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stringclasses 51
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stringclasses 4
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8. Let the unit disk be $D=\left\{(x, y) \in \mathbf{R}^{2} \mid x^{2}+y^{2} \leqslant\right.$ 1\}, and define the width of a strip formed by two parallel lines as the distance between these lines. If the unit disk can be covered by some strips, prove: the sum of the widths of these strips is at least 2.
(19th Iranian Mathematical Olympiad (Second Round))
|
Solution: Consider the unit sphere that contains this unit circle. Let the unit circle belong to plane II. For each strip, replace the two parallel lines that form this strip with two planes that pass through these lines and are perpendicular to plane $I$, then for each strip, we can obtain two parallel planes, which we call a "super box". Each "super box" intersects the unit sphere and forms a region on the sphere's surface. Therefore, the sphere's surface can be covered by these spherical regions, meaning the sum of the areas of these spherical regions is greater than or equal to the area of the sphere's surface. Since the area of a spherical cap is $2 \pi R h$, we have $\sum 2 \pi R \cdot d_{t} \geqslant 4 \pi R^{2}$. Since $R=1$, we have $\sum d_{i} \geqslant 2$.
|
2
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure $3, A C=$ $B C, A C \perp B C$ at point $C, A B=A D=B D$, $C D=C E=D E$. If $A B=\sqrt{2}$, then $B E=$
|
8. 1 .
In $\triangle A D C$ and $\triangle B D C$, given $A D=D B, D C=D C$, $A C=B C$, we can conclude that
$\triangle A D C \cong \triangle B D C, \angle A D C=\angle B D C$.
Since $\angle A D B=60^{\circ}$, it follows that,
$\angle A D C=\angle B D C=\angle E D B=30^{\circ}$.
Therefore, $D B \perp C E, B C=B E$.
Moreover, $\triangle A C B$ is an isosceles right triangle, and $A B=\sqrt{2}$, so, $B C=B E=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 3, $A_{0} A_{1}$ is the diameter of a semicircle, $A_{0} A_{1}=2$, $A_{2}, A_{3}, \cdots, A_{k}$, $A_{k+1}, \cdots$, are points on the semicircle, $\angle A_{0} A_{1} A_{2}=1^{\circ}, \angle A_{1} A_{2} A_{3}=2^{\circ}$, $\angle A_{2} A_{3} A_{4}=3^{\circ}, \cdots, \angle A_{k-1} A_{k} A_{k+1}=k^{\circ}(k$ is a positive integer). If the length of the chord $A_{k} A_{k+1}$ is less than 1, then the minimum value of $k$ is $\qquad$.
|
2. 11 .
As shown in Figure 6, connect
$$
A_{k} O, A_{k+1} O, A_{k} A_{1} \text {, }
$$
then
$$
\begin{array}{l}
\angle O A_{k} A_{k+1}= \\
\angle A_{1} A_{k} A_{k+1}+\angle A_{1} A_{k} O \\
=\angle A_{1} A_{k} A_{k+1}+\angle A_{k} A_{1} O=\left(\frac{k(k+1)}{2}\right)^{\circ} .
\end{array}
$$
Since $A_{k} A_{k+1}60^{\circ}$, that is
$$
\begin{array}{l}
\frac{k(k+1)}{2}>60, \\
\left(k+\frac{1+\sqrt{481}}{2}\right)\left(k-\frac{-1+\sqrt{481}}{2}\right)>0 .
\end{array}
$$
Solving gives $k>\frac{\sqrt{481}-1}{2}$.
Also, $10<\frac{\sqrt{481}-1}{2}<11, k$ is a positive integer, so the minimum value of $k$ is 11.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given that $a, b, c$ are real numbers. The functions are $y_{1} = a x^{2} + b x + c, y_{2} = a x + b (a > 0)$. When $-1 \leqslant x \leqslant 1$, it is given that $-1 \leqslant y_{1} \leqslant 1$ and $y_{2}$ has a maximum value of 2. Try to find the area of the figure formed by connecting in sequence all the lattice points (points with integer coordinates) within and on the boundary of the closed region enclosed by the parabola $y_{1} = a x^{2} + b x + c$ and the line $y_{2} = a x + b$.
|
Three, from $a>0$, we know that $y_{2}$ increases as $x$ increases.
Therefore, $a+b=2$.
When $x=0,1$, $-1 \leqslant c \leqslant 1, -1 \leqslant a+b+c \leqslant 1$, so $-1 \leqslant c=(a+b+c)-2 \leqslant 1-2 \leqslant-1$.
Thus, $c=-1$. Therefore, when $x=0$, $y_{1}=-1$ is the minimum value of $y_{1}=$ $a x^{2}+b x+c$ in the interval $-1 \leqslant x \leqslant 1$.
Hence, $x=0$ is the axis of symmetry of the function $y_{1}=a x^{2}+b x+c$, i.e., $-\frac{b}{2 a}=0, b=0, a=2$.
Thus, $y_{1}=2 x^{2}-1, y_{2}=2 x$.
Solving the system of equations $\left\{\begin{array}{l}y=2 x^{2}-1, \\ y=2 x\end{array}\right.$, we get
$$
\left\{\begin{array}{l}
x_{1}=\frac{1+\sqrt{3}}{2}, \\
y_{1}=1+\sqrt{3};
\end{array} \left\{\begin{array}{l}
x_{2}=\frac{1-\sqrt{3}}{2}, \\
y_{2}=1-\sqrt{3}.
\end{array}\right.\right.
$$
Therefore, the coordinates of the intersection points of the two graphs are $A\left(\frac{1-\sqrt{3}}{2}, 1-\sqrt{3}\right)$ and $B\left(\frac{1+\sqrt{3}}{2}, 1+\sqrt{3}\right)$. The closed figure formed by the two graphs is shown in Figure 8. Since $\frac{1-\sqrt{3}}{2} \leqslant x \leqslant \frac{1+\sqrt{3}}{2}$ has integer solutions $x=0,1$, the corresponding lattice points are $O(0,0), M(0,-1), N(1,1), P(1,2)$. The quadrilateral $O M N P$ is exactly a parallelogram, with area $S_{\text {quadrilateral }}$ MNP $=$ $1 \times 1=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $\lg a<0, \lg b<0, \lg c<0$, and $\lg (a+b+c)=0$. Then the maximum value of $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3. 0 .
Given that $a, b, c$ are positive numbers less than 1, and $a+b+c=1$. We have
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
$\geqslant 9$ (Cauchy-Schwarz inequality).
The equality holds when $a=b=c=\frac{1}{3}$.
Transforming the inequality (1), we get
$$
\begin{array}{l}
9 a b c \leqslant a b+b c+c a \\
=\frac{1}{2}\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right] \\
=\frac{1}{2}\left[1-\left(a^{2}+b^{2}+c^{2}\right)\right],
\end{array}
$$
which implies $a^{2}+b^{2}+c^{2}+18 a b c \leqslant 1$.
Therefore, $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right) \leqslant 0$.
When $a=b=c=\frac{1}{3}$, $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ can achieve its maximum value of 0.
|
0
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. The sum of $n$ consecutive natural numbers starting from the positive integer $m$ is 2004, and $(m, n)>1$ (not coprime). Then the greatest common divisor $(m, n, 2004)=$
|
6. 12 .
According to the problem, $(m, n)>1$ and $m+(m+1)+\cdots+(m+n-1)=2004$, that is,
$$
\frac{(2 m+n-1) n}{2}=2004 \text {. }
$$
And $(2 m+n-1) n$
$$
=1 \times 4008=3 \times 1336=167 \times 24=501 \times 8 \text {, }
$$
where $2 m+n-1$ and $n$ are one odd and one even, and $2 m+n-1$ is greater than $n$. Therefore, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 2 m + n - 1 = 4 0 0 8 , } \\
{ n = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=2004, \\
n=1 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 2 m + n - 1 = 1 3 3 6 , } \\
{ n = 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=667, \\
n=3 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 2 m + n - 1 = 1 6 7 , } \\
{ n = 2 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=72, \\
n=24 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 2 m + n - 1 = 5 0 1 , } \\
{ n = 8 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=247, \\
n=8 .
\end{array}\right.\right.
\end{array}
$$
Given that $m$ and $n$ are not coprime, we can only take
$$
m=72=2^{3} \times 3^{2}, n=24=2^{3} \times 3 \text {. }
$$
Also, $2004=2^{2} \times 3 \times 167$. Therefore, $(m, n, 2004)=12$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let $M$ be a set of $n$ points in the plane, satisfying:
(1) There exist 7 points in $M$ that are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it.
Find the minimum value of $n$.
(Leng Gangsong, provided)
|
Three, prove $n \geqslant 11$.
Consider a convex heptagon $A_{1} A_{2} \cdots A_{7}$ with vertices in $M$, and connect $A_{1} A_{5}$. By condition (2), there is at least one point in $M$ within the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, denoted as $P_{1}$. Connect $P_{1} A_{1}$ and $P_{1} A_{5}$. Then, in the convex pentagon $A_{1} P_{1} A_{5} A_{6} A_{7}$, there is at least one point in $M$, denoted as $P_{2}$, and $P_{2}$ is different from $P_{1}$. Connect $P_{1} P_{2}$, then at least 5 of the vertices $A_{1}, A_{2}, \cdots, A_{7}$ are not on the line $P_{1} P_{2}$. By the pigeonhole principle, on one side of the line $P_{1} P_{2}$, there must be 3 vertices, and these 3 vertices, together with points $P_{1}$ and $P_{2}$, form a convex pentagon that contains at least one point $P_{3}$ in $M$.
Next, draw the lines $P_{1} P_{3}$ and $P_{2} P_{3}$. Let the region $\Pi_{3}$ be the half-plane defined by the line $P_{1} P_{2}$ and on the side opposite to $\triangle P_{1} P_{2} P_{3}$ (excluding the line $P_{1} P_{2}$). Similarly, define regions $\Pi_{1}$ and $\Pi_{2}$. Thus, the regions $\Pi_{1}$, $\Pi_{2}$, and $\Pi_{3}$ cover all points on the plane except for $\triangle P_{1} P_{2} P_{3}$. By the pigeonhole principle, among the 7 vertices $A_{1}, A_{2}, \cdots, A_{7}$, there must be $\left[\frac{7}{3}\right]+1=3$ vertices in the same region (let's say $\Pi_{3}$). These 3 vertices, together with $P_{1}$ and $P_{2}$, form a convex pentagon with vertices in $M$, and its interior contains at least one point $P_{4}$ in $M$. Therefore, $n \geqslant 11$.
Next, construct an example to show that $n=11$ is possible.
As shown in Figure 3, the heptagon $A_{1} A_{2} \cdots A_{7}$ is an integer-point heptagon, and the point set $M$ consists of the 7 vertices $A_{1}, A_{2}, \cdots, A_{7}$ and 4 integer points inside it. Clearly, this satisfies condition (1). This point set $M$ also satisfies condition (2). The proof is as follows.
Assume there exists an integer-point convex pentagon whose interior does not contain any integer points. Since the area of any integer-point polygon can be expressed as $\frac{n}{2}\left(n \in \mathbf{N}_{+}\right)$, by the principle of the smallest number, there must be a smallest area integer-point convex pentagon $A B C D E$ whose interior does not contain any integer points. Considering the parity of the coordinates of the vertices, there are only 4 cases: (odd, even), (even, odd), (odd, odd), (even, even). Therefore, among the vertices of the pentagon $A B C D E$, there must be two vertices with the same parity for both coordinates. Thus, the midpoint $P$ of the line segment connecting these two vertices is also an integer point. Since $P$ is not inside the convex pentagon $A B C D E$, $P$ must lie on one of the sides of the pentagon. Without loss of generality, assume $P$ lies on side $A B$, then $P$ is the midpoint of $A B$. Connect $P E$, then $P B C D E$ is a smaller area integer-point convex pentagon whose interior does not contain any integer points. This is a contradiction.
In conclusion, the minimum value of $n$ is 11.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2008}+2002(y-2)=1 .
\end{array}\right.
$$
Then $x+y=$
|
7. 3
Construct the function $f(t)=t^{2008}+2002 t$. It is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and it is also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y, x+y=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Let $x, y, z$ be positive real numbers, and $x+y+z=1$. Find the minimum value of the function
$$
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
$$
and provide a proof.
|
14. Consider the function $g(t)=\frac{t}{1+t^{2}}$, we know that $g(t)$ is an odd function. Since when $t>0$, $\frac{1}{t}+t$ is decreasing in $(0,1)$, it is easy to see that $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in $(0,1)$. For $t_{1} \backslash t_{2} \in(0,1)$ and $t_{1}<t_{2}$, we have
$$
\left(t_{1}-t_{2}\right)\left[g\left(t_{1}\right)-g\left(t_{2}\right)\right] \geqslant 0 .
$$
Therefore, for any $x \in(0,1)$, we have
$$
\begin{array}{l}
\left(x-\frac{1}{3}\right)\left(\frac{x}{1+x^{2}}-\frac{3}{10}\right) \geqslant 0, \\
\text { hence } \frac{3 x^{2}-x}{1+x^{2}} \geqslant \frac{3}{10}(3 x-1) .
\end{array}
$$
Similarly, $\frac{3 y^{2}-y}{1+y^{2}} \geqslant \frac{3}{10}(3 y-1)$;
$$
\frac{3 z^{2}-z}{1+z^{2}} \geqslant \frac{3}{10}(3 z-1) \text {. }
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}} \\
\geqslant \frac{3}{10}[3(x+y+z)-3]=0 .
\end{array}
$$
When $x=y=z=\frac{1}{3}$, $f(x, y, z)=0$, hence the minimum value is 0.
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=$ $b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is $\qquad$
|
8. 2 .
Let the inradius of $\triangle ABC$ be $r$. Since $a \cos A = b \cos B$, by the Law of Sines, we get $b \sin A = a \sin B$. Therefore, $\sin 2A = \sin 2B$.
Since $A \neq B$, then $A + B = 90^{\circ}$. Thus, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}, a^2 + b^2 = c^2$.
Also, since $c = 10$, and $a, b, c$ form an arithmetic sequence, we have $a = 6, b = 8, (8 - r) + (6 - r) = 10$.
Solving this, we get $r = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $x, y, z$ satisfy $x=y+\sqrt{2}, 2 x y+$ $2 \sqrt{2} z^{2}+1=0$. Then the value of $x+y+z$ is $\qquad$
|
$=1.0$.
Since $x=y+\sqrt{2}$, we have,
$$
\begin{array}{l}
(x-y)^{2}=2, (x+y)^{2}-4 x y=2, \\
2 x y=\frac{1}{2}(x+y)^{2}-1 .
\end{array}
$$
Substituting this into $2 x y+2 \sqrt{2} z^{2}+1=0$ yields
$$
\frac{1}{2}(x+y)^{2}+2 \sqrt{2} z^{2}=0 \text {. }
$$
|
1.0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Points $A(-4,0)$ and $B(2,0)$ are two fixed points on the $x O y$ plane, and $C$ is a moving point on the graph of $y=-\frac{1}{2} x+2$. How many right triangles $\triangle A B C$ can be drawn that satisfy the above conditions?
|
3. 4 .
As shown in Figure 6, draw perpendiculars from $A$ and $B$ to the $x$-axis, intersecting the line $y=-\frac{1}{2} x + 2$ at points $C_{1}$ and $C_{2}$, respectively; with $AB$ as the diameter, draw a semicircle intersecting the line
$$
y=-\frac{1}{2} x + 2 \text { at }
$$
points $C_{3}$ and $C_{4}$. Then,
$\triangle A B C_{1} γ \triangle A B C_{2} γ \triangle A B C_{3} γ \triangle A B C_{4}$ are the required triangles.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $p>0, q>0$, and satisfy $2 p+\sqrt{p q}-$ $q+\sqrt{p}+\sqrt{q}=0$. Then $(2 \sqrt{p}-\sqrt{q}+2)^{3}=$ $\qquad$
|
2. 1 .
The original expression can be transformed into
$$
2(\sqrt{p})^{2}+\sqrt{p q}-(\sqrt{q})^{2}+(\sqrt{p}+\sqrt{q})=0 \text {, }
$$
which is $(\sqrt{p}+\sqrt{q})(2 \sqrt{p}-\sqrt{q}+1)=0$.
Since $p>0, q>0$, then $\sqrt{p}+\sqrt{q}>0$. Therefore, $2 \sqrt{p}-\sqrt{q}+1=0$, which means $2 \sqrt{p}-\sqrt{q}+2=1$.
Thus, $(2 \sqrt{p}-\sqrt{q}+2)^{3}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) A research study group from Class 1, Grade 3 of Yuhong Middle School conducted a survey on students' lunch time at the school canteen. It was found that within a unit of time, the number of people buying lunch at each window and the number of people choosing to eat outside due to unwillingness to wait for a long time are both fixed numbers. Furthermore, it was found that if only 1 window is opened, it takes $45 \mathrm{~min}$ to ensure that all waiting students can buy lunch; if 2 windows are opened simultaneously, it takes $30 \mathrm{~min}$. It was also discovered that if all waiting students can buy lunch within $25 \mathrm{~min}$, the number of people eating outside can be reduced by $80 \%$ within a unit of time. Assuming the total number of students in the school remains constant and everyone needs to have lunch, to facilitate student dining, the research group suggests that the school canteen should sell out lunch within $20 \mathrm{~min}$. How many windows should be opened simultaneously at least?
|
Let each window sell lunch to $x$ people per minute, and $y$ people go out to eat per minute, with the total number of students being $z$ people, and assume that at least $n$ windows need to be open simultaneously. According to the problem, we have
$$
\left\{\begin{array}{l}
45 x=z-45 y, \\
2 \times 30 x=z-30 y, \\
20 n x \geqslant z-20(1-80 \%) y .
\end{array}\right.
$$
From (1) and (2), we get $y=x, z=90 x$. Substituting into (3) yields
$$
20 n x \geqslant 90 x-4 x, n \geqslant 4.3 \text{. }
$$
Therefore, at least 5 windows need to be open simultaneously.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a tetrahedron $ABCD$ with volume 12, points $E$, $F$, and $G$ are on edges $AB$, $BC$, and $AD$ respectively, such that $AE = 2EB$, $BF = FC$, and $AG = 2GD$. A plane through points $E$, $F$, and $G$ intersects the tetrahedron in a section $EFGH$, and the distance from point $C$ to this section is 1. The area of this section is $\qquad$
|
2. 7 .
As shown in Figure 3, it is easy to know that $G E / /$ $D B$. Since $D B / /$ plane $E F H G$, then $H F$ // BD. Therefore, $H$ is also the midpoint of $D C$.
Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{a}, h_{b}, h_{c}, h_{d}$, respectively. We have
$$
\begin{array}{l}
h_{b}=h_{d}, h_{r}=h_{b}, \\
h_{a}=2 h_{b} .
\end{array}
$$
Since $h_{c}=1$, then $h_{b}=h_{d}=1, h_{a}=2$.
Since $S_{i n k}=\frac{1}{6} S_{\text {Mec }}$, then $V_{D-B E F}=\frac{1}{6} V_{\text {Mec }}$.
Since $S_{\triangle A F:}=\frac{1}{2} S_{\triangle A B C}$, then $V_{H-\triangle F C}=-\frac{1}{4} V_{\text {Hexa } A B C C D}=3$.
Let the area of the section $E F H G$ be $x$, then $V_{A-E F H G}=\frac{2}{3} x, V_{D-E F H G}=\frac{1}{3} x$.
As shown in Figure 3, we have $12=-\frac{2}{3},-3+2+\frac{1}{3} x$.
Therefore, $x=7$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $f(x)=a x^{2}+b x+c$ be a quadratic trinomial with integer coefficients. If integers $m, n$ satisfy $f(m)-f(n)=1$. Then $|m-n|=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
4. 1 .
$$
\begin{array}{l}
f(m)-f(n)=a\left(m^{2}-n^{2}\right)+b(m-n) \\
=(m-n)[a(m+n)+b],
\end{array}
$$
then $(m-n)(a m+a n+b)=1$.
Since $m-n$ and $a m+a n+b$ are both integers, it must be that
$$
|m-n|=1 \text{. }
$$
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given real numbers $a, b, x, y$ satisfy $a+b=x+y=2$, $a x+b y=5$. Then the value of $\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right)$ is $\qquad$ .
|
8. -5 .
Given $a+b=x+y=2$, we have
$$
(a+b)(x+y)=a x+b y+a y+b x=4 \text{. }
$$
Since $a x+b y=5$, then,
$$
\begin{array}{l}
a y+b x=-1 . \\
\text{ Thus, }\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right) \\
=(a y+b x)(a x+b y)=-5 .
\end{array}
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. $n$ is a positive integer, $f(n)=\sin \frac{n \pi}{2}$. Then
$$
f(1991)+f(1992)+\cdots+f(2003)=
$$
$\qquad$
|
7. -1 .
It is easy to know that $f(1991)=-1, f(1992)=0, f(1993)=1$, $f(1994)=0, \cdots, f(2003)=-1$. Therefore, $f(1991)+f(1992)+\cdots+f(2003)=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The function defined on the set of positive integers is
$$
f(x)=\left\{\begin{array}{ll}
3 x-1, & x \text { is odd, } \\
\frac{x}{2}, & x \text { is even. }
\end{array}\right.
$$
Let $x_{1}=12, x_{n+1}=f\left(x_{n}\right), n \in \mathbf{N}$, then the number of elements in the set $\left\{x \mid x=x_{n}\right.$, $n \in \mathbf{N}\}$ is $\qquad$.
|
3. 7 .
From $x_{1}=12$ we get $x_{2}=6$, then $x_{3}=3, x_{4}=8, x_{5}=4$, $x_{6}=2, x_{7}=1, x_{8}=2, x_{9}=1, \cdots$, and thereafter it is a cycle of $2,1,2,1$, $\cdots$. Therefore, $x_{n}$ takes a total of 7 different values, i.e., the set $|x| x$ $\left.=x_{n}, n \in \mathbf{N}\right\}$ contains 7 elements.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{49}-\frac{1}{50}\right) \div\left(\frac{1}{20}+\frac{1}{77}+\cdots+\frac{1}{50}\right) \\
= \\
\end{array}
$$
|
$=γ 1.1$
$$
\begin{array}{l}
\text { Original expression }=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{49}+\frac{1}{50}\right)-\right. \\
\left.\quad\left(1+\frac{1}{2}+\cdots+\frac{1}{25}\right)\right] \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \\
=\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right)=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of real roots of the equation $x^{2}|x|-5 x|x|+2 x=0$ is $\qquad$.
The equation $x^{2}|x|-5 x|x|+2 x=0$ has $\qquad$ real roots.
|
3.4 .
From the given, we have $x(x|x|-5|x|+2)=0$.
Thus, $x=0$ or $x|x|-5|x|+2=0$.
(1) When $x>0$, $x^{2}-5 x+2=0$, which has two distinct positive real roots;
(2) When $x<0$, $-x^{2}+5 x+2=0$, which has one negative real root.
Therefore, the original equation has 4 real roots.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) The integers $x_{0}, x_{1}, \cdots, x_{2000}$ satisfy the conditions: $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|$, $\cdots,\left|x_{2004}\right|=\left|x_{2003}+1\right|$. Find the minimum value of $\mid x_{1}+x_{2}+\cdots+$ $x_{2004} \mid$.
|
Three, from the known we can get
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{0}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots . \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$
Thus, $x_{2004}^{2}=x_{0}^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$.
Given $x_{0}=0$, then
$$
\begin{array}{l}
2\left(x_{1}+x_{2}+\cdots+x_{2004}\right) \\
=x_{2004}^{2}+2 x_{2004}-2004=\left(x_{2004}+1\right)^{2}-2005,
\end{array}
$$
which means $\left.\left|x_{1}+x_{2}+\cdots+x_{2001}\right|=\frac{1}{2} \right\rvert\,\left(x_{2001}+1\right)^{2}-20051$.
Since $x_{1}+x_{2}+\cdots+x_{2004}$ is an integer, $x_{2004}+1$ must be an odd number. By comparing $143^{2}-20051$ and $145^{2}-20051$, we get
$$
\left|x_{1}+x_{2}+\cdots+x_{2004}\right| \geqslant \frac{1}{2}\left|45^{2}-2005\right|=10 \text {. }
$$
When $x_{0}=x_{2}=\cdots=x_{1900}=0, x_{1}=x_{3}=\cdots=x_{1999}=$ $-1, x_{11}=1, x_{1962}=2, x_{193}=3, \cdots, x_{2004}=44$, the equality holds.
Therefore, the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$ is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The four sides of the quadrilateral pyramid are isosceles triangles with a leg length of $\sqrt{7}$ and a base length of 2. Then the maximum possible volume of this quadrilateral pyramid is
|
5. 3 .
There are three scenarios for the quadrilateral pyramid that meet the requirements.
(1) As shown in Figure 4, all four lateral edges are $\sqrt{7}$, and the volume is calculated as $V_{1}=\frac{4}{3} \sqrt{5}$.
(2) As shown in Figure 5, two lateral edges are $\sqrt{7}$.
Construct $O E \perp$ plane $A B C D$, and let $A E$ $=y, O E=x$. Thus,
$$
x^{2}+y^{2}=7 \text {, and } 4-x^{2}=7-y^{2} \text {. }
$$
Solving these equations, we get $x=\sqrt{2}, y=\sqrt{5}$. At this point,
$$
V_{2}=\frac{1}{3} \times \sqrt{2} \times 2 \sqrt{5} \times \sqrt{2}=\frac{4}{3} \sqrt{5}=V_{1} \text {. }
$$
(3) As shown in Figure 6, three lateral edges $(O B, O C, O D)$ are $\sqrt{7}$, and one lateral edge $O A=2$.
Let $A D$ and $B C$ intersect at point $E$, and let $B E=x$.
Since $A B=A C=\sqrt{7}, B D=D C=2$, we have
$$
D E^{2}=4-x^{2}, O E^{2}=7-x^{2}, A E^{2}=7-x^{2} \text {. }
$$
Because $\cos \angle O E D=-\cos \angle O E A$, we have
$$
\frac{O E^{2}+E D^{2}-O D^{2}}{20 E \cdot E D}=-\frac{O E^{2}+E A^{2}-O A^{2}}{20 E \cdot E A} \text {. }
$$
Substituting the previous results into the above equation, we solve to get $x=\sqrt{3}$ or $2 \sqrt{2}$. Clearly, $x \neq 2 \sqrt{2}$, so $x=\sqrt{3}$. At this point, $V_{3}=\frac{1}{3} \sqrt{3} \times 3 \sqrt{3}=3$.
In summary, $V_{3}>V_{2}=V_{1}$. Therefore, $V_{\text {max }}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3: In front of each number in $1, 2, 3, \cdots, 1989$, add a β+β or β-β sign to make their algebraic sum the smallest non-negative number, and write out the equation.
(1989, All-Russian Mathematical Olympiad)
|
Proof: First, we prove that the algebraic sum is an odd number.
Consider the simplest case:
If all are filled with β+β, then at this moment
$$
1+2+\cdots+1989=995 \times 1989
$$
is an odd number.
For the general case, it only requires adjusting some β+β to β-β.
Since $a+b$ and $a-b$ have the same parity, the parity of the algebraic sum remains unchanged with each adjustment, meaning the total sum is always odd. Therefore,
$$
1+(2-3-4+5)+(6-7-8+9)+\cdots+
$$
$$
(1986-1987-1988+1989)=1 \text {, }
$$
Thus, the minimum value is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Then the value of $\alpha^{4}+3 \beta$ is $\qquad$
(2003, National Junior High School Mathematics Competition, Tianjin Preliminary Contest)
|
Explanation: $\alpha^{4}+3 \beta$ is not a symmetric expression of the two roots of the equation, and it is obviously impossible to directly substitute it using Vieta's formulas. We can construct the dual expression $\beta^{4}+3 \alpha$ of $\alpha^{4}+3 \beta$, and calculate the sum and difference of the two expressions, then solve for the value through a system of equations. Because
$$
\begin{array}{l}
\alpha^{4}+3 \beta+\beta^{4}+3 \alpha \\
=\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-2(\alpha \beta)^{2}+3(\alpha+\beta) \\
=10, \\
\left(\alpha^{4}+3 \beta\right)-\left(\beta^{4}+3 \alpha\right) \\
=(\alpha-\beta)\left\{(\alpha+\beta)\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]-3\right\} \\
=0,
\end{array}
$$
Therefore, $\alpha^{4}+3 \beta=\frac{10+0}{2}=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a b c \neq 0$, and $a+b+c=0$. Then the value of the algebraic expression $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
|
$\begin{array}{l}\text { I. 1.A. } \\ \text { Original expression }=\frac{-(b+c) a}{b c}+\frac{-(a+c) b}{a c}+\frac{-(a+b) c}{a b} \\ =-\left(\frac{a}{b}+\frac{a}{c}\right)-\left(\frac{b}{a}+\frac{b}{c}\right)-\left(\frac{c}{a}+\frac{c}{b}\right) \\ =\frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3 .\end{array}$
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. The minimum value of the function $f(x)=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$ is $\qquad$ .
|
2. 5 .
Obviously, if $xf(-x)$.
Therefore, when $f(x)$ takes its minimum value, there must be $x \geqslant 0$.
As shown in Figure 6, draw a line segment $A B=4$, $A C \perp A B, D B \perp A B$, and $A C=$ $1, B D=2$. For any point $O$ on $A B$, let $O A=x$, then
$$
\begin{array}{l}
O C=\sqrt{x^{2}+1}, \\
O D=\sqrt{(4-x)^{2}+4} .
\end{array}
$$
Thus, the problem is to find a point $O$ on $A B$ such that $O C+O D$ is minimized.
Let the symmetric point of $C$ with respect to $A B$ be $E$, then the intersection of $D E$ and $A B$ is point $O$. At this point, $O C+O D=O E+O D=D E$. Draw $E F / / A B$ intersecting the extension of $D B$ at $F$.
In the right triangle $\triangle D E F$, it is easy to know that $E F=A B=4, D F=3$, so, $D E=5$.
Therefore, the function
$$
f(x)=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}
$$
has a minimum value of 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$2 . a γ b γ c$ are non-zero real numbers, and $a+b+c \neq 0$. If $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$, then $\frac{(a+b)(b+c)(c+a)}{a b c}$ equals ( ).
(A) 8
(B) 4
(C) 2
(D) 1
|
2.A.
Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$, then $a+b+c=k(a+b+c)$.
Since $a+b+c \neq 0$, we have $k=1$.
Therefore, the original expression $=\frac{2 c \cdot 2 a \cdot 2 b}{a b c}=8$.
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 1. If } n \text { satisfies }(n-2003)^{2}+(2004-n)^{2} \\ =1 \text {, then }(2004-n)(n-2003)=\end{array}$
|
$\begin{array}{l} \text { II.1.0. } \\ \text { From }(n-2003)^{2}+2(n-2003)(2004-n)+ \\ (2004-n)^{2}-2(n-2003)(2004-n) \\ =(n-2003+2004-n)^{2}-2(n-2003)(2004-n) \\ =1-2(n-2003)(2004-n)=1, \\ \text { we get }(n-2003)(2004-n)=0 .\end{array}$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If real numbers $x, y$ satisfy $x \geqslant 0$, and
$$
\max \{1-x, x-1\} \leqslant y \leqslant x+2 \text {, }
$$
then the minimum value of the bivariate function $u(x, y)=2 x+y$ is
$\qquad$ .
|
6.1.
From the given information, we have
$$
\begin{array}{l}
u(x, y)=2 x+y \geqslant 2 x+\max \{1-x, x-1\} \\
=\max \{2 x+(1-x), 2 x+(x-1)\} \\
=\max \{x+1,3 x-1\} \\
\geqslant \max \{1,-1\}=1,
\end{array}
$$
i.e., $u(x, y) \geqslant 1$, with equality holding if and only if $x=0, y=1$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text {. }
$$
Find the minimum value of $b^{2}-4 a c$.
(2004, "TRULY ${ }^{\circledR}$ Xinli Cup" National Junior High School Mathematics Competition)
|
Let $y=a x^{2}+b x+c$.
From $a<0$, we know $\Delta=b^{2}-4 a c>0$.
Therefore, the graph of this quadratic function is a parabola opening downwards, and it intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.
Since $x_{1} x_{2}=\frac{c}{a}<0$, let's assume $x_{1}<x_{2}$.
Then $x_{1}<0<x_{2}$, and the axis of symmetry is $x=-\frac{b}{2 a} \leqslant 0$.
Figure 4
The graph is shown in Figure 4.
Thus, $\left|x_{2}\right|=\left|\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\right|$
$$
=\frac{b-\sqrt{b^{2}-4 a c}}{2 a}=c \text {. }
$$
Therefore, $\frac{4 a c-b^{2}}{4 a} \geqslant c=\frac{b-\sqrt{b^{2}-4 a c}}{2 a} \geqslant-\frac{\sqrt{b^{2}-4 a c}}{2 a}$.
So, $b^{2}-4 a c \geqslant 4$.
When $a=-1, b=0, c=1$, the equality holds.
Thus, the minimum value of $b^{2}-4 a c$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $S=\{1,2,3,4\}$, and the sequence $a_{1}, a_{2}, \cdots, a_{n}$ has the following property: for any non-empty subset $B$ of $S$, there are consecutive $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shanghai High School Mathematics Competition)
|
(Since a binary subset containing a fixed element of $S$ has 3 elements, any element of $S$ appears at least twice in the sequence. Therefore, the minimum value of $n$ is estimated to be 8. On the other hand, an 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the minimum value of $n$ is 8.)
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 Unload several boxes from a cargo ship, with a total weight of $10 \mathrm{t}$, and the weight of each box does not exceed $1 \mathrm{t}$. To ensure that these boxes can be transported away in one go, how many trucks with a carrying capacity of $3 \mathrm{t}$ are needed at least?
(1990, Jiangsu Province Junior High School Mathematics Competition)
|
First, note that the weight of each box does not exceed $1 \mathrm{t}$, so the weight of the boxes that each vehicle can carry at one time will not be less than $2 \mathrm{t}$; otherwise, another box can be added.
Let $n$ be the number of trucks needed, and the weights of the boxes they carry be $a_{1}, a_{2}, \cdots, a_{n}$. Then,
$$
2 \leqslant a_{i} \leqslant 3(i=1,2, \cdots, n) .
$$
Let the total weight of all the goods transported be $S$. Then,
$$
2 n \leqslant S=a_{1}+a_{2}+\cdots+a_{n} \leqslant 3 n,
$$
which means $2 n \leqslant 10 \leqslant 3 n$.
Thus, $\frac{10}{3} \leqslant n \leqslant 5$, i.e., $n=4$ or 5.
In fact, 4 trucks are not enough. This can be illustrated by the following special case. Suppose there are 13 boxes, each weighing $\frac{10}{13} \mathrm{t}$. Since $\frac{10}{13} \times 33$, each truck can only carry 3 boxes. Therefore, 4 trucks cannot carry all the boxes, so at least 5 trucks are needed to transport all the boxes at once.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (16 points) Given real numbers $a, b, c$ satisfy $a+b+c=2$, $abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
|
(1) Let's assume $a=\max \{a, b, c\}$. From the given conditions, we have
$$
a>0, b+c=2-a, bc=\frac{4}{a} \text{. }
$$
Therefore, $b$ and $c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$. Thus,
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0 \\
\Leftrightarrow a^{3}-4a^{2}+4a-16 \geqslant 0 \\
\Leftrightarrow (a^{2}+4)(a-4) \geqslant 0 \Leftrightarrow a \geqslant 4 .
\end{array}
$$
When $a=4, b=c=-1$, the conditions are satisfied, so the minimum value sought is 4.
(2) From $abc=4>0$, we know that $a, b, c$ are all greater than 0 or one is positive and two are negative.
If $a, b, c$ are all greater than 0, then $a, b, c$ all belong to the interval $(0, 2)$, which contradicts the conclusion in (1). Therefore, $a, b, c$ can only be one positive and two negative.
By symmetry, let's assume $a>0, b<0, c<0$. Thus,
$$
\begin{array}{l}
|a|+|b|+|c|=a-(b+c) \\
=a-(2-a)=2a-2 \geqslant 2 \times 4-2=6 .
\end{array}
$$
When $a=4, b=c=-1$, the equality holds.
Therefore, the minimum value of $|a|+|b|+|c|$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $k$ is an integer. If the quadratic equation $k x^{2}+(2 k+3) x+1$ $=0$ has rational roots, then the value of $k$ is $\qquad$
|
2. -2 .
From the given, $\Delta_{1}=(2 k+3)^{2}-4 k$ is a perfect square. Let $(2 k+3)^{2}-4 k=m^{2}(m$ be a positive integer), that is,
$$
4 k^{2}+8 k+9-m^{2}=0 \text {. }
$$
Considering equation (1) as a quadratic equation in $k$, by the problem's condition, it has integer roots, so the discriminant of equation (1)
$$
\Delta_{2}=64-16\left(9-m^{2}\right)=16\left(m^{2}-5\right)
$$
should be a perfect square.
Let $m^{2}-5=n^{2}(n$ be a positive integer, and $m>n)$, then we have $(m+n)(m-n)=5$.
Thus, $\left\{\begin{array}{l}m+n=5, \\ m-n=1 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}m=3, \\ n=2 .\end{array}\right.$
Substituting $m=3$ into equation (1) yields $k=-2$ or $k=0$ (discard).
Therefore, $k=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .
(2003, National Junior High School Mathematics League)
|
(Solution: $m=4$ )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5,5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$ such that for all integers $n$, $a_{n} = b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. Then $b+c+d$ equals ( ).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(1980, American Mathematics Competition)
|
Explanation: Divide the sequence $\{1,3,3,3,5,5,5,5,5, \cdots\}$ into groups:
(1), $(3,3,3),(5,5,5,5,5), \cdots,(2 k-1, 2 k-1, \cdots, 2 k-1), \cdots$,
where the $k$-th group consists of $2 k-1$ occurrences of $2 k-1$. If $a_{n}$ is in the $k$-th group, then $a_{n}=2 k-1$. Therefore,
$$
\begin{array}{l}
1+3+\cdots+(2 k-3)+1 \\
\leqslant n0$, solving this gives
$$
\sqrt{n-1}<k \leqslant \sqrt{n-1}+1 \text {. }
$$
Since $k \in \mathbf{Z}_{+}$, and
$$
\sqrt{n-1}+1-\sqrt{n-1}=1 \text {, }
$$
we have $k=[\sqrt{n-1}+1]=[\sqrt{n-1}]+1$.
Thus, $a_{n}=2 k-1=2[\sqrt{n-1}]+1$.
Hence, $b=2, c=-1, d=1$.
Therefore, $b+c+d=2+(-1)+1=2$.
Note: The 1981 Austrian Mathematical Competition, the second part of the second question, is another form of this problem:
Given $\left\{a_{1}, a_{2}, \cdots\right\}=\{1,3,3,3,5,5,5,5, 5, \cdots\}$, i.e., each odd number $k$ appears $k$ times. Prove: There exist integers $b, c, d$ such that $a_{n}=b[\sqrt{n+c}]+d(n \geqslant 1)$, and there is only one set of values for $b, c, d$ that satisfies the equation.
|
2
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find all positive integers $k$ such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$.
(First China Girls Mathematical Olympiad)
Analysis: To find $k$, we can first determine the upper and lower bounds of $k$, and then argue that the $k$ found satisfies the conditions of the problem.
|
Solution: It is easy to know that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Therefore, the required $k>5$.
Take $a, b, c$ that do not form a triangle ($a=1, b=1, c=2$), then we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right).
$$
This gives $k \leqslant 6$.
Therefore, if $55\left(a^{2}+b^{2}+c^{2}\right),
$$
then $a, b, c$ must form a triangle.
Assume without loss of generality that $0c$.
Construct the function
$$
f(x)=5 x^{2}-6(a+b) x+5 a^{2}+5 b^{2}-6 a b,
$$
then $f(c)f(c)$.
Assume $a+b \leqslant c$, by
$$
a+b, c \in\left[\frac{3}{5}(a+b),+\infty\right)
$$
we know $f(a+b) \leqslant f(c)$.
Contradiction (since $f(x)$ is monotonically increasing to the right of the vertex).
Therefore, $a+b>c$, i.e., $a, b, c$ form a triangle.
Note: To compare the sizes of $a+b$ and $c$, one can first compare the sizes of $f(a+b)$ and $f(c)$, and then deduce the sizes of $a+b$ and $c$ based on the monotonicity of $f(x)$. Of course, this is predicated on constructing an appropriate function $f(x)$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics League) This problem has a resemblance to a 1979 college entrance exam question.
|
Example $\mathbf{3}^{\prime}$ If $(z-x)^{2}-4(x-y)(y-z)=0$, prove that $x, y, z$ form an arithmetic sequence.
In Example 3, taking $a=b=c$ can guess the answer to be 2. But this poses a risk of "root reduction," because the condition is a quadratic expression, while the conclusion
$$
\frac{b+c}{a}=2 \Leftrightarrow b+c=2 a
$$
is a linear expression in the letters. Transitioning from quadratic to linear, could it result in two solutions? There are many safer methods (see [4] P60), one of which is a method of constructing equations: from the given condition, we transform to get
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$
This transformation suggests a discriminant of 0; from the discriminant, we construct the equation
$$
(a-b) x^{2}+(b-c) x+(c-a)=0,
$$
which has equal roots $x_{1}=x_{2}$.
Since the sum of the coefficients of equation (10) is 0, we get
$$
x_{1}=x_{2}=1 \text {. }
$$
Substituting into the relationship between roots and coefficients, we get
$$
1=x_{1} x_{2}=\frac{c-a}{a-b} \text {, }
$$
which means $a-b=c-a$.
Therefore, $\frac{b+c}{a}=2$.
This idea is creative, but a good idea was wasted due to a closed mindset.
(1) Equation (10) is a quadratic equation, which requires $a-b \neq 0$ to ensure, and it is necessary to supplement the discussion when $a=b$ to make the proposition complete.
(2) The entire process uses the discriminant of a quadratic equation, the method of finding roots by observation, and the relationship between roots and coefficients, making the process somewhat convoluted.
The reason is that subconsciously, it is assumed that the discriminant (2) corresponds to a unique equation (1). Eliminating this mental obstacle, a more sensible equation can be constructed. The idea is: from equation (9), think of the discriminant of a quadratic equation being 0, and from the equation, think of equal roots, thus constructing an equation with roots $x_{1}=a-b, x_{2}=c-a$.
Solution: The given condition indicates that the quadratic equation with roots $a-b, c-a$
$$
\begin{array}{l}
{[x-(a-b)][x-(c-a)]=0} \\
\Leftrightarrow x^{2}+(b-c) x+(a-b)(c-a)=0
\end{array}
$$
has a discriminant of 0, i.e.,
$$
\begin{array}{l}
\Delta=(b-c)^{2}-4(a-b)(c-a) \\
=4\left[\frac{1}{4}(b-c)^{2}-(a-b)(c-a)\right]=0 .
\end{array}
$$
Thus, the two roots of the equation are equal, i.e.,
$$
a-b=c-a \text {. }
$$
Therefore, $\frac{b+c}{a}=2$.
Note: If we let $\alpha=a-b, \beta=c-a$, then
$$
b-c=-(\alpha+\beta) \text {. }
$$
The problem can be transformed into
$$
[-(\alpha+\beta)]^{2}-4 \alpha \beta=0 \Rightarrow \alpha=\beta \text {. }
$$
This can be directly derived from $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$. And this is exactly the equation background in equation (6).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In the Cartesian coordinate system $x 0 y$, given two points $M(-1,2)$ and $N(1,4)$, point $P$ moves on the $x$-axis. When $\angle M P N$ takes the maximum value, the x-coordinate of point $P$ is $\qquad$ $-$.
|
12.1.
The center of the circle passing through points $M$ and $N$ lies on the perpendicular bisector of line segment $MN$, which is $y=3-x$. Let the center of the circle be $S(a, 3-a)$, then the equation of circle $S$ is $(x-a)^{2}+(y-3+a)^{2}=2\left(1+a^{2}\right)$.
For a fixed-length chord, the inscribed angle subtended by the major arc increases as the radius of the circle decreases. Therefore, when $\angle M P N$ is at its maximum, the circle $S$ passing through points $M$, $N$, and $P$ must be tangent to the $x$-axis at point $P$. This means that the value of $a$ in the equation of circle $S$ must satisfy $2\left(1+a^{2}\right)=(a-3)^{2}$. Solving this, we get $a=1$ or $a=-7$.
Thus, the corresponding points of tangency are $P(1,0)$ and $P^{\prime}(-7,0)$. The radius of the circle passing through points $M$, $N$, and $P$ is greater than the radius of the circle passing through points $M$, $N$, and $P^{\prime}$, so $\angle M P N > \angle M P^{\prime} N$. Therefore, the point $P(1,0)$ is the solution, meaning the x-coordinate of point $P$ is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $a, b, c \in \mathbf{N}_{+}$, and the parabola $f(x) = ax^{2} + bx + c$ intersects the $x$-axis at two different points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a + b + c$.
(1996, National Junior High School Mathematics Competition)
|
Let $f(x)=a x^{2}+b x+c, A\left(x_{1}, 0\right)$, and $B\left(x_{2}, 0\right)$. According to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
f(1)=a+b+c>0, \\
f(-1)=a-b+c>0 .
\end{array}\right.
$$
From (1), we get $b>2 \sqrt{a c}$; from (3), we get $a+c>b$.
$$
\begin{array}{l}
a+c>b \Rightarrow a+c \geqslant b+1>2 \sqrt{a c}+1 \\
\Rightarrow(\sqrt{a}-\sqrt{c})^{2}>1 .
\end{array}
$$
Since $\frac{c}{a}=x_{1} x_{2}c$, we have
$$
\begin{array}{l}
\sqrt{a}>\sqrt{c}+1 \\
\Rightarrow a>(\sqrt{c}+1)^{2} \geqslant(1+1)^{2}=4 .
\end{array}
$$
Thus, $a \geqslant 5, b>2 \sqrt{5}>4$.
So, $b \geqslant 5$.
Therefore, $a+b+c \geqslant 11$.
Taking $a=5, b=5, c=1$, we get
$$
\begin{array}{l}
f(x)=5 x^{2}+5 x+1=0, \\
-1<x_{1,2}=\frac{-5 \pm \sqrt{5}}{10}<1
\end{array}
$$
which satisfies the conditions. Therefore, the minimum value of $a+b+c$ is 11.
The above text illustrates the method for solving the distribution of roots of a quadratic equation. Of course, for a specific problem, there can be multiple methods to solve it.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a, b$ are the real roots of the equation $x^{2}+2 p x+3=0$, $c, d$ are the real roots of the equation $x^{2}+2 q x+3=0$, and $e, v$ are the real roots of the equation $x^{2}+2\left(p^{2}-q^{2}\right) x+3=0$, where $p, q$ are real numbers. Then, the value of $\frac{(a-c)(b-c)(a+d)(b+d)}{e+v}$ is $\qquad$ .
|
3.6 .
From the relationship between roots and coefficients, we get
$$
\begin{array}{l}
a+b=-2 p, a b=3 ; c+d=-2 q, c d=3 ; \\
e+f=-2\left(p^{2}-q^{2}\right), e f=3 . \\
\text { Therefore } \frac{(a-c)(b-c)(a+d)(b+d)}{e+f} \\
=\frac{\left[a b-(a+b) c+c^{2}\right]\left[a b+(a+b) d+d^{2}\right]}{e+f} \\
=\frac{\left(c d+2 p c+c^{2}\right)\left(c d-2 p d+d^{2}\right)}{-2\left(p^{2}-q^{2}\right)} \\
=\frac{c d(d+2 p+c)(c-2 p+d)}{-2\left(p^{2}-q^{2}\right)} \\
=\frac{3\left[(c+d)^{2}-4 p^{2}\right]}{-2\left(p^{2}-q^{2}\right)}=\frac{3\left(4 q^{2}-4 p^{2}\right)}{-2\left(p^{2}-q^{2}\right)}=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζε¦δΈοΌ
```
One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.
```
|
Given the equation $a x^{2}+b x+c=0$ has two roots $x_{1} γ x_{2}(0<x_{1}<x_{2})$, and $x_{1} x_{2}=\frac{c}{a}=\frac{3}{5}$, we have $a x_{1} x_{2}=3$.
Since $x_{1}+x_{2}=\frac{11}{10}$, we have $a(x_{1}+x_{2})=\frac{11}{10}a>4.4$.
Given $a>4$, and $a$ is a positive integer, so, $a \geqslant 5$.
Taking $f(x)=5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)=5 x^{2}-\frac{11}{2} x+ \frac{3}{2}$, which satisfies the conditions, hence the minimum value of $a$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Try to determine, for any $n$ positive integers, the smallest positive integer $n$ such that at least 2 of these numbers have a sum or difference that is divisible by 21.
|
Three, let $A_{i}$ represent all numbers that leave a remainder of $i$ when divided by 21; $i=0,1$, $\cdots, 20$; let $B_{j}=A_{j} \cup A_{21-j}, j=1,2, \cdots, 10, B_{11}=A_{0}$.
By taking 12 numbers, there will be at least 2 numbers belonging to the same category among the 11 categories $B_{1}, B_{2}$, $\cdots, B_{11}$. Therefore, their sum or difference must be a multiple of 21.
Thus, the smallest positive integer $n$ that satisfies the condition is 12.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Consider the binomial coefficients $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ as a sequence. When $n \leqslant 2004\left(n \in \mathbf{N}_{+}\right)$, the number of sequences in which all terms are odd is $\qquad$ groups.
|
2.10.
From $\mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}+C_{n-1}^{r}$, we know that the 2004 numbers $n=1,2, \cdots, 2004$ can be arranged in the shape of Pascal's Triangle, and the even and odd numbers can be represented by 0 and 1, respectively, as shown in Figure 3.
The pattern of the changes in the parity of each term is:
When and only when $n=2^{k}-1\left(k \in \mathbf{N}_{+}\right)$, all terms are odd. Solving $2^{k}-1 \leqslant 2004$, we get
$$
k=1,2,3,4,5,6,7,8,9,10 .
$$
Therefore, there are 10 sequences where all terms are odd.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n}^{2}+2 a_{n}+3, n \in \mathbf{N}_{+}$, try to find
$$
M=a_{m}^{2}+b_{n}^{2}+m^{2}+n^{2}-2\left(a_{m} b_{n}+m n\right)
$$
$\left(m γ n \in \mathbf{N}_{+}\right)$ the minimum value.
|
(1) It is easy to know that when $n \geqslant 2, n \in \mathbf{N}_{+}$, we have
$4\left(S_{n}-S_{n-1}\right)=\left(a_{n}+1\right)^{2}-\left(a_{n-1}+1\right)^{2}$.
Thus, $\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-2\right)=0$.
Since $a_{n}+a_{n-1}>0$, it follows that,
$$
a_{n}-a_{n-1}=2 \text {. }
$$
It is easy to see that $a_{1}=1, a_{2}=3$, so $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of 2. Therefore,
$$
a_{n}=1+2(n-1)=2 n-1 .
$$
(2) Since $a_{n}=2 n-1$, we have,
$$
b_{n}=(2 n-1)^{2}+2(2 n-1)+3=4 n^{2}+2 \text {. }
$$
The distance from a point $\left(x_{0}, 4 x_{0}^{2}+2\right)\left(x_{0} \geqslant 1\right)$ on the parabola $y=4 x^{2}+2$ to the line $y=2 x-1$ is
$$
\begin{array}{l}
d=\frac{\left|2 x_{0}-\left(4 x_{0}^{2}+2\right)-1\right|}{\sqrt{5}}=\frac{\left(2 x_{0}-\frac{1}{2}\right)^{2}+\frac{11}{4}}{\sqrt{5}} \\
\geqslant \frac{\left(2 \times 1-\frac{1}{2}\right)^{2}+\frac{11}{4}}{\sqrt{5}}=\sqrt{5},
\end{array}
$$
When $x_{0}=1$, $d_{\text {min }}=\sqrt{5}$.
It is easy to see that $M=\left(a_{m}-b_{n}\right)^{2}+(m-n)^{2}$ represents the square of the distance between the point $\left(n, b_{n}\right)$ on the parabola $y=4 x^{2}+2$ and the point $\left(m, a_{m}\right)$ on the line $y=2 x-1$. Clearly, $M \geqslant d_{\text {min }}^{2}=5$.
Considering that when $x_{0}=1$, $y_{0}=4 x_{0}^{2}+2=6$, the point $(1,6)$ lies on the parabola, which is also the point corresponding to the first term of the sequence $\left\{b_{n}\right\}$. Drawing a perpendicular from the point $(1,6)$ to the line $y=2 x-1$, let the foot of the perpendicular be $(x, 2 x-1)$. We have
$$
(x-1)^{2}+(2 x-1-6)^{2}=d_{\min }^{2}=5 .
$$
Solving this, we get $x=3$.
Thus, the foot of the perpendicular is $(3,5)$, which corresponds to $a_{3}=5$ in the sequence $\left\{a_{n}\right\}$.
Therefore, when $m=3, a_{m}=5, n=1, b_{n}=6$,
$$
M_{\text {min }}=d_{\text {min }}^{2}=5 \text {. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text{. }
$$
Find the minimum value of $b^{2}-4 a c$.
|
Solution 1: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a c}=-1(a c \neq 0),
$$
This indicates that the quadratic equation
$$
a c x^{2}+b x+1=0
$$
has a real root $x=-1$. Substituting into equation (1) yields
$$
a c=b-1 \text {. }
$$
Transform equation (3) into the overall structure of the discriminant
$$
b^{2}-4 a c=(2 a c x+b)^{2},
$$
Substitute $x=-1, a c=b-1$, and note that $b \leqslant 0$, we get
$$
\begin{array}{l}
b^{2}-4 a c=[2(b-1)(-1)+b]^{2} \\
=(2-b)^{2} \geqslant 4 .
\end{array}
$$
When $b=0$, thus $a c=-1$, $b^{2}-4 a c$ reaches its minimum value of 4.
Note 1: In the above simplified proof, equation (4) plays a crucial role, indicating that the explicit representation of the essential relationship is a source of "ingenious and clever solutions." The entire solution can be completed mentally.
Note 2: The above solution does not use the conditions $a<0, c<0$, but only uses $a c \neq 0$, so the conditions can be weakened.
Solution 2: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2}=-a c \text {. }
$$
This indicates that the quadratic equation
$$
x^{2}+b x+a c=0
$$
has a real root $x_{1}=-a c$. Therefore,
$$
x_{2}=\frac{a c}{x_{1}}=-1 \text { (using } a c \neq 0 \text { ). }
$$
Substituting $x_{2}=-1$ into equation @ yields
$$
1-b+a c=0 \text {. }
$$
At this point, by $b \leqslant 0$ we know
$$
a c=b-1
$$
$\leqslant-1$ (ensured by $b \leqslant 0$ that $a c<0$ ),
so,
$$
1-a c \geqslant 2 \text {. }
$$
Substituting equations (1) and (9) into the discriminant, we have
$$
\begin{array}{l}
b^{2}-4 a c=(1+a c)^{2}-4 a c \\
=(1-a c)^{2} \geqslant 2^{2} .
\end{array}
$$
From equation (9), we know that when $a c=-1, b=0$, $b^{2}-4 a c$ can reach its minimum value of 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the decimal parts of $7+\sqrt{7}$ and $7-\sqrt{7}$ are $a$ and $b$ respectively, then $a b-3 a+2 b+1=$ $\qquad$ .
|
2.0.
Since $2<\sqrt{7}<3$, therefore, the decimal part of $7+\sqrt{7}$ is $a=\sqrt{7}-2$. Also, because $-3<-\sqrt{7}<-2$, then $0<3-\sqrt{7}<1$. Therefore, the decimal part of $7-\sqrt{7}$ is $b=3-\sqrt{7}$. Hence
$$
\begin{array}{l}
a b-3 a+2 b+1 \\
=(a+2)(b-3)+7=\sqrt{7} \times(-\sqrt{7})+7=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A bookstore is holding a summer book fair, where book buyers can enjoy the following discounts:
(1) For a single purchase not exceeding 50 yuan, no discount is given;
(2) For a single purchase exceeding 50 yuan but not exceeding 200 yuan, a 10% discount is given on the marked price;
(3) For a single purchase exceeding 200 yuan, the first 200 yuan is discounted according to rule (2), and the amount exceeding 200 yuan is given an 80% discount.
A student made two purchases, paying 81 yuan and 126 yuan respectively. If he had bought the same books in one go, he would have saved $\qquad$ yuan.
|
4.3.
Obviously, 81 and 126 are both the prices after the discount. If not discounted, the combined price should be
$$
(81+126) \div 0.9=230 \text { (yuan). }
$$
Since 230 yuan exceeds 200 yuan, according to (3), it should be
$$
200 \times 0.9+30 \times 0.8=204 \text { (yuan) } \text {. }
$$
Thus, we have $81+126-204=3$ (yuan).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $\{x\}$ denote the fractional part of the real number $x$. If $a=$ $(5 \sqrt{13}+18)^{2005}$, then $a \cdot\{a\}=$ $\qquad$
|
5.1.
Let $b=(5 \sqrt{13}-18)^{2005}$.
Since $0<5 \sqrt{13}-18<1$, we have $0<b<1$.
By the binomial theorem, we know
$$
\begin{aligned}
a= & (5 \sqrt{13})^{2005}+\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots+18^{2005}, \\
b= & (5 \sqrt{13})^{2005}-\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& (-1)^{r} \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots- \\
& 18^{2005} .
\end{aligned}
$$
Thus, $a-b=2\left[\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\mathrm{C}_{2005}^{3} \times\right.$ $\left.(5 \sqrt{13})^{2002} \times 18^{3}+\cdots+\mathrm{C}_{2005}^{2005} \times 18^{2005}\right]$ is a positive integer.
Since $0<b<1$, we have $\{a\}=b$.
Therefore, $a \cdot\{a\}=a \cdot b$
$$
\begin{array}{l}
=(5 \sqrt{13}+18)^{2005}(5 \sqrt{13}-18)^{2005} \\
=\left[(5 \sqrt{13})^{2}-18^{2}\right]^{2005}=1^{2005}=1 .
\end{array}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $a \neq b$, and
$$
a^{2}-4 a+1=0, b^{2}-4 b+1=0 \text {. }
$$
Find the value of $\frac{1}{a+1}+\frac{1}{b+1}$.
|
Solution 1: Since
$$
\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1},
$$
we only need to find the values of $a b$ and $a+b$.
From the given conditions, $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so
$$
a+b=4, a b=1 \text {. }
$$
Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1}$
$$
=\frac{4+2}{1+4+1}=1 \text {. }
$$
This solution has been published in many newspapers.
Analysis: Although the analysis for solving the problem is quite reasonable, and it is necessary to find $a+b$ and $a b$, the conclusion is entirely correct. However, from the review of the solution, to get
$$
\frac{(a+b)+2}{a b+(a+b)+1}=1,
$$
having $a b=1$ is sufficient, and finding $a+b$ is an extra step. Therefore, the purpose of proposing Example 6 is:
(1) Emphasize that only $a b=1$ is needed to derive
$$
\frac{1}{a+1}+\frac{1}{b+1}=1 .
$$
However, this is somewhat "shallow and old";
(2) Originating from the "pairing" solution method of this problem.
Solution 2: The given conditions indicate that $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so $a b=1$.
Let $M=\frac{1}{a+1}+\frac{1}{b+1}$, and pair
$$
\begin{array}{l}
N=\frac{a}{a+1}+\frac{b}{b+1} . \\
\text { Then } M+N=\frac{a+1}{a+1}+\frac{b+1}{b+1}=2, \\
M-N=\frac{1-a}{a+1}+\frac{1-b}{b+1} \\
=\frac{(1-a)(1+b)+(1+a)(1-b)}{(a+1)(b+1)} \\
=\frac{2(1-a b)}{(a+1)(b+1)}=0 .
\end{array}
$$
Then $M+N=\frac{a+1}{a+1}+\frac{b+1}{b+1}=2$,
so $M=N=1$.
By incorporating the idea of pairing, the original form of the problem is hidden (only retaining $a b=1$), thus forming the test question.
3.3 Variations
After the test question was released, literature [1], [3-5] discussed its variations, generalizations, and applications.
(1) Variations
Let $a=\frac{x}{m}, b=\frac{y}{m}$, then we have the following Example 8.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 As shown in Figure 14, villages $A$, $B$, and $C$ are located along an east-west oriented highway, with $AB=2 \text{ km}$, $BC=3 \text{ km}$. To the due north of village $B$ is village $D$, and it is measured that $\angle ADC=45^{\circ}$. Now the $\triangle ADC$ area is planned to be a development zone, except for a pond of $4 \text{ km}^2$, the rest is designated for construction or green space. Try to find the area of the construction or green space in this development zone.
|
Solution: As shown in Figure 15, construct the axisymmetric figure of Rt $\triangle A D B$ with respect to the line containing $D A$, which is Rt $\triangle A D B_{1}$. It is easy to know that
Rt $\triangle A D B \cong$
Rt $\triangle A D B_{1}$.
Construct the axisymmetric figure of Rt $\triangle C D E$ with respect to the line containing $D C$, which is Rt $\triangle C D B_{2}$. It is easy to know that
Rt $\triangle C D B \cong \mathrm{Rt} \triangle C D B_{2}$.
Extend $B_{1} A$ and $B_{2} C$ to intersect at $E$, then $B_{1} D B_{2} E$ is a square.
Let $B D=x$, then
\[
\begin{array}{l}
B_{1} D=D B_{2}=B_{2} E=E B_{1}=x . \\
A B_{1}=A B=2, C B_{2}=C B=3, A C=5 .
\end{array}
\]
Therefore, $A E=x-2, C E=x-3$.
In Rt $\triangle A E C$, $A E^{2}+C E^{2}=A C^{2}$, i.e., $(x-2)^{2}+(x-3)^{2}=(2+3)^{2}$.
Rearranging and factoring, we get
\[
(x-6)(x+1)=0 \text{. }
\]
Since $x>0$, then $x+1>0$.
Therefore, $x=6$, i.e., $D B=6 \mathrm{~km}$.
Thus, $S_{\triangle A C D}=\frac{1}{2} \times 5 \times 6=15\left(\mathrm{~km}^{2}\right)$.
Given that there is a pond of $4 \mathrm{~km}^{2}$ in the development zone, the area for construction and greening in this development zone is
\[
15-4=11\left(\mathrm{~km}^{2}\right)
\]
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.2. There are 12 people in the room,
some of whom always lie, while the rest always tell the truth. One of them says: βThere are no honest people hereβ; another says: βThere is at most 1 honest person hereβ; the third person says: βThere are at most 2 honest people hereβ; and so on, until the 12th person says: βThere are at most 11 honest people hereβ. How many honest people are in the room?
|
7.2. There are 6 honest people in the room.
Obviously, the first few people are not honest. If the 6th person is telling the truth, then there are no more than 5 honest people in the room. On the other hand, from him onwards, the next 7 people are telling the truth, and thus, they should all be honest, which is impossible. Therefore, the 6th person is not honest. Through similar analysis, it can be known that the 6 people starting from the 7th person are all honest.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.2. Does there exist a number of the form $1000 \cdots 001$ that can be divided by a number of the form $111 \cdots 11$?
|
7.2. Only for 11
In fact, $1000 \cdots 001=99 \cdots 9+2$, and the remainder of $99 \cdots 9$ divided by $111 \cdots 11$ is a number of the form $99 \cdots 9$ with fewer digits than the divisor. Therefore, if $99 \cdots 9+2$ can be divided by $111 \cdots 11$, then $99 \cdots 9+2$ must equal $111 \cdots 11$. This is only possible for 11.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $k$ be a real number, and the quadratic equation $x^{2}+k x+k+1=0$ has two real roots $x_{1}$ and $x_{2}$. If $x_{1}+2 x_{2}^{2}=k$, then $k$ equals $\qquad$ .
|
2.5.
From $\Delta \geqslant 0$ we get
$$
k^{2}-4 k-4 \geqslant 0 \text {. }
$$
Since $x_{2}^{2}=-k x_{2}-k-1$, then,
$$
\begin{array}{l}
x_{1}+2 x_{2}^{2}=k \Rightarrow x_{1}-2 k x_{2}=3 k+2 \\
\Rightarrow\left(x_{1}+x_{2}\right)-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-k-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-(2 k+1) x_{2}=2(2 k+1) \\
\Rightarrow(2 k+1)\left(x_{2}+2\right)=0 .
\end{array}
$$
Since $2 k+1=0$, i.e., $k=-\frac{1}{2}$ does not satisfy (1), thus, $x_{2}=-2$. Substituting into the original equation yields $k=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of integers $x$ that make $x^{2}+x-2$ a perfect square is $\qquad$.
Make the above text in English, please keep the original text's line breaks and format, and output the translation result directly.
|
3.4 .
Let $x^{2}+x-2=y^{2}, y \in \mathbf{N}$, then $4 x^{2}+4 x-8=(2 y)^{2}$,
i.e., $(2 x-2 y+1)(2 x+2 y+1)=3^{2}$.
Therefore, we have $\left\{\begin{array}{l}2 x-2 y+1=1, \\ 2 x+2 y+1=9 ;\end{array}\left\{\begin{array}{l}2 x-2 y+1=3, \\ 2 x+2 y+1=3 ;\end{array}\right.\right.$ $\left\{\begin{array}{l}2 x-2 y+1=-3, \\ 2 x+2 y+1=-3 ;\end{array}\left\{\begin{array}{l}2 x-2 y+1=-9, \\ 2 x+2 y+1=-1 .\end{array}\right.\right.$
Solving, we get $\left\{\begin{array}{l}x=2, \\ y=2 ;\end{array}\left\{\begin{array}{l}x=1, \\ y=0 ;\end{array}\left\{\begin{array}{l}x=-2, \\ y=0 ;\end{array}\left\{\begin{array}{l}x=-3, \\ y=2 .\end{array}\right.\right.\right.\right.$ Therefore, there are 4 integers $x$ that satisfy the given conditions. Note: This problem can also be solved using the "discriminant method".
|
4
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Five, (20 points) Find the smallest natural number $k$, such that for any $x \in [0,1]$ and $n \in \mathbf{N}_{+}$, the inequality
$$
x^{k}(1-x)^{n}<\frac{1}{(1+n)^{3}}
$$
always holds.
|
Given $x \in [0,1]$, by the AM-GM inequality, we have
$$
\begin{array}{l}
x^{k}(1-x)^{n} \\
=\underbrace{x \cdots \cdots x}_{k \uparrow} x \cdot \underbrace{(1-x) \cdot(1-x) \cdots \cdots(1-x)}_{n \uparrow} \\
=\left(\frac{n}{k}\right)^{n} \underbrace{x \cdots \cdots x}_{k \uparrow} \cdot \underbrace{\frac{k(1-x)}{n} \cdots \cdots \cdot \frac{k(1-x)}{n}}_{n \uparrow} \\
\leqslant \frac{n^{n}}{k^{n}}\left[\frac{k x+k(1-x)}{n+k}\right]^{n+k} \\
=\frac{n^{n} k^{k}}{(n+k)^{n+k}} .
\end{array}
$$
When $x=\frac{k}{n+k}$, $x^{k}(1-x)^{n}$ reaches its maximum value of
$$
\frac{n^{n} k^{k}}{(n+k)^{n+k}}.
$$
Thus, the problem is equivalent to finding the smallest natural number $k$ such that
$$
\frac{(k+n)^{k+n}}{k^{k} n^{n}}>(n+1)^{3} \quad\left(n \in \mathbf{N}_{+}\right)
$$
holds for all $n \in \mathbf{N}_{+}$. Testing $(k, n)=(1,1),(2,1),(3,3)$, we find that $k=1,2,3$ do not satisfy the condition.
When $k=4$, if $n=1$, inequality (1) clearly holds.
If $n \geqslant 2$, then
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\left[\frac{(n-2) n+2 \times 2 n+3(2 n+2)+2^{3}}{n+4}\right]^{n+4} \\
=\left(\frac{n^{2}+8 n+14}{n+4}\right)^{n+4}<\left(\frac{n^{2}+8 n+16}{n+4}\right)^{n+4} \\
=(n+4)^{n+4} .
\end{array}
$$
Therefore, inequality (1) holds. Hence, $k=4$ is the solution.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}$ $=p$. Then $p^{3}+p^{2}+p=$ $\qquad$
|
4.1 .
From the given, we have
$$
\begin{array}{l}
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y}=\frac{x+y-z}{x+y+z}, \\
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z} .
\end{array}
$$
Then \( p^{3}+p^{2}+p=1 \).
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd "Hope Cup" National Mathematics Invitational Competition)
|
Explanation: The original formula is transformed to
$$
u(x, y)=\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}+(x-y)^{2}-2 \text {. }
$$
Consider the points $P_{1}\left(x, \frac{9}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$. When $x \in \mathbf{R}(x \neq 0)$, point $P_{1}$ lies on a hyperbola with the coordinate axes as asymptotes, and point $P_{2}$ lies on the semicircle $x^{2}+y^{2}=2$ $(y \leqslant 0)$, then
$$
\left|P_{1} P_{2}\right| \geqslant 3 \sqrt{2}-\sqrt{2}=2 \sqrt{2} \text {. }
$$
Therefore, $u(x, y)=\left|P_{1} P_{2}\right|^{2}-2 \geqslant 6$.
When and only when $x=-3, y=-1$, $u(x, y)=6$. Thus, $u(x, y)_{\text {min }}=6$.
Note: The key to this solution is to ingeniously convert the extremum of a bivariate function into the shortest distance between a point on a hyperbola and a point on a semicircle.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.5. If $a, b, c, d, e, f, g, h, k$ are all 1 or -1, try to find the maximum possible value of
$$
a e k - a f h + b f g - b d k + c d h - c e g
$$
|
7.5. Since each term in the expression $a e k - a f h + b f g - b d k + c d h - c e g$ is either 1 or -1, the value of the expression is even. However, the expression cannot equal 6. If it did, then $a e k$, $b f g$, and $c d h$ would all have to be 1, making their product 1, while $a f h$, $b d k$, and $c e g$ would all have to be -1, making their product -1. But these two products should be equal, which is a contradiction. The value of the expression can be 4, for example, when $a=b=c=e=h=k=1$ and $d=f=g=-1$, the value of the expression is 4. Therefore, the maximum possible value is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y$ satisfy $x^{3}+y^{3}=2$. Then the maximum value of $x+y$ is $\qquad$ .
|
Ni.1.2.
Let $x+y=k$, it is easy to know that $k>0$.
From $x^{3}+y^{3}=2$, we get
$$
(x+y)\left(x^{2}-x y+y^{2}\right)=2 \text {. }
$$
Thus, $x y=\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)$.
From this, we know that $x, y$ are the two real roots of the equation about $t$
$$
t^{2}-k t+\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)=0
$$
Therefore, we have
$$
\Delta=k^{2}-\frac{4}{3}\left(k^{2}-\frac{2}{k}\right) \geqslant 0 \text {. }
$$
Solving this, we get $k \leqslant 2$.
Hence, the maximum value of $x+y$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three piles of stones have the numbers 21, 10, and 11, respectively. Now, the following operation is performed: each time, 1 stone is taken from any two piles, and then these 2 stones are added to the other pile. The question is:
(1) Can the number of stones in the three piles be 4, 14, and 24 after several such operations?
(2) Can the number of stones in the three piles all be 14 after several such operations?
If the requirements can be met, please complete it with the minimum number of operations; if they cannot be met, please explain the reason.
|
(1) It can be achieved.
The minimum number of operations required is 6, for example:
$$
\begin{array}{l}
(21,10,11) \rightarrow(23,9,10) \rightarrow(22,8,12) \\
\rightarrow(24,7,11) \rightarrow(23,6,13) \rightarrow(25,5,12) \\
\rightarrow(24,4,14) .
\end{array}
$$
Since the minimum number of stones in one pile decreases from 10 to 4, at least 6 operations are required, so it is impossible to achieve in fewer than 6 operations.
(2) It cannot be achieved.
Since after each operation, the number of stones in each pile either decreases by 1 or increases by 2, we can write it as
$$
(a, b, c) \rightarrow(a-1, b-1, c+2) \text {. }
$$
If the remainders of $a$, $b$, and $c$ when divided by 3 are all different, then the remainders of $a-1$, $b-1$, and $c+2$ when divided by 3 will also be different. Since 21, 10, and 11 have remainders of 0, 1, and 2 when divided by 3, and 14, 14, and 14 all have a remainder of 2 when divided by 3, it is impossible to achieve.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} .
\end{array}
$$
Then $\lim _{n \rightarrow \infty}\left(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}\right)=$ ( ).
(A) 2
(B) 3
(C) 4
(D) 5
|
6.C.
$$
\begin{array}{l}
\text { Because } a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} \\
=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}, \\
\text { so } \frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right] \\
=4\left(1-\frac{1}{n+1}\right) \rightarrow 4(n \rightarrow \infty) .
\end{array}
$$
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) An old man divides his savings of $m$ gold coins among his $n$ children ($m, n$ are positive integers greater than 1). First, he gives the eldest child 1 gold coin and $\frac{1}{7}$ of the remainder; then, from the remaining coins, he gives the second child 2 gold coins and $\frac{1}{7}$ of the remainder; and so on, the $n$-th child receives $n$ gold coins and $\frac{1}{7}$ of the remainder, until the youngest child receives the last $n$ gold coins. Does the old man give each child the same number of gold coins?
|
Let the number of gold coins left after giving to the $k$-th child be $a_{k}$, then $a_{0}=m, a_{n-1}=n$,
$$
a_{k}=a_{k-1}-\left(k+\frac{a_{k-1}-k}{7}\right)=\frac{6}{7}\left(a_{k-1}-k\right) \text {. }
$$
Thus, $a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right]$.
This indicates that the sequence $b_{k}=a_{k}+6 k-36$ is a geometric sequence with common ratio $q=\frac{6}{7}$, where the first term and the last term are
$b_{0}=a_{0}-36=m-36$,
$b_{n-1}=a_{n-1}+6(n-1)-36=7 n-42$.
Substituting into the general term formula $b_{n-1}=b_{0} q^{n-1}$ gives
$7 n-42=(m-36)\left(\frac{6}{7}\right)^{n-1}$,
which means $m=36+\frac{7^{n}(n-6)}{6^{n-1}}$.
Since $m$ is a positive integer, $6^{n-1} \mid 7^{n}(n-6)$.
Since $7^{n}$ and $6^{n-1}$ are coprime, it follows that $6^{n-1} \mid(n-6)$.
But $-6^{n-1}+6 \leqslant 01)$.
Solving this yields $n=6$. Thus, $m=36$.
Therefore,
the eldest child gets $1+\frac{36-1}{7}=6$ coins,
the second child gets $2+\frac{30-2}{7}=6$ coins,
the third child gets $3+\frac{24-3}{7}=6$ coins,
the fourth child gets $4+\frac{18-4}{7}=6$ coins,
the fifth child gets $5+\frac{12-5}{7}=6$ coins,
and the youngest child gets the remaining 6 coins.
So, the old man gives each child the same number of gold coins.
Note: This problem provides a prototype for the 6th question of the 9th IMO.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$1 . m$ is what integer when the equation
$$
\left(m^{2}-1\right) x^{2}-6(3 m-1) x+72=0
$$
has two distinct positive integer roots?
|
(Tip: $m^{2}-1 \neq 0, m \neq \pm 1$. Since $\Delta=36(m-3)^{2}>$ 0, hence $m \neq 3$. Using the quadratic formula, we get $x_{1}=\frac{6}{m-1}, x_{2}=$ $\frac{12}{m+1}$. Therefore, $(m-1)|6,(m+1)| 12$. So $m=2$, at this point, $x_{1}=6, x_{2}=4$.)
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. From 3 boys and $n$ girls, any 3 people are selected to participate in a competition, given that the probability of having at least 1 girl among the 3 people is $\frac{34}{35}$. Then $n=$ $\qquad$ .
|
13.4. From the condition, $1-\frac{\mathrm{C}_{3}^{3}}{\mathrm{C}_{n+3}^{3}}=\frac{34}{35}$, solving for $n$ yields $n=4$.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. There are 10 table tennis players participating in a round-robin tournament. The match results show that there are no draws, and among any 5 players, there is 1 player who wins against the other 4, and 1 player who loses to the other 4. Then the number of players who won exactly two matches is $\qquad$.
|
14.1.
It can be proven that under the given conditions, no two players have the same number of wins. Therefore, the number of wins for 10 players are 10 different numbers: $0,1, \cdots, 9$. Hence, the number of players who win exactly two games is 1.
If not, suppose there exist $A$ and $B$ with the same number of wins, and without loss of generality, assume $A$ beats $B$. Then, among the players who lose to $B$, there must be a player $C$ who beats $A$, otherwise, any player who loses to $B$ would also lose to $A$, and $A$ would have at least one more win than $B$ (the game where $A$ beats $B$), which contradicts the assumption that $A$ and $B$ have the same number of wins.
Thus, we find three players $A$, $B$, and $C$ such that $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
For $A$, $B$, and $C$, we can add 2 more players, and among these 5 players, there must be one player who loses to the other 4, and this player cannot be any of $A$, $B$, or $C$, denoted as $D$.
Similarly, for $A$, $B$, and $C$, we add 2 more players (excluding $D$), and we can find one player who loses to the other 4, and this player cannot be any of $A$, $B$, $C$, or $D$, denoted as $E$.
Thus, among the 5 different players $A$, $B$, $C$, $D$, and $E$, no one wins against the other 4, which contradicts the given condition.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $m$ be a non-zero integer, and the quadratic equation in $x$, $m x^{2}-(m-1) x+1=0$, has rational roots. Find the value of $m$.
|
(Given: Let $\Delta=(m-1)^{2}-4 m=n^{2}, n$ be a non-negative integer, then $(m-3)^{2}-n^{2}=8$, i.e., $(m-3-n)(m-3+n)=$ 8. Following Example 2, we get $\left\{\begin{array}{l}m=6, \\ n=1\end{array}\right.$ or $\left\{\begin{array}{l}m=0, \\ n=1\end{array}\right.$ (discard). Therefore, $m=6$, and the two roots of the equation are $\frac{1}{2}$ and $\frac{1}{3}$.)
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.4. There is a 36-digit number, in which the digits $1,2, \cdots, 9$ each appear 4 times, and except for 9, all other digits are less than the digit that follows them. It is known that the first digit of the number is 9. What is the last digit of the number? Please provide all possible answers and prove that there are no other answers.
|
6.4. The last digit of this number is 8.
According to the problem, only 9 can follow 8. If all four 8s are located within the number, then each of them is followed by one 9, and adding the one 9 at the beginning, there are a total of 5 nines.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.2. Person A and Person B perform division with remainder on the same number. A divides it by 8, and B divides it by 9. It is known that the sum of the quotient obtained by A and the remainder obtained by B equals 13. Try to find the remainder obtained by A.
|
7.2. Let $a$ and $b$ represent the quotient and remainder obtained by Jia, and let $c$ and $d$ represent the quotient and remainder obtained by Yi. Thus, we have
$$
8 a+b=9 c+d, a+d=13 \text{. }
$$
Substituting $d=13-a$ into the first equation, we get $9(a-c)=$ $13-b$, so $13-b$ is divisible by 9. Since $b$ can only be an integer from $0 \sim 7$, we know that $b=4$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.3. Several boys and 5 girls are sitting around a round table, and there are 30 pieces of bread on the plate on the table. Each girl takes 1 piece of bread from the plate for each boy she knows, and then each boy takes 1 piece of bread from the plate for each girl he does not know, at which point the bread on the plate is all gone. How many boys are there?
|
6.3. There are $n$ boys, so there are $5 n$ different βboy-girlβ pairs. In each such βpairβ, 1 piece of bread was taken, because if the 2 people in the pair know each other, the girl took 1 piece of bread for the boy; and if they do not know each other, the boy took 1 piece of bread for the girl. Therefore, $5 n=$ $30, n=6$.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given that $f(x)$ is defined on $(-1,1)$, $f\left(\frac{1}{2}\right)=-1$, and satisfies $x, y \in (-1,1)$, we have $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.
(1) The sequence $\{x_n\}$ satisfies
$$
x_1 = \frac{1}{2}, \quad x_{n+1} = \frac{2x_n}{1 + x_n^2}.
$$
Let $a_n = f(x_n)$. Find the general term formula for $\{a_n\}$;
(2) Let $b_n = n^2 + 3n + 1$. Find
$$
1 + f\left(\frac{1}{b_1}\right) + f\left(\frac{1}{b_2}\right) + \cdots + f\left(\frac{1}{b_{2002}}\right) + f\left(\frac{1}{2004}\right)
$$
the value of the above expression.
|
$$
\begin{array}{l}
\text { Therefore, } 1+f\left(\frac{1}{b_{1}}\right)+f\left(\frac{1}{b_{2}}\right)+\cdots+f\left(\frac{1}{b_{2000}}\right)+f\left(\frac{1}{2004}\right) \\
=1+\left[f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)\right]+\left[f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)\right]+ \\
\quad \cdots+\left[f\left(\frac{1}{2003}\right)-f\left(\frac{1}{2004}\right)\right]+f\left(\frac{1}{2004}\right) \\
=1+f\left(\frac{1}{2}\right)=0 .
\end{array}
$$
Three, 13. (1) Let $y=x$, we get $2 f(x)=f\left(\frac{2 x}{1+x^{2}}\right)$.
Since $x_{n+1}=\frac{2 x_{n}}{1+x_{n}^{2}}$, then $f\left(x_{n+1}\right)=2 f\left(x_{n}\right)$. Hence, $a_{n+1}=2 a_{n}$.
Also, $a_{1}=f\left(\frac{1}{2}\right)=-1$, so $\left\{a_{n}\right\}$ is a geometric sequence with the first term -1 and common ratio 2.
Therefore, $a_{n}=-2^{n-1}$.
(2) Let $x=y=0$, we get $f(0)+f(0)=f(0)$.
Then $f(0)=0$.
Let $y=-x$, we get $f(x)+f(-x)=f(0)=0$.
Then $f(-x)=-f(x)$.
$$
\begin{aligned}
& \text { And } f\left(\frac{1}{b_{n}}\right)=f\left(\frac{1}{n^{2}+3 n+1}\right)=f\left(\frac{1}{(n+1)(n+2)-1}\right) \\
= & f\left(\frac{\frac{1}{(n+1)(n+2)}}{1-\frac{1}{(n+1)(n+2)}}\right)=f\left(\frac{\frac{1}{n+1}-\frac{1}{n+2}}{1-\frac{1}{(n+1)(n+2)}}\right) \\
= & f\left(\frac{1}{n+1}\right)+f\left(-\frac{1}{n+2}\right)=f\left(\frac{1}{n+1}\right)-f\left(\frac{1}{n+2}\right),
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is
|
From $2 a+b+c=5, b-c=1$, we can obtain
$$
b=3-a, c=2-a \text {. }
$$
Therefore, $a b+b c+c a$
$$
\begin{array}{l}
=a(3-a)+(3-a)(2-a)+a(2-a) \\
=-a^{2}+6 \leqslant 6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, with $BD$ being the diameter of $\odot O$, and $\overparen{AB}=\overparen{AD}$. If $BC + CD = 4$, then the area of quadrilateral $ABCD$ is $\qquad$ .
|
3.4 .
From $\overparen{A B}=\overparen{A D}$, we know $A B=A D$.
Also, $B D$ is the diameter of $\odot 0$, so $\angle A=\angle C=90^{\circ}$.
Let $A B=A D=x, B C=y, C D=z$.
By the Pythagorean theorem, we have
$$
x^{2}+x^{2}=y^{2}+z^{2} \text {. }
$$
Thus, $2 x^{2}=y^{2}+z^{2}=(y+z)^{2}-2 y z$.
Also, $S_{\text {quadrilateral } A B C D}=\frac{1}{2} A B \cdot A D+\frac{1}{2} B C \cdot C D$ $=\frac{1}{2} x^{2}+\frac{1}{2} y z$,
From equation (1), we know $S_{\text {quadrilateral } A B C D}=\frac{1}{4}(y+z)^{2}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given that $a$ is a positive integer not greater than 2005, and $b, c$ are integers, the parabola $y=a x^{2}+b x+c$ is above the $x$-axis and passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. Find the minimum value of $a-b+c$.
The parabola $y=a x^{2}+b x+c$ passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$.
Substituting these points into the equation of the parabola, we get:
1. For point $A(-1,4 a+7)$:
\[ 4a + 7 = a(-1)^2 + b(-1) + c \]
\[ 4a + 7 = a - b + c \]
\[ 3a + 7 = -b + c \]
\[ c - b = 3a + 7 \quad \text{(1)} \]
2. For point $B(3,4 a-1)$:
\[ 4a - 1 = a(3)^2 + b(3) + c \]
\[ 4a - 1 = 9a + 3b + c \]
\[ -5a - 1 = 3b + c \]
\[ c + 3b = -5a - 1 \quad \text{(2)} \]
We now have the system of linear equations:
\[ c - b = 3a + 7 \quad \text{(1)} \]
\[ c + 3b = -5a - 1 \quad \text{(2)} \]
Subtract equation (1) from equation (2):
\[ (c + 3b) - (c - b) = (-5a - 1) - (3a + 7) \]
\[ 4b = -8a - 8 \]
\[ b = -2a - 2 \]
Substitute \( b = -2a - 2 \) into equation (1):
\[ c - (-2a - 2) = 3a + 7 \]
\[ c + 2a + 2 = 3a + 7 \]
\[ c = a + 5 \]
Now, we need to find \( a - b + c \):
\[ a - b + c = a - (-2a - 2) + (a + 5) \]
\[ a + 2a + 2 + a + 5 \]
\[ 4a + 7 \]
Since \( a \) is a positive integer not greater than 2005, the minimum value of \( 4a + 7 \) occurs when \( a = 1 \):
\[ 4(1) + 7 = 11 \]
Thus, the minimum value of \( a - b + c \) is:
\[ \boxed{11} \]
|
Three, since $a$ is a positive integer, the parabola opens upwards.
The parabola $y=a x^{2}+b x+c$ is above the $x$-axis, so,
$$
b^{2}-4 a c<0.
$$
Given $b=-2-2a$ and $c=a+5$, we have
$$
b^{2}-4 a c=(-2-2a)^{2}-4a(a+5)=4+8a+4a^{2}-4a^{2}-20a=4-12a<0.
$$
This implies
$$
a>\frac{1}{3}.
$$
Since $a$ is a positive integer no greater than 2005, we have $a \geqslant 1$.
Thus, $b=-2-2a \leqslant-4, -b \geqslant 4, c=a+5 \geqslant 6$.
When $a=1, b=-4, c=6$, the parabola $y=x^{2}-4 x+6$ satisfies the given conditions.
Therefore, the minimum value of $a-b+c$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A cube with an edge length of a certain integer is cut into 99 smaller cubes, 98 of which are unit cubes with an edge length of 1, and the other cube also has an integer edge length. Then its edge length is $\qquad$
|
4.3.
Let the edge length of the original cube be $a$, and the edge length of the other cube after cutting be $b\left(a, b \in \mathbf{Z}_{+}, a>b\right)$. Then
$$
a^{3}-98=b^{3} \text {, }
$$
i.e., $(a-b)\left(a^{2}+a b+b^{2}\right)=98$.
Thus, $\left\{\begin{array}{l}a-b=1, \\ a^{2}+a b+b^{2}=98 ;\end{array}\right.$
$$
\left\{\begin{array} { l }
{ a - b = 2 , } \\
{ a ^ { 2 } + a b + b ^ { 2 } = 4 9 ; }
\end{array} \left\{\begin{array}{l}
a-b=7, \\
a^{2}+a b+b^{2}=14 .
\end{array}\right.\right.
$$
Upon inspection, we find that $a=5, b=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given a non-constant sequence $\left\{a_{i}\right\}$ satisfies
$$
a_{i}^{2}-a_{i-1} a_{i}+a_{i-1}^{2}=0 \text {, }
$$
and $a_{i+1} \neq a_{i-1}, i=1,2, \cdots, n$.
For a given positive integer $n, a_{1}=a_{n+1}$. Then
$$
\sum_{i=0}^{n-1} a_{i}=
$$
|
6.0 .
Given $a_{1}^{2}-a_{1-1} a_{i}+a_{1-1}^{2}=0$, so,
$$
a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}=0 \text {. }
$$
Subtracting the two equations, we get
$$
\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}\right)-a_{i}\left(a_{i+1}-a_{i-1}\right)=0 .
$$
That is, $\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}-a_{i}\right)=0$.
Since $a_{i+1} \neq a_{i-1}$, we have,
$$
a_{i+1}+a_{i-1}-a_{i}=0 .
$$
Therefore, $a_{0}=a_{1}-a_{2}, a_{1}=a_{2}-a_{3}, \cdots$
$$
a_{n-2}=a_{n-1}-a_{n}, a_{n-1}=a_{n}-a_{n+1} .
$$
Thus, $\sum_{i=0}^{n-1} a_{i}=a_{1}-a_{n+1}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4: On a circular road in a certain town, there are five primary schools in sequence: No.1 Primary School, No.2 Primary School, No.3 Primary School, No.4 Primary School, and No.5 Primary School. They have 15, 7, 11, 3, and 14 computers respectively. To make the number of computers in each school the same, how many computers should be transferred to the neighboring school: No.1 to No.2, No.2 to No.3, No.3 to No.4, No.4 to No.5, and No.5 to No.1? If School A gives -3 computers to School B, it means School B gives 3 computers to School A. To minimize the total number of computers moved, what arrangement should be made?
|
Solution: As shown in Figure 4, let $A, B, C, D, E$ represent the first to fifth schools, respectively, and let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the number of computers transferred to the neighboring schools. Let the total number of transfers be $y$. According to the problem, we have
$$
\begin{array}{l}
7+x_{1}-x_{2}=11+x_{2}-x_{3}=3+x_{3}-x_{4} \\
=14+x_{4}-x_{5}=15+x_{5}-x_{1}=10 .
\end{array}
$$
Thus, we have $x_{2}=x_{1}-3, x_{3}=x_{1}-2$,
$$
x_{4}=x_{1}-9, x_{5}=x_{1}-5 \text {. }
$$
According to the problem, we have
$$
\begin{aligned}
y= & \left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|+\left|x_{4}\right|+\left|x_{5}\right| \\
= & \left|x_{1}\right|+\left|x_{1}-3\right|+\left|x_{1}-2\right|+\left|x_{1}-9\right|+ \\
& \left|x_{1}-5\right| \\
= & \left|x_{1}\right|+\left|x_{1}-2\right|+\left|x_{1}-3\right|+\left|x_{1}-5\right|+ \\
& \left|x_{1}-9\right| .
\end{aligned}
$$
By the geometric meaning of absolute value, when $x_{1}=3$, $y$ has the minimum value, which is 12. At this time, $x_{2}=0$, $x_{3}=1, x_{4}=-6, x_{5}=-2$.
Therefore, the transfer plan is: School 1 transfers 3 computers to School 2, School 3 transfers 1 computer to School 4, School 5 transfers 6 computers to School 4, and School 1 transfers 2 computers to School 5. This way, the total number of computers transferred is minimized, with the minimum number being 12.
Note: Using the geometric meaning of absolute value, the following conclusion can be explained:
Let $a_{1}, a_{2}, \cdots, a_{n}$ be the real numbers represented by points on the number line arranged in sequence.
When $n$ is odd, if $x=a_{\frac{n+1}{2}}$, then
$$
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|
$$
has the minimum value;
When $n$ is even, if $a_{\frac{n}{2}} \leqslant x \leqslant a_{\frac{n}{2}+1}$, then
$$
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|
$$
has the minimum value.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$13.2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5$ can achieve the minimum value of $\qquad$ .
|
13. -10 .
Original expression $=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10$. When $x=2, y=-1$, it has the minimum value -10.
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $a=1+2+\cdots+2004$. Then the remainder when $a$ is divided by 17 is
|
8.1.
$$
\begin{array}{l}
a=1+2+\cdots+2004 \\
=\frac{2004 \times 2005}{2}=2009010 .
\end{array}
$$
Then 2009010 divided by 17 gives a quotient of 118177, with a remainder of 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given $f(x)=x^{2}+x-1$. If $a b^{2} \neq 1$, and $f\left(a^{-1}\right)=f\left(b^{2}\right)=0$, then $\frac{a}{1+a b^{2}}=$ $\qquad$ .
|
9. -1 .
Given $f(x)=x^{2}+x-1, f\left(a^{-1}\right)=f\left(b^{2}\right)=0, a b^{2} \neq 1$, we know that $a^{-1}$ and $b^{2}$ are the two real roots of $f(x)=x^{2}+x-1$.
By Vieta's formulas, we get $\frac{1}{a}+b^{2}=-1, \frac{b^{2}}{a}=-1$. Thus, $\frac{1}{a}+b^{2}=\frac{b^{2}}{a}=-1$.
Therefore, we have $1+a b^{2}=-a$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A test paper has 4 multiple-choice questions, each with three options (A), (B), (C). Several students take the exam, and after grading, it is found that: any 3 students have 1 question where their answers are all different. How many students can take the exam at most?
|
(ζη€Ί: If 10 people take the exam, then for the 1st question, at least 7 people choose two options: for the 2nd question, at least 5 people among these 7 choose two options; for the 3rd question, at least 4 people among these 5 choose two options; for the 4th question, at least 3 people among these 4 choose two options. Therefore, these 3 people choose only two options for each question, which contradicts the problem statement. Thus, the maximum number of students taking the exam is 9. Additionally, it is easy to construct a scenario where 9 people take the exam.)
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 1, there is a rectangular piece of paper $A B C D, A B=8$, $A D=6$. The paper is folded so that the edge $A D$ lies on the edge $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$ to the right, and the intersection of $A E$ and $B C$ is point $F$. The area of $\triangle C E F$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
|
- .1.A.
From the folding process, we know $D E=A D=6, \angle D A E=\angle C E F=$ $45^{\circ}$. Therefore, $\triangle C E F$ is an isosceles right triangle, and $E C=8-6$ $=2$. Hence, $S_{\triangle C E F}=2$.
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. In the Cartesian coordinate system, the parabola
$$
y=x^{2}+m x-\frac{3}{4} m^{2}(m>0)
$$
intersects the $x$-axis at points $A$ and $B$. If the distances from points $A$ and $B$ to the origin are $O A$ and $O B$, respectively, and satisfy $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, then the value of $m$ is $\qquad$.
|
7. 2 .
Let the roots of the equation $x^{2}+m x-\frac{3}{4} m^{2}=0$ be $x_{1}$ and $x_{2}$, and $x_{1}<0<x_{2}$.
From $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, we know $O A>O B$.
Also, since $m>0$, the axis of symmetry of the parabola is to the left of the $y$-axis, thus,
$$
O A=\left|x_{1}\right|=-x_{1}, O B=x_{2} .
$$
Therefore, $\frac{1}{x_{2}}+\frac{1}{x_{1}}=\frac{2}{3}$,
$$
\text { then } \frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{2}{3}=\frac{-m}{-\frac{3}{4} m^{2}} \text {. }
$$
Solving for $m$ yields $m=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) Given $a, b, c \in \mathbf{R}_{+}$, and
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1 \text{. }
$$
Prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant 12$.
|
Given, we have
$$
\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{1}{b}}+\frac{1}{1+\frac{1}{c}}=1 \text {. }
$$
Let $x=\frac{1}{1+\frac{1}{a}}, y=\frac{1}{1+\frac{1}{b}}, z=\frac{1}{1+\frac{1}{c}}$. Then
$$
x+y+z=1 \text {. }
$$
From $\frac{1}{a}=\frac{1}{x}-1, \frac{1}{b}=\frac{1}{y}-1, \frac{1}{c}=\frac{1}{z}-1$, we get
$$
\begin{array}{l}
\frac{1}{a b c}=\frac{1-x}{x} \cdot \frac{1-y}{y} \cdot \frac{1-z}{z} \\
=\frac{y+z}{x} \cdot \frac{x+z}{y} \cdot \frac{x+y}{z} \\
\geqslant \frac{2 \sqrt{y z}}{x} \cdot \frac{2 \sqrt{x z}}{y} \cdot \frac{2 \sqrt{x y}}{z}=2^{3} .
\end{array}
$$
Thus, $a b c \leqslant 2^{-3}$.
Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{3}{\sqrt[3]{a^{2} b^{2} c^{2}}} \geqslant \frac{3}{\sqrt[3]{2^{-6}}}=12$.
|
12
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. (24 points)(1) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals 2005, find the minimum value of $n$ and explain the reason;
(2) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals $2002^{2000}$, find the minimum value of $n$ and explain the reason.
|
4. (1) Since $2005=1728+125+125+27=12^{3}+$ $5^{3}+5^{3}+3^{3}$, therefore, there exists $n=4$, such that $n_{\min } \leqslant 4$.
Also, $10^{3}=1000,11^{3}=1331,12^{3}=1728,13^{3}=$ 2197, and $12^{3}3 \times 8^{3}$, so the edge length of the largest cube can only be $9,10, 11$ or 12. And
$$
\begin{array}{l}
20050,
\end{array}
$$
thus $x \neq 9$;
$$
\begin{array}{l}
2005-10^{3}-10^{3}=5,2005-10^{3}-9^{3}=276, \\
2005-10^{3}-8^{3}=493,2005-10^{3}-7^{3}-7^{3}>0,
\end{array}
$$
thus $x \neq 10$;
$2005-11^{3}-9^{3}0$,
thus $x \neq 11$;
$2005-12^{3}-7^{3}0$,
thus $x \neq 12$.
Therefore, $n=3$ is not possible.
In conclusion, $n_{\min }=4$.
(2) Let the edge lengths of $n$ cubes be $x_{1}, x_{2}, \cdots, x_{n}$, then
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2005} \text {. }
$$
By $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9)$, we get
$$
\begin{array}{l}
2002^{2005} \equiv 4^{2005} \equiv 4^{668 \times 3+1} \\
\equiv\left(4^{3}\right)^{668} \times 4 \equiv 4(\bmod 9)
\end{array}
$$
Also, when $x \in \mathbf{N}_{+}$, $x^{3} \equiv 0, \pm 1(\bmod 9)$, so,
$$
\begin{array}{l}
x_{1}^{3} \not \equiv 4(\bmod 9), x_{1}^{3}+x_{2}^{3} \not \equiv 4(\bmod 9), \\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \not \equiv 4(\bmod 9) .
\end{array}
$$
Taking equation (1) modulo 9, and from equations (2) and (3), we know $n \geqslant 4$.
Since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, then
$$
\begin{array}{l}
2002^{2005}=2002^{2004} \times\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
=\left(2002^{668}\right)^{3} \times\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
=\left(2002^{668} \times 10\right)^{3}+\left(2002^{668} \times 10\right)^{3}+ \\
\left(2002^{668}\right)^{3}+\left(2002^{668}\right)^{3} .
\end{array}
$$
Therefore, $n=4$ is the required minimum value.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then, $(2+a)(2+b)+b^{2}$
is equal to ( ).
(A) 4
(B) -4
(C) 2
(D) -2
|
1. A.
From the given, we have $(1+a)(1+b)=(1-a)(1-b)$, simplifying to $a+b=0$, which means $a=-b$.
Therefore, $(2+a)(2+b)+b^{2}=(2-b)(2+b)+b^{2}=4$.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Place the natural numbers $1,2, \cdots, 2 n$ randomly on a circle. It is found that among all sets of three consecutive numbers, there are $a$ sets where all three numbers are odd, $b$ sets where exactly two numbers are odd, $c$ sets where only one number is odd, and $d$ sets where all three numbers are even. If $a \neq d$, then the value of $\frac{b-c}{a-d}$ is $\qquad$ .
|
4. -3 .
If each number is counted 3 times, a total of $6 n$ numbers are counted ($3 n$ odd numbers, $3 n$ even numbers), then the system of equations can be set up as
$$
\left\{\begin{array}{l}
3 a+2 b+c=3 n, \\
2 c+b+3 d=3 n .
\end{array}\right.
$$
By eliminating $n$ from equations (1) and (2), we get
$$
\frac{b-c}{a-d}=-3 .
$$
|
-3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given a triangle $\triangle A B C$ with side lengths $4,5,6$ respectively, the circumcircle of $\triangle A B C$ is a great circle of sphere $O$, and $P$ is a point on the sphere. If the distances from point $P$ to the three vertices of $\triangle A B C$ are all equal, then the volume of the tetrahedron $P-A B C$ is $\qquad$ .
|
1.10 .
Since $P A=P B=P C$, the projection of point $P$ on the plane $A B C$ is the circumcenter $O$ of $\triangle A B C$, i.e., $P O \perp$ plane $A B C$, and $P O$ equals the radius $R$ of sphere $O$. Therefore,
$$
V_{P-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot P O=\frac{1}{3} \cdot \frac{a b c}{4 R} \cdot R=10 \text {. }
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Three positive integers $a$, $b$, and $c$ satisfy the conditions:
(1) $a<b<c<30$;
(2) For some positive integer base, the logarithms of $a(2 b-a)$ and $c^{2}+$ $60 b-11 a$ are 9 and 11, respectively.
Then the value of $a-2 b+c$ is $\qquad$.
|
3. -4 .
Let the base be a positive integer $n(n \geqslant 2)$, then from condition (2) we get
$$
\left\{\begin{array}{l}
a(2 b-a)=n^{9} . \\
c^{2}+60 b-11 a=n^{11} .
\end{array}\right.
$$
When $n \geqslant 3$, from equation (1) we get
$$
3^{9} \leqslant n^{9}=a(2 b-a)2^{9}$, i.e., $b \geqslant 23$.
$$
Thus, we have
$$
\left\{\begin{array}{l}
a(2 b-a)=2^{9}, \\
c^{2}+60 b-11 a=2^{11} .
\end{array}\right.
$$
Combining that $a, b, c$ are positive integers, and $a<b<c<30, b \geqslant 23$, we get $a=16, b=24, c=28$.
Therefore, $a-2 b+c=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $\alpha, \beta, \gamma \in \mathbf{R}$,
$$
\begin{aligned}
u= & \sin (\alpha-\beta)+\sin (\beta-\gamma)+ \\
& \sin (\gamma-\alpha) .
\end{aligned}
$$
Then $u_{\text {max }}+u_{\text {min }}=$
|
6.0.
$$
\begin{aligned}
u & =\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \gamma+\sin \gamma \cdot \cos \alpha- \\
& =\left|\begin{array}{lll}
\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \gamma-\cos \gamma \cdot \sin \alpha \\
\sin \alpha & \cos \alpha & 1 \\
\sin \beta & \cos \beta & 1 \\
\sin \gamma & \cos \gamma & 1
\end{array}\right| .
\end{aligned}
$$
Let \( A(\sin \alpha, \cos \alpha) \), \( B(\sin \beta, \cos \beta) \), and \( C(\sin \gamma, \cos \gamma) \). Then \( |u| = 2 S_{\triangle ABC} \).
Clearly, \( A \), \( B \), and \( C \) are any three points on the unit circle \( x^2 + y^2 = 1 \). Since the maximum area of \( \triangle ABC \) is \( \frac{3 \sqrt{3}}{4} \) when and only when \( \triangle ABC \) is an equilateral triangle, we have \( |u| \leq \frac{3 \sqrt{3}}{2} \).
Thus, \( u_{\text{max}} = \frac{3 \sqrt{3}}{2} \) and \( u_{\text{min}} = -\frac{3 \sqrt{3}}{2} \). Therefore, \( u_{\text{max}} + u_{\text{min}} = 0 \).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 (1) In a $4 \times 4$ grid paper, some small squares are colored red, and then two rows and two columns are crossed out. If no matter how they are crossed out, there is at least one red small square that is not crossed out, how many small squares must be colored at least?
(2) If the β$4 \times 4$β grid paper in (1) is changed to β$n \times n$β $(n \geqslant 5)$ grid paper, with other conditions unchanged, how many small squares must be colored at least?
|
Solution: (1) If the number of colored small squares is less than or equal to 4, then we can appropriately strike out two rows and two columns to remove all the colored small squares.
If the number of colored small squares is 5, then by the pigeonhole principle, there must be at least one row with 2 or more colored small squares. Striking out this row, the remaining number of colored small squares does not exceed 3. Then, striking out one more row and two columns can remove all the colored small squares.
If the number of colored small squares is 6, then there must be at least one row with 3 colored small squares or two rows each with 2 colored small squares. Therefore, striking out two rows can remove at least 4 colored small squares, leaving no more than 2 colored small squares. Then, striking out two columns can remove them all.
Thus, the number of colored small squares must be greater than or equal to 7.
Furthermore, if the squares are colored as shown in Figure 1, then striking out any two rows and two columns cannot remove all the colored small squares.
Therefore, at least 7 small squares must be colored.
(2) If the number of colored small squares is less than or equal to 4, then striking out two rows and two columns can definitely remove all of them.
Furthermore, if the squares are colored as shown in Figure 2, then striking out any two rows and two columns cannot remove all the colored small squares.
Therefore, at least 5 small squares must be colored.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $P(x)$ be the product of the digits in the decimal representation of $x$. Try to find all positive integers $x$ such that
$$
P(x)=x^{2}-10 x-22
$$
holds.
|
Solution: Let $x=\overline{a_{n} a_{n-1} \cdots a_{1}}$.
When $n \geqslant 2$, according to (1) and (2), we have the estimate
$$
x \geqslant 10^{n-1} \cdot a_{n}>9^{n-1} \cdot a_{n} \geqslant P(x) \text {, }
$$
thus $x^{2}-10 x-22<x$.
Solving this, we get $0<x<13$.
Upon verification, $x=12$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 On a plane, there are 7 points, and some line segments can be connected between them, so that any three points among the 7 points must have two points connected by a line segment. How many line segments are needed at least? Prove your conclusion.
|
Proof: First, to meet the requirements of the problem, the number of line segments to be connected must be greater than or equal to 9.
Below, we discuss in 4 cases.
(1) If among the 7 points, there exists 1 point not connected to any other points, by the problem's condition, the remaining 6 points must each be connected to every other point, in this case, at least $\frac{6 \times 5}{2}=15$ line segments need to be connected.
(2) If among the 7 points, there is 1 point that only connects to 1 line segment, then the remaining 5 points must each be connected to every other point, in this case, there are at least $\frac{5 \times 4}{2}+1=11$ line segments.
(3) If each point connects to at least 2 line segments, and there is 1 point that exactly connects to 2 line segments, let's assume this point is $A$, and the 2 line segments it connects to are $A B$ and $A C$, then the 4 points not connected to point $A$ must each be connected to every other point, requiring $\frac{4 \times 3}{2}=6$ line segments. And point $B$ must connect to at least 2 line segments, so apart from $B A$, there must be at least 1 more, thus, in this case, at least $6+2+1=9$ line segments need to be connected.
(4) If each point connects to at least 3 line segments, then at least $7 \times 3 \div 2>10$ line segments need to be connected.
In summary, the number of line segments in the graph is greater than or equal to 9.
As shown in Figure 3, an example of construction is given, which indicates that 9 line segments are sufficient.
In conclusion, the minimum number of line segments to be connected is 9.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. (25 points) $a$, $b$, $c$ are positive integers, and $a^{2}+b^{3}=$ $c^{4}$. Find the minimum value of $c$.
δΏηδΊζΊζζ¬ηζ’θ‘εζ ΌεΌγ
|
3. Clearly, $c>1$. From $b^{3}=\left(c^{2}-a\right)\left(c^{2}+a\right)$, if we take $c^{2}-a=b, c^{2}+a=b^{2}$, then at this time
$$
c^{2}=\frac{b(b+1)}{2} \text {. }
$$
Examining $b$ from small to large to make the right side a perfect square, we find that when $b=8$, $c=6$, and thus, $a=28$.
Next, we show that there is no smaller positive integer solution for $c$, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|}
\hline$c$ & $c^{4}$ & Cubes less than $c^{4}$ \\
\hline 2 & 16 & 1,8 \\
\hline 3 & 81 & $1,8,27,64$ \\
\hline 4 & 256 & $1,8,27,64,125,216$ \\
\hline 5 & 625 & $1,8,27,64,125,216,343,512$ \\
\hline
\end{tabular}
In each row of Table 1, $c^{4}-x^{3}$ is never a perfect square, therefore, the smallest value of $c$ is 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and its smallest positive period is 2. Then the value of $f(-1)$ is $\qquad$ .
|
1.0.
Since $f(x+2)=f(x)$, let $x=-1$, then we have $f(-1+2)=f(-1)$,
which means $f(1)=f(-1)=-f(1)$.
Therefore, $f(-1)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find $\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}$.
Analysis: Considering the use of identity (III), we can obtain
$$
\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \text {. }
$$
Let $l=2 n-k-1$, and combining with equation (III) again yields $\frac{l+1}{\mathrm{C}_{2 n}^{l}}$, creating a cyclic phenomenon, which leads to the solution of the problem.
|
Let the original expression be $y_{n}$, from the analysis we have
$$
y_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{C_{2 n}^{2 n-k-1}} \text {. }
$$
By making a change of the summation index, set $l=2 n-$ $k-1$, when $k$ varies from 0 to $2 n-1$, $l$ varies from $2 n-1$ to 0. Thus,
$$
\begin{array}{l}
y_{n}=\sum_{l=0}^{2 n-1}(-1)^{2 n-l} \frac{l+1}{\mathbf{C}_{2 n}^{l}} \\
=\sum_{l=0}^{2 n-1}(-1)^{l} \frac{l+1}{\mathbf{C}_{2 n}^{l}}=-y_{n} .
\end{array}
$$
Therefore, $y_{n}=0$, that is,
$$
\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}=0 .
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 1, in rectangle $A B C D$, $A B=2, B C=$ $\sqrt{3}, E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Line segments $D E$, $B F$, and $E F$ intersect diagonal $A C$ at points $M$, $N$, and $P$ respectively. The number of right triangles formed by the line segments in Figure 1 is ( ).
(A) 8
(B) 12
(C) 16
(D) 20
|
4.B.
Let the notation $[A]$ represent the number of right-angled triangles with $A$ as the right-angle vertex, then $[A]=2$,
Similarly, $[B]=1,[C]=2,[D]=1,[E]=3$,
$$
[F]=3,[M]=[N]=[P]=0 \text {. }
$$
Therefore, the number of right-angled triangles is
$$
2+1+2+1+3+3=12 \text {. }
$$
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Fill in $n$ distinct numbers on a circle so that for every three consecutive numbers, the middle number is equal to the product of the two numbers on its sides. Then $n=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2.6.
According to the problem, $n \geqslant 3$. Take any two adjacent numbers $a, b(a \neq b)$, and by the problem's condition, the numbers on the circle can be written in sequence as
$$
a, b, \frac{b}{a}, \frac{1}{a}, \frac{1}{b}, \frac{a}{b}, a, b, \cdots
$$
Starting from the 7th number, the sequence of numbers has already entered a cycle, so $n \leqslant 6$.
Next, we show that $n$ cannot be $3, 4, 5$.
If $n=3$, then there are only 3 numbers on the circle, so in the above sequence, we should have $a=\frac{1}{a}, b=\frac{1}{b}$, thus,
$$
a= \pm 1, b= \pm 1, \frac{b}{a}= \pm 1 \text {. }
$$
Therefore, $a, b, \frac{b}{a}$ must have two that are equal, which is a contradiction.
If $n=4$, then there are only 4 numbers on the circle (although the sequence of numbers can be formally written, the subsequent numbers actually repeat the first four numbers), so $a=\frac{1}{b}$, i.e., $b=\frac{1}{a}$. Therefore, the second and fourth numbers among the first four numbers are already equal, which is a contradiction.
If $n=5$, then there are 5 numbers on the circle, so $a=\frac{a}{b}$, i.e., $b=1$. Therefore, the $b$ and $\frac{1}{b}$ among the first five numbers are both equal to 1, which is a contradiction.
Therefore, only $n=6$.
On the other hand, when $n=6$, we can specifically give 6 numbers:
$$
2,3, \frac{3}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3} \text {. }
$$
|
6
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
10.1. Try to find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers.
|
10.1.11.
From the problem, we have
$$
\begin{array}{l}
1=\frac{4-2}{4-2}, 3=\frac{8-2}{4-2}, 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}, \\
7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}, \\
2=2 \times 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \times 5=\frac{2^{6}-2^{2}}{2^{3}-2} .
\end{array}
$$
Assume $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$.
Without loss of generality, we can assume $a>b, c>d$. Let $m=a-b$, $n=c-d$, $k=b-d$. Then, we have
$$
11\left(2^{n}-1\right)=2^{k}\left(2^{m}-1\right) \text {. }
$$
The left side of the above equation is odd, so $k=0$.
It is easy to see that $n=1$ cannot satisfy the equation. If $m>n>1$, then $2^{n}-1$ and $2^{m}-1$ both have a remainder of 3 when divided by 4, thus, the left side of the equation has a remainder of 1 when divided by 4, while the right side has a remainder of 3. This is a contradiction.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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