problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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4. Given a complex number $z \neq 0$. Provide the following propositions:
(1) $z+\bar{z}=0$;
(2) $z^{2}=-a^{2}(a \in \mathbf{R}, a \neq 0)$;
(3) $z^{2}=z^{2}$;
(4) $|z+a|=|z-a|(a \in \mathbf{R}, a \neq 0)$.
Among the above propositions, which one is a sufficient and necessary condition for $z$ to be a pure imaginary nu... | 4.(1)、(2)、(4).
$z+\bar{z}=0 \Leftrightarrow \operatorname{Re}(z)=0 \Leftrightarrow z$ is a pure imaginary number.
$=-a i \Leftrightarrow z$ will be purely imaginary.
$$
\begin{array}{c}
|z+a|=|z-a| \Leftrightarrow \sqrt{(x+a)^{2}+y^{2}}= \\
\sqrt{(x-a)^{2}+y^{2}} \Leftrightarrow(x+a)^{2}=(x-a)^{2} \Leftrightarrow 2 a x... | 4.(1)、(2)、(4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,151 |
5. Given that $[x]$ represents the greatest integer not exceeding $x$. Then the number of solutions to the equation
$$
3^{2 x}-\left[10 \times 3^{x+1}\right]+\sqrt{3^{2 x}-10 \times 3^{x+1}+82}=-80
$$
is | 5.2.
The original equation can be transformed into
$$
\begin{array}{l}
3^{2 x}-\left[10 \times 3^{x+1}\right]+82 \\
+\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-2=0 . \\
\therefore\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}+2\right) \\
\quad \cdot\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-1... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,152 |
Є. Let $x_{1}, x_{2}, y_{1}, y_{2} \in \mathbf{R}^{+}, a=\sqrt{x_{1} x_{2}}+$ $\sqrt{y_{1} y_{2}}, b=\sqrt{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)}$. Then the relationship between $a, b$ is $\qquad$ . | $\begin{array}{l}6 . a \leqslant b . \\ \begin{aligned} \frac{a}{b}= & \sqrt{\frac{x_{1} x_{2}}{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)}} \\ & +\sqrt{\frac{y_{1} y_{2}}{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)}} \\ \leqslant & \frac{1}{2}\left(\frac{x_{1}}{x_{1}+y_{1}}+\frac{x_{2}}{x_{2}+y_{2}}\right) ... | a \leqslant b | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,153 |
Four. (Full marks 20 points) The line $y=k x+m$ intersects the hyperbola $\frac{x^{2}}{a^{2}} -\frac{y^{2}}{b^{2}}=1$ and its asymptotes at points $A, B, C, D$. Prove that $|A C|=|B D|$. | Due to the asymptote equations of the hyperbola being $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$, we can set $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=\lambda .(\lambda$ $=0$ or 1 )
(i. $\left.k^{2}\right) e^{2}-2 k m a^{2} x-a^{2} m^{2}-\lambda a^{2} b^{2}=0$. This equation clearly has real solutions.
By Vieta's form... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,154 |
Five. (Full marks 20 points) Let the constant $a>1>b>0$. Then, under what relationship between $a$ and $b$ is the solution set of $\lg \left(a^{x}-b^{x}\right)>0$ $(1, +\infty)$? | Five, $\because a^{\prime}-b^{\prime}>0$, i.e., $\left(\frac{a}{\dot{v}}\right)^{\prime}>1$, and $\frac{a}{b}>1$,
$\therefore$ when $x \in(0,+\infty)$, to make the solution set of the inequality $(1,+\infty)$, it is only necessary that $f(x)=\lg \left(a^{x}-b^{x}\right)$ is an increasing function on $(0,+\infty)$, and ... | a=b+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,155 |
Six. (Full marks 20 points) The integer sequence $\left\{a_{n}\right\}(n \in \mathbf{N})$ satisfies: $a_{1}=$ $2, a_{2}=7$, and
$$
-\frac{1}{2}<a_{n+1}-\frac{a_{n}^{2}}{a_{n-1}} \leqslant \frac{1}{2} \text {. }
$$
Prove: When $n \geqslant 2$, $a_{n}$ is odd. | When $n=2$, $-\frac{1}{2}<a_{3}-\frac{49}{2} \leqslant \frac{1}{2}$, we have $24<a_{3} \leqslant 25$.
$\because a_{n}$ is an integer, $\therefore a_{3}=25=3 a_{2}+2 a_{1}$.
Thus, we conjecture $a_{n+1}=3 a_{n}+2 a_{n-1}$.
When $n=2$, the proposition is obviously true.
Assume that when $n=k$, the proposition holds, i.e.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,156 |
Three, (Full marks 20 points) Prove:
$\operatorname{tg} 70^{\circ}=\operatorname{tg} 20^{\circ}+2 \operatorname{tg} 40^{\circ}+4 \operatorname{tg} 10^{\circ}$. | \begin{aligned} \text { Three, with sides = } & \operatorname{tg} 20^{\circ}+2\left(\operatorname{tg} 10^{\circ}+\operatorname{tg} 40^{\circ}\right)+2 \operatorname{tg} 10^{\circ} \\ = & \operatorname{tg} 20^{\circ}+2 \operatorname{tg} 50^{\circ}\left(1-\operatorname{tg} 40^{\circ} \operatorname{tg} 10^{\circ}\right) \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,157 |
Example 6 In 45 km away from city $A$, there is a metal mine at location $B$. It is known that there is a straight railway $A X$ from $A$ to a certain direction, and the distance from $B$ to this railway is 27 km. To transport materials between $A$ and $B$, it is planned to build a road from a point $C$ on the railway ... | Solution: First, draw
a diagram (Fig.
4), the distance from $B$ to the railway
$A X$ is $B D$
$=27 \mathrm{~km}$, so
$A D^{2}=A B^{2}-$
$B D^{2}, A D=36$
$\mathrm{km}$. Let $A C=x \mathrm{~km}$, then
$$
B C=\sqrt{(36-x)^{2}+27^{2}}(\mathrm{~km}) \text {. }
$$
Assume the railway ton-kilometer freight rate is 1 unit, an... | 20.41 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,158 |
1. In the geometric sequence $\left\{a_{n}\right\}$, let $S_{n}=a_{1}+a_{2}+\cdots+$ $a_{n}$, it is known that $a_{5}=2 S_{4}+3, a_{6}=2 S_{5}+3$. Then the common ratio $q$ of this sequence is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | $\begin{array}{c}-\sqrt{-1}\} \\ a_{6}-a_{5}=\left(2 S_{5}+3\right)-\left(2 S_{4}+3\right)=2\left(S_{5}-S_{4}\right) \\ =2 a_{5} \text {. Therefore, } a_{6}=3 a_{5} \text {, so } q=3 .\end{array}$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,159 |
3. The diameter of the base of a cylinder is $4 R$, and its height is $42 R$. The maximum number of spheres with a radius of $R$ that it can hold is ( ).
(A) 28
(B) 56
(C) 58
(D) 60 | 3. C.
The arrangement is with two on the left and right sides, and two higher ones in the front, placed in layers that cross each other. The distance between successive layers equals the distance between two opposite edges of a regular tetrahedron with the four ball centers as vertices, which is \( d = \sqrt{2} R \). ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,161 |
4. As shown in Figure 2, given the function $y=2 x^{2}$ on $[a, b]$ $(a<b)$, the range of values is $[0,2]$. Then the trajectory of point $(a, b)$ is ( ).
(A) line segments $A B, B C$
(B) line segments $A B, O C$
(C) line segments $O A, B C$
(D) line segments $O A, O C$ | 4. A.
From $y=2 x^{2} \geqslant 0$, we know that when $y=0$, $x=0$. Therefore, $0 \in [a, b]$. Also, when $y=2$, $x=1$ or $x=-1$. There are two possibilities:
(1) If $1 \in [a, b]$, then by $a<b$ and the monotonicity of the function on $\mathbf{R}^{+}$, we have $b=1$, and $-1 \leqslant a \leqslant 0$. The correspondin... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,162 |
5. Given the set $A_{n}=\left\{x \mid 2^{n}<x<2^{n+1}\right.$, and $x=7 m$ $+1, n, m \in \mathbf{N}\}$. Then the sum of all elements in $A_{6}$ is ( ).
(A) 792
(B) 890
(C) 891
(D) 990 | 5. C.
Since $64=2^{6}<x<2^{7}=128$, and $x=7m+1$, thus $a_{1}$ $=71, a_{n}=127$. Given the common difference $d=7$, we find $n=9$. Therefore, $S_{y}=$ $\frac{a_{1}+a_{9}}{2} \times 9=891$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,163 |
6. Let the integer part and the fractional part of $(5 \sqrt{2}+7)^{2 n+1}(n \in \mathbf{N})$ be $I$ and $F$, respectively. Then the value of $F(I+F)$ is ( ).
(A) 1
(B) 2
(C) 4
(D) a number related to $n$ | 6. A.
Since $5 \sqrt{2}-7 \in(0,1)$ and
$(5 \sqrt{2}+7)^{2 n+1}(5 \sqrt{2}-7)^{2 n+1}=1$,
also $(5 \sqrt{2}+7)^{2 n+1}=I+F$.
Therefore, $(5 \sqrt{2}-7)^{2 n+1}=\frac{1}{I+F}$.
By the binomial theorem, $(5 \sqrt{2}+7)^{2 n+1}-(5 \sqrt{2}-7)$
is an integer. That is, $I+F-\frac{1}{I+F}$ is an integer, and since $I$ is an... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,164 |
1. Let $[\operatorname{tg} x]$ denote the greatest integer not exceeding the real number $\operatorname{tg} x$. Then the solution to the equation $[\operatorname{tg} x]=2 \cos ^{2} x$ is | $=1 . x=k \pi+\frac{\pi}{4}, k \in \mathbf{Z}$.
Since $0 \leqslant 2 \cos ^{2} x \leqslant 2$, the value that $[\operatorname{tg} x]$ can take can only be 0, 1, 2. When $[\operatorname{tg} x]=0$, $\cos x=0$, at this time $\operatorname{tg} x$ is undefined. When $[\operatorname{tg} x]=2$, $\cos ^{2} x=1$, at this time $... | x=k \pi+\frac{\pi}{4}, k \in \mathbf{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,165 |
4. Let $f(x)=\log _{2} \frac{1+2^{x}+4^{x} a}{3}$, where $a \in \mathbf{R}$. If $f(x)$ is defined for $x \in(-\infty, 1]$, then the range of values for $a$ is | $$
\text { 4. } a>-\frac{3}{4} \text {. }
$$
According to the problem, $1+2^{x}+4^{x} a>0, x \in(-\infty, 1]$. Therefore,
$$
a>-\left[\left(\frac{1}{4}\right)^{x}+\left(\frac{1}{2}\right)^{x}\right](x \leqslant 1) \text {. }
$$
Since $y=\left(\frac{1}{4}\right)^{x}$ and $y=\left(\frac{1}{2}\right)^{x}$ are decreasing... | a>-\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,168 |
1. A worker: To transport 17 concrete power poles from the company to a road 11 m away for planting, with one pole planted every 0.1 km, the total distance traveled by the car, denoted as $y$, is considered from the company's departure to the completion of the task and return to the company. Due to the limited load cap... | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,169 |
$$
\begin{array}{r}
\text { 5. Given a sequence } z_{0}, z_{1}, \cdots, z_{n}, \cdots \text { satisfying } z_{0}=0, z_{1} \\
=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2,
\end{array}
$$
5. Given a sequence of complex numbers $z_{0}, z_{1}, \cdots, z_{n}, \cdots$ satisfying $z_... | 5.5.
Since $z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right)=\alpha^{2}\left(z_{n-1}-z_{n-2}\right)$ $=\cdots=\alpha^{n}\left(z_{1}-z_{0}\right)=\alpha^{n}$, therefore,
$$
z_{n}-z_{n-1}=\alpha^{n-1}, \cdots, z_{1}-z_{0}=\alpha^{0}=1 \text {. }
$$
Adding the above $n$ equations, we get
$$
z_{n}=\alpha^{n-1}+\cdots+\alph... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,170 |
6. Let the set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}, B_{2}, \cdots, B_{k}\right.$ be a family of non-empty subsets of $A$, and when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Then the maximum value of $k$ is $\qquad$ | 6.175 .
It is not hard to see that the family of all subsets of $A$ containing at most three elements meets the conditions of the problem, where the number of subsets is $\mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{e}+\mathrm{C}_{10}^{3}=175$.
Suppose there is another family of subsets $C$ that meets the conditions, and the... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,171 |
Three. (Full marks 20 points) For every pair of real numbers $x, y$, the function $f(t)$ satisfies $f(x+y)=f(x)+f(y)+xy+1$. If $f(-2)=-2$, find the number of integer solutions $a$ that satisfy $f(a)=a$.
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | Let $x=y=0$, we get $f(0)=-1$; let $x=y=-1$, from $f(-2)=-2$ we get $f(-1)=-2$. Also, let $x=1$, $y=-1$ to get $f(1)=1$. Then let $x=1$, we have
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Therefore, $f(y+1)-f(y)=y+2$, which means when $y$ is a positive integer, $f(y+1)-f(y)>0$.
From $f(1)=1$, we know that for all positive inte... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,172 |
Four, (Full marks 20 points) In $\triangle ABC$, does there exist a point $P$ such that any line passing through point $P$ will divide $\triangle ABC$ into two parts with equal area? Why? | As shown in Figure 5, assume there exists a point $P$ satisfying the conditions. Extend $A^{D}$ to intersect $B C$ at $D$, and connect to. The intersection with $A C$ is at $E$. Then, by $S_{\triangle A B D} = S_{\triangle A C D}$, we get $B D = C D$.
Similarly, we have
$$
A E = C E \text{. }
$$
Thus, $P$ is the centr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,173 |
Five. (Full marks 20 points) A county is located in a desert area, and people have been engaged in a tenacious struggle against nature for a long time. By the end of 1998, the county's greening rate had reached $30 \%$. Starting from 1999, the following situation will occur every year: $16 \%$ of the original desert ar... | (1) Let the current desert area be $b_{1}$, and after $n$ years, the desert area be $b_{n+1}$. Thus, $a_{1}+b_{1}=1, a_{n}+b_{n}=1$.
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} a_{n}$, which is $\frac{96}{1... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,174 |
1. The number of real solutions to the equation $\sqrt{x+19}+\sqrt[3]{x+95}=12$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | $-1 . B$.
Let $\sqrt{x+19}=m, \sqrt[3]{x+95}=n$, then $m+n=12$.
$$
\begin{array}{l}
\therefore m=12-n . \\
\text { Also } n^{3}-m^{2}=76 .
\end{array}
$$
Substituting equation (1) into equation (2) and rearranging, we get
$$
\begin{array}{l}
n^{3}-n^{2}+24 n-220=0, \\
(n-5)\left(n^{2}+4 n+44\right)=0 .
\end{array}
$$
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,175 |
2. $A B$ is the diameter of the semicircle $O$, $\odot O_{1}$ is internally tangent to the semicircle at $C_{1}$, and tangent to $A B$ at $D_{1}$; $\odot O_{2}$ is internally tangent to the semicircle at $C_{2}$, and tangent to $A B$ at $D_{2}$ (as shown in Figure 1). The size relationship between $\angle A C_{1} D_{1}... | 2.C.
The same size, both are $45^{\circ}$. Reason: Obviously, $O 、 O_{1} 、 C_{1}$ are collinear, $O 、 O_{2} 、 C_{2}$ are collinear; also, $O_{1} D_{1} \perp A B, O_{2} D_{2} \perp A B$, so
$$
\begin{aligned}
\angle A C_{1} D_{1} & =\angle A C_{1} O-\angle O C_{1} D_{1} \\
& =\frac{1}{2}\left(\angle C_{1} O B-\angle O ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,176 |
4. As shown in Figure 3, from a known
point $A$ on the circle, two
(A) $\frac{S_{1}}{S_{2}} \geqslant \frac{a}{b}$
(c) $\frac{S_{1}}{S_{2}}=\frac{a}{b}$
(C) $\frac{S_{1}}{S_{2}}>\frac{a}{b}$
(D) $\frac{S_{1}}{S_{2}} \leqslant \frac{a}{b}$
chords $A B$ and $A C$ are drawn with lengths $a$ and $b$ respectively, and $a<... | 4.C.
Connect $B D, D C, \angle B A D=\angle C A D$, we know, $\frac{S_{\triangle A B D}}{S_{\triangle A C D}}=\frac{a}{b}$. We can set $S_{\triangle \triangle B D}=a k, S_{\triangle A C D}=b k$.
It is easy to prove $S_{3}$ lipD $=S_{\text {Зemm. }}$ (set as $S_{0}$ ), then we have
$$
\begin{array}{l}
S_{1}=a k+S_{0}, ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,178 |
5. Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha>\beta$. Then the value of $\frac{2}{\alpha}+3 \beta^{2}$ is ( ).
(A) $\frac{1}{8}(403-85 \sqrt{17})$
(B) $\frac{1}{4}(403-85 \sqrt{17})$
(C) 92
(D) $\sqrt{17}$ | 5. A.
From the relationship between roots and coefficients, i.e., $\alpha+\beta=7, \alpha \beta=8$, we get $\alpha^{2}+\beta^{2}=33, (\alpha-\beta)^{2}=17$.
Given $a>\beta$, we have $\alpha-\beta=\sqrt{17}, \beta-\alpha=-\sqrt{17}$.
Let $A=\frac{2}{\alpha}+3 \beta^{2}, B=\frac{2}{\beta}+3 \alpha^{2}$, then,
$$
\begin{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,179 |
2. A hotel has 100 beds. When the charge is 10 yuan per bed per night, all beds can be rented out. If the charge is increased by 2 yuan per bed per night, then 10 fewer beds will be rented out; if the charge is increased by another 2 yuan per bed per night, then another 10 fewer beds will be rented out. If the charge i... | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,180 |
6. The number of positive roots of the equation $2 x-x^{2}=\frac{2}{x}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 6. A.
When $x>2$, the left side $=(2-x) x < 0$, the original equation has no solution;
When $x=2$, the left side $=0$, the right side $=1$, the equation has no solution.
When $0<x<2$, the left side $2x-x^2$ is a downward-opening parabola, and the right side $\frac{2}{x}$ is a decreasing function. Since the left side i... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,181 |
1. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}$ $+b x^{16}+1$, then $a=$ $\qquad$ | $=1.987$.
Let $p, q$ be the roots of the equation $x^{2}-x-1=0$, then
$$
p+q=1, pq=-1 \text{. }
$$
Given that $p, q$ are also the roots of the equation $a x^{17}+b x^{16}+1=0$, we have $\left\{\begin{array}{l}a p^{17}+b p^{16}+1=0, \\ a q^{17}+b q^{16}+1=0 .\end{array}\right.$
$$
\begin{array}{l}
\text{(1) } \times q^... | 987 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,182 |
2. Let the incircle $\odot O$ of $\triangle A B C$ touch $B C$ at point $D$, and draw the diameter $D E$ through $D$. Connect $A E$ and extend it to intersect $B C$ at point $F$. If $B F+C D=1998$, then $B F+2 C D=$ | 2. 2997.
As shown in Figure 6, let
$\odot O$ be tangent to $AB$ and
$AC$ at points $M$ and $N$, respectively. Draw
a line $GH \parallel BC$ through
point $E$, intersecting
$AB$ and $AC$ at
points $G$ and $H$, respectively. Then $GH$ is tangent
to $\odot O$ at point $E$, and
$\triangle A G E \backsim \triangle A B F, \... | 2997 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,183 |
3. As shown in Figure 4, the side of square $ABCD$ is $AB=1$, and $\overparen{BD}$ and $\overparen{AC}$ are both circular arcs with a radius of 1. Then the difference in the areas of the two unshaded parts is $\qquad$ | 3. $\frac{\pi}{2}-1$.
From Figure 4, we know $S_{1}+S=\frac{\pi}{4}$. Therefore, $S_{1}=\frac{\pi}{4}-S$. Also, $S_{2}+S=1-\frac{\pi}{4}$, so $S_{2}=1-S-\frac{\pi}{4}$. Thus, $S_{1}-S_{2}=\left(\frac{\pi}{4}-S\right)-\left(1-S-\frac{\pi}{4}\right)$ $=\frac{\pi}{2}-1$. | \frac{\pi}{2}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,184 |
4. Given three real
numbers $x_{1}, x_{2}, x_{3}$, any one of them plus five times the product of the other two equals 6. The number of such triples $\left(x_{1}, x_{2}\right.$, $x_{3}$ ) is. | 4.5
From the problem, we have
$$
\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{1} x_{3}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}
$$
(1) - (2) gives $\left(x_{1}-x_{2}\right)\left(5 x_{3}-1\right)=0$.
Thus, $x_{1}=x_{2}$ or $x_{3}=\frac{1}{5}$.
Similarly, $x_{2}=x_{3}$ or $x_{1}=\frac{1}{5}$, $x_{3}=x_{1}$ or... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,185 |
One. (20 points) As shown in Figure 5, in rectangle $ABCD$, $DE = BG$, and $\angle BEC = 90^{\circ}$, $\frac{S_{\text{AEC}}}{S_{\text{EFCH}}} = n$. $ABCD$ is the area of the rectangle (the same below), $\frac{BC}{AB} = \lambda$, and it is known that $n$ is a natural number, and $\lambda$ is a rational number. Prove tha... | I. It is easy to prove that quadrilateral $B G D E$ is a parallelogram.
Let $A B=a, B G=x, S_{\triangle F C C}=S_{1}, S_{E E C H}=S$. Then $B C=\lambda a, G C=\lambda a-x$.
$\because$ Rt $\triangle A B G \backsim R \mathrm{Rt} \triangle G C D$, we have $\frac{A B}{G C}=\frac{B G}{C D}$.
$\therefore(\lambda a-x) x=a^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,186 |
II. (25 points) Does there exist a real number $k$ such that the quadratic equation $x^{2}+(2 k-1) x-(3 k+2)=0$ has two real roots, both of which lie between 2 and 4? If so, determine the range of values for $k$; if not, briefly explain the reason.
Translate the above text into English, please retain the original text... | Let the function $y=f(x)=x^{2}+(2 k-1) x-(3 k+2)$, then its graph is a parabola opening upwards, with its vertex below the $x$-axis, and between $x=2$ and $x=4$. The parabola intersects the $x$-axis at points also between $x=2$ and $x=4$. Therefore, the value of $k$ that meets the requirements should satisfy the follow... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,187 |
Three. (25 points) Let $S$ be a set of numbers composed of some numbers from $1,2,3, \cdots, 50$ (a set of numbers), such that the sum of any two numbers in $S$ cannot be divisible by 7. How many numbers from $1,2,3, \cdots, 50$ can $S$ contain at most (the maximum number of elements in $S$)? Prove your conclusion. | Three, divide the 50 numbers $1,2,3, \cdots, 50$ into seven different sets $F_{0}, F_{1}, \cdots, F_{6}$ based on their remainders when divided by 7:
$F_{0}$ consists of $7,14,21, \cdots, 49$ (divisible by 7), $F_{1}$ consists of $1,8,15, \cdots, 50$ (remainder 1 when divided by 7), $\qquad$
$F_{6}$ consists of $6,13,2... | 23 | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,188 |
1. Let the decimal number 1999 be written as a three-digit number $\overline{x y z}$ in base $b$, and $x+y+z=1+9+9+9$. Then the value of $b$ is ( ).
(A)11
(B) 15
(C) 18
(D) 28 | $-1 . \mathrm{D}$.
[Solution 1]: From the problem, we have
$$
x b^{2}+y b+z=1999 .(x \geqslant 1)
$$
Thus, $b^{3}>1999, b^{2} \leqslant 1999$.
Therefore, $12<b<45$.
Also, $x+y+z=28$,
Subtracting (2) from (1) gives
$$
(b-1)(b x+x+y)=1971 \text {. }
$$
So, $(b-1) \mid 1971=9 \times 3 \times 73$.
$$
\begin{array}{l}
\be... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,189 |
3. Let $O$ be a point inside $\triangle A B C$, and let $D$, $E$, $F$ be the projections of $O$ onto $B C$, $C A$, $A B$ respectively. When $\frac{B C}{O D}+\frac{C A}{O E}+\frac{A B}{O F}$ is minimized, $O$ is the ( ) of $\triangle A B C$.
(A) Centroid
(B) Centroid
(C) Incenter
(D) Circumcenter | 3.C.
Let the area and perimeter of $\triangle A B C$ be denoted as $S$ and $l$, respectively. By Cauchy's inequality, we have
$$
\begin{array}{l}
\left(\frac{B C}{O D}+\frac{C A}{O E}+\frac{A B}{O F}\right)(B C \cdot O D+C A \cdot O E+A B \cdot O F) \\
\geqslant(B C+C A+A B)^{2}=l^{2} . \\
\text { Also, } B C \cdot O ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,192 |
4. Let three real numbers $x_{1}, x_{2}, x_{3}$ be such that the sum of any one of them and the product of the other two is 1. Then the number of ordered triples $\left(x_{1}, x_{2}, x_{3}\right)$ that satisfy this condition is ( ).
(A)1
(B) 3
(C) 5
(D) more than 5
Translate the above text into English, please keep th... | 4.C.
From the problem, we know
$$
\left\{\begin{array}{l}
x_{1}+x_{2} x_{3}=1, \\
x_{2}+x_{1} x_{3}=1, \\
x_{3}+x_{2} x_{1}=1 .
\end{array}\right.
$$
I. If one of $x_{1}, x_{2}, x_{3}$ is 0, then the other two are 1, i.e., $(0,1,1), (1,0,1), (1,1,0)$ all satisfy the problem.
II. If $x_{1} x_{2} x_{3} \neq 0$, from (1)... | C | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 711,193 |
5. Given that $n$ is a natural number, $3^{1999}-3^{1998}-3^{1997}+$ $3^{1996}+3^{n+(-1)^{n}}$ is a perfect square. Then the value of $n$ is ( ).
(A) 2000
(B) 1999
(C) 1998
(D) 1997 | 5.B.
Let the perfect square be $m^{2}$ (where $m$ is a natural number), then the original equation simplifies to
$$
\begin{array}{l}
16 \times 3^{1996} + 3^{n+(-1)^{n}} = m^{2}. \\
\therefore\left(m + 4 \times 3^{998}\right)\left(m - 4 \times 3^{998}\right) = 3^{n+(-1)^{n}}. \\
\text { Let } m + 4 \times 3^{998} = 3^{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,194 |
2. Given the function $f(x+1)=\frac{1999^{2 x+1}}{1999^{2 x+1}-1}$. Then the value of the sum $\sum_{i=1}^{4000} f\left(\frac{i}{4001}\right)$ is $\qquad$ | $$
\begin{array}{l}
=\frac{1999^{2 x-1}}{1999^{2 x-1}-1}+\frac{1999^{1-2 x}}{1999^{1-2 x}-1}=1 \\
\therefore \sum_{i=1}^{4000} f\left(\frac{i}{4001}\right)=\sum_{i=1}^{2000}\left[f\left(\frac{i}{4001}\right)+f\left(\frac{4001-i}{4001}\right)\right] \\
=\sum_{i=1}^{2000}\left[f\left(\frac{i}{4001}\right)+f\left(1-\frac{... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,197 |
3. The solution set of the inequality $\frac{3 x^{2}}{(1-\sqrt[3]{3 x+1})^{2}} \leqslant x+2+$ $\sqrt[3]{3 x+1}$ is $\qquad$ | 3. $-\frac{2}{3} \leqslant x<0$.
Let $y=\sqrt[3]{3 x+1}$. Then $3= \pm i=y^{3}, 3 x=y^{3}$, the original inequality becomes
$$
\begin{array}{l}
\frac{(3 x)^{2}}{1-\sqrt[3]{3}-x+1)^{2}} \leqslant 3 x+6+3 \sqrt[3]{3 x+1}, \\
\frac{\left(y^{3}-1\right)^{2}}{(1-y)^{2}} \leqslant y^{3}+5+3 y, \\
\left(y^{2}+y+1\right)^{2} ... | -\frac{2}{3} \leqslant x < 0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,198 |
4. Given that $a, b, c$ are not all zero. Then the maximum value of $\frac{ab+2bc}{a^2+b^2+c^2}$ is $\qquad$ | 4. $\frac{\sqrt{5}}{2}$.
Let $x, y \in \mathbf{R}^{+}$, then $x^{2} a^{2}+b^{2} \geqslant 2 x a b, y^{2} b^{2}+c^{2} \geqslant 2 y b c$,
i.e., $\square$
$$
\begin{array}{l}
\frac{x}{2} a^{2}+\frac{b^{2}}{2 x} \geqslant a b, \\
y b^{2}+\frac{1}{y} c^{2} \geqslant 2 b c .
\end{array}
$$
(1) + (2) gives
$$
\begin{array}{... | \frac{\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,199 |
5. Given the function $f(x)$ with domain $\left(-\frac{\sqrt{3}}{3}\right.$, $\left.\frac{\sqrt{3}}{3}\right)$, $g(x)=f\left(\frac{\cos x}{2+\sin x}\right)$. Then the domain of the function $g(x)$ is | 5. $\left\{x \mid x \in \mathbf{R}\right.$, and $\left.x \neq 2 k \pi \pm \frac{\pi}{6}, k \in \mathbf{Z}\right\}$.
Let $\frac{\cos x}{2+\sin x}=t, \sin \alpha=\frac{1}{\sqrt{1+t^{2}}}$,
$\cos \alpha=\frac{t}{\sqrt{1+t^{2}}}$. Then
$\cos x-t \sin x=2 t$,
$\sin \alpha \cos x-\cos \alpha \sin x=\frac{2 t}{\sqrt{1+t^{2}}... | \left\{x \mid x \in \mathbf{R}\right., \text{ and } \left.x \neq 2 k \pi \pm \frac{\pi}{6}, k \in \mathbf{Z}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,200 |
6. The positive integer solutions of the equation $x^{2}+y^{2}=2000$ are $\left(x_{0}, y_{0}\right)$. Then the value of $x_{0}+y_{0}$ is $\qquad$ | 6.52 or 60.
Since $2=|1 \pm i|^{2}, 5=|2 \pm i|^{2}$, therefore, $2000=2^{4} \cdot 5^{3}=|1 \pm i|^{8} \cdot|2+i|^{6}$ or $|1 \pm i|^{8} \cdot|2-i|^{6}$ or $|1 \pm i|^{8} \cdot|2-i|^{4} \cdot|2+i|^{2}$ or $|1 \pm i|^{8} \cdot|2+i|^{2} \cdot|2-i|^{4}=|8+$ $44i|^{2}$ or $|40+20i|^{2}$ or $|40-20i|^{2}$ or $|40+20i|^{2}=... | 52 \text{ or } 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,201 |
4. There are $m$ identical machines working together, requiring $m$ hours to complete a task. (1) Suppose the same task is completed by $x$ machines ($x$ is a positive integer not greater than $m$), find the functional relationship between the required time $y$ (hours) and the number of machines $x$; (2) Draw the graph... | 4.(1) $y=\frac{m^{2}}{x}, x$ is a positive integer not greater than $m$ (2) omitted | y=\frac{m^{2}}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,202 |
Three, given $\theta_{i}(i=1,2, \cdots, n)$ are real numbers, and satisfy $\sum_{i=1}^{n}\left|\cos \theta_{i}\right| \leqslant \frac{2}{n+1}(n>1)$.
Prove: $\left|\sum_{i=1}^{n} i \cos \theta_{i}\right| \leqslant\left[\frac{n^{2}}{4}\right]+1$. | When $n=2 k(k \in \mathbf{N})$,
$$
\begin{array}{l}
{\left[\frac{n^{2}}{4}\right]+1-\left|\sum_{i=1}^{n} i \cos \theta_{i}\right| } \\
= k^{2}+1-\left|\sum_{i=1}^{n} i \cos \theta_{i}\right| \\
\geqslant k^{2}+1-\sum_{i=1}^{n} i\left|\cos \theta_{i}\right| \\
= n+(n-1)+\cdots+(k+1)-k-(k-1) \\
-\cdots-1+1-\left(\left|\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,203 |
Four, use $1,2,3,4,5$ to form all $n$-digit numbers where repetition is allowed, and the absolute difference between any two adjacent digits does not exceed 1. How many such numbers are there? | Let the number of $n$-digit numbers with the unit digit being 1 be $a_{n}$, the number of $n$-digit numbers with the unit digit being 2 be $b_{n}$, and the number of $n$-digit numbers with the unit digit being 3 be $c_{n}$. Then, the number of $n$-digit numbers with the unit digit being 4 is $b_{n}$, and the number of ... | \frac{1}{12}(1+\sqrt{3})^{n+3}+\frac{1}{12}(1-\sqrt{3})^{n+3}+\frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,204 |
Five, let $f(x)$ be a function that satisfies the following conditions:
(1) If $x>y$ and $f(x)+x \geqslant w \geqslant f(y)+y$, then there exists a real number $z \in[y, x]$, such that $f(z)=w-z$;
(2) The equation $f(x)=0$ has at least one solution, and among these solutions, there is one that is not greater than all t... | Let $F(x)=f(x)+x$, then $F(0)=1$.
Let $u$ be the smallest root of $f(x)=0$, then $F(u)=u$.
If $u0$
For any real number $x$, by (5) we have
$$
\begin{aligned}
0 & =f(x) f(u)=f(x f(u)+u f(x)+x u) \\
& =f(u f(x)+u u) .
\end{aligned}
$$
$\therefore u f(x)+x u$ is a root of $f(x)=0$.
$\because u$ is the smallest root of $f(... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,205 |
One TV station has $n$ ad breaks in a day, during which a total of $m$ ads were broadcast. In the first ad break, one ad and $\frac{1}{8}$ of the remaining $(m-1)$ ads were broadcast. In the second ad break, 2 ads and $\frac{1}{8}$ of the remaining ads were broadcast. This pattern continued for each subsequent ad break... | Let's denote the number of ads left to be broadcast after $k$ broadcasts as $a_{k}$. Then, the number of ads broadcast at the $k$-th time is:
$$
k+\frac{1}{8}\left(a_{k-1}-k\right)=\frac{1}{8} a_{k-1}+\frac{7}{8} k .
$$
Thus, $a_{k}=a_{k-1}-\left(\frac{1}{8} a_{k-1}+\frac{7}{8} k\right)=\frac{7}{8} a_{k-1}-\frac{7}{8}... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,206 |
II. Given an isosceles $\triangle ABC$ with vertex angle $A$ being $\frac{\pi}{7}$, and point $D$ is on the leg $AB$ such that $CD = \sqrt{2} AD$. Prove: $AD = BC$.
保留源文本的换行和格式，直接输出翻译结果。 | Given $a=\frac{\pi}{14}$,
$$
A D=m, A C=n, B C=a .
$$
In $\triangle A C D$, by the cosine rule we have
$$
\begin{array}{l}
(\sqrt{2} m)^{2}=m^{2}+n^{2} \\
-2 m n \cos 2 \alpha,
\end{array}
$$
which gives $\cos 2 \alpha=\frac{n^{2}-m^{2}}{2 m n}$.
In the isosceles $\triangle A B C$,
$$
\begin{array}{l}
\sin \alpha=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,207 |
Three, it is known that the natural number $N$ has 20 positive integer factors (including 1 and itself), which are sequentially denoted as $d_{1}, d_{2}, d_{3}, \cdots, d_{21}$, and the factor with index $d_{8}$ is $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$. Find the natural number $N$. | Three, because $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$ is a factor of $N$, so, $d_{1}+d_{4}+d_{6}$ and $d_{4}+d_{8}+d_{13}$ are factors of $N$. Therefore,
$$
\begin{array}{l}
d_{1}+d_{4}+d_{6} \geqslant d_{7}, d_{4}+d_{8}+d_{13} \geqslant d_{14} . \\
\therefore d_{d 8}=\left(d_{1}+d_{4}+d_{6}\r... | 2000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,208 |
Initial 83. Given the function $f(x)=a x^{2}+b x+c$, when $x \in$ $[-1,1]$, $|f(x)| \leqslant 1$. Prove:
$$
|a|+|b|+|c| \leqslant 3 .
$$ | $\begin{array}{l}\quad \text { Proof: } \because f(1)=a+b+c, f(0)=c, f(-1)=a \\ -b+c, \\ \quad \therefore|a+b|=|f(1)-f(0)| \\ \leqslant|f(1)|+|f(0)| \leqslant 2, \\ |a-b|=|f(-1)-f(0)| \\ \leqslant|f(-1)|+|f(0)| \leqslant 2, \\ \therefore|a|+|b| \leqslant \max |: a+b|,|a-b| \mid \leqslant 2 . \\ \text { Also } \because|... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,209 |
Let $t$ be a natural number. Then the inequality
$$
\frac{\sqrt{8 t-7}-1}{2}<x<\frac{\sqrt{8 t+1}+1}{2}
$$
has only one positive integer solution. | Proof: Let $T_{1}=\frac{\sqrt{8 t-7}-1}{2}, T_{2}=\frac{\sqrt{8 t+1}+1}{2}$.
$$
\text { From }(2 m)^{2}=4 m^{2},(2 m-1)^{2}=4 m(m-1)+1
$$
we know that the remainder of a perfect square when divided by 8 can only be 0, 1, or 4. We will discuss two cases below:
Case $1.8 t-6,8 t-5, \cdots, 8 t-1,8 t$ do not contain any... | proof | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,210 |
Let $(p, q)$ be a positive integer solution of the equation $x^{2}-m y^{2}=1$ ($m$ is a positive integer), and the sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=\frac{1}{2} a_{n}+\frac{m}{2 a_{n}}$ $(n \geqslant 1)$, and $a_{1}=\frac{p+1}{q}$ (or $a_{1}=\frac{p-1}{q}$. Prove that for any $n>1, \frac{2}{\sqrt{a_{n}... | Proof: First, prove the following lemma.
Lemma: For any $n>1$, there exist $p_{n}, q_{n} \in \mathbf{N}$, satisfying $p_{n}^{2}-m q_{n}^{2}=1$, and such that $a_{n}=\frac{p_{n}}{q_{n}}$.
Prove by mathematical induction.
When $n=2$,
$$
\begin{array}{l}
a_{2}=\frac{1}{2} a_{1}+\frac{m}{2 a_{1}}=\frac{p \pm 1}{2 q}+\frac{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,211 |
Given $x_{i}, y_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n$. Prove:
$$
\begin{array}{l}
\frac{x_{1}^{3}}{y_{1}}+\frac{x_{2}^{3}}{y_{2}}+\cdots+\frac{x_{n}^{3}}{y_{n}} \\
\geqslant \frac{\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{3}}{n\left(y_{1}+y_{2}+\cdots+y_{n}\right)} .
\end{array}
$$ | Proof: Let $A=\sum_{i=1}^{n} \frac{x_{i}^{3}}{y_{i}}, B=\sum_{i=1}^{n} y_{i}$, then
$$
\frac{1}{n}+\frac{1}{A} \cdot \frac{x_{i}^{3}}{y_{i}}+\frac{y_{i}}{B} \geqslant 3 \cdot \frac{x_{i}}{(n A B)^{\frac{1}{3}}}, i=1,2, \cdots, n \text {. }
$$
By adding both sides of the above $n$ inequalities, we get
$$
\begin{array}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,212 |
For example, the perimeter of $\angle S A B C$ is $24$, $M$ is the midpoint of $A B$, $M C=M A=5$. Then the area of $\triangle A B C$ is ( ).
(A) 12
(B) 16
(C) 24
(D) 30 | $$
\begin{array}{c}
\because B M=M A= \\
M C=5, \\
\therefore \angle 1=\angle A, \\
\angle 2=\angle B . \\
\text { So } \angle 1+\angle 2 \\
=90^{\circ} . \\
\therefore A C^{2}+B C^{2}=A B^{2}=100 . \\
\text { Also, since } B B+A C+B C=24, \\
\therefore A C+B C=14, \\
(A C+B C)^{2}=14^{2}, \\
2 A C \cdot B C=196-\left(... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,213 |
Example 2 In the acute triangle $\triangle ABC$, $\angle A=30^{\circ}$, a circle is constructed with $BC$ as its diameter, intersecting $AB$ and $AC$ at $D$ and $E$ respectively. Connecting $DE$ divides $\triangle ABC$ into $\triangle ADE$ and quadrilateral $DBCE$, with areas denoted as $S_{1}$ and $S_{2}$ respectively... | : As shown in Figure 2, connect $B E$, then $\angle A E B=90^{\circ}$.
$$
\begin{array}{l}
\therefore \frac{A E}{A B}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} . \\
\because \angle A E D=\angle A B C, \\
\therefore \triangle A E D \operatorname{\triangle o} \triangle A B C . \\
\therefore \frac{S_{\triangle A E D}}{S_{\triang... | 3:1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,214 |
6. Given that the age order of $a$, $b$, and $c$ satisfies:
(1) If $b$ is not the oldest, then $a$ is the youngest;
(2) If $c$ is not the youngest, then $a$ is the oldest.
Then the ages of these three people from oldest to youngest are ( , .
(A) bac
(B) $c b a$
(C) $a b c$
(D) $a c b$ | 6. (A).
Age is represented from oldest to youngest using $1,2,3$, and $a_{i}$ indicates that $a$'s age ranking is $i(1 \leqslant i \leqslant 3)$, with $b_{i}, c_{i}$ having the same meaning.
From condition (1), if $b$ is not the oldest, then $\bar{b}_{1} a_{3}$; or if $b$ is exactly the oldest, then $a$ can take any ... | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 711,216 |
1. The solution set of the inequality $(x-1) \sqrt{x^{2}-x-2} \geqslant 0$ is $\qquad$ . | 2. $\{-1 \mid \cup\{x \mid x \geqslant 2\}$.
Solution 1: Classify by the sign of $x-1$.
(1) When $x-1 \geqslant 0$, it must be that
$$
\sqrt{x^{2}-x-2} \geqslant 0 \text {. }
$$
That is, $\left\{\begin{array}{l}x \geqslant 1, \\ x \leqslant-1\end{array}\right.$ or $x \geqslant 2$.
Thus, $x \geqslant 2$.
(2) When $x-10... | \{-1\} \cup \{x \mid x \geqslant 2\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,217 |
2. The parabola $y=a x^{2}+b x+c(a \neq 0)$ intersects the $x$-axis at points $A$ and $B$, then the equation of the circle with $A B$ as its diameter is $\qquad$ | 2. $a x^{2}+b x+c+a y^{2}=0$.
By the quadratic formula, the diameter of the circle is
$$
|A B|=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{|a|} .
$$
From the axis of symmetry of the parabola, the center of the circle is $\left(-\frac{b}{2 a}, 0\right)$. Therefore, the circle with diameter $A B$ is
$$
\left(x+\... | a x^{2}+b x+c+a y^{2}=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,218 |
3. The slant height of a cone is $l$, and the angle it forms with the base is $\theta$. The edge length of the inscribed cube in this cone is $\qquad$ (the cube has 4 vertices on the base of the cone and 4 vertices on the lateral surface of the cone). | $3 \cdot \frac{2 l \sin \theta}{2+\sqrt{2} \tan \theta}$.
As shown in Figure 7, consider the axial section, we have
$$
\begin{array}{l}
S A=l, \\
\angle S A B=\theta, \\
A O=l \cos \theta .
\end{array}
$$
Let the edge length of the cube be $x$, then $D O=\frac{\sqrt{2}}{2} x, A D=$ $l \cos \theta-\frac{\sqrt{2}}{2} x$... | \frac{2 l \sin \theta}{2+\sqrt{2} \tan \theta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,219 |
6. Set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$, calculate the sums of any two elements in $A$ to form set $B=$ $\{3,4,5,6,7,8,9,10,11,13\}$. Then $A=$ $\qquad$ . | 6. $\{1,2,3,5,8\}$.
Assume $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=3, \\
x_{4}+x_{5}=13 .
\end{array}
$$
Since in the set of binary elements of $A$, each element appears 4 times, the sum of the elements of set $B$ is
$$
\begin{array}{l}
4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right) \\
=3+4... | A=\{1,2,3,5,8\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,222 |
Three. (Full marks 20 points) The two endpoints of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $A(a, 0)$ and $A_{1}(-a, 0)$. A perpendicular line is drawn from a point within the segment $A A_{1}$ intersecting the ellipse at points $C$ and $D$. Connect $A C$ and $A_{1} D$ intersecting at $P$. Fin... | Three, let a perpendicular line be drawn through the point $\left(x_{1}, 0\right)\left(0<\left|x_{1}\right|<a\right)$ on $A A_{1}$, intersecting the ellipse at $C\left(x_{1}, y_{1}\right), D\left(x_{1},-y_{1}\right)$. From the intersection of $A C$ and $A_{1} D$, we know that $0<\left|y_{1}\right|<b$.
The equations of ... | \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,223 |
Four, (Full marks 20 points) In the right triangular prism $A B C$ $A_{1} B_{1} C_{1}$, $B C_{1} \perp A_{1} C, B C_{1} \perp A B_{1}$. Prove: $\triangle A B C$ is an isosceles triangle. | Extend $B A$ to $D$, such that $A D=A B$, and connect $A_{1} D$ (as shown in Figure 8), then we have $A B_{1} / / A_{1} D$. From $B C_{1} \perp A B_{1}$, we get
$$
{ }^{B C} C_{1} \perp A_{1} D \text{. }
$$
Also, $B C_{1} \perp C A_{1}$, so $B C_{1} \perp$ plane $A_{1} C D$, which implies
$$
P C \perp C D \text{. }
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,224 |
Five. (Full marks 20 points) For given angles $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$, discuss whether the equation $x^{n}+x^{n-1} \sin \alpha_{1}+x^{n-2} \sin \alpha_{2}+\cdots+x \sin \alpha_{n-1}+\sin \alpha_{n}=0$ has complex roots with modulus greater than 2.
---
Please note that the translation preserves th... | Five, the answer is negative. If not, assume there exists $x_{0}$ as a complex solution of the equation, and $\left|x_{0}\right|>2$. Then
$$
\begin{array}{l}
x_{0}^{\pi}=-x_{0}^{\prime \prime}{ }^{1} \sin \alpha_{1}-\cdots-x_{0} \sin \alpha_{n-1}-\sin \alpha_{n} \text {. } \\
\therefore\left|x_{0}\right|^{\prime} \leqs... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,225 |
One, (Full marks 50 points) In the acute triangle $\triangle ABC$, the circumradius is $R=1$, and the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$. Prove:
$$
\frac{a}{1-\sin A}+\frac{b}{1-\sin B}+\frac{c}{1-\sin C}>12 .
$$ | First, prove that
$a+b+c>4$ (or $\sin A+\sin B+\sin C>2$). This is essentially the second problem of the 1999 No. 6 issue of "Middle School Mathematics" Mathematical Olympiad High School Training Question (41). Let $A$, $B$, and $C$ represent the three interior angles of $\triangle ABC$. Here, we provide another trigon... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,227 |
II. (Full marks 50 points) Let $S$ be a subset of $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements in $S$ is not divisible by 7. What is the maximum number of elements that $S$ can have? | Divide $\{1,2, \cdots, 50\}$ into 7 classes according to modulo 7:
$$
\begin{array}{l}
k_{1}=\{1,8,15,22,29,36,43,50\}, \\
k_{2}=\{2,9,16,23,30,37,44\}, \\
k_{3}=\{3,10,17,24,31,38,45\}, \\
k_{4}=\{4,11,18,25,32,39,46\}, \\
k_{5}=\{5,12,19,26,33,40,47\}, \\
k_{6}=\{6,13,20,27,34,41,48\}, \\
k_{0}=\{7,14,21,28,35,42,49\... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,228 |
Three, (Full marks 50 points) Some contestants participate in a math competition, some of whom know each other, while others do not. Any two contestants who do not know each other have exactly two common acquaintances. If $A$ and $B$ know each other but have no common acquaintances, prove: $A$ and $B$ have the same num... | Three, using points to represent people, and connecting points with line segments if the people know each other, according to the problem, there is a line segment between $A$ and $B$ (Figure 9).
Since $A$ and $B$ have no common acquaintances, anyone who knows $A$ does not know $B$, and anyone who knows $B$ does not kn... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,229 |
In $\triangle A B C$, $\angle B=50^{\circ}, \angle C=30^{\circ}, D$ is a point inside $\triangle A B C$, satisfying $\angle D B C=\angle D C B=20^{\circ}$. Find the degree measure of $\angle D A C$. | Solution: As shown in the figure, let $E$ be the circumcenter of $\triangle ABC$, and connect $AE$, $BE$, and $CE$. Since $\angle AEB = 2 \angle ACB = 60^{\circ}$, $\triangle ABE$ is an equilateral triangle, and $\angle ECB = \angle EBC = 10^{\circ}$, $\angle BEC = 160^{\circ}$.
On the extension of $BD$, take a point ... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,230 |
Initial 86. $A D, B E, C F$ are the three altitudes of acute $\triangle A B C$, the distance from the circumcenter $O$ to side $B C$ is equal to $d$, and it is given that $B F + C E = B C$. Prove: $\frac{1}{A D} + \frac{1}{B E} + \frac{1}{C F} = \frac{1}{d}$. | Let $B C=a$, $C A=b$, $A B=c$, the circumradius of $\triangle A B C$ be $R$, the inradius be $r$, the semiperimeter be $p$, and the area be $\triangle$.
Then $R=\frac{a b c}{4 \triangle}, r=\frac{\Delta}{p}$.
From $B F+C E=B C$
$$
\begin{array}{l}
\Rightarrow \frac{B F}{B C}+\frac{C E}{B C}=1 \Rightarrow \cos B+\cos C=... | \frac{1}{A D}+\frac{1}{B E}+\frac{1}{C F}=\frac{1}{d} | Geometry | proof | Yes | Yes | cn_contest | false | 711,231 |
85. Given that $a$, $b$, $c$ are positive numbers, and $abc \leqslant 1$. Prove:
$$
\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geqslant a+b+c+Q \text {. }
$$
where $Q=|(1-a)(1-b)(1-c)|$. | Proof: (1) When $a \leqslant 1, b \leqslant 1, c \leqslant 1$,
$$
\begin{aligned}
\because & (1-a)(1-b) \geqslant 0, \\
\therefore & 2-a-b \geqslant 1-a b \geqslant c(1-a b), \\
& 2+a b c \geqslant a+b+c \geqslant a b+b c+a c . \\
\therefore & \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \\
\geqslant & 3 \geqslant a b+b c+a c+1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,232 |
Example 6: In the plane, there is a fixed point $M$ and... a fixed line $l$. Take any three points $A$, $B$, $C$ on $l$, and take three other points $D$, $E$, $F$ in the plane such that $\triangle M A D$, $\triangle M B E$, $\triangle M C F$ are equilateral triangles oriented counterclockwise. Prove that $D$, $E$, and ... | Solution: Taking point $A$ as the origin and the fixed line $l$ as the real axis to establish a complex plane, let $B$ and $C$ correspond to the real numbers $b$ and $c$, respectively, and point $M$ correspond to the complex number $z_{M}$. Then, by the geometric meaning of complex number multiplication, we have
$$
\be... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,234 |
7. In an equilateral triangle $ABC$ with side length 10, two squares (squares 甲 and 乙 overlap on one side, both have one side on $BC$, square 甲 has one vertex on $AB$, and square 乙 has one vertex on $AC$) are inscribed as shown in Figure 15. Find the minimum value of the sum of the areas of these two inscribed squares. | $$
\left.\frac{25}{2}(3-\sqrt{3})^{2}\right)
$$ | \frac{25}{2}(3-\sqrt{3})^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,239 |
Example 1 Given that $F$ is any point on the bisector of $\angle P$, and two lines $A D$ and $B C$ are drawn through $F$ intersecting the sides of $\angle P$ at $A, D, B, C$.
Prove: $\frac{1}{P A}+\frac{1}{P D}=\frac{1}{P B}+\frac{1}{P C}$. | Proof: As shown in Figure 1, with $F$ as the pole and the ray $F P$ as the polar axis, establish a polar coordinate system.
$$
\begin{array}{c}
\text { Let } \angle A P F= \\
\angle C P F=\alpha, \angle A F X \\
=\theta_{1}, \angle B F X=\theta_{2}, \\
P F=\rho_{0} \text {. Then in } \triangle A F P, \\
\frac{A P}{\sin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,240 |
8. As shown in Figure 16, in hexagon $A B C D E F$, the diagonals $A D, B E, C F$ intersect at point $O$, and $A D=B E=C F=2, \angle A O B=\angle C O D=$ $\angle E O F=60^{\circ}, S$ is the sum of the areas of $\triangle A O B$, $\triangle C O D$, and $\triangle E O F$. Then ( )
(A) $S\sqrt{3}$
(D) The relationship bet... | (Answer: (A)) | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,241 |
Example 3 In the plane of the regular pentagon $A B C D E$, there exists a point $P$ such that the area of $\triangle P C D$ is equal to the area of $\triangle B C D$, and $\triangle A B P$ is an isosceles triangle. The number of different points $P$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | Solution: As shown in Figure 3, point $P$ can only be on line $l_{1}$ (i.e., line $B E$) and line $l_{2}$,
where the distance between $l_{2}$ and line $C D$ is equal to the distance between $l_{1}$ and line $C D$. Therefore, in isosceles $\triangle P A B$:
(1) When $A B$ is the base, the perpendicular bisector of $A B... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,243 |
Example 2 In $\triangle ABC$, the internal angle bisector of $\angle A$ is $AT, |BD|=|CE|, D, E$ are on $AB, AC$ respectively, the projections of $B, C$ on the external angle bisector of $\angle A$ are $P, Q$, and the midpoints of $DE, BC$ are $M, N$ respectively. Prove:
(1) $MN \parallel AT$;
(2) $BQ, CP, AT$ intersec... | Proof: (1) As shown in Figure 2, let
$$
\angle \mathrm{A}=2 \alpha,
$$
$$
\begin{array}{l}
|A C|=b,|A B|= \\
c,|B D|=|C F|=
\end{array}
$$
$t$.
From the coordinates of $B, C, D, E$ and the midpoint formula, we easily get
$$
\begin{array}{l}
M_{y}=\frac{D_{y}+E_{y}}{2}=\frac{1}{2}(c-b) \sin \alpha, \\
N_{y}=\frac{B_{y}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,245 |
For example, $\triangle ABC$ is an isosceles triangle, $AB = AC$. Suppose:
(1) $M$ is the midpoint of $BC$, $O$ is on line $AM$ such that $OB$ is perpendicular to $AB$;
(2) $Q$ is any point on segment $BC$ different from $B$ and $C$;
(3) $E$ is on line $AB$, $F$ is on line $AC$, such that $E$, $Q$, $F$ are distinct and... | Proof: As shown in Figure 3, let $BC = 2$, $\angle C = \alpha$, and $\angle FQC = \theta$. Then $OM = \cot \alpha$.
By the Law of Sines, we have
$$
\frac{QF}{\sin \alpha} = \frac{1 + QM}{\sin (\alpha + \theta)}
$$
and $\frac{QE}{\sin \alpha} = \frac{1 - QM}{\sin (\alpha - \theta)}$.
Sufficiency: When $QE = QF$, it is ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,246 |
Example 4 In the circumcircle of rectangle $A B C D$, take a point $M$ on the arc $\overparen{A B}$ different from vertices $A$ and $B$. Points $P$, $Q$, $R$, and $S$ are the projections of $M$ onto lines $A D$, $A B$, $B C$, and $C D$ respectively.
Prove: Lines $P Q$ and $R S$ are perpendicular to each other. | Proof: As shown in Figure 4, let the unit circle $O: x^{2}+y^{2}=1$, $\mathrm{A}(\cos \theta, \sin \theta), \mathrm{M}(\cos \alpha, \sin \alpha)(\theta<\alpha<\pi-$ $\theta$ ). By symmetry, we have
$$
\begin{array}{l}
Q(\cos \alpha, \sin \theta), S(\cos \alpha,-\sin \theta), \\
P(\cos \theta, \sin \alpha), R(-\cos \th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,247 |
Example 5 Let the incircle of the equilateral $\triangle ABC$ touch the sides $AB, BC, CA$ at $D, E, F$ respectively. For any point $P$ on the arc $\overparen{DF}$ of the incircle, the distances from $P$ to the three sides are $d_{1}, d_{2}, d_{3}$. Prove: $\sqrt{d_{1}}+\sqrt{d_{2}}=\sqrt{d_{3}}$. | Proof: As shown in Figure 5, the incenter $O$ is the origin of the coordinate system. Let the radius $r = 1$, and $P(\cos \theta, \sin \theta)$ $\left(30^{\circ} \leqslant \theta \leqslant 150^{\circ}\right)$. Then the equation of $AC$ is
\[ x \cos 30^{\circ} + y \sin 30^{\circ} - 1 = 0, \]
the equation of $AB$ is
\[ x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,248 |
Example 6 In $\triangle ABC$, $AA_{1}$ is a median, $AA_{2}$ is an angle bisector, $K$ is a point on $AA_{1}$ such that $KA_{2} \parallel AC$. Prove that $AA_{2} \perp KC$.
untranslated text remains the same as requested. | Proof: As shown in Figure 6, let
$$
\begin{array}{l}
B C: y=k_{1} x, \text { (1) } \\
A_{2} K: y=-k_{2} x \text { (2) } \\
\left(k_{1}>0, k_{2}>0\right), \\
A A_{2}=a .
\end{array}
$$
Then the equation of $A C$ is
$$
y=-h_{2} x+a \text {, }
$$
The equation of $A S$ is $y=k_{2} x+a$.
Solving (1) and (4) gives $C\left(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,249 |
For example, $8 A C$ and $C E$ are two diagonals of the regular hexagon $A B C D E F$, and points $M$ and $N$ internally divide $A C$ and $C E$ respectively, such that $A M: A C = C N: C E = r$. If points $B$, $M$, and $N$ are collinear, find $r$.
---
The above text translated into English, preserving the original te... | Proof: As shown in Figure 8, let $AC = 2$, then $A(-1, 0), C(1, 0), B(0, -\frac{\sqrt{3}}{3}), E(0, \sqrt{3})$
$$
\begin{array}{l}
\because AM: AC \\
= CN: CE = r,
\end{array}
$$
$\therefore$ By the section formula, we get
$$
M(2r-1, 0), N(1-r, \sqrt{3}r).
$$
Also, $\because B, M, N$ are collinear,
thus $k_{BM} = k_{M... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,251 |
Example 9 Through the midpoint $M$ of the line segment $AB$ between two intersecting lines, draw any two line segments $CD$ and $EF$ $(C, E$ on the same line), and $CF, ED$ intersect $AB$ at $P, Q$ respectively. Prove: $PM = MQ$.
| Proof: As shown in Figure 9, establish a Cartesian coordinate system. Let \( AB = 2a \), then the equations of \( CD \) and \( EF \) are
\[
\left(y - k_{CD} x\right)\left(y - k_{EF} x\right) = 0;
\]
The equations of \( AD \) and \( BC \) are
\[
\begin{array}{l}
\left[y - k_{AD}(x + a)\right]\left[y - k_{BC}(x - a)\righ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,252 |
For example, $10 G H$ and $E D$ are two perpendicular diameters of a circle. From a point $B$ outside the circle, two tangents are drawn to the circle, intersecting the line $G H$ at $A$ and $C$. The lines $B E$ and $B D$ intersect $G H$ at $F$ and $K$. Prove: $A F=K C$. | Proof: Let the equation of $\odot O$ be $x^{2}+y^{2}=r^{2}$, and $B(a, b)(b \neq \pm r)$. Then the equation of the tangent line is
$$
y-b=k(x-
$$
$a)\left(a \neq \pm r\right.$. When $k_{1}$ and $k_{2}$ exist). It is easy to get
$$
\begin{aligned}
k_{1} k_{2} & =\frac{b^{2}-r^{2}}{a^{2}-r^{2}}, \\
k_{1}+k_{2} & =\frac{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,253 |
For example, as shown in Figure 4, in quadrilateral $A B C D$, $E$ is the midpoint of $B C$, and $A E$ intersects $B D$ at $F$. If $D F = B F, A F = 2 E F$, then $S_{\triangle A C D}: S_{\triangle A B C}: S_{\triangle A B D}$ $=$ $\qquad$ . ( $S_{\triangle A C D}$ represents the area of $\triangle A C D$, and so on) | Solution: Connect $D E$.
$$
\begin{array}{l}
\text { Let } S_{\triangle X E F}=S . \\
\because A F=2 E F, \\
\therefore S_{\triangle B A F}=2 S_{\triangle B E F}=2 S . \\
\text { Also, } \because D F=B F, \\
\therefore S_{\triangle D E F}=S_{\triangle B E F}=S .
\end{array}
$$
Similarly, $S_{\triangle D E C}=S_{\trian... | 1: 3: 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,254 |
Example 11 Circle $O_{1}$ and circle $O_{2}$ are contained within circle $O$, and are tangent to circle $O$ at two different points $M$ and $N$. Circle $O_{1}$ passes through the center of circle $O_{2}$, and the line through the two intersection points of circles $O_{1}$ and $O_{2}$ intersects circle $O$ at points $A$... | Proof: As shown in Figure 11, let the radii of circles $O$, $O_{1}$, and $O_{2}$ be $r$, $r_{1}$, and $r_{2}$ respectively, and $\angle O_{2} O_{1} O = \alpha$. Then,
the equation of circle $O_{1}$ is
$$
x^{2} + y^{2} = r_{1}^{2},
$$
the equation of circle $O_{2}$ is
$$
\left(x - r_{1} \cos \alpha\right)^{2} + \left(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,255 |
1. $M$ is the midpoint of the base $AC$ of isosceles $\triangle ABC$, $MH \perp BC$ at $H$, and $P$ is the midpoint of $MH$. Prove that $AH \perp BP$. | (Given: $A C=4$,
$$
\begin{array}{l}
\angle C=\alpha \text{. Then } A(-2,0), \\
B(0,2 \tan \alpha), H\left(2 \sin ^{2} \alpha, 2 \sin \alpha \cdot \cos \alpha\right) \\
\therefore P\left(\sin ^{2} \alpha, \sin \alpha \cdot \cos \alpha\right) . \\
\therefore k_{A H} \cdot k_{B P} \\
=\frac{2 \sin \alpha \cdot \cos \alph... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,256 |
2. Given a square $A B C D$, points $P$ and $Q$ are on $A B$ and $B C$ respectively, and $B P=B Q, B H \perp P C$ at $H$. Prove: $\angle D H Q$ is a right angle. | (Tip: Let the side length of the square be $1, \angle B C P=\alpha$. Then $D(1,1), B P=$ $\operatorname{tg} \alpha . \therefore Q(\operatorname{tg} \alpha, 0)$, $H\left(\sin ^{2} \alpha, \sin \alpha \cdot \cos \alpha\right)$.
$$
\begin{aligned}
\therefore & k_{Q H} \cdot k_{D H} \\
& =\frac{\sin \alpha \cdot \cos \alph... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,257 |
Example 1 If $f(x)=\left(2 x^{5}+2 x^{4}-53 x^{3}-\right.$ $57 x+54)^{1998}$, then $f\left(\frac{\sqrt{1}-\frac{1}{2}-i}{2}\right)=$ $\qquad$ . | Let $\frac{\sqrt{111}-1}{2}=x$, then we have
$$
\begin{array}{l}
2 x^{2}+2 x-55=0 \\
\because 2 x^{5}+2 x^{4}-53 x^{3}-57 x+54 \\
=x^{3}\left(2 x^{2}+2 x-55\right)+x\left(2 x^{2}+2 x\right. \\
\quad-55)-\left(2 x^{2}+2 x-55\right)-1 \\
\therefore f\left(\frac{\sqrt{111}-1}{2}\right)=(-1)^{1998}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,258 |
Example 2 Given $f(x)=\frac{2 x}{1+x}$. Find
$$
\begin{array}{l}
f(i)+f(2)+\cdots+f(100)+f\left(\frac{1}{2}\right) \\
+f\left(\frac{2}{2}\right)+\cdots+f\left(\frac{100}{2}\right)+\cdots+f\left(\frac{1}{100}\right) \\
+f\left(\frac{2}{100}\right)+\cdots+f\left(\frac{100}{100}\right)=
\end{array}
$$ | Given $f(x)=\frac{2 x}{1+x}$, we have
$$
f(x)+f\left(\frac{1}{x}\right)=2, f(1)=1 \text {. }
$$
In the required expression, there are 100 $f(1)$ terms, and the remaining $f(x)$ and $f\left(\frac{1}{x}\right)$ appear in pairs, totaling $\frac{1}{2}\left(100^{2}-\right.$ 100 ) pairs.
Therefore, the required sum is
$$
10... | 10000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,259 |
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
保留源文本的换行和格式，直接输出翻译结果如下：
```
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, ... | Solution: Let $f(x)=\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}\right.$
$$
\begin{aligned}
& -3 x-3)^{2001} \\
= & a_{0}+a_{1} x+a_{2} x^{2}+\cdots \\
& +a_{4001} x^{4001}+a_{4002} x^{4002} .
\end{aligned}
$$
Let $x=1$ and $x=-1$, we get
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+\cdots+a_{4001}+a_{4002} \\
=f(1)=0, \\
a_{... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,260 |
Example 4 Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=1, f(2)=2, f(3)=3$, then the value of $\frac{1}{4}[f(4)+f(0)]$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8 | Solution: Let $g(x)=f(x)-x$, then
$$
\begin{array}{l}
g(1)=g(2)=g(3)=0, \\
f(x)=g(x)+x
\end{array}
$$
Therefore, $g(x)=(x-1)(x-2)(x-3)(x-r)$.
( $r$ is a constant)
$$
\begin{aligned}
\therefore & \frac{1}{4}[f(4)+f(0)] \\
& =\frac{1}{4}[g(4)+4+g(0)] \\
& =\frac{1}{4}[6(4-r)+4+6 r]=7 .
\end{aligned}
$$
Hence, the answe... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,261 |
Example 5 Given $P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+$ $a_{3} x^{2}+a_{4} x+a_{5}$, and when $k=1,2,3,4$, $P(k)$ $=k \times 1$ 997. Then $P(10)-P(-5)=$ $\qquad$ | Solution: Let $Q(x)=P(x)-1997 x$, then when $k=$ $1,2,3,4$, $Q(k)=P(k)-1997 k=0$.
Therefore, $1,2,3,4$ are roots of $Q(x)=0$.
Thus, we can set $Q(x)=(x-1)(x-2)(x-$ $3)(x-4)(x-r)$, then
$$
\begin{aligned}
& P(10)=Q(10)+1997 \times 10 \\
= & 9 \times 8 \times 7 \times 6 \times(10-r)+1997 \times 10, \\
& P(-5)=Q(-5)+1997 ... | 75315 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,262 |
Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+1994^{2}$. | Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of
$$
\underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\cdots+9^{2}}_{\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2}
$$
The unit digits follow a... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,263 |
Example 7 Let $a>0, b>0$, and $\sqrt{a}(\sqrt{a}+2 \sqrt[3]{b})$ $=\sqrt[3]{b}(\sqrt{a}+6 \sqrt[3]{b})$. Then the value of $\frac{2 a^{4}+a^{3} b-128 a b^{2}-64 b^{3}+b^{4}}{a^{4}+2 a^{3} b-64 a b^{2}-128 b^{3}+2 b^{4}}$ is | Given: From the known, taking $a=4$, we get $b=1$.
$$
\begin{aligned}
\therefore \text { Original expression } & =\frac{2 \times 4^{4}+4^{3}-128 \times 4-64+1}{4^{4}+2 \times 4^{3}-64 \times 4-128+2} \\
& =\frac{1}{2} .
\end{aligned}
$$ | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,264 |
Example 5 Given that the area of square $A B C D$ is 35 square centimeters, $E$ and $F$ are points on sides $A B$ and $B C$ respectively, $A F$ and $C E$ intersect at $G$, and the area of $\triangle A B F$ is 5 square centimeters, the area of $\triangle B C E$ is 14 square centimeters. Then, the area of quadrilateral $... | Solution: As shown in Figure 5, connect $A C, B G$, then
$$
\begin{array}{l}
\frac{B F}{B C}=\frac{S_{\triangle A B F}}{S_{\triangle A B C}} \\
=\frac{2}{7} .
\end{array}
$$
Similarly, $\frac{B E}{B A}=\frac{4}{5}$.
Let $S_{\triangle A G E}=a, S_{\triangle E B G}=b, S_{\triangle B G F}=c$, $S_{\triangle A G C}=d$, the... | 4 \frac{20}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,265 |
Example 8 Let
$$
\begin{aligned}
S= & \sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots \\
& +\sqrt{1+\frac{1}{1997^{2}}+\frac{1}{1998^{2}}} .
\end{aligned}
$$
Then the integer closest to $S$ is ( ).
(A) 1997
(B) 1998
(C) 1999
(D) 2000 | $$
\begin{array}{l}
\text { Solution: } \because \sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}\right)^{2}-\frac{2}{n}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(\frac{n+1}{n}\right)^{2}-2 \times \frac{n+1}{n} \times \frac{1}{n+1}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(\frac{n+1}{n}-\frac{1}{n+1}\ri... | 1998 | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,266 |
Example 9 Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms $S_{n}=$ $n^{2}$. Then
$$
\begin{array}{l}
\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots \\
+\frac{1}{\sqrt{a_{998}}+\sqrt{a_{999}}}=(\quad) .
\end{array}
$$
(A) $\frac{\sqrt{1996}-1}{2}$
(B) $\frac{... | $$
\begin{array}{l}
\text { Solution: Original expression }=\frac{1}{2}\left[\left(\sqrt{a_{2}}-\sqrt{a_{1}}\right)+\left(\sqrt{a_{3}}\right.\right. \\
\left.-\sqrt{a_{2}}\right)+\cdots+\left(\sqrt{a_{999}}\right. \\
\left.\left.-\sqrt{a_{998}}\right)\right] \\
=\frac{1}{2}\left(\sqrt{a_{999}}-\sqrt{a_{1}}\right) \\
=\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,267 |
Example 10 Let $1995 x^{3}=1996 y^{3}=1997 z^{3}$,
$$
\begin{array}{l}
x y z>0, \text { and } \sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}} \\
=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} . \\
\quad \text { Then } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=
\end{array}
$$
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$ | Solution: Let $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, obviously $k \neq 0$. Then we have
$$
1995=\frac{k}{x^{3}}, 1996=\frac{k}{y^{3}}, 1997=\frac{k}{z^{3}} .
$$
From the given, we get
$$
\sqrt{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}}>0 .
$$
Thus... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,268 |
Example 11 Given real numbers $a, b$ satisfy $3 a^{4}+2 a^{2}-4=0$ and $b^{4}+b^{2}-3=0$. Then $4 a^{-4}+b^{4}=$ ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | Solution: Transform the equation $3 a^{4}+2 a^{2}-4=0$ into $\frac{4}{a^{4}}-$ $\frac{2}{a^{2}}-3=0$, and it is known that $-\frac{2}{a^{2}}$ and $b^{2}$ are the two distinct real roots of the equation $x^{2}+x-3=0$.
Let $-\frac{2}{a^{2}}=x_{1}, b^{2}=x_{2}$. By Vieta's formulas, we have $x_{1}+x_{2}=-1, x_{1} x_{2}=-3... | 7 | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,269 |
Example 12 If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-x-1=0$, find the value of $\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$. | Given that $x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$, then $\alpha+\beta+\gamma=0, \beta \gamma+\gamma \alpha+\alpha \beta=$ $-1, \alpha \beta \gamma=1$.
$$
\begin{array}{l}
\therefore \frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma} \\
=2\left(\frac{1}{1-\alpha}+\frac{1}{1-\beta}+\frac{1}{1-... | -7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,270 |
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