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1. If the quadratic equation with integer coefficients
$$
x^{2}+(a+3) x+2 a+3=0
$$
has one positive root $x_{1}$ and one negative root $x_{2}$, and $\left|x_{1}\right|<\left|x_{2}\right|$, then
$$
a=
$$
$\qquad$
|
$=、 1 .-2$.
Since the two roots of the equation are not equal, we have $\Delta>0$, that is
$$
(a+3)^{2}>4(2 a+3) \text {. }
$$
Solving this, we get $a>3$ or $a-3, a<-\frac{3}{2}$, which means $-3<a<-\frac{3}{2}$.
Since $a$ is an integer, then $a=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Given the quadratic function
$$
y=x^{2}+2 m x-n^{2} \text {. }
$$
(1) If the graph of this quadratic function passes through the point $(1,1)$, and let the larger of the two numbers $m, n+4$ be $P$, find the minimum value of $P$;
(2) If $m, n$ vary, these functions represent different parabolas. If each parabola intersects the coordinate axes at three distinct points, prove that the circles passing through these two intersection points all pass through the same fixed point, and find the coordinates of this fixed point.
|
(1) From the quadratic function passing through the point $(1,1)$, we get $m=\frac{n^{2}}{2}$.
Notice that
$$
\begin{array}{l}
m-(n+4)=\frac{n^{2}}{2}-(n+4) \\
=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2),
\end{array}
$$
Therefore, $P=\left\{\begin{array}{ll}\frac{n^{2}}{2}, & n \leqslant-2 \text { or } n \geqslant 4 ; \\ n+4, & -2<n<4 .\end{array}\right.$
Using the function graph, we can see that when $n=-2$, $P_{\min }=2$.
(2) The graph intersects the coordinate axes at three different points, which can be set as $A\left(x_{1}, 0\right) 、 B\left(x_{2}, 0\right) 、 C\left(0,-n^{2}\right)$.
Also, $x_{1} x_{2}=-n^{2}$. If $n=0$, it would contradict the condition of three intersection points, so $x_{1} x_{2}=-n^{2}<0$. Therefore, $x_{1} 、 x_{2}$ are on opposite sides of the origin.
Also, $\left|x_{1} x_{2}\right|=n^{2} \times 1$, so there exists a point $P_{0}(0,1)$ such that $|O A| \cdot|O B|=\left|O P_{0}\right| \cdot|O C|$.
Thus, points $A 、 B 、 C 、 P_{0}$ are concyclic, meaning these circles must pass through the fixed point $P_{0}(0,1)$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In pentagon $A B C D E$, $\angle A=\angle C=90^{\circ}$, $A B=B C=D E=A E+C D=3$. Then the area of this pentagon is $(\quad)$.
(A) 9
(B) 10.5
(C) 12
(D) 13.5
|
3. A.
As shown in Figure 3, extend $DC$ to point $F$ such that $CF = AE$, and connect $BE$, $BD$, and $BF$. Then,
$$
DF = DE = 3.
$$
Also, $\triangle BCF \cong \triangle BAE$, so
$$
BE = BF.
$$
Since $BD = BD$, we have
$$
\triangle BED \cong \triangle BFD.
$$
Therefore, $S_{\triangle BRD} = S_{\triangle BFD}$. Hence, $S_{\text{pentagon } 1BCDE} = S_{\text{quadrilateral } BRDF} = 2S_{\triangle BFD}$
$$
= 2 \times \frac{1}{2} DF \cdot BC = 9.
$$
|
9
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given $a+b+c=0, a^{2}+b^{2}+c^{2}=4$. Then, the value of $a^{4}+b^{4}+c^{4}$ is $\qquad$ .
|
11.8.
From the known equations, we get $a+b=-c, a^{2}+b^{2}=4-c^{2}$.
$$
\begin{array}{l}
\text { Also, } a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] \\
=\frac{1}{2}\left[(-c)^{2}-\left(4-c^{2}\right)\right]=c^{2}-2 .
\end{array}
$$
Therefore, $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}$
$$
=\left(4-c^{2}\right)^{2}-2\left(c^{2}-2\right)^{2}=8-c^{4} \text {. }
$$
Thus, $a^{4}+b^{4}+c^{4}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$ and $b$ are non-negative real numbers, and
$$
a^{2005}+b^{2005}=1, a^{2006}+b^{2006}=1 \text {. }
$$
then $a^{2007}+b^{2007}=$ $\qquad$
|
Ni, 1.1.
If $a$ or $b$ is greater than 1, then $a^{2005} + b^{2005} > 1$; if $0 < a, b < 1$, then $a^{2005} > a^{2006}$, $b^{2005} > b^{2006}$. Therefore, $a^{2000} + b^{2000} > a^{2006} + b^{2000}$, which means $1 > 1$, a contradiction. Hence, one of $a$ or $b$ is 0, and the other is 1. Thus, $a^{200} + b^{20 n} = 1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let the function $f(x)=x^{2}+a x+b(a, b \in$
$\mathbf{R})$. If there exists a real number $m$, such that
$$
|f(m)| \leqslant \frac{1}{4} \text {, and }|f(m+1)| \leqslant \frac{1}{4},
$$
find the maximum and minimum values of $\Delta=a^{2}-4 b$.
|
Solution 1: If $\Delta=a^{2}-4 b$
$$
\frac{-a-\sqrt{\Delta+1}}{2} \leqslant x \leqslant \frac{-a-\sqrt{\Delta-1}}{2}
$$
or $\frac{-a+\sqrt{\Delta-1}}{2} \leqslant x \leqslant \frac{-a+\sqrt{\Delta+1}}{2}$.
If $|f(m)| \leqslant \frac{1}{4}$, and $|f(m+1)| \leqslant \frac{1}{4}$, then it must be true that
$$
\frac{-a+\sqrt{\Delta-1}}{2}-\frac{-a-\sqrt{\Delta-1}}{2} \leqslant 1
$$
or $\frac{-a-\sqrt{\Delta-1}}{2}-\frac{-a-\sqrt{\Delta+1}}{2} \geqslant 1$
or $\frac{-a+\sqrt{\Delta+1}}{2}-\frac{-a+\sqrt{\Delta-1}}{2} \geqslant 1$,
which means $\sqrt{\Delta-1} \leqslant 1$ or $\sqrt{\Delta+1}-\sqrt{\Delta-1} \geqslant 1$.
Solving this, we get $\Delta \leqslant 2$.
When $a=1, b=-\frac{1}{4}$,
$|f(0)|=|f(-1)|=\frac{1}{4}$, and $\Delta=2$,
Therefore, the maximum value of $\Delta$ is 2.
Solution 2: Since $1=\left|\left(m+1-x_{0}\right)-\left(m-x_{0}\right)\right|$
$$
\leqslant\left|m+1-x_{0}\right|+\left|m-x_{0}\right|,
$$
thus, $\left|m+1-x_{0}\right|$ and $\left|m-x_{0}\right|$ must have one that is not less than $\frac{1}{2}$.
If $\Delta=a^{2}-4 b<0$, then
$$
\begin{array}{l}
f(m)=\left(m+\frac{a}{2}\right)^{2}-\frac{\Delta}{4}, \\
f(m+1)=\left(m+1+\frac{a}{2}\right)^{2}-\frac{\Delta}{4}
\end{array}
$$
one of them must be greater than $\frac{1}{4}$.
This contradicts the given condition, so, $\Delta \geqslant 0$.
The rest is the same as Solution 1.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given positive numbers $a, b, c$ satisfying $a+b+c=3$.
Prove:
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+ \\
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 .
\end{array}
$$
|
$$
\begin{array}{l}
\text { 5. } \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}} \\
=\frac{1}{3}\left(1+\frac{2 a+6}{a^{2}-2 a+3}\right)=\frac{1}{3}\left[1+\frac{2 a+6}{(a-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{2 a+6}{2}\right)=\frac{1}{3}(4+a) .
\end{array}
$$
Similarly, $\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}} \leqslant \frac{1}{3}(4+b)$,
$$
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant \frac{1}{3}(4+c) \text {. }
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \\
\leqslant \frac{1}{3}(12+a+b+c)=5 .
\end{array}
$$
|
5
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $\angle BAC=90^{\circ}$, $AB=AC$, points $D_{1}$ and $D_{2}$ are on $AC$, and satisfy $AD_{1}=CD_{2}$, $AE_{1} \perp BD_{1}$, $AE_{2} \perp BD_{2}$, intersecting $BC$ at points $E_{1}$ and $E_{2}$, respectively. Prove that: $\frac{CE_{2}}{BE_{2}}+\frac{CE_{1}}{BE_{1}}=1$.
---
The translation retains the original text's line breaks and formatting.
|
$$
\begin{array}{l}
\frac{C E_{1}}{B E_{1}}=\frac{A F}{A B}, \\
\angle E_{1} A C=\angle F C A . \\
\text { Also, } \angle A B D_{1}=90^{\circ}-\angle B A E_{1}=\angle E_{1} A C, \text { then } \\
\angle A B D_{1}=\angle A C F .
\end{array}
$$
Since $\angle B A D_{1}=\angle C A F=90^{\circ}, A B=A C$, therefore, $\triangle A B D_{1} \cong \triangle A C F$, which means $A D_{1}=A F$.
Thus, $\frac{C E_{1}}{B E_{1}}=\frac{A D_{1}}{A B}$.
Similarly, $\frac{C E_{2}}{B E_{2}}=\frac{A D_{2}}{A B}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{C E_{1}}{B E_{1}}+\frac{C E_{2}}{B E_{2}}=\frac{A D_{1}+A D_{2}}{A B} \\
=\frac{A D_{1}+A C-C D_{2}}{A B}=\frac{A C}{A B}=1 .
\end{array}
$$
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
10. There are four numbers, among which the sum of every three numbers is $24, 36, 28, 32$. Then the average of these four numbers is $\qquad$ .
|
10.10 .
The sum of these four numbers is $(24+36+28+32) \div 3=40$, so the average of these four numbers is 10 .
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. If $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, then $a^{2}+b^{2}=$ $\qquad$ .
|
11.3.
Given $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, we have $\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}+b^{2}\right)-6=0$.
Therefore, $a^{2}+b^{2}=3$ or -2.
Since $a^{2}+b^{2} \geqslant 0$, we have $a^{2}+b^{2}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x y+x+y+7=0, \\
3 x+3 y=9+2 x y,
\end{array}\right.
$$
then $x^{2} y+x y^{2}=$
|
13.6.
From $\left\{\begin{array}{l}x y+x+y+7=0, \\ 3 x+3 y=9+2 x y,\end{array}\right.$ we get $x y=-6, x+y=-1$.
Therefore, $x^{2} y+x y^{2}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. A person's 5 trips to work (unit: $\mathrm{min}$) are $a, b, 8, 9, 10$. It is known that the average of this set of data is 9, and the variance is 2. Then the value of $|a-b|$ is $\qquad$.
|
15.4.
Since the average of $a, b, 8, 9, 10$ is 9 and the variance is 2, it follows that $a, b, 8, 9, 10$ are 5 consecutive integers, so $a=7, b=11$ or $a=11, b=7$. Therefore, the value of $|a-b|$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. If the integer $m$ makes the equation
$$
x^{2}-m x+m+2006=0
$$
have non-zero integer roots, then the number of such integers $m$ is
$\qquad$.
|
16.5.
Let the two integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=m, a \beta=m+2006, \\
\text { i.e., } \alpha \beta-(\alpha+\beta)+1=2006+1 \\
=2007=(\alpha-1)(\beta-1) .
\end{array}
$$
Thus, $\alpha-1= \pm 1, \pm 3, \pm 9$;
$$
\beta-1= \pm 2007, \pm 669, \pm 223 \text {. }
$$
Therefore, there are 5 such integers $m$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $x, y$ be real numbers, the algebraic expression
$$
5 x^{2}+4 y^{2}-8 x y+2 x+4
$$
has a minimum value of
|
9.3.
$$
\begin{array}{l}
\text { Since } 5 x^{2}+4 y^{2}-8 x y+2 x+4 \\
=5\left(x-\frac{4}{5} y+\frac{1}{5}\right)^{2}+\frac{4}{5}(y+1)^{2}+3,
\end{array}
$$
Therefore, when $y=-1, x=-1$, the original expression has a minimum value of 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In a $3 \times 3$ grid filled with the numbers $1 \sim 9$, the largest number in each row is colored red, and the smallest number in each row is colored green. Let $M$ be the smallest number among the three red squares, and $m$ be the largest number among the three green squares. Then $M-m$ can have $\qquad$ different values.
|
14.8.
From the conditions, it is known that the values of $m$ and $M$ are positive integers from 3 to 7 (inclusive of 3 and 7).
Obviously, $M \neq m$, so the value of $M-m$ is
$$
1,2,3,4,-1,-2,-3,-4 \text {. }
$$
Therefore, $M-m$ can have 8 different values.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given $\frac{\sin (\alpha+2 \beta)}{\sin \alpha}=3$, and $\beta \neq \frac{1}{2} k \pi$, $\alpha+\beta \neq n \pi+\frac{\pi}{2}(n, k \in \mathbf{Z})$. Then the value of $\frac{\tan (\alpha+\beta)}{\tan \beta}$ is $\qquad$
|
$\begin{array}{l}13.2 \text {. } \\ \frac{\tan (\alpha+\beta)}{\tan \beta}=\frac{\sin (\alpha+\beta) \cdot \cos \beta}{\cos (\alpha+\beta) \cdot \sin \beta} \\ =\frac{\frac{1}{2}[\sin (\alpha+2 \beta)+\sin \alpha]}{\frac{1}{2}[\sin (\alpha+2 \beta)-\sin \alpha]}=\frac{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}+1}{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}-1} \\ =\frac{3+1}{3-1}=2 .\end{array}$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Given a finite set of planar vectors $M$, for any three elements chosen from $M$, there always exist two elements $\boldsymbol{a}, \boldsymbol{b}$ such that $\boldsymbol{a}+\boldsymbol{b} \in M$. Try to find the maximum number of elements in $M$.
|
Three, the maximum number of elements in the set $M$ is 7.
Let points $A, B, C$ be any three points on a plane. Consider the 7-element set $M=\{\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C A}, \boldsymbol{B A}, \boldsymbol{C B}, \boldsymbol{A C}, \mathbf{0}\}$, which clearly satisfies the conditions. We will now prove that the number of elements in $M$ cannot exceed 7.
When all elements in $M$ are collinear, move all the starting points to the same point and draw a line $l$ parallel to all the elements. Project all elements of $M$ onto the line $l$. Thus, all vectors in $M$ correspond to coordinates on a number line with the common starting point of the elements in $M$ projected onto $l$ as the origin and an arbitrary direction on $l$ as the positive direction. Therefore, the problem can be transformed into a number set problem (from two-dimensional to one-dimensional).
Next, we prove that there are at most 7 elements in this number set.
First, we prove that there are at most 3 positive numbers in the number set. Assume that there are at least 4 positive elements, and the largest 4 numbers are $a_{1}, a_{2}, a_{3}, a_{4}$, with $a_{1} < a_{2} < a_{3} < a_{4}$. Then, $a_{1} + a_{2} + a_{4} > a_{1} + a_{4} > a_{4}$, so the sum $a_{i} + a_{4} \notin M (i=1,2,3)$. Since there is only one element greater than $a_{3}$, which is $a_{4}$, but $a_{2} + a_{3} > a_{1} + a_{3} > a_{3}$, in the set $\left\{a_{1}, a_{3}, a_{4}\right\}$ or $\left\{a_{2}, a_{3}, a_{4}\right\}$, at least one set has the property that the sum of any two elements is not in $M$. This contradicts the given conditions, so the number set has at most 3 positive numbers. Similarly, the number set has at most 3 negative numbers. Adding a 0, thus, the number set $M$ has at most 7 elements.
When the elements in $M$ are not all collinear, move all the starting points to the same point $O$. By the finiteness of $M$, we can draw a Cartesian coordinate system $x O y$ such that the elements in $M$ are not parallel to the coordinate axes.
Next, we prove that there are at most 3 elements in the upper half-plane.
First, we prove that all elements in the upper half-plane are not collinear. Assume there exist elements $a_{1}$ and $a_{2}$ in the upper half-plane that are collinear. Then, take the element $b$ with the smallest angle with $a_{1}$ and $a_{2}$. Consider the set $\left\{a_{1}, a_{2}, b\right\}$. By the choice of $b$, $a_{1} + b$ and $a_{2} + b$ are not in $M$ (the sum vector lies between the two vectors). Thus, there exists $a_{3}$ in $M$ such that $a_{3} = a_{1} + a_{2}$, so $a_{3}$ is collinear with $a_{1}$ and $a_{2}$. Consider the set $\left\{a_{1}, a_{3}, b\right\}$, similarly, there exists $a_{4}$ collinear with $a_{1}$ and $a_{2}$. Continuing this discussion, there are infinitely many elements in $M$ collinear with $a_{1}$ and $a_{2}$, which is a contradiction. Therefore, all elements in the upper half-plane are not collinear.
Next, we prove that there are at most 3 elements in the upper half-plane. Assume there are at least 4 elements in the upper half-plane. Take 4 adjacent elements $a, b, c, d$ in the upper half-plane in a counterclockwise direction. Consider the set $\{a, b, c\}$, then $b = a + c$; consider the set $\{b, c, d\}$, then $c = b + d$. Thus, $b = a + c = a + b + d$, which implies $a + d = 0$. This contradicts the fact that $a$ and $d$ are in the upper half-plane, so there are at most 3 elements in the upper half-plane. Similarly, there are at most 3 elements in the lower half-plane. Adding the zero vector, thus, the set $M$ has at most 7 elements.
In conclusion, the maximum number of elements in the set $M$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of positive integers $n$ such that $n+1$ divides $n^{2006}+2006$ is $\qquad$.
Makes the positive integer $n$ such that $n+1$ can divide $n^{2006}+2006$ total $\qquad$.
Note: The second sentence seems to be a repetition or a different phrasing of the first. If it's meant to be a different statement, please clarify. Otherwise, I will assume the first translation is sufficient.
|
5. 5 .
From the problem, we have
$$
\begin{array}{l}
n^{2006}+2006 \equiv(-1)^{2006}+2006=2007 \\
\equiv 0(\bmod (n+1)) .
\end{array}
$$
Since $2007=3 \times 3 \times 223$, then
$$
n+1=3,9,223,669,2007 \text {. }
$$
Therefore, $n=2,8,222,668,2006$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given $n(n>1)$ integers (which can be the same) $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2007 .
$$
Then the minimum value of $n$ is $\qquad$
|
9. 5 .
Given $a_{1} a_{2} \cdots a_{n}=2007$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. Also, $a_{1}+a_{2}+\cdots+a_{n}=2007$ is an odd number, so $n$ is odd.
If $n=3$, i.e., $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2007$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, then
$$
a_{1} \geqslant \frac{a_{1}+a_{2}+a_{3}}{3}=669, a_{2} a_{3} \leqslant \frac{2007}{a_{1}} \leqslant 3 \text {. }
$$
If $a_{1}=669$, then $a_{2} a_{3}=3$. Thus,
$a_{2}+a_{3} \leqslant 4, a_{1}+a_{2}+a_{3} \leqslant 673669$, only $a_{1}=2007, a_{2} a_{3}=1$ and $a_{2}+a_{3}$ $=0$, which is impossible.
Therefore, $n \geqslant 5$.
$$
\begin{array}{l}
\text { Also, } 2007+1+1+(-1)+(-1) \\
=2007 \times 1 \times 1 \times(-1) \times(-1)=2007,
\end{array}
$$
Thus, the minimum value of $n$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.3. There are 19 cards. Can a non-zero digit be written on each card so that these 19 cards can be arranged to form a 19-digit number divisible by 11?
|
8.3. Answer: Yes.
Write a 2 on each of 10 cards, and a 1 on each of the remaining cards. It is well known that a positive decimal integer is divisible by 11 if and only if the difference $S$ between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11. Under all different arrangements of the 19 cards, we have
$$
-7 \leqslant S \leqslant 11,
$$
where only 11 is a multiple of 11, corresponding to placing the cards with 2s in all odd positions and the cards with 1s in all even positions.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of positive integer pairs $(x, y)$ that satisfy the equation
$$
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2006 x}-\sqrt{2006 y}+\sqrt{2006 x y} \\
\quad=2006
\end{array}
$$
|
3.8 .
Given the equation can be transformed into
$$
(\sqrt{x}+\sqrt{y}+\sqrt{2006})(\sqrt{x y}-\sqrt{2006})=0 \text {. }
$$
Since $\sqrt{x}+\sqrt{y}+\sqrt{2006}>0$, we have
$$
\sqrt{x y}=\sqrt{2006} \text {, }
$$
which means $\square$
$$
\begin{array}{l}
x y=2006=1 \times 2006=2 \times 1003 \\
=17 \times 118=34 \times 59 .
\end{array}
$$
Thus, $(x, y)=(1,2006),(2006,1),(2,1003)$, $(1003,2),(17,118),(118,17),(34,59),(59,34)$. Therefore, there are 8 pairs of positive integers $(x, y)$ that satisfy the given equation.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $x!+y!+z!=\overline{x y z}$. Then the value of $x+y+z$ is $\qquad$
|
5.10.
From $\overline{x y z}=x!+y!+z!$, we get
$$
100 x+10 y+z=x!+y!+z!.
$$
It is easy to see that $x, y, z \leqslant 6$. Otherwise,
$$
x!+y!+z!\geqslant 7!>1000.
$$
If $x=6$, then the left side of equation (1) is $700$, which is a contradiction.
Therefore, $x \leqslant 5$.
Similarly, $y \leqslant 5, z \leqslant 5$.
Thus, $x!+y!+z!\leqslant 3 \times 5!=360$. Hence, $x \leqslant 3$.
When $x=3$,
$$
294+10 y+z=y!+z!\leqslant 2 \times 5!=240,
$$
which is a contradiction.
When $x=2$,
$$
198+10 y+z=y!+z!.
$$
If $y \leqslant 4, z \leqslant 4$, then $y!+z!\leqslant 2 \times 4!<198$, which is a contradiction.
Therefore, one of $y, z$ must be 5.
(i) If $y=5,128+z=z$!, which is impossible;
(ii) If $z=5,83+10 y=y$!, then $y \mid 83$, which is also impossible.
When $x=1$, $99+10 y+z=y!+z!$.
If $y \leqslant 4, z \leqslant 4$, then $y!+z!\leqslant 48<99$, which is a contradiction.
Therefore, one of $y, z$ must be 5.
(i) If $y=5,29+z=z$!, which implies $z \mid 29$, which is impossible;
(ii) If $z=5,10 y=y!+16$, by calculation, $y$ can only be 4.
Thus, $x+y+z=10$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If the graph of the inverse proportion function $y=\frac{k}{x}$ intersects the graph of the linear function $y=a x+b$ at points $A(-2, m)$ and $B(5, n)$, then the value of $3 a+b$ is
保留了源文本的换行和格式。
|
Ni.7.0.
Since points $A(-2, m)$ and $B(5, n)$ lie on the graph of an inverse proportion function, we have
$$
\left\{\begin{array}{l}
m=-\frac{k}{2}, \\
n=\frac{k}{5} .
\end{array}\right.
$$
Since points $A(-2, m)$ and $B(5, n)$ also lie on the line $y=a x+b$, we get
$$
\left\{\begin{array}{l}
m=-2 a+b \\
n=5 a+b
\end{array}\right.
$$
Thus,
$$
\left\{\begin{array}{l}
-2 a+b=-\frac{k}{2}, \\
5 a+b=\frac{k}{5} .
\end{array}\right.
$$
Then,
$$
\left\{\begin{array}{l}
a=\frac{k}{10}, \\
b=-\frac{3 k}{10}
\end{array}\right.
$$.
Solving, we get $b=-3 a$.
Therefore, $3 a+b=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given that $a$, $b$, and $c$ are positive integers, and the graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from points $A$ and $B$ to the origin $O$ are both less than 1, find the minimum value of $a+b+c$.
|
14. Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
Given that $a$, $b$, and $c$ are positive integers, we have
$x_{1}+x_{2}=-\frac{b}{a} < 0$.
Thus, the equation $a x^{2}+b x+c=0$ has two negative real roots, i.e., $x_{1} < 0$.
Solving, we get $b>2 \sqrt{a c}$.
Since $|O A|=\left|x_{1}\right| < 0, a+c > b$.
$$
Given that $a$, $b$, and $c$ are positive integers, we have
$$
a+c \geqslant b+1 > 2 \sqrt{a c}+1.
$$
Thus, $a+c > 2 \sqrt{a c}+1$.
Then, $(\sqrt{a}-\sqrt{c})^{2} > 1, \sqrt{a}-\sqrt{c} > 1, \sqrt{a} > \sqrt{c}+1 \geqslant 2$.
Therefore, $a > 4$, i.e., $a \geqslant 5$.
Hence, $b > 2 \sqrt{a c} \geqslant 2 \sqrt{5 \times 1}=2 \sqrt{5}$, i.e., $b \geqslant 5$.
Thus, taking $a=5, b=5, c=1$, the equation $y=5 x^{2}+5 x+1$ satisfies the conditions.
Therefore, the minimum value of $a+b+c$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. At a party, 9 celebrities performed $n$ "trio dance" programs. If in these programs, any two people have collaborated exactly once, then $n=$ $\qquad$
|
12.12.
Consider 9 people as 9 points. If two people have performed in the same group, then a line segment is drawn between the corresponding two points. Thus, every pair of points is connected, resulting in a total of 36 line segments. Each trio dance corresponds to a triangle, which has three sides. When all sides appear and are not reused, the total number of triangles is \( n = \frac{36}{3} = 12 \), i.e., 12 performances. Such 12 trio dances can be arranged, for example:
$$
\begin{array}{l}
(1,2,3),(4,5,6),(7,8,9) ;(1,4,7),(2,5,8),(3,6,9) ; \\
(1,5,9),(2,6,7),(3,4,8) ;(1,6,8),(2,4,9),(3,5,7)
\end{array}
$$
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find the smallest positive real number $k$ such that for any 4 distinct real numbers $a, b, c, d$ not less than $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation $\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)=0$ has 4 distinct real roots. (Feng Zhigang)
|
2. On one hand, if $k4(d-a)>0, c^{2}-4 b>4(c-b)$ $>0$, therefore, equations (1) and (2) both have two distinct real roots.
Secondly, if equations (1) and (2) have a common real root $\beta$, then
$$
\left\{\begin{array}{l}
\beta^{2}+d \beta+a=0, \\
\beta^{2}+\phi \beta+b=0 .
\end{array}\right.
$$
Subtracting the two equations yields $\beta=\frac{b-a}{d-c}>0$, at this time, $\beta^{2}+d \beta+a>$ 0, which is a contradiction.
Therefore, equations (1) and (2) do not have a common real root.
Thus, $k=4$ meets the requirement.
In summary, $k=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a$, $b$, $c$ be positive integers, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=19 \text {. }
$$
Then the minimum value of $a+b+c$ is $\qquad$
|
4. 10 .
By the cyclic symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$.
Let $a-b=m, b-c=n$, then $a-c=m+n$, and
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[m^{2}+n^{2}+(m+n)^{2}\right],
\end{array}
$$
which is $m^{2}+m n+n^{2}-19=0$.
For the equation in $m$,
$$
\Delta=n^{2}-4\left(n^{2}-19\right) \geqslant 0 \text {. }
$$
Thus, $0 \leqslant n \leqslant \frac{2 \sqrt{57}}{3}0)$ has no solution.
Therefore, when $a=6, b=3, c=1$, $a+b+c$ achieves its minimum value, which is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 1, label the six vertices of a regular hexagon with the numbers $0,1,2,3,4,5$ in sequence. An ant starts from vertex 0 and crawls counterclockwise, moving 1 edge on the first move, 2 edges on the second move, $\cdots \cdots$ and $2^{n-1}$ edges on the $n$-th move. After 2005 moves, the value at the vertex where the ant is located is
|
Ni.1.1.
According to the problem, after 2005 movements, the ant has crawled over
$$
1+2+\cdots+2^{2004}=2^{2005}-1
$$
edges.
Let $x=2^{2005}-1$. Consider the remainder of $x$ modulo 6.
Since $x \equiv 1(\bmod 2), x \equiv 1(\bmod 3)$, we have
$$
3 x \equiv 3(\bmod 6), 2 x \equiv 2(\bmod 6), 5 x \equiv 5(\bmod 6) \text {. }
$$
Also, since $(5,6)=1$, it follows that $x \equiv 1(\bmod 6)$.
Since the value at the vertex where the ant is located is equivalent to the remainder of $x$ modulo 6, the value at the vertex where the ant is located is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $x \in\left(0, \frac{\pi}{2}\right), \sin ^{2} x, \sin x \cdot \cos x$, and $\cos ^{2} x$ cannot form a triangle, and the length of the interval of $x$ that satisfies the condition is $\arctan k$. Then $k$ equals $\qquad$ (Note: If $x \in(a, b), b>a$, then the length of such an interval is $b-a$).
|
2.2.
First, find the range of $x$ that can form a triangle.
(1) If $x=\frac{\pi}{4}$, then $\sin ^{2} \frac{\pi}{4} 、 \cos ^{2} \frac{\pi}{4} 、 \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}$ can form a triangle.
(2) If $x \in\left(0, \frac{\pi}{4}\right)$, at this time,
$$
\sin ^{2} x\cos ^{2} x$, that is,
$$
\tan ^{2} x+\tan x-1>0 \text {. }
$$
Solving this, we get $\arctan \frac{\sqrt{5}-1}{2}<x<\frac{\pi}{4}$.
(3) If $x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, similarly, we get
$$
\frac{\pi}{4}<x<\arctan \frac{\sqrt{5}+1}{2} \text {. }
$$
Thus, the range of $x$ for which these three positive numbers can form the side lengths of a triangle is $\left(\arctan \frac{\sqrt{5}-1}{2}, \arctan \frac{\sqrt{5}+1}{2}\right)$. The length of this interval is $\arctan \frac{\sqrt{5}+1}{2}-\arctan \frac{\sqrt{5}-1}{2}=\arctan \frac{1}{2}$. Therefore, the length of the interval of $x$ that satisfies the condition is $\frac{\pi}{2}-\arctan \frac{1}{2}=\arctan k$. It is easy to find that $k=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\frac{1}{2} \sin 2 y+a=0 .
\end{array}\right.
$$
then the value of $\cos (x+2 y)$ is
|
8.1.
Let $f(t)=t^{3}+\sin t$. Then $f(t)$ is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
From the original system of equations, we get $f(x)=f(-2 y)=2 a$, and since $x$ and $-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, hence $x+2 y=0$. Therefore, $\cos (x+2 y)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If the six edges of a tetrahedron are $2,3,4,5$, 6,7, then there are $\qquad$ different shapes (if two tetrahedra can be made to coincide by appropriate placement, they are considered the same shape).
|
6.10.
Let the edge of length $k$ be denoted as $l_{k}(k \in\{2,3,4,5,6,7\})$. Consider $l_{2}$ and $l_{3}$.
(1) If $l_{2}$ and $l_{3}$ are coplanar, then the other side of the plane must be $l_{4}$.
(i) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a clockwise direction (all referring to the direction of the three sides when viewed from inside the shape, the same below), as shown in Figure 6, then side $D A$ cannot be $l_{6}$ (otherwise, it would make the three sides of $\triangle B C D$ be $2,5,7$, which is a contradiction).
If $D A=l_{5},\{D B, D C\}=\left\{l_{6}, l_{7}\right\}$, there are 2 cases;
If $D A=l_{7},\{D B, D C\}=\left\{l_{5}, l_{6}\right\}$, there are also 2 cases.
A total of 4 cases are obtained.
(ii) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a counterclockwise direction, similarly, 4 cases are obtained.
(2) If $l_{2}$ and $l_{3}$ are skew, let $A B=l_{2}, C D=l_{3}$. Then the remaining four edges, each one must be adjacent to $l_{2}$ and $l_{3}$. Therefore, the other side of the plane containing $l_{2}$ and $l_{7}$ must be $l_{6}$.
(i) If $l_{2}$, $l_{6}$, and $l_{7}$ form a triangle in a clockwise direction, without loss of generality, let $A C=l_{6}, B C=l_{7}$, as shown in Figure 7. The remaining two edges, $B D$ cannot be $l_{4}$, so only $B D=l_{5}, A D=l_{4}$, yielding 1 case.
(ii) If $l_{2}$, $l_{6}$, and $l_{7}$ form a triangle in a counterclockwise direction, similarly, 1 case is obtained.
Therefore, in this problem, there are 10 different shapes.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (16 points) Does there exist a smallest positive integer $t$, such that the inequality
$$
(n+t)^{n+t}>(1+n)^{3} n^{n} t^{t}
$$
holds for any positive integer $n$? Prove your conclusion.
|
$$
\text { Three, 13. Take }(t, n)=(1,1),(2,2),(3,3) \text {. }
$$
It is easy to verify that when $t=1,2,3$, none of them meet the requirement.
When $t=4$, if $n=1$, equation (1) obviously holds.
If $n \geqslant 2$, we have
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\left[\frac{(n-2) n+2 \times 2 n+3(2 n+2)+2^{3}}{n+4}\right]^{n+4} \\
=\left(\frac{n^{2}+8 n+14}{n+4}\right)^{n+4}<\left(\frac{n^{2}+8 n+16}{n+4}\right)^{n+4} \\
=(n+4)^{n+4} .
\end{array}
$$
Therefore, equation (1) holds.
Thus, $t=4$ satisfies that for any positive integer $n$, equation (1) always holds.
|
4
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
15. (22 points) Let $A$ and $B$ be the common left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Let $P$ and $Q$ be moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
A P+B P=\lambda(A Q+B Q)(\lambda \in \mathbf{R},|\lambda|>1) \text {. }
$$
Let the slopes of the lines $A P$, $B P$, $A Q$, and $B Q$ be $k_{1}$, $k_{2}$, $k_{3}$, and $k_{4}$, respectively.
(1) Prove that $k_{1}+k_{2}+k_{3}+k_{4}=0$;
(2) Let $F_{1}$ and $F_{2}$ be the right foci of the ellipse and the hyperbola, respectively. If $P F_{2} \parallel Q F_{1}$, find the value of $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}$.
|
15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}.
$$
Similarly,
$$
k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}.
$$
Let $O$ be the origin, then
$$
2 O P=A P+B P=\lambda(A Q+B Q)=2 \lambda O Q .
$$
Therefore, $O P=\lambda O Q$.
Hence, $O, P, Q$ are collinear.
Thus, $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$.
From equations (1) and (2), we get $k_{1}+k_{2}+k_{3}+k_{4}=0$.
(2) Since point $Q$ is on the ellipse, we have $\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1$.
From $O P=\lambda O Q$, we get $\left(x_{1}, y_{1}\right)=\lambda\left(x_{2}, y_{2}\right)$.
Therefore, $x_{2}=\frac{1}{\lambda} x_{1}, y_{2}=\frac{1}{\lambda} y_{1}$.
Thus, $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=\lambda^{2}$.
Since point $P$ is on the hyperbola, we have $\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1$.
From equations (3) and (4), we get $x_{1}^{2}=\frac{\lambda^{2}+1}{2} a^{2}, y_{1}^{2}=\frac{\lambda^{2}-1}{2} b^{2}$.
Since $P F_{2} \parallel Q F_{1}$, we have $\left|O F_{2}\right|=\lambda\left|O F_{1}\right|$.
Therefore, $\lambda^{2}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}, \frac{x_{1}^{2}}{y_{1}^{2}}=\frac{\left(\lambda^{2}+1\right) a^{2}}{\left(\lambda^{2}-1\right) b^{2}}=\frac{a^{4}}{b^{4}}$.
From equation (1), we get $\left(k_{1}+k_{2}\right)^{2}=\frac{4 b^{4}}{a^{4}} \cdot \frac{x_{1}^{2}}{y_{1}^{2}}=4$.
Similarly, $\left(k_{3}+k_{4}\right)^{2}=4$.
On the other hand, $k_{1} k_{2}=\frac{y_{1}}{x_{1}+a} \cdot \frac{y_{1}}{x_{1}-a}=\frac{b^{2}}{a^{2}}$.
Similarly, $k_{3} k_{4}=-\frac{b^{2}}{a^{2}}$.
Therefore,
$$
k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}
=\left(k_{1}+k_{2}\right)^{2}+\left(k_{3}+k_{4}\right)^{2}-2\left(k_{1} k_{2}+k_{3} k_{4}\right)=8.
$$
|
8
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. Points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y=2 x^{2}$ are symmetric with respect to the line $y=x+m$. If $2 x_{1} x_{2}=-1$, then the value of $2 m$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
|
3. A.
$$
\begin{array}{l}
\text { Given } \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1, \\
\frac{y_{1}+y_{2}}{2}=\frac{x_{1}+x_{2}}{2}+m, \\
2 x_{1} x_{2}=-1 \text { and } y_{1}=2 x_{1}^{2}, y_{2}=2 x_{2}^{2},
\end{array}
$$
we get
$$
\begin{array}{l}
x_{2}-x_{1}=y_{1}-y_{2}=2\left(x_{1}^{2}-x_{2}^{2}\right) \Rightarrow x_{1}+x_{2}=-\frac{1}{2}, \\
2 m=\left(y_{1}+y_{2}\right)-\left(x_{1}+x_{2}\right)=2\left(x_{1}^{2}+x_{2}^{2}\right)+\frac{1}{2} \\
=2\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}+\frac{1}{2}=3 .
\end{array}
$$
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then
$$
\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=
$$
|
ニ、7.9.
From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta$, then
$$
\begin{array}{l}
(\cos \theta-1)^{2}=4 \Rightarrow \cos \theta=-1 . \\
\text { Therefore, }\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=9 .
\end{array}
$$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$ (where $[x]$ denotes the greatest integer not greater than $x$), then $a+b+c+d=$ . $\qquad$
|
8.3.
Given $x_{k^{2}+1}=x_{k^{2}+2}=\cdots=x_{(k+1)^{2}}=2 k+1$, that is, when $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, we have $x_{n}=2 k+1(k=[\sqrt{n-1}])$.
Therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Hence, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (12 points) Find the area of the figure formed by the set of points on the right-angle coordinate plane $b O a$
$$
S=\left\{(b, a) \mid f(x)=a x^{3}+b x^{2}-3 x\right.
$$
is a monotonic function on $\mathbf{R}$, and $a \geqslant-1\}$.
|
Three, 15. When $a=0$, by $f(x)$ being monotonic on $\mathbf{R}$, we know $b=0$. $f(x)$ being monotonic on $\mathbf{R}$ $\Leftrightarrow f^{\prime}(x)$ does not change sign on $\mathbf{R}$.
Since $f^{\prime}(x)=3 a x^{2}+2 b x-3$, therefore, by $\Delta=4 b^{2}+36 a \leqslant 0$,
we get, $a \leqslant-\frac{1}{9} b^{2}$.
Thus, the points $(b, a)$ that satisfy this condition form a closed figure in the $b O a$ coordinate plane, bounded by the curve $y=-\frac{1}{9} x^{2}$ and the line $y=-1$, with an area of
$$
6-2 \int_{0}^{3} \frac{1}{9} x^{2} \mathrm{~d} x=4
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of prime pairs $(p, q)$ that satisfy $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ is $\qquad$ .
|
5.2.
(1) When $p=2$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1$, which is not a prime number, contradiction.
(2) When $p=3$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1+1=2$ is a prime number.
(3) When $p=5$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=2+1=3$ is a prime number.
(4) When $p>5$, since $p$ is a prime number, $p=6k+1$ or $6k+5$.
If $p=6k+1$, then $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=3k+2k+k=6k$, which is not a prime number, contradiction;
If $p=6k+5$, then $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=(3k+2)+(2k+1)+k=6k+3$, which is not a prime number, contradiction.
Therefore, the number of prime pairs $(p, q)$ is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f\left(\frac{\pi}{4}\right)=0$, and for any $x, y \in \mathbf{R}$, we have
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) f\left(\frac{x-y}{2}\right) .
$$
Then $f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)$ $=$ . $\qquad$
|
6.0 .
Let $\frac{x-y}{2}=\frac{\pi}{4}$, then
$$
\begin{array}{l}
f(x)+f\left(x-\frac{\pi}{2}\right)=2 f\left(x-\frac{\pi}{4}\right) f\left(\frac{\pi}{4}\right)=0 . \\
\text { Therefore } f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) Find the maximum value of a prime number $p$ with the following property: there exist two permutations (which can be the same) of $1, 2, \cdots, p$, $a_{1}, a_{2}, \cdots, a_{p}$ and $b_{1}, b_{2}, \cdots, b_{p}$, such that the remainders of $a_{1} b_{1}$, $a_{2} b_{2}, \cdots, a_{p} b_{p}$ when divided by $p$ are all distinct.
|
Lemma (Wilson's Theorem): For any prime $p$, we have $(p-1)! \equiv -1 \pmod{p}$.
Proof of the Lemma: When $p=2,3$, equation (1) is obviously true.
When $p>3$, we prove: For any $2 \leqslant k \leqslant p-2$, there must exist $k^{\prime}\left(2 \leqslant k^{\prime} \leqslant p-2, k^{\prime} \neq k\right)$, such that $k k^{\prime} \equiv 1 \pmod{p}$.
In fact, for $2 \leqslant k \leqslant p-2$, we have $(k, p)=1$.
Thus, $k, 2k, \cdots, pk$ form a complete residue system modulo $p$.
Therefore, there must exist $k^{\prime}\left(1 \leqslant k^{\prime} \leqslant p\right)$, such that $k k^{\prime} \equiv 1 \pmod{p}$.
Clearly, $k^{\prime} \neq p$, otherwise $k k^{\prime} \equiv 0 \pmod{p}$, which is a contradiction.
Moreover, if $k^{\prime}=1$, then by $k k^{\prime} \equiv 1 \pmod{p}$, we get $k \equiv 1 \pmod{p}$, which contradicts $2 \leqslant k \leqslant p-2$.
If $k^{\prime}=p-1$, then by $k(p-1) \equiv 1 \pmod{p}$, we get $-k \equiv 1 \pmod{p}$, which contradicts $2 \leqslant k \leqslant p-2$.
If $k^{\prime}=k$, by $k k^{\prime} \equiv 1 \pmod{p}$, we get $k^2 \equiv 1 \pmod{p}$. Therefore, $p \mid (k+1)(k-1)$, but $1 \leqslant k-1 \leqslant p-3$, which contradicts equation (2), so $p \leqslant 2$.
When $p=2$, let
$$
a_{1}=1, a_{2}=2, b_{1}=1, b_{2}=2.
$$
At this point, $a_{1} b_{1}=1=1 \pmod{2}$,
$$
a_{2} b_{2}=4 \equiv 0 \pmod{2}.
$$
Therefore, $p=2$ satisfies the condition. Hence, $p_{\max}=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ are points on the diagonals $A C$ and $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$
|
5.1.
As shown in Figure 2, draw $M P \perp A B$ at $P$, and connect $P N$. It can be proven that $P N \perp A B$. Thus, $\angle M P N=120^{\circ}$.
Let $A M=F N=x$.
Therefore, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$,
which means $M P=\frac{\sqrt{2}}{2} x$.
Hence, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}\left(\frac{P N}{1}=\frac{B N}{\sqrt{2}}=\frac{B F-N F}{\sqrt{2}}=\frac{\sqrt{2}-x}{\sqrt{2}}\right)$.
Then, $M N^{2}=M P^{2}+P N^{2}-2 P M \cdot P N \cos 120^{\circ}$
$$
\begin{array}{l}
=\frac{x^{2}}{2}+\left(\frac{\sqrt{2}-x}{\sqrt{2}}\right)^{2}-2 \frac{\sqrt{2}}{2} x \cdot \frac{\sqrt{2}-x}{\sqrt{2}} \cos 120^{\circ} \\
=\frac{1}{2}\left(x^{2}-\sqrt{2} x+2\right) \\
=\frac{1}{2}\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\frac{3}{4}(0 \leqslant x \leqslant \sqrt{2}) .
\end{array}
$$
When $x=0$ or $x=\sqrt{2}$, $M N_{\text {max }}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $\cos \beta+\sin \beta \cdot \cot \theta=\tan 54^{\circ}$, $\cos \beta-\sin \beta \cdot \cot \theta=\tan 18^{\circ}$.
Then the value of $\tan ^{2} \theta$ is $\qquad$
|
6.1
From the given conditions, we have
$\cos \beta=\frac{1}{2}\left(\tan 54^{\circ}+\tan 18^{\circ}\right)=\frac{1}{2 \cos 54^{\circ}}$,
$\sin \beta=\frac{\tan \theta}{2}\left(\tan 54^{\circ}-\tan 18^{\circ}\right)=\frac{\tan \theta}{2 \cos 18^{\circ}}$.
Since $\sin ^{2} \beta+\cos ^{2} \beta=1$, we have
$\frac{1}{\cos ^{2} 54^{\circ}}+\frac{\tan ^{2} \theta}{\cos ^{2} 18^{\circ}}=4$.
Thus, $\tan ^{2} \theta=\left(4-\frac{1}{\cos ^{2} 54^{\circ}}\right) \cos ^{2} 18^{\circ}$
$$
\begin{array}{l}
=4 \cos ^{2} 18^{\circ}-\frac{\sin ^{2} 72^{\circ}}{\sin ^{2} 36}=4 \cos ^{2} 18^{\circ}-4 \cos ^{2} 36^{\circ} \\
=4\left(\cos 18^{\circ}+\cos 36^{\circ}\right)\left(\cos 18^{\circ}-\cos 36^{\circ}\right)
\end{array}
$$
$$
\begin{array}{l}
=16 \cos 27^{\circ} \cdot \cos 9^{\circ} \cdot \sin 27^{\circ} \cdot \sin 9^{\circ} \\
=\frac{4 \sin 18^{\circ} \cdot \cos 18^{\circ} \cdot \sin 54^{\circ}}{\cos 18^{\circ}}=\frac{2 \sin 36^{\circ} \cdot \cos 36^{\circ}}{\cos 18^{\circ}} \\
=\frac{\sin 72^{\circ}}{\cos 18^{\circ}}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { 2. Let } M=\frac{2 \cos 34^{\circ}-\cos 22^{\circ}}{\cos 14^{\circ}} \text {, } \\
N=\sin 56^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ} \text {. } \\
\text { Then } \frac{M}{N}=
\end{array}
$$
|
2.8.
$$
\begin{array}{l}
M=\frac{2 \cos 34^{\circ}-\sin 68^{\circ}}{\cos 14^{\circ}}=\frac{2 \cos 34^{\circ}\left(1-\sin 34^{\circ}\right)}{\cos 14^{\circ}} \\
=\frac{2 \cos 34^{\circ}\left(1-\cos 56^{\circ}\right)}{\cos 14^{\circ}}=\frac{4 \cos 34^{\circ} \cdot \sin ^{2} 28^{\circ}}{\cos 14^{\circ}} \\
=8 \cos 34^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ}=8 N .
\end{array}
$$
Therefore, $\frac{M}{N}=8$.
Note: A more general conclusion of this problem is
$$
\frac{2 \sin 4 \alpha-\sin 8 \alpha}{\cos \alpha}=8 \sin \alpha \cdot \sin 2 \alpha \cdot \sin 4 \alpha \text{. }
$$
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. From the 6 face diagonals of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, two tetrahedra $A-B_{1} C D_{1}$ and $A_{1}-B C_{1} D$ can be formed. If the combined volume of the two tetrahedra is 1 (overlapping parts are counted only once), then the volume of the rectangular prism is $\qquad$ .
|
3.2.
Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$ respectively, then the volume of the rectangular prism $V_{0}=a b c$. The volumes of the two tetrahedra (as shown in Figure 4) are both
$$
\begin{array}{l}
V_{1}=V_{0}-V_{A_{1}-A B D}-V_{A_{1}-B B C_{1}}-V_{A_{1}-D_{1} D C_{1}}-V_{C_{1}-B C D} \\
=V_{0}-4 \times \frac{1}{6} a b c=\frac{1}{3} V_{0} .
\end{array}
$$
The common part of the two tetrahedra is an octahedron with vertices at the centers of the 6 faces (as shown in Figure 5), and its volume is
$$
V_{2}=\frac{1}{3} \times \frac{1}{2} S_{\text {quadrilateral } \triangle B C D} \cdot h=\frac{1}{6} V_{0} .
$$
Therefore, the total volume occupied by the two tetrahedra is
$$
1=2 V_{1}-V_{2}=\frac{2}{3} V_{0}-\frac{1}{6} V_{0}=\frac{1}{2} V_{0} \Rightarrow V_{0}=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x=\frac{1}{2-\sqrt{5}}$. Then $x^{3}+3 x^{2}-5 x+1=$
|
From $x=\frac{1}{2-\sqrt{5}}=-2-\sqrt{5}$, we get $x+2=-\sqrt{5}$. Therefore, $(x+2)^{2}=5$, which means $x^{2}+4 x=1$.
Then $x^{3}+3 x^{2}-5 x+1=\left(x^{2}+4 x-1\right)(x-1)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let for any natural numbers $m$, $n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Then the maximum value of $\lambda$ is $\qquad$.
|
5.3.
The original inequality $\Leftrightarrow 7 n^{2}-m^{2} \geqslant \lambda$.
Since $7 n^{2} \equiv 0(\bmod 7), m^{2} \equiv 0,1,2,4(\bmod 7)$, then $\left(7 n^{2}-m^{2}\right)_{\text {min }}=3$, so, $\lambda_{\text {max }}=3$.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (20 points) Given the semi-ellipse $\frac{x^{2}}{4}+y^{2}=1(y>0)$, two perpendicular lines are drawn through a fixed point $C(1,0)$ intersecting the ellipse at points $P$ and $Q$, respectively. Here, $O$ is the origin, and $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse.
(1) Find the minimum value of $\left|P F_{1}+P F_{2}\right|$;
(2) Does there exist a line such that the sum of the x-coordinates of $P F_{1}$ and $P F_{2}$ is minimized? If not, explain why.
|
(1) Since $P O$ is the median of $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}+P F_{2}=2 P O, \left|P F_{1}+P F_{2}\right|=2|P O| \text {. }
$$
Therefore, when $P$ is at the vertex of the minor axis, $\left|P F_{1}+P F_{2}\right|$ achieves its minimum value of 2.
(2) From the problem analysis, the slopes of the lines $C P$ and $C Q$ both exist. Let the slopes of the lines $C P$ and $C Q$ be
$$
k_{C P}=k, k_{C Q}=-\frac{1}{k} \text {. }
$$
Then the equations of these two lines are
$$
l_{C P}: y=k(x-1), l_{C Q}: y=-\frac{1}{k}(x-1) \text {. }
$$
Since $P F_{1}+P F_{2}=2 P O, Q F_{1}+Q F_{2}=2 Q O$, it is sufficient to satisfy $O P \perp O Q$.
Then $k_{O P} k_{O Q}=\frac{k\left(x_{P}-1\right)}{x_{P}} \cdot \frac{-\frac{1}{k}\left(x_{Q}-1\right)}{x_{Q}}=-1$
$$
\Rightarrow x_{P}+x_{Q}=1 \text {. }
$$
From the ellipse equation, we get $y_{P}=\frac{\sqrt{4-x_{P}^{2}}}{2}, y_{Q}=\frac{\sqrt{4-x_{Q}^{2}}}{2}$, then
$$
\begin{array}{l}
k_{O P} k_{Q Q}=\frac{\sqrt{4-x_{P}^{2}}}{2 x_{P}} \cdot \frac{\sqrt{4-x_{Q}^{2}}}{2 x_{Q}}=-1 \\
\Rightarrow \sqrt{16-4\left(x_{P}^{2}+x_{Q}^{2}\right)+x_{P}^{2} x_{Q}^{2}}=-4 x_{P} x_{Q} . \\
\text { Let } x_{P} x_{Q}=t \leqslant 0, \text { and } x_{P}+x_{Q}=1 .
\end{array}
$$
Let $x_{P} x_{Q}=t \leqslant 0$, and $x_{P}+x_{Q}=1$.
Thus $\sqrt{16-4(1-2 t)+t^{2}}=-4 t$
$$
\Rightarrow 15 t^{2}-8 t-12=0 \text {. }
$$
Solving, we get $t=-\frac{2}{3}$ or $t=\frac{6}{5}$ (discard).
Then $\left\{\begin{array}{l}x_{p}+x_{Q}=1, \\ x_{p} x_{Q}=-\frac{2}{3}\end{array}\right.$.
Solving, we get $\left\{\begin{array}{l}x_{P}=\frac{3+\sqrt{33}}{6}, \\ x_{Q}=\frac{3-\sqrt{33}}{6} ;\end{array}\left\{\begin{array}{l}x_{P}=\frac{3-\sqrt{33}}{6}, \\ x_{Q}=\frac{3+\sqrt{33}}{6}\end{array}\right.\right.$.
Therefore, there exist lines such that $\boldsymbol{P F} F_{1}+P F_{2}$ and $\boldsymbol{Q} F_{1}+Q F_{2}$ are perpendicular to each other.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
204 Given that $a, b, c$ are positive numbers satisfying $a+b+c=1$. Prove:
$$
\frac{a^{3}+b}{a(b+c)}+\frac{b^{3}+c}{b(c+a)}+\frac{c^{3}+a}{c(a+b)} \geqslant 5 .
$$
|
Prove: $\frac{a^{3}+b}{a(b+c)}=\frac{a^{2}(1-b-c)+b}{a(b+c)}$
$$
\begin{array}{l}
=\frac{a^{2}+b}{a(b+c)}-a=\frac{a(1-b-c)+b}{a(b+c)}-a \\
=\frac{a+b}{a(b+c)}-1-a .
\end{array}
$$
Similarly, $\frac{b^{3}+c}{b(c+a)}=\frac{b+c}{b(c+a)}-1-b$,
$$
\frac{c^{3}+a}{c(a+b)}=\frac{c+a}{c(a+b)}-1-c \text {. }
$$
Adding the above three equations, we get
$$
\begin{array}{l}
\frac{a^{3}+b}{a(b+c)}+\frac{b^{3}+c}{b(c+a)}+\frac{c^{3}+a}{c(a+b)} \\
=\frac{a+b}{a(b+c)}+\frac{b+c}{b(c+a)}+\frac{c+a}{c(a+b)}-4 \\
\geqslant 3 \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{a(b+c) b(c+a) c(a+b)}}-4 \\
=\frac{3}{\sqrt[3]{a b c}}-4 \geqslant \frac{9}{a+b+c}-4=5 .
\end{array}
$$
Therefore, the original inequality holds.
|
5
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 For all $a, b, c \in \mathbf{R}_{+}$, find
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}
$$
the minimum value.
|
Explanation: Make the substitution
$$
\begin{array}{l}
x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, \\
z=\frac{c}{\sqrt{c^{2}+8 a b}} .
\end{array}
$$
Then $x, y, z \in (0, +\infty)$.
Thus, $x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, which means $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, $\frac{1}{y^{2}}-1=\frac{8 a c}{b^{2}}, \frac{1}{z^{2}}-1=\frac{8 a b}{c^{2}}$.
Multiplying the above three equations yields
$$
\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right)\left(\frac{1}{z^{2}}-1\right)=512 \text{. }
$$
If $x+y+z\frac{\prod\left[\left(\sum x\right)^{2}-x^{2}\right]}{x^{2} y^{2} z^{2}} \\
=\frac{\prod[(y+z)(2 x+y+z)]}{x^{2} y^{2} z^{2}} \\
\geqslant \frac{\prod\left(2 \sqrt{y z} \cdot 4 \sqrt[4]{x^{2} y z}\right)}{x^{2} y^{2} z^{2}} \\
=\frac{\prod\left(8 \sqrt{x^{2} y^{3} z^{3}}\right)}{x^{2} y^{2} z^{2}}=512 .
\end{array}
$$
This leads to a contradiction.
Therefore, $x+y+z \geqslant 1$.
Thus, when $a=b=c$, the minimum value is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Find
$$
\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}
$$
the minimum value.
|
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}(x, y, z \in \mathbf{R}_{+})$, then
$$
\begin{array}{l}
\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot} \\
\quad\left(\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}\right) \\
\geqslant(x+y+z)^{2} .
\end{array}
$$
Thus, $\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}$
$$
\begin{array}{l}
\geqslant \frac{(x+y+z)^{2}}{[y(y+2 x)+z(z+2 y)+x(x+2 z)]} \\
=1,
\end{array}
$$
which means $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \geqslant 1$.
The equality holds if and only if $a=b=c=1$. Therefore, the minimum value is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=1$. Find
$$
\frac{3 a^{2}-a}{1+a^{2}}+\frac{3 b^{2}-b}{1+b^{2}}+\frac{3 c^{2}-c}{1+c^{2}}
$$
the minimum value.
|
( Hint: First prove $\frac{3 a^{2}-a}{1+a^{2}} \geqslant \frac{9}{10}\left(a-\frac{1}{3}\right)$. Similarly, obtain the other two inequalities. Then add the three inequalities, to get the minimum value of 0. )
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 The inequality $x^{2}+|2 x-4| \geqslant p$ holds for all real numbers $x$. Find the maximum value of the real number $p$.
保留了原文的换行和格式，如上所示。
|
Explanation 1: Transform the original inequality into
$|x-2| \geqslant-\frac{x^{2}}{2}+\frac{p}{2}$.
Let $y_{1}=|x-2|, y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$.
Thus, solving inequality (1) is equivalent to determining the range of $x$ values for which the graph of the function $y_{1}=|x-2|$ is above the graph of the function $y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$. In this problem, we are given that this range is all real numbers, and we need to find the maximum value of the parameter $p$.
As shown in Figure 5,
draw the graphs of
the functions $y_{1}=|x-2|$
and $y_{2}=-\frac{x^{2}}{2}$
$+\frac{p}{2}$.
It is easy to see that the critical case where the graph of $y_{1}=|x-2|$ is above the graph of $y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$ is when $p=3$. Therefore, the maximum value of $p$ is 3.
Explanation 2: Let $y=x^{2}+|2 x-4|$. We only need to draw the graph of the function $y=x^{2}+|2 x-4|$.
When $x \geqslant 2$,
$$
y=x^{2}+2 x-4 \text {; }
$$
When $x<2$,
$$
y=x^{2}-2 x+4 \text {. }
$$
As shown in Figure 6, it is clear that the maximum value of $p$ is 3.
Note: Using function graphs to combine numerical and geometric insights is another effective approach to discussing parameters, such as the number of real roots of an equation or the solution set of an inequality. It is important to note that although the graphs are rough sketches, the key conditions, key points, and trends must be accurately represented, otherwise, there will be misjudgments.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CA=4$, $P$ is the midpoint of the semicircular arc $\overparen{AC}$, connect $BP$, the line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$.
|
Ni.6.4.
As shown in Figure 8, let $A C$ and $B P$ intersect at point $D$, and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$. The line segment $B P$ divides the figure $A P C B$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\triangle B E P$, which is twice the area of $\triangle B O P$. Thus,
$$
\begin{array}{l}
S_{\triangle B P O}=\frac{1}{2} P O \cdot C O \\
=\frac{1}{2} \times 2 \times 2=2 .
\end{array}
$$
Therefore, the absolute value of the difference in the areas of these two parts is 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12.B. The real numbers $a, b, c$ satisfy $a \leqslant b \leqslant c$, and $ab + bc + ca = 0, abc = 1$. Find the largest real number $k$ such that the inequality $|a+b| \geqslant k|c|$ always holds.
|
12. B. When $a=b=-\sqrt[3]{2}, c=\frac{\sqrt[3]{2}}{2}$, the real numbers $a, b, c$ satisfy the given conditions, at this time, $k \leqslant 4$.
Below is the proof: The inequality $|a+b| \geqslant 4|c|$ holds for all real numbers $a, b, c$ that satisfy the given conditions.
From the given conditions, we know that $a, b, c$ are all not equal to 0, and $c>0$.
Since $ab=\frac{1}{c}>0, a+b=-\frac{1}{c^{2}}<0$, it follows that $a \leqslant b<0$.
By the relationship between the roots and coefficients of a quadratic equation, $a, b$ are the two real roots of the quadratic equation $x^{2}+\frac{1}{c^{2}} x+\frac{1}{c}=0$.
Then $\Delta=\frac{1}{c^{4}}-\frac{4}{c} \geqslant 0 \Rightarrow c^{3} \leqslant \frac{1}{4}$.
Therefore, $|a+b|=-(a+b)=\frac{1}{c^{2}} \geqslant 4 c=4|c|$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11.B. Given the parabolas $C_{1}: y=-x^{2}-3 x+4$ and $C_{2}: y=x^{2}-3 x-4$ intersect at points $A$ and $B$, point $P$ is on parabola $C_{1}$ and lies between points $A$ and $B$; point $Q$ is on parabola $C_{2}$, also lying between points $A$ and $B$.
(1) Find the length of segment $A B$;
(2) When $P Q \parallel y$-axis, find the maximum length of $P Q$.
|
11. B. (1) Solve the system of equations $\left\{\begin{array}{l}y=-x^{2}-3 x+4, \\ y=x^{2}-3 x-4,\end{array}\right.$ to get
$$
\left\{\begin{array} { l }
{ x _ { 1 } = - 2 , } \\
{ y _ { 1 } = 6 , }
\end{array} \left\{\begin{array}{l}
x_{2}=2, \\
y_{2}=-6 .
\end{array}\right.\right.
$$
Therefore, $A(-2,6)$ and $B(2,-6)$.
Thus, $A B=\sqrt{(2+2)^{2}+(-6-6)^{2}}=4 \sqrt{10}$.
(2) As shown in Figure 11, when
$P Q / / y$ axis, set
$$
\begin{array}{l}
P\left(t,-t^{2}-3 t+4\right), \\
Q\left(t, t^{2}-3 t-4\right) \\
(-2<t<2) .
\end{array}
$$
Then $P Q=2\left(4-t^{2}\right)$
$\leqslant 8$.
Equality holds if and only if $t=0$.
Therefore, the maximum length of $P Q$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $a b=1$, and $\frac{1}{1-2^{x} a}+\frac{1}{1-2^{y+1} b}=1$, then the value of $x+y$ is $\qquad$.
|
8. -1 .
Given that both sides of the equation are multiplied by $\left(1-2^{x} a\right)\left(1-2^{y+1} b\right)$, we get
$$
2-2^{x} a-2^{y+1} b=1-2^{x} a-2^{y+1} b+2^{x+y+1} a b \text {, }
$$
which simplifies to $1=2 \times 2^{x+y}$.
Therefore, $x+y=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. As shown in Figure 2, in Pascal's Triangle, the numbers above the diagonal form the sequence: $1,3,6,10, \cdots$, let the sum of the first $n$ terms of this sequence be $S_{n}$. Then, as $n \rightarrow +\infty$, the limit of $\frac{n^{3}}{S(n)}$ is $\qquad$
|
10.6 .
It is known that $S_{n}=\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n+1}^{2}=\mathrm{C}_{\mathrm{n}+2}^{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{n^{3}}{S(n)}=\lim _{n \rightarrow+\infty} \frac{n^{3} \times 6}{(n+2)(n+1) n}=6$.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Given sets $A_{1}, A_{2}, \cdots, A_{n}$ are different subsets of the set $\{1,2, \cdots, n\}$, satisfying the following conditions:
(i) $i \notin A_{i}$ and $\operatorname{Card}\left(A_{i}\right) \geqslant 3, i=1,2, \cdots, n$;
(ii) $i \in A_{j}$ if and only if $j \notin A_{i}(i \neq j$, $i, j=1,2, \cdots, n)$.
Try to answer the following questions:
(1) Find $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)$;
(2) Find the minimum value of $n$.
|
(1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets, and let $r_{i}$ be the number of sets that contain element $i$ for $i=1,2, \cdots, n$. Then,
$$
\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i} \text {. }
$$
Assume the $r_{1}$ sets that contain element 1 are
$$
A_{2}, A_{3}, \cdots, A_{r_{1}+1} \text {. }
$$
By condition (ii), we know that $2 \notin A_{1}, 3 \notin A_{1}, \cdots \cdots r_{1}+1 \notin A_{1}, r_{1}+2 \in A_{1}, r_{1}+3 \in A_{1}, \cdots \cdots n \in A_{1}$.
Also, by condition (i), we know that $1 \notin A_{1}$.
Thus, $\operatorname{Card}\left(A_{1}\right)=n-\left(r_{1}+2\right)+1=n-r_{1}-1$.
Similarly, $\operatorname{Card}\left(A_{i}\right)=n-r_{i}-1$ for $i=1,2, \cdots, n$.
Therefore, $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n}\left(n-r_{i}-1\right)$ $=\sum_{i=1}^{n}(n-1)-\sum_{i=1}^{n} r_{i}=\sum_{i=1}^{n} r_{i}$.
Thus, $\sum_{i=1}^{n} r_{i}=\frac{1}{2} \sum_{i=1}^{n}(n-1)=\frac{1}{2} n(n-1)$.
(2) Since $\operatorname{Card}\left(A_{i}\right) \geqslant 3$, we have $\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right) \geqslant 3 n$.
And since $A_{i} \subseteq\{1,2, \cdots, n\}$ for $i=1,2, \cdots, n$, at least one of the elements $1,2, \cdots, n$ must be in at least 3 sets. Assume this element is 1, i.e., $r_{1} \geqslant 3$.
$$
\begin{array}{l}
\text { Also, } \operatorname{Card}\left(A_{1}\right)=n-r_{1}-1 \geqslant 3, \text { so } n \geqslant r_{1}+4 \geqslant 7 \text {. And } \\
A_{1}=\{2,3,4\}, A_{2}=\{3,4,5\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{5,6,7\}, A_{5}=\{6,7,1\}, A_{6}=\{7,1,2\}, \\
A_{7}=\{1,2,3\}
\end{array}
$$
satisfy the conditions.
Therefore, the minimum value of $n$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $S$ be a subset of $\{1,2, \cdots, 9\}$ such that the sum of any two distinct elements of $S$ is unique. How many elements can $S$ have at most?
(2002, Canadian Mathematical Olympiad)
|
(When $S=\{1,2,3,5,8\}$, $S$ meets the requirements of the problem. If $T \subseteq\{1,2, \cdots, 9\},|T| \geqslant 6$, then since the sum of any two different numbers in $T$ is between 3 and 17, at most 15 different sum numbers can be formed. And choosing any two numbers from $T$, there are at least $\mathrm{C}_{6}^{2}=15$ ways. If $T$ meets the condition, then the sum numbers must include both 3 and 17, meaning that $1,2,8,9$ must all appear in $T$, but in this case $1+9=2+8$, making $T$ not meet the requirement. Therefore, $S$ can have at most 5 elements.)
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x=\frac{1}{\sqrt{2}-1}, a$ be the fractional part of $x$, and $b$ be the fractional part of $-x$. Then $a^{3}+b^{3}+3 a b=$ $\qquad$ .
|
ニ、1. 1 .
Since $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$, and $2<\sqrt{2}+1<3$, therefore,
$$
\begin{array}{l}
a=x-2=\sqrt{2}-1 . \\
\text { Also }-x=-\sqrt{2}-1, \text { and }-3<-\sqrt{2}-1<-2 \text {, so, } \\
b=-x-(-3)=2-\sqrt{2} .
\end{array}
$$
Then $a+b=1$.
Thus $a^{3}+b^{3}+3 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a b$ $=a^{2}-a b+b^{2}+3 a b=(a+b)^{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given a right trapezoid $A B C D$ with side lengths $A B=2, B C=C D=10, A D=6$, a circle is drawn through points $B$ and $D$, intersecting the extension of $B A$ at point $E$ and the extension of $C B$ at point $F$. Then the value of $B E-B F$ is $\qquad$
|
3.4 .
As shown in Figure 3, extend $C D$ to intersect $\odot O$ at point $G$. Let the midpoints of $B E$ and $D G$ be $M$ and $N$, respectively. It is easy to see that $A M = D N$. Since $B C = C D = 10$, by the secant theorem, it is easy to prove that
$$
\begin{aligned}
B F = & D G. \text{ Therefore, } \\
& B E - B F = B E - D G = 2(B M - D N) \\
& = 2(B M - A M) = 2 A B = 4 .
\end{aligned}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$. Prove: $k=5$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
For every $k$ that satisfies the above equation, let $a_{0}, b_{0}$ satisfy $\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=k$, and $a_{0}+b_{0}$ is the smallest pair. Without loss of generality, assume $a_{0} \geqslant b_{0}$.
(1) If $a_{0}=b_{0}$, then
$$
k=\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=2+\frac{2}{a_{0}^{2}-1}.
$$
Since $a_{0}^{2}-1 \neq 1,2$, $k$ is not an integer.
(2) If $a_{0}>b_{0}$, the quadratic equation in $a$
$$
a^{2}-b_{0} k a+b_{0}^{2}+k=0
$$
has one root $a_{0}$. By Vieta's formulas, the other root $a^{\prime}$ is a positive integer.
Thus, by the minimality of $a_{0}+b_{0}$, we have $a^{\prime} \geqslant a_{0}$.
Therefore, $a_{0}^{2} \leqslant a_{0} a^{\prime}=b_{0}^{2}+k$.
Hence, $k \geqslant a_{0}^{2}-b_{0}^{2} \geqslant a_{0}^{2}-\left(a_{0}-1\right)^{2}=2 a_{0}-1$.
When $b_{0}=1$,
$$
k=\frac{a_{0}^{2}+1}{a_{0}-1}=a_{0}+1+\frac{2}{a_{0}-1}.
$$
Thus, $a_{0}-1=1$ or 2. Therefore, $k=5$.
When $b_{0} \geqslant 2$, we have
$$
\begin{array}{l}
2 a_{0}^{2}>a_{0}^{2}+b_{0}^{2}=k\left(a_{0} b_{0}-1\right) \\
\geqslant\left(2 a_{0}-1\right)^{2}=4 a_{0}^{2}-4 a_{0}+1 .
\end{array}
$$
Thus, $2\left(a_{0}-1\right)^{2}<1$, which is impossible.
In conclusion, $k=5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $a$ be a positive integer, the quadratic function $y=x^{2}+(a+17) x+38-a$, and the reciprocal function $y=\frac{56}{x}$. If the intersection points of the two functions are all integer points (points with both coordinates as integers), find the value of $a$.
---
The above text has been translated into English, preserving the original text's line breaks and format.
|
Three, by eliminating $y$ from the two equations, we get
$$
x^{2}+(a+17) x+38-a=\frac{56}{x} \text {, }
$$
which simplifies to $x^{3}+(a+17) x^{2}+(38-a) x-56=0$.
Factoring, we get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Clearly, $x_{1}=1$ is a root of equation (1), and $(1,56)$ is one of the intersection points of the two function graphs.
Since $a$ is a positive integer, the discriminant of the equation
$$
x^{2}+(a+18) x+56=0
$$
is $\Delta=(a+18)^{2}-224>0$, so it must have two distinct real roots.
Since the intersection points of the two function graphs are all integer points, the roots of equation (2) must all be integers. Therefore, its discriminant $\Delta=(a+18)^{2}-224$ should be a perfect square.
Let $(a+18)^{2}-224=k^{2}$ (where $k$ is a non-negative integer), then $(a+18)^{2}-k^{2}=224$,
which simplifies to $(a+18+k)(a+18-k)=224$.
Following the solution method of the third problem in the A paper, we get
$$
\left\{\begin{array} { l }
{ a = 3 9 , } \\
{ k = 5 5 }
\end{array} \text { or } \left\{\begin{array}{l}
a=12, \\
k=26 .
\end{array}\right.\right.
$$
When $a=39$, equation (2) becomes $x^{2}+57 x+56=0$, whose roots are -1 and -56. In this case, the two function graphs have two additional intersection points $(-1,-56)$ and $(-56,-1)$.
When $a=12$, equation (2) becomes $x^{2}+30 x+56=0$, whose roots are -2 and -28. In this case, the two function graphs have two additional intersection points $(-2,-28)$ and $(-28,-2)$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given that for all real numbers $x$, we have
$$
|x+1|+\sqrt{x-1} \geqslant m-|x-2|
$$
always holds. Then the maximum value that $m$ can take is $\qquad$ .
|
When $-1 \leqslant x \leqslant 2$, the minimum value of $|x+1|+|x-2|$ is
3. Since $\sqrt{x-1} \geqslant 0$, when $x=1$,
$$
|x+1|+\sqrt{x-1}+|x-2|
$$
the minimum value is 3, so, $3 \geqslant m$, hence the maximum value of $m$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x$ and $y$ are real numbers, and $x^{2}+x y+y^{2}=3$. Let the maximum and minimum values of $x^{2}-x y+y^{2}$ be $m$ and $n$, respectively. Then the value of $m+n$ is $\qquad$
|
$=, 1.10$.
Given the equation is symmetric about $x, y$, let $x=a+b, y=a-b$, then $3=x^{2}+xy+y^{2}=3a^{2}+b^{2}$.
Therefore, $3a^{2}=3-b^{2}$.
So, $0 \leqslant b^{2} \leqslant 3$.
$$
\begin{array}{l}
\text { Also, } x^{2}-xy+y^{2}=a^{2}+3b^{2}=\frac{1}{3}\left(3a^{2}+b^{2}\right)+\frac{8}{3}b^{2} \\
=\frac{1}{3} \times 3+\frac{8}{3}b^{2}=1+\frac{8}{3}b^{2},
\end{array}
$$
Then $1 \leqslant 1+\frac{8}{3}b^{2} \leqslant 1+\frac{8}{3} \times 3=9$.
Thus, $m=9, n=1$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a$ and $b$ be integers, and one root of the equation $x^{2} + a x + b = 0$ is $\sqrt{4 - 2 \sqrt{3}}$. Then the value of $a + b$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2
|
4.B.
Notice that $\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1$.
According to the problem, we have $(\sqrt{3}-1)^{2}+a(\sqrt{3}-1)+b=0$, which simplifies to $(a-2) \sqrt{3}+4-a+b=0$.
Thus, $a-2=0$ and $4-a+b=0$.
Solving these, we get $a=2, b=-2$. Therefore, $a+b=0$.
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 1, in the right trapezoid $A B C D$, $A B=B C=4$, $M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}$ : $S_{\triangle A B M}=$ $\qquad$ .
|
2.2.
As shown in Figure 6, draw $A E \perp$ $C D$ intersecting the extension of $C D$ at point $E$, then quadrilateral $A B C E$ is a square.
It is easy to prove
$\mathrm{Rt} \triangle A B M \cong \mathrm{Rt} \triangle A E D$.
Therefore, $B M=D E$.
Thus, $C M=C D$.
Let this value be $x$, then
$$
\begin{array}{l}
x^{2}+x^{2}=4^{2}+(4-x)^{2} \Rightarrow x^{2}=32-8 x . \\
\text { Hence, } \frac{S_{\triangle C D M}}{S_{\triangle B M}}=\frac{\frac{1}{2} x^{2}}{\frac{1}{2} \times 4(4-x)}=\frac{x^{2}}{16-4 x}=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The quality requirements of a product are divided into four different levels from low to high, labeled as $1,2,3,4$. If the working hours remain unchanged, the workshop can produce 40 units of the lowest level (i.e., level 1) product per day, with a profit of 16 yuan per unit; if the level is increased by one, the profit per unit increases by 1 yuan, but the daily production decreases by 2 units. Now the workshop plans to produce only one level of product. To maximize profit, the workshop should produce level $\qquad$ product.
|
3.3.
Let the profit obtained from producing products of the $x$-th grade in the workshop be $y$. According to the problem, we have
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$
Therefore, when $x=3$, the profit $y$ is maximized, at 648 yuan.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of all different integer solutions to the equation $2 x^{2}+5 x y+2 y^{2}=2007$ is $\qquad$ groups.
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。
Note: The note above is not part of the translation but is provided to clarify the instruction. The actual translation is above this note.
|
4.4 .
Let's first assume $x \geqslant y$, the original equation can be transformed into $(2 x+y)(x+2 y)=2007$.
Since $2007=2007 \times 1=669 \times 3=223 \times 9$
$$
\begin{array}{l}
=(-1) \times 2007=(-3) \times(-669) \\
=(-9) \times(-223),
\end{array}
$$
Therefore, $\left\{\begin{array}{l}2 x+y=2007, \\ x+2 y=1 .\end{array}\right.$,
Solving gives $3(x+y)=2008$, which has no integer solutions.
Similarly, $\left\{\begin{array}{l}2 x+y=223, \quad-1,-9 \\ x+2 y=9,-2007,-223,\end{array}\right.$ all have no integer solutions;
$\left\{\begin{array}{l}2 x+y=-3, \\ x+2 y=-669\end{array}\right.$ has integer solutions $\left\{\begin{array}{l}x=221, \\ y=-445 .\end{array}\right.$
If $y \geqslant x$, there are two more sets of integer solutions
$$
\left\{\begin{array} { l }
{ x = - 2 2 1 , } \\
{ y = 4 4 5 , }
\end{array} \left\{\begin{array}{l}
x=-445, \\
y=221 .
\end{array}\right.\right.
$$
Therefore, there are a total of 4 sets of integer solutions.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. For any 4 vertices $A, B, C, D$ of a cube that do not lie on the same plane, the number of cosine values of the dihedral angle $A-BC-D$ that are less than $\frac{1}{2}$ is
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
9.4 .
Among the cosines of the dihedral angles with the body diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A_{1} O_{1} C_{1} \\
=\cos 120^{\circ}=-\frac{1}{2} .
\end{array}
$$
Among the cosines of the dihedral angles with the edge as the edge, only 0 is less than $\frac{1}{2}$.
Among the cosines of the dihedral angles with the face diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A O C_{1} \\
=\frac{\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(1+\frac{1}{2}\right)-(\sqrt{3})^{2}}{2 \times \frac{\sqrt{2}}{2} \times \sqrt{1+\frac{1}{2}}}=-\frac{\sqrt{3}}{3}<\frac{1}{2} \\
\cos \angle A_{1} O C_{1}=\frac{\frac{3}{2}+\frac{3}{2}-2}{2 \sqrt{\frac{3}{2}} \times \sqrt{\frac{3}{2}}}=\frac{1}{3}<\frac{1}{2} .
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There are 5 rooms $A, B, C, D, E$ arranged in a circular pattern, with the number of people living in them being $17, 9, 14, 16, 4$ respectively. Now, adjustments are to be made so that the number of people in each room is the same, and it is stipulated that people can only move to the adjacent left or right room. How many people should each room move to the left or right so that the total number of people moved is minimized?
|
(Tip: Let the number of people moved from room $A$ to room $B$ be $x_{B}$, and so on, then we have
$$
\begin{array}{l}
9+x_{B}-x_{C}=14+x_{C}-x_{D}=16+x_{D}-x_{E} \\
=4+x_{E}-x_{A}=17+x_{A}-x_{B} \\
=\frac{1}{5}(17+9+14+16+4)=12 .
\end{array}
$$
This is transformed into finding
$$
\begin{array}{l}
y=\left|x_{B}-5\right|+\left|x_{B}\right|+\left|x_{B}-3\right|+ \\
\left|x_{B}-1\right|+\left|x_{B}+3\right|
\end{array}
$$
The minimum value of $y$. By the shortest distance between two points is a straight line, we get $x_{B}=1$. Thus, $x_{A}=$ $-4, x_{B}=1, x_{C}=-2, x_{D}=0, x_{E}=4$. The minimum total number of people moved is $4+1+2+0+4=11$.)
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of triangles with integer side lengths and a perimeter of 20 is $\qquad$ .
|
2.8.
Let the three sides of a triangle be $a, b, c$, and $a \geqslant b \geqslant c$, $a+b+c=20$, then $a \geqslant 7$.
Also, from $b+c>a$, we get $2 a<a+b+c=20 \Rightarrow a<10$. Therefore, $7 \leqslant a \leqslant 9$. We can list
$$
\begin{array}{l}
(a, b, c)=(9,9,2),(9,8,3),(9,7,4),(9,6,5), \\
(8,8,4),(8,7,5),(8,6,6),(7,7,6) .
\end{array}
$$
There are 8 sets in total.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange the four numbers $1, 2, 3, 4$ to form a four-digit number, such that this number is a multiple of 11. Then the number of such four-digit numbers is $\qquad$.
|
4.8.
Since $1+4=2+3$, we can place 1 and 4 in the even positions, and 2 and 3 in the odd positions, which gives us four arrangements; placing 2 and 3 in the even positions, and 1 and 4 in the odd positions also gives us four arrangements.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}, \cdots .
\end{array}
$$
Let $S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}$. Then the integer closest to $S_{2006}$ is $(\quad)$.
(A) 2
(B) 3
(C) 4
(D) 5
|
3.C.
Since $a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}=\frac{n}{2}$, we have
$$
\begin{array}{l}
S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right] \\
=4\left(1-\frac{1}{n+1}\right) .
\end{array}
$$
Therefore, $S_{2000}=4\left(1-\frac{1}{2007}\right)$ is closest to the integer 4.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$. Then
$$
f(1)+f(2)+\cdots+f(2006)=
$$
$\qquad$
|
2.0.
Given that the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$, we know that $f(x)=f(1-x)$.
Also, since $f(x)$ is an odd function defined on $\mathbf{R}$, it follows that $f(1-x)=-f(x-1)$.
Therefore, $f(x)+f(x-1)=0$.
Thus, $f(1)+f(2)+\cdots+f(2006)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $\triangle A B C$ with three sides $A B=\sqrt{34}, B C$ $=5 \sqrt{10}, C A=2 \sqrt{26}$. Then the area of $\triangle A B C$ is $\qquad$
|
5.10.
As shown in Figure 4, with $BC$ as the hypotenuse, construct a right triangle $\triangle BCD$ on one side of $\triangle ABC$ such that $\angle BDC = 90^\circ$, $BD = 5$, and $CD = 15$.
Then construct a rectangle $DEA'F$ such that $DE = 2$ and $DF = 5$. Thus, $BE = 3$ and $CF = 10$. At this point,
$$
\begin{array}{l}
A'C = \sqrt{104} = AC, \\
A'B = \sqrt{34} = AB.
\end{array}
$$
Therefore, point $A$ coincides with point $A'$.
Hence,
$$
\begin{array}{l}
S_{\triangle ABC} = S_{\triangle BCD} - S_{\triangle ABE} - \\
\quad S_{\triangle ACF} - S_{\text{rect } EAF} \\
= \frac{1}{2} \times 5 \times 15 - \frac{1}{2} \times 3 \times 5 - \frac{1}{2} \times 2 \times 10 - 2 \times 5 \\
= 10.
\end{array}
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The area of the figure enclosed by the two curves $y=x^{3}$ and $x=y^{3}$ on the Cartesian plane is $\qquad$ .
|
5.1.
Since the two curves are symmetric with respect to the origin, it is only necessary to calculate the area $A$ of the figure enclosed by the two curves in the first quadrant.
When $x>1$, $x^{3}>\sqrt[3]{x}$;
When $0<x<1$, $x^{3}<\sqrt[3]{x}$.
Therefore, the two curves have a unique intersection point $(1,1)$ in the first quadrant.
$$
\begin{array}{l}
\text { Also, } A=\int_{0}^{1}\left(\sqrt[3]{x}-x^{3}\right) \mathrm{d} x \\
=\left.\left(\frac{3}{4} x^{\frac{4}{3}}-\frac{x^{4}}{4}\right)\right|_{0} ^{1}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2},
\end{array}
$$
Therefore, the area of the figure enclosed by the two curves is $2 A=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.3. On the side $BC$ of the rhombus $ABCD$, take a point $M$. Draw perpendiculars from $M$ to the diagonals $BD$ and $AC$, intersecting the line $AD$ at points $P$ and $Q$. If the lines $PB$ and $QC$ intersect $AM$ at the same point, find the ratio $\frac{BM}{MC}$.
|
8.3. As shown in Figure 1, let the intersection of lines $P B$, $Q C$, and $A M$ be $R$.
From the given conditions, $P M \parallel A C$ and $M Q \parallel B D$, thus quadrilaterals $P M C A$ and $Q M B D$ are both parallelograms. Therefore,
$$
\begin{aligned}
M C &= P A, B M = D Q, \text{ and } \\
P Q &= P A + A D + D Q = M C + A D + B M = 2 B C .
\end{aligned}
$$
Since $B C \parallel P Q$ and $B C = \frac{1}{2} P Q$, $B C$ is the midline of $\triangle P R Q$.
This implies that $B M$ is the midline of $\triangle A R P$.
$$
\text{Thus, } M C = P A = 2 B M \text{. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.7. For a natural number $n(n>3)$, we use “$n$ ?" to denote the product of all prime numbers less than $n$. Solve the equation
$$
n ?=2 n+16
$$
|
8.7. The given equation is
$$
n ?-32=2(n-8) \text {. }
$$
Since $n$ ? cannot be divisible by 4, it follows from equation (1) that $n-8$ is odd.
Assume $n>9$, then $n-8$ has an odd prime factor $p$.
Also, $p2 \times 9+16$.
When $n=7$, it is clearly a root of the equation:
However, when $n=5$, we have $n ?=6<16$.
Therefore, the equation has a unique root $n=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{2}+a_{4}+a_{6}+$ $a_{8}+a_{10}=80$, then $a_{7}-\frac{1}{2} a_{8}=(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10
|
4.C.
Since $a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{6}=80$, therefore, $a_{6}=16$.
Thus, $a_{7}-\frac{1}{2} a_{8}=a_{6}+d-\frac{1}{2}\left(a_{6}+2 d\right)=\frac{1}{2} a_{6}=8$.
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. If the three medians $A D$, $B E$, $C F$ of $\triangle A B C$ intersect at point $M$, then $M A+M B+M C=$
|
Ni, 11.0.
Diagram (omitted). Let the midpoint of $A B$ be $D$. By the parallelogram rule, we have
$$
\begin{array}{l}
M A+M B=2 M D=-M C \text {. } \\
\text { Therefore, } M A+M B+M C=0 \text {. } \\
\end{array}
$$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n}+a_{n+1}=1$ $\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then
$$
\mathrm{S}_{2007}-2 \mathrm{~S}_{2006}+\mathrm{S}_{2000}=
$$
$\qquad$
|
13.3.
When $n$ is even, we have
$$
a_{1}+a_{2}=a_{3}+a_{4}=\cdots=a_{n-1}+a_{n}=1 .
$$
Thus, $S_{n}=\frac{n}{2}$.
When $n$ is odd, we have
$$
a_{1}=2, a_{2}+a_{3}=a_{4}+a_{5}=\cdots=a_{n-1}+a_{n}=1 \text {. }
$$
Thus, $S_{n}=2+\frac{n-1}{2}=\frac{n+3}{2}$.
Therefore, $S_{2007}-2 S_{2006}+S_{2005}$
$$
=1005-2 \times 1003+1004=3 \text {. }
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (12 points) In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$.
|
Three, 15. From the given,
$\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B$.
Then, $\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C$ $=3 \sin B$.
Thus, $\sin A+\sin C+\sin (A+C)=3 \sin B$, which means $\sin A+\sin C=2 \sin B$.
Therefore, $2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cdot \cos \frac{B}{2}$.
So, $\cos \frac{A-C}{2}=2 \sin \frac{B}{2}$, which is $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $\triangle ABC$ is an equilateral triangle with side length 8, $M$ is a point on side $AB$, $MP \perp AC$ at point $P$, $MQ \perp BC$ at point $Q$, and connect $PQ$.
(1) Find the minimum length of $PQ$;
(2) Find the maximum area of $\triangle CPQ$.
|
Solution: (1) Let the height of $\triangle ABC$ be $h$, then $h=4 \sqrt{3}$.
From $S_{\triangle C M}+S_{\triangle B C M}=S_{\triangle B B C}$, we get
$$
M P+M O=h=4 \sqrt{3} \text {. }
$$
As shown in Figure 11, draw perpendiculars from points $P$ and $Q$ to side $AB$, with the feet of the perpendiculars being $P_{1}$ and $Q_{1}$, respectively. Since
$$
\begin{array}{l}
\angle P M A=\angle Q M B \\
=30^{\circ},
\end{array}
$$
we have
$$
\begin{array}{l}
P_{1} M=P M \cos 30^{\circ}=\frac{\sqrt{3}}{2} P M, \\
Q_{1} M=Q M \cos 30^{\circ}=\frac{\sqrt{3}}{2} Q M, \\
P Q \geqslant P_{1} Q_{1}=P_{1} M+M Q_{1} \\
=\frac{\sqrt{3}}{2}(P M+Q M)=6 .
\end{array}
$$
When $M$ is the midpoint of $AB$, the equality holds. Therefore, the minimum value of $P Q$ is 6.
(2) Since $\angle P M A=\angle Q M B=30^{\circ}$, we have
$$
\begin{array}{l}
A P+B Q=\frac{1}{2} A M+\frac{1}{2} B M=\frac{1}{2} A B=4, \\
C P+C Q=16-(A P+B Q)=12 .
\end{array}
$$
Thus, $S_{\triangle C P Q}=\frac{1}{2} C P \cdot C Q \sin C=\frac{\sqrt{3}}{4} C P \cdot C Q$
$$
\leqslant \frac{\sqrt{3}}{4} \cdot \frac{(C P+C Q)^{2}}{4}=9 \sqrt{3} \text {. }
$$
When $M$ is the midpoint of $AB$, the equality holds. Therefore, the maximum area of $\triangle C P Q$ is $9 \sqrt{3}$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2-1 Given that $x, y, z$ are positive numbers, and $xyz(x+y+z)=1$.
Find the minimum value of $(x+y)(y+z)$.
(1989, All-Soviet Union Mathematical Competition)
|
Solution 1: $(x+y)(y+z)$
$$
\begin{array}{l}
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2 .
\end{array}
$$
When $x=z=1, y=\sqrt{2}-1$, $y(x+y+z) = xz$, $(x+y)(y+z)$ takes the minimum value 2.
Solution 2: As shown in Figure 1, construct
$\triangle ABC$, with side lengths
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then the area of the triangle is
$$
\begin{array}{l}
S=\sqrt{p(p-a)(p-b)(p-c)} \\
=\sqrt{x y z(x+y+z)}=1 \text { (constant). } \\
\text { Also, }(x+y)(y+z)=a b=\frac{2 S}{\sin C} \geqslant 2, \text { when and }
\end{array}
$$
only when $\sin C=1\left(\angle C=90^{\circ}\right)$, $(x+y)(y+z)$ takes the minimum value 2.
This problem's integration of algebra and geometry has sparked our interest in exploration, providing some insights.
(1) Figure 1 and the corresponding transformation (1) serve as a bridge between the positive numbers $x, y, z$ and $\triangle ABC$, representing a stable approach to integrating algebra and geometry;
(2) Solution 2 tells us: Among triangles with a constant area, the product of two sides is minimized when their included angle is a right angle (right-angled triangle);
(3) Solution 1 tells us: $(x+y)(y+z)$ takes the minimum value under the condition
$$
\begin{array}{l}
x z=y(x+y+z) \\
\Leftrightarrow(z+x)^{2}=(x+y)^{2}+(y+z)^{2} .
\end{array}
$$
Combining this with the converse of the Pythagorean theorem, we see that $\triangle ABC$ is a right-angled triangle, which is consistent with the geometric observation. However, it also provides new insights, namely that from the constant area
$$
\begin{array}{c}
\sqrt{x y z(x+y+z)}=S \text { and equation (2), we get } \\
x z=y(x+y+z)=S .
\end{array}
$$
In this right-angled triangle, $y$ equals the radius of the inscribed circle, so $y(x+y+z)=S$ means that "the area of a right-angled triangle is equal to the product of the semiperimeter and the radius of the inscribed circle."
And $x z=S$ is the following conclusion.
Conclusion: The area of a right-angled triangle is equal to the product of the segments into which the hypotenuse is divided by the points of tangency of the inscribed circle.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2-3 Let quadrilateral $A B C D$ be a rectangle with an area of 2, $P$ a point on side $C D$, and $Q$ the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant 2 B C$;
(2) Find the value of $A Q \cdot B Q$.
(2001, China Western Mathematical Olympiad)
|
Solution 1: (1) From the given information,
$S_{\triangle P A B}=\frac{1}{2} S_{\text {rectangle } A B C D}=1$.
Thus, $P A \cdot P B=\frac{2 S_{\triangle P A B}}{\sin \angle A P B} \geqslant 2$.
Equality holds if and only if $\angle A P B=90^{\circ}$. At this time, point $P$ lies on the circle with $A B$ as its diameter, meaning this circle intersects $C D$ at point $P$.
By the property that "in a right triangle, the median to the hypotenuse is not less than the altitude to the hypotenuse," when $P A \cdot P B$ reaches its minimum value, we have
$$
B C \leqslant \frac{A B}{2} \Rightarrow A B \geqslant 2 B C \text {. }
$$
(2) In the right triangle $\triangle P A B$, by the property that "the lengths of the tangents from a point outside a circle to the circle are equal," we get
$$
|P A-P B|=|A Q-B Q| .
$$
Squaring both sides, we obtain
$$
P A^{2}+P B^{2}-2 P A \cdot P B=A B^{2}-4 A Q \cdot B Q \text {. }
$$
Thus, $A Q \cdot B Q=\frac{1}{2} P A \cdot P B=1$.
Solution 2: As shown in Figure 3, let the incircle of $\triangle P A B$ touch $P A$ and $P B$ at points $M$ and $N$, respectively. Let
$$
\begin{array}{l}
A Q=A M=z, \\
B Q=B N=x, \\
P N=P M=y .
\end{array}
$$
Then $S_{\triangle P A B}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{x y z(x+y+z)} \\
=\frac{1}{2} S_{\text {rectangle } A B C D}=1 .
\end{array}
$$
Thus, $x y z(x+y+z)=1$.
$$
\begin{array}{l}
\text { Also, } P A \cdot P B=(y+z)(x+y) \\
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2,
\end{array}
$$
Therefore, when $x z=y(x+y+z)$, $P A \cdot P B$ reaches its minimum value of 2.
(1) Multiplying both sides of $x z=y(x+y+z)$ by 2 and adding $x^{2}+z^{2}$, we get
$$
(x+z)^{2}=(y+z)^{2}+(x+y)^{2},
$$
which means $P A^{2}+P B^{2}=A B^{2}$.
Thus, $\triangle P A B$ is a right triangle, and $B C$ is the altitude to the hypotenuse of the right triangle. Hence,
$$
2 B C=\frac{2 P A \cdot P B}{A B} \leqslant \frac{P A^{2}+P B^{2}}{A B}=A B .
$$
(2) Adding $x z$ to both sides of $x z=y(x+y+z)$, we get
$$
2 x z=(x+y)(y+z) \text {, }
$$
which means $2 A Q \cdot B Q=P A \cdot P B=2$.
Thus, $A Q \cdot B Q=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $p$ be a positive odd number. Then the remainder of $p^{2}$ divided by 8 is $\qquad$ .
|
Because $p$ is an odd positive integer, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, so
$$
p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1 \text{. }
$$
Since $(k-1)k$ is even, therefore, $p^{2}$ leaves a remainder of 1 when divided by 8.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given an isosceles triangle $\triangle A B C$ with side lengths $a$, $b$, and $c$ all being integers, and satisfying $a+b c+b+c a=24$. Then the number of such triangles is $\qquad$.
|
4.3.
$$
\begin{array}{l}
\text { Since } a+b c+b+a a \\
=(a+b)(c+1)=24=12 \times 2=8 \times 3=6 \times 4, \text { and }
\end{array}
$$
$\triangle A B C$ is an isosceles triangle, so the length of the base can only be $c$.
Thus, there are 3 triangles that satisfy the conditions:
$$
c=1, a=b=6 ; c=2, a=b=4 ; c=3, a=b=3 \text {. }
$$
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2), B(-1, \sqrt{3})$, and $C(c, 2-c)$. Find the value of $a-b+c$.
---
The above text has been translated into English, preserving the original text's line breaks and format.
|
Three, from $\left\{\begin{array}{l}\sqrt{3}+2=\sqrt{3} a+b, \\ \sqrt{3}=-a+b\end{array} \Rightarrow\left\{\begin{array}{l}a=\sqrt{3}-1, \\ b=2 \sqrt{3}-1 .\end{array}\right.\right.$
Therefore, $2-c=a c+b=(\sqrt{3}-1) c+(2 \sqrt{3}-1)$.
Solving for $c$ gives $c=\sqrt{3}-2$.
Thus, $a-b+c=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points)
As shown in Figure 3, 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 to be removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally) in sequence. What is the minimum number of chess pieces that need to be removed to meet this requirement? Explain your reasoning.
|
Second, at least 11 chess pieces must be taken out to possibly meet the requirement.
The reason is as follows:
If a square is in the $i$th row and the $j$th column, then this square is denoted as $(i, j)$.
Step 1 Proof: If any 10 chess pieces are taken, then the remaining chess pieces must have a five-in-a-row, i.e., five chess pieces are consecutively connected in a straight line (horizontally, vertically, or diagonally).
Proof by contradiction.
Assume that 10 chess pieces can be taken such that the remaining chess pieces do not have a five-in-a-row. As shown in Figure 8, one chess piece must be taken from each of the first five squares in every row, and one chess piece must be taken from each of the first five squares in the last three columns. Thus, the 10 taken chess pieces will not be distributed in the shaded area in the lower right corner of Figure 8. By symmetry, they will not be distributed in the shaded areas in the other corners.
The 1st and 2nd rows must each have 1 chess piece taken, and these can only be distributed in the squares $(1,4)$, $(1,5)$, $(2,4)$, and $(2,5)$.
Similarly, at least 2 chess pieces must be taken from the squares $(6,4)$, $(6,5)$, $(7,4)$, and $(7,5)$.
In the 1st, 2nd, and 3rd columns, at least 1 chess piece must be taken from each column, distributed in the squares $(3,1)$, $(3,2)$, $(3,3)$, $(4,1)$, $(4,2)$, $(4,3)$, $(5,1)$, $(5,2)$, and $(5,3)$.
Similarly, at least 3 chess pieces must be taken from the squares $(3,6)$, $(3,7)$, $(3,8)$, $(4,6)$, $(4,7)$, $(4,8)$, $(5,6)$, $(5,7)$, and $(5,8)$.
Thus, at least 10 chess pieces must be taken from these areas.
Therefore, no chess pieces can be taken from the central shaded area.
Since at most 2 of the 4 chess pieces (1), (2), (3), and (4) can be taken, there will inevitably be a five-in-a-row from the diagonal direction. This is a contradiction.
Step 2 Construction of a method:
Take a total of 11
chess pieces, and the
remaining chess pieces
will not have a five-in-a-row.
As shown in Figure 9, as long as the chess pieces in the marked positions are taken,
the remaining chess pieces will not
have a five-in-a-row.
In summary, at least 11 chess pieces must be taken to possibly meet the requirement.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Let $x, y, a, m, n$ be positive integers, and $x+y=a^{m}, x^{2}+y^{2}=a^{n}$. Find how many digits $a^{30}$ has.
保留源文本的换行和格式，直接输出翻译结果。
|
Three, from the known we get
$$
a^{2 m}=x^{2}+y^{2}+2 x y=a^{n}+2 x y \text {. }
$$
From the problem and equation (1), we know that $a^{2 m}>a^{n}$. Therefore, $2 m>n$.
Dividing both sides of equation (1) by $a^{n}$, we get
$$
a^{2 m-n}=1+\frac{2 x y}{a^{n}} \text {. }
$$
Since the left side of equation (2) is a positive integer, $a^{n}$ must divide $2 x y$.
Thus, $2 x y \geqslant a^{n}=x^{2}+y^{2}$, which means $(x-y)^{2} \leqslant 0$.
Solving this, we get $x=y$.
Therefore, $2 x=a^{m}, 2 x^{2}=a^{n}$. So, $a^{2 m-n}=2$.
Considering that $a$ and $2 m-n$ are both positive integers, we have $a=2$, at this time,
$$
2 m-n=1 \text {. }
$$
Next, we need to find how many digits $2^{30}$ has.
Since $2^{30}=\left(2^{10}\right)^{3} = 1024^{3} > \left(10^{3}\right)^{3} = 10^{9}$, so, $2^{30}$ is a number with no fewer than 10 digits.
$$
\begin{array}{l}
\text { Also, } \frac{2^{30}}{10^{9}}=\frac{1}{10} \times \frac{1024^{3}}{1000^{3}}<\frac{1}{10} \times\left[\frac{1025}{1000}\right]^{3} \\
=\frac{1}{10} \times\left[\frac{41}{40}\right]^{3}<\frac{1}{10} \times\left[\frac{41}{40} \times \frac{40}{39} \times \cdots \times \frac{12}{11}\right] \\
=\frac{1}{10} \times \frac{41}{11}=\frac{41}{110}<1,
\end{array}
$$
Therefore, $2^{30}<10^{9}$.
Hence, $2^{30}$ is a number with fewer than 10 digits.
Thus, we can conclude that $2^{30}$ is a 9-digit number.
(Li Ming, Wang Chengyong, Wuhu County No.3 Middle School, Anhui Province, 233300)
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 5, there are 12 equally spaced points on the circumference of a clock face, marked with the numbers $1, 2, \cdots, 12$ in sequence. Please use these equally spaced points as vertices to form 4 triangles (dividing these 12 equally spaced points into 4 groups) such that the following conditions are met:
(1) No two triangles share a common vertex;
(2) In each triangle, one of the numbers marked on the vertices is equal to the sum of the other two numbers.
Find all different grouping schemes.
(A triangle with vertices $a, b, c$ can be represented as $(a, b, c)$.)
|
Three, let 4 triangles be $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, 3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}$.
Let $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, that is
$$
\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10,12) \text {. }
$$
From $8=b_{1}+c_{1}, 9=b_{2}+c_{2}, 10=b_{3}+c_{3}, 12=b_{4}+c_{4}$, it must be that $b_{4}=11, c_{4}=1$, yielding two cases:
$$
\begin{array}{l}
12=11+1,10=7+3,9=5+4,8=6+2 ; \\
12=11+1,10=6+4,9=7+2,8=5+3 .
\end{array}
$$
Corresponding to two divisions:
$$
\begin{array}{l}
(12,11,1),(10,7,3),(9,5,4),(8,6,2) ; \\
(12,11,1),(10,6,4),(9,7,2),(8,5,3) .
\end{array}
$$
If $a_{3}=11$, then $a_{1}+a_{2}=16$.
Thus, $8<a_{2}<11$, yielding respectively
$$
\left(a_{1}, a_{2}\right)=(6,10),(7,9) \text {. }
$$
For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(6,10,11,12)$, we get three divisions:
$(12,8,4),(11,9,2),(10,7,3),(6,5,1)$; $(12,9,3),(11,7,4),(10,8,2),(6,5,1)$; $(12,7,5),(11,8,3),(10,9,1),(6,4,2)$.
For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(7,9,11,12)$, we also get three divisions:
$(12,8,4),(11,10,1),(9,6,3),(7,5,2)$; $(12,10,2),(11,6,5),(9,8,1),(7,4,3)$; $(12,10,2),(11,8,3),(9,5,4),(7,6,1)$.
Therefore, there are eight grouping schemes in total.
(Tao Pingsheng, Department of Mathematics and Computer Science, Jiangxi Science and Technology Normal University, 330013)
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a $3 \times 3$ grid, the numbers $1,2,3,4,5,6,7,8,9$ are filled in, with one number per cell. Now, the cells containing the maximum number in each row are colored red, and the cells containing the minimum number in each row are colored green. Let $M$ be the smallest number in the red cells, and $m$ be the largest number in the green cells. Then $M-m$ can take on different values.
|
2.8
Obviously, $3 \leqslant m, M \leqslant 7$, and $m \neq M$. Therefore,
$$
M-m \in\{-4,-3,-2,-1,1,2,3,4\} \text {. }
$$
As shown in Figure 7, all 8 values can be obtained
\begin{tabular}{|l|l|l|}
\hline 7 & 9 & 8 \\
\hline 6 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 6 & 9 & 8 \\
\hline 7 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 5 & 9 & 8 \\
\hline 7 & 6 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 5 & 9 & 8 \\
\hline 7 & 6 & 1 \\
\hline 4 & 3 & 2 \\
\hline
\end{tabular}
$M=3, m=7$
$M=3, m=6$
$M=3, m=5$
$M=4, m=5$
\begin{tabular}{|l|l|l|} \hline 9 & 8 & 5 \\ \hline 7 & 4 & 2 \\ \hline 6 & 3 & 1 \\ \hline \end{tabular}
\begin{tabular}{|l|l|l|} \hline 9 & 6 & 5 \\ \hline 8 & 4 & 2 \\ \hline 7 & 3 & 1 \\ \hline \end{tabular}
\begin{tabular}{|l|l|l|}
\hline 9 & 6 & 4 \\
\hline 8 & 5 & 2 \\
\hline 7 & 3 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 9 & 6 & 3 \\
\hline 8 & 4 & 2 \\
\hline 7 & 5 & 1 \\
\hline
\end{tabular}
$M=6, m=5$
$M=7, m=5$
$M=7, m=4$
$M=7, m=3$
Figure 7
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of positive integers $m$ that make $m^{2}+m+7$ a perfect square is $\qquad$ .
|
3.2.
It has been verified that when $m=1$, $m^{2}+m+7=9$ is a perfect square;
when $m=2,3,4,5$, $m^{2}+m+7$ are not perfect squares;
when $m=6$, $m^{2}+m+7=49$ is a perfect square.
When $m>6$, $m^{2}+m+7$ are not perfect squares.
Therefore, there are only 2 positive integers $m$ that meet the condition.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) For any real numbers $x, y$, we have
$$
|x-2|+|x-4| \geqslant m\left(-y^{2}+2 y\right)
$$
Determine the maximum value of the real number $m$.
|
Three, by the geometric meaning of absolute value, $|x-2|+|x-4|$ has a minimum value of 2 when $x \in [2,4]$.
And $-y^{2}+2y=-(y-1)^{2}+1$ has a maximum value of 1 when $y=1$.
From the condition, $2 \geqslant m \times 1$, then $m \leqslant 2$.
Therefore, the maximum value of $m$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (25 points) Given the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=p, \\
3 x y+p(x-y)=p^{2}
\end{array}\right.
$$
has integer solutions $(x, y)$. Find the prime number $p$ that satisfies the condition.
|
Five, from $p=x^{2}-y^{2}=(x-y)(x+y)$ and $p$ being a prime number, we have
$\left\{\begin{array}{l}x+y=p, \\ x-y=1\end{array}\right.$ or $\quad\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$
or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p .\end{array}\right.$
Substituting these into $3 x y+p(x-y)=p^{2}$, we get $\frac{3}{4}\left(p^{2}-1\right)+p=p^{2}$,
which simplifies to $p^{2}-4 p+3=0$.
Solving this, we get $p=3$ or $p=1$ (discard).
(2) When $\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$ or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p\end{array}\right.$, it is found through calculation that there are no prime numbers $p$ that satisfy the conditions.
Therefore, the prime number $p$ that satisfies the conditions is $p=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Add the same integer $a(a>0)$ to the numerator and denominator of $\frac{2008}{3}$, making the fraction an integer. Then the integer $a$ added has $\qquad$ solutions.
|
$=1.3$.
From the problem, we know that $\frac{2008+a}{3+a}$ should be an integer, which means $\frac{2008+a}{3+a}=\frac{2005}{3+a}+1$ should be an integer. Therefore, $(3+a) \mid 2005=5 \times 401$.
Since 2005 has 4 divisors, at this point, $a$ can take the values $2002$, $398$, $2$, and one divisor corresponds to a value of $a$ that is less than 0. Therefore, there are 3 values of $a$ that meet the problem's requirements.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $|a|>1$, simplify
$$
\left(a+\sqrt{a^{2}-1}\right)^{4}+2\left(1-2 a^{2}\right)\left(a+\sqrt{a^{2}-1}\right)^{2}+3
$$
the result is $\qquad$ .
|
Let $x_{0}=a+\sqrt{a^{2}-1}$. Clearly, $x_{0}$ is a root of the equation $x^{2}-2 a x+1=0$, and thus it is also a root of the equation
$$
\left(x^{2}-2 a x+1\right)\left(x^{2}+2 a x+1\right)=0
$$
Then $\left(x_{0}^{2}+1\right)^{2}-4 a^{2} x_{0}^{2}=0$, which means
$$
x_{0}^{4}+2\left(1-2 a^{2}\right) x_{0}^{2}+1=0 \text {. }
$$
Therefore, the original expression $=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given
$$
\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}=p \text {. }
$$
Then $p^{3}+p^{2}+p=$ $\qquad$ .
|
Notice
$$
\begin{array}{l}
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z}, \\
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y} \\
=\frac{x+y-z}{x+y+z}, \\
p^{3}+p^{2}+p \\
=\frac{x+y-z}{x+y+z}+\frac{z+x-y}{x+y+z}+\frac{y+z-x}{x+y+z} \\
=\frac{x+y+z}{x+y+z}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the two roots of the equation $x^{2}+x-1=0$ are $\alpha, \beta$. Then the value of $\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}$ is $\qquad$
|
Let $A=\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}, B=\frac{\alpha^{3}}{\alpha}+\frac{\beta^{3}}{\beta}=\alpha^{2}+\beta^{2}$.
From the given information,
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 . \\
\text { Hence } B=(\alpha+\beta)^{2}-2 \alpha \beta=1+2=3 . \\
\text { Also, } \alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)=-4, \\
\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-1}{-1}=1,
\end{array}
$$
Then $A+B=\alpha^{3}\left(\frac{1}{\beta}+\frac{1}{\alpha}\right)+\beta^{3}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$
$$
=\left(\alpha^{3}+\beta^{3}\right)\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=-4 \text {. }
$$
From equations (1) and (2), we get $A=-4-3=-7$.
|
-7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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