problem
stringlengths 15
4.7k
| solution
stringlengths 2
11.9k
| answer
stringclasses 51
values | problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
14. (15 points) Let
$$
A=x^{4}+2 x^{3}-x^{2}-5 x+34 \text {. }
$$
Find the integer values of \( x \) for which \( A \) is a perfect square.
|
14. Notice that,
$$
A=\left(x^{2}+x-1\right)^{2}-3(x-11) \text {. }
$$
So, when $x=11$, $A=131^{2}$ is a perfect square.
Next, we prove: there are no other integer $x$ that satisfy the condition.
(1) When $x>11$, we have $A>0$, thus, $A>\left(x^{2}+x-2\right)^{2}$.
Therefore, $\left(x^{2}+x-2\right)^{2}<A<\left(x^{2}+x-1\right)^{2}$.
Let $A=y^{2}(y \in \mathbf{Z})$. Then
$$
\begin{array}{l}
|y|>\left|x^{2}+x-1\right| \\
\Rightarrow|y|-1 \geqslant\left|x^{2}+x-1\right| \\
\Rightarrow y^{2}-2|y|+1 \geqslant\left(x^{2}+x-1\right)^{2} \\
\Rightarrow-3(x-11)-2\left|x^{2}+x-1\right|+1 \geqslant 0 .
\end{array}
$$
Solving this inequality yields the integer values of $x$ as $\pm 3, \pm 2, \pm 1, 0, -4, -5$, but the corresponding $A$ for these values are not perfect squares.
In summary, the integer value of $x$ that makes $A$ a perfect square is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
381 A chess piece starts from one corner of an $8 \times 8$ chessboard, and moving one square in the row direction or column direction is called a step. After several steps, it passes through every square without repetition and returns to the starting point.
(1) Prove: the number of steps in the row direction is different from the number of steps in the column direction;
(2) Question: what is the minimum difference between the number of steps in the row direction and the number of steps in the column direction?
|
(1) Construct an $8 \times 8$ grid, and connect every adjacent point with a line segment to form a graph. Assume the distance between adjacent points is 1, and call a square with an area of 1 and vertices on the grid points a "cell."
The original problem is equivalent to: Prove that in every Hamiltonian cycle of the graph that meets the requirements, the number of steps in the row direction is not equal to the number of steps in the column direction.
We can determine every Hamiltonian cycle that meets the conditions through the following operations.
First, draw a cycle as shown in Figure 3. Then, starting from the first operation, each time, select a cell inside the previous cycle such that exactly two of its four vertices are on the edge of the previous cycle. Remove the edge of the cell on the cycle and keep the other three edges, thus obtaining the next cycle. Continue this process until a Hamiltonian cycle that meets the conditions is formed.
Conversely, every Hamiltonian cycle that meets the conditions can be determined through the above operations.
Starting from the first operation, each operation will increase the perimeter of the cycle by 2. Since the perimeter of the cycle in Figure 3 is 28, and the perimeter of a Hamiltonian cycle that meets the conditions is 64, a total of 18 operations are required.
It is also easy to see that starting from the first operation, each operation will increase the number of pairs of adjacent points in the $6 \times 6$ grid in the middle of the $8 \times 8$ grid by 1 (no two pairs of points share a point). During each operation:
(i) If the two points are connected by a horizontal line, the operation increases the number of steps in the column direction by 2, while the number of steps in the row direction remains unchanged;
(ii) If the two points are connected by a vertical line, the operation increases the number of steps in the row direction by 2, while the number of steps in the column direction remains unchanged.
Thus, it is sufficient to prove the following proposition:
For a $6 \times 6$ grid, pair the 36 points such that each pair consists of two points that are adjacent in the row direction or the column direction. Then the number of pairs in the row direction and the number of pairs in the column direction (no two pairs share a point) must be different.
In fact, assume that there are 9 pairs in both the row direction and the column direction. Color the $6 \times 6$ grid as shown in Figure 4. It is easy to see that each pair in the column direction includes one black point and one white point, and each pair in the row direction includes two points of the same color. Thus, the 9 pairs in the column direction include 9 black points and 9 white points.
Therefore, the 9 pairs in the row direction also include 9 black points and 9 white points. But this is impossible.
In conclusion, the original proposition is proved.
(2) From (1), we know that the difference is at least 4 (since the number of steps in the row direction and the column direction are both even), and Figure 5 provides an example.
|
4
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$.
Then the integer part of $\sum_{n=0}^{201} \frac{1}{a_{n}+1}$ is $\qquad$.
(2011, National High School Mathematics League Gansu Province Preliminary)
|
Given: $\frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$
$$
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\frac{1}{a_{0}}-\frac{1}{a_{2012}}=4-\frac{1}{a_{2012}} \text {. }
$$
Obviously, the sequence $\left\{a_{n}\right\}$ is monotonically increasing.
$$
\begin{array}{l}
\text { Also, } a_{1}=\frac{5}{16}, a_{2}>\frac{5}{16}+\frac{5}{16} \times \frac{1}{4}=\frac{25}{64}, \\
a_{3}>\frac{25}{64}+\frac{25}{64} \times \frac{1}{3}=\frac{25}{48}>\frac{1}{2}, \\
a_{4}>\frac{1}{2}+\frac{1}{4}=\frac{3}{4}, a_{5}=\frac{3}{4}+\frac{9}{16}>1,
\end{array}
$$
Thus, $a_{2012}>1$.
Therefore, $0<\frac{1}{a_{2012}}<1, \sum_{n=0}^{2011} \frac{1}{a_{n}+1} \in(3,4)$.
Hence, the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (18 points) Given
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) .
$$
(1) If there exist positive numbers $a, b, c$ such that inequality (1) holds, prove: $k>5$;
(2) If there exist positive numbers $a, b, c$ such that inequality (1) holds, and any set of positive numbers $a, b, c$ that satisfy inequality (1) are the sides of some triangle, find the integer $k$ that satisfies the condition.
|
11. (1) Since $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, therefore, from inequality (1) we know
$$
k(a b+b c+c a)>5(a b+b c+c a) .
$$
Noting that $a, b, c$ are positive numbers.
Thus, $a b+b c+c a>0$.
Hence $k>5$.
(2) From (1) we know $k>5$.
Since $k$ is an integer, then $k \geqslant 6$.
Let $a=1, b=1, c=2$. It is easy to see that $a, b, c$ are not the lengths of the three sides of a triangle.
By the problem, we know that the inequality (1) cannot hold, i.e.,
$$
k(1+2+2) \leqslant 5(1+1+4) \Rightarrow k \leqslant 6 \text {. }
$$
Thus, $k=6$.
Next, we prove: $k=6$.
At this point, inequality (1) is
$$
6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) \text {. }
$$
Noting that $a=1, b=1, c=\frac{3}{2}$ satisfies inequality (2).
Assume $a, b, c$ satisfy inequality (2).
By symmetry, without loss of generality, assume $a \leqslant b \leqslant c$.
Transform inequality (2) into
$$
\begin{array}{l}
5 c^{2}-6 c(a+b)+5 a^{2}+5 b^{2}-6 a b<0 . \\
\text { Hence } \Delta=[6(a+b)]^{2}-20\left(5 a^{2}+5 b^{2}-6 a b\right) \\
=-64(a-b)^{2}+64 a b \\
\leqslant 64 a b \leqslant 16(a+b)^{2} .
\end{array}
$$
Thus, from inequality (3) we get
$$
\begin{aligned}
c & <\frac{6(a+b)+\sqrt{\Delta}}{10} \leqslant \frac{6(a+b)+4(a+b)}{10} \\
& =a+b .
\end{aligned}
$$
Therefore, $a, b, c$ are the lengths of the three sides of a triangle.
In summary, the integer $k=6$.
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) In the sequence $\left\{x_{n}\right\}$,
$$
x_{n}=p^{n}+q^{n}, x_{1}=1, x_{3}=4 \text {. }
$$
(1) Prove: $x_{n+2}=x_{n+1}+x_{n}$;
(2) Determine the last digit of $x_{2011}$, but no proof is required.
|
11. (1) Notice,
$$
\begin{array}{l}
p+q=1, p^{3}+q^{3}=4, \\
p^{3}+q^{3}=(p+q)\left(p^{2}-p q+q^{2}\right) .
\end{array}
$$
Thus, $p^{2}-p q+q^{2}=4, p^{2}+2 p q+q^{2}=1$.
$$
\begin{array}{l}
\text { Then } x_{2}=p^{2}+q^{2}=3 \\
\Rightarrow p^{2}+(1-p)^{2}=3 \\
\Rightarrow p^{2}=p+1, q^{2}=q+1 \text {. }
\end{array}
$$
Therefore, $x_{n+2}=p^{n+2}+q^{n+2}=p^{n} p^{2}+q^{n} q^{2}$
$$
\begin{array}{l}
=p^{n}(p+1)+q^{n}(q+1) \\
=p^{n+1}+q^{n+1}+p^{n}+q^{n}=x_{n+1}+x_{n} .
\end{array}
$$
(2) The first few terms of the sequence $\left\{x_{n}\right\}$ are:
$$
\begin{array}{l}
1,3,4,7,11,18,29,47,76,123,199,322, \\
521,843,1364,2207,3571,5778,9349, \\
15127,24476,39603,64079,103682, \\
\quad \ldots
\end{array}
$$
Observing, starting from the first term, every 11 terms, the last digit repeats, i.e., if
$$
x_{n} \equiv r_{n}(\bmod 10)(r \in\{0,1, \cdots, 9\}),
$$
then the sequence $\left\{r_{n}\right\}$ has a period of 12.
Since $2013=167 \times 12+9$, we have
$$
r_{2013}=r_{9}=6 .
$$
|
6
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that for any real number $x$ we have $a \cos x + b \cos 2x \geqslant -1$.
Then the maximum value of $a + b$ is $\qquad$
|
2. 2 .
Let $x=\frac{2 \pi}{3}$, then $a+b \leqslant 2$.
When $a=\frac{4}{3}, b=\frac{2}{3}$,
$$
\begin{array}{l}
a \cos x+b \cos 2 x=\frac{4}{3} \cos x+\frac{2}{3} \cos 2 x \\
=\frac{4}{3} \cos ^{2} x+\frac{4}{3} \cos x-\frac{2}{3} \\
=\frac{4}{3}\left(\cos x+\frac{1}{2}\right)^{2}-1 \\
\geqslant-1 .
\end{array}
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the function
$$
f(x)=x^{3}-6 x^{2}+17 x-5 \text {, }
$$
real numbers $a, b$ satisfy $f(a)=3, f(b)=23$. Then $a+b=$ $\qquad$
|
4.4 .
Notice,
$$
\begin{array}{l}
f(x)=x^{3}-6 x^{2}+17 x-5 \\
=(x-2)^{3}+5(x-2)+13 \text {. } \\
\text { Let } g(y)=y^{3}+5 y .
\end{array}
$$
Then $g(y)$ is an odd function and monotonically increasing.
And $f(a)=(a-2)^{3}+5(a-2)+13=3$, so, $g(a-2)=-10$.
Also, $f(b)=(b-2)^{3}+5(b-2)+13=23$, then $g(b-2)=10$.
Thus $g(a-2)=-g(b-2)=g(2-b)$
$$
\Rightarrow a-2=2-b \Rightarrow a+b=4
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The positive integer solutions of the equation $x+y^{2}+(x, y)^{3}=x y(x, y)$ are $\qquad$ groups $((x, y)$ represents the greatest common divisor of integers $x, y)$.
|
7.4.
Let $(x, y)=d$. Then $d^{2} \mid x$.
Let $x=a d^{2}, y=b d$. Then $(a d, b)=1$.
Thus, the original equation becomes
$$
a+b^{2}+d=a b d^{2} \text {. }
$$
Therefore, $b \mid(a+d) \Rightarrow b \leqslant a+d$.
Then $a+b^{2}+d=a b d^{2}$
$$
\begin{array}{l}
=(a+d) b+(a+d) b+b\left(a d^{2}-2 a-2 d\right) \\
\geqslant a+d+b^{2}+b\left(a d^{2}-2 a-2 d\right) .
\end{array}
$$
Thus, $a d^{2}-2 a-2 d \leqslant 0$.
If $d \geqslant 3$, then
$$
\begin{aligned}
a d^{2}- & 2 a-2 d \geqslant 3 a d-2 a-2 d \\
& =2 a d+a d-2 a-2 d \\
& \geqslant 2 d+3 a-2 a-2 d>0,
\end{aligned}
$$
which is a contradiction.
Therefore, $d \leqslant 2$.
(1) $d=2$.
From $a d^{2}-2 a-2 d \leqslant 0$, we get $a \leqslant 2$.
When $a=2$, from equation (1) we get $b^{2}-8 b+4=0$, in which case, $b$ has no integer solutions;
When $a=1$, from equation (1) we get
$$
b^{2}-4 b+3=0 \Rightarrow b=1 \text { or } 3 \text {, }
$$
In this case, the solutions to the original equation are
$$
(x, y)=(4,2) \text { or }(4,6) \text {. }
$$
(2) $d=1$.
From equation (1) we get
$$
\begin{array}{l}
2=(a-b-1)(b-1) \\
\Rightarrow(a-b-1, b-1)=(2,1) \text { or }(1,2) \\
\Rightarrow(a, b)=(5,2) \text { or }(5,3),
\end{array}
$$
In this case, the solutions to the original equation are
$$
(x, y)=(5,2) \text { or }(5,3) \text {. }
$$
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 1, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then $m+n=$
|
3. 2 .
Solution 1 Notice that,
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .
$$
Since points $M, O, N$ are collinear, we have,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$
Solution 2 Since points $M, O, N$ are on the sides or their extensions of $\triangle A B C$, and the three points are collinear, by Menelaus' theorem we get
$$
\begin{array}{l}
\frac{A M}{M B} \cdot \frac{B O}{O C} \cdot \frac{C N}{N A}=1 . \\
\text { Hence } \frac{n-1}{1-m}=1 \Rightarrow m+n=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (15 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram.
|
12. Given that $\left|F_{1} F_{2}\right|=2$.
As shown in Figure 5, the inscribed quadrilateral $\square A B C D$ of the ellipse has a pair of opposite sides $B C$ and $A D$ passing through the foci $F_{1}$ and $F_{2}$, respectively.
Obviously, the center of symmetry of $\square A B C D$ is the origin.
Figure 5
Let the points be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$.
Then $S_{\square A B C D}=2 S_{\triangle A F, D}=\left|F_{1} F_{2}\right|\left|y_{1}-y_{4}\right|$ $=2\left|y_{1}-y_{4}\right|$.
If the slope of line $A D$ exists, let its equation be
$$
y=k(x-1) \text {. }
$$
Substituting into the ellipse equation and simplifying, we get
$$
\left(3+4 k^{2}\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 \text {. }
$$
By Vieta's formulas, we have
$$
x_{1}+x_{4}=\frac{8 k^{2}}{3+4 k^{2}}, x_{1} x_{4}=\frac{4 k^{2}-12}{3+4 k^{2}} \text {. }
$$
Then $\left|y_{1}-y_{4}\right|^{2}=k^{2}\left|x_{1}-x_{4}\right|^{2}$
$$
\begin{array}{l}
=k^{2}\left[\left(x_{1}+x_{4}\right)^{2}-4 x_{1} x_{4}\right] \\
=\frac{k^{2}}{\left(3+4 k^{2}\right)^{2}}\left[64 k^{4}-16\left(3+4 k^{2}\right)\left(k^{2}-3\right)\right] \\
=9 \times \frac{16 k^{2}\left(k^{2}+1\right)}{\left(3+4 k^{2}\right)^{2}} .
\end{array}
$$
Thus, $S_{\triangle \triangle B C D}=2\left|y_{1}-y_{4}\right|$
$$
=6 \sqrt{\frac{16 k^{2}\left(1+k^{2}\right)}{\left(3+4 k^{2}\right)^{2}}}<6 \text {. }
$$
If the slope of $A D$ does not exist, i.e., $A D \perp x$ axis.
It is easy to get the points $A\left(1, \frac{3}{2}\right), D\left(1,-\frac{3}{2}\right)$.
Thus, $S_{\text {ОАBCD }}=2\left|y_{1}-y_{4}\right|=6$.
In summary, the maximum area of $\square A B C D$ is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the arithmetic mean of $\sin \theta$ and $\cos \theta$ is $\sin \alpha$, and the geometric mean is $\sin \beta$. Then $\cos 2 \alpha-\frac{1}{2} \cos 2 \beta=$ $\qquad$ .
|
4. 0 .
From the given, we have
$$
\begin{array}{l}
\sin \alpha=\frac{\sin \theta+\cos \theta}{2}, \sin ^{2} \beta=\sin \theta \cdot \cos \theta. \\
\text { Therefore, } \cos 2 \alpha-\frac{1}{2} \cos 2 \beta \\
=1-2 \sin ^{2} \alpha-\frac{1}{2}\left(1-2 \sin ^{2} \beta\right) \\
=\frac{1}{2}-2 \sin ^{2} \alpha+\sin ^{2} \beta \\
=\frac{1}{2}-2\left(\frac{\sin \theta+\cos \theta}{2}\right)^{2}+\sin \theta \cdot \cos \theta \\
=\frac{1}{2}-\frac{1+2 \sin \theta \cdot \cos \theta}{2}+\sin \theta \cdot \cos \theta=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If the edge length of the cube $A_{1} A_{2} A_{3} A_{4}-B_{1} B_{2} B_{3} B_{4}$ is 1, then the number of elements in the set
$$
\left\{x \mid x=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}, i, j \in\{1,2,3,4\}\right\}
$$
is $\qquad$
|
5. 1 .
Solution 1 Note that,
$$
\begin{array}{l}
\overrightarrow{A_{1} B_{1}} \perp \overrightarrow{A_{i} A_{1}}, \overrightarrow{A_{1} B_{1}} \perp \overrightarrow{B_{1} B_{j}}(i, j \in\{2,3,4\}) \text {. } \\
\text { Hence } \overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=\overrightarrow{A_{1} B_{1}} \cdot\left(\overrightarrow{A_{i} A_{1}}+\overrightarrow{A_{1} B_{1}}+\overrightarrow{B_{1} B_{j}}\right) \\
=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} A_{1}}+{\overrightarrow{A_{1} B_{1}}}^{2}+\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{B_{1} B_{j}}={\overrightarrow{A_{1} B_{1}}}^{2}=1 \text {. } \\
\end{array}
$$
Solution 2 As shown in Figure 4, for any $i, j \in\{1,2,3,4\}$, the projection of $\overrightarrow{A_{i} B_{j}}$ onto $\overrightarrow{A_{1} B_{1}}$ is $\overrightarrow{A_{1} B_{1}}$, hence
$$
\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}^{2}=a_{n+1} a_{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{1}=\sqrt{2} \text {. }
$$
Then the natural number closest to $\sqrt{a_{2014}}$ is $\qquad$
|
6.8.
From the given, we have
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}} \Rightarrow a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}} \\
\Rightarrow a_{n+1}^{2}=a_{1}^{2}+2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} .
\end{array}
$$
Thus, $a_{2014}^{2}=2+2 \times 2013+\sum_{i=1}^{2013} \frac{1}{a_{i}^{2}}$
$$
>2+2 \times 2013=4028>3969=63^{2} \text {. }
$$
Also, $a_{1}=\sqrt{2}, a_{n+1}=a_{n}+\frac{1}{a_{n}}>2\left(n \in \mathbf{Z}_{+}\right)$, so
$$
\begin{array}{l}
a_{2014}^{2}=2+2 \times 2013+\frac{1}{2}+\sum_{i=2}^{2013} \frac{1}{a_{i}^{2}} \\
<4028+\frac{1}{2}+\frac{1}{4} \times 2012=4531.5 \\
<4624=68^{2} .
\end{array}
$$
Therefore, $9 \approx \sqrt{63}<\sqrt{a_{2014}}<\sqrt{68} \approx 8.3$.
Thus, the natural number closest to $\sqrt{a_{2014}}$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=6$. Then the maximum value of $x+\sqrt{x y}+\sqrt[3]{x y z}$ is $\qquad$ .
|
7.8.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{1}{4} x+y+4 z \geqslant 3 \sqrt[3]{x y z}, \\
\frac{3}{4} x+3 y \geqslant 3 \sqrt{x y} .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
x+4 y+4 z \geqslant 3 \sqrt{x y}+3 \sqrt[3]{x y z} \\
\Rightarrow x+\sqrt{x y}+\sqrt[3]{x y z} \leqslant \frac{4}{3}(x+y+z)=8 .
\end{array}
$$
When $x: y: z=16: 4: 1$, the equality holds. Therefore, the maximum value is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$, $b$, and $c$ are constants, and for any real number $x$ we have
$$
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \text {. }
$$
then $a b c=$ . $\qquad$
|
$$
\text { Two, 1. }-9 \text {. }
$$
From the problem, we know
$$
\begin{array}{c}
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \\
=x^{3}+(a+1) x^{2}+(a+b) x+b .
\end{array}
$$
Thus, $a+1=0, a+b=2, b=c$
$$
\begin{array}{l}
\Rightarrow a=-1, b=c=3 \\
\Rightarrow a b c=-9
\end{array}
$$
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $S_{\triangle M B C}=4, 3 A B=2 B C$, draw the perpendicular from point $C$ to the angle bisector $B E$ of $\angle A B C$, and let the foot of the perpendicular be $D$. Then $S_{\triangle B D C}=$ $\qquad$
|
4.3.
Construct the reflection of point $C$ about $BD$ as point $F$, thus, $BC=BF$. Connect $FA$ and $FD$, then point $F$ lies on the extension of $BA$, and points $C$, $D$, and $F$ are collinear.
$$
\text{Therefore, } \frac{S_{\triangle ABC}}{S_{\triangle FBC}}=\frac{AB}{FB}=\frac{AB}{BC}=\frac{2}{3} \text{.}
$$
Since $S_{\triangle FBC}=2 S_{\triangle DBC}$, we have $\frac{S_{\triangle ABC}}{S_{\triangle DBC}}=\frac{4}{3}$. Thus, $S_{\triangle BDC}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the function $y=f(x)$ is defined on $\mathbf{R}$, with a period of 3, and Figure 1 shows the graph of the function in the interval $[-2,1]$. Then $\frac{f(2014)}{f(5) f(15)}=$ $\qquad$ .
|
$$
\text { II,6. }-2 \text {. }
$$
From the problem statement and combining with the graph, we know that
$$
\frac{f(2014)}{f(5) f(15)}=\frac{f(1)}{f(-1) f(0)}=\frac{2}{(-1) \times 1}=-2 \text {. }
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In $\triangle A B C$, it is known that
$$
\begin{array}{l}
|\overrightarrow{A B}|=\sqrt{3},|\overrightarrow{B C}|=1, \\
|\overrightarrow{A C}| \cos B=|\overrightarrow{B C}| \cos A \text {. } \\
\text { Then } \overrightarrow{A C} \cdot \overrightarrow{A B}=
\end{array}
$$
|
7.2.
Let the three sides of $\triangle A B C$ be $a, b, c$. From the given condition and using the Law of Sines, we have
$$
\begin{array}{l}
\sin B \cdot \cos B=\sin A \cdot \cos A \\
\Rightarrow \sin 2 B=\sin 2 A \\
\Rightarrow \angle B=\angle A \text { or } \angle B+\angle A=90^{\circ} .
\end{array}
$$
We will discuss two cases.
(1) If $\angle B=\angle A$, then
$$
b=a=1, c=\sqrt{3}, \angle A=30^{\circ} \text {. }
$$
Thus, $\overrightarrow{A C} \cdot \overrightarrow{A B}=b c \cos A=1 \times \sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}$.
(2) If $\angle B+\angle A=90^{\circ}$, then
$$
b=\sqrt{2}, c=\sqrt{3}, \cos A=\frac{\sqrt{2}}{\sqrt{3}} \text {. }
$$
Thus, $\overrightarrow{A C} \cdot \overrightarrow{A B}=b c \cos A=\sqrt{2} \times \sqrt{3} \times \frac{\sqrt{2}}{\sqrt{3}}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given integers $a, b, c, d$. If the roots of the equation
$$
z^{4}+a z^{3}+b z^{2}+c z+d=0
$$
correspond to four points $A, B, C, D$ forming the vertices of a square in the complex plane, then the minimum value of the area of square $A B C D$ is $\qquad$
|
7. 2 .
Let the complex number corresponding to the center of the square be $m$. Then, after translating the origin of the complex plane to $m$, the vertices of the square are distributed on a circle, i.e., they are the solutions to the equation $(z-m)^{4}=n$ (where $n$ is a complex number).
From $z^{4}+a z^{3}+b z^{2}+c z+d=(z-m)^{4}-n$, comparing coefficients we know that $m=-\frac{a}{4}$ is a rational number.
Furthermore, from $-4 m^{3}=c$, we know that $m$ is an integer.
Then, from $d=m^{4}-n$, we know that $n$ is an integer.
Thus, the roots of the equation $(z-m)^{4}=n$ are
$$
z_{k}=m+\sqrt[4]{|n|}\left(\cos \frac{k \pi}{2}+\mathrm{i} \sin \frac{k \pi}{2}\right) .
$$
Therefore, the length of the diagonal of the square is $2 \sqrt[4]{|n|}$, and its area is
$$
2 \sqrt{|n|} \geqslant 2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{Z}_{+}\right)$. Find the smallest positive real number $\lambda$, such that for any $n \geqslant 2$, we have
$$
a_{n}^{2}<\lambda \sum_{k=1}^{n} \frac{a_{k}}{k} .
$$
|
Sure, here is the translated text:
```
9. Notice that,
\[
\begin{array}{l}
a_{k}^{2}-a_{k-1}^{2}=\left(a_{k}-a_{k-1}\right)\left(a_{k}+a_{k-1}\right) \\
=\frac{1}{k}\left(2 a_{k}-\frac{1}{k}\right)(k \geqslant 2) .
\end{array}
\]
Then \(a_{n}^{2}-a_{1}^{2}=\sum_{k=2}^{n}\left(a_{k}^{2}-a_{k-1}^{2}\right)=2 \sum_{k=2}^{n} \frac{a_{k}}{k}-\sum_{k=2}^{n} \frac{1}{k^{2}}\)
\[
\begin{array}{l}
\Rightarrow a_{n}^{2}=2 \sum_{k=1}^{n} \frac{a_{k}}{k}-\sum_{k=1}^{n} \frac{1}{k^{2}} \\
\Rightarrow a_{n}^{2}\sum_{k=1}^{n} \frac{1}{k}.
\]
Taking \(n=2^{m}\), thus, when \(m\) is sufficiently large, we have
\[
\begin{array}{l}
\sum_{k=1}^{n} \frac{1}{k}=1+\sum_{i=0}^{m-1} \sum_{i=1}^{2^{i}} \frac{1}{2^{t}+s} \\
>1+\sum_{i=0}^{m-1} 2^{t} \times \frac{1}{2^{t+1}} \\
=1+\frac{m}{2}>\frac{2}{2-\lambda} .
\end{array}
\]
Therefore, equation (1) does not hold.
Thus, \(\lambda \geqslant 2\).
In conclusion, \(\lambda_{\min }=2\).
```
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given that $AB$ is the major axis of the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $CD$ is a chord of the ellipse $\Gamma$. The tangents at points $C$ and $D$ intersect at point $P$, the extension of $AD$ intersects the extension of $CB$ at point $E$, and the extension of $AC$ intersects the extension of $DB$ at point $F$. If $E$, $P$, and $F$ are collinear, find $\frac{EP}{PF}$.
|
10. As shown in Figure 3, let the center of the ellipse be $O$. Connect $O P$, intersecting $C D$ at point $M$.
Let $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right), P\left(x_{0}, y_{0}\right)$.
Then the equation of line $C D$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
From $\left\{\begin{array}{l}\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1, \\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\end{array}\right.$ we get
$$
\left(\frac{x_{0}^{2}}{a^{4}}+\frac{y_{0}^{2}}{a^{2} b^{2}}\right) x^{2}-\frac{2 x_{0}}{a^{2}} x+1-\frac{y_{0}^{2}}{b^{2}}=0 \text {. }
$$
By Vieta's formulas, the coordinates of the midpoint $N$ of $C D$ are
$$
x_{N}=\frac{x_{1}+x_{2}}{2}=\frac{x_{0}}{\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}}, y_{N}=\frac{y_{0}}{\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}} .
$$
Thus, $k_{O N}=\frac{y_{0}}{x_{0}}=k_{O P}$.
Therefore, points $O, N, P$ are collinear.
Hence, point $N$ coincides with $M$.
By Newton's theorem, $\frac{E P}{P F}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For $I$ being the excenter of $\triangle ABC$ with respect to $\angle A$, prove:
$$
\frac{A I^{2}}{C A \cdot A B}-\frac{B I^{2}}{A B \cdot B C}-\frac{C I^{2}}{B C \cdot C A}=1 .
$$
|
Proof As shown in Figure 5, let $\odot I$ be tangent to $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively.
$$
\begin{array}{l}
\text { Let } \angle A I F=\alpha, \\
\angle B I D=\beta, \\
\angle C I E=\gamma .
\end{array}
$$
Then $\alpha=\beta+\gamma$.
Assume $I D=I E=I F=1$. Then $A I^{2}=\frac{1}{\cos ^{2} \alpha}$,
$$
\begin{array}{l}
A B=A F-B F=\tan \alpha-\tan \beta, \\
A C=A E-C E=\tan \alpha-\tan \gamma . \\
\text { Therefore, } \frac{A I^{2}}{C A \cdot A B}=\frac{1}{\cos ^{2} \alpha(\tan \alpha-\tan \beta)(\tan \alpha-\tan \gamma)} \\
=\cos ^{2} \alpha \cdot \frac{\sin (\alpha-\beta)}{\cos \alpha \cdot \cos \beta} \cdot \frac{\sin (\alpha-\gamma)}{\cos \alpha \cdot \cos \gamma} \\
=\frac{1}{\tan \beta \cdot \tan \gamma} .
\end{array}
$$
Similarly, $\frac{B I^{2}}{A B \cdot B C}=\frac{1}{\tan \gamma \cdot \tan \alpha}$,
$$
\begin{array}{l}
\frac{C I^{2}}{B C \cdot C A}=\frac{1}{\tan \alpha \cdot \tan \beta} . \\
\text { Therefore, } \frac{A I^{2}}{C A \cdot A B}-\frac{B I^{2}}{A B \cdot B C}-\frac{C I^{2}}{B C \cdot C A} \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{1}{\tan \gamma \cdot \tan \alpha}-\frac{1}{\tan \alpha \cdot \tan \beta} \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{1}{\tan \alpha} \cdot \frac{\tan \beta+\tan \gamma}{\tan \beta \cdot \tan \gamma} \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{\tan (\beta+\gamma)}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma}(1-\tan \beta \cdot \tan \gamma) \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{1}{\tan \beta \cdot \tan \gamma}(1-\tan \beta \cdot \tan \gamma) \\
=1 .
\end{array}
$$
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. Try to determine the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$
|
4. 3 .
Notice that,
$$
\begin{array}{l}
\frac{\sqrt{14}+2}{\sqrt{14}-2}=\frac{(\sqrt{14}+2)^{2}}{(\sqrt{14}-2)(\sqrt{14}+2)} \\
=\frac{14+4+4 \sqrt{14}}{14-4}=\frac{18+4 \sqrt{14}}{10} .
\end{array}
$$
And $3<\sqrt{14}<4$, thus, $30<18+4 \sqrt{14}<34$.
Therefore, $3<\frac{18+4 \sqrt{14}}{10}<3.4$.
Hence, the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$ is $=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given that vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ are two mutually perpendicular unit vectors in a plane, and
$$
(3 \alpha-\gamma) \cdot(4 \beta-\gamma)=0 .
$$
Then the maximum value of $|\boldsymbol{\gamma}|$ is . $\qquad$
|
11.5.
As shown in Figure 2, let
$$
\begin{array}{l}
\overrightarrow{O A}=3 \alpha, \\
\overrightarrow{O B}=4 \beta, \\
\overrightarrow{O C}=\gamma .
\end{array}
$$
From the given information, $\overrightarrow{A C} \perp \overrightarrow{B C}$. Therefore, point $C$ lies on the circle with $A B$ as its diameter, and this circle passes through the origin.
Thus, the maximum value of $|\overrightarrow{O C}|$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}\left(n \in \mathbf{Z}_{+}\right)$, and define
$$
T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right| \text {. }
$$
If $T_{k} \in \mathbf{Z}_{+}$, find all possible values of $k$.
|
15. From the problem, we know
Also, $a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3$,
$$
a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n-1}+3(n \geqslant 2) \text {, }
$$
Thus, $a_{n+1}-a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) a_{n}$
$\Rightarrow a_{n+1}=3^{\frac{2}{2 k-1}} a_{n}$
$$
\Rightarrow a_{n}=a_{2}\left(3^{\frac{2}{2 k-1}}\right)^{n-2}=3^{\frac{2 n+2 k-3}{2 k-1}}(n \geqslant 2) \text {. }
$$
Clearly, $n=1$ also fits.
Therefore, the general term formula for the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=3^{\frac{2 n+2 k-3}{2 k-1}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Thus, $b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}$
$$
\begin{array}{l}
=\frac{1}{n} \cdot \frac{1}{2 k-1} \sum_{i=1}^{n}(2 i+2 k-3) \\
=1+\frac{n-1}{2 k-1} .
\end{array}
$$
Therefore, $b_{n}-\frac{3}{2}=\frac{n-\left(k+\frac{1}{2}\right)}{2 k-1}$.
Thus, when $n \leqslant k$, $b_{n}-\frac{3}{2}<0$; when $n \geqslant k+1$, $b_{n}-\frac{3}{2}>0$.
$$
\begin{array}{l}
\text { Then } T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right| \\
=\sum_{i=1}^{k}\left(\frac{3}{2}-b_{i}\right)+\sum_{i=k+1}^{2 k}\left(b_{i}-\frac{3}{2}\right) \\
=\frac{k^{2}}{2 k-1} .
\end{array}
$$
Since $T_{k} \in \mathbf{Z}_{+}$, i.e., $(2 k-1) \mid k^{2}$, we have
$$
(2 k-1) \mid\left(4 k^{2}-1+1\right) \text {. }
$$
Thus, $(2 k-1) \mid 1$.
Hence, $2 k-1=1 \Rightarrow k=1$.
Therefore, the only possible value of $k$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The value of the complex number $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{6 n}\left(n \in \mathbf{Z}_{+}\right)$ is
|
3. 1 .
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{6 n}=\left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{3}\right]^{2 n} \\
=\left(\frac{1}{8}+\frac{3}{4} \times \frac{\sqrt{3}}{2} i-\frac{3}{2} \times \frac{3}{4}-\frac{3 \sqrt{3}}{8} i\right) \\
=(-1)^{2 n}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $(\sqrt{2}+1)^{21}=a+b \sqrt{2}$, where $a, b$ are positive integers. Then $(b, 27)=$ $\qquad$
|
8. 1 .
Notice,
$$
\begin{array}{l}
(\sqrt{2}+1)^{21} \\
=(\sqrt{2})^{21}+\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}+\cdots+ \\
\mathrm{C}_{21}^{20} \sqrt{2}+\mathrm{C}_{21}^{21}, \\
(\sqrt{2}-1)^{21} \\
=(\sqrt{2})^{21}-\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}-\cdots+ \\
\quad \mathrm{C}_{2 \mathrm{~N}}^{20}-\mathrm{C}_{21}^{21} .
\end{array}
$$
Therefore, $(\sqrt{2}-1)^{21}=-a+b \sqrt{2}$.
Thus, $2 b^{2}-a^{2}=1$.
If $b \equiv 0(\bmod 3)$, then $a^{2} \equiv-1(\bmod 3)$.
This contradicts $a^{2} \equiv 0,1(\bmod 3)$.
Therefore, $(27, b)=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a+b=\sqrt{5}$, then
$$
\frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b=(\quad) \text {. }
$$
(A) 5
(B) $\frac{3 \sqrt{5}}{2}$
(C) $2 \sqrt{5}$
(D) $\frac{5 \sqrt{5}}{2}$
|
$\begin{array}{l}\text {-1. A. } \\ \frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b \\ =\frac{\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{a^{2}+a b+b^{2}}+3 a b \\ =\left(a^{2}-a b+b^{2}\right)+3 a b \\ =(a+b)^{2}=5 .\end{array}$
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $M$ is the least common multiple of 15 consecutive natural numbers $1,2, \cdots, 15$. If a divisor of $M$ is divisible by exactly 14 of these 15 natural numbers, it is called a "good number" of $M$. Then the number of good numbers of $M$ is $\qquad$.
|
4.4.
It is known that $M=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$.
Since $2 \times 11, 2 \times 13$ are both greater than 15, therefore,
$$
\begin{array}{l}
\frac{M}{11}=2^{3} \times 3^{2} \times 5 \times 7 \times 13, \\
\frac{M}{13}=2^{3} \times 3^{2} \times 5 \times 7 \times 11,
\end{array}
$$
each is a divisor of $M$, and is exactly divisible by the other 14 numbers.
If the power of 2 in $M$ is reduced by 1, then
$$
\frac{M}{2}=2^{2} \times 3^{2} \times 5 \times 7 \times 11 \times 13
$$
is a divisor of $M$, and is exactly divisible by the 14 numbers except 8.
Similarly, if the power of 3 is reduced by 1, then
$$
\frac{M}{3}=2^{3} \times 3 \times 5 \times 7 \times 11 \times 13
$$
is a divisor of $M$, and is exactly divisible by the 14 numbers except 9;
if the power of 5 in $M$ is reduced by 1, then $\frac{M}{5}$ is a divisor of $M$ and is exactly divisible by the 12 numbers except 5, 10, and 15; if 7 is removed, then $\frac{M}{7}$ is a divisor of $M$ and is exactly divisible by the 13 numbers except 7 and 14.
In summary, there are four divisors of $M$ that are exactly divisible by 14 of the 15 natural numbers, namely $\frac{M}{2}, \frac{M}{3}, \frac{M}{11}, \frac{M}{13}$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Find all positive integers $n$, such that there exists an integer-coefficient polynomial $P(x)$, satisfying $P(d)=\left(\frac{n}{d}\right)^{2}$, where, $d$ is each divisor of $n$. [3]
|
For a polynomial $P(x)$ with integer coefficients, and positive integers $a, b (a \neq b)$, we always have
$$
(a-b) \mid (P(a)-P(b)).
$$
This is an invariant.
If $n=1$, then $P(1)=1$. Hence, we can take the polynomial $P(x)=x$.
If $n$ is a prime number, then $n$ has only two factors, 1 and $n$. The polynomial $P(x)$ must satisfy
$$
P(1)=n^{2}, P(n)=1.
$$
Let $P(x)=a x+b$.
From $P(1)=n^{2}, P(n)=1$, we get
$$
P(x)=-(n+1) x+n^{2}+n+1.
$$
If $n=k l(1<k \leqslant l)$ is a composite number, then
$$
\begin{array}{l}
P(1)=n^{2}, P(l)=k^{2}, P(k)=l^{2}, P(n)=1. \\
\text { By }(n-k) \mid (P(n)-P(k)) \\
\Rightarrow k(l-1) \mid (1-l)(1+l) \\
\Rightarrow k \mid (1+l).
\end{array}
$$
Similarly, by
$$
(n-l) \mid (P(n)-P(l)) \Rightarrow l \mid (1+k).
$$
Thus, $k l \mid (1+k)(1+l) \Rightarrow k l \mid (1+k+l)$
$$
\Rightarrow k l \leqslant 1+k+l \Rightarrow (k-1)(l-1) \leqslant 2.
$$
Solving this, we get $k=2, l=3$, i.e., $n=6$.
Then $P(1)=36, P(2)=9, P(3)=4, P(6)=1$.
Let $Q(x)=P(x)-1$.
Then $Q(1)=35, Q(2)=8, Q(3)=3, Q(6)=0$.
Let $Q(x)=(x-6) R(x)$. Then
$$
R(1)=-7, R(2)=-2, R(3)=-1.
$$
Let $R(x)=-1+(x-3) S(x)$. Then
$$
S(1)=3, S(2)=1.
$$
Thus, we find $S(x)=5-2 x$.
Therefore, $P(x)=1+(x-6)[-1+(x-3)(5-2 x)]$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If positive real numbers $a, b$ satisfy
$$
\log _{8} a+\log _{4} b^{2}=5, \log _{8} b+\log _{4} a^{2}=7 \text {, }
$$
then $\log _{4} a+\log _{8} b=$ $\qquad$
|
$=1.4$.
Let $a=2^{x}, b=2^{y}$. Then $\frac{x}{3}+y=5, \frac{y}{3}+x=7$.
Thus, $x=6, y=3$.
Therefore, $\log _{4} a+\log _{8} b=\frac{x}{2}+\frac{y}{3}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the unit digit of $(2+\sqrt{3})^{2013}$.
|
Notice,
$$
\begin{array}{l}
{\left[(2+\sqrt{3})^{2013}\right]=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}-1 .} \\
\text { Let } a_{n}=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}=4 a_{n-1}-a_{n-2}, \\
a_{0}=2, a_{1}=4 .
\end{array}
$$
Then $2 \mid a_{n}$,
$$
\begin{array}{l}
a_{n} \equiv 2,4,4,2,4,4,2, \cdots(\bmod 5) \\
\Rightarrow a_{2013} \equiv 2(\bmod 5) .
\end{array}
$$
Therefore, the last digit is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $a$ and $b$ are two different positive integers. Ask:
$$
\begin{array}{l}
a(a+2), a b, a(b+2), (a+2) b, \\
(a+2)(b+2), b(b+2)
\end{array}
$$
Among these six numbers, what is the maximum number of perfect squares?
|
2. At most two numbers are perfect squares (such as $a=2, b=16$). Notice that,
$$
\begin{array}{l}
a(a+2)=(a+1)^{2}-1, \\
b(b+2)=(b+1)^{2}-1
\end{array}
$$
cannot be perfect squares;
$$
a b \cdot a(b+2)=a^{2}\left(b^{2}+2 b\right)
$$
is not a perfect square, so at most one of $a b$ and $a(b+2)$ is a perfect square.
Similarly, at most one of $(a+2) b$ and $(a+2)(b+2)$ is a perfect square.
Thus, at most two numbers are perfect squares.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 There are three piles of stones. Each time, A moves one stone from one pile to another, and A can receive a reward from B for each move, which is equal to the difference between the number of stones in the pile to which A moves the stone and the number of stones in the pile from which A moves the stone. If this difference is negative, A should return this amount of reward to B (if unable to pay, A can owe it temporarily). It is known that at a certain moment, all the stones are in their initial piles. Try to find the maximum reward A can earn at this moment.
---
The translation maintains the original text's line breaks and format.
|
【Analysis】Due to the uncertainty of A's operations, it is necessary to start from the whole and establish a "substitute" for the increase or decrease of A's reward each time, which must be simple to calculate.
Solution A's reward is 0.
In fact, the three piles of stones can be imagined as three complete graphs (each stone in the same pile is connected by a line segment, and stones in different piles are not connected).
Each time A operates, a vertex $u$ is moved out from a complete graph $A_{1}$ and into another complete graph $A_{2}$, forming two new complete graphs $A_{1}^{\prime}$ and $A_{2}^{\prime}$. The reward is the difference in the number of vertices between $A_{2}^{\prime}$ and $A_{1}$, which is the difference in the sum of the number of edges between $A_{1}^{\prime}$ and $A_{2}^{\prime}$ and the sum of the number of edges between $A_{1}$ and $A_{2}$.
Thus, A's reward for each operation is the change in the number of edges in the graph.
Since it eventually returns to the initial state, A's reward is 0.
【Comment】By establishing the above substitute, the conclusion becomes clear at a glance.
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $n(n \geqslant 4)$ be a positive integer. $n$ players each play a table tennis match against every other player (each match has a winner and a loser). Find the minimum value of $n$ such that after all the matches, there always exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, satisfying that when $1 \leqslant i<j \leqslant 4$, player $A_{i}$ defeats player $A_{j}$.
(Supplied by He Yixie)
|
2. First, prove: when $n=8$, there always exists an ordered quartet that satisfies the problem's conditions.
Since 8 players have played a total of $\mathrm{C}_{8}^{2}=28$ matches, there must be at least one player who has won at least $\left\lceil\frac{28}{8}\right\rceil=4$ matches (where $\left\lceil x \right\rceil$ denotes the smallest integer not less than the real number $x$).
Assume player $A_{1}$ has defeated $a_{1}, a_{2}, a_{3}, a_{4}$.
Among players $a_{1}, a_{2}, a_{3}, a_{4}$, a total of six matches have been played, so at least one player must have won at least $\left\lceil\frac{6}{4}\right\rceil=2$ matches. Assume player $a_{1}$ has defeated players $a_{2}$ and $a_{3}$, and assume player $a_{2}$ has defeated player $a_{3}$. Then, set players $A_{2}, A_{3}, A_{4}$ to be $a_{1}, a_{2}, a_{3}$, respectively. Therefore, the ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$ satisfies the conditions.
Next, prove: when $n \leqslant 7$, an ordered quartet that satisfies the conditions does not necessarily exist.
It suffices to disprove the case for $n=7$.
Label the 7 players as $b_{1}, b_{2}, \cdots, b_{7}$, and agree that $b_{7+k}=b_{k}$. Construct the following scenario:
For $i=1,2, \cdots, 7$, let player $b_{i}$ defeat players $b_{i+1}, b_{i+2}, b_{i+4}$, but lose to players $b_{i+3}, b_{i+5}, b_{i+6}$.
This precisely determines the outcome of every match.
Assume there exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$ that meets the conditions.
Since player $A_{1}$ must be one of the $b_{1}$, players $A_{2}, A_{3}, A_{4}$ can only be a permutation of $b_{i+1}, b_{i+2}, b_{i+4}$. However, because player $b_{i+1}$ defeats player $b_{i+2}$, player $b_{i+2}$ defeats player $b_{i+4}$, and player $b_{i+4}$ defeats player $b_{i+1}$, none of the players $b_{i+1}, b_{i+2}, b_{i+4}$ can serve as player $A_{4}$, leading to a contradiction.
In conclusion, the smallest value of $n$ that satisfies the conditions is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given distinct complex numbers $a$ and $b$ satisfying $a b \neq 0$, the set $\{a, b\}=\left\{a^{2}, b^{2}\right\}$. Then $a+b=$ $\qquad$ .
|
$-1 .-1$.
If $a=a^{2}, b=b^{2}$, by $a b \neq 0$, we get $a=b=1$, which is a contradiction. If $a=b^{2}, b=a^{2}$, by $a b \neq 0$, we get $a^{3}=1$.
Clearly, $a \neq 1$.
Thus, $a^{2}+a+1=0 \Rightarrow a=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
Similarly, $b=-\frac{1}{2} \mp \frac{\sqrt{3}}{2} \mathrm{i}$.
Therefore, $a+b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a positive integer $a$ such that the function
$$
f(x)=x+\sqrt{13-2 a x}
$$
has a maximum value that is also a positive integer. Then the maximum value of the function is $\qquad$ .
|
2.7.
Let $t=\sqrt{13-2 a x} \geqslant 0$. Then
$$
\begin{aligned}
y= & f(x)=\frac{13-t^{2}}{2 a}+t \\
& =-\frac{1}{2 a}(t-a)^{2}+\frac{1}{2}\left(a+\frac{13}{a}\right) .
\end{aligned}
$$
Since $a$ is a positive integer, $y_{\max }=\frac{1}{2}\left(a+\frac{13}{a}\right)$ is also a positive integer, so, $y_{\max }=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the positive geometric sequence $\left\{a_{n}\right\}$,
$$
a_{5}=\frac{1}{2}, a_{6}+a_{7}=3 \text {. }
$$
Then the maximum positive integer $n$ that satisfies $a_{1}+a_{2}+\cdots+a_{n}>a_{1} a_{2} \cdots a_{n}$ is $\qquad$
|
3. 12 .
According to the problem, $\frac{a_{6}+a_{7}}{a_{5}}=q+q^{2}=6$.
Since $a_{n}>0$, we have $q=2, a_{n}=2^{n-6}$.
Thus, $2^{-5}\left(2^{n}-1\right)>2^{\frac{n(n-11)}{2}} \Rightarrow 2^{n}-1>2^{\frac{n(n-11)}{2}+5}$.
Estimating $n>\frac{n(n-11)}{2}+5$, we get $n_{\max }=12$.
Upon verification, $n=12$ meets the requirement.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure 1, given a regular tetrahedron $P-A B C$ with all edge lengths equal to 4, points $D, E, F$ are on edges $P A, P B, P C$ respectively. Then the number of $\triangle D E F$ that satisfy $D E = E F = 3, D F = 2$ is $\qquad$.
|
7.3.
Let $P D=x, P E=y, P F=z$. Then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-x y=9, \\
y^{2}+z^{2}-y z=9, \\
z^{2}+x^{2}-z x=4 .
\end{array}\right.
$$
(1) - (2) gives $x=z$ or $x+z=y$.
When $x=z$, we get $x=z=2, y=1+\sqrt{6}$;
When $x+z=y$, $x z=\frac{5}{2}, x^{2}+z^{2}=\frac{13}{2}$, there are two sets of solutions.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function
$$
f(x)=a \sin x+b \cos x \quad(a, b \in \mathbf{Z}),
$$
and it satisfies
$$
\{x \mid f(x)=0\}=\{x \mid f(f(x))=0\} .
$$
Then the maximum value of $a$ is . $\qquad$
|
2.3.
Let $A=\{x \mid f(x)=0\}, B=\{x \mid f(f(x))=0\}$.
Obviously, set $A$ is non-empty.
Take $x_{0} \in A$, i.e., $x_{0} \in B$, hence
$$
b=f(0)=f\left(f\left(x_{0}\right)\right)=0 \text {. }
$$
Thus, $f(x)=a \sin x(a \in \mathbf{Z})$.
When $a=0$, obviously, $A=B$.
Now assume $a \neq 0$, in this case,
$$
\begin{array}{l}
A=\{x \mid a \sin x=0\}, \\
B=\{x \mid a \sin (a \sin x)=0\} \\
=\{x \mid a \sin x=k \pi, k \in \mathbf{Z}\} .
\end{array}
$$
It is easy to see that $A=B$ if and only if for any $x \in \mathbf{R}$, $a \sin x \neq k \pi(k \in \mathbf{Z}, k \neq 0)$,
i.e., $|a|<\pi$.
Therefore, the maximum value of the integer $a$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let any real numbers $a>b>c>d>0$. To make
$$
\begin{array}{l}
\log _{\frac{b}{a}} 2014+\log _{\frac{d}{b}} 2014+\log _{\frac{d}{c}} 2014 \\
\geqslant m \log _{\frac{d}{a}} 2014
\end{array}
$$
always hold, then the minimum value of $m$ is $\qquad$.
|
5.9.
$$
\begin{array}{l}
\text { Let } x_{1}=-\log _{2014} \frac{b}{a}, x_{2}=-\log _{2014} \frac{c}{b}, \\
x_{3}=-\log _{2014} \frac{d}{c} .
\end{array}
$$
Since $a>b>c>d>0$, we have
$$
x_{1}>0, x_{2}>0, x_{3}>0 \text {. }
$$
Thus, the given inequality can be transformed into
$$
\begin{array}{l}
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}} \leqslant m \cdot \frac{1}{x_{1}+x_{2}+x_{3}} \\
\Rightarrow m \geqslant\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right) \geqslant 9 .
\end{array}
$$
When $x_{1}=x_{2}=x_{3}$, i.e., $a, b, c, d$ form a geometric sequence, the equality holds.
Therefore, the minimum value of $m$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let \( x, y, z \) all be positive real numbers, and
\[
x+y+z=1 \text{. }
\]
Find the minimum value of the function
\[
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
\]
and provide a proof.
|
11. Notice that, $\frac{3 x^{2}-x}{1+x^{2}}=\frac{x(3 x-1)}{1+x^{2}}$.
Consider the function $g(t)=\frac{t}{1+t^{2}}$.
It is easy to see that $g(t)$ is an odd function.
Since when $t>0$, $\frac{1}{t}+t$ is decreasing in the interval $(0,1)$, hence $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in the interval $(0,1)$.
Therefore, for $t_{1}, t_{2} \in(0,1)$, we have
$$
\left(t_{1}-t_{2}\right)\left(g\left(t_{1}\right)-g\left(t_{2}\right)\right) \geqslant 0 \text {. }
$$
Thus, for any $x \in(0,1)$, we have
$$
\begin{array}{l}
\left(x-\frac{1}{3}\right)\left(\frac{x}{1+x^{2}}-\frac{3}{10}\right) \geqslant 0 \\
\Rightarrow \frac{3 x^{2}-x}{1+x^{2}} \geqslant \frac{3}{10}(3 x-1) .
\end{array}
$$
Similarly, $\frac{3 y^{2}-y}{1+y^{2}} \geqslant \frac{3}{10}(3 y-1)$,
$$
\frac{3 z^{2}-z}{1+z^{2}} \geqslant \frac{3}{10}(3 z-1) \text {. }
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}} \\
\geqslant \frac{3}{10}[3(x+y+z)-3]=0 .
\end{array}
$$
When $x=y=z=\frac{1}{3}$, $f(x, y, z)=0$.
Therefore, the minimum value sought is 0.
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: Twelve acrobats numbered $1, 2, \cdots, 12$ are divided into two groups, $A$ and $B$, each with six people. Let the actors in group $A$ form a circle, and each actor in group $B$ stand on the shoulders of two adjacent actors in the $A$ circle. If the number of each actor in the $B$ circle is equal to the sum of the numbers of the two actors beneath him, then such a combination is called a "tower". How many structurally different towers can be built?
[Note] The six people in group $A$, $a_{1}, a_{2}, \cdots, a_{6}$, forming a circle in this order in a clockwise or counterclockwise direction can be considered the same structure. A circle is also considered the same structure after rotation. For convenience, draw a circle, fill in the numbers of the bottom actors inside the circle, and fill in the numbers of the top actors outside the circle. Figure 20 is a tower of 14 people.
|
Let the sum of the elements in groups $A$ and $B$ be denoted as $x$ and $y$ respectively.
Then $y = 2x$.
Therefore, $3x = x + y = 1 + 2 + \cdots + 12 = 78 \Rightarrow x = 26$.
Clearly, $1, 2 \in A, 11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$.
If $d \leq 7$, then
$$
a + b + c + d \leq 4 + 5 + 6 + 7 = 22.
$$
Thus, $a + b + c + d = 23$, and $a \geq 3, 8 \leq d \leq 10$.
(1) $d = 8$.
Then $A = \{1, 2, a, b, c, 8\}, c \leq 7, a + b + c = 15$.
Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, i.e.,
$A = \{1, 2, 3, 5, 7, 8\}$ or $A = \{1, 2, 4, 5, 6, 8\}$.
(i) If $A = \{1, 2, 3, 5, 7, 8\}$, then
$B = \{4, 6, 9, 10, 11, 12\}$.
Since set $B$ contains $4, 6, 11, 12$, set $A$ must have 1 adjacent to 3, 1 adjacent to 5, 5 adjacent to 7, and 8 adjacent to 3.
In this case, there is only one arrangement, resulting in the tower in Figure 21.
Use 21
(ii) If $A = \{1, 2, 4, 5, 6, 8\}$, then
$$
B = \{3, 7, 9, 10, 11, 12\}.
$$
Similarly, set $A$ must have 1 adjacent to 2, 5 adjacent to 6, and 4 adjacent to 8. In this case, there are two arrangements, resulting in the two towers in Figure 22.
(2) $d = 9$.
Then $A = \{1, 2, a, b, c, 9\}, c \leq 8, a + b + c = 14$.
Thus, $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, i.e.,
$A = \{1, 2, 3, 5, 6, 9\}$ or $A = \{1, 2, 3, 4, 7, 9\}$.
(i) If $A = \{1, 2, 3, 5, 6, 9\}$, then
$B = \{4, 7, 8, 10, 11, 12\}$.
To get $4, 10, 12$ in set $B$, set $A$ must have 1, 3, 9 all adjacent, which is impossible.
(ii) If $A = \{1, 2, 3, 4, 7, 9\}$, then
$B = \{5, 6, 8, 10, 11, 12\}$.
To get $6, 8, 12$ in set $B$, set $A$ must have 2 adjacent to 4, 1 adjacent to 7, and 9 adjacent to 3. Thus, there are two arrangements, resulting in the two towers in Figure 23.
(3) $d = 10$.
Then $A = \{1, 2, a, b, c, 10\}, c \leq 9, a + b + c = 13$.
Thus, $(a, b, c) = (3, 4, 6)$, i.e.,
$$
\begin{array}{l}
A = \{1, 2, 3, 4, 6, 10\}, \\
B = \{5, 7, 8, 9, 11, 12\}.
\end{array}
$$
To get $8, 9, 11, 12$ in set $B$, set $A$ must have 6 adjacent to 2, 6 adjacent to 3, 10 adjacent to 1, and 10 adjacent to 2. Thus, there is only one arrangement, resulting in the tower in Figure 24.
Therefore, there are 6 structurally different towers, i.e.,
$T(6) = 6$.
As the value of $n$ increases, the number of such arrangements $T(n)$ increases rapidly, making the situation more complex.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x, y$ be positive real numbers. Find the minimum value of
$$
x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}
$$
(Gu Bin, Jin Aiguo)
|
1. Let $f(x, y)=x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.
If $x \geqslant 1, y \geqslant 1$, then
$$
f(x, y) \geqslant x+y \geqslant 2 \text {; }
$$
If $0<x<1, 0<y<1$, then
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{1-y}{x} = \left(x + \frac{1-x}{y}\right) + \left(y + \frac{1-y}{x}\right) \geqslant 2.
$$
For $0<x<1, y \geqslant 1$, we have
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{y-1}{x} \geqslant x + y + \frac{1-x}{y} \geqslant 2.
$$
For $x \geqslant 1, 0<y<1$, we have
$$
f(x, y) = x + y + \frac{x-1}{y} + \frac{1-y}{x} \geqslant x + y + \frac{1-y}{x} \geqslant 2.
$$
For $x>0, y>0$, we have $f(x, y) \geqslant 2$. Also, $f(1,1)=2$, hence the minimum value is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, in an election, there are 12 candidates, and each member of the electoral committee casts 6 votes. It is known that any two members' votes have at most 2 candidates in common. Find the maximum number of members in the committee.
(Proposed by the Problem Committee)
|
The maximum number of committee members is 4.
Let the number of committee members be $k$, and the candidates be represented by $1,2, \cdots, 12$. Each person's vote is a set $A_{i}(1 \leqslant i \leqslant k)$, and the number of votes each candidate receives is $m_{i}(1 \leqslant i \leqslant 12)$. Then,
$$
\sum_{i=1}^{12} m_{i}=\sum_{i=1}^{k} 6=6 k .
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
2 \sum_{1 \leqslant i<j \leqslant k}\left|A_{i} \cap A_{j}\right|=2 \sum_{i=1}^{12} \mathrm{C}_{m_{i}}^{2}=\sum_{i=1}^{12} m_{i}^{2}-\sum_{i=1}^{12} m_{i} \\
\geqslant \frac{\left(\sum_{i=1}^{12} m_{i}\right)^{2}}{\sum_{i=1}^{12} 1^{2}}-\sum_{i=1}^{12} m_{i}=\frac{(6 k)^{2}}{12}-6 k \\
=3 k^{2}-6 k .
\end{array}
$$
Notice that,
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant k}\left|A_{i} \cap A_{j}\right| \leqslant \sum_{1 \leqslant i<j \leqslant k} 2=2 \mathrm{C}_{k}^{2}=k^{2}-k \\
\Rightarrow & 3 k^{2}-6 k \leqslant 2\left(k^{2}-k\right) \Rightarrow k \leqslant 4 .
\end{aligned}
$$
When $k=4$, a construction that satisfies the conditions is
$$
\begin{array}{l}
A_{1}=\{1,2,3,4,5,6\}, \\
A_{2}=\{1,2,7,8,9,10\}, \\
A_{3}=\{3,4,7,8,11,12\}, \\
A_{4}=\{5,6,9,10,11,12\} .
\end{array}
$$
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Rational numbers $a, b$ have decimal expansions that are repeating decimals with the smallest period of 30. It is known that the decimal expansions of $a-b$ and $a+kb$ have the smallest period of 15. Find the smallest possible value of the positive integer $k$.
|
3. $k_{\min }=6$.
Notice that, $a, b, a-b, a+k b$ can simultaneously become pure repeating decimals by multiplying them by a power of 10. Therefore, assume they are all pure repeating decimals.
Since the decimal part of a rational number is a pure repeating decimal with a period of $T$ if and only if it can be written in the form $\frac{m}{10^{T}-1}(m \in \mathbf{Z})$, we set
$$
a=\frac{m}{10^{30}-1}, b=\frac{n}{10^{30}-1}.
$$
By $a-b=\frac{m-n}{10^{30}-1}, a+k b=\frac{m+k n}{10^{30}-1}$, both are repeating decimals with a period of 15, subtracting them gives
$$
(k+1) b=\frac{(k+1) n}{10^{30}-1}
$$
is a repeating decimal with a period of 15, hence
$$
\left(10^{15}+1\right) \mid(k+1) n.
$$
Since $n$ is not a multiple of $10^{15}+1$ (otherwise, the period length of $b$ would be 15), $k+1$ must be a multiple of some prime factor of $10^{15}+1$.
Also, $10^{15}+1$ is not a multiple of $2,3,5$, thus,
$k+1 \geqslant 7 \Rightarrow k \geqslant 6$.
Next, provide an example for $k=6$.
Take $a=\frac{8}{7\left(10^{15}-1\right)}, b=\frac{1}{7\left(10^{15}-1\right)}$.
Notice that, $10^{3}+1=7 \times 143$, so the smallest period of $a, b$ is 30, $a-b=\frac{1}{10^{15}-1}, a+6 b=\frac{2}{10^{15}-1}$ has the smallest period of 15.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
|
Three, let the equation $x^{2}+p x+k p-1=0$ have integer roots $x_{1}$ and another root $x_{2}$.
By the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-p, x_{1} x_{2}=k p-1 \text {. }
$$
Thus, $x_{2}$ must also be an integer.
Assume $k>1$.
Notice,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right) \\
=x_{1} x_{2}+\left(x_{1}+x_{2}\right)+1=(k-1) p .
\end{array}
$$
Since $p$ is a prime number and $k-1>0$, it follows that $p \mid\left(x_{1}+1\right)$ or $p \mid\left(x_{2}+1\right)$.
Without loss of generality, assume $p \mid\left(x_{1}+1\right)$. Then
$$
x_{1}+1= \pm m p, x_{2}+1= \pm \frac{k-1}{m} \text {, }
$$
where $m$ is a positive integer, and $m \mid(k-1)$.
Adding the two equations, we get
$$
x_{1}+x_{2}+2= \pm\left(m p+\frac{k-1}{m}\right) \text {, }
$$
which means $-p+2= \pm\left(m p+\frac{k-1}{m}\right)$.
If the right side of equation (1) is positive, then
$$
(m+1) p+\frac{k-1}{m}=2 \text {. }
$$
Clearly, the left side of equation (1) is greater than 2, which is a contradiction.
If the right side of equation (1) is negative, then
$$
(m-1) p+2+\frac{k-1}{m}=0 \text {. }
$$
Clearly, the left side of equation (1) is greater than 0, which is a contradiction.
Therefore, the assumption is false.
Since $k$ is a positive integer, the only possible value for $k$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
|
Notice,
$$
\begin{array}{l}
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}-1 \\
=\left(\frac{b^{2}+c^{2}-a^{2}+2 b c}{2 b c}\right)\left(\frac{b^{2}+c^{2}-a^{2}-2 b c}{2 b c}\right) \\
=\frac{(b+c+a)(b+c-a)(b-c+a)(b-c-a)}{4 b^{2} c^{2}} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}-1 \\
=\frac{(c+a+b)(c+a-b)(c-a+b)(c-a-b)}{4 c^{2} a^{2}}, \\
\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}-1 \\
=\frac{(a+b+c)(a+b-c)(a-b+c)(a-b-c)}{4 a^{2} b^{2}} . \\
\text { Therefore, }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}-3 \\
=-(a+b+c)(a+b-c)(b+c-a)(a+c-b) . \\
\left(\frac{1}{4 b^{2} c^{2}}+\frac{1}{4 c^{2} a^{2}}+\frac{1}{4 a^{2} b^{2}}\right) \\
=0 .
\end{array}
$$
Since \(a, b, c\) are positive numbers, therefore,
$$
\begin{array}{l}
a+b+c>0, \\
\frac{1}{4 b^{2} c^{2}}+\frac{1}{4 c^{2} a^{2}}+\frac{1}{4 a^{2} b^{2}}>0 .
\end{array}
$$
Then \((a+b-c)(b+c-a)(a+c-b)=0\).
Hence, among \(a, b, c\), there must be two numbers whose sum equals the third number.
Without loss of generality, let \(a+b=c\). Then
$$
\begin{array}{l}
b^{2}+c^{2}-a^{2}=b^{2}+c^{2}-(c-b)^{2}=2 b c, \\
c^{2}+a^{2}-b^{2}=c^{2}+a^{2}-(c-a)^{2}=2 a c, \\
a^{2}+b^{2}-c^{2}=a^{2}+b^{2}-(a+b)^{2}=-2 a b . \\
\text { Therefore, } \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
=1+1-1=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given real numbers $x, y, z$ satisfy
$$
x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+\frac{1}{x}=\frac{7}{3} \text {. }
$$
Find the value of $x y z$.
|
Multiplying the three conditional expressions yields
$$
\frac{28}{3}=x y z+\frac{1}{x y z}+\frac{22}{3} \Rightarrow x y z=1 .
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $x, y, z$ satisfy
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0 \text {. }
$$
Find the value of the algebraic expression $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$.
|
Notice,
$$
\begin{array}{l}
x+y+z \\
=\left(\frac{x^{2}}{y+z}+x\right)+\left(\frac{y^{2}}{z+x}+y\right)+\left(\frac{z^{2}}{x+y}+z\right) \\
=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right) .
\end{array}
$$
Thus, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=-3$ or 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2$. Find the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$.
|
Solve: Regarding $a$ and $b$ as the main elements, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=2-c, \\
a^{2}+b^{2}=2-c^{2} .
\end{array}\right. \\
\text { Hence } a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=c^{2}-2 c+1=(c-1)^{2} \\
\Rightarrow \frac{(1-c)^{2}}{a b}=1 .
\end{array}
$$
Similarly, $\frac{(1-a)^{2}}{b c}=1, \frac{(1-b)^{2}}{c a}=1$.
Thus, $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{3}-x y z=-5, \\
y^{3}-x y z=2, \\
z^{3}-x y z=21 .
\end{array}\right.
$$
Find the value of $x+y+z$.
|
Let $x y z=k$. Then
$$
\left\{\begin{array}{l}
x^{3}=k-5, \\
y^{3}=k+2, \\
z^{3}=k+21 .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
\begin{array}{l}
k^{3}=(x y z)^{3}=(k-5)(k+2)(k+21) \\
\Rightarrow 18 k^{2}-73 k-210=0 \\
\Rightarrow k_{1}=6, k_{2}=-\frac{35}{18} \text { (not suitable, discard). }
\end{array}
$$
Substituting $k=6$ into the system of equations (1), we solve to get
$$
x=1, y=2, z=3 \text {. }
$$
Therefore, $x+y+z=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The arithmetic sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}+a_{2}+\cdots+a_{14}=77 \text {, and } a_{1} 、 a_{11} \in \mathbf{Z}_{+} \text {. }
$$
Then $a_{18}=$ $\qquad$
|
2. -5 .
From the formula for the sum of an arithmetic sequence, we get
$$
\begin{array}{l}
a_{1}+a_{14}=11 \Rightarrow 2 a_{1}+13 \times \frac{a_{11}-a_{1}}{10}=11 \\
\Rightarrow 7 a_{1}+13 a_{11}=110 \\
\Rightarrow a_{1}=12(\bmod 13), a_{11}=2(\bmod 7) \\
\Rightarrow\left(a_{1}, a_{11}\right)=(12,2) \\
\Rightarrow a_{18}=a_{11}+7 \times \frac{a_{11}-a_{1}}{10}=-5 .
\end{array}
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the function
$$
f(x)=x \log _{2} x+(a-x) \log _{2}(a-x)
$$
be symmetric about the line $x=\frac{1}{2}$. Then for any real numbers $x_{i} \in(0,1)(1 \leqslant i \leqslant 4)$ satisfying $\sum_{i=1}^{4} x_{i}=1$, the minimum value of $s=\sum_{i=1}^{4} x_{i} \log _{2} x_{i}$ is . $\qquad$
|
5. -2 .
From the problem, we know that the midpoint of the interval $(0, a)$ is $\frac{1}{2}$.
Thus, $a=1$.
Then, $f(x)=x \log _{2} x+(1-x) \log _{2}(1-x)$
$\Rightarrow f^{\prime}(x)=\log _{2} \frac{x}{1-x}$.
Let $f^{\prime}(x)=0$, we get $x=\frac{1}{2}$.
For any $x \in\left(0, \frac{1}{2}\right)$, $f^{\prime}(x) < 0$, and for any $x \in\left(\frac{1}{2}, 1\right)$, $f^{\prime}(x) > 0$, so
$$
f(x)_{\text {min }}=f\left(\frac{1}{2}\right)=-1 .
$$
Let $x_{1}+x_{2}=x \in(0,1)$. Then
$$
x_{3}+x_{4}=1-x \in(0,1) \text {. }
$$
From $\frac{x_{1}}{x}+\frac{x_{2}}{x}=1$, we get
$$
\frac{x_{1}}{x} \log _{2} \frac{x_{1}}{x}+\frac{x_{2}}{x} \log _{2} \frac{x_{2}}{x}=f\left(\frac{x_{1}}{x}\right) \geqslant-1,
$$
which means $x_{1} \log _{2} x_{1}+x_{2} \log _{2} x_{2} \geqslant-x+x \log _{2} x$.
Similarly, from the above, we get
$$
\begin{array}{l}
x_{3} \log _{2} x_{3}+x_{4} \log _{2} x_{4} \\
\geqslant-(1-x)+(1-x) \log _{2}(1-x) .
\end{array}
$$
Adding the two inequalities, we get
$$
\sum_{i=1}^{4} x_{i} \log _{2} x_{i} \geqslant-1+f(x) \geqslant-2 \text {. }
$$
When $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{4}$, the equality holds.
Thus, $s_{\min }=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given
$$
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} \text {. }
$$
Find the value of $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}$.
|
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{x+y+z}{x(y+z)}=\frac{1}{2} \\
\Rightarrow \frac{2}{x}=\frac{y+z}{x+y+z} .
\end{array}
$$
Similarly, $\frac{3}{y}=\frac{z+x}{x+y+z}, \frac{4}{z}=\frac{x+y}{x+y+z}$.
Adding the three equations yields $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=1, \\
\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1 .
\end{array}
$$
Then $a b c=$
|
$$
\begin{aligned}
& \frac{1}{1-2 c}+\frac{1}{1-2 a}+\frac{1}{1-2 b}=1 \\
\Rightarrow & (1-2 a)(1-2 b)+(1-2 b)(1-2 c)+(1-2 a)(1-2 c) \\
& =(1-2 a)(1-2 b)(1-2 c) \\
\Rightarrow & 2-2(a+b+c)=8 a b c \\
\Rightarrow & a b c=0 .
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
|
Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$
Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.
By verifying $n=2,4,6,8$ one by one, we find that $N$ is not a square number in any of these cases.
(2) When $n \geqslant 9$, $N=2^{8}\left(9+2^{n-8}\right)$.
For $N$ to be a square number, $9+2^{n-8}$ must be the square of an odd number.
Assume $9+2^{n-8}=(2 k+1)^{2}$. Hence,
$$
2^{n-10}=(k-1)(k+2) \text {. }
$$
Since $k-1$ and $k+2$ are one odd and one even, the left side of the equation is a power of 2, so it must be that $k-1=1$, i.e., $k=2$.
From $2^{n-10}=2^{2}$, we know that $n=12$ is the solution.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given a regular tetrahedron $P-ABC$, points $P, A, B, C$ are all on a sphere with radius $\sqrt{3}$. If $PA, PB, PC$ are mutually perpendicular, then the distance from the center of the sphere to the plane $ABC$ is $\qquad$.
|
$=13 \cdot \frac{\sqrt{3}}{3}$.
Let $P A=P B=P C=x$. Then $A B=\sqrt{2} x$.
Let the centroid of $\triangle A B C$ be $M$, and the center of the circumscribed sphere of the regular tetrahedron be $O, O M=y$. Then
$$
\left.\begin{array}{l}
A M=\frac{2 \sqrt{3}}{3} \times \frac{\sqrt{2}}{2} x=\frac{\sqrt{6}}{3} x .
\end{array}\right\}
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In the right trapezoid $A B C D$, it is known that $A D \perp A B$, $A B / / D C, A B=4, A D=D C=2$. Let $N$ be the midpoint of side $D C$, and $M$ be a moving point within or on the boundary of trapezoid $A B C D$. Then the maximum value of $\overrightarrow{A M} \cdot \overrightarrow{A N}$ is
|
8. 6 .
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. ${ }^{[4]}$
(2013, National High School Mathematics League Shandong Province Preliminary Contest)
|
Solve the general ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
Corresponding problems.
With $F_{1}$ as the pole and $F_{1} x$ as the polar axis, the equation of the ellipse is transformed into
$$
\rho=\frac{e p}{1+e \cos \theta}=\frac{b^{2}}{a-c \cos \theta} .
$$
Therefore, the length of the chord passing through point $F_{1}$
$$
\begin{array}{l}
A B=\frac{b^{2}}{a-c \cos \theta}+\frac{b^{2}}{a+c \cos \theta} \\
=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta},
\end{array}
$$
The area of $\square A B C D$
$$
\begin{aligned}
S & =A B \cdot F_{1} F_{2} \sin \theta=\frac{4 a b^{2} c \sin \theta}{a^{2}-c^{2} \cos ^{2} \theta} \\
& =\frac{4 a b^{2} c \sin \theta}{b^{2}+c^{2} \sin ^{2} \theta}=\frac{4 a b^{2} c}{c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}} . \\
& \text { Let } f(\theta)=c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}\left(0<\theta<\pi\right) . \text { Then } f^{\prime}(\theta)=c^{2} \cos \theta-\frac{b^{2} \cos \theta}{\sin ^{2} \theta}=\frac{\cos \theta\left(c^{2} \sin ^{2} \theta-b^{2}\right)}{\sin ^{2} \theta} .
\end{aligned}
$$
Let $f^{\prime}(\theta)=0$, we get $\sin \theta=\frac{b}{c}$. When $\theta \in\left(0, \frac{b}{c}\right)$, $f^{\prime}(\theta)<0$; when $\theta \in\left(\frac{b}{c}, \pi\right)$, $f^{\prime}(\theta)>0$. Therefore, $f(\theta)$ is monotonically decreasing in the interval $\left(0, \frac{b}{c}\right]$ and monotonically increasing in the interval $\left[\frac{b}{c}, \pi\right)$. Hence,
(1) When $\frac{b}{c} \geqslant 1$, i.e., $b\sqrt{2} b$, $g(t)$ is monotonically decreasing in the interval $\left(0, \frac{b}{c}\right]$ and monotonically increasing in the interval $\left[\frac{b}{c}, 1\right)$. Therefore, the minimum value of $g(t)$ is $g\left(\frac{b}{c}\right)=2 b c$. At this point, $S$ reaches its maximum value $\frac{4 a b^{2} c}{2 b c}=2 a b$.
In summary, when $b\sqrt{2} b$, $S$ reaches its maximum value $2 a b$.
For this problem, $a=2, b=\sqrt{3}, c=1, S$ reaches its maximum value $\frac{4 b^{2} c}{a}=6$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is . $\qquad$
|
4. 2 .
Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$
Combining with $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P F_{1} \perp P F_{2}$.
Therefore, the inradius of $\triangle P F_{1} F_{2}$ is
$$
r=\frac{6+8-10}{2}=2 \text {. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given points $A(1,-1), B(4,0), C(2,2)$, the plane region $D$ consists of all points $P(x, y)$ that satisfy
$$
\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(1<\lambda \leqslant a, 1<\mu \leqslant b)
$$
If the area of region $D$ is 8, then the minimum value of $a+b$ is . $\qquad$
|
8. 4 .
As shown in Figure 3, extend $A B$ to point $N$, and extend $A C$ to point $M$, such that
$$
|A N|=a|A B|,|A M|=b|A C| .
$$
Construct $\square A B E C$ and $\square A N G M$. Then quadrilateral $E H G F$ is a parallelogram.
From the conditions, the region $D$ composed of points $P(x, y)$ is the shaded area in the figure, i.e., $\square E H G F$ (excluding the boundaries $E H$ and $E F$).
Notice that,
$$
\overrightarrow{A B}=(3,1), \overrightarrow{A C}=(1,3), \overrightarrow{B C}=(-2,2) \text {. }
$$
Thus, $|A B|=|A C|=\sqrt{10},|B C|=2 \sqrt{2}$.
Then $\cos \angle C A B=\frac{10+10-8}{2 \times \sqrt{10} \times \sqrt{10}}=\frac{3}{5}$
$$
\Rightarrow \sin \angle C A B=\frac{4}{5} \text {. }
$$
Therefore, $S_{\text {DEHGF }}$
$$
\begin{aligned}
= & (a-1) \sqrt{10}(b-1) \sqrt{10} \times \frac{4}{5}=8 \\
\Rightarrow & (a-1)(b-1)=1 \\
\Rightarrow & a+b=a+\left(\frac{1}{a-1}+1\right) \\
= & (a-1)+\frac{1}{a-1}+2 .
\end{aligned}
$$
Given $a>1, b>1$, we know that when and only when $a-1=1$, i.e., $a=b=2$, $a+b$ reaches its minimum value of 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A certain meeting has 30 participants, each of whom knows at most five others; among any five people, at least two are not acquaintances. Find the largest positive integer $k$, such that in any group of 30 people satisfying the above conditions, there always exists a group of $k$ people, none of whom are acquaintances.
(Cold Gangsong, Contributed)
|
5, hence we have
$30-|X| \leqslant 5|X|$.
Therefore, $|X| \geqslant 5$.
If $|X|=5$, then by equation (1) the equality is achieved, which means the 25 edges are distributed among the five vertices in set $X$, i.e., the neighborhood of each vertex in set $X$ is a set of five points in set $V \backslash X$.
Since $|V \backslash X|=25$, the neighborhoods of any two vertices in set $X$ do not intersect.
Let $X=\{a, b, c, d, e\}$.
Now consider the neighborhood of $a$, denoted as
$$
Y_{a}=\left\{y_{1}, y_{2}, y_{3}, y_{4}, y_{5}\right\}.
$$
By (ii), we know that there are two points in set $Y_{a}$ that are not connected, let them be $y_{1}$ and $y_{2}$.
Since the neighborhoods of any two vertices in set $X$ do not intersect, $y_{1}$ and $y_{2}$ cannot be neighbors of any of $b, c, d, e$.
Thus, $\left\{y_{1}, y_{2}, b, c, d, e\right\}$ is an independent set of graph $G$ with more than 5 elements, which is a contradiction.
This proves that $|X| \geqslant 6$.
(2) Next, we prove: there exists a graph $G$ that meets the conditions, and the maximum independent set has no more than 6 elements.
Partition the set $V$ into point sets $V_{1}, V_{2}, V_{3}$, such that
$\left|V_{i}\right|=10(i=1,2,3)$.
Let $V_{1}=\left\{A_{1}, A_{2}, \cdots, A_{5}, B_{1}, B_{2}, \cdots, B_{5}\right\}$, connect the points in set $V_{1}$ as shown in Figure 3, i.e.,
(i) $A_{i} A_{i+1}(i=1,2, \cdots, 5)$ are connected;
(ii) $B_{i} B_{i+1}(i=1,2, \cdots, 5)$ are connected;
(iii) $A_{i} B_{i}, A_{i} B_{i+1}, A_{i} B_{i-1}(i=1,2, \cdots, 5)$ are connected, where $A_{6}=A_{1}, B_{6}=B_{1}, B_{0}=B_{5}$.
The connection method for point sets $V_{2}$ and $V_{3}$ is the same as for set $V_{1}$, and for any $1 \leqslant i<j \leqslant 3$, there are no edges between point sets $V_{i}$ and $V_{j}$. Then, the degree of any vertex in graph $G$ is 5, and in graph $G$, among any five points, there are always two points that are not connected.
Now, take any maximum independent set $X$ of graph $G$.
We only need to prove: $\left|V_{1} \cap X\right| \leqslant 2$.
In fact, since $A_{i} 、 A_{i+1}(i=1,2, \cdots, 5)$ are adjacent, at most two of $A_{1}, A_{2}, \cdots, A_{5}$ belong to set $X$.
Similarly, at most two of $B_{1}, B_{2}, \cdots, B_{5}$ belong to set $X$.
If exactly two of $A_{1}, A_{2}, \cdots, A_{5}$ belong to set $X$, let them be $\left\{A_{1}, A_{3}\right\}$.
Notice that, the union of the neighborhoods of $A_{1}$ and $A_{3}$ is exactly $\left\{B_{1}, B_{2}, \cdots, B_{5}\right\}$. Hence, $B_{1}, B_{2}, \cdots, B_{5}$ do not belong to set $X$.
Similarly, if exactly two of $B_{1}, B_{2}, \cdots, B_{5}$ belong to set $X$, then $A_{1}, A_{2}, \cdots, A_{5}$ do not belong to set $X$. This proves that $\left|V_{1} \cap X\right| \leqslant 2$.
Similarly, $\left|V_{2} \cap X\right| \leqslant 2,\left|V_{3} \cap X\right| \leqslant 2$.
Thus, $|X|=|V \cap X|$
$$
=\left|V_{1} \cap X\right|+\left|V_{2} \cap X\right|+\left|V_{3} \cap X\right| \leqslant 6.
$$
Therefore, graph $G$ meets the requirements.
Combining (1) and (2), we know that the value of $k$ is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. }
$$
Then the minimum value of $x+y$ is $\qquad$.
|
2. 2 .
Notice,
$$
\begin{array}{l}
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\
\Rightarrow 2 x+\sqrt{4 x^{2}+1} \\
\geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} .
\end{array}
$$
When $x=\frac{1}{y}$,
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right)=1 .
$$
And $f(x)=2 x+\sqrt{4 x^{2}+1}$ is monotonically increasing, so $x \geqslant \frac{1}{y}$.
Thus, $x+y \geqslant y+\frac{1}{y} \geqslant 2$.
When $x=y=1$, the equality holds.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 2. If } \frac{y}{x}+\frac{x}{z}=a, \frac{z}{y}+\frac{y}{x}=b, \frac{x}{z}+\frac{z}{y}=c, \\ \text { then }(b+c-a)(c+a-b)(a+b-c)=\end{array}$
|
2.8.
Notice that,
$$
\begin{array}{l}
b+c-a=\left(\frac{z}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}\right)-\left(\frac{y}{x}+\frac{x}{z}\right) \\
=\frac{2 z}{y} .
\end{array}
$$
Similarly, $c+a-b=\frac{2 x}{z}, a+b-c=\frac{2 y}{x}$.
Therefore, $(b+c-a)(c+a-b)(a+b-c)$ $=\frac{2 z}{y} \cdot \frac{2 x}{z} \cdot \frac{2 y}{x}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate plane $x O y$, the area of the region determined by the system of inequalities
$$
\left\{\begin{array}{l}
|x| \leqslant 2, \\
|y| \leqslant 2, \\
|| x|-| y|| \leqslant 1
\end{array}\right.
$$
is $\qquad$
|
6. 12 .
Obviously, the region is symmetric with respect to both the $x$-axis and the $y$-axis. Therefore, we can assume $x \geqslant 0, y \geqslant 0$.
From $\left\{\begin{array}{l}0 \leqslant x \leqslant 2, \\ 0 \leqslant y \leqslant 2, \\ -1 \leqslant x-y \leqslant 1\end{array}\right.$, draw the part of the region in the first quadrant (including the coordinate axes), and then symmetrically draw the entire region, as shown in Figure 4.
The area of the region sought is the area of a square with side length 4 minus the area of eight isosceles right triangles with leg length 1, i.e., $4^{2}-8 \times \frac{1}{2}=12$.
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given
$$
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \text {. }
$$
Then $f(20)+f(14)=$ $\qquad$ .
|
$$
\begin{array}{l}
\text { II.7.0. } \\
\text { Let } g(x)=x^{2}-53 x+196 \\
=(x-4)(x-49)
\end{array}
$$
Then when $x<4$ or $x>49$, $g(x)>0$;
when $4 \leqslant x \leqslant 49$, $g(x) \leqslant 0$.
Thus, when $x=4,5, \cdots, 49$,
$$
f(x)=g(x)+|g(x)|=g(x)-g(x)=0 \text {. }
$$
Therefore, $f(20)+f(14)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { 4. If } x_{1}>x_{2}>x_{3}>x_{4}>0, \text { and the inequality } \\
\log _{\frac{x_{1}}{2}} 2014+\log _{\frac{x_{2}}{x_{3}}} 2014+\log _{x_{4} \frac{1}{4}} 2014 \\
\geqslant k \log _{\frac{x_{1}}{}} 2014
\end{array}
$$
always holds, then the maximum value of the real number $k$ is . $\qquad$
|
4.9.
From the given, we have
$$
\begin{array}{l}
\frac{\ln 2014}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln 2014}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln 2014}{\ln \frac{x_{3}}{x_{4}}} \\
\geqslant k \frac{\ln 2014}{\ln \frac{x_{1}}{x_{4}}}.
\end{array}
$$
Since $x_{1}>x_{2}>x_{3}>x_{4}>0$, we have
$$
\begin{array}{l}
\ln \frac{x_{1}}{x_{2}}>0, \ln \frac{x_{2}}{x_{3}}>0, \ln \frac{x_{3}}{x_{4}}>0, \\
\ln \frac{x_{1}}{x_{4}}=\ln \frac{x_{1}}{x_{2}}+\ln \frac{x_{2}}{x_{3}}+\ln \frac{x_{3}}{x_{4}}>0. \\
\text{Therefore, } k \leqslant \frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{3}}{x_{4}}}. \\
\text{And } \frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{3}}{x_{4}}} \\
=\left(\ln \frac{x_{1}}{x_{2}}+\ln \frac{x_{2}}{x_{3}}+\ln \frac{x_{3}}{x_{4}}\right) \cdot \\
\left(\frac{1}{\ln \frac{x_{1}}{x_{2}}}+\frac{1}{\ln \frac{x_{2}}{x_{3}}}+\frac{1}{\ln \frac{x_{3}}{x_{4}}}\right) \\
\geqslant 9,
\end{array}
$$
with equality holding if and only if $\ln \frac{x_{1}}{x_{2}}=\ln \frac{x_{2}}{x_{3}}=\ln \frac{x_{3}}{x_{4}}$, i.e., $\frac{x_{1}}{x_{2}}=\frac{x_{2}}{x_{3}}=\frac{x_{3}}{x_{4}}$.
Therefore, the maximum value of the real number $k$ is 9.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$
then $k=$
|
4. 4 .
Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\right)=0 \\
\Rightarrow k=4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) If $x, y \in [0,1]$, try to find the maximum value of
$$
x \sqrt{1-y} + y \sqrt{1-x}
$$
|
$$
\begin{array}{l}
\text { I. Let } s=\sqrt{1-x}, t=\sqrt{1-y}, \\
z=x \sqrt{1-y}+y \sqrt{1-x} .
\end{array}
$$
Then $z=\left(1-s^{2}\right) t+\left(1-t^{2}\right) s=(s+t)(1-s t)$.
Also, $(1-s)(1-t)=1-(s+t)+s t \geqslant 0$, which means $s+t \leqslant 1+s t$,
thus $z \leqslant(1+s t)(1-s t)=1-s^{2} t^{2} \leqslant 1$.
The maximum value of 1 is achieved when $x=1, y=0$ or $x=0, y=1$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$
|
$=, 7 .-1$. Therefore $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4028} x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4026}-a_{4007}-a_{4028}
$$
is $\qquad$
|
6.2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$.
Then $x^{2}+x+2=1$.
Therefore, $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{0}+a_{1} \omega^{2}+a_{2} \omega+a_{3}+a_{4} \omega^{2}+\cdots+ \\
a_{4026}+a_{4027} \omega^{2}+a_{4028} \omega=1 .
\end{array}
$$
Adding the above two equations, we get
$$
\begin{array}{l}
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+ \\
2 a_{405}-a_{4027}-a_{4028}=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A line $l$ is drawn through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersecting the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly three lines $l$, then $\lambda=$ $\qquad$.
|
2.4.
Since there are an odd number of lines that satisfy the condition, by symmetry, the line perpendicular to the $x$-axis satisfies the condition, at this time,
$$
x=\sqrt{3}, y= \pm 2, \lambda=|A B|=4 \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In an All-Star basketball game, 27 players participate, each wearing a jersey with their favorite number, which is a non-negative integer. After the game, they line up in a 3-row, 9-column formation for fans to take photos. An eccentric fan only takes photos where the players in the frame form a rectangle of $a$ rows and $b$ columns $(1 \leqslant a \leqslant 3, 1 \leqslant b \leqslant 9)$ (with the rows and columns aligned with the original formation), and the sum of the jersey numbers of the players in the frame (for a single player, it is just their jersey number) is a multiple of 10. As a result, this fan only takes photos of $s$ players. Find the minimum possible value of $s$.
（Based on a problem from the 2011 National High School Mathematics Competition）
|
Prompt: $s_{\text {min }}=2$.
Assume at most one player is photographed.
By symmetry, without loss of generality, assume no players in the first row are photographed.
For $i=1,2, \cdots, 9$, let the players in the $i$-th column of the 1st, 2nd, and 3rd rows have jersey numbers $a_{i}, b_{i}, c_{i}$, respectively, and denote
$$
S_{k}=\sum_{i=1}^{k} a_{i}, T_{k}=\sum_{i=1}^{k}\left(b_{i}+c_{i}\right),
$$
where $k=0,1, \cdots, 9$.
Considering the 1st row, the 2nd to 3rd rows, and the 1st to 3rd rows, respectively, it can be proven that: $S_{0}, S_{1}, \cdots, S_{9}, T_{0}, T_{1}, \cdots, T_{9}$, and $S_{0}+T_{0}, S_{1}+T_{1}, \cdots, S_{9}+T_{9}$ all form a complete residue system modulo 10, which will lead to a contradiction.
On the other hand, suppose the players choose the numbers as shown in Table 1, it can be verified that no player with a non-zero number can be photographed. In this case, the fan exactly photographs 2 players.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 & 0 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 2 \\
\hline
\end{tabular}
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \). ${ }^{[4]}$
(2011, Hebei Province High School Mathematics Competition)
|
【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.
First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-3 x^{3}+3 x^{2}\right)-\left(y^{4}-3 y^{3}+3 y^{2}\right) .
\end{array}
$$
At this point, it is easy to construct the function
$$
g(x)=x^{4}-3 x^{3}+3 x^{2} \text {. }
$$
Equation (1) can be rewritten as
$$
f(x, y)=\frac{g(x)-g(y)}{x-y} \text {. }
$$
Thus, $f(x, y)$ can be viewed as the slope of the line connecting two points on the graph of $g(x)$.
Furthermore, when $x=y$,
$$
f(x, y)=4 x^{3}-9 x^{2}+6 x \text {. }
$$
Therefore, we only need to find the minimum value of the derivative function $h(x)=4 x^{3}-9 x^{2}+6 x$ of $g(x)$ for $x \geqslant \frac{1}{2}$.
It is easy to find that when $x \geqslant \frac{1}{2}$, the minimum value of $h(x)=4 x^{3}-9 x^{2}+6 x$ is 1.
Thus, the minimum value of $f(x, y)$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$
Determine the periodicity of the sequence $\left\{a_{n}\right\}$.
|
Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$
which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13}{3 a_{n}-5}, \\
a_{n+3}=\frac{5 a_{n+2}-13}{3 a_{n+2}-7}=a_{n},
\end{array}
$$
which means $a_{n+3}=a_{n}(n=1,2, \cdots)$.
In summary, the sequence $\left\{a_{n}\right\}$ is a purely periodic sequence, with the smallest positive period being 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2 . \\
\text { Then } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=
\end{array}
$$
|
7.3.
From the given, we have $a b+b c+c a=1$.
Then $b c=1-a b-a c=1-a(b+c)$ $=1-a(2-a)=(1-a)^{2}$.
Thus, the desired value is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
435 Given a convex polygon $F$, consider all the figures that are positively homothetic to the convex polygon $F$ and smaller than $F$. Let $n(F)$ be the minimum number of such figures (allowing translation but not rotation) needed to cover the convex polygon $F$. Find the value of $n(F)$.
|
(1) On the one hand, let the convex polygon $F$ be the parallelogram $ABCD$. It is easy to see that the convex polygon $F$ can be covered by four smaller parallelograms that are similar to the convex polygon $F$ (as shown in Figure 3).
On the other hand, for any smaller parallelogram $F_1$ that is similar to the convex polygon $F$, if $F_1$ contains point $A$, then $F_1$ cannot contain any other vertices of the convex polygon $F$ (as shown in Figure 4). This indicates that the convex polygon $F$ cannot be covered by fewer than four smaller and similar figures.
Therefore, $n(F)=4$.
(2) Let the polygon $F$ not be a parallelogram. Then $n(F) \geqslant 3$.
In fact, if the convex polygon $F$ can be covered by two smaller and similar figures, let's assume these two figures are of equal size; otherwise, the smaller one can be enlarged. Establish a Cartesian coordinate system such that the corresponding vertices of these two figures have the same y-coordinates. Consider the vertices of the convex polygon $F$ with the maximum and minimum y-coordinates, which cannot be covered by these two figures simultaneously. Hence, $n(F) \geqslant 3$.
Next, we prove that three figures are sufficient.
First, we prove a lemma.
Lemma There exists a triangle $T$ that contains the convex polygon $F$, and each side of $T$ contains a side of the convex polygon $F$.
Proof If the convex polygon $F$ is a triangle, then let $T=F$.
If the convex polygon $F$ is not a triangle, then there exist two sides of the convex polygon $F$ that are neither parallel nor intersecting. Extend these two sides until they intersect (as shown in Figure 5), resulting in a polygon $F_1$ that contains the convex polygon $F$, and the number of sides of $F_1$ is less than that of the convex polygon $F$, with each side of $F_1$ containing a side of the convex polygon $F$.
Repeating this process, the final figure $F'$ will inevitably be a parallelogram or a triangle (only these two figures do not have sides that are neither parallel nor intersecting), and $F'$ contains the convex polygon $F$, with each side of $F'$ containing a side of the convex polygon $F$.
If $F'$ is a triangle, then let $T=F'$.
If $F'$ is a parallelogram $ABCD$, since the convex polygon $F$ is not a parallelogram, $F'$ must have at least one vertex that is not a vertex of the convex polygon $F$, let's say point $A$. Let point $P$ be the point on line segment $AD$ closest to point $A$. Then $P$ is a vertex of the convex polygon $F$, and one of its sides lies on line segment $AD$. Let the other side of the convex polygon $F$ from point $P$ intersect line segment $AB$ at point $Q$. Then the triangle $T$ formed by lines $PQ$, $BC$, and $CD$ contains the convex polygon $F$, and each side of $T$ contains a side of the convex polygon $F$ (as shown in Figure 6).
Back to the original problem.
Using the lemma, assume the convex polygon $F$ is contained in $\triangle ABC$, and the sides $A_1A_2$, $B_1B_2$, and $C_1C_2$ of the convex polygon $F$ lie on segments $BC$, $AC$, and $AB$ respectively (as shown in Figure 7).
Choose any point $O$ inside the convex polygon $F$ and internal points $X$, $Y$, and $Z$ on sides $A_1A_2$, $B_1B_2$, and $C_1C_2$ respectively. Then segments $OX$, $OY$, and $OZ$ divide the convex polygon $F$ into three polygons $F_1$, $F_2$, and $F_3$. Assume polygon $F_1$ is inside quadrilateral $AZOY$.
By the selection of points $O$, $Y$, and $Z$, if $0 < k < 1$ and $k$ is infinitely close to 1, then the homothety $\varphi_1$ with center $A$ and ratio $k$ satisfies $\varphi_1$ (convex polygon $F \cup$ quadrilateral $AZOY$) contains quadrilateral $AZOY$.
Note that the part of quadrilateral $AZOY$ outside the convex polygon $F$ becomes a subset of itself after the homothety, so $\varphi_1(F)$ covers $F_1$.
Similarly, there exist two homotheties $\varphi_2$ and $\varphi_3$ with ratios less than 1 such that $\varphi_2(F)$ covers $F_2$ and $\varphi_3(F)$ covers $F_3$.
Thus, the convex polygon $F$ is covered by the union of $\varphi_1(F)$, $\varphi_2(F)$, and $\varphi_3(F)$.
Therefore, $n(F)=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Add a “+” or “-” in front of each number in $1,2, \cdots, 1989$. Find the minimum non-negative algebraic sum, and write down the equation.
|
First, prove that the algebraic sum is odd.
Consider the simplest case: all filled with " + ", at this time,
$$
1+2+\cdots+1989=995 \times 1989
$$
is odd.
For the general case, it is only necessary to adjust some " + " to " - ".
Since $a+b$ and $a-b$ have the same parity, each adjustment does not change the parity of the algebraic sum, i.e., the total sum remains odd.
$$
\begin{array}{l}
\text { and } 1+(2-3-4+5)+(6-7-8+9)+ \\
\cdots+(1986-1987-1988+1989)=1 .
\end{array}
$$
Therefore, this minimum value is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum_{c y c} \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$
|
To prove that since the inequality to be proved is homogeneous with respect to $a, b, c$, we may assume without loss of generality that $a+b+c=3$.
The original inequality is then $\sum_{\text {cyc }} \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant 8$.
Consider the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{2}}(0<x<3)$.
Notice that,
$$
\begin{array}{l}
f(x)=\frac{x^{2}+6 x+9}{3\left(x^{2}-2 x+3\right)}=\frac{1}{3}\left[1+\frac{8 x+6}{(x-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{8 x+6}{2}\right)=\frac{4 x+4}{3} . \\
\text { Therefore, } \sum_{\text {cyc }} f(a) \leqslant \sum_{\text {cyc }} \frac{4 a+4}{3} \\
=\frac{4(a+b+c)+12}{3}=8 .
\end{array}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \\
\leqslant 60 .
\end{array}
$$
Then $x+y-z=$ . $\qquad$
|
8. 0 .
Original expression
$$
\begin{array}{l}
=\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \\
\leqslant 60 \\
\Rightarrow x=-2, y=5, z=3 \\
\Rightarrow x+y-z=0 .
\end{array}
$$
|
0
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $S_{n}=1+2+\cdots+n$. Then among $S_{1}, S_{2}$, $\cdots, S_{2015}$, there are. $\qquad$ that are multiples of 2015.
|
10.8 .
Obviously, $S_{n}=\frac{1}{2} n(n+1)$.
For any positive divisor $d$ of 2015, it is easy to see that in the range $1 \sim 2015$, there is exactly one $n$ that satisfies $n$ being a multiple of $d$ and $n+1$ being a multiple of $\frac{2015}{d}$. Therefore, each divisor of 2015 will generate one $n$ such that $S_{n}$ is a multiple of 2015.
Since $2015=5 \times 13 \times 31$, 2015 has a total of eight divisors.
Thus, the desired result is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$
then the value of $x+\frac{1}{x}$ is
|
2.3.
Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{array}{l}
x^{2}+\frac{1}{x^{2}}=b^{2}-2, \\
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right) \\
=b\left(b^{2}-3\right)=b^{3}-3 b .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x^{5}+\frac{1}{x^{5}}\right)+\left(x+\frac{1}{x}\right) . \\
\text { Hence }\left(b^{2}-2\right)\left(b^{3}-3 b\right)=123+b \\
\Rightarrow b^{5}-5 b^{3}+5 b-123=0 .
\end{array}
$$
Since $123=3 \times 41$, when $b=3$, the left and right sides of equation (1) are equal. Therefore, $b=3$ is a solution to equation (1).
Dividing the left side of equation (1) by $b-3$, we get the quotient
$$
b^{4}+3 b^{3}+16 b+41 \text {. }
$$
Since this quotient is always greater than 0, we know $b=3$, i.e., $x+\frac{1}{x}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2011}{d}}$ is $\qquad$ .
|
9.7.
Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array}
$$
From $G\left(3^{4}\right)=1$, we know
$$
\begin{aligned}
G\left(13^{155}\right) & =G\left(3^{155}\right)=G\left(3^{4 \times 38+3}\right) \\
=G\left(3^{3}\right) & =7, \\
G\left(403^{5}\right) & =G\left(3^{5}\right)=3 .
\end{aligned}
$$
Thus, the maximum value of the unit digit of $d^{\frac{20115}{d}}$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given sets
$$
A=\{1,3,5,7,9\}, B=\{2,4,6,8,10\} \text {. }
$$
If set $C=\{x \mid x=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is $\qquad$ .
|
,- 1.9 .
Since $a$ is odd and $b$ is even, all elements in set $C$ are odd.
Also, the minimum value of $a+b$ is 3, and the maximum value is 19, and all odd numbers between 3 and 19 can be obtained, thus, set $C$ contains 9 elements.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function
$$
f(x)=\left\{\begin{array}{ll}
0, & x<0, \\
1, & x \geqslant 0 .
\end{array}\right.
$$
Then $f(f(x))=$ $\qquad$
|
2. 1 .
Since for any $x \in \mathbf{R}$, we have $f(x) \geqslant 0$, therefore, $f(f(x))=1$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given
$$
\sin \alpha+\sqrt{3} \sin \beta=1, \cos \alpha+\sqrt{3} \cos \beta=\sqrt{3} \text {. }
$$
Then the value of $\cos (\alpha-\beta)$ is $\qquad$ .
|
3. 0 .
Squaring and adding the two known equations, we get
$$
\begin{array}{l}
4+2 \sqrt{3}(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=4 \\
\Rightarrow \cos (\alpha-\beta)=0
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For a sequence of real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define its "value" as $\max _{1 \leqslant i \leqslant n}\left\{\left|x_{1}+x_{2}+\cdots+x_{i}\right|\right\}$. Given $n$ real numbers, David and George want to arrange these $n$ numbers into a sequence with low value. On one hand, diligent David examines all possible ways to find the minimum possible value $D$. On the other hand, greedy George chooses $x_{1}$ such that $\left|x_{1}\right|$ is as small as possible. From the remaining numbers, he chooses $x_{2}$ such that $\left|x_{1}+x_{2}\right|$ is as small as possible, $\cdots \cdots$ At the $i$-th step, he chooses $x_{i}$ from the remaining numbers such that $\left|x_{1}+x_{2}+\cdots+x_{i}\right|$ is as small as possible. At each step, if there are multiple numbers that give the same minimum absolute sum, George arbitrarily chooses one. Finally, the value of the sequence he gets is $G$. Find the smallest constant $c$, such that for every positive integer $n$, every array of $n$ real numbers, and every sequence George can obtain, we have
$$
G \leqslant c D \text {. }
$$
|
3. $c=2$.
If the initial numbers given are $1, -1, 2, -2$, then David arranges these four numbers as $1, -2, 2, -1$, and George can get the sequence $1, -1, 2, -2$.
Thus, $D=1, G=2$.
Therefore, $c \geqslant 2$.
We now prove: $G \leqslant 2 D$.
Let the $n$ real numbers be $x_{1}, x_{2}, \cdots, x_{n}$, and assume David and George have already obtained their arrangements. Suppose David and George get the sequences $d_{1}, d_{2}, \cdots, d_{n}$ and $g_{1}, g_{2}, \cdots, g_{n}$, respectively.
$$
\begin{array}{c}
\text { Let } M=\max _{1<i \leqslant n}\left\{\left|x_{i}\right|\right\}, \\
S=\left|x_{1}+x_{2}+\cdots+x_{n}\right|, \\
N=\max \{M, S\} .
\end{array}
$$
Then $D \geqslant S$,
$D \geqslant \frac{M}{2}$,
$G \leqslant N=\max \{M, S\}$.
From equations (1), (2), and (3), we get
$$
G \leqslant \max \{M, S\} \leqslant \max \{M, 2 S\} \leqslant 2 D \text {. }
$$
In fact, by the definition of value, equation (1) holds.
For equation (2), consider an index $i$ such that $\left|d_{i}\right|=M$.
Thus, $M=\left|d_{i}\right|$
$=\left|\left(d_{1}+d_{2}+\cdots+d_{i}\right)-\left(d_{1}+d_{2}+\cdots+d_{i-1}\right)\right|$
$\leqslant\left|d_{1}+d_{2}+\cdots+d_{i}\right|+\left|d_{1}+d_{2}+\cdots+d_{i-1}\right|$
$\leqslant 2$.
Therefore, equation (2) holds.
Finally, we prove that equation (3) holds.
Let $h_{i}=g_{1}+g_{2}+\cdots+g_{i}$.
We use mathematical induction on $i$ to prove $\left|h_{i}\right| \leqslant N$.
When $i=1$, $\left|h_{1}\right|=\left|g_{1}\right| \leqslant M \leqslant N$, so the conclusion holds.
Notice that, $\left|h_{n}\right|=S \leqslant N$, so the conclusion also holds.
Assume $\left|h_{i-1}\right| \leqslant N$, and consider two cases.
(1) Assume that $g_{i}, g_{i+1}, \cdots, g_{n}$ do not have two terms with opposite signs. Without loss of generality, assume they are all non-negative.
Then $h_{i-1} \leqslant h_{i} \leqslant \cdots \leqslant h_{n}$.
Thus, $\left|h_{i}\right| \leqslant \max \left\{\left|h_{i-1}\right|,\left|h_{n}\right|\right\} \leqslant N$.
(2) In $g_{i}, g_{i+1}, \cdots, g_{n}$, there are positive and negative numbers, so there exists an index $j \geqslant i$ such that $h_{i-1} g_{j} \leqslant 0$.
By the definition of the sequence George gets,
$$
\begin{array}{l}
\left|h_{i}\right|=\left|h_{i-1}+g_{i}\right| \leqslant\left|h_{i-1}+g_{j}\right| \\
\leqslant \max \left\{\left|h_{i-1}\right|,\left|g_{j}\right|\right\} \leqslant N .
\end{array}
$$
Thus, the conclusion holds.
This shows that equation (3) also holds.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the set $S=\{1,2,3, \cdots, 16\}$ is arbitrarily divided into $n$ subsets, then there must exist a subset in which there are elements $a, b, c$ (which can be the same), satisfying $a+b=c$. Find the maximum value of $n$.
【Note】If the subsets $A_{1}, A_{2}, \cdots, A_{n}$ of set $S$ satisfy the following conditions:
(1) $A_{i} \neq \varnothing(i=1,2, \cdots, n)$;
(2) $A_{i} \cap A_{j}=\varnothing$;
(3) $\bigcup_{i=1}^{n} A_{i}=S$,
then $A_{1}, A_{2}, \cdots, A_{n}$ is called a partition of set $S$.
（Proposed by the Problem Committee）
|
4. First, when $n=3$, assume there exists a partition of the set that does not satisfy the condition. Thus, there must be a subset with at least six elements, without loss of generality, let
$$
A=\left\{x_{1}, x_{2}, \cdots, x_{6}\right\}\left(x_{1}<x_{2}<\cdots<x_{6}\right) \text {. }
$$
Then $x_{6}-x_{1}, x_{6}-x_{2}, \cdots, x_{6}-x_{5} \notin A$. Among these, there must be three elements belonging to another subset, let
$$
x_{6}-x_{i}, x_{6}-x_{j}, x_{6}-x_{k} \in B(1 \leqslant i<j<k \leqslant 5) \text {. }
$$
Then $x_{j}-x_{i}, x_{k}-x_{j}, x_{k}-x_{i} \notin A \cup B$, but they cannot all belong to the third subset, leading to a contradiction.
Therefore, $n=3$ satisfies the condition.
Second, when $n=4$, the partition
$$
\{1\},\{2,3\},\{4,5,6,7,16\},\{8,9, \cdots, 15\}
$$
does not satisfy the condition.
In summary, $n_{\max }=3$.
[Note] $\{1,4,7,10,13,16\},\{2,3,8,9,14,15\}$, $\{5,6\},\{11,12\}$, etc., are also valid.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A shape obtained by removing one unit square from a $2 \times 2$ grid is called an "L-shape". In an $8 \times 8$ grid, place $k$ L-shapes, each of which can be rotated, but each L-shape must cover exactly three unit squares of the grid, and any two L-shapes must not overlap. Additionally, no other L-shape can be placed besides these $k$ L-shapes. Find the minimum value of $k$.
(Zhang Yuezhu, Contributed)
|
6. On one hand, using six straight lines to divide the large square grid into 16 smaller $2 \times 2$ grids, each $2 \times 2$ grid must have at least two cells covered by an L-shape. Therefore, the number of cells covered is no less than 32. Thus, the number of L-shapes is no less than $\left[\frac{32}{3}\right]+1$ $=11$, meaning no fewer than 11 L-shapes.
On the other hand, placing 11 L-shapes as shown in Figure 4 can meet the requirements of the problem.
In conclusion, $k_{\min }=11$.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$, both satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$
|
2. 2 .
From the problem, we know that for any $A(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking
$$
(x, y)=(1,0),(0,1),(-1,-1) \text {, }
$$
we get that the fixed point $B(a, b)$ satisfies the necessary conditions
$$
\left\{\begin{array}{l}
a \leqslant 1, \\
b \leqslant 1, \\
a+b \geqslant-1
\end{array} \quad \Rightarrow a+b \leqslant 2\right. \text {. }
$$
For the fixed point $B(1,1)$, for any $A(x, y) \in D$, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=(1,1) \cdot(x, y)=x+y \leqslant 1 \text {. }
$$
When $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$, the equality holds.
Thus, the point $B(1,1)$ satisfies the conditions of the problem, and in this case, $a+b=2$.
Therefore, $(a+b)_{\max }=2$.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Define the function on $\mathbf{R}$
$$
f(x)=\left\{\begin{array}{ll}
\log _{2}(1-x), & x \leqslant 0 ; \\
f(x-1)-f(x-2), & x>0 .
\end{array}\right.
$$
Then $f(2014)=$
|
3. 1.
First, consider the method for finding the function values on the interval $(0,+\infty)$.
If $x>3$, then
$$
\begin{array}{l}
f(x)=f(x-1)-f(x-2) \\
=f(x-1)-(f(x-1)+f(x-3)) \\
=-f(x-3) .
\end{array}
$$
Thus, for $x>6$, we have
$$
f(x)=-f(x-3)=f(x-6) \text {. }
$$
Therefore, $f(x)$ is a periodic function with a period of 6 on the interval $(0,+\infty)$.
$$
\begin{array}{l}
\text { Hence } f(2014)=f(4)=f(3)-f(2) \\
=-f(0)+f(-1)=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The set $X \backslash Y=\{a \mid a \in X, a \notin Y\}$ is called the difference set of set $X$ and set $Y$. Define the symmetric difference of sets $A$ and $B$ as
$$
A \Delta B=(A \backslash B) \cup(B \backslash A) \text {. }
$$
If two non-empty finite sets $S$ and $T$ satisfy $|S \Delta T|=1$, then the minimum value of $k=|S|+|T|$ is $\qquad$
|
6. 3 .
If $|S \Delta T|=1$, without loss of generality, assume
$$
\left\{\begin{array}{l}
|S \backslash T|=1, \\
|T \backslash S|=0,
\end{array}\right.
$$
i.e., $T$ is a proper subset of $S$, thus,
$$
\begin{array}{l}
|S|>|T| \geqslant 1 . \\
\text { Hence } k \geqslant|T|+1+|T| \geqslant 3 .
\end{array}
$$
$$
\text { Construct sets } S=\{1,2\}, T=\{2\} \text {. }
$$
Clearly, the condition $|S \Delta T|=1$ is satisfied, and
$$
\begin{array}{l}
k=|S|+|T|=3 . \\
\text { Therefore, } k_{\min }=3 .
\end{array}
$$
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there always exists a row where the cells intersecting with these two columns are both filled with 1. Let $x=\max _{1 \leqslant i \leqslant m}\left\{x_{i}\right\}$. Then $x_{\text {min }}=$ $\qquad$
|
8. 5 .
Construct an application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."
Let the $i(i=1,2, \cdots, m)$-th kind of book be purchased by $x_{i}$ people, $x=\max _{1 \leqslant i \leqslant m} x_{i}$. Then
$$
\sum_{i=1}^{m} x_{i}=30 \text {, and } x_{i} \leqslant x(i=1,2, \cdots, m) \text {. }
$$
Considering the total number of "same book pairs" between any two people, we get the inequality
$$
\begin{array}{l}
\sum_{i=1}^{m} \mathrm{C}_{x_{i}}^{2} \geqslant \mathrm{C}_{10}^{2} \\
\Rightarrow \sum_{i=1}^{m} \frac{x_{i}\left(x_{i}-1\right)}{2} \geqslant 45 .
\end{array}
$$
Thus, from conclusion (1) and equation (2), we have
$$
\begin{array}{l}
45 \leqslant \frac{1}{2}(x-1) \sum_{i=1}^{m} x_{i}=15(x-1) \\
\Rightarrow x \geqslant 4 .
\end{array}
$$
If $x=4$, then all $x_{i}=4(i=1,2, \cdots, m)$, which contradicts $\sum_{i=1}^{m} x_{i}=30$. Therefore, $x \geqslant 5$.
Below, we construct a scenario where $x=5$, and the book purchased by the most people is bought by exactly 5 people.
Let the books be $B_{i}$, and the book purchasing situation for the 10 people is as follows:
$$
\begin{array}{l}
B_{1} B_{2} B_{3}, B_{3} B_{4} B_{5}, B_{1} B_{5} B_{6}, B_{1} B_{3} B_{5}, \\
B_{1} B_{2} B_{4}, B_{1} B_{4} B_{6}, B_{2} B_{4} B_{5}, B_{2} B_{5} B_{6}, \\
B_{2} B_{3} B_{6}, B_{3} B_{4} B_{6} .
\end{array}
$$
Therefore, the book purchased by the most people is bought by at least 5 people.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a, b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is
|
From the given, we have $a b=-1, a+b=1$,
$$
\begin{array}{l}
a^{2}=a+1, b^{2}=b+1 . \\
\text { Therefore, } 3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2} \\
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11 .
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a$ and $b$ be distinct real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ .
|
$$
-, 1.4
$$
From the given conditions and the symmetry of the quadratic function graph, we get
$$
\begin{array}{l}
\frac{a+b}{2}=-\frac{a}{2} \Rightarrow 2 a+b=0 \\
\Rightarrow f(2)=4+2 a+b=4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then $\frac{1}{\sin \alpha}+\cos ^{4} \alpha=$ $\qquad$
|
2. 2 .
From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$.
$$
\begin{array}{l}
\text { Then } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha \\
=(1+\sin \alpha)+\left(1-\cos ^{2} \alpha\right) \\
=2+\sin \alpha-\cos ^{2} \alpha=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Then the sum of the squares of all natural numbers $m$ that satisfy
$$
[(2 m+1)\{\sqrt{2 m+1}\}]=m
$$
is $\qquad$ .
|
8.0.
Let $2 m+1=n$. Then $2[n\{\sqrt{n}\}]=n-1$. Note that,
$$
\begin{array}{l}
2 n\{\sqrt{n}\}-21$ when, by $n=2[n\{\sqrt{n}\}]+1$ being odd, we know $n \geqslant 3$.
Thus, $\left|4 n-(2[\sqrt{n}]+1)^{2}\right| \leqslant \frac{4}{\sqrt{n}}+\frac{1}{n}<3$.
Therefore, $4 n-(2[\sqrt{n}]+1)^{2}= \pm 1$.
This leads to
$$
2 n-1=2[\sqrt{n}]([\sqrt{n}]+1)
$$
or $n=[\sqrt{n}]([\sqrt{n}]+1)$,
which is a contradiction.
In summary, the only positive integer $n$ that satisfies the condition is 1.
Thus, $m=0$. Hence, $m^{2}=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Color each cell of a $12 \times 12$ grid either black or white, such that any $3 \times 4$ or $4 \times 3$ subgrid contains at least one black cell. Find the minimum number of black cells.
（Liang Yingde provided the problem）
|
3. The minimum number of black cells required is $n=12$.
First, prove that $n \geqslant 12$.
Since a $12 \times 12$ grid can be divided into $\frac{12 \times 12}{3 \times 4}=12$ non-overlapping $3 \times 4$ sub-grids (excluding the boundaries), according to the problem statement, there must be at least 12 black cells.
To prove that $n=12$, it is sufficient to construct an example that meets the conditions. See Figure 3.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of distinct points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is $\qquad$
|
2. 4 .
Assume the vertices corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$ are arranged in a counterclockwise direction.
Let $z_{1}=\mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\begin{array}{l}
z_{k+1}=\mathrm{e}^{\mathrm{i}\left(\theta+\frac{k}{10}\right)}(k=0,1, \cdots, 19), \\
z_{k+1}^{2015}=\mathrm{e}^{\mathrm{i}\left(2015 \theta+\frac{20115 k}{10}\right)}=\mathrm{e}^{\mathrm{i}\left(2015 \theta-\frac{k}{2}\right)} .
\end{array}
$$
Thus, $\left\{z_{k+1}^{2015}\right\}$ is a periodic sequence with a period of 4. Therefore, the number of different points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.