problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and
$$
\begin{array}{l}
f(x-y)=f(x) g(y)-g(x) f(y), f(-2) \\
=f(1) \neq 0, \text { then } g(1)+g(-1)=\ldots
\end{array}
$$
(Answer with a number). | 4. -1 .
$$
\begin{array}{l}
\because f(x-y)=f(x) g(y)-g(x) f(y), \\
\therefore f(y-x)=f(y) g(x)-g(y) f(x) \\
\quad=-[f(x) g(y)-g(x) f(y)]
\end{array}
$$
There is $f(x-y)=-f(y-x)$
$$
=-f[-(x-y)] \text {. }
$$
Then $f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.
$$
\text { Hence } \begin{aligned}
f(1) & =f(-2)=f(-1-1) ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,729 |
5. Let $\sin \theta+\cos \theta=\frac{\sqrt{2}}{3}, \frac{\pi}{2}<\theta<\pi$. Then the value of $\operatorname{tg} \theta -\operatorname{ctg} \theta$ is $\qquad$ . | 5. $-\frac{8 \sqrt{2}}{7}$.
From $\sin \theta+\cos \theta=\frac{\sqrt{2}}{3}$, we get $\sin \theta \cos \theta=-\frac{7}{18}$.
Also, $\frac{\pi}{2}<\theta<\pi$, so $\sin \theta>0, \cos \theta<0$,
$\sin \theta-\cos \theta$
$=\sqrt{(\sin \theta+\cos \theta)^{2}-4 \sin \theta \cos \theta}=\frac{4}{3}$.
$\therefore \opera... | -\frac{8 \sqrt{2}}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,730 |
6. Let $x, y \in \mathbf{R}^{+}$, and $\frac{19}{x}+\frac{98}{y}=1$. Then the minimum value of $x+y$ is $\qquad$ | \begin{array}{l}\text { 6. } 117+14 \sqrt{38} \text {. } \\ \text { Let } \sin ^{2} \alpha=\frac{19}{x}, \cos ^{2} \alpha=\frac{98}{y} . \\ \text { Then } x=\frac{19}{\sin ^{2} \alpha}, y=\frac{98}{\cos ^{2} \alpha} \text {. } \\ \begin{aligned} \therefore x+y=\frac{19}{\sin ^{2} \alpha}+\frac{98}{\cos ^{2} \alpha} \\ ... | 117+14 \sqrt{38} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,731 |
Three, (Full marks 10 points) In $\triangle A B C$, the three sides $a$, $b$, and $c$ form an arithmetic sequence. Find the value of $5 \cos A - 4 \cos A \cos C + 5 \cos C$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, from the conditions, we know that $2 b=a+c$,
$\therefore 2 \sin B=\sin A+\sin C$.
Also, $\angle A+\angle B+\angle C=\pi$, then
$2 \sin (A+C)=\sin A+\sin C$.
$\therefore 4 \sin \frac{A+C}{2} \cos \frac{A+C}{2}=2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}$.
Since $\sin \frac{A+C}{2} \neq 0$, then $2 \cos -\frac{A... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,732 |
Four, (Full marks 12 points) A gas station needs to manufacture a cylindrical oil storage tank with a volume of $20 \pi \mathrm{m}^{3}$. It is known that the iron plate used for the bottom costs 40 yuan per square meter, and the iron plate used for the side costs 32 yuan per square meter. If the manufacturing loss is n... | Let the radius of the base of a cylindrical oil tank be $r \mathrm{~m}$, the height be $h \mathrm{~m}$, and the cost be $y \overline{\text { yuan }}$.
Given $\pi r^{2} h=20 \pi$, then $h=\frac{20 \pi}{\pi r^{2}}=\frac{20}{r^{2}}$.
$$
\begin{aligned}
\therefore y & =2 \pi r^{2} \cdot 40+2 \pi r h \cdot 32=80 \pi r^{2}+6... | r=2, h=5 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 710,733 |
Five. (Full marks 12 points) As shown in the figure, $AB$ is the diameter of the top base $\odot O$ of a frustum, $C$ is a point on $\odot O$ different from $A$ and $B$, $A'$ is a point on the bottom base $\odot O'$, the section through $A', A, C$ is perpendicular to the bottom base, $M$ is the midpoint of $A'C$. $AC =... | Five, (1) $A C=A A^{\prime}=2, \angle A^{\prime} A C=120^{\circ}$, so $\Lambda^{\prime} C=2 \sqrt{3}$.
$\because M$ is the midpoint of $\Lambda^{\prime} C$,
$\therefore \Lambda M=1$, and $A M \perp A^{\prime} C$.
Also, $\because$ plane $A \prime A C \perp$ the lower base,
$\therefore$ plane $A^{\prime} \triangle C \per... | \operatorname{arctg} \frac{2 \sqrt{3}}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 710,734 |
Six, (Full score 1.2 outside) The ellipse $C_{1}: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{2 a^{2}}=1(a>0)$, the parabola $C_{2}$ has its vertex at the origin $O$, and the focus of $C_{2}$ is the left focus $F_{1}$ of $C_{1}$.
(1) Prove that $C_{1}$ and $C_{2}$ always have two different intersection points;
(2) Does there ex... | $$
=\frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}} .
$$
The distance from the origin to the line $AB$ is $h=\frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}}$.
$$
\begin{aligned}
\therefore S_{\triangle A O B} & =\frac{1}{2} \cdot \frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}} \cdot \frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}} \\
& =6 a^{... | 6 a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,735 |
Seven, (Full marks 12 points) Given a sequence where each term is 1 or 2, the first term is 1, and there are $2^{k-1}$ twos between the $k$-th 1 and the $(k+1)$-th 1, i.e., $1,2,1,2,2,1,2,2,2,2,1,2,2,2,2,2,2,2,2,1, \cdots$
(1) Find the sum of the first 1998 terms of the sequence, $S_{1998}$;
(2) Does there exist a posi... | For the $k$-th segment, we take the $k$-th 1 and then follow it with $2^{k-1}$ 2's, making the $2^{k-1}+i$-th term the $k$-th segment of the sequence.
Suppose the 1958th term is in the $k$-th segment. Then $k$ is the smallest positive integer satisfying
$$
\dot{\kappa}+\left(1+2+2^{2}+\cdots+2^{k-1}\right)=2^{k}+k-1 \g... | 3985 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,736 |
1. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{4}+a_{6}+a_{8}+a_{10}+$ $a_{12}=120$, then the value of $2 a_{9}-a_{10}$ is ( ).
(A) 20
(B) 22
(C) 24
(D) 28 | \begin{array}{l}-1 .(\mathrm{C}) \\ \because a_{4}+a_{6}+a_{8}+a_{10}+a_{12} \\ =5 a_{1}+(3+5+7+9+11) d \\ =5\left(a_{1}+7 d\right)=120 \Rightarrow a_{1}+7 d=24, \\ \therefore 2 a_{9}-a_{10}=2\left(a_{1}+8 d\right)-\left(a_{1}+9 d\right) \\ =a_{1}+7 d=24 .\end{array} | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,737 |
2. Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with two foci $F_{1}$ and $F_{2}$, a chord $A B$ is drawn through the right focus $F_{2}$ with an inclination angle of $\frac{\pi}{4}$. Then the area of $\triangle A B F_{1}$ is ( ).
(A) $\frac{2 \sqrt{2}}{3}$
(B) $\frac{4}{3}$
(C) $\frac{4 \sqrt{2}}{3}$
(D) $\frac{4 \sqrt... | $$
\begin{array}{c}
\because F_{2}(1,0), \\
B(0,-1), \\
F_{1}(-1,0), \\
\therefore \angle A B F_{1}=90^{\circ} . \\
\text { Also, }\left|F_{1} A\right|+ \\
\left|F_{2} A\right|=2 a=2 \sqrt{2}, \\
\therefore\left|F_{1} A\right|=2 \sqrt{2}-\left|F_{2} A\right| . \\
\text { Also, }\left|F_{1} A\right|^{2}=\left|F_{1} B\ri... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,738 |
Example 2 Proof: The smallest integer greater than $(1+\sqrt{3})^{2 n}$ is divisible by $2^{n+1}$ $(n \in \mathbf{N})$.
(6th Putnam Mathematical Competition)
Analysis: From $(1+\sqrt{3})^{2 n} \xrightarrow{\text { association }}(1-\sqrt{3})^{2 n} \in(0$, 1), consider their sum. | Proof: Note that $0<(1-\sqrt{3})^{2 n}<1$, combining with the binomial theorem we have
$$
\begin{array}{l}
(1+\sqrt{3})^{2 n}+(1-\sqrt{3})^{2 n} \\
=2\left(3^{n}+3^{n-1} \mathrm{C}_{2 n}^{2}+3^{n-2} \mathrm{C}_{2 n}^{4}+\cdots\right) \\
\triangle 2 k \in \mathbf{N} .
\end{array}
$$
Then the smallest integer greater th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,739 |
$3.4 \sin 40^{\circ}-\operatorname{tg} 40^{\circ}$ 的值是 $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{3}}{3}$
(D) $\sqrt{3}$
The value of $3.4 \sin 40^{\circ}-\tan 40^{\circ}$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{3}}{3}$
(D) $\sqrt{3}$ | 3. (D).
$$
\begin{aligned}
\text { Original expression } & =\frac{4 \sin 40^{\circ} \cos 40^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}} \\
& =\frac{\sin 80^{\circ}-\sin 40^{\circ}+\sin 80^{\circ}}{\cos 40^{\circ}} \\
& =\frac{\sin 20^{\circ}+\sin 80^{\circ}}{\cos 40^{\circ}}=\frac{2 \sin 50^{\circ} \cos 30^{\circ}}{\cos ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,740 |
6. The function $f(x)$ is defined on $\mathbf{R}$ as an odd function, and $f(2)$ $=0$. For any $x \in \mathbf{R}$, it holds that $f(x+4)=f(x)+f(2)$. Then $f(1998)=(\quad)$.
(A) 3996
(B) 1998
(C) 1997
(D) 0 | 6. (D).
Given $x=-2$, then we have $f(2)=f(-2)+f(4)$, which means $f(2) + f(2) = f(4)$.
$\therefore f(4) = 2 f(2) = 0 \Rightarrow f(x+4) = f(x)$, which means $f(x)$ is a periodic function with a period of 4.
$$
\therefore f(1998) = f(499 \times 4 + 2) = f(2) = 0.
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,743 |
1. Given $\alpha+\beta=15^{\circ}$. Then $\frac{1-\operatorname{tg} \alpha-\operatorname{tg} \beta-\operatorname{tg} \alpha \operatorname{tg} \beta}{1+\operatorname{tg} \alpha+\operatorname{tg} \beta-\operatorname{tg} \alpha \operatorname{tg} \beta}$ | \[
\begin{array}{l}
=1 \cdot \frac{\sqrt{3}}{3} \text {. } \\
\text { Original expression }=\frac{1-\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}}{1+\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}}=\frac{1-\tan(\alpha+\beta)}{1+\tan(\alpha+\beta)} \\
=\frac{1-\tan 15^{\circ}}{1+\tan 15^{\circ}}=\tan... | \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,744 |
2. If the complex number $\frac{z+\frac{1}{3}}{z-\frac{1}{3}}$ is purely imaginary, then $|z|=$ | 2. $\frac{1}{3}$.
Let $z=a+b i$. Then
$$
\begin{aligned}
\text { Original expression }= & \frac{3 z+1}{3 z-1}=1+\frac{2}{3 z-1}=\frac{2}{(3 a-1)+3 b i}+1 \\
& =\frac{(3 a-1)^{2}+9 b^{2}+2(3 a-1)}{(3 a-1)^{2}+9 b^{2}} \\
& -\frac{6 b i}{(3 a-1)^{2}+9 b^{2}}
\end{aligned}
$$
is a pure imaginary number.
$$
\begin{array}... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,745 |
3. Given $a-b=1$. Then the minimum value of $(a+1)^{2}+(b+1)^{2}$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. $\frac{1}{2}$.
$\because$ Point $(a, b)$ lies on the line $x-y=1$, and point $(-1,-1)$ is outside the line $x-y=1$,
$$
\begin{array}{l}
\therefore a=\frac{|-1 \cdot(-1)-1|}{\sqrt{2}}=\frac{1}{\sqrt{2}}, \\
\text { and } \sqrt{(a+1)^{2}+(b+1)^{2}} \geqslant d=\frac{1}{\sqrt{2}}, \\
\therefore(a+1)^{2}+(b+1)^{2} \geqs... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,746 |
5. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=\frac{4}{3}, P$ is a moving point on the incircle of $\triangle A B C$, and $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$, then $d_{\t... | 5. 160.
From the given information, we have $\angle C=90^{\circ}, a=$ $6, b=8, c=10$, and the inradius $r$ $=2$. Establish a rectangular coordinate system as shown, and let $P(x, y)$. Clearly, $(x-2)^{2}+(y-$ $2)^{2}=4$. Therefore, we have
$$
\begin{aligned}
d^{2} & =x^{2}+(y-8)^{2}+(x \\
& -6)^{2}+y^{2}+x^{2}+y^{2} \... | 160 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,748 |
6. The positive integer solution of the system of equations $\left\{\begin{array}{l}\frac{z^{y}}{y}=y^{2 x}, \\ 2^{z}=2 \cdot 4^{x}, \\ x+y+z=12\end{array}\right.$ is $\qquad$ | $6 . x=2, y=5, z=5$.
From $2^{z}=2^{2 x+1} \Rightarrow z=2 x+1$. Substituting into $x+y+z=12 \Rightarrow 3 x+y=11$.
$\because x, y$ are positive integers
then $\left\{\begin{array}{l}x=1, \\ y=8\end{array}\right.$ or $\left\{\begin{array}{l}x=2, \\ y=5\end{array}\right.$ or $\left\{\begin{array}{l}x=3, \\ y=2 .\end{arr... | x=2, y=5, z=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,749 |
Example 3 Let $m=4 l+1, l$ be a non-negative integer. Prove: $a=\mathrm{C}_{n}^{1}+m \mathrm{C}_{n}^{3}+m^{2} \mathrm{C}_{n}^{5}+\cdots+m^{\frac{n-1}{2}} \mathrm{C}_{n}^{n}$ $(n=2 k+1, k \in \mathbf{N})$ is divisible by $2^{n-1}$.
Analysis: Considering the structure of the binomial expansion, construct the binomial ex... | Proof: By the binomial theorem, we know
$$
\begin{aligned}
a= & \frac{1}{2 \sqrt{m}}\left[(1+\sqrt{m})^{n}-(1-\sqrt{m})^{n}\right] \\
= & 2^{n-1} \cdot \frac{1}{\sqrt{m}}\left(\left(\frac{1+\sqrt{m}}{2}\right)^{n}\right. \\
& \left.-\left(\frac{1-\sqrt{m}}{2}\right)^{n}\right) .
\end{aligned}
$$
To prove the original ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,750 |
Three. (Full marks 20 points) Let the complex numbers $z_{1}, z_{2}, \cdots, z_{10}$ form a geometric sequence. Given that $z_{1} \neq 1, z_{2}=z_{10}=1$. Find the modulus and argument of $z_{1}$.
保留源文本的换行和格式，翻译结果如下：
Three. (Full marks 20 points) Let the complex numbers $z_{1}, z_{2}, \cdots, z_{10}$ form a geometric... | $\begin{array}{l}\text { Three, } \because z_{1} \neq 1 \text {, let } z_{1}=r(\cos \theta+i \sin \theta) \text {. } \\ \text { Also, } z_{2}=1, \therefore q=\frac{1}{z_{1}}=\frac{1}{r}(\cos \theta-i \sin \theta) \text {. } \\ \because z_{10}=z_{2} \cdot q^{9}=\frac{1}{r^{8}}(\cos \theta-i \sin \theta)^{8} \\ =\frac{1}... | \theta=\frac{k \pi}{4}, k \in \mathbb{Z}, k \neq 8 m, m \in \mathbb{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,751 |
Five. (Full marks 20 points) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e=2+\sqrt{6}-\sqrt{3}-\sqrt{2}$, a line $l$ passing through its right focus $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. Find the size of $\angle A F_{1} F_... | $$
\begin{array}{l}
\text { Five, as shown in the figure, let } \angle A F_{1} F_{2} \\
=\alpha, \angle F_{1} A_{2} F_{2}=\beta .
\end{array}
$$
Five, as shown in the figure, let $\angle A F_{1} F_{2}$
then $\alpha+\beta=90^{\circ}$.
By the Law of Sines, we have
$$
\begin{array}{l}
\frac{\left|A F_{2}\right|}{\sin \... | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,753 |
Six. (Full marks 20 points) If $a_{1}+a_{2}+\cdots+a_{n}=1$, prove:
$$
\begin{array}{l}
\frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}} \\
+\cdots+\frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^{2}+a_{1}^{3}} \geqslant \... | $$
\begin{aligned}
\text { Six, let } A= & \frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}} \\
& +\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}} \\
& +\cdots+\frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^{2}+a_{1}^{3}}, \\
B= & \frac{a_{2}^{4}}{a_{1}^{3}+a_{1}^{2} a_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,754 |
1. In an acute triangle $\triangle A B C$, let $x=\sin A+\sin B+$ $\sin C, y=\cos A+\cos B+\cos C$. Then the relationship between $x$ and $y$ is $(\quad)$.
(A) $x>y$
(B) $x=y$
(C) $x<y$
(D) The relationship between $x$ and $y$ is uncertain | $$
\begin{array}{l}
-1 .(\Lambda) . \\
\angle A+\angle B=180^{\circ}-\angle C>90^{\circ} \\
\Rightarrow \sin A>\sin \left(90^{\circ}-B\right)=\cos B .
\end{array}
$$
Similarly, $\sin B>\cos C, \sin C>\cos A$.
$$
\therefore \sin A+\sin B+\sin C>\cos A+\cos B+\cos C \text {. }
$$
That is, $x>y$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,755 |
2. Given that the two real roots of the equation $a x^{2}+b x+c=0$ are $a$ and $c (a c \neq 0)$. Then the root situation of the equation $9 c x^{2}+3 b x+a=0$ is ( ).
(A) One root must be $\frac{1}{3}$
(B) One root must be $\frac{1}{9}$
(C) The two roots are $\frac{1}{3},-\frac{1}{3}$
(D) One root must be $\frac{1}{3}$... | 2. (D).
By Vieta's formulas, we get $a+c=-\frac{b}{a}, a c=\frac{c}{a}$, then $a^{2} c=$ c. Since $a c \neq 0$, we have $a= \pm 1$. The equation $9 c x^{2}+3 b x+a=0(a c$ $\neq 0$ ) has no zero roots, so,
$$
c+\frac{b}{3 x}+\frac{a}{9 x^{2}}=0,
$$
which means $a \cdot\left(\frac{1}{3 x}\right)^{2}+b \cdot\left(\frac{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,756 |
3. In circle $\odot O$, the radius $r=5 \mathrm{~cm}$, $A B$ and $C D$ are two parallel chords, and $A B=8 \mathrm{~cm}, C D=6 \mathrm{~cm}$. Then the length of $A C$ has ( ).
(A) 1 solution
(B) 2 solutions
(C) 3 solutions
(D) 4 solutions | 3. (c).
According to the theorem and the Pythagorean theorem, the distance between $AB$ and $CD$ can be found to be $7 \mathrm{~cm}$ or $1 \mathrm{~cm}$. As shown in Figure 2, we solve for four cases respectively:
$$
\begin{array}{c}
A C_{1}=\sqrt{7^{2}+1^{2}} \\
=5 \sqrt{2}, \\
A C_{2}=\sqrt{7^{2}+7^{2}}=7 \sqrt{2}, ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,757 |
4. The fractional equation about $x$ $\frac{1}{x+2}-\frac{k}{x-2}=1$ $-\frac{4 x}{x^{2}-4}$ has two real roots. Then, $k$ should satisfy ( ).
(A) $k^{2}-18 k+33>0$
(B) $k^{2}-18 k+33>0$ and $k \neq 2$
(C) $k \neq 2$
(D) None of the above answers is correct | 4. (B).
Eliminating the denominator and rearranging gives
$$
x^{2}+(k-5) x+(2 k-2)=0 \text {. }
$$
According to the problem, $k$ should satisfy
$$
\left\{\begin{array}{l}
\Delta=(k-5)^{2}-4(2 k-2)=k^{2}-18 k+33>0, \\
2^{2}+2(k-5)+2 k-2 \neq 0, \\
(-2)^{2}-2(k-5)+2 k-2 \neq 0 .
\end{array}\right.
$$
That is, $k^{2}-1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,758 |
Example 1 In the donation activity of a school to the Hope Project, the total donation amount of $m$ boys and 11 girls in Class A is equal to the total donation amount of 9 boys and $n$ girls in Class B, which is $(m n+9 m+11 n+145)$ yuan. It is known that the donation amount per person is the same, and it is an intege... | Solution: Let the donation amount per person be $x$ yuan. Since the total donation amounts of the two classes are equal and the donation amount per person is the same, the number of people in the two classes is also equal, denoted as $y$. Then,
$$
y=m+11=n+9 \text{, }
$$
and $\square D \square D \square D \square$
$$
... | 25 \text{ or } 47 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,761 |
Example 2: In a certain year, the total coal production of a coal mine, apart from a certain amount of coal used for civilian, export, and other non-industrial purposes each year, the rest is reserved for industrial use. According to the standard of industrial coal consumption of a certain industrial city in that year,... | Solution: Let the total coal production of the mine for the year be $x$, the annual non-industrial coal quota be $y$, and the industrial coal consumption of each industrial city for the year be $z$. Let $p$ be the number of years the coal can supply only one city. According to the problem, we have the system of equatio... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,762 |
6. A soldier has to check for mines in a triangular region. The effective range of his detection instrument is equal to half the height of the equilateral triangle. If the soldier starts from one vertex of the triangle, what is the shortest path he can take to complete his mission?
(15th IMO) | 6. When the soldier passes through the path $A D E$, he can inspect the entire $\triangle A B C$.
Analysis: Connect $E F$, and construct a circle through points $B, C, E$ intersecting $E F$ at $H$. Connect accordingly,
$$
\begin{array}{l}
E K^{2}=E M^{2} \\
=E C \cdot E D \\
=E H \cdot E F .
\end{array}
$$ | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,763 |
Three. (Full marks 20 points) As shown in the figure above, in quadrilateral $ABCD$, $AC$ intersects $BD$ at point $O$. Line $l$ is parallel to $BD$ and intersects the extensions of $AB$, $DC$, $BC$, $AD$, and $AC$ at points $M$, $V$, $R$, $S$, and $P$ respectively. Prove that $PM \cdot PN = PR \cdot PS$.
---
Note: T... | $\equiv 、 \because B D \| l, \therefore \angle O D C=\angle P N C$.
If $\triangle O U C \sim \triangle P N C$, then $\frac{P N}{O D}=\frac{C P}{O C}$.
By analogy, $\frac{P h}{\partial P}=\frac{C P}{(Q)}$.
Therefore, $\frac{P N}{P K}=\frac{O D}{Q B}$.
Also, $\because B D \| l$,
$\therefore \triangle A B O \sim \triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,768 |
Four, (Full marks 20 points) For what rational values of $x$ is the algebraic expression $9 x^{2}+23 x-2$ the product of two consecutive positive integers?
When $x$ is what rational number, the algebraic expression $9 x^{2}+23 x-2$ is the product of two consecutive positive integers? | Let two consecutive positive even numbers be $k, k+2$. Then we have
$$
9 x^{2}+23 x-2=k(k+2) \text {, }
$$
which simplifies to $9 x^{2}+23 x-\left(2+k^{2}+2 k\right)=0$. Since $x$ is a rational number, the discriminant must be a perfect square, i.e.,
$$
\begin{array}{l}
\Delta=23^{2}+4 \times 9\left(k^{2}+2 k+1+1\rig... | x=2 \text{ or } x=-17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,769 |
1. Among the following groups of quadratic radicals, the ones that belong to the same type of quadratic radicals are ( ).
(A) $\sqrt{\frac{24}{9}}, \sqrt{\frac{147}{4}}$
(B) $\sqrt{\frac{5}{18}}, \sqrt{\frac{216}{49}}$
(C) $\sqrt{\frac{24}{9}}, \sqrt{\frac{216}{49}}$
(D) $\sqrt{\frac{5}{18}}, \sqrt{\frac{147}{4}}$ | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,771 |
2. Given as shown, $A D$ is the diameter of $\odot O$, $A D^{\prime} \perp B C, A B, A C$ intersect the circle at $E, F$ respectively. Then, which of the following equations must be true?
$$
\begin{array}{l}
\text { (A) } A E \cdot B E=A F \cdot C F \\
\text { (B) } A E \cdot A B=A O \cdot A D^{\prime} \\
\text { (C) }... | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,772 |
Example 1 In a cyclic quadrilateral $ABCD$, extend $AB$ and $DC$ to meet at $E$, and extend $AD$ and $BC$ to meet at $F$. $EM$ and $FN$ are tangents to the circle, and arcs are drawn with $E$ and $F$ as centers and $EM$ and $FN$ as radii, respectively, intersecting at $K$.
Prove: $EK \perp FK$.
(1997, Taiyuan City Juni... | $F K^{2}=F N^{2}=F C \cdot F B=F H \cdot F E$. Therefore, $E K^{2}+F K^{2}=E F^{2}$, which means $E K \perp F K$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,774 |
4. As shown in the figure, the radius of $\odot O$ is $2$, and the distance from point $P$ to the center of $\odot O$ is $1$. The chord $AB$ passing through point $P$ intersects the minor arc. Then the minimum value of the area of the resulting segment is ( ).
(A) $\frac{2 \pi}{3}+\sqrt{3}$
(B) $\frac{4 \pi}{3} \div \s... | 4. C
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 | 710,775 |
7. Among the natural numbers starting from 1, list in ascending order those that can be expressed as the difference of squares of two integers. What is the 1998th number in this sequence?
(A) 2662
(B) 2664
(C) 2665
(D) 2666 | 7. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,778 |
8. Ten distinct rational numbers, the sum of any nine of which is an "irreducible proper fraction with a denominator of 22 (a fraction where the numerator and denominator have no common divisor other than 1)." Then the sum of these ten rational numbers is ( ).
(A) $\frac{1}{2}$
(B) $\frac{11}{18}$
(C) $\frac{7}{6}$
(D)... | 8.D
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 | 710,779 |
Example 2 Let $\triangle ABC$ be a triangle, and let a circle with center $O$ pass through points $A$ and $C$, and intersect segments $AB$ and $BC$ at points $K$ and $N$ respectively, where $K$ and $N$ are distinct. The circumcircle of $\triangle ABC$ and the circumcircle of $\triangle KBN$ intersect at $B$ and another... | Analysis: By the root axis property, we know that $B M$, $K N$, and $A C$ intersect at a point $D$. Since $\angle D C N = \angle A K N = \angle B M N$, it follows that $D$, $C$, $N$, and $M$ are concyclic, thus
$$
\begin{array}{l}
B M \cdot B D = B N \cdot B C = B O^{2} - r^{2}, \\
D M \cdot D B = D N \cdot D K = D O^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,785 |
6. Irrational equation
$$
2 x^{2}-15 x-\sqrt{2 x^{2}-15 x+1998}=-18
$$
The solution is $\qquad$ | 6.
$$
x=-\frac{3}{2} \text { or } x=9
$$ | x=-\frac{3}{2} \text{ or } x=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,786 |
Three, (Full marks 12 points) At the foot of the mountain is a pond, the scene: a steady flow (i.e., the same amount of water flows into the pond from the river per unit time) continuously flows into the pond. The pond contains a certain depth of water. If one Type A water pump is used, it will take exactly 1 hour to p... | Three, let the amount of water flowing into the pond from the spring every minute be $x$ m$^{3}$, each water pump extracts $y$ m$^{3}$ of water per minute, and the pond originally contains $z$ m$^{3}$ of water. It takes $t$ minutes for three water pumps to drain the pond. According to the problem, we have
$$
\left\{\be... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,789 |
Four, (Full marks 12 points) Given that $\triangle ABC$ is an equilateral triangle, $E$ is any point on the extension of $AC$, choose a point $D$ such that $\triangle CDE$ is an equilateral triangle. If $M$ is the midpoint of segment $AD$, and $N$ is the midpoint of segment $BE$, prove that $\triangle CMN$ is an equila... | Four, $\because \triangle A C D \cong \triangle B C E(S 、 A 、 S)$,
$$
\therefore A D=B E, A M=B N \text {. }
$$
In $\triangle A M C$ and $\triangle B N C$,
$$
\because A C=B C, A M=B N, \angle C A M=\angle C B N \text {, }
$$
$\therefore \triangle A M C \cong \triangle B N C$.
$$
\therefore C M=C N, \angle A C M=\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,790 |
Five, (Full marks 12 points) Find all values of $k$ such that the roots of the equation $k x^{2}+(k+1) x+(k-1)=0$ are integers.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | When $k=0$, the given equation is a linear equation $x-1=0$, which has an integer root 1.
When $k \neq 0$, the given equation is a quadratic equation.
Let the two integer roots be $x_{1}$ and $x_{2}$, then we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{k+1}{k}=-1-\frac{1}{k}, \\
x_{1} x_{2}=\frac{k-1}{b}=1-\frac... | k=0, k=-\frac{1}{7}, k=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,791 |
Six. (Full marks 12 points) As shown in the figure, $AB$ is the diameter of a semicircle, $AC \perp AB, AC=AB$, and any point $D$ is taken on the semicircle. Construct $DE \perp CD$, intersecting the line $AB$ at point $E$, and $BF \perp AB$, intersecting the extension of line segment $AD$ at point $F$.
(1) If the arc ... | $$
\text { Six, (1) } 0<x<90 \text {. }
$$
(2) Special positions and the observation of the trend of motion changes can be used to exclude impossible situations from the opposite side, or to conjecture $B E = B F$ from the positive side. The proof is given below.
Connect $B D$.
$\because A B$ is the diameter of the sem... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,792 |
Seven, (Full marks 12 points) There are several table tennis teams, and players from different teams all played one match against each other, while players from the same team did not play against each other. The match statistician recorded the following: A total of 10 players participated in this competition, and a tot... | If a team has at least 7 members, then at least $7 \times 6 \div 2=21$ matches must be reduced, so each team has at most 6 members.
If each team has at most 5 members, then the possible number of members in each team (which will not reduce the number of matches) is as follows: it can be seen that there must be a team ... | 3 \text{ teams with } 6, 3, \text{ and } 1 \text{ members respectively} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,793 |
Example 3 In an acute triangle $\triangle ABC$, $O$ is the circumcenter. A circle passing through points $A$, $B$, and $O$ intersects $AC$ and $BC$ at points $E$ and $F$. Prove: $CO \perp EF$.
---
The translation maintains the original text's line breaks and format. | Analysis: As shown in the figure, connect $\dot{E} O$ and extend it to intersect $B C$ at $M$. Since $\angle B O M = \angle A = \frac{1}{2} \angle B O C = \angle C O M$, it follows that $O M \perp B C$, i.e., $E M \perp L F C$. Similarly, $F N \perp E C$. Therefore, $O$ is the orthocenter of $\triangle C E F$, which me... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,796 |
2. Regarding fractions, which of the following four statements is correct? ( ).
(A) An algebraic expression containing a denominator is called a fraction
(B) Multiplying (or dividing) the denominator and numerator of a fraction by $2 a+3$, the value of the fraction remains unchanged
(C) When $x=2$, the value of the fra... | 2. D
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 | 710,804 |
3. As shown in the figure, the side length of the equilateral $\triangle AEF$ is equal to the side length of the rhombus $ABCD$, with points $E$ and $F$ on $BC$ and $CD$ respectively. Then the measure of $\angle B$ is ( ).
(A) $70^{\circ}$
(B) $75^{\circ}$
(C) $80^{\circ}$
(D) $95^{\circ}$ | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,805 |
Example 4 As shown in the figure, $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ are internally tangent to $\odot O$ at $A$ and $B$, respectively. Their common tangent line touches $\odot O_{1}$ and $\odot O_{2}$ at points $C$ and $D$. Let the intersection of $A C$ and $B D$ be $E$. Prove: $O E \perp C D$. | Analysis: To prove $O E \perp C D$, it is only necessary to prove $O E \parallel O_{1} C$, which only requires proving $\angle A C O_{1}=\angle A=\angle A E O$, i.e., proving that $E$ is on $\odot O$. Therefore, it is only necessary to prove $\angle A E B=\frac{1}{2} \angle A O B$ or $180^{\circ}-\frac{1}{2} \angle A O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,807 |
8. Using 3 ones, 2 twos, and 1 three, the number of different 6-digit numbers that can be formed is ( ).
(A) 45
(B) 50
(C) 60
(D) 70 | 8.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,811 |
Four, (Full marks 15 points) The unit digit of an $n$-digit number is 6. Moving the 6 to the front of the number while keeping the other digits in place results in a new $n$-digit number, which is 4 times the original $n$-digit number. Find the smallest positive number that satisfies the above conditions.
---
untran... | From the problem, we know that the first digit must be 1.
Expression:
$$
\begin{array}{r}
1 \cdots E D C B A 6 \\
\hline 6 \cdots E D C B A
\end{array}
$$
We get $A=4, B=8, C=3, D=5, F=1$. $\therefore$ The smallest number is 15384. | 15384 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,813 |
Five. (Full marks 15 points) Expand new numbers according to the following rules:
Given two numbers $a$ and $b$, a new number can be expanded according to the rule $c = ab + a + b$. Any two numbers among $a$, $b$, and $c$ can then be used to expand another new number according to the rule, and so on. Each expansion of ... | (1) The first time, you can only get $1 \times 4+4+1=9$.
Since the goal is to get the maximum new number, the $\cdots$ time, take 4 and 9, to get $4 \times 9+4+9=49$.
Similarly, the second time, take 9 and 49, to get
$9 \times 49+9+49=441$.
(2) $\because a=a \cdot a+b=(a+1)(b+1)-1$,
$$
\therefore a+1=(a+1)(b+1).
$$
Ta... | 441 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,814 |
2. Three lines divide a regular hexagon into six congruent figures, the number of ways to do this ( ).
(A) is exactly 1
(B) is 2
(C) is 6
(D) is infinitely many | 2. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,817 |
For example, in $\triangle ABC$, the incenter is $I$, and the incircle touches $BC$, $CA$ at points $D$, $E$ respectively. If $BI$ intersects $DE$ at point $G$, prove that $AG = BG$.
(1994, Mathematics Linke) | Analysis: Inverse ID and $A I$, it is easy to know
$$
\angle A I B=\angle B D G=90^{\circ}+\frac{1}{2} \angle A C B \text {, }
$$
Then $\triangle A B I \sim \triangle G B D$.
$$
\begin{array}{l}
\therefore \frac{A B}{B I}=\frac{B G}{B D} \\
\therefore \triangle A B G \backsim \triangle I B D
\end{array}
$$
Since $I D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,818 |
3. Between 9:00 AM and 10:00 AM, the hour hand and the minute hand of a clock will overlap once. The time segment for this overlap is ( ).
(A) $9: 48 \sim 9: 49$
(B) $9: 49 \sim 9: 50$
(C) $9: 50 \sim 9: 51$
(D) $9: 51 \sim 9: 52$ | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 710,819 |
4. In a right trapezoid $A B C D$, the upper base $A D=\sqrt{3}$, the lower base $B C=3 \sqrt{3}$, and the leg $A B=6$ perpendicular to the bases. Choose a point $P$ on $A B$ such that $\triangle P A D$ and $\triangle P B C$ are similar. The number of such points $P(\quad)$.
(A) is 1
(B) is 2
(C) is 3
(D) does not exis... | 4. B
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 | 710,820 |
6. Given a convex quadrilateral $A B C D$ with the lengths of its four sides $A B, B C, A D, D C$ being $1, 9, 9, 8$ respectively, and $\cos D=\frac{7}{18}$. Consider the following statements:
(1) Quadrilateral $A B C D$ is a trapezoid;
(2) The area of quadrilateral $A B C D$ is $\frac{45 \sqrt{11}}{4}$;
(3) If $M$ is ... | 6. C
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 | 710,822 |
7. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1998(x-1)=\cdots 1, \\
(y-1)^{3}+1998(y-1)=1 .
\end{array}\right.
$$
Then $x+y=(\quad)$.
(A) 1
(B) -1
(C) 2
(D) -2 | 7.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. The text "7.C" does not require translation as it is already in a form that is the same in both Chinese and ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,823 |
8. Let $b, c$ be integers. When $x$ takes the values $1, 3, 6, 11$ respectively, a student calculates the values of the polynomial $x^{2} + b x + c$ to be $3, 5, 21, 93$. Upon verification, only one of these results is incorrect. The incorrect result is ( ).
(A) When $x=1$, $x^{2} + b x + c = 3$
(B) When $x=3$, $x^{2} ... | 8. 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 | 710,824 |
12. If $a>0, b<0$, then the range of $x$ that satisfies
$$
|x-a|+|x-b|=a-b
$$
is $\qquad$ | 12. $b \leqslant x \leqslant a$ | b \leqslant x \leqslant a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,828 |
For example, let the circumcircle of $\triangle ABC$ be $\odot O$, $\angle C = 60^{\circ}$, $N$ is the midpoint of $\overparen{A B}$, and $H$ is the orthocenter.
Prove: $C N \perp O H$.
(1994, Bulgarian Mathematical Olympiad) | Analysis: As shown in the figure, add auxiliary lines.
From $\angle C=60^{\circ}$, we know $\angle C A H^{\prime}=30^{\circ}$.
Therefore, $\triangle O C H^{\prime}$ is an equilateral triangle.
From the orthocenter $H$, it is easy to get $\mathrm{CH}=\mathrm{CH}^{\prime}$;
From $N$ being the midpoint of $\bar{A} B$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,829 |
15. The distance between locations A and B is 70 kilometers. Two cars start from the two locations simultaneously and continuously travel back and forth between A and B. The first car departs from A, traveling at 30 kilometers per hour, and the second car departs from B, traveling at 40 kilometers per hour. When the fi... | $15.150,200$ | 150,200 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,832 |
17. Given the quadratic equation in $x$, $x^{2} + b x + c = 0$.
(1) If $b$ and $c$ are two different roots of this equation, find the values of $b$ and $c$;
(2) Let $a$ and $\beta$ be the two roots of this equation, factorize
$$
\left[x^{2} + (b+1) x + c\right]^{2} + b\left[x^{2} + (b+1) x + c\right] + c \text{. }
$$ | (1) From the relationship between the roots and coefficients of a quadratic equation, we have
$$
\left\{\begin{array}{l}
b+c=-b, \\
b c=c .
\end{array}\right.
$$
If $c=0$, then $b=0$, which contradicts the condition that the roots are distinct, hence $c \neq 0$.
Solving this, we get $b=1, c=-2$.
$$
\text { (2) } \begi... | (x-\alpha)(x-\beta)(x-\alpha+1)(x-\beta+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,834 |
18. In the right triangle $\triangle ABC$, $CD$ is the altitude on the hypotenuse $AB$, and $O$, $O_{1}$, $O_{2}$ are the points of intersection of the angle bisectors of $\triangle ABC$, $\triangle ACD$, and $\triangle BCD$ respectively. Prove:
(1) $O_{1} O \perp C O_{2}$;
(2) $O C=O_{1} O_{2}$. | 18. (1) Given that $O_{1}$ and $O$ are both on the angle bisector of $\angle A$, let this bisector intersect $\mathrm{CO}_{2}$ at $E$.
$$
\begin{array}{l}
\because \angle A=\angle D C B, \\
\therefore \angle E A C \\
\quad=\angle O_{2} C B, \\
\therefore \angle E A C+\angle A C E \\
\quad=\angle O_{2} C B+\angle A C E=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,835 |
2. Given the function $f(x)=\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right) \cos x$ $+b \sin x+6$ (where $a$ and $b$ are constants, and $a>1$), $f\left(\lg \log _{8} 1000\right)=8$. Then the value of $f(\lg \lg 2)$ is ( ).
(A) 8
(B) 4
(C) -4
(D) depends on the values of $a$ and $b$ | 2. (B).
Let $F(x)=f(x)-6$.
It is not hard to prove that $F(-x)=-F(x)$, i.e., $F(x)$ is an odd function.
Then
$$
\begin{array}{l}
F\left(-\lg \log _{8} 1000\right)=-F\left(\operatorname{lglog}_{8} 1000\right), \\
F(\operatorname{lglg} 2)=-F(\operatorname{lglog} 8000), \\
f(\lg 22)-6=-\left[f\left(\operatorname{lglog}_{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,837 |
4. Given $f(x)=\sum_{k=0}^{9} x^{k}$. Then the remainder of the polynomial $f\left(x^{1999}\right)$ divided by $f(x)$ is ( ).
(A) 0
(B) 1
(C) $f(x)-x^{9}$
(D) $f(x)-x^{9}-1$ | 4. (A).
$\because f(x)=\frac{x^{10}-1}{x-1}$,
$\therefore \omega$ is a tenth root of unity of $f(x)=0$.
That is, $\omega^{10}=1$, and $\omega \neq 1$.
Furthermore, $\omega^{n}$ is a root of $f(x)=0$, where $n=1,2, \cdots, 9$.
Also, $f\left(\left(\omega^{n}\right)^{1099}\right)=\frac{\left[\left(\omega^{n}\right)^{1990}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,839 |
Example 8 As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, $O_{1}$ lies on the circumference of $\odot O_{2}$, and the chord $AC$ of $\odot O_{1}$ intersects $\odot O_{2}$ at point $D$. Prove: The line segment $O_{1} D$ is perpendicular to $BC$. | Analysis: As shown in the figure, draw auxiliary lines.
From $\angle B O_{1} D=\angle B A C=\frac{1}{2} \angle B O_{1} C$,
we know that in the isosceles $\triangle O_{1} B C$, the angle bisector $O_{1} D$ is perpendicular to the base $B C$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,840 |
5. As shown in Figure 1, there are 1999 circles each tangent to both sides of $\angle M O N$, and these circles are sequentially tangent to each other. If the radii of the largest and smallest circles are 1998 and 222 respectively, then the radius of the circle in the middle is $(\quad)$.
(A) 1110
(B) 1776
(C) 666
(D) ... | 5. (C).
As shown in Figure 3. Let the radii of the externally tangent circles $\odot O_{n}, \odot O_{n+1}$ be $R_{n}$ and $R_{n+1}$, and $\angle M O N=2 \alpha$ (where $\alpha$ is a constant). Construct $O_{n} A \perp O M$ at $A$, $O_{n+1} B \perp O M$ at $B$, and $O_{n} C \perp O_{n+1} B$ at $C$.
Then $O_{n+1} C=O_{n... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,841 |
6. As shown in Figure 2,
A particle moves in the first quadrant. In the first second, it moves from the origin to $(0,1)$, and then it continues to move back and forth in directions parallel to the $x$-axis and $y$-axis, moving one unit length each time. Therefore, after 1999 seconds, the position of the particle is (... | 6. (D).
Since $1+3+5+\cdots+(2 n-1)=n^{2}$.
When $n$ is odd, the particle's position after $n^{2}$ seconds is $(0, n)$, moving to the right.
When $n$ is even, the particle's position after $n^{2}$ seconds is $(n, 0)$, moving to.
$$
\because 1+3+\cdots+(2 \times 44-1)=c, 4^{2}=1936 \text {, }
$$
$\therefore 1936$ seco... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,842 |
1. Calculate: $\sin ^{2} 35^{\circ}+\sin ^{2} 85^{\circ}-\cos 5^{\circ} \cos 55^{\circ}=$ | $\begin{array}{l}=1 . \frac{3}{4} . \\ \text { Original expression }=\sin ^{2} 35^{\circ}+\sin ^{2} 85^{\circ}-\sin 85^{\circ} \sin 35^{\circ} \\ =\left(\sin 35^{\circ}-\sin 85^{\circ}\right)^{2}+\sin 35^{\circ} \sin 85^{\circ} \\ =\left(2 \sin 25^{\circ} \cos 60^{\circ}\right)^{2}+\frac{1}{2}\left(\cos 50^{\circ}-\cos... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,843 |
$2.1999^{2000}$ divided by $10^{10}$, the remainder is $\qquad$ | $$
\begin{array}{l}
2.5996000001 . \\
1999^{2000}=(1-2000)^{2000} \\
=1-2000 \times 2000+\frac{1}{2} \times 2000 \times 1999 \times 2000^{2} \\
\quad-\frac{1}{16} \times 2000 \times 1999 \times 1998 \times 2000^{3}+\cdots \\
\quad+2000^{2000} .
\end{array}
$$
In the expression on the right side of the equation above, ... | 5996000001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,844 |
3. Given $x \in \mathbf{C}, \arg \left(x^{2}-2\right)=\frac{3}{4} \pi, \arg \left(x^{2}\right.$ $+2 \sqrt{3})=\frac{\pi}{6}$. Then the value of $x$ is $\qquad$ . | 3. $\pm(1+i)$.
Let $x^{2}-2=\rho_{1}\left(\cos \frac{3}{4} \pi+i \sin \frac{3}{4} \pi\right)$,
$$
x^{2}+2 \sqrt{3}=\rho_{2}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \text {. }
$$
(2) - (1) gives
$2+2 \sqrt{3}=\left(\frac{\sqrt{3}}{2} \rho_{2}+\frac{\sqrt{2}}{2} \rho_{1}\right)+i\left(\frac{1}{2} \rho_{2}-\f... | \pm(1+i) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,845 |
5. Given two points $A(0,1), B(6,9)$. If there is an integer point $C$ (Note: A point with both coordinates as integers is called an integer point), such that the area of $\triangle A B C$ is minimized. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ | 5.1.
Since the slope of line $AB$ is $k=\frac{3}{4}$, the line passing through point $C$ and parallel to $AB$ can be expressed as $3 y-4 x=m$. Let the coordinates of point $C$ be $\left(x_{0}, y_{0}\right)$, where $x_{0}, y_{0}$ are integers. Therefore, we have
$$
3 y_{0}-4 x_{0}=m \text {. }
$$
Thus, $m$ is an integ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,847 |
6. Given in a Cartesian coordinate system there are two moving points $P\left(\sec ^{2} \alpha, \operatorname{tg} \alpha\right), Q(\sin \beta, \cos \beta+5)$, where $\alpha, \beta$ are any real numbers. Then the shortest distance between $P$ and $Q$ is $\qquad$
Translate the above text into English, please retain the ... | $6.2 \sqrt{5}-1$.
The trajectory equation of $P$ is $y^{2}=x-1$, and the trajectory equation of $Q$ is $x^{2}+(y-5)^{2}=1$.
To find the shortest distance between $P$ and $Q$, it is only necessary to find the shortest distance from a point on $P$ to $(0,5)$.
Let the point on $P$ be $(x, y)$, and the distance to $(0,5)$ ... | 2 \sqrt{5}-1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,848 |
Three. (Full marks 20 points) Given a sequence of positive numbers $\left\{a_{n}\right\}$ satisfying
$$
\begin{array}{l}
\sqrt{a_{n} a_{n+1}+a_{n} a_{n+2}} \\
=3 \sqrt{a_{n} a_{n+1}+a_{n+1}^{2}}+2 \sqrt{a_{n} a_{n+1}},
\end{array}
$$
and $a_{1}=1, a_{2}=3$. Find the general term formula for $\left\{a_{n}\right\}$. | Divide the original equation by $\sqrt[a]{a}, \overline{a_{n}+1}$, we get
$$
\sqrt{1+\frac{a_{n}+2}{a_{n+1}}}:=3 \sqrt{1+\frac{a_{n+1}}{a_{n}}}+2 \text {. }
$$
Let $b_{n}=\sqrt{1+\frac{a_{n}+1}{a_{n}}}+\mathrm{k}$, then $b_{1}=3$.
Transform (1) into
$$
\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}+1=3\left(\sqrt{1+\frac{a_{n+1}}{a... | a_{n}=\left\{\begin{array}{l}1,(n=1) \\ \prod_{k=1}^{n-1}\left[\left(3^{k}-1\right)^{2}-1\right] .(n>1)\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,849 |
Four, (Full marks 20 points) The sum of \( m \) different positive even numbers divisible by 5 and \( n \) different positive odd numbers divisible by 3 is \( M \). For all such \( m \) and \( n \), the maximum value of \( 5m + 3n \) is 123. What is the maximum value of \( M \)? Please prove your conclusion.
---
The ... | From the problem, we know
$$
\begin{aligned}
M \geqslant & 5(2+4+\cdots+2 m) \\
& +3[1+3+5+\cdots+(2 n-1)] \\
= & 5 m(m+1)+3 n^{2} .
\end{aligned}
$$
That is, \( 5\left(m^{2}+m\right)+3 n^{2} \leqslant M \),
$$
5\left(m+\frac{1}{2}\right)^{2}+3 n^{2} \leqslant M+\frac{5}{4} \text {. }
$$
By the Cauchy-Schwarz inequal... | 1998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,850 |
Five. (Full marks 20 points) Given $0<\alpha_{i}<\frac{\pi}{4}(i=1$, $2,3,4)$, and $\sum_{i=1}^{4} \sin ^{2} \alpha_{i}=1$. Prove: $\sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 \alpha_{i}} \geqslant$ 2. | Let $a_{i}=\sin ^{2} \alpha_{i}$, then $0<a_{i}<\frac{1}{2}$,
$$
\begin{array}{l}
\cos 2 \alpha_{i}=1-2 \sin ^{2} \alpha_{i}=1-2 a_{i} . \\
\frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}}=\frac{a_{i}}{1-2 a_{i}}=2\left(\frac{a_{i}^{2}}{1-2 a_{i}}+\frac{a_{i}}{2}\right) \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\r... | 2 | Inequalities | proof | Yes | Yes | cn_contest | false | 710,852 |
One, (Full marks $5(1$ points) Among the natural numbers greater than 2000, arbitrarily select 601 numbers. Then among these 601 numbers, there must exist two numbers whose difference is 3 or 4 or 7. | Divide the natural numbers not exceeding 2000 into 200 groups, with ten consecutive natural numbers forming one group. Each group is $10 k+1 \sim 10 k+10$, where $k=0,1,2, \cdots, 199$.
Since $\left[\frac{601}{200}\right]+i==4$, by the pigeonhole principle, at least one group must contain at least 4 numbers. Let's ass... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,853 |
II. (Full marks 50 points) Given two points $A$ and $B$ on the same side of a line $l$, with a distance of $15 \mathrm{~cm}$ between them, and the distances from $A$ and $B$ to the line $l$ are $4 \mathrm{~cm}$ and $16 \mathrm{~cm}$, respectively. Can a circle be constructed that passes through points $A$ and $B$ and i... | $$
\begin{array}{l}
\text { Question: } \odot \mathrm{O}_{2} \text { is also a circle that meets the conditions of the problem, } \\
\text { Solution 1: As shown in Figure 5. } \\
\text { Draw perpendiculars from } A \text { and } B \text { to } l \text {, } \\
\text { with feet at } C \text { and } D \text {, respecti... | \frac{65}{8} \text{ cm or } \frac{185}{8} \text{ cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,854 |
Three, (Full marks 50 points) If the hundreds digit of an $n$-digit natural number $N$ is 9, and the sum of its digits is $M$, where $n>3$, when the value of $\frac{N}{M}$ is the smallest, what is $N$?
---
Please note that the translation retains the original formatting and structure of the text. | $$
\begin{array}{l}
\text { Three, (1) When } n=4 \text {, let } N=\overline{a 9 b c} \text {. Then } \\
\frac{N}{M}=\frac{1000 a+900+10 b+c}{a+9+b+c} \\
=1+\frac{999 a+891+9 b}{a+9+b+c} \\
\geqslant 1+\frac{999 a+891+9 b}{a+9+b+9} \text { (equality holds when } c=9 \text {) } \\
=1+\frac{9(a+b+18)+990 a+729}{a+b+18} \... | 1999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,855 |
Initial 77. Given that $a$, $b$, and $c$ are real numbers in $[0,1]$. Prove: $(a+b+c)(1-abc) \leqslant 2$.
保留源文本的换行和格式，直接输出翻译结果。
---
Given that $a$, $b$, and $c$ are real numbers in $[0,1]$. Prove: $(a+b+c)(1-abc) \leqslant 2$. | Proof: Let $l=a+b+c$. When $l<2$, the inequality obviously holds.
Below, we discuss the case where $2 \leqslant l \leqslant 3$.
$$
\begin{array}{l}
\because(1-a)(1-b) \geqslant 0, \\
\therefore a b \geqslant a+b-1, \\
\therefore-a b c \leqslant c(1-a-b) \leqslant 1-a-b \\
\quad \leqslant 2-a-b-c,
\end{array}
$$
That i... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,856 |
Example 10 In the circumcircle of rectangle $ABCD$, take a point $M$ on the arc $AB$ different from vertices $A, B$. Points $P, Q, R, S$ are the projections of $M$ onto lines $AD, AB, BC$, and $CD$ respectively. Prove that lines $PQ$ and $RS$ are perpendicular.
(1983, Yugoslav Mathematical Olympiad) | Analysis: As shown in the figure, let the unit circle $O$ be:
$$
\begin{array}{l}
x^{2}+y^{2}=1 . \\
A(\cos \theta, \sin \theta), \\
M(\cos \alpha, \sin \alpha) \\
(\theta<\alpha<\pi-\theta) .
\end{array}
$$
By symmetry, it is easy to know that
$$
\begin{array}{l}
Q(\cos \alpha,-\sin \theta), S(\cos \alpha,-\sin \thet... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,857 |
Example $3$ City $A$, City $B$, and City $C$ have 10, 10, and 8 units of a certain machine, respectively. Now it is decided to allocate these machines to City $D$ (18 units) and City $E$ (10 units). It is known that the cost to transport one unit from City $A$ to City $D$ and City $E$ is 200 yuan and 800 yuan, respecti... | Solution: (1) According to the problem, the number of machines sent from City A, City B, and City C to City D are $x$, $x$, and $18-2x$; the number of machines sent to City E are $10-x$, $10-x$, and $2x-10$. Therefore,
$$
\begin{aligned}
w= & 200 x+300 x+400(18-2 x) \\
& +800(10-x)+700(10-x) \\
& +500(2 x-10) \\
= & -8... | 9800 \text{ (yuan) and } 14200 \text{ (yuan)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,858 |
Example 11 Construct a semicircle with diameter $PC$ of $\triangle ABC$, intersecting $AB$ and $AC$ at points $D$ and $E$, respectively. Draw perpendiculars from $D$ and $E$ to $BC$, meeting $BC$ at points $F$ and $G$, respectively. The line segments $DG$ and $EF$ intersect at point $M$. Prove that $AM \perp BC$.
(37th... | Analysis: With the center of the circle as the origin, establish a coordinate system as shown in the figure. Without loss of generality, let $BC=2$, $\angle EBC=\alpha$, $\angle DCB=\beta$. Then
$$
\begin{array}{l}
BD: y=\operatorname{ctg} \beta \cdot(x+1), \\
CE: y=-\operatorname{ctg} \alpha \cdot(x-1).
\end{array}
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,859 |
Example 12 Let $A B C D$ be a convex quadrilateral, $A C = B D$. On its sides $A B, B C, C D, D A$, construct equilateral triangles with centers $O_{1}, O_{2}, O_{3}, O_{4}$ respectively. Prove that $\mathrm{O}_{1} \mathrm{O}_{3} \perp \mathrm{O}_{2} \mathrm{O}_{4}$.
(35th IMO Preliminary Problem) | Analysis: Given the geometric configuration, let $\omega = \frac{1}{2} + \frac{\sqrt{3}}{2} \mathrm{i}$, then $\bar{\omega} = \frac{1}{2} - \frac{\sqrt{3}}{2} \mathrm{i}$, which is a complex cube root of -1, and has the properties:
$$
\begin{array}{l}
\omega \bar{\omega} = 1, \omega^{2} = -\bar{\omega}, \omega + \bar{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,860 |
1. $M$ is the midpoint of the base $A C$ of isosceles $\triangle A B C$, $M H \perp B C$ at $H$, and $I$ is the midpoint of $M H$. Prove: $A H \perp B P$. | (Tip one: Take the midpoint $N$ of $HC$, then $MN // AH$, and $PN // AC$, it is easy to know that $P$ is the orthocenter of $\triangle BMN$, so $BP \perp MN$, thus $BP \perp AH$.
(Tip two: From $\triangle ABM \sim \triangle CMH$, it can be deduced that $\triangle BMP \sim \triangle AHC$, then $A, B, C, M$ are concyclic... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,862 |
3. As shown in the right figure, in $\triangle A B C$, $A B>A C, \angle A$ has an external angle bisector intersecting the circumcircle of $\triangle A B C$ at $E, F$ is on $A B$ and $2 A F=A B-A C$. Prove: $E F \perp A B$. | (Tip: Take $B G=A C$, then $A F=G F$. From $\triangle B E G \simeq \triangle E A$ we get $E G=E A$. Then $E F \perp A B$.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,864 |
5. Given that circle $O$ and circle $O^{\prime}$ intersect at points $A$ and $B$, a line through point $A$ intersects circle $O$ and circle $O^{\prime}$ at points $P$ and $Q$ respectively, such that $HA P = A Q$. Also, $M$ is the midpoint of $PB$, and $N$ is the midpoint of $QB$. Prove that $MN \perp AB$. | (提示：如图由 $\triangle A C P \backsim \triangle A B M \backsim \triangle B C M$ 得 $M B^{2}=M A \cdot M C$
及 $P A \cdot A B=M A \cdot$
$A C$. 故 $M A^{2}-$
$$
\begin{array}{l}
M B^{2}=M A \cdot(M A \\
-M C)=P A \cdot A B .
\end{array}
$$
From the figure, since $\triangle A C P \sim \triangle A B M \sim \triangle B C M$, we ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,865 |
6. As shown in the left figure, Rt $\triangle O A B$ and Rt $\triangle O C D$ share $\angle O$, $C D \perp O A, A B \perp O C$, a perpendicular line is drawn from point $B$ to $O A$, and a perpendicular line is drawn from point $D$ to $O C$, with the feet of the perpendiculars being $M$ and $N$ respectively. If $A B$ i... | ( Hint: Connect $A C$, obviously $P$ is the orthocenter of $\triangle A O C$, prove $M N / / A C$ using cyclic quadrilateral properties twice. ) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,866 |
1. The sum of the interior angles of a convex $n$-sided polygon is less than $1999^{\circ}$. Then, the maximum value of $n$ is ( ).
(A) 11
(B) 12
(C) 13
(D) 14 | $-1 .(\mathrm{C})$.
Since the sum of the interior angles of a convex $n$-sided polygon is $(n-2) \cdot 180^{\circ}$, we have $(n-2) \cdot 180^{\circ}<1999^{\circ}, n-2<12, n<14$.
Moreover, the sum of the interior angles of a convex 13-sided polygon is
$$
(13-2) \times 180^{\circ}=1980^{\circ}<1999^{\circ} \text {, }
$$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,868 |
Example 4 A store sells goods that cost 10 yuan each at 18 yuan each, and can sell 60 of them per day. After conducting a market survey, the store manager found that if the selling price of the goods (based on 18 yuan each) is increased by 1 yuan, the daily sales volume will decrease by 5; if the selling price of the g... | Solution: Let the selling price of each item be $x$ yuan, and the daily profit be $s$ yuan.
$$
\begin{array}{l}
\text { When } x \geqslant 18 \text {, we have } \\
s=[60-5(x-18)](x-10) \\
=-5(x-20)^{2}+500,
\end{array}
$$
That is, when the selling price of the item is increased to $x=20$ yuan, the daily profit $s$ is ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,869 |
2. A city charges monthly gas fees according to the following rules: If the gas usage does not exceed 60 cubic meters, it is charged at 0.8 yuan per cubic meter; if it exceeds 60 cubic meters, the excess part is charged at 1.2 yuan per cubic meter. It is known that a user's average gas fee per cubic meter in April was ... | 2. (B).
Since $0.88>0.8$, the coal used this month exceeded 60 cubic meters. Let $x$ be the cubic meters used. Then, we have
$$
60 \div 0.8 + 1.2(x - 60) = 0.88x,
$$
Solving for $x$ gives $x = 75$.
$$
75 \times 0.88 = 66(\text{ yuan }).
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,870 |
3. Given $\frac{1}{a}-|a|=1$, what is the value of $\frac{1}{a}+|a|$?
(A) $\frac{\sqrt{5}}{2}$
(B) $-\frac{\sqrt{5}}{2}$
(C) $-\sqrt{5}$
(D) $\sqrt{5}$ | 3. (D).
From the given, we can obtain that $a, \frac{1}{a},|a|$ are all positive, so
$$
\begin{array}{l}
\left(\frac{1}{a}-|a|\right)^{2}=1, \frac{1}{a^{2}}+|a|^{2}=3 \cdot\left(\frac{1}{a}+|a|\right)^{2}=5, \\
\frac{1}{a}+|a|=\sqrt{5} .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,871 |
4. In $\triangle A B C$, $D$ is a point on side $B C$, it is known that $A C$ $=5, A D=6, B D=10, C D=5$. Then, the area of $\triangle A B C$ is ( ).
(A) 30
(B) 36
(C) 72
(D) 125 | 4. (13).
As shown in the figure, draw $CH \perp AD$ at $H$. Since $\triangle ACD$ is an isosceles right triangle, in the right triangle $\triangle ACH$, $AC=5, AH=3$. Therefore, $CH=4$. Thus,
$$
\begin{array}{l}
S_{\text {AUT }}=\frac{1}{2} A \cdot CH=\frac{1}{2} \times 6 \times 4=12, \\
S_{\triangle ABC}=\frac{BC}{DC... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,872 |
5. If the parabola $y=x^{2}-(k-1) x-k-1$ intersects the $x$-axis at points $A$ and $B$, and the vertex is $C$, then the minimum value of the area of $\triangle A B C$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5. (A).
First, $\Delta=(k-1)^{2}+4(k+1)=k^{2}+2 k+5=$ $(k+1)^{2}+4>0$, so for any value of $k$, the parabola always intersects the $x$-axis at two points. Let the $x$-coordinates of the intersection points of the parabola with the $x$-axis be $x_{1}$ and $x_{2}$. Then,
$$
\begin{aligned}
|A B| & =\sqrt{\left(x_{1}-x_{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,873 |
6. 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 | 6. (D).
As shown in the figure, point $P$ can only be on line $l_{1}$ (line BE) and line $l_{2}$, where the distance between $l_{2}$ and line $CD$ is the same as the distance between $l_{1}$ and line $(D)$. Therefore, in isosceles $\triangle PAB$, when $AB$ is the base, the perpendicular bisector of $AB$ intersects $l... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,874 |
7. Given $x=\frac{1}{\sqrt{3}+\sqrt{2}}, y=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then, $x^{2}+y^{2}$ is
untranslated part: 轩隹
Note: The term "轩隹" does not have a clear meaning in this context and has been left untranslated. If you can provide more context or clarify the term, I can attempt to translate it accurately. | $=, 7.10$.
Let $x=\sqrt{3}-\sqrt{2}, y=\sqrt{3}+\sqrt{2}$, so,
$$
x^{2}+y^{2}=(\sqrt{3}-\sqrt{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}=10 .
$$ | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,875 |
8. Square $A B C D$ has a side length of $10 \mathrm{~cm}$, point $E$ is on the extension of side $C B$, and $E B=10$ $\mathrm{cm}$, point $P$ moves on side $D C$, and the intersection of $E P$ and $A B$ is point $F$. Let $D P=x \mathrm{~cm}$, and the sum of the areas of $\triangle E F B$ and quadrilateral $A F P D$ is... | $\begin{array}{l}\text { S. } y=5 x+50) \text {. } \\ \text { H } I P=x \text { pending } P C=10 \quad x, J B=\frac{1}{2}(10 \quad x) \text {. } \\ \text { M. } y=\frac{1}{2} \times 10 \times \frac{1}{2}(10-x) \\ +\frac{1}{2}\left[10 \cdots \frac{1}{2}(10-x)+x\right] \times 10 \\ =5 x+50 \\\end{array}$ | y=5x+50 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,876 |
10. As shown in Figure 2, given a square $O A B C$ with side length 1 in the Cartesian coordinate system, points $A$ and $B$ are in the first quadrant, and $O A$ makes an angle of $30^{\circ}$ with the $x$-axis. Therefore, the coordinates of point $B$ are $\qquad$ | 10. $\left(-\frac{1+\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$.
Draw $A D \perp x$-axis at $D$, $B E \perp e$-axis at $E$, $A F \perp B E \Psi$
$F$ is
$$
\begin{array}{l}
F B=\text { (A) } F A=E D=I A \\
\because O A=1 . \angle I O A=30 \\
\therefore F A=E D=\frac{1}{2} .
\end{array}
$$
Also, $O D=\sqrt{O A^{2}-D A^{... | \left(-\frac{1+\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,878 |
11. Consider an equilateral triangle with a side length of 1, denoted as $A_{1}$ (as shown in Figure 3).
Trisect each side of $A_{1}$, and construct an equilateral triangle outward on the middle segment, then remove the middle segment to obtain the figure denoted as $A_{2}$ (as shown in Figure 4); trisect each side of ... | 11. $\frac{64}{9}$.
Starting from $A_{1}$, each operation results in a figure whose perimeter is $\frac{4}{3}$ times the perimeter of the original figure. Therefore,
the perimeter of $A_{2}$ is $\frac{4}{3} \times 3=4$,
the perimeter of $A_{3}$ is $\frac{4}{3} \times 4=\frac{16}{3}$,
the perimeter of $A_{4}$ is $\frac... | \frac{64}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,879 |
Example 5 A public bus company, after two technological innovations, adjusted the seating from New Year's Day 1957, allowing each bus to carry 6 more people, so that the number of people each 5 trips could carry exceeded 270. From New Year's Day 1958, a trailer was added, allowing each bus to carry 98 more people than ... | Solution: Let the number of people each car can carry before New Year's Day 1957 be $x$, then on New Year's Day 1957 and 1958, each car can carry $x+6$ and $x+98$ people, respectively. According to the problem, we have $\left\{\begin{array}{l}5(x+6)>270, \\ 3(x+98)>8(x+6) .\end{array}\right.$
Solving, we get $\left\{\b... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,880 |
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