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1. Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha$ $>\beta$. Without solving the equation, use the relationship between roots and coefficients to find the value of $\frac{2}{\alpha}+3 \beta^{2}$.
(8th "Zu Chongzhi Cup" Mathematics Invitational Competition) | (Tip: Construct a conjugate. Answer: $\frac{1}{8}(403-85 \sqrt{17})$ ) | \frac{1}{8}(403-85 \sqrt{17}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,680 |
2. In the plane $\alpha$ there is $\triangle A B C, \angle A B C=60^{\circ}, A C=\sqrt{3}$. On both sides of the plane $\alpha$ there are points $S$ and $T$, satisfying $S A=S B=S C$ $=2, T A=T B=T C=3$. Then the length of $S T$ is $\qquad$ | $2 \cdot \sqrt{3}+2 \sqrt{2}$ | 2 \cdot \sqrt{3}+2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,682 |
4. The function $f(x)=\left|x^{2}-a\right|$ has a maximum value $M(a)=$ $\qquad$ on the interval $[-1,1]$ | 4. $\left\{\begin{array}{ll}1-a, & \text { when } a \leqslant \frac{1}{2} \text {, } \\ a, & \text { when } a>\frac{1}{2} \text { }\end{array}\right.$ | \left\{\begin{array}{ll}1-a, & \text { when } a \leqslant \frac{1}{2} \text {, } \\ a, & \text { when } a>\frac{1}{2} \text { }\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,684 |
5. Given that $\triangle A B C$ is an equilateral triangle with side length 5, and $P$ is a point inside it such that $P A=4, P B=3$. Then the length of $P C$ is $\qquad$. | 5. $\sqrt{25-12 \sqrt{3}}$ | \sqrt{25-12 \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,685 |
10. If for all positive real numbers $x, y$, we always have
$$
\frac{x y}{\sqrt{\left(x^{2}+y^{2}\right)\left(3 x^{2}+y^{2}\right)}} \leqslant \frac{1}{k} \text {. }
$$
then the maximum value of $k$ is $\qquad$ | $10.1+\sqrt{3}$ | 1+\sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,690 |
2. Given that $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{1}{3} y^{2}=25, \\
z^{2}+\frac{1}{3} y^{2}=9, \\
x^{2}+z x+z^{2}=16 .
\end{array}\right.
$$
Find the value of $x y+2 y z+3 x z$.
(17th All-Russian High School Mathematics Competition) | (Hint: Construct a triangle. Answer: $24 \sqrt{3}$ ) | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,691 |
II. (16 points) Given that $a$ and $b$ are real numbers, and $\mathrm{i}$ is the imaginary unit, the quadratic equation in $z$ is
$$
4 z^{2}+(2 a+\mathrm{i}) z-8 b(9 a+4)-2(a+2 b) \mathrm{i}=0
$$
has at least one real root. Find the maximum value of this real root. | Let $x$ be the real root of a known quadratic equation, then
$$
\begin{array}{l}
4 x^{2}+2 a x-8 b(9 a+4)+[x-2(a+2 b)] \mathrm{i}=0 \\
\Leftrightarrow\left\{\begin{array}{l}
4 x^{2}+2 a x-8 b(9 a+4)=0, \\
x-2(a+2 b)=0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
5 a^{2}+16\left(b-\frac{1}{4}\right)^{2}=... | \frac{3 \sqrt{5}}{5} + 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,692 |
Three, (16 points) Given real numbers $a, b, c$ satisfy $a+b+c=2$, $abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$. | (1) Let's assume $a=\max \{a, b, c\}$. From the given conditions, we have
$$
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^{... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,693 |
1. If $\left[x+\frac{1}{100}\right]+\left[x+\frac{2}{100}\right]+\cdots+\left[x+\frac{99}{100}\right]+$ $[x+1]=356$, where $[x]$ denotes the greatest integer not greater than $x$, then $x$ can take the value ( ).
(A) 3.54
(B) 3.55
(C)3.44
(D) 3.45 | 1.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "1.B" is already in English and does not require translation, here is the same text in English:
1.B | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,695 |
3. Positive integers are arranged in ascending order: $1, 2, 3, \cdots$. After removing the perfect squares and perfect cubes from the sequence, a new sequence is formed without changing the order. What is the 200th number in the new sequence?
(A) 216
(B) 217
(C) 218
(D) 219 | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,697 |
4. Given $x+y+x y+1=0, x^{2} y+x y^{2}+2=0$. Then $(x-y)^{2}$ equals ( ).
(A) 1 or 8
(B) 0 or 9
(C) 4 or 8
(D) 4 or 9 | 4.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,698 |
5. As shown in Figure 2, in $\triangle A B C$, $E$ and $D$ are points on sides $A B$ and $A C$ respectively, $B D$ and $C E$ intersect at $F$, and the extension of $A F$ intersects $B C$ at point $H$. If $\angle 1=\angle 2, A E=$ $A D$, then the number of pairs of congruent triangles in Figure 2 is ( ).
(A) 3
(B) 5
(C)... | 5.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,699 |
3. Given $a \neq 0$, and $a^{2}+5 a=\frac{1}{b}+\frac{5}{\sqrt{b}}=1$. Then the value of the algebraic expression $\frac{a^{3} b \sqrt{b}+1}{b \sqrt{b}}$ is $\qquad$ .
(2003, Shanghai Junior High School Mathematics Competition) | (Tip: Formulate an equation. Answer: -140 )
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,702 |
1. Observe the following expressions:
$$
97^{2}=9409,997^{2}=994009,9997^{2}=99940009 \text {. }
$$
Conjecture $999998^{2}=$ $\qquad$ | $1.999996000004$ | 999996000004 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,703 |
1. (16 points) Given that $m, n$ are integers, and satisfy $2 m^{2}+n^{2}+$ $3 m+n-1=0$. Find $m, n$.
| Three, 1. Taking $m$ as the variable, we get the quadratic equation in $m$: $2 m^{2}+3 m+n^{2}+n-1=0$.
Since $m$ has integer solutions, we have
$$
\Delta=9-8\left(n^{2}+n-1\right)=17-8 n(n+1) \geqslant 0
$$
and it must be a perfect square.
Also, since $n$ is an integer, we consider the following cases:
(1) When $n=0$ ... | m=-1, n=1 \text{ or } m=-1, n=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,710 |
2. (16 points) In a school, the ratio of male to female students in the first year of junior high school is $8: 7$. After one year, 40 new students are transferred in, and the ratio of male to female students becomes 17:15. By the third year of junior high school, some students have transferred out, and some new studen... | 2. Let the total number of new students in Grade 7 be $15 a$ people, the total number of students in Grade 8 be $32 b$ people, and the total number of students in Grade 9 be $13 c$ people, where $a$, $b$, and $c$ are integers. From the problem, we have
$$
\left\{\begin{array}{l}
15 a+40=32 b, \\
15 a+50=13 c .
\end{arr... | 320 \text{ boys, } 280 \text{ girls} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,711 |
3. (18 points) As shown in Figure 7, in $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$, $C D$ and $C_{1} D_{1}$ are the angle bisectors of $\angle A C B$ and $\angle A_{1} C_{1} B_{1}$, respectively, and $C D=C_{1} D_{1}, A B=A_{1} B_{1}, \angle A D C=\angle A_{1} D_{1} C_{1}$. Can you determine whether $\triangle... | 3. $\triangle A B C$ can be congruent to $\triangle A_{1} B_{1} C_{1}$.
As shown in Figure 8, extend $D A$ to $E$ such that $A E=B D$, and extend $D_{1} A_{1}$ to $E_{1}$ such that $A_{1} E_{1}=B_{1} D_{1}$.
Since $A B=D E=A_{1} B_{1}=D_{1} E_{1}$,
$\angle A D C=\angle A_{1} D_{1} C_{1}, D C=D_{1} C_{1}$,
therefore, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,712 |
4. If $S=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\cdots+\frac{1}{2001}}$, then the integer part of $S$ is $\qquad$
(2001, Shandong Province Junior High School Mathematics Competition) | (Tip: Construct an inequality. Answer: 90)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,713 |
1. If $0<a<1,-2<b<-1$, then the value of $\frac{|a-1|}{a-1}-\frac{|b+2|}{b+2}+\frac{|a+b|}{a+b}$ is ( ).
(A) 0
(B) -1
(C) -2
(D) -3 | - I. D. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,714 |
2. Let $\alpha, \beta$ be the two real roots of the equation $2 x^{2}-3|x|-2=0$. Then the value of $\frac{\alpha \beta}{|\alpha|+|\beta|}$ is ( ).
(A) -1
(B) 1
(C) $-\frac{2}{3}$
(D) $\frac{2}{3}$ | 2. A.
The original equation is $(2|x|+1)(|x|-2)=0$.
Since $2|x|+1>0,|x|-2=0$, therefore, $x= \pm 2$. Hence, $\frac{a \beta}{|\alpha|+|\beta|}=\frac{2 \times(-2)}{|2|+|-2|}=-1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,715 |
3. As shown in Figure 1, in $\triangle A B C$, $\angle A=60^{\circ}, \angle C=80^{\circ}$, the angle bisector of $\angle C$ intersects the external angle bisector of $\angle A$ at point $D$, and $B D$ is connected. Then the value of $\tan \angle B D C$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) $\sqrt{3}$
(D) $\frac{\sqrt{3}}... | 3.D.
It is known that $D$ lies on the external angle bisector of $\angle B$.
Given $\angle A B C=40^{\circ}$, we have $\angle D B A=70^{\circ}, \angle D B C=110^{\circ}$.
Also, $\angle B C D=40^{\circ}, \angle B D C=30^{\circ}$, so $\tan \angle B D C=\frac{\sqrt{3}}{3}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,716 |
4. There is a kind of football made of 32 pieces of black and white cowhide, with black pieces being regular pentagons and white pieces being regular hexagons, and all sides being of equal length (as shown in Figure 2). Then the number of white pieces is ( ).
(A) 22
(B) 20
(C) 18
(D) 16 | 4.B.
Let the number of white pieces be $x$, then the number of black pieces is $32-x$.
Since the black pieces are pentagons, the total number of edges of the black pieces is $5(32-x)$.
Also, because each white piece has 3 edges connected to black pieces, the total number of edges of the black pieces can also be expre... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,717 |
5. As shown in Figure 3, in $\triangle A B C$, $\angle C=90^{\circ}, A C=\sqrt{11}$, $B C=5$, a circle is drawn with $C$ as the center and $B C$ as the radius, intersecting the extension of $B A$ at point $D$. Then the length of $A D$ is ( ).
(A) $\frac{3}{7}$
(B) $\frac{5}{7}$
(C) $\frac{7}{3}$
(D) $\frac{5}{3}$ | 5.C.
As shown in Figure 8, extend $AC$ to intersect the circle at $E$ and $F$, then
$$
\begin{array}{l}
A F=5+\sqrt{11}, \\
A E=5-\sqrt{11} .
\end{array}
$$
It is easy to calculate that $A B=6$. By the intersecting chords theorem $A D \cdot A B=A E \cdot A F$
we get
$$
A D=\frac{A E \cdot A F}{A B}=\frac{(5-\sqrt{11}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,718 |
6. Given the equation $\sqrt{x}+3 \sqrt{y}=\sqrt{300}$. The number of positive integer solutions to this equation is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 6.C.
Since $\sqrt{300}=10 \sqrt{3}, x, y$ are positive integers, thus $\sqrt{x}, \sqrt{y}$ when simplified to their simplest radical form should be of the same type as $\sqrt{3}$.
There can only be the following three cases: | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,719 |
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)... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,721 |
3. In trapezoid $A B C D$; $A D / / B C, A C 、 B D$ intersect at point $O, \triangle A O D$ and $\triangle A O B$ have areas of 9 and 12, respectively. Then the area of trapezoid $A B C D$ is $\qquad$ | 3. 49.
As shown in Figure 10, since
$$
\begin{array}{l}
\frac{S_{\triangle A O D}}{S_{\triangle A U B}}=\frac{O D}{O B}=\frac{O A}{O C} \\
=\frac{S_{\triangle A O B}}{S_{\triangle B C C}},
\end{array}
$$
thus $S_{\triangle A B B}^{2}=S_{\triangle M D} \cdot S_{\triangle B O C}$,
which means $S_{\triangle B O C}=\frac... | 49 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,722 |
4. Let the quadratic function $y=x^{2}+2 a x+\frac{a^{2}}{2}(a<0)$ have its vertex at $A$, and its intersections with the $x$-axis at $B$ and $C$. When $\triangle A B C$ is an equilateral triangle, the value of $a$ is . $\qquad$ | 4. $-\sqrt{6}$.
The graph of the quadratic function $y=x^{2}+2 a x +\frac{a^{2}}{2}(a<0)$ is shown in Figure 11. Let $A\left(-a,-\frac{a^{2}}{2}\right)$, $B\left(x_{1}, 0\right)$, $C\left(x_{2}, 0\right)$. Denote the intersection of the axis of symmetry with the $x$-axis as $D$, then
$$
\begin{array}{l}
|B C|=\left|x_... | -\sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,723 |
Example 1 Let $k \in \mathbf{N}_{+}, I_{k}=(2 k-1,2 k+1]$. Find the range of $a$ such that the equation $(x-2 k)^{2}=a x$ has two distinct real roots in $I_{k}$. | Solution: Rearrange the equation $(x-2 k)^{2}=a x$, we get
$$
x^{2}-(4 k+a) x+4 k^{2}=0 \text {. }
$$
Since the equation has two distinct real roots in $I_{k}$, there is the system of inequalities
$$
\left\{\begin{array}{l}
\Delta=(4 k+a)^{2}-16 k^{2}>0, \\
x_{1}=\frac{4 k+a+\sqrt{\Delta}}{2} \leqslant 2 k+1, \\
2 k-1... | 0<a \leqslant \frac{1}{2 k+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,724 |
Example 2 When does $a$ satisfy the condition that the equation
$$
\left(a^{2}-1\right) x^{2}-6(3 a-1) x+72=0
$$
has roots of opposite signs and the absolute value of the negative root is larger? | The equation $\left(a^{2}-1\right) x^{2}-6(3 a-1) x+72=0$ has roots of opposite signs with the negative root having a larger absolute value under the conditions:
$$
\left\{\begin{array}{l}
a^{2}-1 \neq 0, \\
\Delta=36(3 a-1)^{2}-4\left(a^{2}-1\right) \times 72>0, \\
x_{1}+x_{2}=\frac{6(3 a-1)}{a^{2}-1}<0, \\
x_{1} x_{2... | \frac{1}{3}<a<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,725 |
4. For the equation $x^{2}+(2 m-1) x+(m-6)=0$, one root is no greater than -1, and the other root is no less than 1. Try to find:
(1) the range of values for the parameter $m$;
(2) the maximum and minimum values of the sum of the squares of the roots.
(9th Junior High School Mathematics Competition, Jiangsu Province) | (Hint: (2) Convert $x_{1}^{2}+x_{2}^{2}$ into a quadratic function of $m$, then find the maximum and minimum values. Answer: (1) $-4 \leqslant m \leqslant 2$ (2) Maximum value 101, Minimum value $\frac{43}{4}$ ) | -4 \leqslant m \leqslant 2, \text{ Maximum value } 101, \text{ Minimum value } \frac{43}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,726 |
3. If $a>b>0$, then the minimum value of $a^{3}+\frac{1}{b(a-b)}$ is $\qquad$ .
| $$
\begin{array}{l}
\text { 3. } \frac{5}{3} \sqrt[5]{144} . \\
a^{3}+\frac{1}{b(a-b)} \geqslant a^{3}+\frac{4}{a^{2}}=2 \times \frac{a^{3}}{2}+3 \times \frac{4}{3 a^{2}} \\
\geqslant 5 \sqrt[5]{\left(\frac{1}{2}\right)^{2} \times\left(\frac{4}{3}\right)^{3}}=\frac{5}{3} \sqrt[5]{144} .
\end{array}
$$
The equality hol... | \frac{5}{3} \sqrt[5]{144} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,727 |
4. If the sum of the squares of two pairs of opposite sides of a spatial quadrilateral are equal, then, the angle formed by its two diagonals is | 4. $90^{\circ}$.
In Figure 2, $ABCD$ is a spatial quadrilateral. $AC$ and $BD$ are its diagonals. The midpoints of $AB$, $BC$, $CD$, $DA$, $BD$, and $AC$ are $M$, $N$, $G$, $P$, $K$, and $H$, respectively. Using the property that the sum of the squares of the sides of a parallelogram equals the sum of the squares of i... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,728 |
5. Point $P$ is equidistant from point $A(1,0)$, $B(a, 2)$, and the line $x=-1$. If such a point exists and is unique, then the value of $a$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. $\pm 1$.
Let the coordinates of point $P$ be $P(x, y)$. Given that the distance from $P$ to $A(1,0)$ and the line $x=-1$ are equal, the locus of point $P$ is $y^{2}=4 x$.
Also, $|P A|=|P B|$, thus
$$
\sqrt{(x-1)^{2}+y^{2}}=\sqrt{(x-a)^{2}+(y-2)^{2}},
$$
which simplifies to $2(1-a) x-4 y+3+a^{2}=0$.
The problem is ... | \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,729 |
6. In the Cartesian coordinate system, the set of points $(x, y)$ that satisfy $x<y$, $|x|<3$, $|y|<3$, and make the equation
$$
\left(x^{3}-y^{3}\right) t^{4}+(3 x+y) t^{2}+\frac{1}{x-y}=0
$$
have no real roots is denoted as $N$. Then the area of the region formed by the point set $N$ is $\qquad$ . | 6. $\frac{81}{5}$.
Let $u=t^{2}$, the original equation transforms into
$$
\begin{array}{l}
\left(x^{3}-y^{3}\right) u^{2}+(3 x+y) u+\frac{1}{x-y}=0 . \\
\Delta=(3 x+y)^{2}-4\left(x^{3}-y^{3}\right) \cdot \frac{1}{x-y} \\
=5 x^{2}+2 x y-3 y^{2}=(5 x-3 y)(x+y) .
\end{array}
$$
The given equation has no real roots if a... | \frac{81}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,730 |
Three. (20 points) Given the equation $17 x^{2}-16 x y+4 y^{2}$ $-34 x+16 y+13=0$ represents an ellipse in the $xy$ plane. Find its center of symmetry and axes of symmetry.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, it is known that points $A(1,1), B(1,-1)$ are on the curve.
Let the center of symmetry of the ellipse be $(a, b)$, then points $A^{\prime}(2 a-1, 2 b-1), B^{\prime}(2 a-1,2 b+1)$ are also on the curve. Therefore, we have
$$
\begin{array}{l}
17(2 a-1)^{2}-16(2 a-1)(2 b-1)+4(2 b-1)^{2}- \\
34(2 a-1)+16(2 b-1)+13=0... | y=\frac{13 \pm 5 \sqrt{17}}{16}(x-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,731 |
Four. (20 points) Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a rectangular prism, the perpendiculars from point $A$ to the lines $A_{1} B$, $A_{1} C$, and $A_{1} D$ intersect the lines $A_{1} B_{1}$, $A_{1} C_{1}$, and $A_{1} D_{1}$ at points $M$, $N$, and $P$, with the feet of the perpendiculars being $E$, $G$, a... | Solution 1: Establish a spatial rectangular coordinate system as shown in Figure 4. Let $A D=a$, $A B=b$, and $A A_{1}=c$. Then the coordinates of the vertices of the rectangular prism are:
$$
A(a, 0,0) \text {, }
$$
$B(a, b, 0)$, $C(0, b, 0)$, $D(0,0,0)$, $A_{1}(a, 0, c)$, $B_{1}(a, b, c)$, $C_{1}(0, b, c)$, $D(0,0, c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,732 |
Five. (20 points) Given
$$
f(x)=\frac{x^{4}+k x^{2}+1}{x^{4}+x^{2}+1}, k, x \in \mathbf{R} \text {. }
$$
(1) Find the maximum and minimum values of $f(x)$;
(2) Find all real numbers $k$ such that for any three real numbers $a, b, c$, there exists a triangle with side lengths $f(a)$, $f(b)$, and $f(c)$. | Five, $f(x)=1+\frac{(k-1) x^{2}}{x^{4}+x^{2}+1}$.
(1) Since $x^{4}+1 \geqslant 2 x^{2}$. Then $x^{4}+x^{2}+1 \geqslant 3 x^{2}$. Therefore,
$0 \leqslant \frac{x^{2}}{x^{4}+x^{2}+1} \leqslant \frac{1}{3}$.
When $k \geqslant 1$, $f(x)_{\text {max }}=\frac{k+2}{3}, f(x)_{\text {min }}=1$;
When $k<1$, $f(x)_{\text {max }}=... | -\frac{1}{2}<k<4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,733 |
One, (50 points) The incircle $\odot O$ of $\triangle ABC$ touches $AB$ at $D, S_{\triangle ABC} = AD \cdot DB$ if and only if $\angle C = 90^{\circ}$. | Given the figure $6, \odot 0$ is tangent to
$AB$ at $D$, let $AB=c, BC=a, CA=b$. It is easy to know that
$$
\begin{aligned}
& AD=\frac{b+c-a}{2}, \\
& DB=\frac{a+c-b}{2}. \\
& \text{Therefore, } S_{\triangle ABC}=AD \cdot DB \\
\Leftrightarrow & \frac{1}{4} \sqrt{(a+b+c)(a+b-c)(b+c-a)(a+c-b)} \\
= & \frac{1}{4}(b+c-a)(... | c^{2}=a^{2}+b^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 714,734 |
II. (50 points) Find the maximum value of the area of an inscribed triangle in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. | As shown in Figure 7, let \( A\left(a \cos \theta_{1}, b \sin \theta_{1}\right) \), \( B\left(a \cos \theta_{2}, b \sin \theta_{2}\right) \), \( C\left(a \cos \theta_{3}, b \sin \theta_{3}\right) \), where \( 0 \leqslant \theta_{1} < \theta_{2} < \theta_{3} < 2 \pi \). Other cases are similar.
Draw the line \( l: y = -... | \frac{3 \sqrt{3}}{4} a b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,735 |
Three, (50 points) The sequence of integers $a_{0}, a_{1}, a_{2}, a_{3}, \cdots$ satisfies $a_{0}=1$, and for each positive integer $n$, there is $a_{n+1} a_{n-1}$ $=k a_{n}$, where $k$ is a positive integer. If $a_{2000}=2000$, find all possible values of $k$.
保留源文本的换行和格式，直接输出翻译结果如下：
```
Three, (50 points) The seque... | Let $a_{1}=a$, from the problem we know $a_{2}=k a_{1}=k a$.
If $a=0$, then $a_{2}=0$. Also, $a_{2} a_{4}=k a_{3}$, so $a_{3}=0$. By induction, we can get
$a_{n}=0\left(n \in \mathbf{N}_{+}\right)$. This contradicts $a_{2000}=2000$.
Therefore, $a \neq 0$. From the problem, we have
$$
a_{2}=k a, a_{3}=k^{2}, a_{4}=\frac... | 100,200,400,500,1000,2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,736 |
5. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .
(2003, National Junior High School Mathematics League) | (Solution: $m=4$ ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,737 |
As shown in Figure 1, given that $\triangle ABC$ is inscribed in $\odot O$, $AB > AC$, and chord $EF \parallel BC$. The extensions of $FA$ and $EA$ intersect the extension of $BC$ at points $P$ and $Q$ respectively, and $AD$ is the angle bisector of $\angle BAC$. Prove:
$$
\frac{1}{CD} - \frac{1}{BD} = \frac{1}{DP} + \... | Proof: As shown in Figure 2, extend $BA$ to $G$, and draw the tangent line $MN$ of $\odot O$ through point $A$.
Obviously, $\angle GAQ = \angle EAB$.
Since $\angle EAB = \frac{1}{2} \overparen{EB^{\circ}}$, therefore, $\angle GAQ = \frac{1}{2} \overparen{EB^{\circ}}$.
By the tangent-chord angle theorem, we have $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,738 |
As shown in Figure 3, $\angle A B C=\angle A E F=90^{\circ}$, $\angle B A C=\angle E A F$ and points $A$, $C$, and $F$ are collinear. $B F$ intersects ray $E C$ at point $P$. Prove that $A P \perp B E$. | Proof: As shown in Figure 4, ray $BC$ intersects $EF$ at $M$. Through $F$, draw lines parallel to $AB$ and $AE$ intersecting ray $AP$ and $AM$ at $K$ and $L$, respectively. Through $B$, draw a perpendicular to $AC$ intersecting $AC$, $AM$, and $AE$ at $S$, $Y$, and $G$, respectively. $AF$ intersects $BE$ at $X$, and $A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,739 |
Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$, we have
$$
f(x f(y))=(1-y) f(x y)+x^{2} y^{2} f(y) .
$$ | Solution: Clearly, $f(x) \equiv 0$ satisfies equation (1). Now assume $f(x) \neq 0$.
Substitute $y=1$ into (1), we get
$$
f(x f(1))=x^{2} f(1).
$$
Thus, $f(0)=0$.
Assume $f(1) \neq 0$, take $x=\frac{1}{f(1)}$, then $f(1)=\frac{1}{f(1)}$.
Solving this, we get $f(1)=1$ or $f(1)=-1$.
Therefore, $f(x)=x^{2}$ or $f(x)=-x^{... | f(x) \equiv 0 \text{ or } f(x)=x-x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,740 |
142 As shown in Figure 5, from a point $P$ outside a circle, two secants $PAB$ and $PCD$ are drawn, intersecting the circle at $A, B, C, D$ respectively. Chords $AD$ and $BC$ intersect at point $Q$. A secant $PEF$ passing through point $Q$ intersects the circle at $E, F$. Prove:
$$
\frac{1}{P E}+\frac{1}{P F}=\frac{2}{... | Proof: As shown in Figure 6, draw tangents $PM$ and $PN$ from $P$ to the circle, with $M$ and $N$ as the points of tangency. Let $MN$ intersect $AB$ at point $S$, $BN$ intersect $AD$ at point $H$, and $AN$ intersect $BC$ at point $K$. Then,
Similarly, $\frac{AK}{KN} = \frac{AC \sin \angle ACK}{NC \sin \angle NCK}, \fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,741 |
Example 1 Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the first term $a_{1}$ and common difference $d$. The sequence $\left\{a_{n}\right\}$ is grouped according to the rule of $3k$ numbers in the $k$-th group as follows:
$$
\left(a_{1}, a_{2}, a_{3}\right),\left(a_{4}, a_{5}, \cdots, a_{9}\right),\left(a_... | Explanation: If $a_{n}$ is in the $k$-th group, then the first $k-1$ groups have
$$
3+6+9+\cdots+3(k-1)=\frac{3 k(k-1)}{2}
$$
terms. Therefore,
$$
\frac{3 k(k-1)}{2}+1 \leqslant n0 .\end{array}\right.$$
Solving this system of equations yields
$$
-\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}<k \leqslant \frac{1}{2}+\frac{\sqrt... | k a_{1}+\frac{3 k\left(k^{2}+2 k-1\right) d}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,742 |
Example 2 Let $\left\{a_{n}\right\}$ be the sequence of all numbers in the set $\left\{2^{s}+2^{t} \mid 10 \leqslant s<t\right.$ and $s, t \in \mathbf{Z}\}$ arranged in ascending order, i.e., $\square$
$$
\begin{aligned}
& a_{1}=3, a_{2}=5, a_{3}=6, a_{4}=9, a_{5}=10, \\
a_{6}= & 12, \cdots .
\end{aligned}
$$
Arrange ... | (1) Arrange the above triangular number table into a right-angled triangular number table:
$$
\begin{array}{l}
a_{1}=2^{0}+2^{1} \\
a_{2}=2^{0}+2^{2} \quad a_{3}=2^{1}+2^{2} \\
a_{4}=2^{0}+2^{3} \quad a_{5}=2^{1}+2^{3} \quad a_{6}=2^{2}+2^{3} \\
a_{7}=2^{0}+2^{4} \quad a_{8}=2^{1}+2^{4} \quad a_{9}=2^{2}+2^{4} \quad a_... | 16640 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,743 |
Example 3 The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=a_{n}+$ $\frac{1}{\left[a_{n}\right]}$, where $\left[a_{n}\right]$ denotes the greatest integer not exceeding $a_{n}$. Find the general term formula of the sequence $\left\{a_{n}\right\}$. | This is a problem I devised in 2002. At first glance, it seems unrelated to grouped sequences, but if you write out the first few terms and arrange them in a right-angled triangular table, you will find it similar to Example 2.
$$
\begin{array}{l}
a_{1}=\frac{0}{1}+1 \\
a_{2}=\frac{0}{2}+2 \quad a_{3}=\frac{1}{2}+2 \\
... | a_{n}=\frac{n-1}{k}+\frac{k+1}{2}, \text{ where } k=\left[\frac{1+\sqrt{8 n-7}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,744 |
Example 4 Figure 1 is a chessboard that can extend infinitely to the right and downward, with horizontal rows and vertical columns. Fill in the squares with positive integers according to the pattern shown in the $4 \times 4$ grid already filled.
(1) Find the number in the square located at the 3rd row and 8th column;
... | Explanation: Grouping diagonally from top-right to bottom-left:
(1), (2,3), (4,5,6), ,
The $k$-th group has $k$ numbers, and the column of the first number in this group is the $k$-th column. The first $k-1$ groups have
$$
1+2+3+\cdots+(k-1)=\frac{k(k-1)}{2}
$$
numbers.
Let the number in the $i$-th row and $j$-th col... | 48, (21, 5), 2n^2-2n+1, \frac{n(2n^2+1)}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,745 |
Example 5 Delete all perfect squares from the sequence of positive integers $1,2,3, \cdots$, to get a new sequence $\left\{a_{n}\right\}$. Then the 2003rd term of $\left\{a_{n}\right\}$ is ( ).
(A) 2046
(B) 2047
(C) 2048
(D) 2049
(2003, National High School Mathematics Competition) | Explanation: We solve a more general problem, finding the general term formula of the new sequence $\left\{a_{n}\right\}$.
Since $(k+1)^{2}-k^{2}-1=2 k, k \in \mathbf{Z}_{+}$, in the sequence of positive integers $1,2,3, \cdots$, there are $2 k$ numbers between $k^{2}$ and $(k+1)^{2}$, all of which are terms of $\left... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,746 |
Example 6 In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5,5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$ such that for all integers $n$, $a_{n} = b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$.... | Explanation: Divide the sequence $\{1,3,3,3,5,5,5,5,5, \cdots\}$ into groups:
(1), $(3,3,3),(5,5,5,5,5), \cdots,(2 k-1, 2 k-1, \cdots, 2 k-1), \cdots$,
where the $k$-th group consists of $2 k-1$ occurrences of $2 k-1$. If $a_{n}$ is in the $k$-th group, then $a_{n}=2 k-1$. Therefore,
$$
\begin{array}{l}
1+3+\cdots+(2 k... | 2 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,747 |
Example 7 If $n$ runs through all positive integers, prove that $f(n)$ $=n+\left[\sqrt{\frac{n}{3}}+\frac{1}{2}\right]$ runs through all positive integers, except for the terms of the sequence $a_{n}=3 n^{2}-2 n$.
(29th IMO Shortlist) | Explanation: Let the sequence of positive integers, excluding all terms of the sequence $\left\{a_{n}\right\}$, be arranged in ascending order to form the sequence $\left\{b_{n}\right\}$. We only need to prove that
$$
b_{n}=f(n)=n+\left[\sqrt{\frac{n}{3}}+\frac{1}{2}\right] \text {. }
$$
Since $a_{1}=1, a_{k+1}-a_{k}-... | b_{n}=n+\left[\frac{1}{2}+\sqrt{\frac{n}{3}}\right] | Number Theory | proof | Yes | Yes | cn_contest | false | 714,748 |
1. Let the sequence $\left\{a_{n}\right\}=\{1,2,2,3,3,3, \cdots, k, k, \cdots, k, \cdots\}$, where each positive integer $k$ appears $k$ times. Find the general term formula for $\left\{a_{n}\right\}$ and the sum of the first $n$ terms $S_{n}$. | $\begin{array}{l}\left(\text { Answer: } a_{\mathrm{n}}=\left[\frac{1+\sqrt{8 n-7}]}{2}\right], S_{\mathrm{n}}=k n-\frac{1}{6} k(k\right. \\ \left.+1)(k-1), k=\left[\frac{1+\sqrt{8 n-7}}{2}\right]\right)\end{array}$ | a_{\mathrm{n}}=\left[\frac{1+\sqrt{8 n-7}}{2}\right], S_{\mathrm{n}}=k n-\frac{1}{6} k(k+1)(k-1), k=\left[\frac{1+\sqrt{8 n-7}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,749 |
Example 3 Consider the equation in $x$
$$
a x^{2}+(a+2) x+9 a=0
$$
has two distinct real roots $x_{1}, x_{2}$, and $x_{1}<\frac{2}{5}<x_{2}$.
(A) $a>-\frac{2}{5}$
(B) $a<-\frac{2}{5}$
(C) $a<-\frac{2}{7}$
(D) $-\frac{2}{11}<a<0$
(2002, National Junior High School Mathematics Competition) | Solution: Let $t=x-1$, then the original equation becomes
$$
a t^{2}+(3 a+2) t+11 a+2=0 \text {. }
$$
Thus, the condition that one root of the original equation is greater than 1 and the other is less than 1 is transformed into the condition that one root of equation (1) is greater than 0 and the other is less than 0.... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,750 |
2. The sequence $1,1,2,2,2,2,2, \cdots, k, k, \cdots, k, \cdots$, where each positive integer $k$ appears $3 k-1$ times. Then the 2004th term of this sequence is $\qquad$ | (Answer: 37) | 37 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,751 |
3. Given the sequence $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}$, $\frac{1}{4}, \cdots$. Then, $\frac{n}{m}$ is the nth term of this sequence? | (Answer: the $\frac{(m+n-1)(m+n)}{2}-n+1$ term) | \frac{(m+n-1)(m+n)}{2}-n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,752 |
4. Arrange the positive odd numbers as shown in Figure 2, and denote the element in the $i$-th row and $j$-th column as $a_{4}$.
(1) Find the sum of all $a_{\eta}$ that satisfy $i+j=n+1\left(n \in \mathbf{N}_{+}\right)$;
(2) Find the sum of all $a_{4}$ that satisfy $i+j \leqslant n+1\left(n \in \mathbf{N}_{+}\right)$;
... | (1) $n^{3}$
(2) $\left[\frac{n(n+1)}{2}\right]^{2}$
(3) $i=13 \quad j$ = 33) | n^{3}, \left[\frac{n(n+1)}{2}\right]^{2}, i=13, j=33 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,753 |
5. Given the sequence of positive integers excluding all terms of the sequence $12 n^{2}-n:\left(n \in \mathbf{N}_{+}\right)$, the remaining terms arranged in ascending order form the sequence $\left\{a_{n}\right\}$. Try to find the general term formula for $a_{n}$. | (Answer: $a_{n}=2 k^{2}-k+[n-2 k(k-1)]=n+k$, where $\left.k=\left[\frac{1+\sqrt{2 n-1}}{2}\right]\right)$ | a_{n}=n+k, \text{ where } k=\left[\frac{1+\sqrt{2 n-1}}{2}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,754 |
Example 1 Let positive real numbers $x, y$ satisfy $xy=1$. Find the range of the function
$$
f(x, y)=\frac{x+y}{[x][y]+[x]+[y]+1}
$$
(where $[x]$ denotes the greatest integer not exceeding $x$). | Solution: Without loss of generality, let $x \geqslant y$, then $x^{2} \geqslant 1, x \geqslant 1$.
(1) When $x=1$, $y=1$, at this time $f(x, y)=\frac{1}{2}$.
(2) When $x>1$, let $[x]=n,\{x\}=x-[x]=\alpha$, then $x=n+\alpha, 0 \leqslant \alphaa_{2}=a_{3}, a_{3}b_{2}>\cdots>b_{n}>\cdots .
\end{array}
$$
Thus, when $x>1... | \left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,755 |
Example 2 Find the largest real number $m$, such that as long as the quadratic function $f(x)=a x^{2}+b x+c$ satisfies the conditions
(1) $a>0$;
(2) $|f(x)| \leqslant 1$, when $x \in[-1,1]$;
(3) $x_{0}=-\frac{b}{2 a} \in[-2,-1) \cup(1,2]$,
it must be true that $f(x) \geqslant m(x \in[-2,2])$.
Analysis: $m_{\text {max ... | Solution: Let $f(x)=a\left(x-x_{0}\right)^{2}+d$. Without loss of generality, assume $x_{0}=-\frac{b}{2 a} \in(1,2]$. From (2), we have
$$
\left\{\begin{array}{l}
1 \geqslant f(-1)=a\left(1+x_{0}\right)^{2}+d, \\
-1 \leqslant f(1)=a\left(1-x_{0}\right)^{2}+d .
\end{array}\right.
$$
Thus, $1-a\left(1+x_{0}\right)^{2} \... | -\frac{5}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,756 |
Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
\left(1+4^{2 x-y}\right) 5^{1-2 x+3}=1+2^{2 x-y+1}, \\
y^{3}+4 x+1+\ln \left(y^{2}+2 x\right)=0 .
\end{array}\right.
$$
(1999, Vietnam Mathematical Olympiad) | Solution: In (1), let $u=2 x-y$, then
$$
f(u)=5 \times\left(\frac{1}{5}\right)^{u}+5 \times\left(\frac{4}{5}\right)^{u}-2 \times 2^{u}-1=0 \text {. }
$$
Since $f(u)$ is monotonically decreasing on $\mathbf{R}$, and $f(1)=0$, thus, $u=1$, i.e., $2 x=y+1$.
Substituting the above equation into (2) yields
$$
\begin{array}... | x=0, y=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,757 |
Example 4 Let the function $y=f(x)$ have the domain $\mathbf{R}$, and when $x>0$, we have $f(x)>1$, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x) f(y)$. Solve the inequality
$$
f(x) \leqslant \frac{1}{f(x+1)} .
$$
Analysis: This problem involves an abstract function, and we can find a prototype function $f(x... | (1) Proof that $f(x)>0$ always holds.
From $f(x)=f\left(\frac{x}{2}+\frac{x}{2}\right)=f^{2}\left(\frac{x}{2}\right) \geqslant 0$, if there exists $x_{0} \in \mathbf{R}$, such that $f\left(x_{0}\right)=0$, then for any real number $x$, we have
$$
f(x)=f\left(x_{0}+x-x_{0}\right)=f\left(x_{0}\right) f\left(x-x_{0}\right... | x \leqslant -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,758 |
Example 5 Find all positive integers $k$ such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$.
(First China Girls Mathematical Olympiad)
Analysis: To find $k$, we can first determine the upper... | Solution: It is easy to know that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Therefore, the required $k>5$.
Take $a, b, c$ that do not form a triangle ($a=1, b=1, c=2$), then we have
$$
k(1 \times 1+1 \times 2+2 \times 1) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right).
$$
This gives $k \leqslant 6$.
Therefore, if $55\left(... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,759 |
Example 6 As shown in Figure 1, the sides of the equilateral $\triangle ABC$ with side length 1 are each divided into $n$ equal parts, and lines parallel to the other two sides are drawn through the division points, dividing the triangle into smaller equilateral triangles. The vertices of these smaller triangles are ca... | Solution: (1) Consider the figure composed of three small triangles as shown in Figure 2 (there are two rhombuses in the figure), and the numbers placed at each vertex are $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. According to the problem, we have
$$
a_{1}+a_{5}=a_{2}+a_{4}, a_{2}+a_{5}=a_{3}+a_{4}.
$$
Subtracting the two ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,760 |
Example 4 The equation $7 x^{2}-(k+13) x+k^{2}-k-2$ $=0$ (where $k$ is a constant) has two real roots $\alpha, \beta$, and $0<\alpha<1$, $1<\beta<2$. Then, the range of values for $k$ is ( ).
(A) $3<k<4$
(B) $-2<k<-1$
(C) $-2<k<-1$ or $3<k<4$
(D) No solution
(1990, National Junior High School Mathematics League) | Solution: Let $f(x)=7 x^{2}-(k+13) x+k^{2}-k-2$.
Since $7>0$, the graph of $f(x)$ opens upwards. To make the two real roots $\alpha$ and $\beta$ satisfy $0<\alpha<1$ and $0<\beta<1$, we need:
$$
\left\{\begin{array}{l}
\Delta=(k+13)^{2}-4 \cdot 7 \cdot\left(k^{2}-k-2\right)>0, \\
0<\frac{k+13}{14}<1, \\
f(0)=k^{2}-k-2>... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,761 |
Let the lengths of the three sides of a triangle be integers $l$, $m$, $n$, and $l > m > n$. It is known that
$$
\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\},
$$
where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the min... | 1 Another Solution to the Test Question
Solution: From the given, we have
$$
\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 10^{4}\right) . \\
\text { Equation (1) } \Leftrightarrow\left\{\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 2^{4}\right), \\
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod... | 3003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,762 |
2 Promotion of Test Questions
Let the three sides of a triangle be integers $l, m, n$, and $l > m > n$. Given that $p$ is an odd prime, $M$ is a positive integer,
$$
\begin{array}{l}
(p, M)=1, \text{ and } \\
\left\{\frac{p^{l}}{M}\right\}=\left\{\frac{p^{m}}{M}\right\}=\left\{\frac{p^{n}}{M}\right\},
\end{array}
$$
w... | Given, we have $p^{l} \equiv p^{m} \equiv p^{n}(\bmod M)$.
$$
p^{l} \equiv p^{m} \equiv p^{n}\left(\bmod p_{i}^{a_{i}}\right), i=1,2, \cdots, s .
$$
From equation (1), we get $p^{l-n} \equiv p^{m-n} \equiv 1\left(\bmod p_{i}^{o_{i}}\right)$.
Since $\left(p, p_{i}^{o_{i}}\right)=1$, according to Euler's theorem, we hav... | k^{\prime}(u+v)+3 n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,763 |
Example 1 If $x_{0}$ is a root of the quadratic equation $a x^{2}+b x+$ $c=0(a \neq 0)$, then the relationship between the discriminant $\Delta=b^{2}-4 a c$ and the quadratic form $M=\left(2 a x_{0}+b\right)^{2}$ is ( ).
(A) $\Delta>M$
(B) $\Delta=M$
(C) $\Delta<M$
(D) cannot be determined
(1992, National Junior High S... | There are many methods to solve this problem, but no matter which method is used, the final result we emphasize (i.e., equation (2)) must be obtained: the discriminant is not an isolated $b^{2}-4 a c$, it is related to the quadratic equation and is expressed as a perfect square. The solution method that best reflects t... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,764 |
Example 2 Find all real roots of the equation $x^{2}-2 x \sin \frac{\pi x}{2}+1=0$.
(1956, Beijing Mathematical Competition) | Text [3] P73 Consider the equation as a quadratic equation with a parameter in the linear term, and directly use the discriminant to solve it. Later, many journals published the solution method using the discriminant for this problem, and then made a "correction" to the "ingenious solution," the main reason being that ... | x=1, -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,765 |
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+... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,766 |
Title As shown in Figure 1, on a plane, there are three circles, where the two external common tangents of each pair of circles intersect at a point. Try to prove that the three intersection points lie on a straight line.
This is a famous problem proposed by the renowned engineer and educator Sweet, which has been hail... | Proof: (1) As shown in Figure 3, let the radii of $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ be $r_{1}$, $r_{2}$, and $r_{3}$, respectively. Let $P_{1}$ and $P_{2}$ be the intersection points of the two internal common tangents of $\odot O_{1}$ and $\odot O_{2}$, and $\odot O_{1}$ and $\odot O_{3}$, respectively. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,767 |
Proposition Let the three sides, circumradius, and inradius of $\triangle A B C$ be $a, b, c, R, r$ respectively. Then we have
$$
\frac{b^{2}+c^{2}}{2 b c} \leqslant \frac{R}{2 r} \text {. }
$$ | Proof: Let the area of $\triangle ABC$ be $S$. From $abc = 4RS$ and $S = \frac{1}{2} r(a + b + c)$, we know that equation (1) is equivalent to
$$
\frac{b^2 + c^2}{2bc} \leq \frac{abc(a + b + c)}{16S^2}.
$$
By Heron's formula, we have
$$
\begin{aligned}
16S^2 = & (a + b + c)(b + c - a) \\
& (c + a - b)(a + b - c).
\end... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,768 |
1. Let acute angle $\theta$ make the equation about $x$, $x^{2}+4 x \cos \theta+\cot \theta$ $=0$, have a repeated root. Then the radian measure of $\theta$ is $($ ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{12}$ or $\frac{5 \pi}{12}$
(C) $\frac{\pi}{6}$ or $\frac{5 \pi}{12}$
(D) $\frac{\pi}{12}$ | $-1 . B$
Since the equation has a repeated root, we have $\Delta=16 \cos ^{2} \theta-4 \cot \theta=0$.
Given $0<\theta<\frac{\pi}{2}$, thus $4 \cos \theta(2 \sin 2 \theta-1)=0$, which gives $\sin 2 \theta=\frac{1}{2}$. Therefore, $\theta=\frac{\pi}{12}$ or $\theta=\frac{5 \pi}{12}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,769 |
2. Given $M=\left\{(x, y)\left|x^{2}+2 y^{2}=3\right|, N=\{(x, y) \mid\right.$ $y=m x+b$. If for all $m \in \mathbf{R}$, there is $M \cap N \neq \varnothing$, then the range of $b$ is ( ).
(A) $\left[-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right]$
(B) $\left(-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right)$
(C) $\left(-\... | 2. A.
$M \cap N \neq \varnothing$ is equivalent to the point $(0, b)$ being on or inside the ellipse $\frac{x^{2}}{3}+\frac{2 y^{2}}{3}=1$, so $\frac{2 b^{2}}{3} \leqslant 1$. Therefore, $-\frac{\sqrt{6}}{2} \leqslant b \leqslant \frac{\sqrt{6}}{2}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,770 |
3. The solution set of the inequality $\sqrt{\log _{2} x-1}+\frac{1}{2} \log _{\frac{1}{2}} x^{3}+2>0$ is ( ).
(A) $[2,3)$
(B) $(2,3]$
(C) $[2,4)$
(D) $(2,4]$ | 3.C.
The original inequality is equivalent to
$$
\left\{\begin{array}{l}
\sqrt{\log _{2} x-1}-\frac{3}{2} \log _{2} x+\frac{3}{2}+\frac{1}{2}>0 \\
\log _{2} x-1 \geqslant 0
\end{array}\right.
$$
Let $\sqrt{\log _{2} x-1}=t$, then we have
$$
\left\{\begin{array}{l}
t-\frac{3}{2} t^{2}+\frac{1}{2}>0, \\
t \geqslant 0
\... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,771 |
Example 5 Find the range of $k$ such that the equation
$$
x^{2}-k x-k+3=0
$$
has two roots in the open intervals $(0,1)$ and $(1,2)$, respectively. | Solution: The original equation can be transformed into $x^{2}+3=k(x+1)$. Let $y_{1}=x^{2}+3, y_{2}=k(x+1)$.
Draw the graphs of $y_{1}$ and $y_{2}$, as shown in Figure 1. The intersection points of the parabola and the family of lines are on $\overparen{A B}$ and $\overparen{B C}$ (excluding points $A, B, C$) when the... | 2<k<\frac{7}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,772 |
4. As shown in Figure 1, let point $O$ be inside $\triangle A B C$, and $O A + 2 O B + 3 O C = 0$. Then the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$ is ( ).
(A) 2
(B) $\frac{3}{2}$
(C) 3
(D) $\frac{5}{3}$ | 4.C.
Let $D$ and $E$ be the midpoints of sides $AC$ and $BC$, respectively, then
$$
\begin{array}{l}
O A+O C=2 O D, \\
2(O B+O C)=4 O E .
\end{array}
$$
From (1) and (2), we get
$$
O A+2 O B+3 O C=2(O D+2 O E)=0,
$$
which means $O D$ and $O E$ are collinear, and $|O D|=2|O E|$.
Therefore, $\frac{S_{\triangle A B C}}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,773 |
5. Let the three-digit number $n=\overline{a b c}$. If the lengths of the sides of a triangle can be formed by $a, b, c$ to form an isosceles (including equilateral) triangle, then the number of such three-digit numbers $n$ is ( ) .
(A) 45
(B) 81
(C) 165
(D) 216 | 5.C.
$a, b, c$ must be able to form the sides of a triangle, clearly none of them can be 0, i.e., $a, b, c \in\{1,2, \cdots, 9\}$.
(1) If they form an equilateral triangle, let the number of such three-digit numbers be $n_{1}$. Since all three digits in the three-digit number are the same, then $n_{1}=\mathrm{C}_{9}^{1... | 165 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,774 |
6. As shown in Figure 2, the axis section of a cone with vertex $P$ is an isosceles right triangle. $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, with the foot of the perpendicular at $B$, $O H \perp P B$, with the f... | 6.D.
Since $A B \perp O B, A B \perp O P$, then $A B \perp P B$.
Since $O H \perp P B$, the plane $P A B \perp$ plane $P O B$. Therefore, $O H \perp H C, O H \perp P A$. Since $C$ is the midpoint of $P A$, then $O C \perp P A$. Thus, $P C$ is the height of the tetrahedron $P-H O C$, and $P C=2$.
In the right triangle... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,775 |
7. In the Cartesian coordinate system $x O y$, the graph of the function $f(x)=$ $a \sin a x+\cos a x(a>0)$ over an interval of the smallest positive period length and the graph of the function $g(x)=\sqrt{a^{2}+1}$ enclose a closed figure whose area is $\qquad$ | $$
\text { Ni. } 7 . \frac{2 \pi}{a} \sqrt{a^{2}+1} .
$$
$f(x)=\sqrt{a^{2}+1} \sin (a x+\varphi)$, where $\varphi=\arctan \frac{1}{a}$, its smallest positive period is $\frac{2 \pi}{a}$, and the amplitude is $\sqrt{a^{2}+1}$. Due to the symmetry of the closed figure formed by the graphs of $f(x)$ and $g(x)$, the figure... | \frac{2 \pi}{a} \sqrt{a^{2}+1} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 714,776 |
8. Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfy $f(0)=1$, and for any $x, y$ $\in \mathbf{R}$, there is $f(x y+1)=f(x) f(y)-f(y)-x+2$. Then $f(x)=$ $\qquad$ . | $$
\text { 8. } x+1 \text {. }
$$
For any $x, y \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x y+1)=f(x) f(y)-f(y)-x+2, \\
f(x y+1)=f(y) f(x)-f(x)-y+2 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } f(x) f(y)-f(y)-x+2 \\
=f(y) f(x)-f(x)-y+2,
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
f(x)+y=f(y)+x... | x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,777 |
11. The sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$, satisfies the relation (3$\left.a_{n+1}\right)\left(6+a_{n}\right)=18$, and $a_{0}=3$. Then $\sum_{i=0}^{n} \frac{1}{a_{i}}=$ $\qquad$ . | 11. $\frac{1}{3}\left(2^{n+2}-n-3\right)$.
Let $b_{n}=\frac{1}{a_{n}}, n=0,1,2, \cdots$.
From the given condition, we have $3 b_{n+1}-6 b_{n}-1=0$. Therefore,
$$
b_{n+1}+\frac{1}{3}=2\left(b_{n}+\frac{1}{3}\right) \text {. }
$$
Thus, the sequence $\left\{b_{n}+\frac{1}{3}\right\}$ is a geometric sequence with a commo... | \frac{1}{3}\left(2^{n+2}-n-3\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,779 |
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 subt... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,780 |
13. The rules of a "level-passing game" stipulate: on the $n$-th level, a die must be rolled $n$ times. If the sum of the points from these $n$ rolls is greater than $2^{n}$, the level is considered passed. The questions are:
(1) What is the maximum number of levels a person can pass in this game?
(2) What is the proba... | Three, 13. Since the die is a uniform cube, the possibility of each point number appearing after throwing is the same.
(1) Because the maximum point number that the die can show is 6, and $6 \times 4 > 2^{4}$, $6 \times 5 < 2^{5}$, therefore, when $n \geqslant 5$, the sum of the point numbers appearing $n$ times being ... | \frac{100}{243} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,781 |
Example 6 If the parabola $y=x^{2}+m x+2$ intersects the line segment $MN$ (including points $M$ and $N$) connecting points $M(0,1)$ and $N(2,3)$ at two distinct points, find the range of values for $m$. | Solution: The function expression of line segment $MN$ is $y=x+1$. Therefore, the original problem is equivalent to the equation $x^{2}+m x+2=x+1$ having two distinct real roots in the interval $[0,2]$. Organizing, we get
$$
\begin{array}{l}
x^{2}+(m-1) x+1=0 . \\
\text { Let } f(x)=x^{2}+(m-1) x+1 .
\end{array}
$$
To... | -\frac{3}{2} \leqslant m < -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,782 |
15. Given that $\alpha, \beta$ are the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the function $f(x)=\frac{2 x-t}{x^{2}+1}$ has the domain $[\alpha, \beta]$.
(1) Find $g(t)=\max f(x)-\min f(x)$;
(2) Prove: For $u_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3)$, if $\sin u_{1}$
$$
... | 15. (1) Let $\alpha \leqslant x_{1} < x_{2} \leqslant \beta$.
$$
\begin{array}{l}
\text{Then } f\left(x_{2}\right)-f\left(x_{1}\right)=\frac{2 x_{2}-t}{x_{2}^{2}+1}-\frac{2 x_{1}-t}{x_{1}^{2}+1} \\
=\frac{\left(x_{2}-x_{1}\right)\left[t\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}+2\right]}{\left(x_{2}^{2}+1\right)\left(x_{1... | \frac{3}{4} \sqrt{6} | Algebra | proof | Yes | Yes | cn_contest | false | 714,783 |
One, (50 points) As shown in Figure 4, in the acute triangle $\triangle ABC$, the altitude $CE$ from $A B$ intersects with the altitude $BD$ from $A C$ at point $H$. The circle with diameter $DE$ intersects $AB$ and $AC$ at points $F$ and $G$, respectively. $FG$ intersects $AH$ at point $K$. Given that $BC=25, BD=20, B... | Given that $\angle A D B=\angle A E C=90^{\circ}$. Therefore,
$\triangle A D B \backsim \triangle A E C$.
Thus, $\frac{A D}{A E}=\frac{B D}{C E}=\frac{A B}{A C}$.
From the given, $C D=15, C E=24$. From (1) we have
$\left\{\begin{array}{l}\frac{A D}{A E}=\frac{5}{6}, \\ \frac{A E+7}{A D+15}=\frac{5}{6}\end{array} \Right... | 8.64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,784 |
II. (50 points) In the Cartesian coordinate system $x O y$, the sequence of points $\left\{A_{n}\right\}$ on the positive $y$-axis and the sequence of points $\left\{B_{n}\right\}$ on the curve $y=\sqrt{2 x} (x \geqslant 0)$ satisfy $\left|O A_{n}\right|=\left|O B_{n}\right|=\frac{1}{n}$. The intercept of the line $A_{... | (1) It is known that $A_{n}\left(0, \frac{1}{n}\right), B_{n}\left(b_{n}, \sqrt{2 b_{n}}\right)\left(b_{n}>0\right)$.
From $\left|O B_{n}\right|=\frac{1}{n}$, we get $b_{n}^{2}+2 b_{n}=\left(\frac{1}{n}\right)^{2}$, hence
$$
b_{n}=\sqrt{\left(\frac{1}{n}\right)^{2}+1}-1, n \in \mathbf{N}_{+} \text {. }
$$
Furthermore,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,785 |
For the integer $n(n \geqslant 4)$, find the smallest integer $f(n)$, such that for any positive integer $m$, any $f(n)$-element subset of the set $\{m, m+1, \cdots, m+n-1\}$ contains at least 3 pairwise coprime elements. | When $n \geqslant 4$, for the set $M=\{m, m+1, m+2, \cdots, m+n-1\}$, if $2 \mid m$, then $m+1, m+2, m+3$ are pairwise coprime; if $2 \nmid m$, then $m, m+1, m+2$ are pairwise coprime.
Thus, in all $n$-element subsets of $M$, there are at least 3 elements that are pairwise coprime. Therefore, $f(n)$ exists, and $f(n) ... | f(n) \leqslant n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,786 |
1. Let $m$ be a fixed integer greater than 1, and the sequence $x_{0}, x_{1}$, $x_{2}, \cdots$ is defined as follows:
$$
x_{i}=\left\{\begin{array}{ll}
2^{i}, & 0 \leqslant 1 \leqslant m-1, \\
\sum_{j=1}^{m} x_{i-j}, & i \geqslant m .
\end{array}\right.
$$
Find the maximum value of $k$ such that there are $k$ consecut... | Solution: Let $r_{i}$ be the remainder of $x_{i}$ modulo $m$. In the sequence, divide the terms into blocks of $m$ consecutive terms. Then, there are at most $m^{m}$ different cases for the remainders. By the pigeonhole principle, one type of case will appear repeatedly. Since the defined recurrence relation can be use... | m-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,787 |
2. For every positive integer $a$, the number $d=d(a)$ is obtained through the following process:
(1) Move the last digit of $a$ to the first position to get the number $b$;
(2) Square $b$ to get the number $c$;
(3) Move the first digit of $c$ to the last position to get the number $d$.
For example, if $a=2003$, then $... | Let the positive integer $a$ satisfy $d=d(a)=a^{2}$, and $a$ has $n+1$ digits, $n \geqslant 0$. Let the last digit of $a$ be $s$, and the first digit of $c$ be $f$. Because
$$
\begin{array}{l}
(* \cdots * s)^{2}=a^{2}=d=* \cdots * f, \\
(s * \cdots *)^{2}=b^{2}=c=f * \cdots *,
\end{array}
$$
where * represents a singl... | a=3, a=2, a=\underbrace{2 \cdots 2}_{n \hat{1}} 1, n \geqslant 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,788 |
4. Let $b$ be an integer greater than 5. For each positive integer $n$, consider the number $x_{n}=\underbrace{11 \cdots 122 \cdots 2}_{n-1 \uparrow} \underbrace{}_{n \uparrow}$ in base $b$.
Prove that the statement "there exists a positive integer $M$ such that for any integer $n$ greater than $M$, the number $x_{n}$... | Proof: For $b=6,7,8,9$, classifying $x$ modulo $b$, it can be directly verified that $x^{2} \equiv 5(\bmod b)$ has no solutions.
Since $x_{n} \equiv 5(\bmod b)$, $x_{n}$ is not a perfect square.
For $b=10$, direct calculation yields
$$
x_{n}=\frac{1}{b-1}\left(b^{2 n}+b^{n+1}+3 b-5\right)=\left(\frac{10^{n}+5}{3}\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,789 |
5. An integer $n$ is called a "good" number if $|n|$ is not a perfect square. Find all integers $m$ such that $m$ can be expressed as the sum of three distinct "good" numbers in infinitely many ways, and the product of these three "good" numbers is a perfect square of an odd number.
(Korea provided) | Assume that $m$ can be expressed as $m=u+v+w$, and $u w w$ is the square of an odd number. Therefore, $u, v, w$ are all odd numbers, and $u w \equiv 1(\bmod 4)$. Thus, either two of $u, v, w$ are congruent to 3 modulo 4, or none of them are. In either case, we have
$$
m=u+v+w \equiv 3(\bmod 4).
$$
Next, we prove that ... | m=4k+3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,790 |
7. The sequence $a_{0}, a_{1}, a_{2}, \cdots$ is defined as follows: for all $k$ $(k \geqslant 0), a_{0}=2, a_{k+1}=2 a_{k}^{2}-1$. Prove: If an odd prime $p$ divides $a_{n}$, then $2^{n+3}$ divides $p^{2}-1$.
(Brazil provided) | Proof: By mathematical induction, we can prove
$$
a_{n}=\frac{(2+\sqrt{3})^{2^{n}}+(2-\sqrt{3})^{2^{n}}}{2} \text {. }
$$
If $x^{2} \equiv 3(\bmod p)$ has integer solutions, let integer $m$ satisfy $m^{2} = 3(\bmod p)$. Given $p \mid a_{n}$, we have
$$
(2+\sqrt{3})^{2^{n}}+(2-\sqrt{3})^{2^{n}} \equiv 0(\bmod p) \text ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,791 |
Example 7 Given $a, b, c \in \mathbf{N}_{+}$, and the parabola $f(x) = ax^{2} + bx + c$ intersects the $x$-axis at two different points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a + b + c$.
(1996, National Junior High School Mathematics Competition) | Let $f(x)=a x^{2}+b x+c, A\left(x_{1}, 0\right)$, and $B\left(x_{2}, 0\right)$. According to the problem, we have
$$
\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... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,792 |
8. Let $p$ be a prime number, and $A$ be a set of positive integers, satisfying the following conditions:
(1) The set of prime factors of the elements in $A$ contains $p-1$ elements;
(2) For any non-empty subset of $A$, the product of its elements is not a $p$-th power of an integer.
Find the maximum number of elements... | Solution: The maximum value is $(p-1)^{2}$.
Let $r=p-1$, and assume the distinct primes are $p_{1}$, $p_{2}, \cdots, p_{r}$, define
$$
B_{i}=\left\{p_{i}, p_{i}^{p+1}, p_{i}^{2 p+1}, \cdots, p_{i}^{(r-1) p+1}\right\} .
$$
Let $B=\bigcup_{i=1}^{\prime} B_{i}$, then $B$ has $r^{2}$ elements, and satisfies conditions (1)... | (p-1)^2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,793 |
2. Let $D_{1}, D_{2}, \cdots, D_{n}$ be closed disks in the plane, and assume that each point in the plane belongs to at most 2003 disks $D_{i}$. Prove: There exists a disk $D_{k}$ such that $D_{k}$ intersects at most $7 \times 2003-1$ other disks $D$.
(Georgia provided) | Proof: Let the disk $S$ have the smallest radius $s$. As shown in Figure 9, divide the plane into 7 regions: $S$ is one region, and the other 6 regions are 6 congruent regions outside $S$.
Since $s$ is the smallest radius, any disk with its center in $S$ and different from $S$ contains the center $O$ of disk $S$, and ... | 7 \times 2003-1 | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,794 |
3. Let integer $n(n \geqslant 5)$. Find the maximum integer $k$, such that there exists an $n$-sided polygon (convex or concave, as long as the boundary does not intersect itself) with $k$ right angles.
(Lithuania provided) | Solution: We prove that when $n=5$, the satisfying $k$ equals 3; when $n \geqslant 6$, $k$ equals $\left[\frac{2 n}{3}\right]+1$ (where $[x]$ represents the greatest integer not exceeding $x$).
Assume there exists an $n$-sided polygon with $k$ right angles, since all other angles are less than $360^{\circ}$, we have
$... | k \leqslant \left[\frac{2 n}{3}\right]+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,795 |
4. Let $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$ be real numbers, and the matrix $A=\left(a_{i j}\right)_{1 \leqslant i, j \leqslant n}$ satisfies
$$
a_{i j}=\left\{\begin{array}{ll}
1, & \text { if } x_{i}+y_{j} \geqslant 0, \\
0, & \text { if } x_{i}+y_{j}<0.
\end{array}\right.
$$
If the $n \ti... | Proof: Let $B=\left(b_{i j}\right)_{1 \leqslant i, j \leqslant n}$. Define
$$
S=\sum_{1 \leqslant i, j \leqslant n}\left(x_{i}+y_{j}\right)\left(a_{i j}-b_{i j}\right) \text {. }
$$
On one hand, we have
$$
\begin{array}{l}
S=\sum_{i=1}^{n} x_{i}\left(\sum_{j=1}^{n} a_{i j}-\sum_{j=1}^{n} b_{i j}\right)+\sum_{j=1}^{n} ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,796 |
5. It is known that each integer point on the Cartesian plane is the center of a disk with a radius of $\frac{1}{1000}$. Prove:
(1) There exists an equilateral triangle whose three vertices are in three different disks;
(2) The side length of every equilateral triangle whose three vertices are in three different disks ... | Proof: (1) Define $f: \mathbf{Z} \rightarrow[0,1)$ as
$$
f(x)=\sqrt{3} x-[\sqrt{3} x] \text {. }
$$
By the pigeonhole principle, there must exist two different integers $x_{1}$ and $x_{2}$, such that
$$
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<0.001 \text {. }
$$
Let $a=\left|x_{1}-x_{2}\right|$. Then eit... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,797 |
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