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1. Given $S=1^{2}-2^{2}+3^{2}-4^{2}+\cdots-100^{2}+$ $101^{2}$. Then the remainder when $S$ is divided by 103 is
|
$=.1 .1$.
Notice that
$$
\begin{array}{l}
S=1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(101^{2}-100^{2}\right) \\
=1+2+3+\cdots+100+101 \\
=\frac{101 \times 102}{2}=5151=50 \times 103+1 .
\end{array}
$$
Therefore, the required remainder is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$
|
2. -1 .
Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$. Since $\triangle ABC$ is a right triangle, it is known that $x_{1}$ and $x_{2}$ must have opposite signs, so $x_{1} x_{2}=\frac{c}{a}<0$. By the projection theorem, we know that $|O C|^{2}=|A O| \cdot|B O|$, which means $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\left|\frac{c}{a}\right|$.
Therefore, $|a c|=1, a c=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y, z$ satisfy
$$
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \leqslant 60 \text {. }
$$
Then the value of $x+y-z$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
|
-.1.D.
From the given equation, we have
$$
\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \leqslant 60 \text {. }
$$
Therefore, $x=-2, y=5, z=3$.
Thus, $x+y-z=-2+5-3=0$.
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given integers $a, b, c, d$ satisfy
$$
27\left(3^{a}+3^{b}+3^{c}+3^{d}\right)=6888 \text {. }
$$
Then the value of $a+b+c+d$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
|
4.C.
Assume $a \leqslant b \leqslant c \leqslant d$. From the given equation, we have
$$
3^{a+3}\left(1+3^{b-a}+3^{c-a}+3^{d-a}\right)=3 \times 2296 \text {. }
$$
(1) If $3^{a+3}=1$, i.e., $a=-3$, then
$$
31\left(1+3^{b+3}+3^{c+3}+3^{d+3}\right)=3 \times 2296 \text {. }
$$
Since $3^{b+3} \leqslant 3^{c+3} \leqslant 3^{d+3}$, we have
$$
3^{b+3}+3^{c+3}=2 \text {. }
$$
This implies $b=c=-3$. Thus, $3^{d+2}=2295$.
This case has no solution.
$$
\begin{array}{l}
\text { (2) If } 3^{a+3}=3 \text {, i.e., } a=-2 \text {, then } \\
1+3^{b+2}+3^{c+2}+3^{d+2}=2296 \\
\Rightarrow 3^{b}\left(1+3^{c-b}+3^{d-b}\right)=3 \times 85 \\
\Rightarrow b=1 \Rightarrow 1+3^{c-1}+3^{d-1}=85 \\
\Rightarrow 3^{c}\left(1+3^{d-c}\right)=3^{2} \times 28 \Rightarrow c=2, d=5 .
\end{array}
$$
Therefore, $a+b+c+d=-2+1+2+5=6$.
|
6
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $x$ is a real number. Then the maximum value of $\sqrt{2000-x}+$ $\sqrt{x-2000}$ is $\qquad$ .
|
4.4.
$$
\begin{array}{l}
\text { Let } t_{1}=\sqrt{2008-x}+\sqrt{x-2000} \text {. Then } \\
t_{1}^{2}=8+2 \sqrt{(2008-x)(x-2000)} \\
\leqslant 8+8=16 .
\end{array}
$$
Therefore, $t \leqslant 4$, i.e., $t_{\max }=4$. At this point, $x=2004$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 2, each segment of the broken line $A-B-C-D$ is parallel to the sides of the rectangle, and it divides the rectangle into two equal areas. Point $E$ is on the side of the rectangle such that segment $A E$ also bisects the area of the rectangle.
Given that segment $A B=30, B C=$
$24, C D=10$. Then $D E=$
$\qquad$
|
8.12.
Solution 1: Let line segment $A E$ intersect $B C$ at point $M$, and draw $E P \perp B C$, with the foot of the perpendicular being $P$.
Let $E D=x, B M=y$. Then $M P=24-x-y$.
By the problem, $S_{\triangle B B Y}=S_{\text {quadrilateral } U C D E}$, then
$15 y=120-5 y+5 x$.
Thus, $4 y=24+x$.
Also, $\triangle A M B \backsim \triangle E M P$, so $\frac{A B}{E P}=\frac{B M}{M P}$.
Therefore, $\frac{30}{10}=\frac{y}{24-x-y}$.
Hence, $4 y=72-3 x$.
From equations (1) and (2), we get $x=12$.
Solution 2: On both sides of the original rectangle, subtract equal-area smaller rectangles from each part so that $C D$ lies on the edge of the remaining rectangle (as shown in Figure 10). At this point, the broken line $A-B-C$ also divides this smaller rectangle into two equal-area parts. When both are further reduced by the area of the shaded part, they are still equal, i.e.,
$$
30 \times 24-10 \times 24=A G \times 40 \Rightarrow A G=12 .
$$
Since $A E$ also divides this smaller rectangle into two equal-area parts, we have $A G=E D=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (40 points) A rectangular box with dimensions $a_{1} \times b_{1} \times c_{1}$ can fit into another rectangular box with dimensions $a_{2} \times b_{2} \times c_{2}$ if and only if $a_{1} \leqslant a_{2} 、 b_{1} \leqslant b_{2} 、 c_{1} \leqslant c_{2}$. Therefore, among the rectangular boxes with dimensions $a \times b \times c$ (where $a 、 b 、 c$ are integers and $1 \leqslant a \leqslant b \leqslant c \leqslant 5$), what is the maximum number of boxes that can be selected such that no box can fit into another?
|
4. Among the boxes that meet the conditions, there are 5 types of boxes that are cubes, and 20 types of boxes that are square-based but not cubic prisms, because they have five different heights and four different choices for the square side length.
In addition, there are 10 types of boxes where the length, width, and height are all different, because three can be chosen from five lengths, resulting in $\frac{5 \times 4 \times 3}{3 \times 2}$ types. Table 2 categorizes these 35 types of boxes based on their length + width + height values. Within the same category, boxes with smaller length + width + height values can fit into boxes with larger length + width + height values.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline Length+Width+Height & First Class Second Class Third Class Fourth Class Fifth Class \\
\hline 15 & 555 & & & & \\
\hline 14 & 455 & & & & \\
\hline 13 & 355 & 445 & & & \\
\hline 12 & 255 & 345 & & 444 & \\
\hline 11 & 155 & 245 & 335 & 344 & \\
\hline 10 & 145 & 244 & 235 & 334 & \\
\hline 9 & 135 & 144 & 225 & 234 & 333 \\
\hline 8 & & 134 & 125 & 224 & 233 \\
\hline 7 & & 124 & 115 & 223 & 133 \\
\hline 6 & & & 114 & 222 & 123 \\
\hline 5 & & & 113 & & 122 \\
\hline 4 & & & & & 112 \\
\hline 3 & & & & & 111 \\
\hline
\end{tabular}
One box with a length + width + height value of 9 can be selected from each category, i.e., $(1,3,5)$, $(1,4,4)$, $(2,2,5)$, $(2,3,4)$, and $(3,3,3)$. Each of these boxes has at least one dimension that is larger than the others, making it impossible for them to fit inside each other. Therefore, the maximum number of boxes that can be selected is 5.
Note: Giving an example with 4 boxes earns 10 points, giving an example with 5 boxes earns 20 points. Simply stating 5 boxes without giving an example earns 0 points. Proving the maximum of 5 boxes earns 20 points.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.(40 points) As shown in Figure 7, quadrilateral $ABCD$ is inscribed in a circle, $AB=AD$, and its diagonals intersect at point $E$. Point $F$ lies on segment $AC$ such that $\angle BFC = \angle BAD$. If $\angle BAD = 2 \angle DFC$, find the value of $\frac{BE}{DE}$.
保留源文本的换行和格式，直接输出翻译结果。
|
7. From $AB=AD$, we know $\angle ABD=\angle ADB=\theta$. By the property of equal arcs subtending equal angles at the circumference, we have
$$
\angle ACD=\angle ACB=\theta.
$$
Let $\angle DFC=\varphi$. Then
$$
\angle BAD=\angle BFC=2\varphi.
$$
Therefore, $\angle ABD + \angle ADB + \angle BAD$
$$
=\theta + \theta + 2\varphi = 180^{\circ}.
$$
Thus, $\theta + \varphi = 90^{\circ}$, and $\angle CDF = 90^{\circ}$.
On the other hand, from
$$
\begin{array}{l}
\angle FBC = 180^{\circ} - \theta - 2\varphi = \theta = \angle FCB \\
\Rightarrow FB = FC.
\end{array}
$$
Let $M$ be the midpoint of side $BC$, and connect $FM$. It is easy to see that $\triangle FCD \cong \triangle FBM$, and $BC = 2CD$.
Since $AC$ is the angle bisector of $\angle BCD$, by the Angle Bisector Theorem, we have $\frac{BE}{DE} = \frac{BC}{CD} = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A sequence of numbers, the first three numbers are $1, 9, 9$, and each subsequent number is the remainder of the sum of the three preceding numbers divided by 3. What is the 1999th number in this sequence?
|
(Tip: Apart from the first three numbers $1,9,9$, this sequence repeats every 13 terms (i.e., $1,1,2,1,1,1,0,2,0,2,1,0,0$). Since $1999-3=13 \times 153+7$, the 1999th number is the 7th number in the 154th cycle, which is exactly 0.)
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Several 1s and 2s are arranged in a row
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$
The rule is: the 1st number is 1, the 2nd number is 2, the 3rd number is 1, ... Generally, first write a row of 1s, then insert $k$ 2s between the $k$th 1 and the $(k+1)$th 1 ($k=1$, $2, \cdots$). Try to answer:
(1) Is the 2005th number 1 or 2?
|
Explanation: Clearly, the position of 1 is somewhat special. For the convenience of calculation, we might as well divide this sequence of numbers into $n$ groups: the 1st group has 1 number, the 2nd group has 2 numbers, the 3rd group has 3 numbers, ..., the $n$th group has $n$ numbers, and the last number of each group is 1, while the rest are 2, i.e.,
(1), $(2,1),(2,2,1), \cdots,(\underbrace{2,2, \cdots, 2,1}_{n-1})$.
Thus, to determine whether the 2005th number is 1 or 2, we just need to use the Gauss summation formula to find out which group the 2005th number is in.
Below, we will use the trial method to estimate.
These $n$ groups have a total of $1+2+\cdots+n=\frac{n(n+1)}{2}$ numbers.
When $n=62$, there are 1953 numbers;
When $n=63$, there are 2016 numbers.
From this, we can see that the 2005th number is in the 63rd group, and it is not the last number.
Therefore, the 2005th number is 2.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure $2, \odot O$ is internally tangent to $\odot O^{\prime}$ at point $P, \odot O$'s chord $A B$ is tangent to $\odot O^{\prime}$ at point $C$, and $A B / / O O^{\prime}$. If the area of the shaded part is $4 \pi$, then the length of $A B$ is $\qquad$
|
4.4.
As shown in Figure 10, connect $O^{\prime} C$ and $O A$. Draw $O D \perp A B$ at $D$. Then quadrilateral $O O^{\prime} C D$ is a rectangle.
Therefore, $O^{\prime} C=O D$.
Thus, $S_{\text {fill }}$
$=\pi O A^{2}-\pi O^{\prime} C^{2}$ $=\pi\left(O A^{2}-O D^{2}\right)=\pi A D^{2}=4 \pi$.
So, $A D=2$.
By the Perpendicular Diameter Theorem, $A B=2 A D=2 \times 2=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The digit at the 2007th position after the decimal point of the irrational number $0.2342343423434342343434342 \cdots$ is $\qquad$ .
|
(Observation: Note the position of the digit 2 after the decimal point. The $n$th 2 is at the $n^{2}$th position after the decimal point. Since $44^{2}<2007<45^{2}$, the digit at the 2007th position after the decimal point is between the 44th and 45th 2, with 44 instances of 34 in between. The digit 3 is in the odd positions, and $2007-1936=71$, so the digit at the 2007th position is 3.)
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find four distinct natural numbers such that the sum of any two of them can be divided by their difference. If the sum of the largest and smallest of these four numbers is to be minimized, what is the sum of the middle two numbers?
(3rd Hua Luogeng Cup)
|
Analysis: Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four numbers that meet the conditions, and $a_{1}8$, which contradicts the condition that $a_{1}+a_{4}$ is the smallest.
In summary, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(2,3,4,6)$, the sum of the middle two numbers is 7. .
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If a natural number $N$ is appended to the right of any natural number, the resulting number can be divided by $N$ (for example, 2 appended to 35 results in 352, which is divisible by 2), then $N$ is called a "magic number". Among the natural numbers less than 130, how many magic numbers are there?
|
Analysis: To calculate how many magic numbers there are, we first need to clarify what constitutes a magic number. Although the problem provides the definition of a magic number, it is not convenient to directly use this definition to determine whether a number is a magic number. Therefore, we need to explore a simplified and equivalent condition.
Obviously, the key to simplification is how to accurately describe in mathematical language "a number appended to the right of a natural number." Imagine $P$ appended to the right of $Q$; what is the resulting number? How can this number be expressed in terms of $P$ and $Q$? This clearly depends on how many digits $P$ has. If $P$ is an $m$-digit number, then the number obtained by appending $P$ to the right of $Q$ is $Q \cdot 10^{m} + P$. Using this, we can derive a simplified and equivalent condition based on the definition of a magic number.
Solution: Let $P$ be a magic number, and assume it is an $m$-digit number.
According to the condition, for any natural number $Q$, we have
$$
P \mid\left(Q \cdot 10^{m} + P\right),
$$
which implies $P \mid Q \cdot 10^{m}$ (for any $Q \in \mathbf{N}$).
Taking $Q=1$, we get $P \mid 10^{m}$.
Conversely, if $P \mid 10^{m}$, then for any $Q \in \mathbf{N}$, we have $P \mid Q \cdot 10^{m}$.
Thus, $P \mid\left(Q \cdot 10^{m} + P\right)$.
Therefore, $P$ is an $m$-digit magic number if and only if $P \mid 10^{m}$.
(1) When $m=1$, $P \mid 10$, and $P$ is a one-digit number, so $P=1,2,5$.
(2) When $m=2$, $P \mid 10^{2}$, i.e., $P \mid\left(2^{2} \times 5^{2}\right)$, and $P$ is a two-digit number, so,
$$
P=2^{0} \times 5^{2}, 2^{1} \times 5^{1}, 2^{1} \times 5^{2}, 2^{2} \times 5^{1}.
$$
(3) When $m=3$, $P \mid 10^{3}$, i.e., $P \mid\left(2^{3} \times 5^{3}\right)$, and $P$ is a three-digit number, $P<130$, so,
$$
P=2^{0} \times 5^{3}, 2^{2} \times 5^{2}.
$$
In summary, there are $3+4+2=9$ magic numbers that meet the conditions.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
244 Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=ab+ac+ad+bc+bd+cd=3 \text{. }
$$
Find the maximum real number $k$, such that the inequality
$$
a+b+c+2ab+2bc+2ca \geqslant k d
$$
always holds.
|
Given:
$$
\begin{aligned}
& (a+b+c+d)^{2}=3^{2} \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 a c+2 a d+ \\
& 2 b c+2 b d+2 c d=9 \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}=3 .
\end{aligned}
$$
Let \( y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \).
Then \( y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \).
Since \( y \geqslant 0 \) and the coefficient of \( x^{2} \) is \( 3 > 0 \), we have:
$$
\Delta=[-2(a+b+c)]^{2}-4 \times 3\left(a^{2}+b^{2}+c^{2}\right)
$$
\(\leqslant 0\).
Thus, \( 4(3-d)^{2}-4 \times 3\left(3-d^{2}\right) \leqslant 0 \), which simplifies to:
$$
2 d^{2}-3 d \leqslant 0 \text {. }
$$
Solving this, we get \( 0 \leqslant d \leqslant \frac{3}{2} \) (when \( a=b=c=1 \), \( d=0 \); when \( a=b=c=\frac{1}{2} \), \( d=\frac{3}{2} \)).
From the given, \( a+b+c=3-d \),
$$
\begin{array}{l}
a b+b c+c a=3-d(a+b+c) \\
=3-d(3-d)=3-3 d+d^{2} .
\end{array}
$$
Then \( a+b+c+2 a b+2 b c+2 c a \)
$$
\begin{array}{l}
=3-d+2\left(3-3 d+d^{2}\right):=2 d^{2}-7 d+9 \\
=2\left(\frac{3}{2}-d\right)^{2}+3\left(\frac{3}{2}-d\right)+2 d
\end{array}
$$
Since \( 0 \leqslant d \leqslant \frac{3}{2} \), we have:
$$
2\left(\frac{3}{2}-d\right)^{2}+3\left(\frac{3}{2}-d\right) \geqslant 0 \text {. }
$$
Therefore, \( a+b+c+2 a b+2 b c+2 c a \geqslant 2 d \).
When \( a=b=c=\frac{1}{2}, d=\frac{3}{2} \), the equality holds.
Thus, the largest real number \( k \) such that the inequality
$$
a+b+c+2 a b+2 b c+2 c a \geqslant k d
$$
always holds is \( k=2 \).
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given the function $f(x)=x+\frac{t}{x}(t>0)$ and the point $P(1,0)$, draw two tangent lines $P M$ and $P N$ from $P$ to the curve $y=f(x)$, with the points of tangency being $M$ and $N$.
(1) Let $|M N|=g(t)$, find the expression for the function $g(t)$.
(2) Does there exist a $t$ such that $M$, $N$, and $A(0,1)$ are collinear? If so, find the value of $t$; if not, explain the reason.
(3) Under the condition in (1), if for any positive integer $n$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m}, a_{m+1}$ in the interval $\left[2, n+\frac{64}{n}\right]$ such that the inequality
$$
g\left(a_{1}\right)+g\left(a_{2}\right)+\cdots+g\left(a_{m}\right)<g\left(a_{m+1}\right)
$$
holds, find the maximum value of $m$.
|
Solution: (1) The equation of the chord of tangents $MN$ passing through the external point $P(1,0)$ of the curve $y=x+\frac{t}{x}(t>0)$ is
$$
y=2 x+2 t \text {. }
$$
Substituting into $y=x+\frac{t}{x}(t>0)$, we get $x^{2}+2 t x-t=0$.
Thus, $x_{1}+x_{2}=-2 t, x_{1} x_{2}=-t \ldots$
Therefore, $g(t)=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}+\frac{t}{x_{1}}-x_{2}-\frac{t}{x_{2}}\right)^{2}}$
$$
=\sqrt{5}\left|x_{1}-x_{2}\right|=2 \sqrt{5 t^{2}+5 t}(t>0) \text {. }
$$
(2) When $M, N$ and $A(0,1)$ are collinear, substituting $A(0,1)$ into $y=2 x+2 t$, we get $t=\frac{1}{2}$.
(3) It is easy to see that $g(t)$ is an increasing function on the interval $\left[2, n+\frac{64}{n}\right]$, then
$$
\begin{array}{l}
m g(2) \leqslant g\left(a_{1}\right)+g\left(a_{2}\right)+\cdots+g\left(a_{m}\right) \\
<g\left(a_{m+1}\right) \leqslant g\left(n+\frac{64}{n}\right)
\end{array}
$$
holds for all positive integers $n$, i.e.,
$$
m<\sqrt{\frac{1}{6}\left[\left(n+\frac{64}{n}\right)^{2}+\left(n+\frac{64}{n}\right)\right]}
$$
holds for all positive integers $n$.
Since $n+\frac{64}{n} \geqslant 16$, we have
$$
\sqrt{\frac{1}{6}\left[\left(n+\frac{64}{n}\right)^{2}+\left(n+\frac{64}{n}\right)\right]} \geqslant \sqrt{45 \frac{1}{3}} \text {. }
$$
Since $m$ is a positive integer, we take $m=\left[\sqrt{45 \frac{1}{3}}\right]=6$.
When $m=6$, there exist
$$
a_{1}=a_{2}=\cdots=a_{m}=2, a_{m+1}=16 \text {, }
$$
satisfying the condition for all $n$, thus, the maximum value of $m$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Choose $n$ numbers from $1,2, \cdots, 9$, among which there must be several numbers (at least one, or all), the sum of which can be divisible by 10. Find the minimum value of $n$.
(2008, National Junior High School Mathematics Competition)
|
Solution: When $n \leqslant 4$, since no combination of $n$ numbers from $1,3,5,8$ can sum to a multiple of 10, we have $n \geqslant 5$.
Below, we prove that $n=5$ meets the requirement, i.e., we prove that from $1,2, \cdots, 9$, any 5 numbers chosen will always include some numbers whose sum is a multiple of 10.
We use proof by contradiction.
Assume that 5 numbers can be chosen such that no combination of these numbers sums to a multiple of 10. Divide $1,2, \cdots, 9$ into the following 5 groups:
$$
\{1,9\},\{2,8\},\{3,7\},\{4,6\},\{5\},
$$
Then the chosen numbers can only contain one number from each group, thus 5 must be chosen.
(1) If 1 is chosen from the first group, then 9 is not chosen. Since $1+5+4=10$, 4 is not chosen, thus 6 is chosen. Since $1+6+3=10$, 3 is not chosen, thus 7 is chosen. Since $1+7+2=10$, 2 is not chosen, thus 8 is chosen. But $5+7+8=20$ is a multiple of 10, a contradiction. (2) If 9 is chosen from the first group, then 1 is not chosen. Since $9+5+6=20$, 6 is not chosen, thus 4 is chosen. Since $9+4+7=20$, 7 is not chosen, thus 3 is chosen. Since $9+3+8=20$, 8 is not chosen, thus 2 is chosen. But $5+3+2=10$ is a multiple of 10, a contradiction. In summary, the minimum value of $n$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. As shown in Figure 5, in quadrilateral $A B C D$, $\angle A=$ $\angle B C D=90^{\circ}, B C=$ $C D, E$ is a point on the extension of $A D$. If $D E=A B$ $=3, C E=4 \sqrt{2}$, then the length of $A D$ is
|
11.5.
In Figure 5, connect $A C$. It is easy to prove that
$\triangle C D E \cong \triangle C B A, \angle A C E=90^{\circ}$.
Since $C A=C E=4 \sqrt{2}$, therefore, $A E=8$.
Thus, $A D=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Xiao Wang walks along the street at a constant speed, and he notices that a No. 18 bus passes him from behind every $6 \mathrm{~min}$, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals, then, the interval between dispatches is $\qquad$ min.
|
13.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, and the distance between two consecutive buses traveling in the same direction be $s \mathrm{~m}$. From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s .
\end{array}\right.
$$
Solving these equations, we get $s=4 x$, i.e., $\frac{s}{x}=4$.
Therefore, the interval time for bus No. 18 to depart from the terminal is $4 \mathrm{~min}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $n$ is an integer, and the quadratic equation in $x$
$$
(n-1)^{2} x^{2}-5 n(n-1) x+\left(6 n^{2}-n-1\right)=0
$$
has at least one integer root. Then the sum of all possible values of $n$ is $\qquad$
|
$$
[(n-1) x-(2 n-1)][(n-1) x-(3 n+1)]=0 .
$$
Since $n \neq 1$, then
$$
\begin{array}{l}
x_{1}=\frac{2 n-1}{n-1}=2+\frac{1}{n-1}, \\
x_{2}=\frac{3 n+1}{n-1}=3+\frac{4}{n-1} .
\end{array}
$$
Because the original equation has at least one integer root, and $n$ is an integer, so, $n-1= \pm 1, \pm 2, \pm 4$.
Solving gives $n=2,0,3,-1,5,-3$.
Therefore, the sum of all $n$ values is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9}$, $x \in[0,1]$. Provide the following conclusions:
(1) $f(x)>0$;
(2) $f(x)<0$;
(3) There exists $x_{0} \in[0,1)$, such that $f\left(x_{0}\right)=0$;
(4) There exists $x_{0} \in[0,1]$, such that $f\left(x_{0}\right)<0$.
Among them, the correct conclusion numbers are $\qquad$.
|
2. (1)
$$
\begin{array}{l}
f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9} \\
=x^{3}(1-x)+3 x^{3}-3 x^{2}+(x-1)^{2}+\frac{4}{9} \\
=x^{3}(1-x)+(x-1)^{2}+\frac{3}{2} x^{3}+\frac{3}{2} x^{3}+\frac{4}{9}-3 x^{2} \\
\geqslant x^{3}(1-x)+(x-1)^{2}+3 \sqrt[3]{\frac{3}{2} x^{3} \cdot \frac{3}{2} x^{3} \cdot \frac{4}{9}}-3 x^{2} \\
=x^{3}(1-x)+(x-1)^{2} \geqslant 0 .
\end{array}
$$
The first equality holds when $x=\frac{2}{3}$, and the second equality holds when $x=1$.
Therefore, the equalities cannot hold simultaneously.
Thus, $f(x)>0$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $f(x)=x^{2}+2 x+1$, there exists a real number $t$ such that when $x \in[1, m]$, $f(x+t) \leqslant x$ always holds, then the maximum value of $m$ is $\qquad$ .
|
2.4.
Translate the graph of $f(x)$ to the right by $-t$ units, and from the graphical analysis, the maximum value of $m$ is the larger of the x-coordinates of the two intersection points of $\left\{\begin{array}{l}y=x, \\ y=f(x+t)\end{array}\right.$
From $f(1+t)=1$, we get $t=-1, t=-3$.
Then, from $f(x-3)=x$, we have
$x=1$ (discard), $x=4$.
Therefore, the maximum value of $m$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $n$ be a positive integer. If
$$
A(n)=1986^{n}+1987^{n}+1988^{n}+1989^{n}
$$
is not divisible by 5, find $n$.
|
(Tip: Only consider the unit digit $G(A(n))$ of $A(n)$. When $n=4 k\left(k \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{4}+7^{4}+8^{4}+9^{4}\right) \\
=G(6+1+6+1)=4,
\end{array}
$$
at this time, 5 does not divide $A(n)$;
$$
\begin{array}{l}
\text { When } n=4 k+r(k \in \mathbf{N}, 1 \leqslant r \leqslant 3) \text {, } \\
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{r}+7^{\prime}+8^{r}+9^{r}\right)=0,
\end{array}
$$
contradiction.)
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There are 10 positive integers arranged from smallest to largest: $1, 4, 8$, $10, 16, 19, 21, 25, 30, 43$. How many groups of consecutive numbers have a sum that is divisible by 11?
|
Let 10 positive integers be $a_{1}, a_{2}$, $\cdots, a_{10}$, and let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$$
a_{i+1}+a_{i+2}+\cdots+a_{j}=S_{j}-S_{i} .
$$
The sequence $a_{n}(n=1,2, \cdots, 10)$ has remainders modulo 11 of $1,4,-3,-1,5,-3,-1,3,-3,-1$; the sum of the first $n$ terms $S_{n}$ has remainders modulo 11 of $1,5,2$, $1,6,3,2,5,2,1$. Therefore,
$$
\begin{array}{l}
S_{1} \equiv S_{4} \equiv S_{10} \equiv 1(\bmod 11), \\
S_{2} \equiv S_{8} \equiv 5(\bmod 11), \\
S_{3} \equiv S_{7} \equiv S_{9} \equiv 2(\bmod 11) .
\end{array}
$$
Thus, there are $3+1+3=7$ pairs $(i, j)$ such that $11 \mid \left(S_{j}-S_{i}\right)$, i.e., $11 \mid \left(a_{i+1}+a_{i+2}+\cdots+a_{j}\right)$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If
$$
\frac{4^{5}+4^{5}+4^{5}+4^{5}}{3^{5}+3^{5}+3^{5}} \times \frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5}}{2^{5}+2^{5}}=2^{n},
$$
then, $n=$ . $\qquad$
|
$-1.12$
The left side of the original equation can be transformed into
$$
\frac{4 \times 4^{5}}{3 \times 3^{5}} \times \frac{6 \times 6^{5}}{2 \times 2^{5}}=\frac{4^{6}}{3^{6}} \times \frac{6^{6}}{2^{6}}=4^{6}=2^{12} .
$$
Thus, $2^{12}=2^{n}$. Therefore, $n=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let the integer $a$ divided by 7 leave a remainder of 3, and the integer $b$ divided by 7 leave a remainder of 5. If $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7.
(1994, Tianjin City Junior High School Mathematics Competition)
|
Solution: Let $a=7 m+3, b=7 n+5$. Then
$$
\begin{array}{l}
a^{2}-4 b=(7 m+3)^{2}-4(7 n+5) \\
=49 m^{2}+42 m+9-28 n-20 \\
=7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{array}
$$
Therefore, the remainder when $a^{2}-4 b$ is divided by 7 is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $a, b, c$ simultaneously satisfy $a-7b+8c=4$ and $8a+4b-c=7$. Then, $a^{2}-b^{2}+c^{2}=$ $\qquad$ .
|
4.1.
According to the conditions $a+8c=4+7b, 8a-c=7-4b$. Squaring both sides of the two equations and then adding them yields
$$
\begin{array}{l}
(a+8c)^{2}+(8a-c)^{2} \\
=(7+4b)^{2}+(7-4b)^{2} .
\end{array}
$$
Simplifying and organizing, we get $65\left(a^{2}+c^{2}\right)=65\left(1+b^{2}\right)$.
Therefore, $a^{2}-b^{2}+c^{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In an equilateral triangle $\triangle A B C$ with side length $2 \sqrt{7}$, $M$ is the midpoint of side $B C$, and $P$ is a point on $A C$.
(1) For what value of $P C$ is $B P+P M$ minimized?
(2) Find the minimum value of $B P+P M$.
|
7. As shown in Figure 8, connect $A M$, and flip $\triangle A B C$ along the axis $A C$ by $180^{\circ}$ to get $\triangle A D C$. Let $N$ be the symmetric point of $M$ (about $A C$). At this point, we always have $P M = P N$. Since $B$ and $N$ are fixed points, $B P + P N \geqslant B N$, with equality holding if and only if point $P$ lies on the side $B N$.
To find the value of $P C$, extend $B C$ to $E$ such that $C E = C N$, forming an equilateral $\triangle C E N$. By the construction, we have
$$
P C \parallel N E, \frac{P C}{N E} = \frac{B C}{B E}, \text{ i.e., } \frac{P C}{\sqrt{7}} = \frac{2 \sqrt{7}}{3 \sqrt{7}}.
$$
Thus, when $P C = \frac{2 \sqrt{7}}{3}$, $B P + P M$ achieves its minimum value.
Draw $N F \perp C E$, with $F$ as the foot of the perpendicular. Then $N F = \frac{\sqrt{21}}{2}$, $B F = \frac{5 \sqrt{7}}{2}$,
$$
B N = \sqrt{B F^{2} + F N^{2}} = \sqrt{\frac{175}{4} + \frac{21}{4}} = 7
$$
Therefore, the minimum value of $B P + P M$ is 7.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 There is a sequence of numbers: $1,3,4,7,11,18, \cdots$, starting from the third number, each number is the sum of the two preceding numbers.
(1) What is the remainder when the 1991st number is divided by 6?
(2) Group the above sequence as follows:
$(1),(3,4),(7,11,18), \cdots$,
where the $n$-th group has exactly $n$ numbers. What is the remainder when the sum of the numbers in the 1991st group is divided by 6?
(6th Spring Festival Cup Mathematics Competition)
|
Solution: Let the $n$-th number of the sequence be $a_{n}$. Then
$$
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
$$
Through experimentation, it is known that the remainders of $a_{1}, a_{2}, \cdots, a_{26}$ when divided by 6 are
$$
\begin{array}{l}
1,3,4,1,5,0,5,5,4,3,1,4,5, \\
3,2,5,1,0,1,1,2,3,5,2,1,3 .
\end{array}
$$
Notice that $a_{25} \equiv a_{1}(\bmod 6), a_{26} \equiv a_{2}(\bmod 6)$, and using $a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$, we get
$$
a_{27}=a_{26}+a_{25} \equiv a_{2}+a_{1}=a_{3}(\bmod 6) .
$$
Continuing this way, when $n>24$, we have
$$
a_{n} \equiv a_{n-24}(\bmod 6) \text {. }
$$
(1) Since $1991=24 \times 82+23$, we have
$$
a_{1991} \equiv a_{23} \equiv 5(\bmod 6) \text {. }
$$
Thus, the remainder when the 1991st number is divided by 6 is 5.
(2) When $n>24$, $a_{n} \equiv a_{n-24}(\bmod 6)$, so for any natural number $n$, we have
$$
\begin{array}{l}
a_{n+1}+a_{n+2}+\cdots+a_{n+x} \equiv a_{1}+a_{2}+\cdots+a_{21} \\
\equiv 1+3+4+1+5+0+5+5+4+ \\
3+1+4+5+3+2+5+1+0+ \\
1+1+2+3+5+2 \\
=66 \equiv 0(\bmod 6) .
\end{array}
$$
Since the first 1990 groups contain a total of $1+2+\cdots+1990$ $=1981045$ (let this be $m$) numbers, the sum of the numbers in the 1991st group is
$$
\begin{array}{l}
a_{m+1}+a_{m+2}+\cdots+a_{m+1991} \\
\equiv\left(a_{m+1}+a_{m+2}+\cdots+a_{m+1991}+a_{m+192}\right)- \\
\quad a_{m+192} \\
\equiv 83 \times 0-a_{m+1922} \equiv-a_{13} \equiv-5 \equiv 1(\bmod 6) .
\end{array}
$$
Thus, the remainder when the sum of the numbers in the 1991st group is divided by 6 is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (15 points) The function $f(x)$ defined on the interval $[0,1]$ satisfies $f(0)=f(1)=0$, and for any $x_{1}, x_{2} \in [0,1]$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant f\left(x_{1}\right)+f\left(x_{2}\right) \text {. }
$$
(1) Prove: For any $x \in [0,1]$, we have
$$
f(x) \geqslant 0 \text {; }
$$
(2) Find the value of $f\left(\frac{3}{4}\right)$;
(3) Calculate $f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{2^{k}}\right)+$
$$
\cdots+f\left(\frac{1}{2^{2 \alpha B}}\right) .
$$
|
(1) Take $x_{1}=x_{2}=x \in[0,1]$, then
$$
f\left(\frac{2 x}{2}\right) \leqslant f(x)+f(x),
$$
i.e., $f(x) \leqslant 2 f(x)$.
Thus, $f(x) \geqslant 0$.
Therefore, for any $x \in[0,1]$, $f(x) \geqslant 0$.
(2) From $f(0)=f(1)=0$, we get
$$
f\left(\frac{0+1}{2}\right) \leqslant f(0)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{1}{2}\right) \leqslant 0$.
But from (1), we know $f\left(\frac{1}{2}\right) \geqslant 0$, so $f\left(\frac{1}{2}\right)=0$.
From $f\left(\frac{1}{2}\right)=0, f(1)=0$, we get
$$
f\left(\frac{\frac{1}{2}+1}{2}\right) \leqslant f\left(\frac{1}{2}\right)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{3}{4}\right) \leqslant 0$.
From (1), we know $f\left(\frac{3}{4}\right) \geqslant 0$, so $f\left(\frac{3}{4}\right)=0$.
(3) From $f(0)=0, f\left(\frac{1}{2}\right)=0$, we get
$$
f\left(\frac{0+\frac{1}{2}}{2}\right) \leqslant f(0)+f\left(\frac{1}{2}\right)=0+0=0.
$$
Thus, $f\left(\frac{1}{4}\right) \leqslant 0$.
But from (1), we know $f\left(\frac{1}{4}\right) \geqslant 0$, so
$$
f\left(\frac{1}{4}\right)=f\left(\frac{1}{2^{2}}\right)=0.
$$
Continuing this process, we get
$$
f\left(\frac{1}{2^{k}}\right)=0(k=1,2, \cdots, 2008).
$$
Therefore, $f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{2^{k}}\right)+\cdots$
$$
+f\left(\frac{1}{2^{2008}}\right)=0.
$$
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) For a set $S\left(S \subseteq \mathbf{N}_{+}\right)$, if for any $x \in S, \dot{x}$ cannot divide the sum of elements of any non-empty subset of $S \backslash\{x\}$, then $S$ is called a "good set" ($S \backslash\{x\}$ represents the set $S$ after removing the element $x$).
(1) If $\{3,4, n\}$ is a good set, find the minimum value $n_{0}$ of $n$;
(2) Prove: the set $\left\{3,4, n_{0}, m\right\}$ is not a good set, where $m \in \mathbf{N}_{+}, m \neq 3,4, n_{0}$.
|
Three, (1) Obviously, $n>4$.
If $n=5$, then $4 \mid (3+5)$; if $n=6$, then $3 \mid 6$; if $n=7$, then $7 \mid (3+4)$; if $n=8$, then $4 \mid 8$; if $n=9$, then $3 \mid 9$.
When $n=10$, it is verified that $\{3,4,10\}$ is a good set.
Therefore, the minimum value of $n$ is $n_{0}=10$.
(2) If $\{3,4,10, m\}$ is a good set, then $m$ is not a multiple of 3, and 3 cannot divide $10+m$, i.e., $m$ is not of the form $3k+2$. 3 also cannot divide $10+4+m$, i.e., $m$ is not of the form $3k+1$. Therefore, for any positive integer $m (m \neq 3,4,10)$, the set $\{3,4,10, m\}$ is not a good set.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Given Theorem: "If three prime numbers $a, b, c$ greater than 3 satisfy the equation $2a + 5b = c$, then $a + b + c$ is a multiple of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
(1997, National Junior High School Mathematics League)
|
Solution: Let $a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}$. Then
$$
\begin{array}{l}
a+b+c=3(a+2 b) \\
=3\left(3 k_{1}+r_{1}+6 k_{2}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) .
\end{array}
$$
Since $a, b$ are both primes greater than 3, we have
$$
r_{1} r_{2} \neq 0.
$$
If $r_{1} \neq r_{2}$, then $r_{1}=1, r_{2}=2$, or $r_{1}=2, r_{2}=1$. In this case,
$$
\begin{aligned}
c & =2 a+5 b=6 k_{1}+2 r_{1}+15 k_{2}+5 r_{2} \\
& =3\left(2 k_{1}+5 k_{2}+r_{2}\right)+2\left(r_{1}+r_{2}\right) \\
& =3\left(2 k_{1}+5 k_{2}+r_{2}\right)+6,
\end{aligned}
$$
which contradicts the fact that $c$ is a prime. Therefore, $r_{1}=r_{2}$.
$$
\begin{array}{l}
\text { Hence } a+b+c=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}+r_{1}\right) .
\end{array}
$$
Thus, $a+b+c$ is a multiple of 9, i.e., $n=9$ when the conclusion holds.
Next, we prove: $n \leqslant 9$.
Take $a=11, b=5$, then $c=2 a+5 b=47$. In this case, $a+b+c=63$, so $n \mid 63$.
Take $a=13, b=7$, then $c=2 a+5 b=61$. In this case, $a+b+c=81$, so $n \mid 81$.
Therefore, $n$ is a common divisor of 63 and 81.
Hence $n \leqslant(63,81)=9$.
In conclusion, the maximum value of $n$ is 9.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. Find the remainder when $47^{37^{2}}$ is divided by 7.
Try to find the remainder of $47^{37^{2}}$ when divided by 7.
|
(Note that $47^{6} \equiv(-2)^{6} \equiv 2^{6} \equiv 8^{2} \equiv 1^{2}$ $\equiv 1(\bmod 7)$. Also, $37^{23} \equiv 1^{23} \equiv 1(\bmod 6)$, let $37^{23}$ $=6k+1$. Then $47^{37^{23}}=47^{k+1}=\left(47^{6}\right)^{k} \times 47 \equiv$ $\left.1^{k} \times 47 \equiv 47 \equiv 5(\bmod 7).\right)$
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
7.6.
Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
-1<\beta<0,1<\alpha<2 \text {, }
$$
and $\alpha+\beta=1, \alpha \beta=-1$.
Therefore, $\alpha, \beta$ are the two distinct real roots of $x^{2}-x-1=0$.
Let $A_{n}=\alpha^{n}+\beta^{n}$. Then $A_{n+2}=A_{n+1}+A_{n}$.
By $A_{1}=\alpha+\beta=1$,
$$
A_{2}=\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3 \text {, }
$$
we know that $\left\{A_{n}\right\}$ is an integer sequence, and the sequence of remainders when divided by 7 is 1, $3,4,0,4,4,1,5,6,4,3,0,3,3,6,2,1,3,4, \cdots$, which is a periodic sequence with a period of 16.
Since $2009=16 \times 125+9$, we know that the remainder when $A_{2000}$ is divided by 7 is 6.
$$
\text { Since }-1<\left(\frac{1-\sqrt{5}}{2}\right)^{200}<0 \text {, hence }\left[\left(\frac{1+\sqrt{5}}{2}\right)^{209}\right]
$$
is 6 when divided by 7.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that the three non-zero real roots of the equation $x^{3}+a x^{2}+b x+c$ $=0$ form a geometric progression. Then $a^{3} c-b^{3}$ $=$ . $\qquad$
|
8.0.
Let the three roots be $d$, $d q$, and $d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a, \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b, \\
d^{3} q^{3}=-c .
\end{array}\right.
$$
Dividing (2) by (1) gives $d q=-\frac{b}{a}$.
Substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$. Therefore, $a^{3} c-b^{3}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (15 points) Given two lines
$$
\begin{array}{l}
l_{1}: 3 x+4 y-25=0, \\
l_{2}: 117 x-44 y-175=0,
\end{array}
$$
Point $A$ has projections $B$ and $C$ on $l_{1}$ and $l_{2}$, respectively.
(1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$;
(2) If $\odot T:\left(x-\frac{39}{5}\right)^{2}+\left(y-\frac{27}{5}\right)^{2}=r^{2}$ intersects the curve $\Gamma$ at exactly 7 points, find the value of $r$.
|
10. (1) Let $A(x, y)$. It is easy to know
$$
\begin{array}{l}
A B=\frac{|3 x+4 y-25|}{\sqrt{3^{2}+4^{2}}}=\frac{|3(x-3)+4(y-4)|}{5}, \\
A C=\frac{|117 x-44 y-175|}{\sqrt{117^{2}+44^{2}}} \\
=\frac{|117(x-3)-44(y-4)|}{125} .
\end{array}
$$
Let the angles of inclination of $l_{1}$ and $l_{2}$ be $\dot{\alpha}_{1}$ and $\alpha_{2}$, respectively, and the angle between $l_{1}$ and $l_{2}$ be $\beta=\alpha_{1}-\alpha_{2}$. Then $\sin \beta=\frac{24}{25}$. Therefore,
$$
\begin{aligned}
& S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \beta \\
= & \frac{12}{5}\left|351(x-3)^{2}-176(y-4)^{2}+336(x-3)(y-4)\right| \\
= & \frac{1728}{625} \\
\Rightarrow & \mid 351(x-3)^{2}-176(y-4)^{2}+ \\
& 336(x-3)(y-4) \mid=3600 .
\end{aligned}
$$
Equation (1) is the trajectory equation of point $A$.
(2) It is easy to know $T\left(\frac{39}{5}, \frac{27}{5}\right)$, and the intersection point of $l_{1}$ and $l_{2}$ is $D(3,4)$.
Translate $\odot T$ and curve $\Gamma$ so that point $D$ is moved to the origin, and then perform a rotation transformation centered at point $D^{\prime}$ to make the angle bisector of $l_{1}$ and $l_{2}$ parallel to the coordinate axes, obtaining $\odot T^{\prime}$ and curve $\Gamma^{\prime}$, i.e.,
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{\prime}=\frac{24}{25}(x-3)+\frac{7}{25}(y-4), \\
y^{\prime}=\frac{24}{25}(y-4)-\frac{7}{25}(x-3)
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x-3=\frac{24 x^{\prime}-7 y^{\prime}}{25}, \\
y-4=\frac{7 x^{\prime}+24 y^{\prime}}{25} .
\end{array}\right.
\end{array}
$$
Substitute into equation (1) and simplify to get
$$
\left|\left(4 x^{\prime}\right)^{2}-\left(3 y^{\prime}\right)^{2}\right|=144 \text {. }
$$
Then curve $\Gamma^{\prime}$ is two sets of hyperbolas:
$$
\Gamma_{1}: \frac{x^{2}}{9}-\frac{y^{2}}{16}=1, \Gamma_{2}: \frac{x^{2}}{9}-\frac{y^{2}}{16}=-1 \text {. }
$$
Also, $T^{\prime}(5,0)$, so $\odot T^{\prime}:(x-5)^{2}+y^{2}=r^{2}$.
Since $\odot T$ intersects curve $\Gamma$ at exactly 7 points, i.e., $\odot T^{\prime}$ intersects $\Gamma^{\prime}$ at exactly 7 points, and both $\odot T$ and $\Gamma^{\prime}$ are symmetric about the $x$-axis, $\odot T$ and $\Gamma^{\prime}$ must have one intersection point on the $x$-axis, i.e., $(-3,0)$ or $(3,0)$ is on $\odot T^{\prime}$. Therefore,
$r=2$ or 8.
If $r=2$, since the distance from $T^{\prime}(5,0)$ to the asymptotes of $\Gamma^{\prime}$ is
$$
\frac{|4 \times 5 \pm 3 \times 0|}{\sqrt{4^{2}+3^{2}}}=4>r,
$$
so, $\odot T^{\prime}$ does not intersect $\Gamma_{2}$. And $\odot T^{\prime}$ intersects $\Gamma_{1}$ at most 4 points, which contradicts the fact that $\odot T^{\prime}$ intersects $\Gamma^{\prime}$ at 7 points. Therefore, $r=8$.
It is easy to see that when $r=8$, the conditions are satisfied.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange $1, 2, 3$ in a cyclic order from left to right, forming a 2009-digit number: $x=123123 \cdots 12312$. Find the remainder when $x$ is divided by 101.
|
(Tip: Since $100 \equiv -1 \pmod{101}$, we have
$$
\begin{array}{l}
123123 = 123 \times 100 + 123 \\
\equiv -123 + 123 \equiv 0 \pmod{101}.
\end{array}
$$
Therefore, $123123 \cdots 12312 = 123123 \cdots 12300 + 12$
$$
\equiv 0 + 12 \equiv 12 \pmod{101}.
$$
)
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find the unit digit of the natural number $2^{100}+3^{101}+4^{100}$.
|
Solution: Notice that
$$
\begin{array}{l}
2^{100} \equiv 2^{1 \times 25} \equiv\left(2^{4}\right)^{25} \equiv 16^{25} \\
\equiv 6^{25} \equiv 6(\bmod 10), \\
3^{101} \equiv 3^{4 \times 25+1} \equiv\left(3^{4}\right)^{25} \times 3^{1} \\
\equiv 1^{25} \times 3^{1} \equiv 3(\bmod 10), \\
4^{102} \equiv 4^{2 \times 51} \equiv\left(4^{2}\right)^{51} \equiv 16^{51} \equiv 6(\bmod 10) . \\
\text { Therefore, } 2^{100}+3^{101}+4^{102} \equiv 6+3+6 \equiv 5(\bmod 10),
\end{array}
$$
Thus, the unit digit of the natural number $2^{100}+3^{101}+4^{102}$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The inequality $x^{2}+|2 x-6| \geqslant a$ holds for all real numbers $x$. Then the maximum value of the real number $a$ is $\qquad$
|
2.5.
When $x \geqslant 3$,
left side $=x^{2}+2 x-6=(x+1)^{2}-7 \geqslant 9$;
When $x<3$,
left side $=x^{2}-2 x+6=(x-1)^{2}+5 \geqslant 5$.
Therefore, for the given inequality to hold for all real numbers $x$, the maximum value of $a$ should be 5.
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find the largest positive number $\lambda$, such that for any real numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality
$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2}$ holds.
(Zhang Zhengjie)
|
5. Notice that
$$
\begin{array}{l}
1=x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{array}
$$
and when $y=\frac{\sqrt{2}}{2}, x=\frac{\sqrt{2} \lambda}{2 \sqrt{\lambda^{2}+1}}, z=\frac{\sqrt{2}}{2 \sqrt{\lambda^{2}+1}}$, both equalities can be achieved simultaneously.
Therefore, $\frac{\sqrt{1+\lambda^{2}}}{2}$ is the maximum value of $|\lambda x y+y z|$.
Let $\frac{\sqrt{1+\lambda^{2}}}{2}=\frac{\sqrt{5}}{2}$, then $\lambda=2$.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $n$ be a positive integer, and let $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq$ $a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, the signs of $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ are opposite.
(1) Find the value of $f(10)$;
(2) Determine the remainder when $f(2008)$ is divided by 13.
|
And the $n$-digit waveform number $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfies $a_{1}>a_{2}$ is called a "B-type number". According to symmetry, when $n \geqslant 2$, the number of B-type numbers is also $g(n)$. Therefore, $f(n)=2 g(n)$.
Next, we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit A-type waveform numbers with the last digit being $i$ $(i=1,2,3,4)$. Then $g(n)=\sum_{i=1}^{4} m_{n}(i)$. Since $a_{2 k-1}a_{2 k+1}$, then 1) when $k$ is even,
$m_{k+1}(4)=0, m_{k+1}(3)=m_{k}(4)$,
$m_{k+1}(2)=m_{k}(4)+m_{k}(3)$,
$m_{k+1}(1)=m_{k}(4)+m_{k}(3)+m_{k}(2)$.
2) When $k$ is odd,
$m_{k+1}(1)=0, m_{k+1}(2)=m_{k}(1)$,
$m_{k+1}(3)=m_{k}(1)+m_{k}(2)$,
$m_{k+1}(4)=m_{k}(1)+m_{k}(2)+m_{k}(3)$. (2)
It is easy to see that $m_{2}(1)=0, m_{2}(2)=1, m_{2}(3)=2$, $m_{2}(4)=3$, then $g(2)=6$. Therefore,
$$
\begin{array}{l}
m_{3}(1)=m_{2}(2)+m_{2}(3)+m_{2}(4)=6, \\
m_{3}(2)=m_{2}(3)+m_{2}(4)=5, \\
m_{3}(3)=m_{2}(4)=3, \\
m_{3}(4)=0 .
\end{array}
$$
So, $g(3)=\sum_{i=1}^{4} m_{3}(i)=14$.
And by $m_{4}(1)=0, m_{4}(2)=m_{3}(1)=6$,
$m_{4}(3)=m_{3}(1)+m_{3}(2)=11$,
$$
m_{4}(4)=m_{3}(1)+m_{3}(2)+m_{3}(3)=14,
$$
So, $g(4)=\sum_{i=1}^{4} m_{4}(i)=31$.
Similarly, we get
$g(5)=70, g(6)=157, g(7)=353$,
$g(8)=793, \cdots$.
Generally, when $n \geqslant 5$,
$g(n)=2 g(n-1)+g(n-2)-g(n-3)$.
Now we prove formula (3). By induction on $n$.
When $n=5,6,7,8$, it has been verified.
Assume formula (3) holds up to $n$. Consider the case for $n+1$.
When $n$ is even, according to 1) and 2), we have
$m_{n+1}(4)=0$,
$m_{n+1}(3)=m_{n}(4)$,
$m_{n+1}(2)=m_{n}(4)+m_{n}(3)$,
$m_{n+1}(1)=m_{n}(4)+m_{n}(3)+m_{n}(2)$.
And $m_{n}(1)=0$, then
$g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$
$=2\left(\sum_{i=1}^{4} m_{n}(i)\right)+m_{n}(4)-m_{n}(2)$
$=2 g(n)+m_{n}(4)-m_{n}(2)$.
Also, $m_{n}(4)$
$=m_{n-1}(1)+m_{n-1}(2)+m_{n-1}(3)+0$
$=\sum_{i=1}^{4} m_{n-1}(i)=g(n-1)$,
$m_{n}(2)=m_{n-1}(1)$
$=m_{n-2}(4)+m_{n-2}(3)+m_{n-2}(2)+0$
$=g(n-2)$,
At this point, we have
$g(n+1)=2 g(n)+g(n-1)-g(n-2)$.
When $n$ is odd, $g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$.
And $m_{n+1}(1)=0, m_{n+1}(2)=m_{n}(1)$,
$m_{n+1}(3)=m_{n}(1)+m_{n}(2)$,
$m_{n+1}(4)=m_{n}(1)+m_{n}(2)+m_{n}(3)$,
Then $g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$
$=2 \sum_{i=1}^{4} m_{n}(i)+m_{n}(1)-m_{n}(3)$
$=2 g(n)+m_{n}(1)-m_{n}(3)$.
Also, $m_{n}(1)$
$=m_{n-1}(4)+m_{n-1}(3)+m_{n-1}(2)+0$
$=g(n-1)$,
$m_{n}(3)=m_{n-1}(4)$
$=m_{n-2}(1)+m_{n-2}(2)+m_{n-2}(3)+0$
$=g(n-2)$,
At this point, we also have
$g(n+1)=2 g(n)+g(n-1)-g(n-2)$.
Therefore, by induction, for all $n \geqslant 5$, formula (3) holds.
According to formula (3),
$g(9)=2 g(8)+g(7)-g(6)=1782$,
$g(10)=2 g(9)+g(8)-g(7)=4004$,
So, $f(10)=2 g(10)=8008$.
Now consider the sequence $\{g(n) \mid$ modulo 13.
Using formula (3), it is easy to calculate that when $n=2,3, \cdots, 14,15$,
$16, \cdots$, the remainders of $g(n)$ when divided by 13 are:
$6,1,5,5,1,2,0,1,0,1,1,3 ; 6,1,5,5, \cdots$
Therefore, when $n \geqslant 2$, the sequence of remainders of $\{g(n)\}$ when divided by 13 is a periodic sequence, with the smallest period being 12.
Since $2008=12 \times 167+4$, then
$g(2008) \equiv 5(\bmod 13)$.
Therefore, $f(2008) \equiv 10(\bmod 13)$.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given that positive integers $p, q$ are both prime numbers, and $7p+q, pq+11$ are also prime numbers. Find $p+q$.
保留源文本的换行和格式，直接输出翻译结果。
|
When $p \equiv q \equiv 1(\bmod 2)$,
$$
7 p+q \equiv 7+1 \equiv 0(\bmod 2) \text{. }
$$
But $7 p+q \geqslant 7>2$, which contradicts the fact that $7 p+q$ is a prime number. Therefore, one of $p, q$ must be even.
Since $p, q$ are both primes, then $p=2$ or $q=2$.
(1) When $p=2$, $14+q$ and $2 q+11$ are both primes.
If $q \equiv 1(\bmod 3)$, then
$$
14+q \equiv 14+1 \equiv 0(\bmod 3) ; \quad \div i
$$
Contradiction.
If $q \equiv-1(\bmod 3)$, then
$$
2 q+11 \equiv-2+11 \equiv 0(\bmod 3),
$$
Contradiction.
Therefore, $q \equiv 0(\bmod 3)$.
But $q$ is a prime, so $q=3$.
(2) When $q=2$, $7 p+2$ and $2 p+11$ are both primes.
If $p \equiv 1(\bmod 3)$, then
$$
7 p+2 \equiv 7+2 \equiv 0(\bmod 3),
$$
Contradiction.
If $p \equiv-1(\bmod 3)$, then
$$
2 p+11 \equiv-2+11 \equiv 0(\bmod 3),
$$
Contradiction.
Therefore, $p \equiv 0(\bmod 3)$.
But $p$ is a prime, so $p=3$.
Upon verification, $p=2, q=3$ or $p=3, q=2$ satisfy the conditions. Hence $p+q=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ .
|
8. -1 .
Let $z$ be a root satisfying the condition. Then the original equation is equivalent to
$$
z^{2008}(z+1)=-1 \text {. }
$$
Taking the modulus on both sides, we get $|z+1|=1$.
Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle I: with center $(-1,0)$ and radius 1, and the circle with center $(0,0)$ and radius 1 in the complex plane. Therefore, $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
: Upon verification, both roots are roots of the original equation. Thus, the sum of all roots with modulus 1 is -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. There are 10 players $A_{1}, A_{2}, \cdots, A_{10}$, whose points are $9,8,7,6,5,4,3,2,1,0$, and their rankings are 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th. Now a round-robin tournament (i.e., each pair of players plays exactly one match) is held, and each match must have a winner. If the player with a higher ranking beats the player with a lower ranking, the winner gets 1 point, and the loser gets 0 points; if the player with a lower ranking beats the player with a higher ranking, the winner gets 2 points, and the loser gets 0 points. After all the matches, the total points of each player (i.e., the sum of the points obtained in this round-robin tournament and the previous points) are calculated, and the players are re-ranked based on the total points. Find the minimum value of the new champion's total points (ties in ranking are allowed).
|
15. The minimum cumulative score of the new champion is 12.
If the new champion's score does not exceed 11 points, then
$A_{1}$ can win at most 2 games; $A_{2}$ can win at most 3 games;
$A_{3}$ can win at most 4 games; $A_{4}$ can win at most 5 games.
$A_{5}$ can increase by at most 6 points, but there are only 5 players with fewer points than him at the start. Therefore, if he increases by 6 points, he must win at least 1 game against players ranked higher than him, meaning he can win at most 4 games against players ranked lower. Thus, he can win at most 5 games.
$A_{6}$ can increase by at most 7 points, but there are only 4 players with fewer points than him at the start. Therefore, if he increases by 7 points, he must win at least 2 games against players ranked higher than him, meaning he can win at most 3 games against players ranked lower. Thus, he can win at most 5 games.
$A_{7}$ can increase by at most 8 points, but there are only 3 players with fewer points than him at the start. Therefore, if he increases by 8 points, he must win at least 3 games against players ranked higher than him, meaning he can win at most 2 games against players ranked lower. Thus, he can win at most 5 games.
$A_{8}$ can increase by at most 9 points, but there are only 2 players with fewer points than him at the start. Therefore, if he increases by 9 points, he must win at least 4 games against players ranked higher than him, meaning he can win at most 1 game against players ranked lower. Thus, he can win at most 5 games.
$A_{9}$ can increase by at most 10 points, but there is only 1 player with fewer points than him at the start. Therefore, if he increases by 10 points, he must win at least 5 games against players ranked higher than him. Thus, he can win at most 5 games.
$A_{10}$ can increase by at most 11 points, and he can win at most 5 games against players ranked higher than him. Thus, he can win at most 5 games.
In summary, the maximum number of games won by all players is $2+3+4+5 \times 7=44$, but each match between two players results in one win, totaling $\mathrm{C}_{10}^{2}=45$ wins, which is a contradiction.
The following example shows that the new champion's cumulative score can be 12 points.
$A_{1}$ wins against $A_{2}, A_{3}, A_{4}$, loses to $A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}$, with a cumulative score of $9+3=12$;
$A_{2}$ wins against $A_{3}, A_{4}, A_{5}, A_{6}$, loses to $A_{7}, A_{8}, A_{9}, A_{10}$, with a cumulative score of $8+4=12$;
$A_{3}$ wins against $A_{4}, A_{5}, A_{6}, A_{7}$, loses to $A_{8}, A_{9}, A_{10}$, with a cumulative score of $7+4=11$;
$A_{4}$ wins against $A_{5}, A_{6}, A_{7}, A_{8}$, loses to $A_{9}, A_{10}$, with a cumulative score of $6+4=10$;
$A_{5}$ wins against $A_{6}, A_{7}, A_{8}, A_{9}$, loses to $A_{10}$, with a cumulative score of $5+2+4=11$;
$A_{6}$ wins against $A_{7}, A_{8}, A_{9}, A_{10}$, with a cumulative score of $4+2+4=10$;
$A_{7}$ wins against $A_{8}, A_{9}, A_{10}$, with a cumulative score of $3+2 \times 2+3=10$;
$A_{8}$ wins against $A_{9}, A_{10}$, with a cumulative score of $2+2 \times 3+2=10$;
$A_{9}$ wins against $A_{10}$, with a cumulative score of $1+2 \times 4+1=10$; $A_{10}$ has a cumulative score of $0+2 \times 5=10$.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $x$ and $y$ are integers, $y=\sqrt{x+2003}-$ $\sqrt{x-2009}$. Then the minimum value of $y$ is $\qquad$ .
|
3.2 .
Let $\sqrt{x+2003}=a, \sqrt{x-2009}=b$.
Since $x, y$ are integers,
$y=a-b=\frac{a^{2}-b^{2}}{a+b}=\frac{4012}{a+b} \in \mathbf{Z}_{+}$.
Thus, $a-b \in \mathbf{Q}, a+b \in \mathbf{Q}$
$\Rightarrow a \in \mathbf{Q}, b \in \mathbf{Q} \Rightarrow a, b \in \mathbf{Z}$.
Then $(a+b)(a-b)=4012=2006 \times 2$.
Since $a+b$ and $a-b$ have the same parity, the minimum value of $a-b$ is 2.
Therefore, the minimum value of $y$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, on the hypotenuse $A B$ respectively intercept $A D=$
$$
A C, B E=B C, D E=6 \text{, }
$$
$O$ is the circumcenter of $\triangle C D E$, as shown in Figure 2. Then the sum of the distances from $O$ to the three sides of $\triangle A B C$ is . $\qquad$
|
4.9.
As shown in Figure 8, connect
$$
O A, O B, O C, O D \text {, and }
$$
$O E$.
Since $O C=O D$,
we know that point $O$ lies on the perpendicular bisector of segment $C D$.
Also, $A D=A C$, so $O A$ bisects $\angle C A D$.
Similarly, $O B$ bisects $\angle C B D$.
Therefore, $O$ is the incenter of $\triangle A B C$.
Thus, the distances from point $O$ to the three sides of $\triangle A B C$ are all equal to the radius of the incircle of the right triangle $\triangle A B C$. This radius is
$$
\begin{array}{l}
\frac{1}{2}(A C+B C-A B) \\
=\frac{1}{2}(A D+B E-A B)=\frac{1}{2} D E=3 .
\end{array}
$$
Therefore, the required sum is $3 \times 3=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sum of $2 n+1\left(n \in \mathbf{N}_{+}\right)$ consecutive positive integers is $a$, and the difference between the sum of the squares of the last $n$ numbers and the sum of the squares of the first $n$ numbers is $b$. If $\frac{a}{b}=\frac{11}{60}$, then the value of $n$ is
|
2.5
Let the $(n+1)$-th number be $m$. Then
$$
\begin{array}{l}
a=(2 n+1) m, \\
b=\sum_{i=1}^{n}(m+i)^{2}-\sum_{i=1}^{n}(m-i)^{2} \\
=2 m \sum_{i=1}^{n} 2 i=2 m n(n+1) . \\
\text { Hence } \frac{a}{b}=\frac{2 n+1}{2 n(n+1)}=\frac{11}{60} .
\end{array}
$$
Solving gives $n=5$ or $-\frac{6}{11}$ (discard).
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. From the arithmetic sequence $2,5,8,11, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$
|
2.8 .
First, take $2,5,8,11,20,41,110,1640$, it is easy to see that the sum of their reciprocals is 1, i.e., $k=8$ satisfies the requirement.
Second, suppose we take $x_{1}, x_{2}, \cdots, x_{k}$ from the sequence, such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1$.
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
Since $x_{n} \equiv 2(\bmod 3)$, taking the modulus 3 of both sides of equation (1) gives
$k \cdot 2^{k-1} \equiv 2^{k}(\bmod 3)$, i.e., $k \equiv 2(\bmod 3)$.
When $k=2$, $\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{2}+\frac{1}{5}<1$;
When $k=5$,
$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{5}} \leqslant \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}<1$.
Therefore, $k \geqslant 8$. Hence, the minimum value of $k$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are any two points (which can coincide) on the graph of the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{2 x}{1-2 x}, & x \neq \frac{1}{2} \\
-1, & x=\frac{1}{2}
\end{array}\right.
$$
Point $M$ lies on the line $x=\frac{1}{2}$, and $\overrightarrow{A M}=\overrightarrow{M B}$. Then the value of $y_{1}+y_{2}$ is
|
4. -2 .
Given that point $M$ is on the line $x=\frac{1}{2}$, let $M\left(\frac{1}{2}, y_{M}\right)$.
Also, $\overrightarrow{A M}=\overrightarrow{M B}$, that is,
$$
\begin{array}{l}
\overrightarrow{A M}=\left(\frac{1}{2}-x_{1}, y_{M}-y_{1}\right), \\
\overrightarrow{M B}=\left(x_{2}-\frac{1}{2}, y_{2}-y_{M}\right) .
\end{array}
$$
Thus, $x_{1}+x_{2}=1$.
(1) When $x_{1}=\frac{1}{2}$, $x_{2}=\frac{1}{2}$, then
$$
y_{1}+y_{2}=f\left(x_{1}\right)+f\left(x_{2}\right)=-1-1=-2 \text {; }
$$
(2) When $x_{1} \neq \frac{1}{2}$, $x_{2} \neq \frac{1}{2}$, then
$$
\begin{array}{l}
y_{1}+y_{2}=\frac{2 x_{1}}{1-2 x_{1}}+\frac{2 x_{2}}{1-2 x_{2}} \\
=\frac{2 x_{1}\left(1-2 x_{2}\right)+2 x_{2}\left(1-2 x_{1}\right)}{\left(1-2 x_{1}\right)\left(1-2 x_{2}\right)} \\
=\frac{2\left(x_{1}+x_{2}\right)-8 x_{1} x_{2}}{1-2\left(x_{1}+x_{2}\right)+4 x_{1} x_{2}}=-2 .
\end{array}
$$
In summary, $y_{1}+y_{2}=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $n \in \mathbf{N}$, and $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$ is a positive integer, then $n=$ $\qquad$
|
6.5.
Notice that $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$
$$
=\frac{33}{\sqrt{n^{2}+24}+\sqrt{n^{2}-9}},
$$
thus $\sqrt{n^{2}+24}+\sqrt{n^{2}-9}$ could be $1,3,11,33$. Therefore, the solution is $n=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a, b \in \mathbf{N}_{+}$, when $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$, and the remainder is $r$. Then the number of pairs $(a, b)$ that satisfy $q^{2}+r=2009$ is $\qquad$ pairs.
|
-1.0 .
From $2\left(a^{2}+b^{2}\right) \geqslant(a+b)^{2}$, we get $\frac{a^{2}+b^{2}}{a+b} \geqslant \frac{a+b}{2}$.
Then $q+1>q+\frac{r}{a+b}=\frac{a^{2}+b^{2}}{a+b}$ $\geqslant \frac{a+b}{2} \geqslant \frac{r+1}{2}$.
Thus, $r<2 q+1$.
From $q^{2}+r=2009<(q+1)^{2}$, we get $q \geqslant 44$. From $q^{2}+r=2009 \geqslant q^{2}$, we get $q \leqslant 44$. Therefore, $q=44 \Rightarrow r=2009-44^{2}=73$.
Thus, $a^{2}+b^{2}=44(a+b)+73$, which means $(a-22)^{2}+(b-22)^{2}=1041$.
Therefore, the equation has no positive integer solutions.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $a, b, c, d \in \mathbf{Z}$,
$$
f(x)=a x^{3}+b x^{2}+c x+d \text {. }
$$
Then among $A(1,1)$, $B(2009,1)$, $C(-6020,2)$, $D(2,-6020)$, at most $\qquad$ of them can be on the curve $y=f(x)$.
|
4.3.
If $A$ and $C$ are both on $y=f(x)$, then $f(1)=1, f(-6020)=2$.
It is easy to see that $\left(x_{1}-x_{2}\right) \mid \left(f\left(x_{1}\right)-f\left(x_{2}\right)\right)$. Therefore, $[1-(-6020)] \mid (f(1)-f(-6020))=-1$, which is a contradiction.
Thus, $A$ and $C$ cannot both be on $y=f(x)$.
Similarly, $B$ and $C$ cannot both be on $y=f(x)$.
Therefore, at most two points $(A, B)$ out of $A, B, C$ can be on $y=f(x)$.
In this case, $(x-1)(x-2009) \mid (f(x)-1)$.
Let $f(x)=a(x-1)(x-2009)(x-t)+1$.
Substituting $D(2,-6020)$ into equation (1) yields
$a(2-t)=3$.
Thus, when $a=1, t=-1$, $A, B, D$ are on the $y=f(x)$ defined by equation (1).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) On the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. Find the smallest positive integer $k$ such that: for every circle that contains $k$ rational points on its circumference, the circle must contain infinitely many rational points on its circumference.
|
12. First, prove: If a circle's circumference contains 3 rational points, then the circumference must contain infinitely many rational points.
Let $\odot C_{0}$ in the plane have 3 rational points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ on its circumference, with the center $C_{0}\left(x_{0}, y_{0}\right)$.
Since the perpendicular bisectors of segments $P_{1} P_{2}$ and $P_{1} P_{3}$ pass through the center $C_{0}$, we have
$$
\left\{\begin{array}{l}
\left(y_{2}-y_{1}\right)\left(y_{0}-\frac{y_{1}+y_{2}}{2}\right)+\left(x_{2}-x_{1}\right)\left(x_{0}-\frac{x_{1}+x_{2}}{2}\right)=0, \\
\left(y_{3}-y_{1}\right)\left(y_{0}-\frac{y_{1}+y_{3}}{2}\right)+\left(x_{3}-x_{1}\right)\left(x_{0}-\frac{x_{1}+x_{3}}{2}\right)=0 .
\end{array}\right.
$$
Since $x_{i} 、 y_{i}(i=1,2,3)$ are all rational numbers, the solution $\left(x_{0}, y_{0}\right)$ of the above system of linear equations in $x_{0} 、 y_{0}$ is also a rational number, i.e., $C_{0}$ is a rational point.
Suppose the coordinates of the rational point $P_{n}\left(x_{n}, y_{n}\right)$ are
$$
\begin{array}{l}
\qquad\left\{\begin{array}{l}
x_{n}=x_{0}+a_{n}\left(x_{3}-x_{0}\right)-b_{n}\left(y_{3}-y_{0}\right), \\
y_{n}=y_{0}+b_{n}\left(x_{3}-x_{0}\right)+a_{n}\left(y_{3}-y_{0}\right),
\end{array}\right. \\
\text { where, } a_{n}=\frac{n^{2}-1}{n^{2}+1}, b_{n}=\frac{2 n}{n^{2}+1}(n=4,5, \cdots) .
\end{array}
$$
Then $\left|P_{n} C_{0}\right|^{2}$
$$
\begin{aligned}
= & {\left[a_{n}\left(x_{3}-x_{0}\right)-b_{n}\left(y_{3}-y_{0}\right)\right]^{2}+} \\
& {\left[b_{n}\left(x_{3}-x_{0}\right)+a_{n}\left(y_{3}-y_{0}\right)\right]^{2} } \\
= & \left(a_{n}^{2}+b_{n}^{2}\right)\left[\left(x_{3}-x_{0}\right)^{2}+\left(y_{3}-y_{0}\right)^{2}\right] \\
= & \left(x_{3}-x_{0}\right)^{2}+\left(y_{3}-y_{0}\right)^{2}=\left|P_{3} C_{0}\right|^{2} .
\end{aligned}
$$
Hence, the points $P_{n}(n=4,5, \cdots)$ are all on the circumference of $\odot C_{0}$, i.e., the circumference of $\odot C_{0}$ contains infinitely many rational points.
Second, construct an example of a circle whose circumference contains only two rational points.
$$
C:(x-\sqrt{2})^{2}+(y-\sqrt{2})^{2}=6 .
$$
It is easy to verify that $P_{1}(-1,1)$ and $P_{2}(1,-1)$ are on the circumference of $C$. If the circumference of $C$ contains another rational point $P_{3}\left(x_{3}, y_{3}\right)$ different from $P_{1}$ and $P_{2}$, then
$$
\left(x_{3}-\sqrt{2}\right)^{2}+\left(y_{3}-\sqrt{2}\right)^{2}=6 \text {, }
$$
i.e., $x_{3}^{2}+y_{3}^{2}-2=2 \sqrt{2}\left(x_{3}+y_{3}\right)$.
Since the left side is a rational number and $\sqrt{2}$ is an irrational number, we have $x_{3}+y_{3}=0$. Thus, $x_{3}^{2}=1$. Hence, $x_{3}= \pm 1, y_{3}=$ $\mp 1$. This contradicts the assumption that $P_{3}$ is different from $P_{1}$ and $P_{2}$.
In conclusion, the minimum value of $k$ is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $a, b, c \geqslant 0$, and $a b+b c+c a=\frac{1}{3}$. Prove:
$$
S=\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-a c+1}+\frac{1}{c^{2}-a b+1} \leqslant 3 .
$$
|
Explanation: When $a=b=c=\frac{1}{3}$, $S=3$.
Also, when $a=0, b=\frac{\sqrt{3}}{3}, c=\frac{\sqrt{3}}{3}$, $S=3$ as well.
Therefore, the maximum value 3 is both conventional and unconventional, which is quite rare. Moreover, $S$ has no other extremum (the lower bound of $S$ is $\frac{5}{2}$, $S>\frac{5}{2}$ but cannot take the equality).
Below, we prove this using the graphical method.
Let $a+b+c=M$, it is easy to see that $M \geqslant 1$.
Substituting $a b+b c+a c=\frac{1}{3}$, we get
$$
\begin{aligned}
S & =\frac{1}{a^{2}+a b+a c-\frac{1}{3}+1}+ \\
& \frac{1}{b^{2}+b c+b a-\frac{1}{3}+1}+ \\
& \frac{1}{c^{2}+c a+c b-\frac{1}{3}+1} \\
& =\frac{1}{M a+\frac{2}{3}}+\frac{1}{M b+\frac{2}{3}}+\frac{1}{M c+\frac{2}{3}} .
\end{aligned}
$$
As shown in Figure 3, draw
$$
y=\frac{1}{M x+\frac{2}{3}}(x>
$$
0 ) (a part of a hyperbola).
By the properties of convex functions and decreasing nature, for any $k$ and $\varepsilon>0$ we have
$$
f(k)+f(0)>f(\varepsilon)+f(k-\varepsilon) .
$$
The left side of equation (1) is twice the length of the midline of a trapezoid with bases $f(k)$ and $f(0)$, and the right side is twice the length of the midline of a trapezoid with bases $f(\varepsilon)$ and $f(k-\varepsilon)$. Since the two midlines coincide and the left side is longer, equation (1) holds.
This indicates that when $S$ takes its maximum value, one of the variables must be 0 (let $a=0$). Then $b c=\frac{1}{3}$. Substituting into the sum, we get
$$
S=\frac{3}{2}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1} \text {. }
$$
Let $y=\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}$
$$
\begin{array}{l}
=\frac{1}{b^{2}+1}+\frac{1}{\left(\frac{1}{3 b}\right)^{2}+1} \\
=\frac{9 b^{4}+18 b^{2}+1}{9 b^{4}+10 b^{2}+1} .
\end{array}
$$
Then $(9 y-9) b^{4}+(10 y-18) b^{2}+y-1=0$.
Since $b^{2} \in \mathbf{R}$, we have
$$
\Delta=(10 y-18)^{2}-4 \times 9(y-1)^{2} \geqslant 0 \text {. }
$$
Solving this, we get $y \leqslant \frac{3}{2}(y \geqslant 3$ is discarded $)$.
Thus, $S=\frac{3}{2}+y \leqslant 3$.
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
(1) Find the minimum value of the smallest side length of a peculiar triangle;
(2) Prove that there are infinitely many isosceles peculiar triangles;
(3) How many non-isosceles peculiar triangles are there?
|
(1) Let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of a peculiar triangle. Then, by Heron's formula, we have
$$
\begin{aligned}
16 \Delta^{2}= & (a+b+c)(a+b-c) . \\
& (a-b+c)(-a+b+c) .
\end{aligned}
$$
Since $(a, b, c)=1$, at least one of $a, b, c$ must be odd. If there is an odd number of odd numbers among $a, b, c$, then $a+b+c, a+b-c, a-b+c, -a+b+c$ are all odd, which contradicts equation (1).
Therefore, exactly two of $a, b, c$ are odd.
If $a=1$, by $c1$, it leads to a contradiction.
When $c=b+1$, $b, c$ are one odd and one even.
Thus, $a, b, c$ have exactly one odd number, which is a contradiction.
If $a=4$, then $b, c$ are both odd.
By $cb-\Delta_{1}$, we have $b+\Delta_{1}=4, b-\Delta_{1}=1$, hence $2b=5$, which is a contradiction.
When $c=b+2$,
$$
p=\frac{1}{2}(a+b+c)=b+3, \\
\Delta^{2}=(b+3)(b-1) \times 3 \times 1,
$$
$$
3 \Delta_{1}^{2}=(b+3)(b-1).
$$
If $3 \mid (b+3)$, then $3 \mid b$, which contradicts the peculiar triangle. If $3 \mid (b-1)$, then $3 \mid (b+2)=c$, which also contradicts the peculiar triangle.
In summary, $a \geqslant 5$.
Moreover, $(5,5,8)$ is a peculiar triangle, so the minimum value of the smallest side length of a peculiar triangle is 5.
(2) If $m, n \in \mathbf{N}_{+}, m>n, 3 \mid m \dot{n}, (m, n)=1, m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, m^{2}+n^{2}, 2 m^{2}-2 n^{2}\right)$ is a peculiar triangle.
In fact, $\Delta=2 m n\left(m^{2}-n^{2}\right)$ is an integer.
Furthermore, since $m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, 2\right)=1$.
Thus, $\left(m^{2}+n^{2}, 2 m^{2}-2 n^{2}\right)$
$=\left(m^{2}+n^{2}, m^{2}-n^{2}\right)=\left(m^{2}+n^{2}, m^{2}\right)$
$=\left(m^{2}, n^{2}\right)=1$.
Finally, because $3 \mid m n$, and $(m, n)=1$, one of $m, n$ is a multiple of 3, so $m^{2}+n^{2}, 2 m^{2}-2 n^{2}$ are not multiples of 3.
In particular, taking $m=6 k+1, n=6$, then
$\left(36 k^{2}+12 k+37,36 k^{2}+12 k+37,72 k^{2}+24 k-70\right)$
is a peculiar triangle.
Similarly, if $m, n \in \mathbf{N}_{+}, m>n, 3 \mid \left(m^{2}-n^{2}\right)$, $(m, n)=1, m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, m^{2}+n^{2}, 4 m n\right)$ is a peculiar triangle.
In particular, taking $m=6 k+1, n=2$, then
$$
\left(36 k^{2}+12 k+5,36 k^{2}+12 k+5,48 k+8\right)
$$
is a peculiar triangle.
(3) There are infinitely many non-isosceles peculiar triangles.
Take $t \equiv 5(\bmod 30)$, and let
$$
\begin{array}{l}
a=5 t^{2}, b=\frac{1}{4}\left(25 t^{4}-6 t^{2}+1\right), \\
c=\frac{1}{4}\left(25 t^{4}+6 t^{2}+1\right) .
\end{array}
$$
Since $t$ is odd, $a, b, c$ are integers, and clearly $a<b<c$.
Also, since $t$ is not a multiple of 3, $a, b, c$ are not multiples of 3.
Finally, since $5 \mid t$, $b, c$ are not multiples of 5, and thus, by $\left(t^{2}, 25 t^{4} \pm 6 t^{2}+1\right)=1$, we have $(a, b, c)=1$.
By calculation, $\Delta=\frac{1}{2} t^{2}\left(25 t^{4}-1\right)$ is an integer.
Therefore, $(a, b, c)$ is a non-isosceles peculiar triangle.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In trapezoid $A B C D$, it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition)
|
(提示: Extend $D A$ to point $F$, complete the trapezoid into a square $C B F D$, then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$. It is easy to see that $\triangle A B E \cong \triangle G B E$. Therefore, $A E$ $=E G=E C+A F=10$. Let $E C=x$, then $A F=10-$ $x, D E=12-x, A D=2+x$. By the Pythagorean theorem, $(12-x)^{2}+(2+x)^{2}=10^{2}$. Solving this, we get $x=4$ or 6. )
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In trapezoid $A B C D$, it is known that $A D / / B C, A D \perp$ $C D, B C=C D=2 A D, E$ is a point on side $C D$, $\angle A B E=45^{\circ}$. Then $\tan \angle A E B=$ $\qquad$
(2007, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
(提示: Extend $D A$ to point $F$, such that $A F=A D$, to get the square $B C D F$. Then extend $A F$ to point $G$, such that $F G=C E$. Then $\triangle B F G \cong \triangle B C E$. It is also easy to prove that $\triangle A B G \cong \triangle A B E$. Let $C E=x, B C=2 a$. Then $F G=x, A F=A D=a, A E=A G=x+a, D E=2 a-x$. From $(x+a)^{2}=(2 a-x)^{2}+a^{2}$, solving gives $x=\frac{2}{3} a$. Therefore, $\tan \angle A E B=\tan \angle B G F=\frac{B F}{F G}=3$ is the answer. )
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
(2008, National High School Mathematics Competition)
|
Explanation: By the symmetry of the parabola, we can assume
$$
P\left(2 t^{2}, 2 t\right)(t>0) \text {. }
$$
Since the equation of the circle is $x^{2}+y^{2}-2 x=0$, the equation of the chord of contact $M N$ is
$$
2 t^{2} x+2 t y-\left(x+2 t^{2}\right)=0,
$$
which simplifies to $\left(2 t^{2}-1\right) x+2 t y-2 t^{2}=0$.
Thus, the double tangents $P B$ and $P C$ are the quadratic curve passing through all common points of the double line
$$
\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0
$$
and the circle, with the equation
$$
(x-1)^{2}+y^{2}-1+\lambda\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0 .
$$
Substituting the coordinates of point $P$ into the above equation gives $\lambda=-\frac{1}{4 t^{4}}$. Therefore, the equation of the double tangents $P B$ and $P C$ can be written as
$$
(x-1)^{2}+y^{2}-1-\frac{1}{4 t^{4}}\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0 \text {. }
$$
Setting $x=0$ in the above equation, we get
$$
y=\frac{1}{t} y-1 \text { or } y=1-\frac{1}{t} y \text {. }
$$
Since when $t=1$, there is only one tangent $P B$ intersecting the axis, and when $01$.
Thus, $y_{B}=\frac{t}{1+t}, y_{C}=\frac{t}{1-t}$,
$$
\begin{array}{l}
B C=\frac{t}{1+t}-\frac{t}{1-t}=\frac{2 t^{2}}{t^{2}-1} . \\
\text { Hence } S_{\triangle P B C}=\frac{1}{2} B C\left|x_{P}\right|=\frac{1}{2} \cdot \frac{2 t^{2}}{t^{2}-1} \cdot 2 t^{2} \\
=\frac{2 t^{4}}{t^{2}-1}=2\left[2+\left(t^{2}-1\right)+\frac{1}{t^{2}-1}\right] \geqslant 8 .
\end{array}
$$
The equality holds if and only if $t=\sqrt{2}$. Therefore, the minimum value sought is 8.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Quadratic Function
$$
f(x)=a x^{2}+b x+c(a, b \in \mathbf{R} \text {, and } a \neq 0)
$$
satisfies the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and
$$
f(x) \geqslant x \text {; }
$$
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find the largest $m(m>1)$, such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, we have $f(x+t) \leqslant x$.
(2002, National High School Mathematics Competition)
|
Explanation: First, derive the analytical expression of $f(x)$ from the given conditions, then classify and discuss $m$ and $t$ to determine the value of $m$.
Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about the line $x=-1$. Therefore,
$$
-\frac{b}{2 a}=-1 \Rightarrow b=2 a \text {. }
$$
From condition (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From conditions (1) and (2), we get $f(1) \geqslant 1, f(1) \leqslant 1$.
Thus, $f(1)=1$, i.e., $a+b+c=1$.
From the above, we can solve for $a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4}$.
Therefore, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Since the graph of the parabola $f(x)$ opens upwards, and the graph of $f(x+t)$ is obtained by translating the graph of $y=f(x)$ by $t$ units, to ensure that on $[1, m]$, the graph of $y=f(x+t)$ is below the graph of $y=x$, and $m$ is maximized, $1$ and $m$ should be the two roots of the equation $\frac{1}{4}(x+t+1)^{2}=x$.
Since $1$ is a root of the equation $\frac{1}{4}(x+t+1)^{2}=x$, we can solve for $t=0$ or $-4$.
Substituting $t=0$ into the original equation gives $x_{1}=x_{2}=1$ (which contradicts $m>1$);
Substituting $t=-4$ into the original equation gives $x_{1}=1, x_{2}=9$.
Therefore, $m=9$.
In summary, the maximum value of $m$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 A $6 \times 6$ chessboard has 36 squares. Placing chess pieces in the squares, if there are 4 pieces of the same color in a straight line (horizontal, vertical, or at a $45^{\circ}$ diagonal), it is called a "four-in-a-row". Player A places white pieces, and Player B places black pieces. If A goes first, what is the minimum number of pieces A needs to place to ensure that B cannot subsequently form a four-in-a-row ${ }^{[1]}$?
|
Solution: At least 10 chess pieces need to be placed.
This is because, to prevent the four-in-a-row formation by the black pieces placed subsequently by player B, player A must place at least one white piece in each $1 \times 4$ rectangle, meaning that at least 8 white pieces must be placed in the $2 \times 4$ rectangles on the four sides as shown in Figure 1.
If at least 2 white pieces are placed in the 4 central squares, then the number of white pieces is no less than 10.
If no white pieces are placed in the 4 central squares, then at least 2 white pieces must be placed in each of the two middle columns, at least 2 white pieces in each of the two middle rows, and at least 2 white pieces on each of the two main diagonals, and these white pieces are distinct, so at least $2 \times 6 = 12 (>10)$ white pieces must be placed.
If exactly 1 white piece is placed in the 4 central squares, by symmetry, assume it is placed at position $C$ in Figure 2, with no white pieces at $A$, $B$, or $D$. Then at least 1 white piece must be placed at $E$ and $F$, at least 1 white piece at $G$ and $H$, at least 2 white pieces on the diagonal of $A$ and $D$, and at least 1 white piece on the diagonal $M Q$, and these white pieces are distinct (at least 5 in total). Thus, by the pigeonhole principle, at least one of the two shaded areas in Figure 2 must contain 3 white pieces (assume the upper left $2 \times 4$ square contains 3 pieces), and by Figure 1, the part outside $A$, $B$, $C$, and $D$ in Figure 2 must contain at least $3 + 3 \times 2 = 9$ pieces, plus $C$, making a total of at least 10 white pieces.
Figure 3 shows one possible placement (shaded areas indicate where white pieces are placed).
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 4, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally). What is the minimum number of chess pieces that need to be removed to meet this requirement? Explain your reasoning $^{\mid 2\rfloor}$.
---
The translation preserves the original text's formatting and structure.
|
Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.
Inspired by Example 1, we will also start from the local perspective.
Solution: In each $1 \times 5$ rectangle, at least one chess piece must be taken. Following Example 1, as shown in Figure 5, the area outside the shaded part must have at least 10 chess pieces taken.
If at least one chess piece is taken from the shaded part in Figure 5, then the total number of chess pieces taken is no less than 11.
If no chess pieces are taken from the shaded part in Figure 5, then at least 2 chess pieces must be taken from each of the middle two columns, and at least 2 chess pieces must be taken from each of the middle three rows. Therefore, at least 10 chess pieces have been taken.
Considering the four squares (1), (2), (3), and (4) in Figure 6.
Looking diagonally, since the shaded area in Figure 6 indicates that no chess pieces have been taken, to avoid a line of five, the four squares (1), (2), (3), and (4) must all have chess pieces taken. Therefore, in Figure 6, 2 chess pieces are taken from the 3rd column and 2 chess pieces are taken from the 6th column; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 7th, and 8th columns, totaling 4 chess pieces; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 6th, and 7th rows, totaling 4 chess pieces. Therefore, 12 chess pieces are taken.
In summary, at least 11 chess pieces must be taken.
Figure 7 provides one possible method.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five distinct integers satisfying the condition
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=9
$$
If $b$ is an integer root of the equation
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)=2009
$$
then the value of $b$ is $\qquad$.
|
8. 10 .
Notice
$$
\left(b-a_{1}\right)\left(b-a_{2}\right)\left(b-a_{3}\right)\left(b-a_{4}\right)\left(b-a_{5}\right)=2009,
$$
and $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five different integers, so all $b-a_{1}, b-a_{2}, b-a_{3}, b-a_{4}, b-a_{5}$ are also five different integers.
$$
\begin{array}{l}
\text { Also, } 2009=1 \times(-1) \times 7 \times(-7) \times 41, \text { then } \\
b-a_{1}+b-a_{2}+b-a_{3}+b-a_{4}+b-a_{5}=41 .
\end{array}
$$
From $a_{4}+a_{2}+a_{3}+a_{4}+a_{5}=9$, we get $b=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A. $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following conditions:
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$.
|
14. A. Let in $a_{1}, a_{2}, \cdots, a_{n}$, after removing $a_{i}(i=1$, $2, \cdots, n)$, the arithmetic mean of the remaining $n-1$ numbers is a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Therefore, for any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
b_{i}-b_{j}=\frac{a_{j}-a_{i}}{n-1} \text {. }
$$
Thus, $(n-1) \mid\left(a_{j}-a_{i}\right)$.
Since $b_{1}-b_{n}=\frac{a_{n}-a_{1}}{n-1}=\frac{2008}{n-1}$ is a positive integer, then $(n-1) \mid 2^{3} \times 251$.
Also, $a_{n}-1$
$$
\begin{array}{l}
=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right) \\
\geqslant(n-1)+(n-1)+\cdots+(n-1) \\
=(n-1)^{2},
\end{array}
$$
then $(n-1)^{2} \leqslant 2008$.
Thus, $n \leqslant 45$.
Combining $(n-1) \mid 2^{3} \times 251$, we know $n \leqslant 9$.
On the other hand, let
$$
\begin{array}{l}
a_{1}=8 \times 0+1, a_{2}=8 \times 1+1, \\
a_{3}=8 \times 2+1, \cdots \cdots \\
a_{8}=8 \times 7+1, a_{9}=8 \times 251+1 .
\end{array}
$$
Then these 9 numbers satisfy the problem's requirements.
In summary, the maximum value of $n$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x_{1}, x_{2}, \cdots, x_{6}$ be positive integers, and they satisfy the relation
$$
\begin{aligned}
x_{6} & =2288, \\
x_{n+3} & =x_{n+2}\left(x_{n+1}+2 x_{n}\right)(n=1,2,3) .
\end{aligned}
$$
Then $x_{1}+x_{2}+x_{3}=$
|
3. 8 .
Substituting $n=1,2,3$ into the relation
$$
\begin{array}{l}
x_{n+3}=x_{n+2}\left(x_{n+1}+2 x_{n}\right) . \\
\text { we get } x_{4}=x_{3}\left(x_{2}+2 x_{1}\right), \\
x_{5}=x_{4}\left(x_{3}+2 x_{2}\right)=x_{3}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right), \\
x_{6}=x_{3}^{2}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right)\left(x_{2}+2 x_{1}+2\right) . \\
\text { Also, } x_{6}=2288=2^{4} \times 11 \times 13, \text { and } x_{1} 、 x_{2} 、 x_{3}
\end{array}
$$
are all positive integers, so
$$
x_{3}+2 x_{2} \geqslant 3, x_{2}+2 x_{1} \geqslant 3 \text {. }
$$
Since $x_{2}+2 x_{1}+2$ is 2 more than $x_{2}+2 x_{1}$, and $2^{4} \times 11 \times 13$ only has 13 and 11 with a difference of 2 and both are factors not less than 3, thus,
$$
x_{2}+2 x_{1}+2=13 \Rightarrow x_{2}+2 x_{1}=11 \text {. }
$$
Therefore, it can only be
$$
x_{3}^{2}\left(x_{3}+2 x_{2}\right)=2^{4}, x_{3}=1 \text { or } 2 \text {. }
$$
When $x_{3}=1$, $1+2 x_{2}=2^{4}, x_{2}$ is not an integer;
When $x_{3}=2$, we get $x_{2}=1, x_{1}=5$ which meets the conditions.
Thus, $x_{1}+x_{2}+x_{3}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
(2003, National Junior High School Mathematics Competition)
|
Solution: (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$.
From the problem, we have $a>0$, and $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
Therefore,
$$
\begin{array}{l}
\Delta=[-(2-a)]^{2}-4 \times \frac{4}{a} \geqslant 0 \\
\Rightarrow a^{3}-4 a^{2}+4 a-16 \geqslant 0 \\
\Rightarrow\left(a^{2}+4\right)(a-4) \geqslant 0 .
\end{array}
$$
Solving this, we get $a \geqslant 4$.
When $a=4, b=c=-1$, the minimum value of the maximum among $a, b, c$ is 4.
(2) From $abc>0$, we know that $a, b, c$ are all greater than 0 or one is positive and two are negative.
(i) If $a, b, c$ are all greater than 0, then from (1) we know that the minimum value of the maximum among $a, b, c$ is 4, which contradicts $a+b+c=2$.
(ii) If $a, b, c$ are one positive and two negative, let $a>0$, $b<0$, $c<0$. Then
$$
\begin{array}{l}
|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2 a-2 .
\end{array}
$$
From (1), we know $a \geqslant 4$, hence $2 a-2 \geqslant 6$.
When $a=4, b=c=-1$, the conditions of the problem are satisfied and the equality holds.
Therefore, the minimum value of $|a|+|b|+|c|$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For any real numbers $x, y$, the inequality
$$
|x-1|+|x-3|+|x-5| \geqslant k(2-|y-9|)
$$
always holds. Then the maximum value of the real number $k$ is $\qquad$
|
3. 2 .
From the geometric meaning of absolute value, we know that the minimum value of $|x-1|+|x-3|+|x-5|$ is 4 (at this time $x=3$), and the maximum value of $2-|y-9|$ is 2 (at this time $y=9$). According to the condition, we get $4 \geqslant k \cdot 2$, so, $k \leqslant 2$. Therefore, the maximum value of $k$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+p_{5}^{2}=p_{6}^{2}$ has $\qquad$ groups of positive prime solutions $\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)$.
|
7.5.
Obviously, $p_{6} \neq 2$.
Since $p^{2} \equiv 1$ or $4(\bmod 8)$ (where $p$ is a prime), and $p^{2} \equiv 4(\bmod 8)$ if and only if $p=2$, it follows that among $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$, there must be 4 twos (let's assume they are $p_{1}$, $p_{2}$, $p_{3}$, and $p_{4}$), and the other one is an odd prime, and $p_{6}$ is also an odd prime.
Thus, $16=p_{6}^{2}-p_{5}^{2}=\left(p_{6}+p_{5}\right)\left(p_{6}-p_{5}\right)$.
Clearly, the only solution is $p_{5}=3, p_{6}=5$.
Therefore, the number of all prime solutions is $\mathrm{C}_{5}^{4}=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $a$, $b$, $c$ are all real numbers, and $a+b+c=0$, $abc=2$, then the minimum value that $|a|+|b|+|c|$ can reach is $\qquad$ .
|
(Tip: Refer to Example 3. Answer: 4. )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find the maximum value of the function $y=\sqrt{2 x^{2}+3 x+1}+\sqrt{7-2 x^{2}-3 x}$.
|
( Hint: Refer to Example 8. Answer: The maximum value is 4 . )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Each point in the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist four points of different colors that are concyclic.
(3rd Northern Mathematical Olympiad Invitational)
|
Explanation: If $n=4$, then a colored plane can be constructed such that: there is exactly one point on a circle with three colors, and the rest of the points are of another color, satisfying the problem's conditions and ensuring no four points of different colors are concyclic. Therefore, $n \geqslant 5$.
When $n=5$, by condition (2), there exists a line $l$ on which exactly two colors of points lie. Without loss of generality, assume that line $l$ has only points of colors 1 and 2. By condition (1), there exist points $A$, $B$, and $C$ of colors 3, 4, and 5, respectively, that are not collinear. Let the circle passing through points $A$, $B$, and $C$ be $\odot O$.
(i) If $\odot O$ intersects line $l$, then there exist four points of different colors that are concyclic.
(ii) If $\odot O$ is separated from line $l$ and $\odot O$ has points of colors 1 and 2, then there exist four points of different colors that are concyclic.
(iii) If $\odot O$ is separated from line $l$ and $\odot O$ has no points of colors 1 and 2, as shown in Figure 11, draw a perpendicular from $O$ to line $l$ intersecting $l$ at point $D$. Assume the color of $D$ is 1, and the perpendicular intersects $\odot O$ at points $E$ and $S$. Assume the color of $E$ is 3. Consider a point $F$ on line $l$ with color 2, and let $FS$ intersect $\odot O$ at point $G$. Since $EG \perp GF$, points $D$, $E$, $G$, and $F$ are concyclic.
If $G$ is not a point of color 3, then there exist four points of different colors that are concyclic;
If $G$ is a point of color 3, and $B$, $C$ must have one point different from $S$ (assume it is $B$), $SB$ intersects line $l$ at point $H$. Since $EB \perp BH$, points $B$, $E$, $D$, and $H$ are concyclic.
If $H$ is a point of color 2, then $B$, $H$, $D$, and $E$ are four points of different colors and are concyclic;
If $H$ is a point of color 1, since
$$
SB \cdot SH = SE \cdot SD = SG \cdot SF,
$$
then $B$, $H$, $F$, and $G$ are four points of different colors and are concyclic.
In summary, when $n=5$, there exist four points of different colors that are concyclic.
Therefore, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the minimum value of $y=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
|
Solution: As shown in Figure 2, construct Rt $\triangle P A C$ and Rt $\triangle P B D$ such that $A C=1, B D=2, C P=x, P D=4-x$. Thus, the original problem is transformed into: finding a point $P$ on line $l$ such that the value of $P A + P B$ is minimized.
Since $y = P A + P B \geqslant A B$, the minimum value of $y = P A + P B$ is $A B$.
Draw $B E \parallel l$, intersecting the extension of $A C$ at point $E$. Then,
$$
\begin{array}{l}
A E = A C + C E \\
= A C + B D = 3, \\
B E = C D \\
= x + 4 - x = 4 .
\end{array}
$$
Therefore, $A B = 5$. Hence, the minimum value of $y$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Consider rectangles in the plane with sides parallel to the coordinate axes (both length and width are greater than 0), and call such a rectangle a "box". If two boxes have a common point (including common points on the interior or boundary of the boxes), then the two boxes are said to "intersect". Find the largest positive integer $n$, such that there exist $n$ boxes $B_{1}, B_{2}, \cdots, B_{n}$, satisfying $B_{i}$ intersects $B_{j}$ if and only if $i \neq j \pm 1(\bmod n)$.
|
1. The maximum value that satisfies the condition is 6.
An example is shown in Figure 1.
Below is the proof: 6 is the maximum value.
Assume the boxes $B_{1}$, $B_{2}, \cdots, B_{n}$ satisfy the condition. Let the closed intervals corresponding to the projections of $B_{k}$ on the $x$-axis and $y$-axis be $I_{k}$ and $J_{k}$ respectively $(1 \leqslant k \leqslant n)$.
If $B_{i}$ and $B_{j}$ intersect, and the common point is $(x, y)$, then $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$. Therefore, $I_{i} \cap I_{j}$ and $J_{i} \cap J_{j}$ are non-empty. Conversely, if there exist real numbers $x$ and $y$ such that $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Thus, $B_{i}$ and $B_{j}$ do not intersect if and only if at least one of the intervals $I_{i}, I_{j}$ and $J_{i}, J_{j}$ do not intersect, i.e., either $I_{i} \cap I_{j}=\varnothing$ or $J_{i} \cap J_{j}=\varnothing$.
For convenience, if the indices of the boxes differ by 1 modulo $n$, then the two boxes or intervals are called "adjacent." Other cases are called non-adjacent.
For each $k=1,2, \cdots, n$, the adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect, so $\left(I_{k}, I_{k+1}\right)$ or $\left(J_{k}, J_{k+1}\right)(1 \leqslant k \leqslant n)$ is a pair of non-intersecting intervals. Therefore, among $\left(I_{1}, I_{2}\right), \cdots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right)$; $\left(J_{1}, J_{2}\right), \cdots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$, there are at least $n$ pairs of non-intersecting intervals.
Since any two non-adjacent boxes intersect, the corresponding intervals on both coordinate axes also intersect.
Proposition: Let $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$ be $n$ closed intervals on the number line, and any two non-adjacent intervals intersect. Then the number of pairs of non-intersecting intervals $\left(\Delta_{k}, \Delta_{k+1}\right)(k=1,2, \cdots, n)$ is at most 3.
Proof of the proposition: Let $\Delta_{k}=\left[a_{k}, b_{k}\right](1 \leqslant k \leqslant n)$, $\alpha=\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be the rightmost left endpoint of $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$, and $\beta=\min \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be the leftmost right endpoint of $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$. Without loss of generality, assume $\alpha=a_{2}$.
If $\alpha \leqslant \beta$, then for all $i(i=1,2, \cdots, n)$, we have $a_{i} \leqslant \alpha \leqslant \beta \leqslant b_{i}$. Each $\Delta_{i}$ contains $\alpha$, so there are no non-intersecting interval pairs $\left(\Delta_{i}, \Delta_{i+1}\right)$.
If $\beta<\alpha$, then there exists $i \in\{1,2, \cdots, n\}$ such that $\beta=b_{i}$, and $a_{i}<b_{i}=\beta<\alpha=a_{2}<b_{2}$. Thus, $\Delta_{2}$ and $\Delta_{i}$ are non-intersecting. Since $\Delta_{2}$ intersects with all other intervals except $\Delta_{1}$ and $\Delta_{3}$, $\Delta_{2}$ and $\Delta_{i}$ are non-intersecting if and only if $i=1$ or $i=3$. By symmetry, assume $i=3$, then $\beta=b_{3}$. Since each of $\Delta_{4}, \Delta_{5}, \cdots, \Delta_{n}$ intersects with $\Delta_{2}$, for $i=4,5, \cdots, n$, we have $a_{i} \leqslant \alpha \leqslant b_{i}$. Therefore, $\alpha \in \Delta_{4} \cap \Delta_{5} \cap \cdots \cap \Delta_{n}$, i.e., $\Delta_{4} \cap \Delta_{5} \cap \cdots \cap \Delta_{n} \neq \varnothing$. Similarly, $\Delta_{5}, \cdots, \Delta_{n}, \Delta_{1}$ all intersect with $\Delta_{3}$. Thus, $\beta \in \Delta_{5} \cap \cdots \cap \Delta_{n} \cap \Delta_{1}$, i.e., $\Delta_{5} \cap \cdots \cap \Delta_{n} \cap \Delta_{1} \neq \varnothing$. Therefore, the remaining interval pairs $(\Delta_{1}, \Delta_{2})$, $(\Delta_{2}, \Delta_{3})$, $(\Delta_{3}, \Delta_{4})$ are the possible non-intersecting interval pairs. Thus, there are at most 3.
By the proposition, the maximum value of the positive integer $n$ that satisfies the condition is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ .
|
ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Find the maximum value of the function $y=\sqrt{5-2 x}+\sqrt{3+2 x}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This part is not translated as it contains instructions for the translation task itself. Here is the requested translation above.
|
Solution: From the given condition, we know $y>0$.
Notice that the variance of the two numbers $\sqrt{5-2 x}$ and $\sqrt{3+2 x}$ is
$$
\begin{array}{l}
S^{2}=\frac{1}{2}\left[(\sqrt{5-2 x})^{2}+(\sqrt{3+2 x})^{2}-2\left(\frac{y}{2}\right)^{2}\right] \\
\geqslant 0 .
\end{array}
$$
Solving this, we get $16-y^{2} \geqslant 0$, which means $y^{2} \leqslant 16$.
Therefore, $y_{\text {max }}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. As shown in Figure 1, given that the circumradius $R$ of the acute triangle $\triangle ABC$ is $1$, $\angle BAC = 60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively. The line segment $OH$ intersects the extension of $BC$ at point $P$. Find:
(1) The area of the concave quadrilateral $ABHC$;
(2) The value of $\mathrm{PO} \cdot \mathrm{OH}$.
|
14. (1) As shown in Figure 2, connect $A H$, and draw $O D \perp B C$ at point D.
Since $O$ is the
circumcenter of
$\triangle A B C$ and $\angle B A C$ $=60^{\circ}$, we have
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
O D=O C \cos 60^{\circ}=\frac{1}{2} .
\end{array}
$$
By the property of the Euler line, $A H=2 O D=1$.
By the Law of Sines, $B C=2 R \sin A=\sqrt{3}$, so the area of the concave quadrilateral $A B H C$ is $\frac{1}{2} A H \cdot B C=\frac{\sqrt{3}}{2}$.
(2) Since $H$ is the orthocenter of $\triangle A B C$,
$$
\angle B H C=180^{\circ}-\angle B A C=120^{\circ}=\angle B O C \text {. }
$$
Therefore, points $B, C, H, O$ are concyclic.
Thus, $P O \cdot P H=P B \cdot P C$.
Since $O$ is the circumcenter of $\triangle A B C$, and by the power of a point theorem, $P B \cdot P C=P O^{2}-R^{2}$. Hence,
$$
P O \cdot P H=P B \cdot P C=P O^{2}-R^{2} .
$$
Therefore, $P O^{2}-P O \cdot P H=1$, which means $P O \cdot O H=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
|
15. When $a=b=c=1$, we can get $k \geqslant 2$.
Below is the proof: the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \cdot \frac{1}{b}}=3 .
$$
Similarly, $b c+\frac{1}{b}+\frac{1}{c} \geqslant 3, c a+\frac{1}{c}+\frac{1}{a} \geqslant 3$.
Adding the above three inequalities, we get
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9 .
$$
In conclusion, the minimum value of $k$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given
$$
x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}=1 \text {. }
$$
then the value of $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}$ is
|
$$
\text { II, 1. }-1 \text {. }
$$
Notice that
$$
\begin{array}{l}
x \sqrt{x^{2}+3 x+18}+x \sqrt{x^{2}-6 x+18} \\
=\frac{x^{2}\left(x^{2}+3 x+18\right)-x^{2}\left(x^{2}-6 x+18\right)}{x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}} \\
=\frac{x^{2} \cdot 9 x}{1}=9 x^{3} .
\end{array}
$$
Subtracting the above equation from the given equation, we get
$$
2 x \sqrt{x^{2}-6 x+18}=9 x^{3}-1 \text {. }
$$
Therefore, $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the three vertices $A$, $B$, and $C$ of the right triangle $\triangle ABC$ are all on the parabola $y=x^{2}$, and the hypotenuse $AB$ is parallel to the $x$-axis. Then the height $h$ from the hypotenuse is $\qquad$.
|
3. 1 .
Let point $A\left(a, a^{2}\right)$ and $C\left(c, c^{2}\right)(|c|<a)$. Since $|AB|=|AC|$, we have $a^{2}-c^{2}=1$.
Therefore, the altitude from $C$ to the hypotenuse $AB$ is $a^{2}-c^{2}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2, in $\triangle A B C$, it is known that $A B=9, B C$ $=8, C A=7, A D$ is the angle bisector, and a circle is drawn with $A D$ as a chord,
touching $B C$ and intersecting $A B$ and $A C$ at points $M$ and $N$, respectively. Then $M N=$
$\qquad$
|
4. 6 .
As shown in Figure 4, connect $D M$.
$$
\begin{array}{l}
\text { By } \angle B D M \\
=\angle B A D \\
=\angle C A D \\
=\angle D M N,
\end{array}
$$
we get $M N \parallel B C$.
Thus, $\triangle A M N \sim \triangle A B C$.
It is easy to know $B D=\frac{9}{9+7} \times 8=\frac{9}{2}, B M \cdot B A=B D^{2}$
$\Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2} \Rightarrow B M=\frac{9}{4}$
$\Rightarrow A M=9-\frac{9}{4}=\frac{3}{4} \times 9$.
Also, $\frac{M N}{B C}=\frac{A M}{A B}=\frac{\frac{3}{4} \times 9}{9}=\frac{3}{4}$, therefore,
$$
M N=\frac{3}{4} \times 8=6
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the function $f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x$ has a range of $[n, M]$ on the interval $[-k, k](k>0)$, then $M+n$ $=$ $\qquad$
|
2. 2 .
Notice that
$$
f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x=1+\frac{2^{x}-1}{2^{x}+1}+\sin x \text {. }
$$
Let $g(x)=f(x)-1$. Then
$$
g(-x)=\frac{2^{-x}-1}{2^{-x}+1}+\sin (-x)=\frac{1-2^{x}}{1+2^{x}}-\sin x=-g(x)
$$
is an odd function.
Let the maximum value of $g(x)$ be $g\left(x_{0}\right)$. By the given information, we have
$$
g\left(x_{0}\right)=M-1 \text {, }
$$
and $g\left(x_{0}\right) \geqslant g(x)(x \in[-k, k])$.
Thus, $-g\left(x_{0}\right) \leqslant-g(x)=g(-x)(x \in[-k, k])$, which means $-g\left(x_{0}\right)$ is the minimum value of $g(x)$. Therefore,
$$
-g\left(x_{0}\right)=n-1 \text {. }
$$
Adding the maximum and minimum values of $g(x)$, we get
$$
0=(M-1)+(n-1) \Rightarrow M+n=2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The 29th Summer Olympic Games were held in Beijing on August 8, 2008, forming a memorable number 20080808. The number of different positive divisors of 20080808 divided by 8 is $\qquad$
|
3. 8 .
From $20080808=2^{3} \times 11 \times 17 \times 31 \times 433$, we know that the number of different positive divisors of 20080808 is
$$
N=(3+1)(1+1)^{4}=8^{2} \text { (divisors). }
$$
Dividing by 8 gives 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A line $l$ passing through the right focus $F$ of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$, and there are exactly 3 such lines, find $\lambda$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
(The polar equation of the hyperbola is
$$
\rho=\frac{2}{1-\sqrt{3} \cos \theta} \text {. }
$$
Let $A B$ be a chord passing through the right focus and intersecting only the right branch.
Then $|A B|=\rho_{1}+\rho_{2}$
$$
\begin{array}{l}
=\frac{2}{1-\sqrt{3} \cos \theta}+\frac{2}{1-\sqrt{3} \cos (\theta+\pi)} \\
=\frac{4}{1-3 \cos ^{2} \theta} \geqslant 4
\end{array}
$$
(when $\theta=\frac{\pi}{2}$, the equality holds), that is, among the chords passing through the right focus of the hyperbola and intersecting two points on the right branch, the minimum length is 4 if and only if the chord is perpendicular to the $x$-axis. Since there are exactly 3 lines that satisfy the condition, we have
(1) There is only one line that intersects both the left and right branches of the hyperbola, which must be the real axis of the hyperbola by symmetry, and there are only two lines that intersect the right branch, which can be verified not to meet the condition;
(2) There are only two lines that intersect both the left and right branches of the hyperbola, and only one line that intersects the right branch, which must be perpendicular to the $x$-axis by symmetry. In this case, $|A B|=\lambda=4$. When $\lambda=4$, it can be proven that there are two lines that intersect both the left and right branches of the hyperbola and have a length of 4, so $\lambda=4$. )
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 2, given that $A, B, C, D$ are four points on a plane that are not concyclic,
$$
\begin{array}{l}
\triangle A B D, \triangle A D C \text {, } \\
\triangle B C D, \triangle A B C
\end{array}
$$
have circumcenters
$$
\text { as } E, F, G, H \text {, }
$$
respectively. The line segments $E G, F H$
intersect at point $I$. If
$$
A I=4, B I=3 \text {, then } C I=
$$
|
8.4.
Since points $E$ and $G$ lie on the perpendicular bisector of $BD$, $EG$ is the perpendicular bisector of segment $BD$. Similarly, $FH$ is the perpendicular bisector of segment $AC$. Since $I$ is the intersection of $EG$ and $FH$, it follows that $CI = AI = 4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (40 points) There is a stack of cards numbered $1 \sim 15$. After arranging them in a certain order, the following two operations are performed: Place the top card on the table, then place the second card at the bottom of the stack. These two operations are repeated until all 15 cards are sequentially placed on the table. If the cards on the table are numbered from top to bottom as $1 \sim 15$ in order, what is the fourth card from the bottom of the original stack?
|
1. Assuming the cards are arranged in descending order from top to bottom as shown from left to right in Table 1. The first card originally becomes the bottom card, so it must be 15. The third card must be 14. And so on.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & $\beta$ & 13 & & 12 & & 11 & & 10 & & 9 & & 8 \\
\hline
\end{tabular}
Since the top 8 cards have been removed, the card marked “ $\alpha$ ” becomes the card after 8 and should be placed at the bottom. Therefore, the card marked “ $\beta$ ” is 7. And so on, we get Table 2.
Table 2
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & 7 & 13 & & 12 & 6 & 11 & & 10 & 5 & 9 & & 8 \\
\hline
\end{tabular}
Similarly, we can determine the original arrangement of the cards (see Table 3).
Table 3
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & 4 & 14 & 7 & 13 & 2 & 12 & 6 & 11 & 3 & 10 & 5 & 9 & 1 & 8 \\
\hline
\end{tabular}
Therefore, the fourth-to-last card is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In a $10 \times 10$ grid, there is a shape composed of $4 n$ $1 \times 1$ small squares, which can be covered by $n$ "田" shaped figures, or by $n$ "世" or "円" shaped figures (which can be rotated). Find the minimum value of the positive integer $n$.
(Zhu Huaiwei, problem contributor)
|
7. Let the given shapes be denoted as type $A$ and type $B$.
First, we prove that $n$ is even.
Color the $10 \times 10$ grid as shown in Figure 6.
Regardless of which 4 squares type $A$ covers, the number of black squares must be even, while for type $B$ it is odd. If $n$ is odd, the number of black squares covered by $n$ type $A$ shapes must be even; while the number of black squares covered by $n$ type $B$ shapes must be odd, leading to a contradiction. Therefore, $n$ must be even.
If $n=2$, the shapes formed by two type $A$ shapes can only be the two cases shown in Figure 7. However, neither of these can be formed by two type $B$ shapes.
Thus, $n \geq 4$.
Figure 8 shows the tiling when $n=4$.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $41^{x}=2009,7^{y}=2009$. Then the value of $\frac{1}{x}+\frac{2}{y}$ is . $\qquad$
|
4. 1 .
From $41^{x}=2009$, we get $41^{x y}=2009^{y}$.
$$
\begin{array}{l}
\text { Also } 7^{y}=2009 \Rightarrow 7^{2 y}=2009^{2} \\
\Rightarrow 49^{y}=2009^{2} \Rightarrow 49^{x y}=2009^{2 x} .
\end{array}
$$
$$
\text { Also } 41 \times 49=2009 \Rightarrow 41^{x y} \times 49^{x y}=2009^{x y} \text {. }
$$
Therefore $2009^{y} \times 2009^{2 x}=2009^{x y}$
$$
\begin{array}{c}
\Rightarrow 2009^{2 x+y}=2009^{x y} \Rightarrow 2 x+y=x y \\
\Rightarrow \frac{2}{y}+\frac{1}{x}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (14 points) Let positive real numbers $x, y, z$ satisfy $xyz=1$. Try to find the maximum value of
$$
f(x, y, z)=(1-yz+z)(1-xz+x)(1-xy+y)
$$
and the values of $x, y, z$ at that time.
|
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 - y z + z < 0 , } \\
{ 1 - x z + x < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+x z<1, \\
y+x y<1
\end{array}\right.\right. \\
\Rightarrow(x+x z)(y+x y)<1 \\
\Leftrightarrow x+x y+x^{2} y<0,
\end{array}
$$
Contradiction. Therefore, among $1-y z+z$, $1-x z+x$, and $1-x y+y$, at most one can be negative.
Assume all three expressions are positive. Then,
$$
\begin{array}{l}
(1-y z+z)(1-x z+x)(1-x y+y) \\
=(x-x y z+x z)(y-x y z+x y)(z-x y z+y z) \\
=(x-1+x z)(y-1+x y)(z-1+y z),
\end{array}
$$
Thus,
$$
\begin{array}{l}
f^{2}(x, y, z) \\
= {[(1-y z+z)(1-x z+x)(1-x y+y)]^{2} } \\
= {[(1-y z+z)(1-x z+x)(1-x y+y)] . } \\
{[(x-1+x z)(y-1+x y)(z-1+y z)] } \\
= {\left[z^{2}-(1-y z)^{2}\right]\left[x^{2}-(1-x z)^{2}\right]\left[y^{2}-(1-x y)^{2}\right] } \\
\leqslant(x y z)^{2}=1 .
\end{array}
$$
When $x=y=z=1$, the equality holds.
Therefore, $f(x, y, z)_{\text {max }}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If $p$ is a prime number, and $p+3$ divides $5p$, then the last digit of $p^{2009}$ is $\qquad$ .
|
8. 2 .
Since the positive divisors of $5p$ are $1, 5, p, 5p$, and $(p+3)$ divides $5p$, therefore, $p=2$.
Thus, $p^{2009}=2^{2009}=\left(2^{4}\right)^{502} \times 2=2 \times 16^{502}$.
Given that the last digit of $16^{502}$ is 6, it follows that the last digit of $p^{2009}$ is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in Figure 2, in quadrilateral $A B C D$, $\angle A C B=$ $\angle B A D=105^{\circ}, \angle A B C=\angle A D C=45^{\circ}$. If $A B$ $=2$, then the length of $C D$ is $\qquad$.
|
9. 2 .
As shown in Figure 6, draw $EA \perp AB$ intersecting the extension of $BC$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle AEB=45^{\circ} \\
\quad=\angle ADC, \\
AE=AB=2 . \\
\text { Also } \angle DAC=\angle DAB-\angle CAB \\
=\angle DAB-\left(180^{\circ}-\angle ABC-\angle ACB\right) \\
=75^{\circ}=180^{\circ}-\angle ACB=\angle ACE,
\end{array}
$$
Therefore, $\triangle ACE \cong \triangle CAD \Rightarrow CD=AE=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given $\frac{x y}{x+y}=2, \frac{x z}{x+z}=3, \frac{y z}{y+z}=4$. Find the value of $7 x+5 y-2 z$.
|
Three, 11. Given
$$
\begin{array}{l}
\frac{1}{2}=\frac{1}{x}+\frac{1}{y}, \frac{1}{3}=\frac{1}{x}+\frac{1}{z}, \\
\frac{1}{4}=\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
Solving simultaneously, we get $x=\frac{24}{7}, y=\frac{24}{5}, z=24$.
Therefore, $7 x+5 y-2 z=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (14 points) Let the line $l: y=k x+m(k, m \in$ Z) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Question: Does there exist a line $l$ such that the vector $\overrightarrow{A C}+\overrightarrow{B D}=\mathbf{0}$? If it exists, indicate how many such lines there are; if not, explain the reason.
|
Given the system of equations:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{16}+\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0.
$$
Let \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then:
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}. \\
\Delta_{1}=(8 k m)^{2}-4\left(3+4 k^{2}\right)\left(4 m^{2}-48\right)>0.
\end{array}
$$
From the system:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{4}-\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3-k^{2}\right) x^{2}-2 k m x-m^{2}-12=0.
$$
Let \( C\left(x_{3}, y_{3}\right) \) and \( D\left(x_{4}, y_{4}\right) \). Then:
$$
x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}.
$$
$$
\Delta_{2}=(-2 k m)^{2}+4\left(3-k^{2}\right)\left(m^{2}+12\right)>0.
$$
Since \( \overrightarrow{A C}+\overrightarrow{B D}=0 \), we have:
$$
\left(x_{4}-x_{2}\right)+\left(x_{3}-x_{1}\right)=0.
$$
Thus, \( \left(y_{4}-y_{2}\right)+\left(y_{3}-y_{1}\right)=0 \).
From \( x_{1}+x_{2}=x_{3}+x_{4} \), we get:
$$
-\frac{8 k m}{3+4 k^{2}}=\frac{2 k m}{3-k^{2}}.
$$
Therefore, \( 2 k m=0 \) or \( -\frac{4}{3+4 k^{2}}=\frac{1}{3-k^{2}} \).
Solving the above, we get \( k=0 \) or \( m=0 \).
When \( k=0 \), from equations (1) and (2), we get:
$$
-2 \sqrt{3}<m<2 \sqrt{3}.
$$
Since \( m \) is an integer, the values of \( m \) are:
$$
-3,-2,-1,0,1,2,3.
$$
When \( m=0 \), from equations (1) and (2), we get \( -\sqrt{3}<k<\sqrt{3} \). Since \( k \) is an integer, the values of \( k \) are:
$$
-1,0,1.
$$
Thus, there are 9 lines that satisfy the conditions.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (15 points) Find the maximum and minimum values of the function
$$
y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}
$$
|
3. The domain of the function is $[0,13]$.
$$
\begin{array}{l}
\text { Given } y=\sqrt{x}+\sqrt{x+27}+\sqrt{13-x} \\
=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \\
\geqslant \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13},
\end{array}
$$
we know that the equality holds when $x=0$.
Thus, the minimum value of $y$ is $3 \sqrt{3}+\sqrt{13}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
y^{2}=(\sqrt{x}+\sqrt{x+27}+\sqrt{13-x})^{2} \\
\leqslant\left(\frac{1}{2}+1+\frac{1}{3}\right)[2 x+(x+27)+3(13-x)] \\
=121 .
\end{array}
$$
Thus, $y \leqslant 11$.
By the condition for equality in the Cauchy-Schwarz inequality, we get
$$
4 x=9(13-x)=x+27 \Rightarrow x=9 \text {. }
$$
Thus, when $x=9$, the equality in (1) holds.
Therefore, the maximum value of $y$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given three non-zero real numbers $a, b, c$, the set $A=$ $\left\{\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b}\right\}$. Let $x$ be the sum of all elements in set $A$, and $y$ be the product of all elements in set $A$. If $x=2 y$, then the value of $x+y$ is $\qquad$
|
$-1 .-6$.
From the problem, we have
$$
x=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}
$$
$$
\begin{aligned}
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-3, \\
y & =\frac{a+b}{c} \cdot \frac{b+c}{a} \cdot \frac{c+a}{b} \\
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-1 .
\end{aligned}
$$
Then $x=y-2$.
Also, $x=2y$, solving this gives $y=-2$.
Thus, $x+y=2y-2=-6$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Let $A=\{i \in \mathbf{N} \mid 1 \leqslant i \leqslant$ $2880\}, B \subseteq A,|B| \geqslant 9$. If all elements in set $A$ can be represented by the sum of no more than 9 different elements from $B$, find $\min |B|$, and construct a set corresponding to the minimum $|B|$.
|
Let $|B|=k \geqslant 9$.
According to the problem, we should have $S_{k}=\sum_{i=1}^{9} \mathrm{C}_{k}^{i} \geqslant 2880$.
Notice that $S_{9}=2^{9}-1, S_{10}=2^{10}-2$,
$$
\begin{array}{l}
S_{11}=2^{11}-2-\mathrm{C}_{11}^{10}=2035, \\
S_{12}=2^{12}-2-\mathrm{C}_{12}^{10}-\mathrm{C}_{12}^{11}=4016>2880 .
\end{array}
$$
Thus, $k \geqslant 12$.
Next, we prove:
$$
\begin{aligned}
B= & \left\{1,2,2^{2}, 2^{3}, 2^{4}, 2^{5}-1,2^{6}-2,2^{7}-4,\right. \\
& \left.2^{8}-8,2^{9}-16,2^{10}-32,2^{11}-80\right\} \\
= & \left\{b_{i} \mid i=0,1, \cdots, 11\right\}
\end{aligned}
$$
satisfies the conditions.
(1) First, we prove by mathematical induction: For any $n \in\left[1, b_{i}-1\right](i=5,6, \cdots, 10)$, $n$ can be expressed as the sum of at most $i-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, i-1\right\}$.
When $i=5$, for any $n \in\left[1, b_{5}-1\right]$, by binary knowledge,
$$
n=\sum_{j=0}^{4} \varepsilon_{j} j^{j}=\sum_{j=0}^{4} \varepsilon_{j} b_{j},
$$
where $\varepsilon_{j}=0$ or $1$, $\varepsilon_{j}$ are not all $1$, $j=0,1, \cdots, 4$.
That is, $n$ can be expressed as the sum of at most 4 different elements from $\left\{b_{j} \mid j=0,1, \cdots, 4\right\}$.
Assume the proposition holds for $i=k \geqslant 5$.
When $i=k+1$, by the induction hypothesis, it is easy to see that the proposition holds for $n \in \left[1, b_{k}\right]$; for $n \in\left[b_{k}+1, b_{k+1}-1\right]$, $n-b_{k} \in\left[1, b_{k}-1\right]$.
By the induction hypothesis, $n-b_{k}$ can be expressed as the sum of at most $k-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}$, hence $n$ can be expressed as the sum of at most $k$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k\right\}$.
(2) For $n \in\left[b_{10}+1, b_{11}-1\right]$, take $k \in \mathbf{N}$ such that
$$
\sum_{j=k}^{10} b_{j}>n \geqslant \sum_{j=k+1}^{10} b_{j} .
$$
If $k \leqslant 3$, then $n \geqslant \sum_{j=4}^{10} b_{j}=b_{11}+1$, which is a contradiction.
If $k=4$, then $n-\sum_{j=5}^{10} b_{j} \in\left[0, b_{4}-2\right]$, by (1) we know $n-\sum_{j=5}^{10} b_{j}$ can be expressed as the sum of at most 3 different elements from $\left\{b_{j} \mid j=0,1,2,3\right\}$. Hence $n$ can be expressed as the sum of at most 9 different elements from $\left\{b_{j} \mid j=0,1, \cdots, 10\}$.
If $k \geqslant 4$, then $n-\sum_{j=k+1}^{10} b_{j} \in\left[0, b_{k}-1\right]$, by (1) we know $n-\sum_{j=k+1}^{10} b_{j}$ can be expressed as the sum of at most $k-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}$. Hence $n$ can be expressed as the sum of at most
$$
(10-k)+(k-1)=9
$$
different elements from $\left\{b_{j} \mid j=0,1, \cdots, 10\}$.
(3) For $n \in\left[b_{11}+1,2880\right]$, then
$n-b_{11} \in\left[1, b_{10}-1\right]$.
Take $k \in \mathbf{N}$ such that $\sum_{j=k}^{9} b_{j}>n-b_{11} \geqslant \sum_{j=k+1}^{9} b_{j}$.
Thus, $n-b_{11}-\sum_{j=k+1}^{9} b_{j} \in\left[0, b_{k}-1\right]$.
By (1) we know $n-b_{11}-\sum_{j=k+1}^{9} b_{j}$ can be expressed as the sum of
$$
\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}
$$
of at most $k-1$ different elements.
Hence $n$ can be expressed as the sum of at most
$$
1+(9-k)+(k-1)=9
$$
different elements from $\left\{b_{j} \mid j=0,1, \cdots, 11\}$.
In summary, $\min |B|=12, B=\{1,2,4,8,16, 31,62,124,248,496,992,1968\}$ is a set that satisfies the conditions.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let point $O$ be outside $\triangle A B C$, and
$$
\overrightarrow{O A}-2 \overrightarrow{O B}-3 \overrightarrow{O C}=0 \text {. }
$$
Then $S_{\triangle A B C}: S_{\triangle O B C}=$
|
$-1.4$.
As shown in Figure 4, let $D$ and $E$ be the midpoints of sides $AB$ and $BC$, respectively, and connect $CD$. Then
$$
\begin{array}{l}
\overrightarrow{OA}+\overrightarrow{OB}=2 \overrightarrow{OD}, \\
\overrightarrow{OB}+\overrightarrow{OC}=2 \overrightarrow{OE} .
\end{array}
$$
(1) - (2) $\times 3$ gives
$$
\begin{array}{l}
0=\overrightarrow{OA}-2 \overrightarrow{OB}-3 \overrightarrow{OC} \\
=2 \overrightarrow{OD}-6 \overrightarrow{OE} . \\
\text { Therefore, } \overrightarrow{OD}=3 \overrightarrow{OE} .
\end{array}
$$
Thus, $\overrightarrow{OD}$ and $\overrightarrow{OE}$ are collinear, and $|\overrightarrow{OD}|=3|\overrightarrow{OE}|$.
Hence, $|\overrightarrow{DE}|=2|\overrightarrow{OE}|$.
$$
\text { Therefore, } \frac{S_{\triangle BCD}}{S_{\triangle OBC}}=\frac{2}{1}=2, \frac{S_{\triangle ABC}}{S_{\triangle OBC}}=\frac{2 S_{\triangle BCD}}{S_{\triangle OBC}}=4 \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (14 points) In the Cartesian coordinate system $x O y$, points with both integer coordinates are called integer points. Given $O(0,0), A(2,1)$, and $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$. If $S_{\triangle O M M}=3$, find the number of integer points $M$ that satisfy this condition.
|
9. Connect $O A$. It is easy to know that there are two integer points $M_{1}(-6,0)$ and $M_{2}(6,0)$ on the $x$-axis inside the ellipse that satisfy the problem.
Draw two lines $l_{1}$ and $l_{2}$ parallel to the line $O A$ through points $M_{1}$ and $M_{2}$, respectively.
According to the principle that triangles with equal bases and heights have equal areas, all integer points $M$ that meet the conditions lie on the lines $l_{1}$ and $l_{2}$. It is easy to know
$$
k_{O A}=\frac{1-0}{2-0}=\frac{1}{2} \text {. }
$$
Therefore, the equations of the lines $l_{1}$ and $l_{2}$ are
$$
y=\frac{1}{2}(x+6), y=\frac{1}{2}(x-6) \text {. }
$$
Given that $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$, we have
$$
\frac{x^{2}}{200}+\frac{y^{2}}{8}<1 \text {. }
$$
Solve $\left\{\begin{array}{l}\frac{x^{2}}{200}+\frac{y^{2}}{8}<1, \\ y=\frac{1}{2}(x+6)\end{array}\right.$ and $\left\{\begin{array}{l}\frac{x^{2}}{200}+\frac{y^{2}}{8}<1, \\ y=\frac{1}{2}(x-6),\end{array}\right.$ to get $-10<x<-\frac{10}{29}, \frac{10}{29}<x<10$.
Since $M$ is an integer point and lies on the lines $l_{1}$ and $l_{2}$, $x$ must be an even number.
Therefore, in $-10<x<-\frac{10}{29}$ and $\frac{10}{29}<x<10$, $x$ has 4 even numbers each.
Thus, the number of integer points that meet the conditions is 8.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]$ equals $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$).
|
2. 6 .
Given $0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$, then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]=0$ or 1. From the problem, we know that 18 of them are equal to 1. Therefore,
$$
\begin{array}{l}
{\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0,} \\
{\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1}
\end{array}
$$
Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$
$\Rightarrow 18 \leqslant 30 a<19 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$
$\Rightarrow[10 a]=6$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=
$$
$\qquad$ .
|
5. 2 .
Notice that
$$
\begin{array}{l}
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
= \sqrt{(x-1)+2 \sqrt{x-1}+1}+ \\
\sqrt{(x-1)-2 \sqrt{x-1}+1} \\
= \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}} \\
=|\sqrt{x-1}+1|+|\sqrt{x-1}-1| .
\end{array}
$$
Since $1 \leqslant x \leqslant 2$, we have $\sqrt{x-1}-1 \leqslant 0$.
$$
\begin{array}{l}
\text { Therefore, } \sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
=\sqrt{x-1}+1-(\sqrt{x-1}-1)=2 \text {. }
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given an integer $n \geqslant 3$. Find the smallest positive integer $k$, such that there exists a $k$-element set $A$ and $n$ pairwise distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1}+x_{2}, x_{2}+x_{3}, \cdots$, $x_{n-1}+x_{n}, x_{n}+x_{1}$ all belong to $A$. (Xiong Bin provided)
|
2. Let $x_{1}+x_{2}=m_{1}, x_{2}+x_{3}=m_{2}, \cdots \cdots$
$x_{n-1}+x_{n}=m_{n-1}, x_{n}+x_{1}=m_{n}$.
First, $m_{1} \neq m_{2}$, otherwise, $x_{1}=x_{3}$, which is a contradiction.
Similarly, $m_{i} \neq m_{i+1}\left(i=1,2, \cdots, n, m_{n+1}=m_{1}\right)$.
Thus, $k \geqslant 2$.
If $k=2$, let $A=\{a, b\}(a \neq b)$, such that
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{2}+x_{3}=b, \\
\cdots \ldots . \\
x_{n-1}+x_{n}=b, \\
x_{n}+x_{1}=a
\end{array} \text { ( } n\right. \text { is odd), } \\
\text { or }\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{2}+x_{3}=b, \\
\cdots \cdots \\
x_{n-1}+x_{n}=a, \\
x_{n}+x_{1}=b
\end{array} \quad(n \text { is even). }\right.
\end{array}
$$
For the system (1), $x_{n}=x_{2}$, which is a contradiction.
For the system (2),
$$
\begin{array}{l}
\frac{n}{2} a=\left(x_{1}+x_{2}\right)+\left(x_{3}+x_{4}\right)+\cdots+\left(x_{n-1}+x_{n}\right) \\
=\left(x_{2}+x_{3}\right)+\left(x_{4}+x_{5}\right)+\cdots+\left(x_{n}+x_{1}\right) \\
=\frac{n}{2} b,
\end{array}
$$
Thus, $a=b$, which is a contradiction.
For $k=3$, an example can be constructed as follows:
$$
\text { Let } x_{2 k-1}=k, x_{2 k}=n+1-k(k=1,2, \cdots) \text {. }
$$
Then, when $n$ is even,
$$
x_{\mathrm{i}}+x_{\mathrm{i}+1}=\left\{\begin{array}{ll}
n+1, & i \text { is odd; } \\
n+2, & i \text { is even and } i<n ; \\
\frac{n}{2}+2, & i=n\left(x_{n+1}=x_{1}\right) .
\end{array}\right.
$$
When $n$ is odd,
$$
x_{i}+x_{i+1}=\left\{\begin{array}{ll}
n+1, & i \text { is odd and } i<n ; \\
n+2, & i \text { is even; } \\
\frac{n-1}{2}+2, & i=n\left(x_{n+1}=x_{1}\right) .
\end{array}\right.
$$
In summary, the minimum value of $k$ is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
|
11. 9 .
Since $(\sqrt{2}+\sqrt{3})^{2010}+(\sqrt{2}-\sqrt{3})^{2010}$ is an integer, therefore, the fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is
$$
\begin{array}{l}
1-(\sqrt{2}-\sqrt{3})^{2010} . \\
\text { Also } 0<(\sqrt{2}-\sqrt{3})^{2010}<0.2^{1005}<(0.008)^{300} \text {, then } \\
0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1 .
\end{array}
$$
It follows that the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9 .
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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