$.
Since the units digit of $16^{7}$ is 6, the units digit of $4 \times 16^{7}$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 4, given that the equilateral $\triangle ABC$ is inscribed in $\odot O, AB$ $=86$. If point $E$ is on side $AB$, and through $E$ a line $DG \parallel BC$ intersects $\odot O$ at points $D, G$, and intersects $AC$ at point $F$, and let $AE=x, DE$ $=y$. If $x, y$ are both positive integers, then $y=$ $\qquad$
|
4.12.
From the problem, we know that $E F=A E=x$.
By the symmetry of the circle, $F G=D E=y$.
By the intersecting chords theorem, we have $A E \cdot E B=D E \cdot E G$, which means $x(86-x)=y(x+y)$.
If $x$ is odd, then $x(86-x)$ is also odd, and in this case, $y(x+y)$ is even. Therefore, $x$ must be even, and $y$ must also be even.
From $x(86-x)=y(x+y)$, we get
$x^{2}+(y-86) x+y^{2}=0$.
Thus, $x=\frac{86-y \pm \sqrt{(y-86)^{2}-4 y^{2}}}{2}$.
By $\Delta=(y-86)^{2}-4 y^{2} \geqslant 0$, we get
$0<y \leqslant 28$.
Since $x$ is a positive integer, $\Delta$ must be a perfect square.
Since $\Delta=(86-3 y)(86+y)$, and $y=2,4$, $\cdots, 28$, it is verified that only when $y=12$, $\Delta$ is a perfect square. At this time, $x=2$ or 72.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 5, in $\odot O$, $AB$ and $CD$ are two perpendicular diameters. Point $E$ is on radius $OA$, and point $F$ is on the extension of radius $OB$, such that $OE = BF$. Lines $CE$ and $CF$ intersect $\odot O$ at points $G$ and $H$, respectively. Lines $AG$ and $AH$ intersect line $CD$ at points $N$ and $M$, respectively. Prove that:
$$
\frac{DM}{MC} - \frac{DN}{NC} = 1.
$$
|
As shown in Figure 7, connect $D G$ and $D H$, and extend $C G$ to point $P$. Since $A B$ and $C D$ are perpendicular diameters,
we have
$$
\overparen{A C}=\overparen{A D}=\overparen{B C}=\overparen{B D}.
$$
Thus,
$$
\begin{array}{l}
\angle A H C=\angle A H D \\
=\angle A G C \\
=\angle P G N=45^{\circ}.
\end{array}
$$
Since $A, H, D, G$ are concyclic,
$$
\angle D G N=\angle A H D=45^{\circ}.
$$
Therefore, $\angle C H M=\angle D H M$, and $\angle P G N=\angle D G N$.
Thus, $H M$ is the angle bisector of $\triangle C H D$, and $G N$ is the external angle bisector of $\triangle C G D$.
So, $\frac{D M}{M C}=\frac{H D}{H C}$, and $\frac{D N}{N C}=\frac{G D}{G C}$.
Hence, $\frac{D M}{M C}-\frac{D N}{N C}=\frac{H D}{H C}-\frac{G D}{G C}$.
By the similarity of Rt $\triangle C D H \sim$ Rt $\triangle C F O$ and Rt $\triangle C D G \sim$ Rt $\triangle C E O$, we get
$$
\frac{H D}{H C}=\frac{O F}{O C}, \frac{G D}{G C}=\frac{O E}{O C}.
$$
Therefore, $\frac{H D}{H C}-\frac{G D}{G C}=\frac{O F}{O C}-\frac{O E}{O C}=\frac{O F-O E}{O C}$.
Since $O E=B F$ and $O F=O B+B F$, we have $O F-O E=O B=O C$.
Thus, $\frac{D M}{M C}-\frac{D N}{N C}=1$.
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the geometric sequence $z_{1}, z_{2}, \cdots, z_{n}, \cdots$ be such that $z_{1}=$ $1, z_{2}=a+b \mathrm{i}, z_{3}=b \mathrm{i}(a, b \in \mathbf{R}, ab>0)$. Then the smallest natural number $n$ for which $z_{1} z_{2} \cdots z_{n}<0$ is $\qquad$ .
|
$Ni, 1.8$.
$$
\begin{array}{l}
\text { Given } z_{2}^{2}=z_{1} z_{3} \Rightarrow(a+b \mathrm{i})^{2}=b \mathrm{i} \\
\Rightarrow a^{2}-b^{2}+2 a b \mathrm{i}=b \mathrm{i} \\
\Rightarrow\left\{\begin{array} { l }
{ a ^ { 2 } = b ^ { 2 } , } \\
{ 2 a b = b }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\frac{1}{2}, \\
b=\frac{1}{2}
\end{array}\right.\right. \\
\Rightarrow \text { common ratio } q=\frac{z_{2}}{z_{1}}=\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+\mathrm{i} \sin \frac{\pi}{4}\right) \\
\Rightarrow z_{1} z_{2} \cdots z_{n}=\left(z_{1}\right)^{n} q^{\frac{n(n-1)}{2}} \\
=\left(\frac{\sqrt{2}}{2}\right)^{\frac{n(n-1)}{2}}\left[\cos \frac{n(n-1)}{8} \pi+i \sin \frac{n(n-1)}{8} \pi\right] \\
<0 \\
\Rightarrow\left\{\begin{array}{l}
\cos \frac{n(n-1)}{8} \pi=-1, \\
\sin \frac{n(n-1)}{8} \pi=0
\end{array}\right. \\
\Rightarrow \frac{n(n-1)}{8}=k \text { (odd number). } \\
\end{array}
$$
The smallest value of $n$ that satisfies the condition is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A has a box, inside there are 4 balls in total, red and white; Person B has a box, inside there are 2 red balls, 1 white ball, and 1 yellow ball. Now, A randomly takes 2 balls from his box, B randomly takes 1 ball from his box. If the 3 balls drawn are all of different colors, then A wins. To ensure A has the highest probability of winning, the number of red balls in A's box should be
|
3.2.
Suppose box A contains $n(n \geqslant 1)$ red balls, then it has $4-n$ white balls.
Therefore, the probability of A winning is
$$
P=\frac{\mathrm{C}_{n}^{1} \mathrm{C}_{4-n}^{1}}{\mathrm{C}_{4}^{2} \mathrm{C}_{4}^{1}}=\frac{1}{24} n(4-n) \text {. }
$$
Since $\sqrt{n(4-n)} \leqslant \frac{n+4-n}{2}=2$, that is,
$$
n(4-n) \leqslant 4 \text {, }
$$
the equality holds if and only if $n=2$, at which $P$ is maximized.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the ellipse $\frac{y^{2}}{a^{2}}+$ $\frac{x^{2}}{b^{2}}=1(a>b>0)$, $m=\left(\frac{x_{1}}{b}, \frac{y_{1}}{a}\right)$, $n$ $=\left(\frac{x_{2}}{b}, \frac{y_{2}}{a}\right)$, and $\boldsymbol{m} \cdot \boldsymbol{n}=0$. The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$, the length of the minor axis is $2$, and $O$ is the origin. Then the area of $\triangle A O B$ is - $\qquad$
|
4.1.
Given $e^{2}=\frac{c^{2}}{a^{2}}=\frac{3}{4}, b=1$, then $a^{2}=b^{2}+c^{2}=4$.
Also, $\boldsymbol{m} \cdot \boldsymbol{n}=0 \Rightarrow y_{1} y_{2}=-4 x_{1} x_{2}$.
Substituting the coordinates of points $A$ and $B$ into the ellipse equation, we get
$$
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{4}+x_{1}^{2}=1 \\
\frac{y_{2}^{2}}{4}+x_{2}^{2}=1 .
\end{array}\right.
$$
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2}|O A| \cdot|O B| \sin \angle A O B \\
=\frac{1}{2}|O A| \cdot|O B| \sqrt{1-\cos ^{2} \angle A O B} \\
=\frac{1}{2}|O A| \cdot|O B| \sqrt{1-\left(\frac{O A \cdot O B}{|O A| \cdot|O B|}\right)^{2}} \\
=\frac{1}{2} \sqrt{|O A|^{2} \cdot|O B|^{2}-(O A \cdot O B)^{2}} \\
=\frac{1}{2} \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)-\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}} .
\end{array}
$$
Substituting equations (1), (2), and (3) into the above equation, we get
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2} \sqrt{\left(4-3 x_{1}^{2}\right)\left(4-3 x_{2}^{2}\right)-\left(-3 x_{1} x_{2}\right)^{2}} \\
=\sqrt{4-3\left(x_{1}^{2}+x_{2}^{2}\right)} .
\end{array}
$$
From (1) $)^{2}$, we easily get $y_{1}^{2} y_{2}^{2}=16 x_{1}^{2} x_{2}^{2}$, i.e.,
$$
16\left(1-x_{1}^{2}\right)\left(1-x_{2}^{2}\right)=16 x_{1}^{2} x_{2}^{2} \text {. }
$$
Thus, $x_{1}^{2}+x_{2}^{2}=1$.
Therefore, $S_{\triangle A O B}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the Cartesian coordinate system, there is a parabola $y=$ $x^{2}-(5 c-3) x-c$ and three points $A\left(-\frac{1}{2} c, \frac{5}{2} c\right)$, $B\left(\frac{1}{2} c, \frac{9}{2} c\right)$, $C(2 c, 0)$, where $c>0$. There exists a point $P$ on the parabola such that the quadrilateral with vertices $A$, $B$, $C$, and $P$ is a parallelogram. Then the number of such points $P$ is $\qquad$.
|
4.3.
(1) If $A B$ is the diagonal, then $P_{1}(-2 c, 7 c)$. For $P_{1}$ to be on the parabola, it must satisfy
$$
7 c=(-2 c)^{2}-(5 c-3)(-2 c)-c \text {. }
$$
Solving this, we get $c_{1}=0$ (discard), $c_{2}=1$.
Thus, $c=1$, and at this point, $P_{1}(-2,7)$.
(2) If $B C$ is the diagonal, then $P_{2}(3 c, 2 c)$. Similarly, we get $c=1$, and at this point, $P_{2}(3,2)$.
(3) If $A C$ is the diagonal, then $P_{3}(c,-2 c)$, we get $c=1$, and at this point, $P_{3}(1,-2)$.
In summary, there exist three points $P_{1}(-2,7)$, $P_{2}(3,2)$, and $P_{3}(1,-2)$, such that the quadrilateral with vertices $A$, $B$, $C$, and $P$ is a parallelogram.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 2, $EF$ intersects the diagonal $AC$ of $\square ABCD$ at point $G$,
intersects $AB$, $AD$ at points
$E$, $F$, and connects $CE$,
$CF$, $BG$. If $\frac{1}{S_{\triangle ACE}}+$
$\frac{1}{S_{\triangle ACF}}=\frac{\lambda}{S_{\triangle ABG}}$, find
the value of $\lambda$.
|
II. As shown in Figure 8, draw $EM \parallel BC$ intersecting $AC$ at $M$, then $\frac{AB}{AE}=\frac{AC}{AM}$.
From $\frac{AD}{AF}=\frac{BC}{AF}$
$=\frac{BC}{EM} \cdot \frac{EM}{AF}$
$=\frac{AC}{AM} \cdot \frac{GM}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AM}+\frac{AC}{AM} \cdot \frac{GM}{AG}$
$=\frac{AC}{AM}\left(1+\frac{GM}{AG}\right)=\frac{AC}{AM} \cdot \frac{AM}{AG}=\frac{AC}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AG}$
$\Rightarrow \frac{AB \cdot AG}{AC \cdot AE}+\frac{AD \cdot AG}{AC \cdot AF}=1$.
Since $\frac{AB \cdot AG}{AC \cdot AE}=\frac{S_{\triangle ABG}}{S_{\triangle ACE}}, \frac{AD \cdot AG}{AC \cdot AF}=\frac{S_{\triangle ADG}}{S_{\triangle ACF}}$, thus
$\frac{S_{\triangle ABG}}{S_{\triangle ACE}}+\frac{S_{\triangle ADG}}{S_{\triangle ACF}}=1$.
It is easy to prove that $\triangle ABC \cong \triangle CDA$, then
$S_{\triangle ABC}=S_{\triangle ADC}$.
Draw $BP \perp AC$, and $DQ \perp AC$, with feet of the perpendiculars at $P$ and $Q$ respectively.
From $S_{\triangle ABC}=\frac{1}{2} AC \cdot BP, S_{\triangle ADC}=\frac{1}{2} AC \cdot DQ$, we get
$BP=DQ$.
Also, $S_{\triangle ABG}=\frac{1}{2} AG \cdot BP, S_{\triangle ADG}=\frac{1}{2} AG \cdot DQ$,
thus $S_{\triangle ABG}=S_{\triangle ADG}$
$\Rightarrow \frac{S_{\triangle ABG}}{S_{\triangle ACE}}+\frac{S_{\triangle ABG}}{S_{\triangle ACF}}=1 \Rightarrow \lambda=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
---
In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
|
Three, as shown in Figure 9, let the sides of $\triangle ABC$ be $BC = a$, $CA = b$, and $AB = c$. Let $CD = h$, $AD = x$, and $BD = y$. The radii of the two incircles are $r_1$ and $r_2$. Then,
$$
\begin{array}{l}
RS = |RD - SD| \\
= |r_1 - r_2|. \\
\text{Since } b = AC = AP + CP = AE + CR \\
= (x - r_1) + (h - r_1),
\end{array}
$$
Thus, $r_1 = \frac{x + h - b}{2}$.
Similarly, $r_2 = \frac{y + h - a}{2}$.
Therefore, $|RS| = |r_1 - r_2|$
$$
\begin{array}{l}
= \left| \frac{x + h - b}{2} - \frac{y + h - a}{2} \right| \\
= \frac{1}{2} |(x - y) + (a - b)|.
\end{array}
$$
Also, $a^2 - y^2 = h^2 = b^2 - x^2$, so $x^2 - y^2 = b^2 - a^2$.
Since $x + y = c$, we have
$$
x - y = \frac{(b + a)(b - a)}{c}.
$$
Substituting equation (2) into equation (1) gives
$$
\begin{array}{l}
RS = \frac{1}{2} \left| \frac{(b + a)(b - a)}{c} + (a - b) \right| \\
= \frac{|b - a|}{2c} |a + b - c|.
\end{array}
$$
Since $a$, $b$, and $c$ are three consecutive natural numbers, let these three consecutive natural numbers be $n$, $n+1$, and $n+2$.
Since equation (3) is symmetric with respect to $a$ and $b$, we can assume without loss of generality that $a > b$. There are three cases:
(1) $a = n+1$, $b = n$, $c = n+2$, in this case, $RS = \frac{1 \times (n-1)}{2(n+2)}$ is not an integer.
(2) $a = n+2$, $b = n$, $c = n+1$, in this case, $RS = \frac{2(n+1)}{2(n+1)} = 1$, there are infinitely many such $n$, so there are infinitely many triangles that satisfy the condition.
(3) $a = n+2$, $b = n+1$, $c = n$, in this case, $RS = \frac{n+3}{2n}$, for $RS$ to be a positive integer, then $n+3 \geq 2n$.
Thus, $n \leq 3$.
Since $n$ is a positive integer, then $n = 1, 2, 3$.
When $n = 1$, $a = 3$, $b = 2$, $c = 1$, in this case, $b + c = a$ cannot form a triangle, so $n \neq 1$.
When $n = 2$, $RS = \frac{5}{4}$ is not an integer.
When $n = 3$, $RS = 1$.
In summary, $RS = 1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 As shown in Figure 5, in Rt $\triangle C A B$, $\angle A=$ $90^{\circ}, \angle B γ \angle C$ are bisected and intersect at $F$, and intersect the opposite sides at points $D γ E$. Find $S_{\text {quadrilateral } B C D E}: S_{\triangle B F C}$.
|
Solution: Let $A B=c, A C=b, B C=a$. By the property of the internal angle bisector, we have
$$
\frac{A E}{E B}=\frac{b}{a} \text {. }
$$
Then $\frac{A E}{c}=\frac{b}{a+b}$, i.e., $A E=\frac{b c}{a+b}$.
Thus, $B E=c-A E=\frac{a c}{a+b}$.
Therefore, $S_{\triangle C B E}=\frac{1}{2} B E \cdot b=\frac{a b c}{2(a+b)}$.
And $\frac{C F}{F E}=\frac{a}{B E}=\frac{a+b}{c}$, so
$$
\frac{C F}{C E}=\frac{a+b}{a+b+c}, \frac{S_{\triangle C B F}}{S_{\triangle C B E}}=\frac{a+b}{a+b+c},
$$
$$
S_{\triangle B P C}=\frac{a b c}{2(a+b+c)} \text {. }
$$
Similarly, we can get $A D=\frac{b c}{a+c}$.
Then $S_{\text {quadrilateral } B C D E}=\frac{1}{2} b c-\frac{1}{2} \cdot \frac{b c}{a+c} \cdot \frac{b c}{a+b}$
$$
=\frac{a b c(a+b+c)}{2(a+b)(a+c)} \text {. }
$$
Noting that $a^{2}=b^{2}+c^{2}$, then
$$
\begin{array}{l}
\frac{S_{\text {quadrilateralBCDE }}}{S_{\triangle B F C}}=\frac{(a+b+c)^{2}}{(a+b)(a+c)} \\
=\frac{a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a}{a^{2}+a b+a c+b c} \\
=\frac{2 a^{2}+2 a b+2 b c+2 c a}{a^{2}+a b+a c+b c}=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given that the equation $x^{3}+3 x^{2}-x+a$ $=0$ has three real roots that form an arithmetic sequence. Then the real number $a=$ $\qquad$
|
12. -3 .
Let these three roots be $b-d$, $b$, and $b+d$. Then
$$
\begin{array}{l}
x^{3}+3 x^{2}-x+a \\
=(x-b+d)(x-b)(x-b-d),
\end{array}
$$
i.e., $3 x^{2}-x+a$
$$
=-3 b x^{2}+\left(3 b^{2}-d^{2}\right) x-b^{3}+b d^{2} \text {. }
$$
Comparing coefficients, we get
$$
-3 b=3,3 b^{2}-d^{2}=-1,-b^{3}+b d^{2}=a \text {. }
$$
Thus, $a=-3$.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The incenter of $\triangle A B C$ is $I$, and the angle bisector of $\angle B$ intersects $A C$ at point $P$. If $A P+A B=B C$, and $A B=3, B C=$ 5, then the value of $A I$ is $\qquad$ .
|
2.2.
As shown in Figure 7, on segment $B C$, take $B A^{\prime}=B A$, extend $A I$ to intersect $B C$ at $Q$, and connect $P A^{\prime}$.
$$
\begin{array}{l}
\text { Given } B A^{\prime}=B A, \\
\angle A B P=\angle A^{\prime} B P, \\
B P=B P,
\end{array}
$$
we have $\triangle A B P \cong \triangle A^{\prime} B P$.
Thus, $A^{\prime} P=A P=B C-B A=B C-B A^{\prime}=A^{\prime} C$.
Therefore, $\angle A^{\prime} P C=\angle C$.
Since $\angle B A C=\angle B A^{\prime} P=\angle C+\angle A^{\prime} P C=$ $2 \angle C$, we have
$$
\begin{array}{l}
\angle A P B=\angle P B C+\angle C \\
=\frac{1}{2} \angle B+\angle C=\frac{1}{2}(\angle B+\angle A). \\
\text { Also, } \angle A I P=\angle I A B+\angle I B A \\
=\frac{1}{2}(\angle A+\angle B),
\end{array}
$$
Thus, $\angle A P B=\angle A I P$, and $A I=A P$.
Since $A P=B C-A B=5-3=2$, we have $A I=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Calculate: $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=$
|
9.4 .
$$
\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=\frac{2 \sin \left(30^{\circ}-10^{\circ}\right)}{\frac{1}{2} \sin 20^{\circ}}=4
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. To cut a rectangular prism into $k$ tetrahedra, the minimum value of $k$ is
|
11.5 .
According to the equivalence, we only need to consider the cutting situation of a unit cube.
On the one hand, first, we need to show that 4 is not enough. If there were 4, since all faces of a tetrahedron are triangles and are not parallel to each other, the top face of the cube would have to be cut into at least two triangles, and the bottom face would also have to be cut into at least two triangles. The area of each triangle is less than or equal to $\frac{1}{2}$, and these four triangles must belong to four different tetrahedrons. The height of a tetrahedron with such a triangle as its base is less than or equal to 1.
Therefore, the sum of the volumes of the four different tetrahedrons is less than or equal to $4\left(\frac{1}{3} \times \frac{1}{2} \times 1\right)=\frac{2}{3}<1$, which does not meet the requirement.
Thus, $k \geqslant 5$.
On the other hand, as shown in Figure 5, the unit cube can be cut into 5 tetrahedrons, for example, by removing a tetrahedron $A_{1} B C_{1} D$ from the center of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, leaving four tetrahedrons at the corners.
In total, there are 5 tetrahedrons.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then, in the plane, the area of the figure formed by all points satisfying $[x]^{2}+[y]^{2}=50$ is
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ
|
5.12.
First, consider the first quadrant.
From $50=1^{2}+7^{2}=7^{2}+1^{2}=5^{2}+5^{2}$, we get $([x],[y])=(1,7),(7,1),(5,5)$.
And from $[x]=1,[y]=7$, we get the unit square $1 \leqslant x<2,7 \leqslant y<8$, whose area is 1.
Similarly, from $([x],[y])=(7,1)$ and $(5,5)$, we also get one unit square each, and these three squares do not overlap.
If we consider all four quadrants, we get a total of 12 non-overlapping unit squares, with a total area of 12.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $x, y, z$ be non-negative real numbers, and $x+y+z=$ 2. Then the sum of the maximum and minimum values of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ is $\qquad$ .
|
2.1.
Since $x, y, z$ are non-negative real numbers, we have
$$
A=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant 0.
$$
When $x=y=0, z=2$, $A=0$, thus the minimum value of $A$ is 0.
Assume $A$ reaches its maximum value at $(x, y, z)$, without loss of generality, let $x \leqslant y \leqslant z$, prove: $x=0$.
In fact, if $x>0$, then let $x^{\prime}=0, y^{\prime}=y+x, z^{\prime}=z$, at this time,
$$
\begin{array}{l}
A^{\prime}=x^{\prime 2} y^{\prime 2}+y^{\prime 2} z^{\prime 2}+z^{\prime 2} x^{\prime 2}=(y+x)^{2} z^{2} \\
=y^{2} z^{2}+z^{2} x^{2}+2 x y z^{2} \\
\geqslant y^{2} z^{2}+z^{2} x^{2}+2 x y(x y) \\
>y^{2} z^{2}+z^{2} x^{2}+x y(x y)=A,
\end{array}
$$
which contradicts the maximality of $A$.
When $x=0$,
$$
y+z=2, A=y^{2} z^{2} \leqslant\left(\frac{y+z}{2}\right)^{4}=1,
$$
the equality holds when $y=z=1, x=0$.
Therefore, the maximum value of $A$ is 1.
Thus, the sum of the maximum and minimum values is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Find the smallest positive integer $t$, such that for any convex $n$-gon $A_{1} A_{2} \cdots A_{n}$, as long as $n \geqslant t$, there must exist three points $A_{i} γ A_{j} γ A_{k}(1 \leqslant i<j<k \leqslant n)$, such that the area of $\triangle A_{i} A_{j} A_{k}$ is no more than $\frac{1}{n}$ of the area of the convex $n$-gon $A_{1} A_{2} \cdots A_{n}$.
|
Three, first prove a lemma.
Lemma For any convex hexagon $A_{1} A_{2} \cdots A_{6}$, there exists $1 \leqslant i\frac{S}{k+1}$, then
$S_{\text {pentagon } A_{1} A_{2} \cdots A_{k}}<S-\frac{S}{k+1}=\frac{k S}{k+1}$.
By the induction hypothesis, there must be $1 \leqslant i<j<r \leqslant n$, such that $S_{\triangle A_{i} A_{j} A_{1}} \leqslant \frac{1}{k} \cdot \frac{k S}{k+1}=\frac{S}{k+1}$.
The conclusion holds.
In summary, the minimum value of $t$ is 6.
(Feng Yuefeng, Shenzhen Senior High School, 518040)
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x^{2}+y^{2} \leqslant 1$. Then the maximum value of the function $z=$ $\frac{\cos x+\cos y}{1+\cos x y}$ is $\qquad$ .
|
2 1.1.
Assume $x \geqslant 0, y \geqslant 0$. When $0 \leqslant x \leqslant 1$, $0 \leqslant x y \leqslant y<1<\pi$, so $\cos x y \geqslant \cos y$.
Therefore, $\cos x+\cos y \leqslant 1+\cos x y$, where the equality holds when $x=y=0$.
Hence $z_{\max }=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (20 points) As shown in Figure 1, $EF$ is a chord of the parabola $\Gamma: y^{2}=2px$. Tangents to $\Gamma$ at points $E$ and $F$ intersect at point $C$. Points $A$ and $B$ are on the rays $EC$ and $CF$ respectively, and $\frac{EC}{CA}=\frac{CF}{FB}=\lambda$.
(1) Prove that the line $AB$ is tangent to $\Gamma$;
(2) Let $AB$ be tangent to $\Gamma$ at point $G$. Find $\frac{S_{\triangle EFG}}{S_{\triangle ABC}}$.
|
(1) Let $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then the equations of the tangents $E C$ and $C F$ are
$$
y_{1} y=p\left(x+x_{1}\right), y_{2} y=p\left(x+x_{2}\right).
$$
By solving the system of equations, we get
$$
C\left(\frac{y_{1} y_{2}}{2 p}, \frac{y_{1}+y_{2}}{2}\right).
$$
From $\frac{E C}{C A}=\frac{C F}{F B}=\lambda$, we have
$$
\frac{\boldsymbol{E} \boldsymbol{A}}{\boldsymbol{A C}}=\frac{\boldsymbol{C B}}{\boldsymbol{B F}}=-(1+\lambda).
$$
Let $A\left(m_{1}, n_{1}\right), B\left(m_{2}, n_{2}\right)$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
m_{1}=\frac{x_{1}-(1+\lambda) \frac{y_{1} y_{2}}{2 p}}{1-(1+\lambda)}=\frac{(1+\lambda) y_{1} y_{2}-y_{1}^{2}}{2 p \lambda}, \\
n_{1}=\frac{y_{1}-(1+\lambda) \frac{y_{1}+y_{2}}{2}}{1-(1+\lambda)}=\frac{(1+\lambda) y_{2}-(1-\lambda) y_{1}}{2 \lambda} ;
\end{array}\right. \\
\left\{\begin{array}{l}
m_{2}=\frac{\frac{y_{1} y_{2}}{2 p}-(1+\lambda) x_{2}}{1-(1+\lambda)}=\frac{(1+\lambda) y_{2}^{2}-y_{1} y_{2}}{2 p \lambda}, \\
n_{2}=\frac{\frac{y_{1}+y_{2}}{2}-(1+\lambda) y_{2}}{1-(1+\lambda)}=\frac{(1+2 \lambda) y_{2}-y_{1}}{2 \lambda} .
\end{array}\right. \\
\end{array}
$$
Thus, $n_{1}-n_{2}=\frac{y_{1}-y_{2}}{2}$,
$m_{1}-m_{2}=\frac{\left(y_{1}-y_{2}\right)\left[(1+\lambda) y_{2}-y_{1}\right]}{2 p \lambda}$.
Hence $\frac{n_{1}-n_{2}}{m_{1}-m_{2}}=\frac{p \lambda}{(1+\lambda) y_{2}-y_{1}}$.
Then $n_{1}-\frac{n_{1}-n_{2}}{m_{1}-m_{2}} \cdot m_{1}$
$$
\begin{aligned}
& =\frac{(1+\lambda) y_{2}-(1-\lambda) y_{1}}{2 \lambda}- \\
& \frac{p \lambda}{(1+\lambda) y_{2}-y_{1}} \cdot \frac{(1+\lambda) y_{1} y_{2}-y_{1}^{2}}{2 p \lambda} \\
= & \frac{(1+\lambda) y_{2}-y_{1}}{2 \lambda} .
\end{aligned}
$$
Therefore, the equation of the line $A B$ is
$$
y=\frac{n_{1}-n_{2}}{m_{1}-m_{2}} x+n_{1}-\frac{n_{1}-n_{2}}{m_{1}-m_{2}} \cdot m_{1}
$$
which can be written as
$$
y=\frac{p \lambda}{(1+\lambda) y_{2}-y_{1}} x+\frac{(1+\lambda) y_{2}-y_{1}}{2 \lambda} .
$$
By combining this with the equation $y^{2}=2 p x$ and eliminating $x$, we get
$$
y^{2}-\frac{2\left[(1+\lambda) y_{2}-y_{1}\right]}{\lambda} y+\frac{\left[(1+\lambda) y_{2}-y_{1}\right]^{2}}{\lambda^{2}}
$$
$=0$.
Since its discriminant
$$
\begin{array}{l}
\Delta=4\left[\frac{(1+\lambda) y_{2}-y_{1}}{\lambda}\right]^{2}-4 \cdot \frac{\left[(1+\lambda) y_{2}-y_{1}\right]^{2}}{\lambda^{2}} \\
=0,
\end{array}
$$
the line $A B$ is tangent to $\Gamma$.
(2) Let the y-coordinate of the tangent point $G$ be $n$. From (1), we have
$$
\begin{array}{l}
n=\frac{(1+\lambda) y_{2}-y_{1}}{\lambda} . \\
\text { Hence } \frac{A B}{B G}=\frac{n_{2}-n_{1}}{n-n_{2}}
\end{array}
$$
$$
\begin{array}{l}
=\frac{\frac{y_{2}-y_{1}}{2}}{\frac{(1+\lambda) y_{2}-y_{1}}{\lambda}-\frac{(1+2 \lambda) y_{2}-y_{1}}{2 \lambda}} \\
=\frac{\frac{y_{2}-y_{1}}{2}}{\frac{y_{2}-y_{1}}{2 \lambda}}=\lambda .
\end{array}
$$
Thus, $\frac{S_{\triangle A B G}}{S_{\triangle A B C}}=\frac{|A E| \cdot|A G|}{|A C| \cdot|A B|}$
$$
=(1+\lambda) \frac{1+\lambda}{\lambda}=\frac{(1+\lambda)^{2}}{\lambda}.
$$
Therefore, $\frac{S_{\text {quadrilateral } B C B C}}{S_{\triangle A B C}}=\frac{(1+\lambda)^{2}-\lambda}{\lambda}=\frac{\lambda^{2}+\lambda+1}{\lambda}$.
$$
\begin{array}{l}
\text { And } S_{\triangle C E F}=\frac{|E C| \cdot|C F|}{|C A| \cdot|C B|} S_{\triangle A B C} \\
=\lambda \cdot \frac{\lambda}{1+\lambda} S_{\triangle A B C}=\frac{\lambda^{2}}{1+\lambda} S_{\triangle A B C}, \\
S_{\triangle B G F}=\frac{|F B| \cdot|B G|}{|C B| \cdot|A B|} S_{\triangle A B C} \\
=\frac{1}{1+\lambda} \cdot \frac{1}{\lambda} S_{\triangle A B C}=\frac{1}{\lambda \cdot(1+\lambda)} S_{\triangle A B C}, \\
\text { Hence } S_{\triangle E F G}=S_{\text {quadrilateral } C C B C}-S_{\triangle C E F}-S_{\triangle B G F} \\
=2 S_{\triangle A B C}, \\
\text { i.e., } \frac{S_{\triangle E F G}}{S_{\triangle A B C}}=2 .
\end{array}
$$
|
2
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) The most recent mathematics competition consisted of 6 problems, with each correct answer scoring 7 points and each incorrect (or unanswered) question scoring 0 points. After the competition, a certain team scored a total of 161 points, and it was found during the score tallying that: any two participants in the team had at most two problems answered correctly in common, and no three participants answered the same two problems correctly. How many participants are there in the team at least?
Please retain the original text's line breaks and format, and output the translation result directly.
|
We find that by moving a point in the first column to another column in the same row, we can reduce the number of columns in Figure 6. For example, by making the move $(6,1) \rightarrow(6,2)$, we can simultaneously make the moves $(4,10) \rightarrow(6,3)$, $(3,9) \rightarrow(6,4)$, and $(5,9) \rightarrow(6,7)$, thus obtaining Figure 7 with 23 red points. Similarly, we can obtain Figure 8. This indicates that the minimum value of $n$ is no more than 7.
Three, suppose the team has $n$ players, denoted as $a_{1}$, $a_{2}$, ..., $a_{n}$, and the 6 problems are numbered as $1, 2, \cdots, 6$. We create a $6 \times n$ grid table, where if player $a_{i} (i=1,2, \cdots, n)$ answers the $j$-th problem $(j=1,2, \cdots, 6)$ correctly, we color the center of the small grid $(j, i)$ in the $j$-th row and $i$-th column red. Our problem is to find the minimum value of $n$ such that in a $6 \times n$ grid table, there are at least 23 red points without any "horizontal" 6 points.
If the first column has 6 red points, then the subsequent columns can have at most 2 red points each. Since $C_{6}^{2}=15>9$, we can take columns 2 to 10, where columns 2 to 9 each have 2 red points, and column 10 has 1 red point (as shown in Figure 6), which satisfies the problem's conditions. This indicates that the minimum value of $n$ is no more than 10.
Next, we prove that the minimum value of $n$ is greater than 6.
For a grid table with exactly 6 columns, by the pigeonhole principle, at least one column must have no fewer than 4 red points. Without loss of generality, assume the first column, and the first 4 rows of the first column have red points. If a column has 2 red points, we call it a "row pair" in that column. Thus, in the first 4 rows, each of the 5 columns other than the first column can have only one row pair. Therefore, the first 4 rows have a total of $C_{4}^{2}+5=11$ row pairs. Consider the last two rows: if the first column has additional red points, then the row with red points cannot have any other red points. If the first column has 2 additional red points, this can add 9 row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+9=20$; if the first column has 1 additional red point, assume the 5th row of the first column has a red point, then even if the 6th row has red points in all other columns except the first column, this can add $C_{4}^{1}+5 \times 2=14$ row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+14=25$; if the first column has no additional red points, then in the last two rows, there can be at most 2 row pairs, which occupy two columns, and in the remaining three columns, each column can have at most 1 red point, thus adding $2 \times 5+3 \times 2=16$ row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+16=27$. This indicates that 27 is the maximum possible total number of row pairs.
Let the number of red points in the $i$-th column be $x_{i} (i=1,2, \cdots, 6)$, and $\sum_{i=1}^{6} x_{i}=k$. Then the total number of row pairs is $\sum_{i=1}^{6} C_{x_{i}}^{2} \leqslant 27$, i.e.,
$$
\sum_{i=1}^{6} x_{i}^{2}-\sum_{i=1}^{6} x_{i} \leqslant 54.
$$
By the Cauchy-Schwarz inequality, we have
$$
\sum_{i=1}^{6} x_{i}^{2} \geqslant \frac{1}{6}\left(\sum_{i=1}^{6} x_{i}\right)^{2}=\frac{1}{6} k^{2}.
$$
Therefore, $\frac{k^{2}}{6} \leqslant k+54$.
Solving this, we get $3-3 \sqrt{37} \leqslant k \leqslant 3+3 \sqrt{37}$.
Since $k$ is a positive integer, $k \leqslant 21$. This indicates that the maximum number of red points in a $6 \times 6$ grid table is 21.
Furthermore, when $n \leqslant 5$, the total number of red points in the grid table is no more than $4 \times 5=20$. This indicates that the minimum value of $n$ is at least 7. In summary, the team must have at least 7 players.
(Liu Shixiong, Zhongshan Affiliated High School of South China Normal University, 528447)
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $f(x)$ is an odd function on the interval $\left[t, t^{2}-2 t-2\right]$, then the value of $t$ is $\qquad$
|
3. -1 .
The domain of an odd function is symmetric about the origin, and the right endpoint of the interval is not less than the left endpoint. Therefore,
$$
\left\{\begin{array} { l }
{ - t = t ^ { 2 } - 2 t - 2 > 0 , } \\
{ t \leqslant t ^ { 2 } - 2 t - 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t^{2}-t-2=0, \\
t<0, \\
t^{2}-3 t-2 \geqslant 0 .
\end{array}\right.\right.
$$
Solving this, we get $t \doteq-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The number of all distinct real solutions of the equation $2 x^{3}+6 x^{2}+12 x+8+x^{3} \sqrt{x^{2}+1}$ $+(x+2)^{3} \sqrt{(x+2)^{2}+1}=0$ is $\qquad$.
|
6.1.
Rewrite the equation as
$$
\begin{array}{l}
x^{3}+x^{3} \sqrt{x^{2}+1}+(x+2)^{3}+ \\
(x+2)^{3} \sqrt{(x+2)^{2}+1}=0 . \\
\text { Let } f(x)=x^{3}+x^{3} \sqrt{x^{2}+1} \\
=x^{3}\left(1+\sqrt{x^{2}+1}\right) .
\end{array}
$$
Then equation (1) can be rewritten as
$$
f(x+2)=-f(x) \text {. }
$$
Clearly, $f(-x)=-f(x)$.
From equations (2) and (3), we get
$$
f(x+2)=f(-x) \text {. }
$$
Clearly, $g(x)=x^{3}$ is an increasing function, and $h(x)=1+\sqrt{x^{2}+1}$ is an increasing function for $x \geqslant 0$ and a decreasing function for $x \leqslant 0$, and $h(x) \geqslant 2(x \in \mathbf{R})$. Therefore, when $x \geqslant 0$, $f(x)=g(x) h(x)$ is an increasing function; when $x \leqslant 0$, $f(x)=g(x) h(x)$ is also an increasing function.
If $x_{2}<0<x_{1}$, then $f\left(x_{2}\right)<f(0)<f\left(x_{1}\right)$. Therefore, $f(x)$ is an increasing function on $\mathbf{R}$.
Thus, from equation (4) we get $x+2=-x$, i.e., $x=-1$.
Clearly, $x=-1$ satisfies equation (1). Hence, the given equation has a unique real root $x=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Xiao Wang walks along the street at a uniform speed and finds that a No. 18 bus passes him from behind every 6 min, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals, then, the interval between dispatches is $\qquad$ $\min$.
|
7.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, and the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, with the distance between two consecutive buses traveling in the same direction being $s \mathrm{~m}$.
From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s
\end{array}\right.
$$
Solving these equations, we get $s=4 x$, i.e., $\frac{s}{x}=4$.
Therefore, the interval time for bus No. 18 to depart from the main station is $4 \mathrm{~min}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 1, in $\triangle A B C$, $A B=7, A C$ $=11, M$ is the midpoint of $B C$, $A D$ is the angle bisector of $\angle B A C$, $M F / / A D$. Then the length of $F C$ is $\qquad$
|
8.9.
As shown in Figure 5, let $N$ be the midpoint of $AC$, and connect $MN$. Then $MN \parallel AB$.
Also, $MF \parallel AD$, so,
$\angle FMN = \angle BAD = \angle DAC = \angle MFN$.
Therefore, $FN = MN = \frac{1}{2} AB$.
Thus, $FC = FN + NC = \frac{1}{2} AB + \frac{1}{2} AC = 9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14.A. Choose $n$ numbers from $1,2, \cdots, 9$. Among them, there must be some numbers (at least one, or possibly all) whose sum is divisible by 10. Find the minimum value of $n$.
|
14. A. When $n=4$, the numbers $1,3,5,8$ do not have any subset of numbers whose sum is divisible by 10.
When $n=5$, let $a_{1}, a_{2}, \cdots, a_{5}$ be five different numbers from $1,2, \cdots, 9$. If the sum of any subset of these numbers cannot be divisible by 10, then $a_{1}, a_{2}, \cdots, a_{5}$ cannot simultaneously contain 1 and 9, 2 and 8, 3 and 7, 4 and 6. Therefore, $a_{1}, a_{2}, \cdots, a_{5}$ must contain the number 5.
If $a_{1}, a_{2}, \cdots, a_{5}$ contains 1, then it does not contain 9. Thus, it does not contain $4 (4+1+5=10)$, so it contains 6; thus, it does not contain $3 (3+6+1=10)$, so it contains 7; thus, it does not contain $2 (2+1+7=10)$, so it contains 8. However, $5+7+8=20$ is a multiple of 10, which is a contradiction.
If $a_{1}, a_{2}, \cdots, a_{5}$ contains 9, then it does not contain 1. Thus, it does not contain $6 (6+9+5=20)$, so it contains 4; thus, it does not contain $7 (7+4+9=20)$, so it contains 3; thus, it does not contain $8 (8+9+3=10)$, so it contains 2. However, $5+3+2=10$ is a multiple of 10, which is a contradiction.
In summary, the minimum value of $n$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a, b$ are positive numbers, and
$$
a^{2009}+b^{2009}=a^{2007}+b^{2007} \text {, }
$$
then the maximum value of $a^{2}+b^{2}$ is $\qquad$
|
Given $a, b$, without loss of generality, assume $a \geqslant b>0$.
Then $a^{2} \geqslant b^{2}, a^{2007} \geqslant b^{2007}$.
$$
\begin{array}{l}
\text { At this point, }\left(a^{2007}-b^{2007}\right)\left(a^{2}-b^{2}\right) \geqslant 0 \\
\Rightarrow a^{2009}+b^{2009} \geqslant a^{2} b^{2007}+a^{2007} b^{2} \\
\Rightarrow\left(a^{2007}+b^{2007}\right)\left(a^{2}+b^{2}\right) \\
\quad=a^{2009}+b^{2009}+a^{2} b^{2007}+a^{2007} b^{2} \\
\quad \leqslant 2\left(a^{2009}+b^{2009}\right) .
\end{array}
$$
Considering $a^{2009}+b^{2009}=a^{2007}+b^{2007}$, and both sides are positive, so $a^{2}+b^{2} \leqslant 2$.
When $a=b=1$, $a^{2}+b^{2}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $P G=O G-O P=\frac{1}{3}(a+b)-\dot{m} a$
$$
\begin{array}{l}
=\left(\frac{1}{3}-m\right) a+\frac{1}{3} b, \\
G Q=O Q-O G=n b-\frac{1}{3}(a+b) \\
=-\frac{1}{3} a+\left(n-\frac{1}{3}\right) b,
\end{array}
$$
Thus, $\left(\frac{1}{3}-m\right) a+\frac{1}{3} \boldsymbol{b}$
$$
=\lambda\left[-\frac{1}{3} \boldsymbol{a}+\left(n-\frac{1}{3}\right) \boldsymbol{b}\right] \text {. }
$$
Since $a, b$ are not collinear, we have
$$
\left\{\begin{array}{l}
\frac{1}{3}-m=-\frac{1}{3} \lambda, \\
\frac{1}{3}=\lambda\left(n-\frac{1}{3}\right) .
\end{array}\right.
$$
Solving these equations, we get $3 m n=m+n$.
Therefore, $\frac{1}{m}+\frac{1}{n}=3$.
|
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
$$
\begin{array}{l}
\frac{2 p-1}{q}=m, \\
\frac{2 q-1}{p}=n .
\end{array}
$$
Then $m$ and $n$ are positive integers, and it is easy to see that $m>n$.
From equation (2), we get
$$
q=\frac{n p+1}{2} \text {. }
$$
Substituting equation (3) into equation (1), we get
$$
2 p-1=m q=m \cdot \frac{n p+1}{2} \text {. }
$$
Simplifying, we get $(4-m n) p=m+2$.
Therefore, $4-m n$ is a positive integer.
Thus, $m n=1$ or 2 or 3.
Since $m>n$, we have $m=2, n=1$ or $m=3, n=1$.
When $m=2, n=1$, from equations (1) and (2), we solve to get $p=2, q=\frac{3}{2}$ (discard);
When $m=3, n=1$, from equations (1) and (2), we solve to get $p=5, q=3$.
In summary, $p+q=8$.
|
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 $f(0)=c, f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$,
$$
\begin{array}{l}
f(1)=a+b+c, \\
f^{\prime}(0)=b \\
=4\left(\frac{a}{4}+\frac{b}{2}+c\right)-(a+b+c)-3 c .
\end{array}
$$
Then $\left.\left|f^{\prime}(0)\right| \leqslant 4\left|f\left(\frac{1}{2}\right)\right|+|f(1)|+3 \right| f(0)$ |
$$
\leqslant 4+1+3=8 \text {. }
$$
Therefore, $A \leqslant 8$.
$$
\text { Also, } f(x)=-8 x^{2}+8 x-1 \text {, when } 0 \leqslant x \leqslant 1
$$
then, $|f(x)| \leqslant 1, f^{\prime}(x)=-16 x+8, f^{\prime}(0)=8$.
Thus, $A \geqslant 8$.
Therefore, $A=8$.
|
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
$$
\alpha^{2005}+\beta^{2005}=\sum_{i, j} c_{i, j} \sigma_{1}^{i} \sigma_{2}^{j},
$$
where $c_{i, j}$ are constants.
$$
\begin{array}{l}
\text { Hence } \sum_{i, j} c_{i, j}=\left.\left(\alpha^{2005}+\beta^{2005}\right)\right|_{\sigma_{1}=\sigma_{2}=1} \\
=2 \operatorname{Re}\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{2005}=2 \operatorname{Re}\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right) \\
=1 .
\end{array}
$$
Therefore, the sum of the coefficients of the required polynomial is 1.
|
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 United States of America Mathematical Olympiad)
|
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 $a_{1}, a_{2}, \cdots, a_{n} (n>3)$ satisfying $a_{1}+a_{2}+\cdots +a_{n} \geqslant n$ and $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$, we have
$\max \left\{\ddot{a}_{1}, a_{2}, \cdots, a_{n}\right\} \geqslant 2$.
We use proof by contradiction.
Assume $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} < 2, i \vee j \geqslant 0, i+j=n$. Then, $\max \left\{x_{1}, x_{2}, \cdots, x_{i}\right\} < 2$,
\[
\begin{array}{l}
2 i=2+2+\cdots+2 > x_{1}+x_{2}+\cdots+x_{i} \\
\geqslant n+y_{1}+y_{2}+\cdots+y_{j} \\
=i+j+y_{1}+y_{2}+\cdots+y_{j}.
\end{array}
\]
Rearranging gives $i-j > y_{1}+y_{2}+\cdots+y_{j}$.
Since $x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2}+\left(-y_{1}\right)^{2}+\left(-y_{2}\right)^{2}+\cdots+\left(-y_{j}\right)^{2} \geqslant n^{2}$, we have
\[
\begin{array}{l}
4 i > x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2} \\
\geqslant n^{2}-\left(y_{1}^{2}+y_{2}^{2}+\cdots+y_{j}^{2}\right) \\
\geqslant n^{2}-\left(y_{1}+y_{2}+\cdots+y_{j}\right)^{2} \\
> n^{2}-(i-j)^{2} \\
= (i+j)^{2}-(i-j)^{2} = 4 i j.
\end{array}
\]
Since $i \geqslant 0$, we have $j a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$.
This leads to a contradiction with $n>3$.
Next, take $a_{1}=a_{2}=\cdots=a_{n}=2$, then $a_{1}+a_{2}+\cdots+a_{n} \geqslant n$ and $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2}$ both hold. At this point,
$\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=2$.
In summary, the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is 2.
|
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 can get $x=y$.
Therefore, $\left(x-\sqrt{x^{2}-2008}\right)^{2}=2008$.
Solving this, we get $x^{2}=2008$.
$$
\begin{array}{l}
\text { Hence } 3 x^{2}-2 y^{2}+3 x-3 y-2007 \\
=3 x^{2}-2 x^{2}+3 x-3 x-2007 \\
=x^{2}-2007=1 \text {. }
\end{array}
$$
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$
|
$$
\begin{array}{l}
\text { Given } a^{2}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\frac{3-\sqrt{5}}{2}=1-a \\
\Rightarrow a^{2}+a=1 . \\
\text { Therefore, } \frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a} \\
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a \cdot a^{2}-a} \\
=\frac{a^{3}-2 a^{3}-1+2}{a(1-a)-a}=\frac{1-a^{3}}{-a^{2}}=-\frac{1-a^{3}}{1-a} \\
=-\left(1+a+a^{2}\right)=-2 .
\end{array}
$$
|
-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, totaling $4 \times$ $68=272$ digits;
$100^{2}$ to $316^{2}$, each result occupies 5 digits, totaling $5 \times 217=1085$ digits;
At this point, we still need
$$
2008-(3+12+66+272+1085)=570
$$
digits.
$317^{2}$ to $411^{2}$, each result occupies 6 digits, totaling $6 \times 95=570$ digits.
Therefore, the digit at the 2008th position is precisely the unit digit of $411^{2}$, which is 1.
|
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 in the $k$-th column are both 1. The number of such triplets $\{i, j, k\}$ in the $k$-th column is $\mathrm{C}_{l_{k}}^{2}$, so the total number of such triplets $\{i, j, k\}$ is $\sum_{k=1}^{n} \mathrm{C}_{l_{k}}^{2}$.
Second, we calculate the total number of such triplets $\{i, j, k\}$ using another method: for any $1 \leqslant i<j \leqslant 6$, let $m_{i, j}$ be the number of columns where the $i$-th row and the $j$-th row are both 1, then
$$
\sum_{1 \leqslant i<j<6} m_{i, j}=\sum_{k=1}^{n} \mathrm{C}_{l_{k}}^{2} \text {. }
$$
Thus, by the condition $m_{i, j} \leqslant 2$, we have
$$
\begin{array}{l}
30=\frac{6(6-1)}{2} \times 2 \geqslant \sum_{1 \leqslant i<j \leqslant 6} m_{i, j} \\
=\sum_{k=1}^{n} \mathrm{C}_{l_{k}}^{2}=\frac{1}{2}\left(\sum_{k=1}^{n} l_{k}^{2}-\sum_{k=1}^{n} l_{k}\right) \\
\geqslant \frac{1}{2}\left[\frac{1}{n}\left(\sum_{k=1}^{n} l_{k}\right)^{2}-\sum_{k=1}^{n} l_{k}\right] \\
=\frac{1}{2 n} \times 30^{2}-\frac{1}{2} \times 30,
\end{array}
$$
which implies $n \geqslant 10$.
For $n=10$, Figure 2 is a matrix that meets the requirements:
$$
\begin{array}{llllllllll}
1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0
\end{array}
$$
Figure 2
Therefore, the minimum value of $n$ is 10.
(Fu Longrang provided)
|
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 $AF = 10 - x$, $DE = 12 - x$, $AD = 2 + x$. By the Pythagorean theorem, $(12 - x)^2 + (2 + x)^2 = 10^2$. Solving this, we get $x = 4$ or 6.)
|
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) is $64 \mathrm{~cm}$, then the side length of square (7) is $\qquad$ $\mathrm{cm}$.
|
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
$$
32 \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{3}=16 \sqrt{2}(\mathrm{~cm}) \text {, }
$$
the side length of square (5) is
$$
16 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{4}=16(\mathrm{~cm}) \text {, }
$$
the side length of square (6) is
$$
16 \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{5}=8 \sqrt{2}(\mathrm{~cm}) \text {, }
$$
the side length of square (7) is
$$
8 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{6}=8(\mathrm{~cm}) \text {. }
$$
Investigation: The side length of square (γ
) is
$$
64 \times\left(\frac{\sqrt{2}}{2}\right)^{n} \mathrm{~cm} .
$$
|
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 $-\frac{-9-\sqrt{156}}{3} \leqslant a \leqslant \frac{-9+\sqrt{156}}{3}$.
The integer $a$ can only take non-zero integer values from $1, -1$, $-2, -3, -4, -5, -6, -7$. From the equation, we have
$$
x=\frac{a+5 \pm \sqrt{52-3(a+3)^{2}}}{2 a} \text{. }
$$
When $a=1$, from equation (1) we get $x=2$ and 4;
When $a=-1$, the equation has no rational roots;
When $a=-2$, from equation (1) we get $x=1$ and $-\frac{5}{2}$;
When $a=-3$, the equation has no rational roots;
When $a=-4$, from equation (1) we get $x=-1$ and $\frac{3}{4}$;
When $a=-5$, the equation has no rational roots;
When $a=-6$, from equation (1) we get $x=\frac{1}{2}$ and $-\frac{1}{3}$;
When $a=-7$, from equation (1) we get $x=0$ and $\frac{2}{7}$.
Therefore, for different values of $a$, the equation has a total of 11 rational roots.
|
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 $f(x)$ is $\qquad$.
|
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:
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, 2 a_{n}=f\left(a_{n-1}\right)+15(n=2,3, \cdots), \\
b_{n}=\frac{1}{2+a_{n}}(n=1,2, \cdots),
\end{array}
$$
The sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is denoted as $S_{n}$, and the product of the first $n$ terms is denoted as $T_{n}$. Prove:
(1) $a=2$, and $b=-15$;
(2) For any positive integer $n, 2^{n+1} T_{n}+S_{n}$ is a constant.
|
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, $f(x)=(x-\alpha)(x-\beta)$
$$
=x^{2}-(\alpha+\beta) x+\alpha \beta=x^{2}+2 x-15 \text {. }
$$
Notice that
$$
\begin{array}{l}
\left|x^{2}+2 x-15\right| \leqslant 2\left|x^{2}+2 x-15\right| \\
=\left|2 x^{2}+4 x-30\right|
\end{array}
$$
holds for all $x \in \mathbf{R}$.
Thus, $a=2, b=-15$ is the unique set of values that satisfy the given conditions.
(2) From (1), we know $f(x)=x^{2}+2 x-15$.
Then $2 a_{n}=a_{n-1}^{2}+2 a_{n-1}$.
Thus, $2 a_{n+1}=a_{n}\left(a_{n}+2\right)$.
Therefore, $b_{n}=\frac{1}{2+a_{n}}=\frac{a_{n}}{2 a_{n+1}}$, and
$$
\begin{array}{l}
b_{n}=\frac{1}{2+a_{n}}=\frac{a_{n}^{2}}{2 a_{n} a_{n+1}} . \\
=\frac{2 a_{n+1}-2 a_{n}}{2 a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} .
\end{array}
$$
Hence, $T_{n}=b_{1} b_{2} \cdots b_{n}$
$$
\begin{array}{l}
=\frac{a_{1}}{2 a_{2}} \cdot \frac{a_{2}}{2 a_{3}} \cdots \cdot \frac{a_{n}}{2 a_{n+1}}=\frac{1}{2^{n+1} a_{n+1}}, \\
S_{n}=b_{1}+b_{2}+\cdots+b_{n} \\
=\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\cdots+\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right) \\
=\frac{1}{a_{1}}-\frac{1}{a_{n+1}}=2-\frac{1}{a_{n+1}} .
\end{array}
$$
Therefore, $2^{n+1} \vec{T}_{n}+S_{n}=\frac{1}{a_{n+1}}+\left(2-\frac{1}{a_{n+1}}\right)=$ 2 is a constant.
|
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
$I=\left\{P\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \mid c_{i}=0 \text{ or } 1, i=1, 2, 3, 4\right\}$.
If for any $n$-element subset $Q=\left\{P_{1}, P_{2}, \cdots, P_{n}\right\}$ of $I$, there exist $P_{i}, P_{j}, P_{k} \in Q$ such that $\triangle P_{i} P_{j} P_{k}$ is an equilateral triangle, i.e., $P_{i} P_{j}=P_{j} P_{k}=P_{k} P_{i}$, find the minimum value of $n$.
|
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, $n_{\min } \geqslant 9$.
When $n=9$, let
$$
S_{m}=\left\{P \mid P \in \dot{I} \text {, and } P P_{m}=1\right\}(m=1,2, \cdots, 9).
$$
Then $\left.\mid S_{m}\right\}=4$.
Therefore, $\sum_{m=1}^{9}\left|S_{m}\right|=4 \times 9=36$.
Since $|I|=2^{4}=16$, there exists $P \in I$ such that $P \in S_{i} \cap S_{j} \cap S_{k}$,
i.e., $P P_{i}=P P_{j}=P P_{k}=1$.
Thus, $P_{i} P_{j}=P_{j} P_{k}=P_{k} P_{i}=\sqrt{2}$.
Therefore, $\triangle P_{i} P_{j} P_{k}$ is an equilateral triangle.
In conclusion, the minimum value of $n$ is 9.
|
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
$$
\begin{aligned}
y^{2}= & (x-5)^{2}+(x-4)^{2}+\cdots+x^{2}+\cdots+ \\
& (x+4)^{2}+(x+5)^{2} \\
= & x^{2}+2\left(x^{2}+1^{2}\right)+2\left(x^{2}+2^{2}\right)+\cdots+2\left(x^{2}+5^{2}\right) \\
= & 11 x^{2}+2\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\right) \\
= & 11\left(x^{2}+10\right) .
\end{aligned}
$$
Since $11\left(x^{2}+10\right)$ is a perfect square, and 11 is a prime number, it follows that 11 divides $\left(x^{2}+10\right)$.
Thus, $x^{2}+10 \geqslant 11$, which means $x^{2} \geqslant 1$.
Therefore, $y^{2}=11\left(x^{2}+10\right) \geqslant 11^{2}$, which implies $|y| \geqslant 11$.
The equality holds when $x^{2}=1$.
Hence, the minimum value of $|y|$ is 11.
Note: This problem is adapted from the 1998 National Junior High School Mathematics Competition. The original problem asked for the minimum value of $y$, but using it as a junior high school mathematics competition problem is questionable. Although it has been shown that $|y| \geqslant 11$, i.e., $y \leqslant-11$ or $y \geqslant 11$, it is not sufficiently justified to conclude that the minimum value of $y$ is -11 (the original answer).
In fact, since $y^{2}=11\left(x^{2}+10\right)$ is a perfect square, it follows that $x^{2}+10=11 k^{2}\left(k \in \mathbf{N}_{+}\right)$, at which point, $y= \pm 11 k$. If the equation $x^{2}+10=11 k^{2}$ has only a finite number of integer solutions, let the solution that makes $k$ the largest be $\left(x_{0}, k_{0}\right)$, then the minimum value of $y$ is $-11 k_{0}$; if the equation $x^{2}+10=11 k^{2}$ has an infinite number of integer solutions, then the minimum value of $y$ does not exist. Clearly, discussing all integer solutions to the equation $x^{2}+10=11 k^{2}$ goes beyond the scope of junior high school mathematics competition knowledge.
Moreover, taking the equation $x^{2}+10=11 k^{2}$ modulo 11, we get $111\left(x^{2}+10\right)$, thus, $111\left(x^{2}-1\right)$.
Therefore, 11 divides $(x-1)$ or 11 divides $(x+1)$.
When 11 divides $(x-1)$, taking $x=23$ satisfies the condition, at which point, $k=7$, and the corresponding $y= \pm 77$.
When 11 divides $(x+1)$, taking $x=43$ satisfies the condition, at which point, $k=13$, and the corresponding $y= \pm 143$.
It is evident that $y_{\text {min }}=-11$ is clearly incorrect.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 3. Given real numbers } a, b, c. \text { If } \\ \frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 c a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1, \\ \text { then }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2008}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2008}+ \\ \left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2008}=\text {. }\end{array}$
|
3.3.
Notice
$$
\begin{array}{l}
\frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 a a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+1+\frac{(c-a)^{2}-b^{2}}{2 c a}+1+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}-1=1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+\frac{(c-a)^{2}-b^{2}}{2 c a}+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}=0 \\
\Rightarrow a\left[(b-c)^{2}-a^{2}\right]+b\left[(c-a)^{2}-b^{2}\right]+ \\
c\left[(a+b)^{2}-c^{2}\right]=0 \\
\Rightarrow a(b-c+a)(b-c-a)+ \\
b(c-a+b)(c-a-b)+ \\
c(a+b+c)(a+b-c)=0 \\
\Rightarrow(a+b-c)\left(a b-a c-a^{2}-b c+a b-\right. \\
\left.b^{2}+a c+b c+c^{2}\right)=0 \\
\Rightarrow(a+b-c)(a+c-b)(b+c-a)=0
\end{array}
$$
$$
\Rightarrow c=a+b \text { or } b=c+a \text { or } a=b+c \text {. }
$$
When $c=a+b$, we have
$$
\begin{array}{l}
\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(b-c)^{2}-a^{2}}{2 b c}+1 \\
=\frac{a^{2}-a^{2}}{2 b c}+1=1 .
\end{array}
$$
Similarly, $\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1, \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-1$.
Therefore, the original expression $=3$.
Similarly, when $b=c+a$ or $a=b+c$, the original expression $=3$.
|
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 $i=1,2, \cdots, m, \operatorname{Card}\left(A_{i}\right)=n$;
(2) For any $1 \leqslant i < j \leqslant m, \operatorname{Card}\left(A_{i} \cap A_{j}\right) \leqslant 10$,
Given $n>10$, find the maximum positive integer $m$, such that there exists a suitable family of sets containing $m$ sets.
|
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 empty set.
Assume the family of sets $\left\{A_{i} \subseteq A\right\}$ contains 5 subsets of $A$. The 5 subsets $A_{i}(i=1,2, \cdots, 5)$ contain a total of $5 n$ elements.
(1) If there is 1 element that appears 5 times, then the other elements appear at most 2 times. Thus, the 5 subsets contain at most $5 + 2(2 n-1) = 4 n + 3$ elements, which is a contradiction.
(2) If there are 2 elements each appearing 4 times, then these 2 elements must appear together in at least 3 subsets, which is a contradiction.
(3) If exactly 1 element appears 4 times, and let the number of elements appearing 3 times be $s$, then
$$
\begin{array}{l}
4 + 3 s + 2(2 n - s - 1) \geq 5 n \\
\Rightarrow s \geq n - 2 .
\end{array}
$$
(i) When $s > n - 2$, at least $n$ elements each appear at least 3 times;
(ii) When $s = n - 2$, there is 1 element appearing 4 times, and $n - 2$ elements each appearing 3 times.
(4) If no element appears 4 times, and let the number of elements appearing 3 times be $s$. Then
$$
3 s + 2(2 n - s) \geq 5 n \Rightarrow s \geq n .
$$
In summary, there are only two cases:
(1) At least $n$ elements each appear at least 3 times;
(2) One element appears 4 times, and $n - 2$ elements each appear 3 times.
For (1), since the intersection of any 3 subsets out of the 5 subsets contains at most 1 element, the number of elements appearing at least 3 times, $n \leq \mathrm{C}_{5}^{3} = 10$, which is a contradiction.
For (2), similarly, by the number of elements appearing at least 3 times, $n - 1 \leq \mathrm{C}_{5}^{3} = 10$, we know $n = 11$ and the intersection of any 3 subsets out of the 5 subsets is exactly a singleton set. These 10 singleton sets are all different, but for the element appearing 4 times, it appears in 4 singleton sets, which is a contradiction.
Therefore, there does not exist a suitable family of sets containing 5 subsets.
In conclusion, the maximum positive integer $m=4$.
|
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 perfect square can only be even 0, even 1, even 4, even 9, 25, odd 6, only $\overline{(a+2)(a+3)}=56$, 89 meet the conditions, i.e., $a=$ 3,6.
When $a=6$, $N=7689$ is not a perfect square, so it is discarded.
When $a=3$, $N=4356=66^{2}$, which meets the conditions.
Therefore, $a=3$ is the solution.
|
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, $b=d=-(a+c)$.
$$
\begin{array}{l}
\text { Also, }\left(a^{2}+a c\right)+\left(c^{2}+a c\right)=2+4=6, \\
\left(a^{2}+a c\right)-\left(c^{2}+a c\right)=2-4=-2,
\end{array}
$$
we get $a+c= \pm \sqrt{6},(a-c)(a+c)=-2$.
When $a+c=\sqrt{6}$, $a-c=-\frac{\sqrt{6}}{3}$.
Solving, we get $a=\frac{\sqrt{6}}{3}, c=\frac{2 \sqrt{6}}{3}$.
When $a+c=-\sqrt{6}$, $a-c=\frac{\sqrt{6}}{3}$.
Solving, we get $a=-\frac{\sqrt{6}}{3}, c=-\frac{2 \sqrt{6}}{3}$.
Therefore, $6 a+2 b+3 c+2 d=6 a+3 c+4 b$
$$
=6 a+3 c-4(a+c)=2 a-c=0 .
$$
|
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., $\frac{9-k}{2} \geqslant 0$, hence $k \leqslant 9$.
Therefore, $a+b= \pm \sqrt{\frac{9-k}{2}}, a b=\frac{3-k}{2}$.
Thus, the real numbers $a$ and $b$ can be considered as the two roots of the quadratic equation $x^{2} \mp \sqrt{\frac{9-k}{2}} x+\frac{3-k}{2}=0$. Then, $\Delta=\left(\sqrt{\frac{9-k}{2}}\right)^{2}-4 \times \frac{3-k}{2}=\frac{3 k-3}{2} \geqslant 0$.
Solving this, we get $k \geqslant 1$.
Therefore, $1 \leqslant k \leqslant 9$.
Thus, the maximum value of $a^{2}-a b+b^{2}$ is $m=9$, and the minimum value is $n=1$.
Hence, $m+n=10$.
|
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
|
$$
\begin{array}{l}
\sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\
= 3(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta), \\
\sin \alpha \cdot \cos \beta = 2 \cos \alpha \cdot \sin \beta. \\
\text{Therefore, } \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \cdot \sin \beta} = 2.
\end{array}
$$
|
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, $(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}=100 \overline{a b}$, which means
$$
\overline{a b}+2 \sqrt{c d}=100 \text {. }
$$
From equation (1), we know that $\overline{c d}$ is a perfect square. Therefore, $c d$ can be $01, 04, 09, 16, 25, 36, 49, 64, 81$, a total of 9 values.
From equation (1), the corresponding $\overline{a b}$ are
$$
98, 96, 94, 92, 90, 88, 86, 84, 82 \text {. }
$$
Thus, the four-digit numbers
$$
\begin{array}{l}
9801, 9604, 9409, 9216, 9025, \\
8836, 8649, 8464, 8281
\end{array}
$$
all satisfy the condition for $\overline{a b c d}$.
Rewriting equation (1) as
$$
\overline{a b}+\sqrt{\overline{c d}}=100-\sqrt{c d} .
$$
In fact,
$$
\begin{array}{l}
\sqrt{8281}=82+\sqrt{81}=100-\sqrt{81} ; \\
\sqrt{8464}=84+\sqrt{64}=100-\sqrt{64} ; \\
\sqrt{8649}=86+\sqrt{49}=100-\sqrt{49} ; \\
\sqrt{8836}=88+\sqrt{36}=100-\sqrt{36} ; \\
\sqrt{9025}=90+\sqrt{25}=100-\sqrt{25} ; \\
\sqrt{9216}=92+\sqrt{16}=100-\sqrt{16} ; \\
\sqrt{9409}=94+\sqrt{09}=100-\sqrt{09} ; \\
\sqrt{9604}=96+\sqrt{04}=100-\sqrt{04} ; \\
\sqrt{9801}=98+\sqrt{01}=100-\sqrt{01} .
\end{array}
$$
In summary, there are 9 four-digit numbers $\overline{a b c d}$ that meet the condition.
|
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 this point $x=-4$.
(2) When $-4 \leqslant x \leqslant 1$,
$$
y=-2(x-1)+(x+4)=-x+6 \text {. }
$$
Therefore, the minimum value of $y$ is 5, at this point $x=1$.
(3) When $1 \leqslant x \leqslant 2$,
$$
y=2(x-1)+(x+4)=3 x+2 \text {. }
$$
Therefore, the minimum value of $y$ is 5, at this point $x=1$.
In summary, the minimum value of $2|x-1|+|x+4|$ is 5, which is achieved when $x=1$.
|
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 the value of $|M O|-|M T|$.
|
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 definition of the hyperbola, we get
$$
\begin{array}{l}
|P F|-\left|P F^{\prime}\right|=6, \\
|F T|=\sqrt{|O F|^{2}-|O T|^{2}}=4 .
\end{array}
$$
Thus, $|M O|-|M T|$
$$
\begin{array}{l}
=\frac{1}{2}\left|P F^{\prime}\right|-|M F|+|F T| \\
=\frac{1}{2}\left(\left|P F^{\prime}\right|-|P F|\right)+|F T|=1 .
\end{array}
$$
|
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 $\max \{x, y\}$ steps, but not in fewer steps. Therefore, the distance between these two squares is $\max \{x, y\}$.
Let $d(A, B)$ denote the distance between squares $A$ and $B$.
Suppose $A$, $B$, and $C$ are three small squares, each pair of which is 100 units apart. Then, each pair of these squares will have either x-coordinates or y-coordinates that differ by 100. Thus, among the three pairs of squares $(A, B)$, $(B, C)$, and $(A, C)$, there must be two pairs such that the two squares in each pair have x-coordinates or y-coordinates that differ by 100. Without loss of generality, assume that $A$ and $B$, as well as $A$ and $C$, have x-coordinates that differ by 100, then the x-coordinates of $B$ and $C$ differ by 0 or 200. Since $d(B, C) = 100$, $B$ and $C$ must have the same x-coordinate. Therefore, their y-coordinates differ by 100. Without loss of generality, let the coordinates of $A$, $B$, and $C$ be $B(0,0)$, $C(0,100)$, and $A(100, x) (0 \leqslant x \leqslant 100)$.
Considering the small square $X$ that is 50 units away from $A$, $B$, and $C$, its x-coordinate must be 50 (otherwise, $d(X, A) > 50$ or $d(X, B) > 50$).
Similarly, its y-coordinate must also be 50 (otherwise, $d(X, C) > 50$ or $d(X, B) > 50$).
Therefore, $X(50,50)$.
Hence, there is a unique small square that is 50 units away from $A$, $B$, and $C$.
|
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 $\left|x^{2}-n\right|=2$.
Solving, we get $x=1, m=1$ or $x^{2}-n= \pm 2$.
When $x=1, m=1$, $n=4$.
$$
\left|x^{2}-n\right|=3 \text {. }
$$
So, $x=1$ satisfies the condition.
When $x^{2}-n=2$,
$$
x^{2}=n+2 \in\{3,4,5,6\} \text {. }
$$
Then $x=2$.
In this case, $n=2, m=3, x^{2}+m=7$.
So, $x=2$ satisfies the condition.
When $x^{2}-n=-2$,
$$
x^{2}=n-2 \in\{-1,0,1,2\} \text {. }
$$
Then $x=1$.
When $x=1$, $n=3, m=2, x^{2}+m=3$ is a prime number.
So, $x=1$ satisfies the condition.
Therefore, there are 2 possible values for $x$.
|
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
\[ \frac{\left|y_{0} - b + x_{0} b\right|}{\sqrt{\left(y_{0} - b\right)^{2} + x_{0}^{2}}} = 1. \]
Thus, \( \left(y_{0} - b\right)^{2} + x_{0}^{2} \)
\[ = \left(y_{0} - b\right)^{2} + 2 x_{0} b \left(y_{0} - b\right) + x_{0}^{2} b^{2}. \]
It is known that \( x_{0} > 2 \). Simplifying the above equation, we get
\[ \left(x_{0} - 2\right) b^{2} + 2 y_{0} b - x_{0} = 0. \]
Similarly, \( \left(x_{0} - 2\right) c^{2} + 2 y_{0} c - x_{0} = 0 \).
Therefore, \( b + c = \frac{-2 y_{0}}{x_{0} - 2} \), \( bc = \frac{-x_{0}}{x_{0} - 2} \).
Thus, \( (b - c)^{2} = \frac{4 x_{0}^{2} + 4 y_{0}^{2} - 8 x_{0}}{\left(x_{0} - 2\right)^{2}} \).
Since \( P\left(x_{0}, y_{0}\right) \) is a point on the parabola, we have \( y_{0}^{2} = 2 x_{0} \), so
\[ (b - c)^{2} = \frac{4 x_{0}^{2}}{\left(x_{0} - 2\right)^{2}}, \quad b - c = \frac{2 x_{0}}{x_{0} - 2}. \]
Therefore, \( S_{\triangle P B C} = \frac{1}{2} (b - c) x_{0} = \frac{x_{0}}{x_{0} - 2} \cdot x_{0} \)
\[ = \left(x_{0} - 2\right) + \frac{4}{x_{0} - 2} + 4 \geqslant 2 \sqrt{4} + 4 = 8. \]
When \( \left(x_{0} - 2\right)^{2} = 4 \), the above inequality holds with equality, at which point,
\[ x_{0} = 4, \quad y_{0} = \pm 2 \sqrt{2}. \]
Therefore, the minimum value of \( S_{\triangle P B C} \) is 8.
|
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 & $\cdots$ \\
\hline 1 & 3 & 5 & 7 & 9 & $\cdots$ \\
\hline 1 & 4 & 7 & 10 & 13 & $\cdots$ \\
\hline 1 & 5 & 9 & 13 & 17 & $\cdots$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline
\end{tabular}
|
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 positive integer solutions are
$$
\begin{aligned}
(m, n)= & (2,100),(4,34),(10,12), \\
& (12,10),(34,4),(100,2) .
\end{aligned}
$$
Therefore, the code 100 appears 6 times in Table 1.
|
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 to identify the coins with weights $a$ and $c$ using the fewest number of weighings. Here is a step-by-step approach:
1. **First Weighing:**
- Divide the nine coins into three groups of three coins each: Group 1, Group 2, and Group 3.
- Weigh Group 1 against Group 2.
**Case 1:** If Group 1 and Group 2 balance, then the coin weighing $a$ and the coin weighing $c$ must be in Group 3.
- Weigh two coins from Group 3 against each other.
- If they balance, the remaining coin in Group 3 is the one weighing $c$.
- If they do not balance, the lighter coin is the one weighing $a$ and the heavier coin is the one weighing $c$.
**Case 2:** If Group 1 and Group 2 do not balance, then the coin weighing $a$ and the coin weighing $c$ are in the lighter and heavier groups, respectively.
- Identify the lighter group (Group 1 or Group 2) and the heavier group (Group 1 or Group 2).
- Weigh two coins from the lighter group against each other.
- If they balance, the remaining coin in the lighter group is the one weighing $a$.
- If they do not balance, the lighter coin is the one weighing $a$.
- Weigh two coins from the heavier group against each other.
- If they balance, the remaining coin in the heavier group is the one weighing $c$.
- If they do not balance, the heavier coin is the one weighing $c$.
In both cases, we can identify the coins weighing $a$ and $c$ in a minimum of 2 weighings.
Thus, the minimum number of weighings required is \(\boxed{2}\).
|
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 $(E, F)$ against $(G, H)$. At this point, there are four possible scenarios:
(1) $(A, B) > (C, D)$ and $(E, F) > (G, H)$. Then the heavy coin is in $(A, B)$, and the light coin can only be in $(G, H)$. For the fourth weighing, place $(A, B, G, H)$ on one side and $(C, D, E, F)$ on the other side. The lightness or heaviness or balance of $(A, B, G, H)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$.
(2) $(A, B) > (C, D)$ and $(E, F) = (G, H)$. Then the heavy coin is in $(A, B)$, and the light coin can only be in $(C, D, I)$.
If the light coin is in $(C, D)$, then it is already known that $a+c > 2b$. For the fourth weighing, place $(A, B, I)$ on one side and $(E, F, G)$ on the other side. The lightness or heaviness of the $(A, B, I)$ side indicates whether $a+c$ is lighter or heavier than $2b$.
(3) $(A, B) = (C, D)$ and $(E, F) > (G, H)$. Then the light coin is in $(G, H)$, and the heavy coin can only be in $(E, F, I)$. If the heavy coin is in $(E, F)$, then it is already known that $a+c < 2b$. For the fourth weighing, place $(A, B, C)$ on one side and $(G, H, I)$ on the other side. The lightness or heaviness or balance of the $(G, H, I)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$.
(4) If both weighings are unbalanced, assume $(A, B) > (C, D)$ and $(E, F) > (G, H)$. For the third weighing, compare $(A, B)$ with $(G, H)$. If they balance, then the fake coins are in $(C, D, E, F)$. For the fourth weighing, compare $(A, B, G, H)$ with $(C, D, E, F)$. The lightness or heaviness or balance of the $(C, D, E, F)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$; if they do not balance, then the fake coins are in $(A, B, G, H)$. For the fourth weighing, compare $(A, B, G, H)$ with $(C, D, E, F)$. The lightness or heaviness or balance of the $(A, B, G, H)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$. (For the third weighing, compare $(A, B)$ with $(E, F)$ (or symmetrically $(C, D)$ with $(G, H))$. The heavier side contains the fake heavy coin, and the other side contains the real coins, thus identifying the location of the light coin.)
In summary, four weighings are sufficient to determine the result.
|
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 make the original equation have integer solutions is 1.
|
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 of 1 when divided by 3;
$B=\{11,17,23\}$, where each number leaves a remainder of 2 when divided by 3.
Suppose the sums $p_{a}, p_{b}, p_{c}$ of the three groups $A, B, C$ are distinct primes.
Since $p_{a}+p_{b}+p_{c}=1+2+\cdots+9=45$ is divisible by 3, the three sums $p_{a}, p_{b}, p_{c}$ must be of the same type. Note that the sum of the three numbers in class $A$ is $7+13+19=39 \neq 45$, which is a contradiction.
Therefore, two of the three sums must be equal.
(2) According to (1), expressing 45 as the sum of three numbers from 7, 11, 13, 17, 19, 23 (with two of the numbers being equal), there are only four cases:
(1) $19+19+7$, (2) $17+17+11$,
(3) $13+13+19$, (4) $11+11+23$.
Since there are 5 odd numbers in $1,2, \cdots, 9$, one of the three groups must consist of three odd numbers, and the other two groups must each have one odd number.
For case (1), the group with a sum of 7 can only be $\{1,2,4\}$. Dividing the remaining six numbers $3,5,6,7,8,9$ into two groups with a sum of 19, and one of these groups must consist of all odd numbers, there is only one unique division: $\{3,7,9\}$ and $\{5,6,8\}$.
For case (2), if the group of three odd numbers is $\{1,7,9\}$, then the other two groups can be $\{4,5,8\},\{2,3,6\}$ or $\{3,6,8\},\{2,4,5\}$;
If the group of three odd numbers is $\{3,5,9\}$, then the other two groups can be $\{2,8,7\},\{1,4,6\}$ or $\{4,6,7\},\{1,2,8\}$;
If the group of three odd numbers is $\{1,3,7\}$, then the other two groups can be $\{2,6,9\},\{4,5,8\}$.
This results in 5 different divisions.
For case (3), if the group of three odd numbers is $\{3,7,9\}$, then the other two groups can be $\{1,4,8\},\{2,5,6\}$;
If the group of three odd numbers is $\{1,3,9\}$, then the other two groups can be $\{2,4,7\},\{5,6,8\}$ or $\{2,5,6\},\{4,7,8\}$;
If the group of three odd numbers is $\{1,5,7\}$, then the other two groups can be $\{3,4,6\},\{2,8,9\}$ or $\{2,3,8\},\{4,6,9\}$.
This results in 5 different divisions.
For case (4), the group with a sum of 23 can only be $\{6,8,9\}$, then the other two groups can be $\{1,3,7\},\{2,4,5\}$.
In total, there are $1+5+5+1=12$ different divisions.
(Tao Pingsheng, Department of Mathematics and Computer Science, Jiangxi Science and Technology Normal University, 330013)
|
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
$\triangle A D F \cong \triangle B G F$
$\Rightarrow B G = A D = 3$
$\Rightarrow \angle A D F = \angle B G F$
$\Rightarrow A D \parallel G B$
$\Rightarrow \angle G B E + \angle A C B = 180^{\circ}$
$\Rightarrow \angle G B E = 90^{\circ}$
$\Rightarrow G E = \sqrt{G B^{2} + E B^{2}} = 5$.
Also, $D F = F G$, and $E F \perp D G$
$\Rightarrow \triangle E F D \cong \triangle E F G$
$\Rightarrow D E = G E = 5$.
|
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 number" (for example, 1 is a representable number, because
$$
+1+2-3-4+5+6-7-8+9
$$
is one way to represent 1).
(1) Prove that 7 is a representable number, and 8 is an unrepresentable number;
(2) Find the number of different ways 25 can be represented.
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ
|
(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 an odd number, and it cannot be 8. Therefore, 8 is a number that cannot be represented.
(2) Let the sum of the numbers with β+β signs be $x$, and the sum of the numbers with β-β signs be $y$. Then $x-y=25$.
$$
\begin{array}{l}
\text { Also, } x+y=1+2+\cdots+9=45 \text {, then } \\
x=35, y=10 .
\end{array}
$$
Since $9<10,1+2+3+4=10$, therefore, the numbers with β-β signs must be at least 2 and at most 4.
Thus, the sum of the numbers with β-β signs is 10.
Next, we just need to place a β-β sign in front of the numbers whose sum is 10, and a β+β sign in front of the other numbers to get one way to represent 25. Therefore, we only need to calculate the number of different ways to select several numbers from 1 to 9 whose sum is 10, to get the number of different ways to represent 25.
(i) 10 equals the sum of two numbers: $10=1+9=2+8=$ $3+7=4+6$, a total of 4 methods;
(ii) 10 equals the sum of three numbers: $10=1+2+7=1+$ $3+6=1+4+5=2+3+5$, a total of 4 methods;
(iii) 10 equals the sum of four numbers: $10=1+2+3+4$, only 1 method.
In summary, there are $4+4+1=9$ different ways to represent 25.
(Provided by the Sichuan Mathematical Competition Committee)
|
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
$-(a+1)<b<0$;
From equation (1), we get $b<-2 \sqrt{a}$.
Therefore, $-(a+1)<b<-2 \sqrt{a}$.
When $a=1,2,3,4$, there are no integers $b$ that satisfy equation (4);
When $a=5, b=-5$, the equation is $5 x^{2}-5 x+1=0$, and the two roots are between 0 and 1.
Thus, the minimum value of $a$ is 5.
|
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
$$
\begin{array}{l}
\{2 m-a, 2 m-b, 2 m-c, 2 m-d, 2 m-e\} \\
=\{-1,1,-7,7,41\}.
\end{array}
$$
Thus, from the sum of the elements
$$
10 m-(a+b+c+d+e)=41,
$$
we get $a+b+c+d+e=10 m-41$, whose last digit is 9.
|
9
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
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