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3. The first 24 digits of $\pi$ are
3. 14159265358979323846264 .
Let $a_{1}, a_{2}, \cdots, a_{24}$ be any permutation of these 24 digits. Then
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right) \\
\equiv \quad(\bmod 2) .
\end{array}
$$
$(\bmod 2)$.
|
3. 0 .
In the first 24 digits of $\pi$, there are 13 odd numbers. When these 13 odd numbers are placed into 12 parentheses, by the pigeonhole principle, there must be two in the same parenthesis, making the difference in this parenthesis even. Thus, the product $\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right)$ is even, and an even number has a remainder of 0 when divided by 2.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a, b, c \in \mathbf{R}$, and $a+b+c=3$. Then the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is $\qquad$
|
3. 9 .
Since the function $y=3^{x}$ is an increasing function on $(-\infty,+\infty)$, by the property of increasing functions, for any $x_{1}$, $x_{2}$, we have $\left(x_{1}-x_{2}\right)\left(3^{x_{1}}-3^{x_{2}}\right) \geqslant 0$, so,
$$
(a-1)\left(3^{a}-3^{1}\right) \geqslant 0,
$$
which means $3^{a} a-3^{\circ} \geqslant 3 a-3$.
Similarly, $3^{b} b-3^{b} \geqslant 3 b-3$, $3^{c} c-3^{c} \geqslant 3 c-3$.
Adding the three inequalities and transforming them, we get
$$
\begin{array}{l}
3^{a} a+3^{b} b+3^{c} c \\
\geqslant 3^{a}+3^{b}+3^{c}+3(a+b+c)-9 \\
=3^{a}+3^{b}+3^{c}+3 \times 3-9=3^{a}+3^{b}+3^{c} . \\
\text { Also, } 3^{a}+3^{b}+3^{c} \geqslant 3 \sqrt[3]{3^{c} \times 3^{b} \times 3^{c}}=3^{2}=9,
\end{array}
$$
with equality holding if and only if $a=b=c=1$.
Therefore, the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
|
From the second equation, we get $z(x+y)=23$, and since 23 is a prime number, we can obtain $\left\{\begin{array}{l}z=1, \\ x+y=23\end{array}\right.$ or $\left\{\begin{array}{l}z=23, \\ x+y=1 \text {. }\end{array}\right.$ According to the problem, $x+y=1$ is not valid, so we discard it. Substituting $z=1$ into the given equations, we get
$$
\left\{\begin{array}{l}
x y+y=63, \\
x+y=23 .
\end{array}\right.
$$
Solving this, we get $(x, y)=(20,3),(2,21)$.
Therefore, the solutions to the original system of equations are
$$
(x, y, z)=(20,3,1),(2,21,1) .
$$
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $P_{n}(k)$ denote the number of permutations of $\{1,2, \cdots, n\}$ with $k$ fixed points. Let $a_{t}=\sum_{k=0}^{n} k^{t} P_{n}(k)$. Then
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=
\end{array}
$$
|
8. 0 .
Notice
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1} \\
=\sum_{k=0}^{n}\left(k^{5}-10 k^{4}+35 k^{3}-50 k^{2}+25 k\right) P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] . \\
\frac{n!}{(n-k)!k!} P_{n-k}(0) \\
=\sum_{k=0}^{n}\left[\frac{n!}{(n-5)!} \cdot \frac{(n-5)!}{(k-5)!(n-k)!} P_{(n-5)-(k-5)}(0)+\right. \\
\left.n \cdot \frac{(n-1)!}{(k-1)!(n-k)!} P_{(n-1)-(k-1)}(0)\right] \\
=\frac{n!}{(n-5)!} \sum_{k=0}^{n-5} P_{n-5}(k)+n \sum_{k=0}^{n-1} P_{n-1}(k) .
\end{array}
$$
Since $a_{0}=\sum_{k=0}^{n} P_{n}(k)=n!$, therefore,
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=\frac{n!}{(n-5)!}(n-5)!+n \cdot(n-1)!-2 \cdot n! \\
=0 .
\end{array}
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
265 It is known that for all positive integers $n$,
$$
\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right) \geqslant \frac{k}{2} \sqrt[3]{19 n+8}
$$
always holds. Try to find the maximum value of $k$.
|
Let $T_{n}=\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right)$. Then
$$
\frac{T_{n+1}}{T_{n}}=1+\frac{1}{3 n+2}=\frac{3 n+3}{3 n+2} \text {. }
$$
Given $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$, we have $k \leqslant \frac{2 T_{n}}{\sqrt[3]{19 n+8}}$.
Let $f(n)=\frac{2 T_{n}}{\sqrt[3]{19 n+8}}$. Then
$$
\begin{array}{l}
f(n+1)=\frac{2 T_{n+1}}{\sqrt[3]{19 n+27}}, \\
\frac{f(n+1)}{f(n)}=\frac{2 T_{n+1}}{\sqrt[3]{19 n+27}} \cdot \frac{\sqrt[3]{19 n+8}}{2 T_{n}} \\
=\frac{T_{n+1}}{T_{n}} \cdot \frac{\sqrt[3]{19 n+8}}{\sqrt[3]{19 n+27}} \\
=\frac{3 n+3}{3 n+2} \sqrt[3]{\frac{19 n+8}{19 n+27}} \\
=\sqrt[3]{\frac{(3 n+3)^{3}(19 n+8)}{(3 n+2)^{3}(19 n+27)}} .
\end{array}
$$
Notice that
$$
\begin{aligned}
& (3 n+3)^{3}(19 n+8)-(3 n+2)^{3}(19 n+27) \\
= & (19 n+8)\left[(3 n+3)^{3}-(3 n+2)^{3}\right]- \\
& 19(3 n+2)^{3} \\
= & (19 n+8)\left(27 n^{2}+45 n+19\right)- \\
& 19\left(27 n^{3}+54 n^{2}+36 n+8\right) \\
= & 45 n^{2}+37 n>0 .
\end{aligned}
$$
Then $(3 n+3)^{3}(19 n+8)$
$$
\begin{array}{l}
>(3 n+2)^{3}(19 n+27) . \\
\text { Hence } \sqrt[3]{\frac{(3 n+3)^{3}(19 n+8)}{(3 n+2)^{3}(19 n+27)}}>1 \\
\Rightarrow \frac{f(n+1)}{f(n)}>1 \Rightarrow f(n)<f(n+1) .
\end{array}
$$
Therefore, $f(n)$ is a monotonically increasing function with respect to $n$.
Since $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$ holds for all positive integers $n$,
$$
\begin{array}{l}
k \leqslant \min \{f(n) \mid n=1,2, \cdots\}=f(1) \\
=\frac{2 T_{1}}{\sqrt[3]{19+8}}=\frac{2\left(1+\frac{1}{2}\right)}{\sqrt[3]{27}}=1 .
\end{array}
$$
Thus, the maximum value of $k$ is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given real numbers $a, b \neq 0$, let
$$
x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{a b}{|a b|} \text {. }
$$
Then the sum of the maximum and minimum values of $x$ is $\qquad$ [1]
|
Question 1 Original Solution ${ }^{[1]}$ According to the number of negative numbers in $a$ and $b$, there are three cases:
(1) If $a$ and $b$ are both positive, then $x=3$;
(2) If $a$ and $b$ are one positive and one negative, then $x=-1$;
(3) If $a$ and $b$ are both negative, then $x=-1$.
In summary, the maximum value of $x$ is 3, and the minimum value is -1.
Therefore, the required answer is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a, b, c$ satisfy
$$
\frac{a b}{a+b}=\frac{1}{3}, \frac{b c}{b+c}=\frac{1}{4}, \frac{c a}{c+a}=\frac{1}{5} \text {. }
$$
then $a b+b c+c a=$ $\qquad$
|
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{c}+\frac{1}{a}=5 \\
\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6 \\
\Rightarrow \frac{1}{c}=3, \frac{1}{a}=2, \frac{1}{b}=1 \\
\Rightarrow a=\frac{1}{2}, b=1, c=\frac{1}{3} \\
\Rightarrow a b+b c+c a=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ are six different positive integers, taking values from $1, 2, 3, 4, 5, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{aligned}
$$
Then the minimum value of $S$ is $\qquad$
|
4. 10 .
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{array}{l}
S \geqslant\left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
\quad\left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots+\left(6-x_{6}\right)\right| \\
= 10 . \\
\text { Also, when } x_{i}=7-i(i=1,2, \cdots, 6) \text {, } S=10 .
\end{array}
$$
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the value of $\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}$.
|
$\begin{array}{l}\text { Solve the original expression }=\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}\left(1+5^{1998}\right)}{7^{1998}\left(1+5^{1998}\right)}} \\ =\left(\frac{7}{3}\right)^{999} \times\left(\frac{3}{7}\right)^{999}=\left(\frac{7}{3} \times \frac{3}{7}\right)^{999}=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A writes down the positive integers $1, 2, \cdots$, 2009 on the blackboard, then turns away from the blackboard, and asks Person B to erase some of these numbers and then add the remainder of the sum of the erased numbers when divided by 7. After several such operations, only two numbers remain on the blackboard, one of which is a single-digit number. A asks B: “What is the larger of the two remaining numbers?” B answers: “100.” Then the single-digit number is
|
3.5.
Since $1+2+\cdots+2009 \equiv 0(\bmod 7)$, therefore, the single digit $a$ satisfies $100+a \equiv 0(\bmod 7)$. Hence $a \equiv 5(\bmod 7)$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of solutions to the equation $\pi^{x-1} x^{2}+\pi^{x^{2}} x-\pi^{x^{2}}=x^{2}+x-1$ is $\qquad$ ( $\pi$ is the ratio of a circle's circumference to its diameter).
|
4. 2 .
The original equation is transformed into
$$
x^{2}\left(\pi^{x-1}-1\right)+(x-1)\left(\pi^{x^{2}}-1\right)=0 \text {. }
$$
When $x \neq 0,1$, $x^{2}\left(\pi^{x-1}-1\right)$ and $(x-1)\left(\pi^{x^{2}}-1\right)$ have the same sign.
Therefore, $x=0,1$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ that satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}(n=2,3, \cdots, m$ is a real parameter $)$. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$.
|
10. Since $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}$,
$$
\begin{array}{l}
a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}, a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, \\
a_{6}=\frac{\frac{a_{0}}{m a_{1}}}{m \frac{1}{m^{2} a_{1}}}=a_{0}, a_{7}=\frac{a_{0}}{m \frac{a_{0}}{m a_{1}}}=a_{1},
\end{array}
$$
Therefore, the sequence $\left\{a_{n}\right\}$ is a periodic sequence with a period of 6.
Thus, $\frac{a_{0}}{a_{1}}=a_{2009}=a_{5}=\frac{a_{0}}{m a_{1}} \Rightarrow m=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $x=2 \frac{1}{5}, y=-\frac{5}{11}, z=-2 \frac{1}{5}$. Then $x^{2}+x z+2 y z+3 x+3 z+4 x y+5=$
|
Solve: From the given, we have
$$
x y=-1, x+z=0, y z=1 \text {. }
$$
Therefore, the expression to be found is
$$
\begin{array}{l}
=x(x+z)+2 y z+3(x+z)+4 x y+5 \\
=2 \frac{1}{5} \times 0+2 \times 1+3 \times 0+4 \times(-1)+5=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (15 points) Divide the sides of an equilateral $\triangle ABC$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points where the sides of $\triangle ABC$ and these parallel lines intersect are called grid points. If $n$ grid points are chosen from these 10 grid points, there will always be three grid points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
|
11. $n_{\min }=5$.
Let the two equal division points on side $AB$ from point $A$ to $B$ be $D$ and $E$, the two equal division points on side $BC$ from point $B$ to $C$ be $F$ and $G$, and the two equal division points on side $CA$ from point $C$ to $A$ be $H$ and $I$, with the central grid point being $K$.
If the minimum value of $n$ is 4, taking the grid points $A, D, E, B$, then there do not exist three grid points that can form an isosceles triangle.
Therefore, $n \geqslant 5$.
We now prove: Any selection of five grid points will definitely contain three grid points that can form an isosceles triangle.
Assume the selected points are red points.
We only need to prove: There must exist an isosceles triangle formed by red points.
If these five red points include grid point $K$, divide the other nine grid points into three sets:
$$
L=\{D, E, F\}, M=\{G, H, I\}, N=\{A, B, C\}.
$$
By the pigeonhole principle, there must be at least one set containing at least two red points. Regardless of which set and which two points are red, they will form an isosceles triangle with red point $K$.
If these five red points do not include grid point $K$, when grid point $A$ is a red point, in
$$
\begin{array}{l}
U=\{D, I\}, V=\{E, H\}, \\
W=\{F, G\}, T=\{B, C\}
\end{array}
$$
if one of these sets contains two red points, the conclusion holds; otherwise, each set contains exactly one red point.
Assume $D$ is a red point, then $I$ is not a red point.
If $B$ is a red point, then $G, C$ are not red points. Thus, $F$ is a red point, and regardless of whether $E$ or $H$ is a red point, they will form an isosceles triangle with $D$ and $F$.
If $B$ is not a red point, then $C$ is a red point, thus $E$ is not a red point, and $H$ is a red point. Regardless of whether $F$ or $G$ is a red point, they will form an isosceles triangle with $D$ and $H$ or $H$ and $C$.
Similarly, when grid point $B$ or $C$ is a red point, the conclusion still holds.
If $K, A, B, C$ are not red points, then among $D, E, F, G, H, I$, there are five red points, and the conclusion is clearly true.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let a tangent line of the circle $x^{2}+y^{2}=1$ intersect the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of $|AB|$ is $\qquad$ .
|
3. 2 .
By symmetry, without loss of generality, assume the point of tangency is
$$
P(\cos \theta, \sin \theta)\left(0<\theta<\frac{\pi}{2}\right) \text {. }
$$
Then $|P A|=\tan \theta,|P B|=\cot \theta$.
Thus, $|A B|=|P A|+|P B|=\tan \theta+\cot \theta \geqslant 2$.
Equality holds if and only if $\tan \theta=\cot \theta$, i.e., $\theta=\frac{\pi}{4}$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $f(x)$ be an odd function defined on $\mathbf{R}$, $f(1)=2$, when $x>0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$. Then the maximum value of the function $f(x)$ on the interval $[-3,-2]$ is
|
5. -4 .
Since $f(x)$ is an odd function and is increasing on $(0,+\infty)$, therefore, $f(x)$ is also increasing on $(-\infty, 0)$. Thus, $f(-3) \leqslant f(x) \leqslant f(-2)$.
$$
\begin{array}{l}
\text { Also, } f(2)=f(1)+f(1)=4, \text { then } \\
f(-2)=-f(2)=-4 .
\end{array}
$$
Therefore, the maximum value of the function $f(x)$ on $[-3,-2]$ is -4.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure $2, A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two moving points on the semicircle, with $C D / / A B$. If the radius of the semicircle $\odot O$ is 1, then the maximum value of the perimeter of trapezoid $A B C D$ is $\qquad$ .
|
7.5.
In Figure 2, connect $A C$, and draw $C H \perp A B$ at point $H$. Let $\angle A B C=\theta\left(0<\theta<\frac{\pi}{2}\right)$. Then $A D=B C=A B \cos \theta=2 \cos \theta$.
Thus, $B H=B C \cos \theta=2 \cos ^{2} \theta$.
Therefore, $C D=A B-2 B H=2-4 \cos ^{2} \theta$.
Hence, the perimeter of trapezoid $A B C D$ is
$$
\begin{aligned}
l & =A B+B C+C D+D A \\
& =4+4 \cos \theta-4 \cos ^{2} \theta \\
& =5-4\left(\cos \theta-\frac{1}{2}\right)^{2} .
\end{aligned}
$$
Therefore, when $\cos \theta=\frac{1}{2}$, i.e., $\theta=\frac{\pi}{3}$, $l_{\max }=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 3, in $\triangle A B C$, $A B=3, A C=5$. If $O$ is the circumcenter of $\triangle A B C$, then the value of $\overrightarrow{A O} \cdot \overrightarrow{B C}$ is
|
8. 8 .
Let $D$ be the midpoint of side $B C$, and connect $O D, A D$. Then $O D \perp B C$. Therefore,
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdot \overrightarrow{B C}=\overrightarrow{A D} \cdot \overrightarrow{B C} \\
=\frac{1}{2}(\overrightarrow{A C}+\overrightarrow{A B}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \\
=\frac{1}{2}\left(|\overrightarrow{A C}|^{2}-|\overrightarrow{A B}|^{2}\right)=8
\end{array}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Calculate:
$$
\frac{2009^{2}-2008^{2}}{19492009^{2}-19492008 \times 19492010+2 \times 2008} .
$$
|
Let $19492009=a, 2008=b$. Then the original expression $=\frac{(b+1)^{2}-b^{2}}{a^{2}-(a-1)(a+1)+2 b}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given real numbers $x, y, z$ satisfy $x-y=8$, $xy+z^{2}=-16$. Then $x+y+z=$ $\qquad$ .
|
Let $x=a+b, y=a-b$. Then $x-y=2b \Rightarrow b=4$.
Since $xy+z^{2}=-16$, we have $(a+b)(a-b)+z^{2}+16=0$. Also, $-b^{2}=-16$, then $a^{2}+z^{2}=0$. By the property of non-negative numbers, we know $a=0, z=0$. Therefore, $x=4, y=-4$. Hence, $x+y+z=4-4+0=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an "integer point". The number of integer points on the graph of the function $y=\frac{x+12}{2 x-1}$ is $\qquad$
|
2. 6 .
Notice that $2 y=1+\frac{25}{2 x-1}$.
Since $y$ is an integer, $\frac{25}{2 x-1}$ must also be an integer. Therefore, $2 x-1= \pm 1, \pm 5, \pm 25$.
Solving for $x$ gives $x=-12,-2,0,1,3,13$.
Correspondingly, $y=0,-2,-12,13,3,1$.
Thus, there are 6 integer points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups.
|
4. 12 .
$$
\begin{array}{r}
\text { Since }(a+1)(b+1) \\
=2009=41 \times 7^{2} \text {, and given }
\end{array}
$$
that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let convex quadrilateral $ABCD$ satisfy $AB=AD=1$, $\angle A=160^{\circ}, \angle C=100^{\circ}$. Then the range of the length of diagonal $AC$ is $\qquad$ .
|
2. $\{1\}$.
Since $\angle C=180^{\circ}-\frac{\angle A}{2}$, therefore, points $B$, $C$, and $D$ lie on a circle with $A$ as the center and a radius of 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Simplify:
$$
\frac{(\sqrt{x}-\sqrt{y})^{3}+2 x \sqrt{x}+y \sqrt{y}}{x \sqrt{x}+y \sqrt{y}}+\frac{3 \sqrt{x y}-3 y}{x-y} .
$$
|
Let $\sqrt{x}=a+b, \sqrt{y}=a-b$. Then
$$
\sqrt{x y}=a^{2}-b^{2}, \sqrt{x}+\sqrt{y}=2 a, \sqrt{x}-\sqrt{y}=2 b \text {. }
$$
Therefore, the original expression is
$$
\begin{aligned}
= & \frac{(2 b)^{3}+2(a+b)^{3}+(a-b)^{3}}{(a+b)^{3}+(a-b)^{3}}+ \\
& \frac{3\left(a^{2}-b^{2}\right)-3(a-b)^{2}}{4 a b} \\
= & \frac{3\left(a^{2}+3 b^{2}\right)(a+b)}{2 a\left(a^{2}+3 b^{2}\right)}+\frac{3(a-b)}{2 a} \\
= & \frac{3 a+3 b+3 a-3 b}{2 a}=\frac{6 a}{2 a}=3 .
\end{aligned}
$$
[Note] This problem uses sum and difference substitution to transform the simplification of radicals into the simplification of fractions, thereby reducing the difficulty and improving the speed of solving the problem.
From the above examples, it can be seen that the essence of sum and difference substitution
is substitution and transformation, the key is to construct and set variables, the purpose is to introduce new variables to make complex calculations and proofs easier to handle.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (22 points) Suppose a set of planar points $S$ has the following properties:
(1) No three points are collinear;
(2) The distance between any two points is unique.
For two points $A$ and $B$ in $S$, if there exists a point $C \in S$ such that $|A C|<|A B|<|B C|$, then $A B$ is called a "middle edge" of $S$. For three points $A$, $B$, and $C$ in $S$, if $A B$, $A C$, and $B C$ are all middle edges of $S$, then $\triangle A B C$ is called a "middle edge triangle" of $S$. Find the smallest $n$ such that any $n$-element planar point set $S$ with properties (1) and (2) must contain a middle edge triangle.
|
4. Color all middle edges of $S$ red, and color other edges blue.
When $n \geqslant 6$, according to Ramsey's theorem, there must exist a monochromatic triangle, which must have a middle edge, and it must be a middle edge triangle.
The following set of five points with properties (1) and (2) does not have a middle edge triangle:
Assume five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are arranged in a counterclockwise order on the circumference of a circle, and
$$
\begin{array}{c}
{\overparen{P_{1} P_{2}}}_{2}^{\circ}=\frac{\pi}{10},{\overparen{P_{2} P_{3}}}^{0}=\frac{3 \pi}{5},{\overparen{P_{3} P_{4}}}^{0}=\frac{3 \pi}{10}, \\
\overparen{P}_{4} P_{5}^{0}=\frac{3 \pi}{4},{\overparen{P_{5} P_{1}}}^{0}=\frac{\pi}{4} .
\end{array}
$$
Then the distances between points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are all different, and $P_{2} P_{3}, P_{3} P_{1}, P_{1} P_{5}, P_{5} P_{4}, P_{4} P_{2}$ are middle edges, but there is no middle edge triangle.
For cases with fewer than five points, simply remove some points from the previous example, and there will still be no middle edge triangle.
(Provided by An Sai)
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The line $l: x+y=t$ intersects the circle $\odot O: x^{2}+y^{2}=20$ at points $A$ and $B$, and $S_{\triangle O A B}$ is an integer. Then the number of all positive integer values of $t$ that satisfy the condition is $\qquad$ .
|
7.2.
Let $\angle A O B=2 \alpha, O C \perp A B$, with the foot of the perpendicular being $C$. Then $O C=\sqrt{20} \cos \alpha=\frac{t}{\sqrt{1+1}}=\frac{t}{\sqrt{2}}$ $\Rightarrow \cos \alpha=\frac{t}{2 \sqrt{10}}$.
Also, $S_{\triangle O A B}=10 \sin 2 \alpha \leqslant 10$, so $\sin 2 \alpha=\frac{k}{10}(k \in\{1,2, \cdots, 10\})$.
Thus, $2 \cos \alpha \sqrt{1-\cos ^{2} \alpha}=\frac{k}{10}$.
Substituting equation (1) into the above equation and simplifying, we get $t^{2}\left(40-t^{2}\right)=4 k^{2}$.
It is easy to see that $t$ is an even number. Let $t=2 t_{0}$. Then $4 t_{0}^{2}\left(10-t_{0}^{2}\right)=k^{2} \Rightarrow t_{0}<\sqrt{10}$. Upon inspection, $t_{0}=1,3$ satisfy the conditions. Therefore, $t=2,6$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If the polynomial $f(x)=x^{3}-6 x^{2}+a x+a$ has three roots $x_{1}, x_{2}, x_{3}$ that satisfy
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 \text {, }
$$
then the value of the real number $a$ is .. $\qquad$
|
6. -9 .
From the problem, we have
$$
\begin{array}{l}
g(t)=f(t+3) \\
=(t+3)^{3}-6(t+3)^{2}+a(t+3)+a \\
=t^{3}+3 t^{2}+(a-9) t+4 a-27
\end{array}
$$
The three roots \( t_{1}, t_{2}, t_{3} \) satisfy \( t_{1}^{3}+t_{2}^{3}+t_{3}^{3}=0 \).
By the relationship between roots and coefficients, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
t_{1}+t_{2}+t_{3}=-3, \\
t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}=a-9, \\
t_{1} t_{2} t_{3}=-(4 a-27) .
\end{array}\right. \\
\text { Substituting into } t_{1}^{3}+t_{2}^{3}+t_{3}^{3}-3 t_{1} t_{2} t_{3} \\
=\left(t_{1}+t_{2}+t_{3}\right)\left[\left(t_{1}+t_{2}+t_{3}\right)^{2}-\right. \\
\left.3\left(t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}\right)\right],
\end{array}
$$
we get \( 3(4 a-27)=-3[9-3(a-9)] \).
Solving for \( a \), we get \( a=-9 \).
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given an ellipse centered at the origin, with foci on the $x$-axis, the length of the major axis is twice the length of the minor axis, and it passes through the point $M(2,1)$. A line $l$ parallel to $OM$ has a $y$-intercept of $m (m<0)$, and intersects the ellipse at two distinct points $A$ and $B$. Find the $x$-coordinate of the incenter $I$ of $\triangle ABM$.
untranslated text remains the same as the source text.
|
10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then $\left\{\begin{array}{l}a=2 b, \\ \frac{4}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=8, \\ b^{2}=2 \text {. }\end{array}\right.\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$.
As shown in Figure 1, because
the line $l$ is parallel
to $O M$, and the y-intercept
is $m$, and $k_{O S}=\frac{1}{2}$,
so, the equation of the line $l$ is $y=\frac{1}{2} x+m$.
From $\left\{\begin{array}{l}y=\frac{1}{2} x+m, \\ \frac{x^{2}}{8}+\frac{y^{2}}{2}=1\end{array} \Rightarrow x^{2}+2 m x+2 m^{2}-4=0\right.$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then
$x_{1}+x_{2}=-2 m, x_{1} x_{2}=2 m^{2}-4$.
Let the slopes of the lines $M A 、 M B$ be $k_{1} 、 k_{2}$. Then
$k_{1}=\frac{y_{1}-1}{x_{1}-2}, k_{2}=\frac{y_{2}-1}{x_{2}-2}$.
Thus, $k_{1}+k_{2}=\frac{y_{1}-1}{x_{1}-2}+\frac{y_{2}-1}{x_{2}-2}$
$=\frac{\left(y_{1}-1\right)\left(x_{2}-2\right)+\left(y_{2}-1\right)\left(x_{1}-2\right)}{\left(x_{1}-2\right)\left(x_{2}-2\right)}$.
The numerator of the above expression
$$
\begin{array}{l}
=\left(\frac{1}{2} x_{1}+m-1\right)\left(x_{2}-2\right)+\left(\frac{1}{2} x_{2}+m-1\right)\left(x_{1}-2\right) \\
=x_{1} x_{2}+(m-2)\left(x_{1}+x_{2}\right)-4(m-1) \\
=2 m^{2}-4+(m-2)(-2 m)-4(m-1) \\
=0 .
\end{array}
$$
Therefore, $k_{1}+k_{2}=0$.
Hence, the angle bisector $M I$ of $\triangle A B M$ is perpendicular to the x-axis. Therefore, the x-coordinate of the incenter $I$ is equal to the x-coordinate of point $M$, which is 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}, x y z>0$,
and
$$
\begin{array}{l}
\sqrt[3]{\frac{2010}{x^{2}}+\frac{2011}{y^{2}}+\frac{2012}{z^{2}}} \\
=\sqrt[3]{2010}+\sqrt[3]{2011}+\sqrt[3]{2012} .
\end{array}
$$
Find the value of $x+y+z$.
|
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}=k$, obviously $k \neq 0$.
Then $2010=x^{3} k, 2011=y^{3} k, 2012=z^{3} k$.
From the given, we have
$$
\sqrt[3]{x k+y k+z k}=\sqrt[3]{x^{3} k}+\sqrt[3]{y^{3} k}+\sqrt[3]{z^{3} k},
$$
which means $\sqrt[3]{k} \sqrt[3]{x+y+z}=\sqrt[3]{k}(x+y+z)$.
Since $k \neq 0$, we have $\sqrt[3]{x+y+z}=x+y+z$.
Given that $x>0, y>0, z>0$.
Therefore, $x+y+z=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find all positive integers $n$, such that $2^{n-1} n+1$ is a perfect square.
(2004, Slovenia IMO National Team Selection Test
|
Let $2^{n-1} n+1=m^{2}\left(m \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=(m+1)(m-1)$.
When $n=1,2,3,4$, $2^{n-1} n+1$ is not a perfect square.
Therefore, $n \geqslant 5, 16 \mid (m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, both $m+1$ and $m-1$ are even, and $m$ is odd.
Let $m=2 k-1\left(k \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=2 k(2 k-2)$.
Thus, $2^{n-3} n=k(k-1)$.
Since $k$ and $k-1$ have different parities, $2^{n-3}$ can only be a divisor of one of them.
Also, $2^{n-3} n=k(k-1) \neq 0$, so
$2^{n-3} \leqslant k$.
Therefore, $n \geqslant k-1$.
Hence, $2^{n-3} \leqslant k \leqslant n+1$.
By the properties of functions or mathematical induction, when $n \geqslant 6$, $2^{n-3}>n+1$.
Thus, $n \leqslant 5$. Since $n \geqslant 5$, we have $n=5$.
At this point, $2^{n-1} n+1=81$ is a perfect square, satisfying the condition.
In summary, the required positive integer $n=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Proof: There exist positive integers $a_{i}(1 \leqslant i \leqslant 8)$, such that
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}}=2 .
\end{array}
$$
|
Prove the construction of the identity:
$$
\begin{array}{l}
\sqrt{(2 i+1)^{2}}-\sqrt{(2 i+1)^{2}-1} \\
=2 i+1-2 \sqrt{i(i+1)} \\
=(\sqrt{i+1}-\sqrt{i})^{2} .
\end{array}
$$
Take \( a_{i}=(2 i+1)^{2} \) for \( i=1,2, \cdots, 8 \), then
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}} \\
=\sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\cdots+\sqrt{8+1}-\sqrt{8} \\
=\sqrt{9}-\sqrt{1}=2 .
\end{array}
$$
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that $a$ is a root of the equation $x^{2}-3 x+1=0$. Then the value of the fraction $\frac{2 a^{6}-6 a^{4}+2 a^{5}-a^{2}-1}{3 a}$ is $\qquad$ -
|
II. 7. -1.
According to the problem, we have $a^{2}-3 a+1=0$.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 a^{3}\left(a^{2}-3 a+1\right)-\left(a^{2}+1\right)}{3 a} \\
=-\frac{a^{2}+1}{3 a}=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Robots A and B simultaneously conduct a $100 \mathrm{~m}$ track test at a uniform speed, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. After calculation, this track is not standard. Then this track is $\qquad$ m longer than $100 \mathrm{~m}$.
|
8. 1 .
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙 }}$, respectively. When 甲 is $1 \mathrm{~m}$ away from the finish line, the time spent is $t$. Then $v_{\text {乙 }} t=x-2$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}$.
Let the time 甲 takes to reach the finish line be $t^{\prime}$. Then $v_{\text {甲 }} t^{\prime}=x, v_{\text {乙 }} t^{\prime}=x-1.01$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x}{x-1.01}$.
We can get the equation $\frac{x-1}{x-2}=\frac{x}{x-1.01} \Rightarrow x=101 \mathrm{~m}$.
Therefore, this track is $1 \mathrm{~m}$ longer than $100 \mathrm{~m}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The graphs of the functions $f(x)=2 x^{2}-2 x-1$ and $g(x)=$ $-5 x^{2}+2 x+3$ intersect at two points. The equation of the line passing through these two points is $y=a x+b$. Find the value of $a-b$.
|
6. The graphs of the functions $f(x)$ and $g(x)$ intersect at two points, which are the solutions to the system of equations $\left\{\begin{array}{l}y=2 x^{2}-2 x-1, \\ y=-5 x^{2}+2 x+3\end{array}\right.$. Therefore,
$$
\begin{array}{l}
7 y=5\left(2 x^{2}-2 x-1\right)+2\left(-5 x^{2}+2 x+3\right) \\
=-6 x+1 .
\end{array}
$$
Thus, the two intersection points lie on the line $y=-\frac{6}{7} x+\frac{1}{7}$, i.e., $a=-\frac{6}{7}, b=\frac{1}{7}$.
Therefore, $a-b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Three. (20 points) (1) Prove: } \\
\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=4 \sin ^{2} 2 x-3(x \in \mathbf{R}) ;
\end{array}
$$
(2) Find the value: $\prod_{t=0}^{2^{8}}\left(4 \sin ^{2} \frac{t \pi}{2^{9}}-3\right)$.
|
$$
\begin{array}{l}
=(1)\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=16 \sin ^{2} x \cdot \cos ^{2} x-12\left(\sin ^{2} x+\cos ^{2} x\right)+9 \\
=4 \sin ^{2} 2 x-3
\end{array}
$$
(2) Let $A_{k}=\prod_{i=0}^{2 k-1}\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)$.
When $k \geqslant 2$, by (1) we have
$$
\begin{aligned}
A_{k}= & \prod_{t=0}^{2^{k-2}-1}\left\{\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)\right. \\
& {\left.\left[4 \sin ^{2} \frac{\left(2^{k-1}-t\right) \pi}{2^{k}}-3\right]\right\} } \\
& =\prod_{t=0}^{2^{k-2}-1}\left(4 \sin ^{2} \frac{2^{k-1} \pi}{2^{k}}-3\right) \\
& \left.=-\sin ^{2} \frac{t \pi}{2^{k-1}}-3\right)\left[-\left(4 \sin ^{2} \frac{2^{k-2} \pi}{2^{k-1}}-3\right)\right]
\end{aligned}
$$
Therefore, $A_{9}=A_{1}=\left(4 \sin ^{2} 0-3\right)\left(4 \sin ^{2} \frac{\pi}{2}-3\right)$ $=-3$.
|
-3
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$ (for example, $f(3)$ is the sum of the digits of $3 \times 3^{2}+3+1=31$, i.e., $f(3)=4$).
(1) Prove that for any positive integer $n, f(n) \neq 1$, and $f(n) \neq 2$;
(2) Try to find a positive integer $n$ such that $f(n)=3$.
|
(1) Since $3 n^{2}+n+1$ is an odd number greater than 3, hence $f(n) \neq 1$.
If $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$ (where $k$ is an integer greater than 1).
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Since $(n, 3 n+1)=1$, we have $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k} \text {. }\end{array}\right.$
Therefore, $3 n+1 \leqslant 4 n=4 \times 2^{k}<5^{k}$. This is a contradiction.
Hence, $f(n) \neq 2$.
(2) When $n=8$, since $3 n^{2}+n+1=201$, thus, $f(8)=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a three-digit number $x y z(1 \leqslant x \leqslant 9,0 \leqslant y, z$ $\leqslant 9)$. If $x y z=x!+y!+z!$, then the value of $x+y+z$ is
|
7. 10 .
Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$.
And $1!=1, 2!=2, 3!=6, 4!=24, 5!=120$.
By observation, we get $x=1, y=4, z=5$. Therefore,
$$
x+y+z=10 \text {. }
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Given that $a$ and $b$ are integers, the equation $a x^{2} + b x + 2 = 0$ has two distinct negative real roots greater than -1. Find the minimum value of $b$.
|
Let the equation $a x^{2}+b x+2=0(a \neq 0)$ have two distinct negative real roots $x_{1} 、 x_{2}\left(x_{1}<x_{2}<0\right)$.
\end{array}\right.
$$
Solving, we get $a>0, b>0$.
Since $a$ and $b$ are both integers, it follows that $a$ and $b$ are both positive integers.
Let $y=a x^{2}+b x+2$. Then this parabola opens upwards, intersects the $x$-axis at two different points, and when $x=-1$, $y>0$. Therefore,
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ \Delta = b ^ { 2 } - 8 a > 0 , } \\
{ a - b + 2 > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
8 a \leqslant b^{2}-1, \\
a \geqslant b-1
\end{array}\right. \\
\Rightarrow b^{2}-1 \geqslant 8 b-8 \\
\Rightarrow(b-1)(b-7) \geqslant 0 .
\end{array}
$$
Since $b-1 \geqslant 0$, then $b-7 \geqslant 0 \Rightarrow b \geqslant 7$.
When $b=7$, substituting into equations (1) and (2) gives $a=6$, at which point the roots of the original equation are $-\frac{1}{2},-\frac{2}{3}$.
Therefore, the minimum value of $b$ is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 2, fold $\triangle A B C$ along the dotted line $D E$ to get a heptagon $A D E C F G H$. If the area ratio of the heptagon to the original triangle is $2: 3$, and the area of the overlapping part after folding is 4, then the area of the original $\triangle A B C$ is
|
3. 12 .
Let the area of the non-overlapping part after folding be $x$. Then the area of the original triangle is $8+x$, and the area of the heptagon is $4+x$. From the given condition, we have
$$
\begin{array}{l}
(8+x):(4+x)=3: 2 \\
\Rightarrow 16+2 x=12+3 x \Rightarrow x=4 .
\end{array}
$$
$$
\text { Hence } S_{\triangle A B C}=12 \text {. }
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ The value is
The value of $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is
|
4. 4 .
Let $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}=x$.
Cubing both sides and simplifying, we get
$$
x^{3}-6 x-40=0 \text {. }
$$
By observation, 4 is a root of the equation. Therefore,
$$
(x-4)\left(x^{2}+4 x+10\right)=0 \text {. }
$$
Since $\Delta=4^{2}-4 \times 10=-24<0$, the equation $x^{2}+4 x+10=0$ has no real roots.
Hence, the equation $x^{3}-6 x-40=0$ has only one real root $x=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+\cdots+b_{n} x^{n},
\end{array}
$$
and $b_{1}+b_{2}+\cdots+b_{n}=1013$.
Then a possible value of the positive integer $n$ is
|
13. 9 .
Let $x=0$, we get $b_{0}=n$.
Let $x=1$, we get
$$
2+2^{2}+\cdots+2^{n}=b_{0}+b_{1}+\cdots+b_{n},
$$
which is $2\left(2^{n}-1\right)=n+1013$.
Solving this, we get $n=9$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 4, given $\angle A O M=60^{\circ}$, there is a point $B$ on ray $O M$ such that the lengths of $A B$ and $O B$ are both integers, thus $B$ is called an "olympic point". If $O A=8$, then the number of olympic points $B$ in Figure 4 is $\qquad$
|
4. 4 .
As shown in Figure 7, draw $A H \perp O M$ at point $H$.
Since $\angle A O M$
$=60^{\circ}$, and $O A$
$=8$, therefore,
$O H=4$,
$A H=4 \sqrt{3}$.
Let $A B=$
$m, H B=n$
($m, n$ are positive integers). Clearly, in the right triangle $\triangle A H B$, we have
$$
m^{2}-n^{2}=(4 \sqrt{3})^{2} \text {, }
$$
which simplifies to $(m+n)(m-n)=48$.
Since $m+n$ and $m-n$ have the same parity, the factorization of 48 can only be the following three:
$$
48=24 \times 2=12 \times 4=8 \times 6 .
$$
Thus, when $(m, n)=(13,11),(8,4),(7,1)$, there are Olympic points $B_{1}, B_{2}, B_{3}$.
Additionally, it is easy to see that the point $B_{3}$ symmetric to the line $A H$ is also an Olympic point, denoted as $B_{3}^{\prime}$.
Therefore, there are four Olympic points: $B_{1}, B_{2}, B_{3}, B_{3}^{\prime}$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are
|
6. 4 .
According to the problem, we know that $(2 n+1) \mid(20 n+1)$. Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$. Therefore, $n \in\{1,4\}$. Upon verification, $n=4$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: There is an electronic flea jumping back and forth on a number line, and it lights up a red light when it jumps to a negative number, but does not light up when it jumps to a positive number. The starting point is at the point representing the number -2 (record one red light), the first step is to jump 1 unit to the left, the second step is to jump 2 units to the right, the third step is to jump 3 units to the left, the fourth step is to jump 4 units to the right, and so on.
(1) When it jumps to the tenth step, how far is the electronic flea from the origin? How many times has the red light been lit?
(2) If the electronic flea needs to light up the red light ten times, what is the minimum number of jumps it needs to make? At this point, how far is it from the origin?
|
(1) Notice
$$
\begin{array}{l}
S=\mid-2-1+2-3+4-5+\cdots-9+101 \\
=1-2+1 \times 5 \mid=3 .
\end{array}
$$
Therefore, the negative numbers appear as $-2,-3,-1,-4,-5$, $-6,-7$, a total of seven times. So, the red light will flash seven times.
(2) In fact, after the electronic flea jumps four times, it returns to the origin, and the red light has already flashed four times (including the initial jump). After that, the red light flashes once for each odd-numbered jump. Therefore, to make the red light flash ten times, the electronic flea needs to jump $4+$ $2 \times 6-1=15$ times. At this point, the distance of the electronic flea from the origin is
$$
\begin{array}{l}
S=|-2-1+2-3+4-5+\cdots-15| \\
=|-2+1 \times 7-15|=10 .
\end{array}
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 If real numbers $a, b$ satisfy the conditions
$$
a^{2}+b^{2}=1,|1-2 a+b|+2 a+1=b^{2}-a^{2} \text {, }
$$
then $a+b=$ $\qquad$
(2009, National Junior High School Mathematics Joint Competition)
|
Given $a^{2}+b^{2}=1$, we have
$$
b^{2}=1-a^{2} \text {, and }-1 \leqslant a \leqslant 1,-1 \leqslant b \leqslant 1 \text {. }
$$
From $|1-2 a+b|+2 a+1=b^{2}-a^{2}$, we get
$$
\begin{array}{l}
|1-2 a+b|=b^{2}-a^{2}-2 a-1 \\
=\left(1-a^{2}\right)-a^{2}-2 a-1=-2 a^{2}-2 a .
\end{array}
$$
Thus, $-2 a^{2}-2 a \geqslant 0 \Rightarrow-1 \leqslant a \leqslant 0$.
Hence, $1-2 a+b \geqslant 0$.
Therefore, $1-2 a+b=-2 a^{2}-2 a$, which means
$$
1+b=-2 a^{2}=-2\left(1-b^{2}\right) \text {. }
$$
Rearranging gives $2 b^{2}-b-3=0$.
Solving this, we get $b=-1$ (the other root $b=\frac{3}{2}$ is discarded).
Substituting $b=-1$ into $1+b=-2 a^{2}$, we get $a=0$.
Thus, $a+b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
|
12. By the Pythagorean theorem, we have
$$
c^{2}=a^{2}+b^{2} \text {. }
$$
Also, $c=\frac{1}{3} a b-(a+b)$, substituting into equation (1) gives
$$
\begin{array}{l}
a^{2}+b^{2} \\
=\frac{1}{9}(a b)^{2}-\frac{2}{3} a b(a+b)+a^{2}+2 a b+b^{2} .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\Rightarrow(a-6)(b-6)=18 .
\end{array}
$$
Since $a, b$ are both positive integers, without loss of generality, let $a<b$. Then
$$
\left\{\begin{array} { l }
{ a - 6 = 1 , } \\
{ b - 6 = 1 8 }
\end{array} \text { or } \left\{\begin{array} { l }
{ a - 6 = 2 , } \\
{ b - 6 = 9 }
\end{array} \text { or } \left\{\begin{array}{l}
a-6=3, \\
b-6=6 .
\end{array}\right.\right.\right.
$$
Solving, we get $(a, b, c)$
$$
=(7,24,25),(8,15,17),(9,12,15) \text {. }
$$
Therefore, there are three right triangles that satisfy the conditions.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In $\triangle A B C$, it is known that $A B=b^{2}-1, B C=$ $a^{2}, C A=2 a$, where $a$ and $b$ are both integers greater than 1. Then the value of $a-b$ is $\qquad$
|
2.0.
Since $a$ is an integer greater than 1, we have $a^{2} \geqslant 2 a$. By the triangle inequality, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
a^{2}+2 a>b^{2}-1, \\
2 a+b^{2}-1>a^{2}
\end{array}\right. \\
\Leftrightarrow(a-1)^{2}<b^{2}<(a+1)^{2} .
\end{array}
$$
Given $a, b \in \mathbf{N}_{+}$, hence $b=a$.
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ .
|
4.5.
Factorizing the left side of the equation, we get
$$
\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 \text {. }
$$
Since $x^{2}+x+2=0$ has no real roots, the equation $x^{2}+x+k=0$ has two real roots $x_{1} 、 x_{2}$.
According to the problem, $x_{1} x_{2}=k=-2$.
$$
\begin{array}{l}
\text { Also, } x_{1}+x_{2}=-1 \text {, so } \\
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=5 .
\end{array}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 The function $y=|x+1|+|x+2|+|x+3|$. When $x=$ $\qquad$, $y$ has its minimum value, and the minimum value is $\qquad$
(2007, National Junior High School Mathematics Competition Zhejiang Regional Finals)
|
It is known that $|x+1|+|x+2|+|x+3|$ represents the sum of the distances (lengths of segments) from point $x$ to points $-1$, $-2$, and $-3$ on the number line. It is easy to see that when $x=-2$, this sum of distances is minimized, i.e., the value of $y$ is the smallest, at which point, $y_{\min }=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, the minimum value of $7 y=2 x^{2}+4|x|-1$ is
$\qquad$
(2007, National Junior High School Mathematics Competition, Zhejiang Province Re-test)
|
Let $t=|x| \geqslant 0$. Then
$$
y=2 t^{2}+4 t-1=2(t+1)^{2}-3(t \geqslant 0) \text{.}
$$
It is easy to see that when $t=|x|=0$, i.e., $x=0$, $y$ is minimized at -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If for any real number $x$, the function
$$
f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4
$$
is always a non-negative real number, then the maximum value of the real number $a$ is
|
7. 1 .
From the conditions, we have $\left\{\begin{array}{l}f(0)=-|1+a|+2 \geqslant 0, \\ f(1)=-|a|+2 \geqslant 0 .\end{array}\right.$
Solving this, we get $-2 \leqslant a \leqslant 1$.
When $a=1$, we have
$$
f(x)=x^{2}-2 x-2|x-2|+4,
$$
which is $f(x)=\left\{\begin{array}{ll}x^{2}, & x \leqslant 2 ; \\ x^{2}-4 x+8, & x>2\end{array}\right.$.
It is easy to see that, for any real number $x$, the value of $f(x)$ is always a non-negative real number.
Therefore, $a=1$ meets the requirement.
Hence, the maximum value of the real number $a$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Find the minimum value of the function
$$
f(x)=\max \left\{|x+1|,\left|x^{2}-5\right| \right\}
$$
and find the value of the independent variable $x$ when $f(x)$ takes the minimum value.
|
Solve As shown in Figure 1, in the same Cartesian coordinate system, draw the graphs of $f_{1}(x)=|x+1|$ and $f_{2}(x)=\left|x^{2}-5\right|$. The two graphs intersect at four points, $A$, $B$, $C$, and $D$, whose x-coordinates can be obtained by solving the equation $|x+1|=\left|x^{2}-5\right|$.
Removing the absolute value signs gives
$$
x+1=x^{2}-5 \text { or } x+1=5-x^{2} \text {. }
$$
Solving these, we get $x_{1}=3, x_{2}=-2$,
$$
x_{3}=\frac{-1+\sqrt{17}}{2}, x_{4}=\frac{-1-\sqrt{17}}{2} \text {. }
$$
From Figure 1, it is clear that the x-coordinates of points $A$, $B$, $C$, and $D$ are $\frac{-1-\sqrt{17}}{2},-2, \frac{-1+\sqrt{17}}{2}, 3$ respectively.
According to the definition of $f(x)$, its graph is shown by the solid line in the figure, and the y-coordinate of point $B$ is the minimum value of the function $f(x)$, at which $f(-2)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $a, b$ satisfy
$$
|a+1|+|a+3|+|b+2|+|b-5|=9 \text {. }
$$
Let the maximum and minimum values of $ab + a + b$ be $m, n$ respectively. Then the value of $m+n$ is ( ).
(A) -12 (B) -10 (C) -14 (D) -8
|
-1. A.
$$
\begin{array}{l}
\text { Since } 9=|a+1|+|a+3|+|b+2|+|b-5| \\
=(|a+1|+|-a-3|)+(|b+2|+|5-b|) \\
\geqslant|1-3|+|2+5|=9 .
\end{array}
$$
Thus $-3 \leqslant a \leqslant-1$, and $-2 \leqslant b \leqslant 5$
$$
\begin{array}{l}
\Rightarrow-2 \leqslant a+1 \leqslant 0, \text { and }-1 \leqslant b+1 \leqslant 6 \\
\Rightarrow-12 \leqslant(a+1)(b+1) \leqslant 2 \\
\Rightarrow-13 \leqslant a b+a+b \leqslant 1 .
\end{array}
$$
Therefore, $m=1, n=-13$.
Thus, $m+n=-12$.
|
-12
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 On the blackboard, there are two positive integers, one is 2002, and the other is a number less than 2002. If the average of these two numbers $m$ is an integer, then the following operation can be performed: one of the numbers is erased and replaced by $m$. How many times can such an operation be performed at most?
|
Without loss of generality, let these two positive integers be $a, b(a>b)$.
Consider the absolute value of the difference between the two numbers.
Initially, it is $|a-b|$. After the first operation, the absolute value of the difference between the two numbers is $\left|a-\frac{a+b}{2}\right|=\left|\frac{a-b}{2}\right|$. Proceeding in this manner, after each operation, the absolute value of the difference between the two numbers becomes half of the original absolute value of the difference, and the average $m$ of the two numbers is an integer. Therefore, the number of allowed operations $n$ should satisfy
$2^{n}<2002$.
Clearly, the maximum value of $n$ is 10, meaning the operation can be performed at most 10 times.
At this point, the initial two positive integers on the blackboard are 2002 and 978.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the equation
$$
n \sin x+(n+1) \cos x=n+2
$$
has two distinct real roots in $0<x<\pi$, then the minimum value of the positive integer $n$ is $\qquad$.
|
4.4.
From the given, we have $\frac{1}{n}=-1-\frac{-1-\sin x}{2-\cos x}$.
And $\frac{-1-\sin x}{2-\cos x}$ represents the slope $k$ of the line connecting a moving point $P(\cos x, \sin x)$ on the upper half of the unit circle (excluding endpoints) and a fixed point $Q(2,-1)$.
To satisfy the problem, the line $PQ$ must intersect the upper half of the unit circle (excluding endpoints) at two distinct points, in which case, $k \in\left(-\frac{4}{3},-1\right)$.
Thus, $\frac{1}{n} \in\left(0, \frac{1}{3}\right)$, which gives $n>3$.
Therefore, the smallest positive integer value of $n$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sum of the x-coordinates of the points where the graph of the function $y=x^{2}-2009|x|+2010$ intersects the x-axis is $\qquad$ .
|
3. 0 .
The original problem can be transformed into finding the sum of all real roots of the equation
$$
x^{2}-2009|x|+2010=0
$$
If a real number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1).
Therefore, the sum of all real roots of the equation is 0, that is, the sum of the x-coordinates of the points where the graph intersects the x-axis is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given
$$
A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n}(n \in \mathbf{N})
$$
can be expressed as the product of $k(k \in \mathbf{N}, k>1)$ consecutive integers. Then $n+k=$ $\qquad$ .
|
4. 2 .
If $k \geqslant 4$, then $81 A$.
$$
\begin{array}{l}
\text { But } A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n} \\
\equiv 1+4-3^{7 n} \equiv 5-3^{n} \not \equiv 0(\bmod 8),
\end{array}
$$
Contradiction.
If $k=3$, let $A=m\left(m^{2}-1\right)(m \in \mathbf{Z})$.
But $A \equiv 2^{2012 n}-2^{4 n}+2 \times(-1)^{7 n}$
$$
\equiv 2 \times(-1)^{n} \equiv \pm 2(\bmod 5),
$$
This contradicts $m\left(m^{2}-1\right) \equiv 0$ or $\pm 1(\bmod 5)$.
If $k=2$, let $A=m(m+1)$. Then
$$
(2 m+1)^{2}=4 A+1 \text {. }
$$
But when $n \geqslant 1$, we have
$$
\left(2 \times 17^{100 \sigma_{n}}\right)^{2}<4 A+1<\left(2 \times 17^{100 \sigma_{n}}+1\right)^{2} \text {, }
$$
Contradiction.
Therefore, $n=0$. In this case, $A=12=3 \times 4$ satisfies the condition.
Thus, $(n, k)=(0,2)$. Hence $n+k=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A photographer took some photos of eight people at a party, with any two people (there are 28 possible combinations) appearing in exactly one photo. Each photo can be a duo or a trio. How many photos did the photographer take at least?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
6. 12 .
Let the number of group photos with three people be $x$, and the number of group photos with two people be $y$. Then $3 x+y=28$.
Thus, when $x$ is maximized, the total number of photos taken is minimized.
When $x>8$, by $\frac{3 x}{8}>3$ we know that there is a person $A$ who appears 4 times, and in the 4 group photos where $A$ appears, there are a total of eight appearances by other people, so there must be a person $B$ who appears with $A$ in two group photos, which is a contradiction.
When $x=8$, $y=4$, it can be done.
Number the eight people as $0,1, \cdots, 7$, the 12 photos taken are
$$
\{012,034,056,713,745,726,146,235 \text {, }
$$
$07,15,24,36\}$.
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is
|
10.9.
Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
n_{0}+1=[2,3, \cdots, k],
$$
where $[2,3, \cdots, k]$ represents the least common multiple of $2,3, \cdots, k$.
$$
\begin{array}{l}
\text { Since }[2,3, \cdots, 8]=840, \\
{[2,3, \cdots, 9]=2520,} \\
{[2,3, \cdots, 10]=2520,} \\
{[2,3, \cdots, 11]=27720,}
\end{array}
$$
Therefore, the smallest positive integer $k$ that satisfies $2000<n_{0}<3000$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. B. Color the five sides and five diagonals of the convex pentagon $A B C D E$, such that any two segments sharing a common vertex are of different colors. Find the minimum number of colors needed.
|
14. B. Since vertex $A$ is the common point of four segments $A B$, $A C$, $A D$, and $A E$, at least four colors are needed.
If only four colors are used, let's assume they are red, yellow, blue, and green. Then, the four segments extending from each vertex must include one each of red, yellow, blue, and green. Therefore, the number of red segments would be $\frac{5}{2}$, which is a contradiction. Thus, at least five colors are needed.
The following example shows that five colors can be used to color the ten segments to meet the conditions: color $A B$ and $C E$ with color 1; color $B C$ and $D A$ with color 2; color $C D$ and $E B$ with color 3; color $D E$ and $A C$ with color 4; color $E A$ and $B D$ with color 5. In this way, any two segments sharing a common vertex will have different colors.
In conclusion, the minimum number of colors required is 5.
(Provided by Ben Guo Min)
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If the equation about $x$
$$
x^{3}+a x^{2}+b x-4=0\left(a 、 b \in \mathbf{N}_{+}\right)
$$
has a positive integer solution, then $|a-b|=$
|
7.1.
Solution 1 Let $m$ be a positive integer solution of the equation.
If $m \geqslant 2$, then $a m^{2}+b m=4-m^{3}<0$, which contradicts that $a$ and $b$ are both positive integers.
Therefore, only $m=1$. Substituting it in, we get $a+b=3$.
Since $a, b \in \mathbf{N}_{+}$, $\{a, b\}=\{1,2\}$.
Thus, $|a-b|=1$.
Solution 2 It is easy to see that the positive integer solution of the equation must be a divisor of 4. The divisors of 4 are $1,2,4$. Substituting them into the original equation, we get
$$
\begin{array}{l}
a+b-3=0, \\
4 a+2 b+4=0, \\
16 a+4 b+60=0 .
\end{array}
$$
It is easy to see that the last two equations have no positive integer solutions.
Therefore, $a+b=3$.
The rest is the same as Solution 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $|x| \leqslant 1,|y| \leqslant 1$, and
$$
k=|x+y|+|y+1|+|2 y-x-4| \text {. }
$$
Then the sum of the maximum and minimum values of $k$ is $\qquad$
|
2. 10 .
From the given conditions, we know
$$
-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1 \text{. }
$$
Thus, $y+1 \geqslant 0, 2y-x-40$ when,
$$
k=x+y+y+1-(2y-x-4)=2x+5 \text{. }
$$
Therefore, $3 \leqslant k \leqslant 7$.
Hence, the maximum value of $k$ is 7, and the minimum value is 3.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $x, y, z$ satisfy $x+y=5$ and $z^{2}=x y+y-9$.
Then $x+2 y+3 z=$ $\qquad$ .
|
Solve: From $x+y=5$, we can set $x=\frac{5}{2}+t, y=\frac{5}{2}-t$.
Substitute into $z^{2}=x y+y-9$ and simplify to get $z^{2}+\left(t+\frac{1}{2}\right)^{2}=0$.
Thus, $z=0, t=-\frac{1}{2}$.
Therefore, $x=2, y=3$.
Hence, $x+2 y+3 z=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Let real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value?
|
Solve: From $x+y=z-1$, we can set
$$
x=\frac{z-1}{2}+t, y=\frac{z-1}{2}-t \text {. }
$$
Substituting into $x y=z^{2}-7 z+14$ and simplifying, we get
$$
3 z^{2}-26 z-55=-4 t^{2} \text {. }
$$
Therefore, $3 z^{2}-26 z-55 \leqslant 0$.
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\text { From } x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(z-1)^{2}-2\left(z^{2}-7 z+14\right) \\
=-(z-6)^{2}+9,
\end{array}
$$
Thus, when $z=5$, the maximum value of $x^{2}+y^{2}$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given an equilateral $\triangle A B C$ with side length $2, P$ is a point inside $\triangle A B C$, and the distances from point $P$ to the three sides $B C, A C, A B$ are $x, y, z$ respectively, and their product is $\frac{\sqrt{3}}{9}$. Then the sum of the squares of $x, y, z$ is
|
2. 1 .
As shown in Figure 4, connect
$$
P A, P B, P C \text{. }
$$
It is easy to see that, in the equilateral $\triangle A B C$, we have
$$
\begin{array}{l}
x+y+z \\
=\frac{\sqrt{3}}{2} A B=\sqrt{3} .
\end{array}
$$
Thus, $\sqrt{3}=x+y+z$
$$
\geqslant 3 \sqrt[3]{x y z}=3 \sqrt[3]{\frac{\sqrt{3}}{9}}=\sqrt{3} \text{. }
$$
Since the equality holds, we have
$$
x=y=z=\frac{\sqrt{3}}{3} \text{. }
$$
Therefore, $x^{2}+y^{2}+z^{2}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the quadratic function
$$
y=3 a x^{2}+2 b x-(a+b) \text {, }
$$
when $x=0$ and $x=1$, the value of $y$ is positive. Then, when $0<x<1$, the parabola intersects the $x$-axis at $\qquad$ points.
|
3. 2 .
From the given, we have $\left\{\begin{array}{l}-(a+b)>0, \\ 3 a+2 b-(a+b)>0 .\end{array}\right.$
Solving, we get $a>0$.
Let $x=\frac{1}{2}$, then
$$
y=\frac{3 a}{4}+b-(a+b)=-\frac{a}{4}<0 .
$$
By the Intermediate Value Theorem, $3 a x^{2}+2 b x-(a+b)=0$ has one root when $0<x<\frac{1}{2}$, and one root when $\frac{1}{2}<x<1$. Therefore, there are two intersection points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x, y, z \in \mathbf{R}_{+}$. Then the minimum value of $\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}$ is $\qquad$ .
|
3.2.
Let $x=y=1, z=\sqrt[3]{2}$, we get
$$
\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}=2 \text {. }
$$
Notice that
$$
\begin{array}{l}
\left(x^{2}+y^{2}\right)^{3}+z^{6} \\
=\left(x^{6}+y^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}\right)+z^{6} \\
=\left(2 x^{4} y^{2}+2 x^{2} y^{4}\right)+\left(x^{6}+y^{6}+x^{4} y^{2}+x^{2} y^{4}\right)+z^{6} \\
\geqslant 4 x^{3} y^{3}+\left(x^{3}+y^{3}\right)^{2}+\left(z^{3}\right)^{2} \\
\geqslant 4 x^{3} y^{3}+2\left(x^{3}+y^{3}\right) z^{3} \\
=4 x^{3} y^{3}+2 x^{3} z^{3}+2 z^{3} y^{3} .
\end{array}
$$
Therefore, the minimum value sought is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Fill $1,2, \cdots, n^{2}$ into an $n \times n$ chessboard, with each cell containing one number, and each row forming an arithmetic sequence with a common difference of 1. If any two of the $n$ numbers on the chessboard are neither in the same row nor in the same column, then the sum of these $n$ numbers is called a “basic sum”. If a basic sum is randomly selected, the probability that this basic sum equals the arithmetic mean of all basic sums is
|
5.1.
The numbers in the $k$-th row of the number table are
$$
(k-1) n+1,(k-1) n+2, \cdots,(k-1) n+n \text {. }
$$
Let the number taken from the $k$-th row be
$$
(k-1) n+a_{k} \text {. }
$$
Since these numbers are from different columns, $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$.
Therefore, the sum of the $n$ numbers taken is
$$
\begin{array}{l}
\sum_{k=1}^{n}\left[(k-1) n+a_{k}\right]=n \sum_{k=1}^{n}(k-1)+\sum_{k=1}^{n} a_{k} \\
=n \frac{n(n-1)}{2}+\frac{n(n+1)}{2}=\frac{n\left(n^{2}+1\right)}{2} .
\end{array}
$$
Thus, any basic sum is equal to the arithmetic mean of all basic sums.
Therefore, the required probability is 1.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 2 of the hopscotch game: a person can only enter the first
square from outside; in the squares, each time they can
jump forward 1 square or 2 squares.
Then, the number of ways a person can jump
from outside to the sixth square is $\qquad$.
|
Solution 1 (Enumeration Method) Transform the problem as follows:
Enter the 6th grid from the 1st grid, walking 5 grids. Represent the number 5 as the sum of several 1s or 2s, where different orders of 1s and 2s represent different methods (e.g., $5=1+1+1+1+1$). For convenience, abbreviate the expression to $5=11111$. Then, decompose 5 in the following ways:
$$
\begin{array}{l}
5=11111=1112=1121=1211 \\
=2111=122=221=212,
\end{array}
$$
There are 8 methods in total.
Solution 2 (Recursive Method) Consider the number of ways to enter the $n$-th grid, denoted as $F_{n}$. It is easy to see that $F_{1}=1, F_{2}=1$ (Note: a person can only enter the first grid from outside the grid). For $n \geqslant 3$, the number of ways to enter the $n$-th grid, $F_{n}$, can be divided into two categories: taking one step from the $(n-1)$-th grid; taking two steps from the $(n-2)$-th grid. Therefore,
$$
F_{n}=F_{n-1}+F_{n-2} \text {, }
$$
This is a Fibonacci sequence. Extracting $F_{6}=8$ gives the answer to this problem.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 The condition for three line segments to form a triangle is: the sum of the lengths of any two line segments is greater than the length of the third line segment. There is a wire of length $144 \mathrm{~cm}$, which is to be cut into $n$ $(n>2)$ small segments, with each segment being no less than $1 \mathrm{~cm}$ in length. If any three of these segments cannot form a triangle, then the maximum value of $n$ is $\qquad$
|
Since the necessary and sufficient condition for forming a triangle is that the sum of any two sides is greater than the third side, the condition for not forming a triangle is that the sum of any two sides is less than or equal to the largest side. The shortest piece of wire is 1, so we can place 2 ones, and the third segment is 2 (to maximize $n$, the remaining wire should be as long as possible, so each segment is always the sum of the previous two segments), in sequence:
$$
1,1,2,3,5,8,13,21,34,55 \text {, }
$$
The sum of the above numbers is 143, which is 1 less than 144. Therefore, the last segment can be 56, at which point $n$ reaches its maximum of 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A certain linear function graph is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through the point $(-1,-25)$. Then on the line segment $AB$ (including $A$ and $B$), the number of points with both integer coordinates is $\qquad$ .
|
3.5.
Let the linear function be $y=\frac{5}{4} x+b$.
Given that it passes through the point $(-1,-25)$, we get $b=-\frac{95}{4}$.
Therefore, $y=\frac{5}{4} x-\frac{95}{4}$.
Thus, $A(19,0)$ and $B\left(0,-\frac{95}{4}\right)$.
From $y=\frac{5}{4} x-\frac{95}{4}(0 \leqslant x \leqslant 19)$, taking $x=3,7,11$,
15,19, $y$ is an integer, so there are 5 points that satisfy the condition.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the function $f(x)$ satisfies for all real numbers $x, y$,
$$
\begin{array}{l}
f(x)+f(2 x+y)=f(3 x-y)+x-2010 \text {. } \\
\text { Then } f(2010)=
\end{array}
$$
|
4.0.
Let $x=2010, y=1005$, then
$$
\begin{array}{l}
f(2010)+f(5025) \\
=f(5025)+2010-2010 .
\end{array}
$$
Therefore, $f(2010)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$
Then the product of the maximum and minimum values of $t$ is $\qquad$
|
8. 1.
Let $a=x+y, b=x-y$. Then
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$
Simplifying, we get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-x^{2}-3 y^{2}=8 x^{2}-3 .
\end{array}
$$
Therefore, $-3 \leqslant t \leqslant-\frac{1}{3}$.
Hence, $-3 \times\left(-\frac{1}{3}\right)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (12 points) As shown in Figure 3, in $\triangle ABC$, it is given that $AB=9, BC=8, AC=7$, and $AD$ is the angle bisector. A circle is drawn with $AD$ as a chord, tangent to $BC$, and intersecting $AB$ and $AC$ at points $M$ and $N$, respectively. Find the length of $MN$.
|
II. As shown in Figure 3, connect DM. Given
$\angle B D M=\angle B A D=\angle C A D=\angle D M N$
$\Rightarrow M N / / B C \Rightarrow \triangle A M N \backsim \triangle A B C$.
It is easy to know that $B D=\frac{9}{2}$.
Then $B M \cdot B A=B D^{2} \Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2}$
$\Rightarrow B M=\frac{9}{4} \Rightarrow A M=\frac{27}{4}$.
Also, $\frac{M N}{B C}=\frac{A M}{A B}=\frac{\frac{27}{4}}{9}=\frac{3}{4}$, thus $M N=6$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a(a \neq 0$, and $a \neq 1)$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\frac{a}{1-a}\left(1-a_{n}\right)$. Let $b_{n}=a_{n} \lg \left|a_{n}\right|\left(n \in \mathbf{N}_{+}\right)$. When $a=-\frac{\sqrt{7}}{3}$, does there exist a positive integer $m$ such that for any positive integer $n$, $b_{n} \geqslant b_{m}$? If it exists, find the value of $m$; if not, explain the reason.
|
2. When $n \geqslant 2$,
$$
S_{n}=\frac{a}{1-a}\left(1-a_{n}\right), S_{n-1}=\frac{a}{1-a}\left(1-a_{n-1}\right) .
$$
Then $a_{n}=S_{n}-S_{n-1}$
$$
\begin{array}{l}
=\frac{a}{1-a}\left[\left(1-a_{n}\right)-\left(1-a_{n-1}\right)\right] \\
=\frac{a}{1-a}\left(a_{n-1}-a_{n}\right),
\end{array}
$$
i.e., $a_{n}=a a_{n-1}$.
Since $a_{1}=a \neq 0$, thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term and common ratio both being $a$.
Therefore, $a_{n}=a^{n}, b_{n}=a_{n} \lg \left|a_{n}\right|=n a^{n} \lg |a|$.
Also, $a=-\frac{\sqrt{7}}{3} \in(-1,0)$, then $\lg |a|<0$.
It can be seen that if there exists a positive integer $m$ that satisfies the condition, then $m$ must be even.
Notice that
$$
\begin{array}{l}
b_{2 k+2}-b_{2 k}=\left[(2 k+2) a^{2 k+2}-2 k a^{2 k}\right] \lg |a| \\
=2 a^{2 k}\left[(k+1) a^{2}-k\right] \lg |a| \\
=2 a^{2 k}\left[(k+1) \times \frac{7}{9}-k\right] \lg |a| \\
=\frac{2 a^{2 k}}{9}(7-2 k) \lg |a|\left(k \in \mathbf{N}_{+}\right) . \\
\text {Also, } \frac{a^{2}}{1-a^{2}}=\frac{7}{2}, \text { thus, }
\end{array}
$$
When $2 k>7$, $b_{2 k+2}>b_{2 k}$, i.e.,
$$
b_{8}<b_{10}<b_{12}<\cdots \text {; }
$$
When $2 k<7$, $b_{2 k+2}<b_{2 k}$, i.e.,
$$
b_{8}<b_{6}<b_{4}<b_{2} \text {. }
$$
Therefore, there exists a positive integer $m=8$, such that for any positive integer $n$, $b_{n} \geqslant b_{m}$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.1. Distribute 24 pencils of four colors (6 pencils of each color) to 6 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$.
|
9.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
In fact, there are 6 pencils of each color, and each student has 4 pencils, so there exists a student who has at least two colors of pencils. Clearly, any other color must be owned by at least one student.
Thus, the conclusion holds.
Next, we provide an example to show: There exists a distribution method such that any two students have at most three colors of pencils.
One student has 4 pencils of the second color, one student has 4 pencils of the third color, one student has 4 pencils of the fourth color, one student has 2 pencils of the first color and 2 pencils of the second color, one student has 2 pencils of the first color and 2 pencils of the third color, one student has 2 pencils of the first color and 2 pencils of the fourth color.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
A4. Find the smallest positive integer $v$, such that for any two-coloring of the edges of the complete graph $K_{v}$, there are two monochromatic triangles that share exactly one vertex.
|
The solution to A4 is $v=9$.
On one hand, take two red complete graphs $K_{4}$, and color all the edges between them blue. The resulting complete graph $K_{8}$ has no two monochromatic triangles with exactly one common vertex.
On the other hand, consider any two-coloring of the complete graph $K_{9}$. By problem $\mathrm{B}$, there is a monochromatic triangle (let $\triangle x_{1} x_{2} x_{3}$ be red). The complete graph $K_{6}$ formed by the other six points, if all edges are red, the conclusion is evident.
Assume the conclusion does not hold. Suppose there is an edge $v_{4} v_{5}$ that is blue. By the same argument as in problem $\mathrm{B}$ (Figure 1), we can assume there is another red $\triangle x_{1} x_{2} x_{4}$.
The complete graph $K_{5}$ formed by the other five points $x_{5} 、 x_{6} 、 x_{7} 、 x_{8} 、 x_{9}$, if all edges are red, the conclusion is evident.
Suppose the edge $x_{8} x_{9}$ is blue (as in Figure 5).
If the conclusion does not hold, $x_{4} x_{8} 、 x_{4} x_{9}$ are not both blue.
Assume $x_{4} x_{8}$ is red (Figure 6). Then $x_{1} x_{8} 、 x_{2} x_{8}$ are blue, $x_{9} x_{1} 、 x_{9} x_{2}$ are red, $x_{9} x_{4} 、 x_{9} x_{3}$ are blue, $x_{3} x_{4}$ is red, $\triangle x_{2} x_{3} x_{4}$ and $\triangle x_{2} x_{1} x_{9}$ are two red triangles with exactly one common vertex.
Thus, the conclusion holds.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A set of $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane, no three of which are collinear, is denoted as $D$. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side" of the triangle; if all three sides of a triangle are middle sides (not necessarily the middle sides of this triangle), then the triangle is called a "middle side triangle" in the set $D$. A line $l$ that does not pass through any point $P_{i}(i=1,2, \cdots, n)$ divides the set $D$ into two subsets $D_{1}$ and $D_{2}$. If, regardless of how these $n$ points are distributed and how $l$ is chosen, there always exists a subset $D_{k}(k \in\{1,2\})$ such that $D_{k}$ contains a middle side triangle. Find the minimum value of $n$.
|
4. The minimum value of $n$ is 11.
When $n \geqslant 11$, regardless of how $l$ is chosen, there always exists a subset, let's assume it is $D_{1}$, such that $D_{1}$ contains at least six points.
Consider all the middle edges of the triangles in $D_{1}$, and color them red, then color the other edges blue.
By Ramsey's theorem, there must exist a monochromatic triangle. Since every triangle has a middle edge, this monochromatic triangle must be red. Therefore, there exists a middle edge triangle in $D_{1}$.
Below is the proof: When $n \leqslant 10$, there exists a point set $D$ and a line $l$, such that neither $D_{1}$ nor $D_{2}$ contains a middle edge triangle.
If $n=10$, consider the two subsets $D_{1}$ and $D_{2}$ on either side of the line $l$, each containing five points, and the distribution is the same.
Assume the five points in $D_{1}$ are $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$, and they are arranged in a counterclockwise order on the circumference of a circle, with
\[
\begin{array}{l}
\overparen{P_{1} P_{2}}=\frac{\pi}{10}, \overparen{P_{2} P_{3}}=\frac{3 \pi}{5}, \overparen{P_{3} P_{4}}=\frac{3 \pi}{10}, \\
\overparen{P_{4} P_{5}}=\frac{3 \pi}{4}, \overparen{P_{5} P_{1}}=\frac{\pi}{4} .
\end{array}
\]
Then the distances between $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are all different, and $P_{2} P_{3}, P_{3} P_{1}, P_{1} P_{5}, P_{5} P_{4}, P_{4} P_{2}$ are middle edges, but there is no middle edge triangle.
If $n<10$, for the cases where $D_{1}$ and $D_{2}$ have fewer than five points, simply remove some points from the previous example, and there will still be no middle edge triangle.
In summary, the minimum value of $n$ is 11.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. 1. Distribute 40 pencils of four colors (10 pencils of each color) to 10 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$.
|
10.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
Since there are 10 pencils of each color, and each student has 4 pencils, there must be a student who has at least two different colors of pencils.
Obviously, any other color of pencils must be owned by at least one student. Therefore, the conclusion holds.
Next, we provide an example to show that there exists a distribution method such that any two students have at most three colors of pencils.
Two students have 4 pencils of the second color, two students have 4 pencils of the third color, two students have 4 pencils of the fourth color, one student has 4 pencils of the first color, one student has 2 pencils of the first color and 2 pencils of the second color, one student has 2 pencils of the first color and 2 pencils of the third color, and one student has 2 pencils of the first color and 2 pencils of the fourth color.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In two boxes, Jia and Yi, each contains the same number of pieces of jewelry. After transferring a piece of jewelry worth 50,000 yuan from Jia to Yi, the average value of the jewelry in Jia decreases by 10,000 yuan, and the average value of the jewelry in Yi increases by 10,000 yuan. Then the total value of the jewelry in both boxes is ( ) ten thousand yuan.
(A) 11
(B) 12
(C) 13
(D) 14
|
4. B.
Let the original number of jewels in box A be $n(n>1)$, with a total value of $x$ million yuan, and the total value in box B be $y$ million yuan. Then $\left\{\begin{array}{l}\frac{x-5}{n-1}+1=\frac{x}{n}, \\ \frac{y+5}{n+1}-1=\frac{y}{n}\end{array} \Rightarrow\left\{\begin{array}{l}x=-n^{2}+6 n, \\ y=-n^{2}+4 n .\end{array}\right.\right.$
Since $x>0, y>0$, we have $0<n<4$. Therefore, $n=2$ or 3.
Thus, $x+y=-2 n^{2}+10 n=12$.
|
12
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. Set $A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}y=\sqrt{1-x}, \\ y=1-x^{2}\end{array}\right\}\right.\right.$ has the number of subsets as $\qquad$
|
3. 8 .
Notice that the number of elements in set $A$ is the number of solutions to the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2}
\end{array}\right.
$$
Substituting equation (1) into equation (2) and simplifying, we get
$$
x(x-1)\left(x^{2}+x-1\right)=0 \text {. }
$$
Solving this, we get $x_{1}=0, x_{2}=1$,
$$
x_{3}=\frac{-1+\sqrt{5}}{2}, x_{4}=\frac{-1-\sqrt{5}}{2} \text {. }
$$
Noting that $\left|x_{4}\right|>1$, we discard it.
Therefore, the system of equations has three solutions
$$
(0,1),(1,0),\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right) \text {. }
$$
Since set $A$ has 3 elements, set $A$ has $2^{3}=8$ subsets.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\text { Three. (50 points) Given the sequence }\left\{a_{n}\right\}: 1,3,5,7, \cdots \text {, }
$$
starting from the 5th term, $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$. Find:
$$
a_{2008}^{2}+a_{2009}^{2}+a_{2010}^{2}+a_{2011}^{2}
$$
Can it be divisible by 4?
|
Consider the sequence $\left\{a_{n}\right\}$ and its modulo 2 residue sequence $\left\{b_{n}\right\}$.
Given that $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$, we have $b_{n+4} \equiv b_{n}+b_{n+3}(\bmod 2)$.
By recursion, we get
$$
\begin{array}{l}
b_{n+10} \equiv b_{n+6}+b_{n+9} \\
\equiv\left(b_{n+2}+b_{n+5}\right)+\left(b_{n+5}+b_{n+8}\right) \\
\equiv b_{n+2}+b_{n+8} \\
\equiv b_{n+2}+\left(b_{n+4}+b_{n+7}\right) \\
\equiv b_{n+2}+\left(b_{n}+b_{n+3}\right)+\left(b_{n+3}+b_{n+6}\right) \\
\equiv b_{n+2}+b_{n}+b_{n+6} \\
\equiv b_{n+2}+b_{n}+\left(b_{n+2}+b_{n+5}\right) \\
\equiv b_{n}+b_{n+5}(\bmod 2) .
\end{array}
$$
Then, $b_{n+15} \equiv b_{(n+5)+10} \equiv b_{n+5}+b_{n+10}$
$$
\begin{array}{l}
\equiv b_{n+5}+\left(b_{n}+b_{n+5}\right) \\
\equiv b_{n}(\bmod 2) .
\end{array}
$$
It is evident that $\left\{b_{n}\right\}$ is a periodic sequence with a period of 15, hence
$$
\begin{array}{l}
a_{2008} \equiv b_{2008} \equiv b_{15 \times 133+13} \equiv b_{13} \equiv 0(\bmod 2), \\
a_{2009} \equiv b_{2009} \equiv b_{15 \times 133+14} \equiv b_{14} \equiv 0(\bmod 2), \\
a_{2010} \equiv b_{2010} \equiv b_{15 \times 134} \equiv b_{15} \equiv 0(\bmod 2), \\
a_{2011} \equiv b_{2011} \equiv b_{15 \times 134+1} \equiv b_{1} \equiv 1(\bmod 2) .
\end{array}
$$
Since the square of an even number is divisible by 4, and the square of an odd number is congruent to 1 modulo 4, we have
$$
S=a_{2008}^{2}+a_{2008}^{2}+a_{2010}^{2}+a_{2011}^{2} \equiv 1(\bmod 4) \text {. }
$$
Therefore, $S$ is not divisible by 4.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
279 Given positive real numbers $x, y, z$ satisfying $(\sqrt{3}+1) x y + 2 \sqrt{3} y z + (\sqrt{3}+1) z x = 1$.
(1) Find the minimum value of $x + y + z$;
(2) Find the minimum value of $\frac{\sqrt{3} x y}{z} + \frac{(8-4 \sqrt{3}) y z}{x} + \frac{\sqrt{3} z x}{y}$.
|
(1) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{(\sqrt{3}+1) x^{2}}{2}+(\sqrt{3}-1) y^{2} \geqslant 2 x y, \\
2 y^{2}+2 z^{2} \geqslant 4 y z, \\
(\sqrt{3}-1) z^{2}+\frac{(\sqrt{3}+1) x^{2}}{2} \geqslant 2 x z .
\end{array}
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
(\sqrt{3}+1)\left(x^{2}+y^{2}+z^{2}\right) \geqslant 2(x y+2 y z+z x) \\
\Rightarrow x^{2}+y^{2}+z^{2} \geqslant(\sqrt{3}-1)(x y+2 y z+z x) .
\end{array}
$$
Equality holds if and only if $\frac{(\sqrt{3}+1) x^{2}}{2}=(\sqrt{3}-1) y^{2}, 2 y^{2}=2 z^{2},(\sqrt{3}-1) z^{2}=\frac{(\sqrt{3}+1) x^{2}}{2}$, i.e., $y=z=\frac{(\sqrt{3}+1) x}{2}$.
$$
\begin{array}{l}
\text { Then }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \\
\geqslant(\sqrt{3}-1)(x y+2 y z+z x)+2(x y+y z+z x) \\
\geqslant(\sqrt{3}+1) x y+2 \sqrt{3} y z+(\sqrt{3}+1) z x=1 .
\end{array}
$$
$$
\text { Thus, } x+y+z \geqslant 1 \text {. }
$$
Equality holds if and only if $y=z=\frac{(\sqrt{3}+1) x}{2}$ and $(\sqrt{3}+1) x y+2 \sqrt{3} y z+(\sqrt{3}+1) z x=1$, i.e., $y=z=\frac{\sqrt{3}-1}{2}$, $x=2-\sqrt{3}$.
Therefore, when $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$, $x+y+z$ attains its minimum value of 1.
(2) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{y}{z}+\frac{z}{y} \geqslant 2, \\
\frac{z}{x}+\frac{(2+\sqrt{3}) x}{2 z} \geqslant \sqrt{3}+1, \\
\frac{y}{x}+\frac{(2+\sqrt{3}) x}{2 y} \geqslant \sqrt{3}+1 .
\end{array}
$$
Equality holds if and only if $y=z=\frac{(\sqrt{3}+1) x}{2}$.
Introducing positive parameters $u, v, w$, we have
$$
\begin{array}{l}
u x\left(\frac{y}{z}+\frac{z}{y}\right)+v y\left[\frac{(2+\sqrt{3}) x}{2 z}+\frac{z}{x}\right]+ \\
w z\left[\frac{(2+\sqrt{3}) x}{2 y}+\frac{y}{x}\right] \\
=\left[u+\frac{(2+\sqrt{3}) v}{2}\right] \frac{x y}{z}+(v+w) \frac{y z}{x}+ \\
{\left[u+\frac{(2+\sqrt{3}) w}{2}\right] \frac{z x}{y} .} \\
\text { Also, } u x\left(\frac{y}{z}+\frac{z}{y}\right)+v y\left[\frac{(2+\sqrt{3}) x}{2 z}+\frac{z}{x}\right]+ \\
w z\left[\frac{(2+\sqrt{3}) x}{2 y}+\frac{y}{x}\right] \\
\geqslant 2 u x+(\sqrt{3}+1) v y+(\sqrt{3}+1) w z \text {, } \\
\text { Hence, }\left[u+\frac{(2+\sqrt{3}) v}{2}\right] \frac{x y}{z}+(v+w) \frac{y z}{x}+ \\
{\left[u+\frac{(2+\sqrt{3}) w}{2}\right] \frac{z x}{y}} \\
\geqslant 2 u x+(\sqrt{3}+1) v y+(\sqrt{3}+1) w z . \\
\text { Let } u=\sqrt{3}+1, v=w=2 \text {. Substituting into the above inequality, we get } \\
(3+2 \sqrt{3}) \frac{x y}{z}+4 \times \frac{y z}{x}+(3+2 \sqrt{3}) \frac{z x}{y} \\
\geqslant 2(\sqrt{3}+1) x+2(\sqrt{3}+1) y+2(\sqrt{3}+1) z \text {, } \\
\text { Then } \frac{\sqrt{3} x y}{z}+\frac{(8-4 \sqrt{3}) y z}{x}+\frac{\sqrt{3} z x}{y} \\
\geqslant 2(\sqrt{3}-1)(x+y+z) \geqslant 2(\sqrt{3}-1) \text {. } \\
\end{array}
$$
Equality holds if and only if $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$.
Therefore, when $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$, $\frac{\sqrt{3} x y}{z}+$ $\frac{(8-4 \sqrt{3}) y z}{x}+\frac{\sqrt{3} z x}{y}$ attains its minimum value of $2(\sqrt{3}-1)$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$
(Fourth Jiangsu Province Junior High School Mathematics Competition)
|
Given that $m$ and $n$ are the roots of the equation $x^{2}=x+1$, i.e., $x^{2}-x-1=0$.
Let $a_{k}=m^{k}+n^{k}$. Then
$$
a_{k}=a_{k-1}+a_{k-2}(k \geqslant 3) \text {. }
$$
By $a_{1}=m+n=1$,
$$
a_{2}=(m+n)^{2}-2 m n=3 \text {, }
$$
we know $a_{3}=a_{2}+a_{1}=4, a_{4}=a_{3}+a_{2}=7$.
Therefore, $m^{5}+n^{5}=a_{5}=a_{4}+a_{3}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $m$ and $n$ are rational numbers, and the equation
$$
x^{2}+m x+n=0
$$
has a root $\sqrt{5}-2$. Then $m+n=$ $\qquad$ .
(2001, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
$$
\begin{array}{l}
(\sqrt{5}-2)^{2}+m(\sqrt{5}-2)+n=0 \\
\Rightarrow(n-2 m+9)=(4-m) \sqrt{5} \\
\Rightarrow\left\{\begin{array} { l }
{ n - 2 m + 9 = 0 , } \\
{ 4 - m = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=4, \\
n=-1
\end{array}\right.\right. \\
\Rightarrow m+n=3 \text {. } \\
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Let positive integers $a, b, c (a \geqslant b \geqslant c)$ be the lengths of the sides of a triangle, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-a c-b c=13 \text {. }
$$
Find the number of triangles that meet the conditions and have a perimeter not exceeding 30.
|
Given the known equation:
$$
(a-b)^{2}+(b-c)^{2}+(a-c)^{2}=26 \text {. }
$$
Let $a-b=m, b-c=n$.
Then $a-c=m+n(m, n$ are natural numbers $)$.
Thus, equation (1) becomes
$$
m^{2}+n^{2}+m n=13 \text {. }
$$
Therefore, the pairs $(m, n)$ that satisfy equation (2) are:
$$
(m, n)=(3,1),(1,3) \text {. }
$$
(1) When $(m, n)=(3,1)$,
$$
b=c+1, a=b+3=c+4 \text {. }
$$
Since $a, b, c$ are the lengths of the sides of a triangle,
$$
b+c>a \Rightarrow(c+1)+c>c+4 \Rightarrow c>3 \text {. }
$$
Also, the perimeter of the triangle does not exceed 30, i.e.,
$$
\begin{array}{l}
a+b+c=(c+4)+(c+1)+c \leqslant 30 \\
\Rightarrow c \leqslant \frac{25}{3} .
\end{array}
$$
Thus, $3<c \leqslant \frac{25}{3}$.
Therefore, $c$ can take the values $4, 5, 6, 7, 8$, corresponding to 5 triangles that meet the conditions.
(2) When $(m, n)=(1,3)$, similarly,
$1<c \leqslant \frac{23}{3}$.
Thus, $c$ can take the values $2, 3, 4, 5, 6, 7$, corresponding to 6 triangles that meet the conditions.
In summary, the number of triangles that meet the conditions and have a perimeter not exceeding 30 is 11.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given the quadratic function $y=x^{2}+b x-c$ whose graph passes through two points $P(1, a)$ and $Q(2,10 a)$.
(1) If $a$, $b$, and $c$ are all integers, and $c<b<8 a$, find the values of $a$, $b$, and $c$;
(2) Let the graph of the quadratic function $y=x^{2}+b x-c$ intersect the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If the roots of the equation $x^{2}+b x-c=0$ are both integers, find the area of $\triangle A B C$.
|
Three, given that points $P(1, a)$ and $Q(2, 10a)$ are on the graph of the quadratic function $y = x^2 + bx - c$, we have
$$
1 + b - c = a, \quad 4 + 2b - c = 10a.
$$
Solving these equations, we get $b = 9a - 3$ and $c = 8a - 2$.
(1) From $c < b < 8a$, we know
$$
8a - 2 < 9a - 3 < 8a \Rightarrow 1 < a < 3.
$$
Since $a$ is an integer, we have
$$
a = 2, \quad b = 9a - 3 = 15, \quad c = 8a - 2 = 14.
$$
(2) Let $m$ and $n$ be the two integer roots of the equation, with $m \leq n$. By the relationship between roots and coefficients, we have
$$
m + n = -b = 3 - 9a, \quad mn = -c = 2 - 8a.
$$
Eliminating $a$, we get $9mn - 8(m + n) = -6$.
Multiplying both sides by 9 and factoring, we get
$$
(9m - 8)(9n - 8) = 10.
$$
Thus, $(9m - 8, 9n - 8)$
$$
= (1, 10), (2, 5), (-10, -1), (-5, -2).
$$
Solving these, we get $(m, n)$
$$
= (1, 2), \left(\frac{10}{9}, \frac{13}{9}\right), \left(-\frac{2}{9}, \frac{7}{9}\right), \left(\frac{1}{3}, \frac{2}{3}\right).
$$
Since $m$ and $n$ are integers, we have $(m, n) = (1, 2)$.
Therefore, $b = -(m + n) = -3$,
$$
c = -mn = -2.
$$
Thus, the quadratic function is
$$
y = x^2 - 3x + 2.
$$
It is easy to find that $A(1, 0)$, $B(2, 0)$, and $C(0, 2)$.
So, $S_{\triangle ABC} = \frac{1}{2} \times (2 - 1) \times 2 = 1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $p$ be a prime number greater than 2, and $k$ be a positive integer. If the graph of the function $y=x^{2}+p x+(k+1) p-4$ intersects the $x$-axis at two points, at least one of which has an integer coordinate, find the value of $k$.
---
The function is given by:
\[ y = x^2 + px + (k+1)p - 4 \]
To find the points where the graph intersects the $x$-axis, we set $y = 0$:
\[ x^2 + px + (k+1)p - 4 = 0 \]
This is a quadratic equation in the form:
\[ x^2 + px + (k+1)p - 4 = 0 \]
The roots of this quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = p \), and \( c = (k+1)p - 4 \).
Substituting these values into the quadratic formula, we get:
\[ x = \frac{-p \pm \sqrt{p^2 - 4 \cdot 1 \cdot ((k+1)p - 4)}}{2 \cdot 1} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4((k+1)p - 4)}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4(k+1)p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4kp - 4p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4kp - 4p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4p(k + 1) + 16}}{2} \]
For the quadratic equation to have at least one integer root, the discriminant must be a perfect square. Let's denote the discriminant by \( D \):
\[ D = p^2 - 4p(k + 1) + 16 \]
We need \( D \) to be a perfect square. Let \( D = m^2 \) for some integer \( m \):
\[ p^2 - 4p(k + 1) + 16 = m^2 \]
Rearranging the equation, we get:
\[ p^2 - 4p(k + 1) + 16 - m^2 = 0 \]
This is a quadratic equation in \( p \):
\[ p^2 - 4p(k + 1) + (16 - m^2) = 0 \]
For \( p \) to be a prime number greater than 2, the discriminant of this quadratic equation must be a perfect square. The discriminant of this quadratic equation is:
\[ \Delta = (4(k + 1))^2 - 4 \cdot 1 \cdot (16 - m^2) \]
\[ \Delta = 16(k + 1)^2 - 4(16 - m^2) \]
\[ \Delta = 16(k + 1)^2 - 64 + 4m^2 \]
\[ \Delta = 16(k + 1)^2 + 4m^2 - 64 \]
For \( \Delta \) to be a perfect square, we need:
\[ 16(k + 1)^2 + 4m^2 - 64 = n^2 \]
for some integer \( n \).
Simplifying, we get:
\[ 4(4(k + 1)^2 + m^2 - 16) = n^2 \]
\[ 4(k + 1)^2 + m^2 - 16 = \left(\frac{n}{2}\right)^2 \]
Let \( \frac{n}{2} = t \), then:
\[ 4(k + 1)^2 + m^2 - 16 = t^2 \]
We need to find integer solutions for \( k \) and \( m \) such that the above equation holds. Testing small values of \( k \):
For \( k = 1 \):
\[ 4(1 + 1)^2 + m^2 - 16 = t^2 \]
\[ 4 \cdot 4 + m^2 - 16 = t^2 \]
\[ 16 + m^2 - 16 = t^2 \]
\[ m^2 = t^2 \]
This is true for \( m = t \). Therefore, \( k = 1 \) is a solution.
Thus, the value of \( k \) is:
\[ \boxed{1} \]
|
From the problem, we know that the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
has at least one integer root among its two roots $x_{1}, x_{2}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=(k+1) p-4 . \\
\text { Hence }\left(x_{1}+2\right)\left(x_{2}+2\right) \\
=x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4 \\
=(k-1) p .
\end{array}
$$
(1) If $k=1$, then the equation becomes
$$
x^{2}+p x+2(p-2)=0 \text {, }
$$
which has two integer roots -2 and $2-p$.
(2) If $k>1$, then $k-1>0$.
Since $x_{1}+x_{2}=-p$ is an integer, and at least one of $x_{1}, x_{2}$ is an integer, it follows that $x_{1}, x_{2}$ are both integers.
Also, since $p$ is a prime number, by equation (1), we know that $p \mid (x_{1}+2)$ or $p \mid (x_{2}+2)$.
Without loss of generality, assume $p \mid (x_{1}+2)$. Then we can set $x_{1}+2=m p$ (where $m$ is a non-zero integer).
Thus, from equation (1), we get $x_{2}+2=\frac{k-1}{m}$.
Therefore, $\left(x_{1}+2\right)+\left(x_{2}+2\right)=m p+\frac{k-1}{m}$, which means
$x_{1}+x_{2}+4=m p+\frac{k-1}{m}$.
Since $x_{1}+x_{2}=-p$, we have
$-p+4=m p+\frac{k-1}{m}$
$\Rightarrow (m+1) p+\frac{k-1}{m}=4$.
If $m$ is a positive integer, then
$(m+1) p \geqslant (1+1) \times 3=6, \frac{k-1}{m}>0$.
Thus, $(m+1) p+\frac{k-1}{m}>6$, which contradicts equation (2).
If $m$ is a negative integer, then
$(m+1) p \leqslant 0, \frac{k-1}{m}<0$.
Thus, $(m+1) p+\frac{k-1}{m}<0$, which also contradicts equation (2).
Therefore, when $k>1$, the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
cannot have integer roots.
In summary, $k=1$.
(Provided by Xu Shenglin)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $a, b$ be positive integers, and satisfy $2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{15}{b}}\right)$ is an integer. Then the number of such ordered pairs $(a, b)$ is $\qquad$ pairs.
$(2009$, National Junior High School Mathematics League)
|
First, guess that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
The following is an attempt to prove this.
Let $\frac{15}{a}=A, \frac{15}{b}=B$. According to the problem, we can assume
$\sqrt{A}+\sqrt{B}=C\left(A, B, C \in \mathbf{Q}_{+}\right)$.
Thus, $\sqrt{A}=C-\sqrt{B} \Rightarrow A=(C-\sqrt{B})^{2}$
$\Rightarrow A=C^{2}-2 C \sqrt{B}+B \Rightarrow \sqrt{B}=\frac{C^{2}+B-A}{2 C}$
$\Rightarrow \sqrt{B}$ is a rational number.
Similarly, $\sqrt{A}$ is a rational number.
Therefore, $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
Further conjecture: When $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are expressed in their simplest fractional form, the numerator is 1.
The following is an attempt to prove this.
Let $\sqrt{\frac{15}{a}}=\frac{q}{p}\left(p, q \in \mathbf{Z}_{+}\right.$, and $p$ and $q$ are coprime $)$.
Then $\left.\frac{15}{a}=\frac{q^{2}}{p^{2}} \Rightarrow a q^{2}=15 p^{2} \Rightarrow q^{2} \right\rvert\, 15 p^{2}$.
Since $p$ and $q$ are coprime, $p^{2}$ and $q^{2}$ are coprime. Hence
$q^{2} \left\lvert\, 15 \Rightarrow q^{2}=1 \Rightarrow q=1 \Rightarrow \sqrt{\frac{15}{a}}=\frac{1}{p}\right.$.
Similarly, when $\sqrt{\frac{15}{b}}$ is expressed in its simplest fractional form, the numerator is 1.
From the previous analysis, we can assume
$\sqrt{\frac{15}{a}}=\frac{1}{x}, \sqrt{\frac{15}{b}}=\frac{1}{y}\left(x, y \in \mathbf{Z}_{+}\right)$. Then $a=15 x^{2}, b=15 y^{2}$.
According to the problem, we can assume $\frac{1}{x}+\frac{1}{y}=\frac{k}{2}\left(k \in \mathbf{Z}_{+}\right)$.
Since $x \geqslant 1, y \geqslant 1$, we have
$$
\frac{k}{2}=\frac{1}{x}+\frac{1}{y} \leqslant 1+1=2 \Rightarrow k \leqslant 4 \text {. }
$$
Therefore, $k$ can only be $1,2,3,4$.
(1) When $k=1$,
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$
$\Rightarrow(x-2)(y-2)=4$
$\Rightarrow(x, y)=(4,4)$ or $(3,6)$ or $(6,3)$;
(2) When $k=2$, $(x, y)=(2,2)$;
(3) When $k=3$, $(x, y)=(2,1)$ or $(1,2)$;
(4) When $k=4$, $(x, y)=(1,1)$.
Therefore, there are 7 such ordered pairs $(a, b)$: $(240,240),(135,540),(540,135)$, $(60,60),(60,15),(15,60),(15,15)$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If real numbers $a, b$ satisfy
$$
\left(a-\sqrt{a^{2}+2010}\right)\left(b+\sqrt{b^{2}+2010}\right)+2010=0 \text {, }
$$
then $a \sqrt{b^{2}+2011}-b \sqrt{a^{2}+2011}=$ $\qquad$
|
From the given, we have
$$
\begin{array}{l}
a-\sqrt{a^{2}+2010}=-\frac{2010}{b+\sqrt{b^{2}+2010}} \\
=b-\sqrt{b^{2}+2010} .
\end{array}
$$
Similarly,
$$
b+\sqrt{b^{2}+2010}=a+\sqrt{a^{2}+2010} \text {. }
$$
Subtracting (2) from (1) and simplifying, we get $a=b$.
Therefore, the answer is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $P$ is a point inside the circle $\odot O$ with radius 15, among all the chords passing through point $P$, 24 chords have integer lengths. Then $O P=$ $\qquad$ .
|
2. 12 .
Among the chords of circle $\odot O$ passing through point $P$, the diameter is the longest chord, and there is only one; the shortest chord is the one perpendicular to $OP$, and there is also only one. Therefore, there is only one chord each with lengths of $30$ and $18$, and two chords each with lengths of $29, 28, \cdots, 19$, making a total of 24 chords.
By the perpendicular diameter theorem and the intersecting chords theorem, we have
$$
(15+O P)(15-O P)=\frac{18}{2} \times \frac{18}{2} .
$$
Thus, $O P=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Prove: $\frac{(2+\sqrt{3})^{2010}+(2-\sqrt{3})^{2010}}{2}$ is an integer, and find its remainder when divided by 4.
|
Notice
$$
\begin{array}{l}
(2+\sqrt{3})+(2-\sqrt{3})=4, \\
(2+\sqrt{3})(2-\sqrt{3})=1 .
\end{array}
$$
Therefore, $2+\sqrt{3}$ and $2-\sqrt{3}$ are the two roots of $x^{2}-4 x+1=0$.
Let $a_{n}=\frac{1}{2}(2+\sqrt{3})^{n}+\frac{1}{2}(2-\sqrt{3})^{n}$. Then
$$
a_{n}-4 a_{n-1}+a_{n-2}=0
$$
That is, $a_{n}=4 a_{n-1}-a_{n-2}(n \geqslant 3)$.
Since $a_{1}=2, a_{2}=7$ are both integers, and combining with equation (1), we know that $a_{n}$ is an integer for all positive integers $n$.
Thus, the original expression $=a_{2010}$ is an integer.
From equation (1), we know that the sequence of remainders when $\left\{a_{n}\right\}$ is divided by 4 is $2,3,2,1 ; 2,3,2,1 ; \cdots \cdots$ with a period of 4.
Therefore, its remainder when divided by 4 is 3.
|
3
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. $F$ is the right focus of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $P$ is a moving point on the ellipse. For the fixed point $A(-2, \sqrt{3}),|P A|+$ $2|P F|$ the minimum value is $\qquad$ .
|
4. 10 .
It is known that the eccentricity of the ellipse is $\frac{1}{2}$, and the equation of the right directrix $l$ is $x=8$. The distance from point $A$ to $l$ is 10.
Let the projection of point $P$ onto $l$ be $H$. Then
$\frac{|P F|}{|P H|}=\frac{1}{2}$.
Thus, $|P A|+2|P F|=|P A|+|P H| \geqslant 10$.
The minimum value is achieved when point $P$ lies on the segment $A H$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a$ ] equals $\qquad$
$(2009$, Beijing Mathematical Competition (Grade 8))
|
Notice that
$0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$.
Then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]$ equals
0 or 1.
By the problem statement, 18 of these are equal to 1. Therefore,
$\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0$,
$\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1$.
Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$.
Therefore, $18 \leqslant 30 a<19 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$.
Hence, $[10 a]=6$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let
$$
\begin{aligned}
S & =\frac{1}{\left[\frac{(10 \times 11-1)^{2}}{10 \times 11}\right]}+\frac{1}{\left[\frac{(11 \times 12-1)^{2}}{11 \times 12}\right]}+\cdots+ \\
& \frac{1}{\left[\frac{(49 \times 50-1)^{2}}{49 \times 50}\right]}
\end{aligned}
$$
Then $[30 S]=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 0
(2002, "Five Sheep Cup" Mathematics Competition (Grade 8))
|
When $n \geqslant 10$, and $n$ is an integer,
$$
\begin{array}{l}
{\left[\frac{[n(n+1)-1]^{2}}{n(n+1)}\right]=\left[n(n+1)-2+\frac{1}{n(n+1)}\right]} \\
=n(n+1)-2=(n-1)(n+2) .
\end{array}
$$
Thus, $S=\frac{1}{9 \times 12}+\frac{1}{10 \times 13}+\cdots+\frac{1}{48 \times 51}$.
Notice that $\frac{1}{k(k+3)}=\frac{1}{3}\left(\frac{1}{k}-\frac{1}{k+3}\right)$.
$$
\begin{array}{l}
\text { Then } 3 S=\left(\frac{1}{9}-\frac{1}{12}\right)+\left(\frac{1}{10}-\frac{1}{13}\right)+\cdots+\left(\frac{1}{48}-\frac{1}{51}\right) \\
=\frac{1}{9}+\frac{1}{10}+\frac{1}{11}-\frac{1}{49}-\frac{1}{50}-\frac{1}{51} . \\
\text { Therefore, } \frac{3}{10}-\frac{3}{40}<3 S<\frac{3}{9}-\frac{3}{60}, \\
2<\frac{9}{4}<30 S<\frac{17}{6}<3 .
\end{array}
$$
Therefore, $[30 S]=2$.
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $n$ ($n<100$) be a positive integer, and there exists a positive integer $k$, such that $1 \leqslant k \leqslant n-1$, satisfying
$$
\frac{4 k+1}{2 n}+\frac{1-2 k^{2}}{n^{2}}=\frac{1}{2} \text {. }
$$
How many values of $n$ satisfy the condition? Prove your conclusion.
|
1. From equation (1), we get $(2 k-n)^{2}=n+2$.
Solving, we get $k=\frac{1}{2}(n \pm \sqrt{n+2})$.
Since $k$ is an integer, therefore, $n=m^{2}-2(m \in \mathbf{N})$.
Thus, $k=\frac{1}{2}\left(m^{2} \pm m-2\right)$.
Also, $1 \leqslant k \leqslant n-1$, so
$$
1 \leqslant \frac{m^{2} \pm m-2}{2} \leqslant m^{2}-3 \text {. }
$$
Solving, we get $m \geqslant 3$.
But $n<100$, hence the $n$ that satisfies the problem are $7,14,23,34,47,62,79,98$.
A total of 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-$ $xy+y^{2}$ is $\qquad$
|
2. 1 .
Notice that
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also }|x \pm y| \leqslant|x|+|y| \leqslant 1 \text {, then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ take 0 and 1 respectively, the equality holds.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In trapezoid $A B C D$, $A D / / B C, E F$ is the midline, the area ratio of quadrilateral $A E F D$ to quadrilateral $E B C F$ is $\frac{\sqrt{3}+1}{3-\sqrt{3}}$, and the area of $\triangle A B D$ is $\sqrt{3}$. Then the area of trapezoid $A B C D$ is
|
6. 2 .
Let $A D=a, B C=b$, and the height be $2 h$. The area of trapezoid $A B C D$ is $S$. Then
$$
\begin{array}{l}
E F=\frac{a+b}{2}, \frac{a}{b}=\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\sqrt{3}}{S-\sqrt{3}} . \\
\text { Also, } \frac{\sqrt{3}+1}{3-\sqrt{3}}=\frac{S_{\text {quadrilateral } A E F D}}{S_{\text {quadrilateral } E B C F}}=\frac{S+2 \sqrt{3}}{3 S-2 \sqrt{3}} .
\end{array}
$$
Applying the ratio theorem, we get $S=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (25 points) Let positive real numbers $a$, $b$, $c$ satisfy
$$
(a+2 b)(b+2 c)=9 \text {. }
$$
Prove: $\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant 3$.
(Zhang Lei)
|
Because $a, b, c > 0$, so,
$$
\begin{array}{l}
\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \\
\geqslant \frac{a+b}{2}+2 \cdot \frac{b+c}{2} \\
=\frac{1}{2}[(a+2 b)+(b+2 c)] \\
\geqslant \sqrt{(a+2 b)(b+2 c)}=3 .
\end{array}
$$
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, let $G$ and $H$ be the centroid and orthocenter of $\triangle ABC$ respectively, $F$ be the midpoint of segment $GH$, and the circumradius of $\triangle ABC$ be $R=1$. Then
$$
\begin{array}{l}
|\overrightarrow{A F}|^{2}+|\overrightarrow{B F}|^{2}+|\overrightarrow{C F}|^{2} \\
= \\
.
\end{array}
$$
|
6.3.
Let the circumcenter $O$ of $\triangle ABC$ be the origin of a Cartesian coordinate system. Then,
$$
\begin{array}{l}
\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}, \\
\overrightarrow{O G}=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
\end{array}
$$
$$
\text { Then } \overrightarrow{O F}=\frac{1}{2}(\overrightarrow{O G}+\overrightarrow{O H})=\frac{2}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \text {. }
$$
Therefore, $|\overrightarrow{A F}|^{2}+|\overrightarrow{B F}|^{2}+|\overrightarrow{C F}|^{2}$
$$
\begin{aligned}
= & (\overrightarrow{O A}-\overrightarrow{O F}) \cdot(\overrightarrow{O A}-\overrightarrow{O F})+ \\
& (\overrightarrow{O B}-\overrightarrow{O F}) \cdot(\overrightarrow{O B}-\overrightarrow{O F})+ \\
& (\overrightarrow{O C}-\overrightarrow{O F}) \cdot(\overrightarrow{O C}-\overrightarrow{O F}) \\
= & |\overrightarrow{O A}|^{2}+|\overrightarrow{O B}|^{2}+|\overrightarrow{O C}|^{2}- \\
& 2(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{O F}+3 \overrightarrow{O F} \cdot \overrightarrow{O F} \\
= & |\overrightarrow{O A}|^{2}+|\overrightarrow{O B}|^{2}+|\overrightarrow{O C}|^{2}=3 .
\end{aligned}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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