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Three, if $\mathrm{a}, \mathrm{b}$ are skew lines, $\mathrm{A}, \mathrm{~B}$ are on line $\mathrm{a}$, $\mathrm{C}, \mathrm{D}$ are on line $\mathrm{E}$, then $\mathrm{AC}$ and $\mathrm{BD}$ are skew lines. | Three, Proof (by contradiction)
Assume that $\mathrm{AC}$ and $\mathrm{BD}$ are in the same plane $\beta$, then $A, B, C, D$ four points are all in plane $\beta$ $\Rightarrow A, B$ are in $\beta$ $\Rightarrow$ line $\mathrm{a}$ is in plane $\beta$.
Similarly, line $\mathrm{b}$ is in plane $\beta$.
Thus, $a, b$ are bo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,682 |
Four, $\mathrm{AC}$ is in the plane $\mathrm{M}$ of the dihedral angle $\mathrm{M}-\mathrm{EF}-\mathrm{N}$, forming a $45^{\circ}$ angle with $\mathrm{EF}$, and $\mathrm{AC}$ forms a $30^{\circ}$ angle with the plane $N$. Find the degree measure of the dihedral angle $M-E F-N$. | Four, Solution: Let $\mathrm{AB}$ be on plane $\mathrm{N}$, connect $\mathrm{BC}$, then $\angle \mathrm{ACB}$
$$
=30^{\circ} \text{. Also, construct BD } \perp \text{ EF, }
$$
then AD is perpendicular to EF. (By the Three Perpendiculars Theorem)
$\therefore \angle \mathrm{ADB}$ is the plane angle of the dihedral angle... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,683 |
Seven, in plane $\alpha$ there is an equilateral $\triangle \mathrm{A}^{\prime} \mathrm{BC}, \mathrm{DE} / / \mathrm{BC}$, fold $\triangle \mathrm{A}^{\prime} \mathrm{DE}$ to the position of $\triangle \mathrm{ADE}$, making plane $\mathrm{ADE}$ perpendicular to plane $\alpha$, ask where DE should be taken so that the d... | Seven, Solution: Let $\mathrm{A}^{\prime} \mathrm{M}$ be the altitude of $\triangle \mathrm{ABC}$ $\Rightarrow \mathrm{AN} \perp \mathrm{DE}$.
$\because$ Plane $\mathrm{ADE}$ is on plane $\alpha, \therefore \mathrm{AN}$ is perpendicular to plane $\alpha$.
Let the side length of the equilateral $\triangle \mathrm{A}^{\p... | \sqrt{\frac{5}{8}} \mathrm{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,686 |
3. The cofactor of the element -1 in the determinant $\left|\begin{array}{rrr}3 & 1 & 0 \\ 5 & -2 & -1 \\ 2 & 4 & 1\end{array}\right|$ is $\qquad$ , and the value of this determinant is $\qquad$ | 3. $-\left|\begin{array}{ll}8 & 1 \\ 2 & 4\end{array}\right|,-21$,
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
However, since the content is a mathematical expression and a number, it remains the same in English:
3. $-\left|\... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,690 |
Approximations:
(1) $\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}+n}-n\right)=$ $\qquad$ .
(2) $\lim _{x \rightarrow 0} \frac{\tan x}{x}=$ $\qquad$ | 4. (1) $\frac{1}{2}$, (2) 1 ; | \frac{1}{2}, 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,691 |
5 . Given the sum of the first $n$ terms of an arithmetic sequence $S_{4}=5 n^{2}+3 n$, then its general term formula $a_{0}=$ $\qquad$ . | $5 . a_{n}=10 n+8$
$5 . a_{n}=10 n+8$
The above text has been translated into English, maintaining the original format and line breaks. | a_{n}=10 n+8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,692 |
Sure, here is the translated text:
```
Solve the equation:
\[
\left|\begin{array}{ccc}
-4 & 2 & -4+3 i \\
3 & -5 & 2-5 i \\
-4+3 i & 2-5 i & x
\end{array}\right|=0
\]
``` | $$
\begin{array}{l}
\because\left|\begin{array}{lcc}
-4 & 2 & -4+3 \mathrm{i} \\
8 & -5 & 2-5 \mathrm{i} \\
-4+3 \mathrm{i} & 2-5 \mathrm{i} & \mathbf{x}
\end{array}\right| \begin{array}{l}
\text { (multiply the first row by } \\
\text { the row above) }
\end{array} \\
\left|\begin{array}{ccc}
-4 & 2 & -4+3 i \\
3 & -5... | x=1+5i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,696 |
Three, Prove: $(-2)^{1084}+(-2)^{-1084}>2$.
| $\begin{array}{l}\text { Three, Proof } \because(-2)^{1084}>0,(-2)^{-1084}>0, \\ \text { and }(-2)^{1084} \neq 1, \\ \therefore \frac{(-2)^{1084}+(-2)^{-1084}}{2}> \\ \sqrt{(-2)^{1084}(-2)^{-1084}}=\sqrt{(-2)^{0}}=1. \\ \text { That is, }(-2)^{1084}+(-2)^{-1084}>2. \\\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,697 |
Seven, given that $a, b, c$ form an arithmetic sequence, $\alpha, \beta, \gamma$ also form an arithmetic sequence, and $\frac{a}{\alpha}, \frac{b}{\beta}, \frac{c}{\gamma}$ form a geometric sequence. Prove that: $\frac{a}{\gamma} + \frac{\gamma}{a} = \frac{a}{c} + \frac{c}{a}$. | $$
\begin{array}{c}
\text{Given that } 2 b=a+c, 2 \beta=\alpha+\gamma. \\
\because \frac{a}{\alpha}, \frac{b}{\beta}, \frac{c}{\gamma} \text { form a geometric sequence, } \\
\therefore \frac{b^{2}}{\beta^{2}}=\frac{a c}{\alpha \gamma}, \text { i.e., } \frac{\beta^{2}}{\alpha \gamma}=\frac{b^{2}}{a c}. \\
\therefore \f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,702 |
Eight, on one side $0 x$ of the acute angle $x o y$, there is a point $\mathrm{A}$, from which a perpendicular line is drawn to the other side, with its foot of the perpendicular being $\mathrm{B}$, then from $\mathrm{B}$ a perpendicular line is drawn to $0 \mathrm{x}$, with the foot of the perpendicular being $\mathrm... | Eight, solve $\because \triangle A B C \backsim \triangle B C D \backsim \triangle C D E \sim \triangle D E F \backsim$
$$
\begin{array}{l}
\therefore \frac{B C}{A B}=\frac{b}{a}=\frac{C D}{B C}=\frac{D E}{C D}=\frac{E F}{D E}=\cdots \\
C D=B C \times \frac{b}{a}=b \times \frac{b}{a}, \\
\mathrm{DE}=\mathrm{CD} \times ... | \frac{a^2}{a-b} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,703 |
4. If $|\mathrm{a}| \geqslant 1$, then the range of $\operatorname{arctg} \frac{1}{a}$ is | 4. $\left[-\frac{\pi}{4}, 0\right) \cup\left(0, \frac{\pi}{4}\right]$, | \left[-\frac{\pi}{4}, 0\right) \cup\left(0, \frac{\pi}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,707 |
5. The domain of the function $y=\frac{1}{\sin x-\cos x}$ is $\qquad$ | $\begin{array}{l}\text { 5. }\{x \mid x \in R \text { but } \\ \left.x \neq n \pi+\frac{\pi}{4}, n \in Z\right\}\end{array}$ | \{x \mid x \in R \text { but } x \neq n \pi+\frac{\pi}{4}, n \in Z\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,708 |
二、 1 . The sum of the first 15 terms of an arithmetic sequence is 225, find the 8th term. | 二、1, Solve $\because 15 \mathrm{a}_{1}+\frac{15 \times 14}{2} \mathrm{~d}=225$, i.e., $15\left(a_{1}+7 d\right)=225, \quad \therefore a_{3}=a_{1}+7 d=\frac{225}{15}=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,711 |
2. The sum of the first term and the second term of a geometric sequence is 30, the sum of the third term and the fourth term is 120, find the sum of the fifth term and the sixth term. | 2. Solution Let the first term be $\mathrm{a}$, and the common ratio be $\mathrm{q}$. Then
$$
\begin{array}{l}
a+a q=30, \\
a q^{2}+a q^{3}=120 .
\end{array}
$$
From (1) and (2), we get $q= \pm 2$. When $q=2$, $a=10$, at this time the sum of the 5th and 6th terms is 480. When $q=-2$, $a=-30$, at this time the sum of t... | 480 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,712 |
$\begin{array}{l}\text { Four, } 1 . \text { Given } \mathrm{a}>\mathrm{b}>\mathrm{c} \text {, prove that } \\ \frac{1}{\mathrm{a}-\mathrm{b}}+\frac{1}{\mathrm{~b}-\mathrm{c}}+\frac{1}{\mathrm{c}-\mathrm{a}}>0 .\end{array}$ | $$
\begin{array}{l}
\text { Four, 1. Prove the left side }=\frac{(a-b)^{2}+(b-c)^{2}+(a-c)^{2}}{2(a-b)(b-c)(a-c)} . \\
\because a>b>c, \therefore a-b>0, b-c>0, a-c>0,
\end{array}
$$
Therefore, the original equality holds. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,715 |
2. If $a>0, b>0$, prove that
$$
\left(\frac{s^{2}}{b}\right)^{\frac{1}{2}}+\left(\frac{b^{2}}{a}\right)^{\frac{1}{2}} \geqslant \sqrt{a}+\sqrt{b} .
$$ | 2. Station Far Right $=\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}-\sqrt{a}-\sqrt{b}$
$$
=\frac{a-b}{\sqrt{b}}+\frac{b-a}{\sqrt{a}}=(a-b)\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a b}}\right) .
$$
$\because a>0, b>0$, and $a-b$ has the same sign as $\sqrt{a}-\sqrt{b}$,
$\therefore(a-b)\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a b}}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,716 |
Five, write the formula for the sum of the first $\mathrm{n}$ terms of a geometric sequence with the first term $a$ and common ratio $q(q \neq 1)$, and prove it using mathematical induction.
Write the formula for the sum of the first $\mathrm{n}$ terms of a geometric sequence with the first term $a$ and common ratio $... | Five, $S_{2}=\frac{a_{1}\left(1-q^{*}\right)}{1-q}$.
(1) When $n=1$, $S_{1}=\frac{a_{1}(1-q)}{1-q}=a_{1}$, the formula is correct.
(2) Assume the formula is correct when $n=k$, i.e.,
$$
\begin{aligned}
S_{k} & =\frac{a_{1}\left(1-q^{k}\right)}{1-q} . \\
\text { When } n & =k+1, \text { } S_{k+1}=S_{k}+a_{k+1} \\
& =\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,717 |
1. Given $P_{1}(2,1) 、 P_{2}(3,-9)$, and point $P$ divides $P_{1} P_{2}$ in the ratio $\lambda=4$, then the coordinates of point $P$ are $\qquad$ 3 | I. $\left(\frac{14}{5},-7\right)$ | \left(\frac{14}{5},-7\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,719 |
Theorem 6 If the last digit of $\mathrm{N}_{\mathbf{r}}$ is only taken from $1,3,7,9$, then $\mathbf{N}_{\mathbf{r}}{ }^{10} 0_{\mathbf{k}+1}$ has the same last three digits as $\mathbf{N}_{\mathbf{z}}$. | To prove, by analogy, it suffices to show that $10^{3} \mathbf{N}_{\mathbf{z}}{ }^{100} \mathbf{k}-1$ holds.
By Theorem 4, taking 8-3, we get $8 \times 5^{3} \mid \mathbf{N}_{\mathbf{r}}{ }^{4 \times 5}{ }^{2} \cdot \mathbf{k}-1$ holds.
Therefore, $10^{3} \mid \mathbf{N}_{\mathbf{z}}{ }^{10} 0_{\mathbf{k}-1}$ holds, an... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,720 |
4. The equation of the tangent line passing through the point $\mathrm{P}\left(-\frac{1}{3}, \frac{2 \sqrt{2}}{3}\right)$ on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ is $\qquad$ , | $4 . x-2 \sqrt{2} y+3=0$ | x-2 \sqrt{2} y+3=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,723 |
6. The curve represented by the equation $x^{2}+y^{2}=1 \quad(0 \leqslant x \leqslant 1)$ is what shape? | 6. The arc of the unit circle in the first quadrant (including endpoints).
| The \ arc \ of \ the \ unit \ circle \ in \ the \ first \ quadrant \ (including \ endpoints) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,725 |
2. Find the equation of the ellipse centered at the origin, with foci on the y-axis, eccentricity equal to 0.8, and focal distance of 4. | 2. Let the equation of the ellipse be $\frac{\mathrm{x}^{2}}{\mathrm{~b}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{a}^{2}}=1$. From the problem, we have: $2 \mathrm{c}=4, \mathrm{c}=2$.
$$
\left\{\begin{array}{l}
a^{2}-b^{2}=4, \\
\frac{2}{a}=0.8 .
\end{array}\right.
$$
Solving these, we get $\mathrm{a}=\frac{5}{2}, \mathrm{... | \frac{4 x^{2}}{9}+\frac{4 y^{2}}{25}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,727 |
Three, Prove that the three altitudes of a triangle are concurrent (using the analytical method).
This is the translation of the provided text, maintaining the original formatting and line breaks. | Three, Proof: Taking the line on which side $\mathrm{AB}_{2}$ of $\triangle \mathrm{ABC}$ lies as the $\mathrm{x}$-axis.
Let the coordinates of points A, B, and C be $(a, 0)$, $(b, 0)$, and $(0, c)$, respectively. Then the equations of the lines on which the three altitudes $\mathrm{AD}$, $\mathrm{BE}$, and $\mathrm{C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,728 |
Four, in isosceles $\triangle A B C$, the base $B C$ lies on the line $x+y=0$, the vertex $\mathrm{A}(2,3)$, and it is known that one of its equal sides $\mathrm{AB}$ is parallel to the line $\mathbf{x}-4 y-6=0$. Find the equation of the other equal side $\mathrm{AC}$. | Let the equation of the line $\mathrm{AB}$ be
$$
\mathbf{x}-4 \mathrm{y}+\mathrm{m}=0 \text{. }
$$
Since point $\mathrm{A}(2,3)$ lies on this line, $\therefore \mathrm{m}=10$.
Solve the system of equations
$$
\left\{\begin{array}{l}
x-4 y+10=0, \\
\mathbf{x}+\mathbf{y}=0 .
\end{array}\right.
$$
to get the coordinates ... | 4 x-y-5=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,729 |
Five, find the length of the chord intercepted by the graph $x^{2}+y^{2}-2 a x \cos \alpha-2 b y \sin \alpha$ $-a^{2} \sin ^{2} \alpha=0$ on the $y$-axis. | Let $\mathbf{x}=0$, the equation of the circle becomes
$$
\mathrm{y}^{2}-2 \mathrm{by} \sin \alpha-\mathrm{a}^{2} \sin ^{2} \alpha=0 \text {. }
$$
By Vieta's formulas, we have
$$
\begin{aligned}
y_{1} & +y_{2}=2 b \sin \alpha, y_{1} y_{2}=-a^{2} \sin ^{2} \alpha, \\
\text { chord length } & =\left|y_{1}-y_{2}\right|=\... | 2|\sin \alpha| \sqrt{a^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,730 |
Example 1. Find the last three digits of $769^{304}$.
untranslated text remains in its original form, preserving the line breaks and format. | Solution: Since $769^{304}=769^{301} \times 769^{3}$, applying Theorem Six we get
$769^{301}=769^{4 \times 8^{2} \times 3+1}$, whose last three digits are 769.
It is not difficult to calculate that the last three digits of $769^{8}$ are 609.
Therefore, the last three digits determined by $769 \times 609$ are 321,
which... | 321 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,731 |
Six, a circle center at the origin, the focus $\mathrm{F}$ is on the $\mathbf{x}$ axis, the line connecting this focus and the vertex $\mathrm{B}(0,2 \sqrt{2})$ has an inclination angle of $135^{\circ}$, find the coordinates of the intersection points of the line passing through $\mathrm{F}$ and perpendicular to $\math... | Let the line passing through $\mathrm{F}$ and perpendicular to $\mathrm{BF}$ be 1.
$\because \mathrm{BF}$ has an inclination angle of $1.35^{\circ}, \therefore 1$ has an inclination angle of $45^{\circ}$.
$\because b=c=2 \sqrt{2}, \quad \therefore a=\sqrt{2} b=4$.
The ellipse equation is $\frac{x^{2}}{16}+\frac{y^{2}}{... | (0,-2 \sqrt{2}) \text{ and } \left(\frac{8 \sqrt{2}}{3}, \frac{2 \sqrt{2}}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,732 |
Seven, draw a chord from the origin to the circle $(x-8)^{2}+y^{2}=64$, find the equation of the locus of one point of the chord.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Seven, Solution: Let the coordinates of the midpoint P of the chord OM be $(x, y)$, and let the coordinates of M be $\left(x_{1}, y_{1}\right)$. Then, by the midpoint formula, we have
$$
\begin{array}{l}
x_{1}=2 x, y_{1}=2 y . \\
\because \text { point } M\left(x_{1}, y_{1}\right)
\end{array}
$$
lies on the circle $(\... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,733 |
Eight, the moving circle is tangent to the circle $x^{2}+y^{2}=1$, and also tangent to the $y$-axis. Find the equation of the locus of the center of the moving circle, and draw the graph.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation resul... | Eight, Solution Let the coordinates of the center $P$ of the moving circle be $(x, y)$.
If the moving circle is externally tangent to the circle $x^{2}+y^{2}=1$, then
$$
\sqrt{x^{2}+y^{2}}-1=|x|
$$
If the moving circle is internally tangent to the circle $x^{2}+y^{2}=1$, then
$$
1-\sqrt{x^{2}+y^{2}}=|x|
$$
From (1), ... | y^{2}=2 x+1 \text { and } y^{2}=-2 x+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,734 |
1. If $a_{n}-a_{n-1}=a_{n-1}-a_{n-2}(n=3,4, \cdots)$, then $\left\{a_{n}\right\}$ is: (A) a geometric sequence, (B) an arithmetic sequence, (C) both arithmetic and geometric, (D) neither arithmetic nor geometric. Answer ( )
| 1. (B)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,735 |
1. Solve the equation $\left|\begin{array}{ccc}x+1 & 2 x & 1 \\ x & 3 x-2 & 2 x \\ 1 & 1 & x\end{array}\right|=0$ ; | Three, 1. Hint: Subtract the first column from the second column, then take $x-1$ from the second column, resulting in the equation $(x-1)\left(x^{2}+4 x-2\right)=0$, solving for $x_{1}=1, x_{2}, 3=-2 \pm \sqrt{6} 3$ | x_1=1, x_2, x_3=-2 \pm \sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,747 |
2. Solve the inequality $\frac{x^{2}-2 x-15}{x-2}<0$ , | 2. $2<x<5$ or $x<-3$,
| 2<x<5 \text{ or } x<-3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,748 |
$\left\{b_{2}\right\}$ also forms an exponential sequence, the first term is $b_{1}$, and the common ratio is $q_{2}$. Try to prove that $\left\{a, b_{2}\right\}$ also forms a geometric sequence, and the common ratio is $q_{1} \cdot q_{2}$. And find the sum of the first $n$ terms of $\left\{a_{0} b_{n}\right\}$. | $3 . \because \frac{a_{1} \cdot b_{n}}{a_{n-1} \cdot b_{n-1}}=\frac{a_{2}}{a_{n-1}} \cdot \frac{b_{n}}{b_{n-1}}=q_{1} \cdot q_{2}$ (constant),
$\therefore\left\{a_{0} b_{0}\right\}$ is a geometric sequence, the first term is $a_{1} b_{1}$, common ratio
is $q_{1} \cdot q_{2}, S_{2}=\frac{a_{1} b_{1}\left[1-\left(q_{1} ... | S_{2}=\frac{a_{1} b_{1}\left[1-\left(q_{1} q_{2}\right)^{n}\right]}{1-q_{1} q_{2}} | Algebra | proof | Yes | Yes | cn_contest | false | 701,751 |
1. Given the sequence $\left\{\frac{n+1}{n+2}\right\}$, for any $e>0$, take $\mathbf{N}=$ $\qquad$ $\sim$, then when $n>N$, it always holds that $\qquad$ $<\varepsilon$, so the limit of the sequence $\left\{\frac{n+1}{n+2}\right\}$ is 1. | 1. The integer part of $\left(\frac{1}{\varepsilon}-2\right)$, $\left|\frac{n+1}{n+2}-1\right|$, | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,755 |
4. If $-\frac{2}{3}<x<0, y=a x^{3}+a x^{2}+b \quad(a \neq 0)$ is a decreasing function, then the range of values for $\mathrm{a}$ is $\qquad$ . | 4. $(0,+\infty)$ | (0,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,758 |
6 . Let $a$, $b$, $c$ form an arithmetic sequence, $m$ is the arithmetic mean of $a$, $b$, $n$ is the arithmetic mean of $b$, $c$, then $\frac{a}{m}+\frac{b}{n}=$ $\qquad$ | 6.2 .
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,760 |
Three, in an arithmetic sequence, if the 11th term is 20, find the sum of the first 21 terms. | Three, Solution From $a_{1}+10 \mathrm{~d}=20$, we have
$$
S_{21}=21 a_{1}+\frac{21 \cdot 20}{2} d=21\left(a_{1}+10 d\right)=420 .
$$ | 420 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,761 |
Five, find the maximum and minimum values of the function $y=\arcsin \frac{x}{2}+\sqrt{4-x^{2}}$ on its domain. | Five, solve. The domain of the function is $[-2,2]$.
$$
\begin{aligned}
y^{\prime} & =\frac{1-x}{\sqrt{4-x^{2}}}, \text { let } y^{\prime}=0 \text { to get the critical point } x=1 . \\
\left.\because \quad y\right|_{x=1} & =\frac{\pi}{6}+\sqrt{3}, y=-\frac{\pi}{2}, \\
y=2 & =\frac{\pi}{2}, \\
\therefore \quad y_{m} y_... | y_{\text{max}} = \frac{\pi}{6} + \sqrt{3}, \, y_{\text{min}} = -\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,763 |
Six, the tangent line to the curve $f(x)=x^{2}$ at the point $(1,1)$ intersects the x-axis at the point $(\alpha, 0)$. Find $\lim _{n \rightarrow \infty} f(\alpha)$. | Six, solve $\because f^{\prime}(x)=n x^{0-1}$,
$\therefore f(x)$ at the point $(1,1)$ has the tangent line equation
$$
y-1=n(x-1) \text {. }
$$
Let $\mathrm{y}=0$, we get $\alpha=1-\frac{1}{\mathrm{n}}$.
Thus, $f(\alpha)=\alpha^{*}=\left(1-\frac{1}{n}\right)^{n}$.
$\therefore \lim _{n \rightarrow \infty} f(\alpha)=\li... | \frac{1}{e} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,765 |
Seven, for a cylindrical can with a fixed volume, when the relationship between the diameter of the base and the height is what, the least amount of iron sheet is used?
When the volume of a cylindrical can is fixed, the relationship between the diameter of the base and the height that results in the least amount of ... | Seven, Solution Let the base radius be $\mathrm{r}$, height be $\mathrm{h}$, total surface area be $\mathrm{S}$, and volume be $\mathbf{V}$.
$$
\begin{array}{l}
\because V=\pi r^{2} h, \quad \therefore h:=\frac{V}{\pi r^{2}} . \\
S=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+\frac{2 V}{r}, \\
S^{\prime}=\frac{4 \pi r^{2}-2 V}{r^... | h=2r | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,766 |
Eight, Buzhi Line
$1_{1:} \mathrm{y}=\mathrm{kx}+1$ ( 0
$<\mathrm{k}<1$ ), $1_{2}$
$y=x$. Through the intersection point $P_{0}$ of $l_{1}$ and the $y$-axis, draw a line parallel to the $x$-axis to intersect $1_{2}$ at point
$Q_{13}$ Draw a line parallel to the $y$-axis through $Q_{1}$ to intersect $I_{1}$ at $P_{1}$,
... | (1) $\because P_{0}(0,1), Q_{1}(1,1)$,
$$
\begin{aligned}
\therefore & a_{1}=P_{0} Q_{1}=1, \\
& \because P_{1}(1, k+1), Q_{2}(k+1, k+1), \\
& \therefore a_{2}=P_{1} Q_{2}=k . \\
& \because P_{2}\left(k+1, k^{2}+k+1\right), \\
& Q_{3}\left(k^{2}+k+1, k^{2}+k+1\right), \\
& \therefore a_{3}=P_{2} Q_{3}=k^{2}, \cdots, \\... | \frac{k}{2\left(1-k^{2}\right)} | Other | math-word-problem | Yes | Yes | cn_contest | false | 701,767 |
Question 3. The sum of several natural numbers is 1983. When is their product the largest, and what is the maximum value?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Since $\frac{1983}{\left[\frac{1983}{e}\right]}$ and $\frac{1983}{\left[\frac{1983}{e}\right]+1}$ are not integers, the solution method from the previous problem cannot be applied. A completely different approach is needed.
Assume that the natural numbers $a_{1}, a_{2}, \cdots, a_{n}$ (where $\mu$ is also unknown) sum... | 3^{661} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,768 |
1. In a geometric sequence, $a_{3}=\frac{8}{2}, S_{3}=\frac{9}{2}$. Find $a_{1}$ and q. | Solving from the given:
$$
\left\{\begin{array}{l}
\frac{9}{2}=\frac{a_{1}\left(1-q^{8}\right)}{1-q} \\
\frac{3}{2}=a_{1} q^{2}
\end{array}\right.
$$
$\Rightarrow\left\{\begin{array}{l}q=-\frac{1}{2} \\ a_{1}=6,\end{array}\right.$ or $\left\{\begin{array}{l}q=1, \\ a_{1}=\frac{3}{2}\end{array}\right.$.
The flaw in this... | a_{1}=\frac{3}{2}, q=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,771 |
1. Prove by mathematical induction that, $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+$
$$
\frac{1}{n(n+1)}=\frac{n}{n+1} \text {. }
$$ | If the proof steps are as follows, then it is incorrect:
(1) When $\mathrm{n}=1$, left $=\frac{1}{1 \cdot 2}=\frac{1}{2}$, right $=\frac{1}{1+1}=\frac{1}{2}$,
left = right.
(2) Assume that the original formula holds when $n=k$, then when $n=k+1$,
$$
\begin{array}{c}
\text { left }=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,775 |
2. Let $B_{1}, B_{2}, B_{3}$ be the algebraic cofactors of $b_{1}, b_{2}, b_{3}$, respectively. Prove that $\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=b_{1} B_{1}+b_{2} B_{2}+b_{3} B_{3}$. | The correct solution applies three definitions: the definition of a third-order determinant, the definition of a second-order determinant, and the definition of an algebraic cofactor. The proof is as follows:
$$
\begin{aligned}
t & =a_{1} b_{2} c_{3}+a_{2} b_{3} c_{1}+a_{3} b_{1} c_{2}-a_{3} b_{2} c_{1} \\
& -a_{2} b_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,776 |
Question 1. Find a prime number that remains prime when 10 or 14 is added to it. | Because $2+10=12,2+14=16$, the prime number 2 does not meet the requirement;
Because $3+10=13,3+14=17$, the prime number 3 meets the requirement,
Because $5+10=15,5+14=19$, the prime number 5 does not meet the requirement;
Because $7+10=17,7+14=21$, the prime number 7 does not meet the requirement;
Because $11+10=2... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,777 |
The chord intersects the axis at a point 9 (as shown in Figure 2). Prove: 1) The length of the segment of this chord inside the parabola is equal to $\frac{2 \mathrm{p}}{\sin ^{2} \theta}$,
2) If this chord is divided into two segments of lengths $m$ and $n$ by the focus, then $\frac{1}{m}+\frac{1}{n}$ $=\frac{2}{\math... | The image characteristic of this problem is that the vertex of the $\theta$ angle is located at the focus, which is suitable for using the polar coordinate equation of a parabola with the focus at the pole:
$$
\rho=\frac{\mathrm{p}}{1-\cos \theta} \text{. }
$$
1) $m=|F M|=\frac{p}{1-\cos \theta}$,
$\mathrm{n}=|\mathrm{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,780 |
Question 4. Let $A$, $B$, $C$ be the three angles of $\triangle ABC$, prove that $\operatorname{ctg}^{2} A+\operatorname{ctg}^{2} B+\operatorname{ctg}^{2} C \geqslant 1$. | In $\triangle ABC$, $\operatorname{ctg} A \cdot \operatorname{ctg} B + \operatorname{ctg} B \cdot \operatorname{ctg} C + \operatorname{ctg} C \cdot \operatorname{ctg} A = 1$.
For this problem, it suffices to prove the following inequality immediately:
$\operatorname{ctg}^{2} A + \operatorname{ctg}^{2} B + \operatornam... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,781 |
Let $A_{i} (i=1,2, \cdots, n)$ be the interior angles of a convex n-gon, and $\sum_{i=1}^{n} \lg \sin A_{i}=0$, then this convex $n$-gon is a rectangle. | Proof $\because 0^{\circ}<A_{i}<180^{\circ}$,
$$
\begin{array}{l}
\therefore \quad 0<\sin A_{i} \leqslant 1, \quad \lg \sin A_{i} \leqslant 0 . \\
\because \quad \sum_{i=1}^{n} \lg \sin A i=0,
\end{array}
$$
$\therefore \lg \sin A_{i}=0$, i.e., $\sin A_{i}=1, A_{i}=90^{\circ}$.
From the formula for the sum of interior ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,782 |
Example 1. Solve the equation: $\frac{3 x}{x^{2}-1}+\frac{x^{2}-1}{3 x}=-\frac{5}{2}$. | Solve the original equation $\stackrel{\text { Proposition } 2 \text { }}{\Longleftrightarrow} \frac{3 x}{x^{2}-1}=2$ or $\frac{3 x}{x^{2}-1}=$ $\frac{1}{2}$, solving it yields $x_{1}=2, x_{2}=-\frac{1}{2}, x_{3}=3+\sqrt{10}$, | x_{1}=2, x_{2}=-\frac{1}{2}, x_{3}=3+\sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,783 |
Example. Regarding the equation about $x$: $x+\frac{1}{x-1}=a+\frac{1}{a-1}$. | Solve the original equation $\stackrel{\text { transformation }}{\Longleftrightarrow} x-1+\frac{1}{x-1}=a-1+\frac{1}{a-1}$ $\stackrel{\text { Proposition 2 }}{\Longleftrightarrow} x-1=a-1$ or $x-1=\frac{1}{a-1}$, solving these gives $x_{1}=a$, $\mathbf{x}_{2}=\frac{\mathrm{a}}{\mathrm{a}-1}$ as the roots of the origina... | x_{1}=a, \mathbf{x}_{2}=\frac{\mathrm{a}}{\mathrm{a}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,784 |
Example 3. Solve the equation for $x$: $\frac{a-x}{b+x}=5-\frac{4(b+x)}{a-x}$. $(a+b \neq 0)$ (Exercise 3 (2), page 117) | Solve the original equation $\stackrel{\text { transformation }}{\Longleftrightarrow} \frac{a-x}{b+x}+\frac{4(b+x)}{a-x}=5 \stackrel{\text { Proposition 1 }}{\Longleftrightarrow}$ $\frac{a-x}{b+x}=1$ or $\frac{a-x}{b+x}=4$, solving these gives $x_{1}=\frac{a-b}{2}, x_{2}=\frac{a-4b}{5}$ as the roots of the original equ... | x_{1}=\frac{a-b}{2}, x_{2}=\frac{a-4b}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,785 |
Example 5. Solve the equation $\mathbf{x}^{2}+\mathbf{x}+1=\frac{\mathbf{2}}{\mathbf{x}^{2}+\mathbf{x}}$ (Page 125, 5.(3)) | Solve the original equation $\stackrel{\text { transformation }}{\Longleftrightarrow} x^{2}+x-\frac{2}{x^{2}+x}=-1$ <Proposition 1> $x^{2}+x=1$ or $x^{2}+x=-2$ (no solution), solving it yields $x_{1}=$ $\frac{-1+\sqrt{5}}{2}, x_{2}=\frac{-1-\sqrt{5}}{2}$ as the roots of the original equation. | x_{1}=\frac{-1+\sqrt{5}}{2}, x_{2}=\frac{-1-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,787 |
Example 6. Solve the equation $\sqrt{\frac{x+2}{x-1}}+\sqrt{\frac{x-1}{x+2}}=\frac{5}{2}$. | It is known that $x>1$ or $x<-2$, then the original equation $\stackrel{\text { Proposition } 2 \text { }}{\leftrightharpoons}$ $\Leftrightarrow$ $\sqrt{\frac{x+2}{x-1}}=2$ or $\sqrt{\frac{x+2}{x-1}}=\frac{1}{2}$, solving these yields $x_{1}=2, x_{2}=-3$ as the roots of the original equation. | x_{1}=2, x_{2}=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,788 |
Example 7. Solve the equation for $x$:
$$
\sqrt{\frac{a-x}{b+x}}+{ }^{n} \sqrt{\frac{b+x}{a-x}}=2.
$$
$$
(a>b>0)
$$ | It is known that $-\mathrm{b}b>0, \therefore-b$ $<\frac{1}{2}(a-b)<a$, so $x=\frac{1}{2}(a-b)$ is a root of the original equation. | x=\frac{1}{2}(a-b) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,789 |
Example 1. Calculate the side length of a square inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. (Problem 14, Page 171) | Solve As shown in Figure 1, let the vertex A of the square inscribed in the ellipse in the first quadrant be
$$
\left\{\begin{array}{l}
x=a \cos \theta, \\
y=b \sin \theta
\end{array}\right.
$$
(obviously $0<\theta<\frac{x}{2}$)
From $x=x$, we have $a \cos \theta=b \sin \theta$, i.e., $\operatorname{tg} \theta=\frac{a}... | \frac{2 a b}{\sqrt{a^{2}+b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,790 |
Example 8. Solve the equation $\sqrt{\frac{x^{2}-2 x+3}{2 x^{2}+4 x+8}}+\sqrt{\frac{x^{2}+2 x+4}{x^{2}-2 x+3}}$
$$
=\frac{2+\sqrt{2}}{2} \text {. }
$$ | Solve the original equation $\stackrel{\text { transformation }}{\Longleftrightarrow} \frac{1}{\sqrt{2}} \cdot \sqrt{\frac{x^{2}-2 x+8}{x^{2}+2 x+4}}+$ $\sqrt{\frac{x^{2}+2 x+4}{x^{2}-2 x+3}}=\frac{1}{\sqrt{2}}+1 \stackrel{\text { Proposition 1 }}{\Longleftrightarrow} \sqrt{\frac{x^{2}-2 x+3}{x^{2}+2 x+4}}$ $=1$ or $\s... | x_{1}=-\frac{1}{4}, x_{2}=-1, x_{3}=-5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,791 |
Example 2. If $z$ is a complex number, $\theta$ is the argument of $z$, and it satisfies $\left|z+\frac{1}{z}\right|=1$. Prove that $z$ $k \pi+\frac{\pi}{3} \leqslant \theta \leqslant k \pi+\frac{2}{3} \pi$.
(k is an integer) | Let $z=r(\cos \theta+i \sin \theta)$,
$$
\frac{1}{z}=\mathbf{r}^{-1}(\cos \theta-i \sin \theta) .
$$
From $\left|z+\frac{1}{z}\right|=1$ we get
$$
\begin{array}{l}
\left|\left(r+r^{-1}\right) \cos \theta+i\left(r-r^{-1}\right) \sin \theta\right|=1, \\
\left(r+r^{-1}\right)^{2} \cos ^{2} \theta+\left(r-r^{-1}\right)^{2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,795 |
Example 3. Given real numbers $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$, $\mathrm{e}$ satisfy the equations
$$
\begin{array}{l}
a+b+c+d+e=8 \\
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16
\end{array}
$$
Prove that $0 \leqslant e \leqslant \frac{16}{5}$. | Proof of 3:
From the known equation, eliminate $a$ to get
$$
(8-b-c-d-e)^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16 \text {. }
$$
Rearrange into a quadratic equation in $b$ to get
$$
\begin{array}{l}
2 b^{2}-2(8-c-d-e) b+(8-c-d-e)^{2} \\
+c^{2}+d^{2}+e^{2}-16=0 .
\end{array}
$$
Since $b$ is a real number, we have
$$
\begin{align... | 0 \leqslant e \leqslant \frac{16}{5} | Algebra | proof | Yes | Yes | cn_contest | false | 701,796 |
Example 1. Given that DE is tangent to the circle with diameter $AB$ at point $C$, $\mathrm{AD} \perp \mathrm{DE}$, $\mathrm{BE} \perp \mathrm{DE}$, $\mathrm{CH} \perp \mathrm{AB}$. Prove:
(1) $\mathrm{CH}=\frac{1}{2} \mathrm{DE}$,
(2) $\mathrm{AD} \cdot \mathrm{BE}=\mathrm{CH}^{2}$. | Prove: As shown in Figure 1, draw a line parallel to $\mathrm{DE}$ through 0, intersecting $\mathrm{AD}$ at $\mathrm{M}$ and the extension of $\mathrm{EB}$ at $\mathrm{N}$. Let the angle parameter $\alpha=\angle \mathrm{COB}$, then $\angle \mathrm{A}=\angle \mathrm{OBN}=\alpha$.
Assume the radius of the circle is $R$,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,799 |
Example 2. Given that $C D$ is the altitude on the hypotenuse of $R t \triangle A B C$, and a circle is constructed with $C D$ as its diameter, intersecting $\mathrm{AC}$ and $\mathrm{BC}$ at $E$ and $F$ respectively. Prove:
$$
\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}^{\mathrm{S}}}{\mathrm{BC}^{3}} \text {. }
... | Proof As shown in Figure 2, let the angle parameter $\alpha = \angle \mathrm{A}$, and connect $\mathrm{DE}$ and $\mathrm{DF}$.
Then
$$
\begin{aligned}
\mathrm{AE} & = \mathrm{AD} \cos \alpha = (\mathrm{AC} \cdot \cos \alpha) \cos \alpha \\
& = (\mathrm{AB} \cdot \cos \alpha) \cos^{2} \alpha = \mathrm{AB} \cos^{3} \alph... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,800 |
Example 2. Find a point on the meadow $\frac{x^{2}}{45}+\frac{y^{2}}{20}=1$ such that the lines connecting it to the two foci are perpendicular. (Problem 13, Page 171) | Let $P$ be the point we are looking for. According to the ellipse equation, we can set $\mathbf{P}$
$$
\begin{array}{l}
(\sqrt{45} \cos \theta, \sqrt{20} \sin \theta) \\
\text { and }|\mathrm{C}_{1} \mathrm{O}|=|\mathrm{C}_{2} \mathrm{O}| \\
=\sqrt{45-20}=5 .
\end{array}
$$
Since $\mathrm{C}_{1} \mathrm{P} \perp \math... | P_{1}(3,4), P_{2}(-3,4), P_{3}(-3,-4), P_{4}(3,-4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,801 |
Example 3, Let CD be the altitude on the hypotenuse of right $\triangle \mathrm{ABC}$. Prove: $\mathrm{AB}+\mathrm{CD}>\mathrm{AC}+\mathrm{BC}$. | Proof As shown in Figure 3, let
angle parameter $\alpha=\angle \mathrm{CAB}, \mathrm{AB}$
$=c$, then $B C=c \sin \alpha$,
$A C=c \cos \alpha, C D=A C \sin \alpha=c \sin \alpha \cos \alpha$,
$\because \alpha$ is an acute angle,
$\therefore 0<\sin \alpha<1, 0<\cos \alpha<1$.
Thus, $\sin \alpha \cos \alpha<\sin \alpha+\co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,802 |
Example 4. Let $\triangle \mathrm{ABC}$ be an equilateral triangle, and $\mathrm{P}$ be any point on its incircle. Prove that: $\mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}$ is a constant. | $$
\begin{array}{l}
\mathrm{PA}^{2}=\mathrm{PO}^{2}+\mathrm{AO}^{2}-2 \mathrm{PO} \cdot \mathrm{AO} \cdot \cos \omega \\
=5 \mathrm{r}^{2}-4 \mathrm{r}^{2} \cos \alpha, \\
\mathrm{PB}^{2}=\mathrm{PC}^{2}+\mathrm{BC}^{2}-2 \mathrm{FO} \cdot \mathrm{BO} \cdot \cos \left(120^{\circ}-\alpha\right) \\
=5 \mathrm{r}^{2} \ldo... | 15r^2 | Geometry | proof | Yes | Yes | cn_contest | false | 701,803 |
Example 5. Let the side lengths of $\triangle \mathrm{ABC}$ be $\mathrm{BC}=\mathrm{a}, \mathrm{CA}=\mathrm{b}, \mathrm{D}$ is any point on the base BC, $\mathrm{DB}=\mathrm{q}, \quad \mathrm{DC}=\mathrm{p}, \mathrm{DA}=$ $x$. Prove:
$$
x=\sqrt{c^{2} \frac{p}{a}+b^{2} \frac{q}{a}-p q}
$$ | Proof: Let the angle parameter $\alpha = \angle \mathrm{ADC}$, then by the cosine rule we have:
$$
\begin{array}{l}
\mathrm{x}^{2} + \mathrm{p}^{2} - \mathrm{b}^{2} = 2 \mathrm{x} \mathrm{p} \cos \alpha, \\
\mathrm{x}^{2} + \mathrm{q}^{2} - \mathrm{c}^{2} = 2 \mathrm{x} \mathrm{q} \cos (180^{\circ} - \alpha).
\end{arra... | x = \sqrt{c^{2} \frac{p}{a} + b^{2} \cdot \frac{q}{a} - p q} | Geometry | proof | Yes | Yes | cn_contest | false | 701,804 |
Example 6. Prove: The sum of the areas of two squares constructed on the two adjacent sides $\mathrm{AB}$ and $\mathrm{AD}$ of a parallelogram $\mathrm{ABCD}$ is not less than the area of the rectangle constructed on the two diagonals $\mathrm{AC}$ and $\mathrm{BD}$, and prove under what conditions the two are equal. | Prove: As shown in Figure 6, let the angle parameter $\alpha = \angle \mathrm{DAB}$, $\mathrm{AB} = \mathrm{a}$, $\mathrm{AD} = \mathrm{b}$, $\mathrm{AC} = \mathrm{p}$, $\mathrm{BD} = \mathrm{q}$, then $\angle \mathrm{ABC} = 180^{\circ} - \alpha$.
By the cosine rule:
In $\triangle \mathrm{ABC}$, $q^{2} = a^{2} + b^{2} ... | a^{2} + b^{2} = \frac{1}{2} (p^{2} + q^{2}) \geqslant pq | Geometry | proof | Yes | Yes | cn_contest | false | 701,805 |
Example 7. Given a square $\mathrm{ABCD}$ with side length $a$ inscribed in $\odot O$, extend $\mathrm{DC}$ to $\mathrm{Z}$ such that $\mathrm{CZ}=\frac{1}{2} \mathrm{a}$, and take $\mathrm{BF}=\frac{1}{3} \mathrm{a}$ on $\mathrm{BC}$. The extension of $\mathrm{AF}$ intersects $\mathrm{BZ}$ at $\mathrm{G}$. Prove that ... | Prove as shown in Figure 7, set the angle parameters
$$
\alpha=\angle \mathrm{BAG}, \beta=\angle \mathrm{CBZ} \text {, }
$$
then $\operatorname{tg} \alpha=\frac{1}{3}, \operatorname{tg} \beta=\frac{1}{2}$,
$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,806 |
Example 8. In $\triangle \mathrm{ABC}$, $\mathrm{AB} = \mathrm{BC}$, $\mathrm{AD}$ and $\mathrm{BE}$ are the altitudes to $\mathrm{BC}$ and $\mathrm{AC}$ respectively, intersecting at $\mathrm{H}$. $\mathrm{N}$ is the midpoint of $\mathrm{AH}$. Draw $\mathrm{EF} \perp \mathrm{BC}$, extend $\mathrm{AD}$ to $\mathrm{M}$,... | Proof As shown in Figure 8, let the angle parameter $2 \alpha = \angle \mathrm{A}, \mathrm{BC} = 2 \mathrm{a}$, then $\angle \mathrm{BAD} = \angle \mathrm{CAD} = \angle \mathrm{HBD} = \alpha, \mathrm{AD} = \operatorname{actg} \alpha$, $\mathrm{HD} = \operatorname{atg} \alpha$. Therefore,
$$
\begin{aligned}
\mathrm{ND} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,807 |
Example 9., In isosceles triangle $\mathrm{ABC}$, take a point $D$ on the extension of the base $\mathrm{BC}$. Draw a perpendicular from $A$ to $AB$ intersecting $\mathrm{AC}$ at $\mathrm{E}$ and $\mathrm{AB}$ at $\mathrm{F}$. If the area of $\triangle \mathrm{AEF}$ is twice the area of $\triangle \mathrm{CDE}$, prove ... | Let $\alpha = \angle \mathrm{CED}$, then $\angle \mathrm{AEF} = \alpha$. In $\triangle \mathrm{AEF}$ and $\triangle \mathrm{CED}$, applying the Law of Sines, we get:
$$
\begin{array}{l}
\mathrm{EF} = \frac{\mathrm{AF} \sin \mathrm{A}}{\sin \alpha}, \mathrm{DE} = \frac{\mathrm{CD} \sin \angle \mathrm{ECD}}{\sin \alpha}.... | \frac{\mathrm{DE}}{\mathrm{EF}} = \frac{\mathrm{AB}}{\mathrm{BC}} | Geometry | proof | Yes | Yes | cn_contest | false | 701,808 |
10. Given that ABCD is a square, $P$ is a point inside the square, and $\angle PAB = \angle PBA = 15^{\circ}$.
Prove: $\triangle PCD$ is an equilateral triangle. | Let $\alpha - \angle \mathrm{PAB}$
$$
= \angle \mathrm{PBA} = 15^{\circ} \text{.}
$$
In $\triangle \mathrm{ABP}$, by the Law of Sines, we have:
$$
\frac{\mathrm{PA}}{\sin 15^{\circ}} = \frac{\mathrm{AB}}{\sin 150^{\circ}} = \frac{\mathrm{AB}}{\sin 30^{\circ}} = 2 \mathrm{AB} \text{.}
$$
Thus, $\mathrm{PA} = 2 \mathrm... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,809 |
Example 11. Given that $\mathrm{ABCD}$ is a square, $\mathrm{BE} / / \mathrm{AC}$, $\mathrm{CE}=\mathrm{AC}$, extend $\mathrm{EC}$ to intersect the extension of $\mathrm{BA}$ at $\mathrm{F}$. Prove: $A E=A F$. | 11
Proof As shown in Figure 11, let the parameter $\alpha-\angle \mathrm{CBE}-\angle \mathrm{ACB}-45^{\circ}$. And $C E=A C=\sqrt{2 \mathrm{BC}}$, in $\triangle \mathrm{BCE}, \frac{\sqrt{2 \mathrm{BC}}}{\sin 45^{\circ}}$
$$
=\frac{\mathrm{BC}}{\sin \angle \mathrm{BEC}} \text {. }
$$
Thus, $\sin \angle \mathrm{BEC}=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,810 |
Example 1. Through a fixed point $A$ on the bisector of a fixed angle $\angle \mathrm{XOY}$, any line is drawn intersecting $O X$ and $O Y$ at $P$ and $Q$ respectively.
Prove: $\frac{1}{O P}+\frac{1}{O Q}$ is a constant. | $$
\begin{array}{l}
\text{Given } CA = 2 \text{ (constant)}, \angle FO\| = \angle QOA = \alpha \text{ (constant)}. \\
\because S \triangle P O = S \Delta r O \Delta + S \Delta \Delta \cdot Q, \\
\therefore \frac{1}{2} O P \cdot O Q \sin 2 \alpha = \frac{1}{2} O P \cdot a \sin \alpha \\
\quad + \frac{1}{2} O Q \cdot a \... | \frac{2 \cos \alpha}{a} | Geometry | proof | Yes | Yes | cn_contest | false | 701,811 |
Example 3. For what value of $m$ does the line $y=-x+m$ intersect the ellipse $\frac{x^{2}}{20}+\frac{y^{2}}{5}=1$
(i) intersect;
(ii) be tangent,
(iii) be
separated? (Problem 13, page 132) | Solve the parametric equations of the ellipse as $\left\{\begin{array}{l}x=\sqrt{20} \cos \theta, \\ y=\sqrt{5} \sin \theta .\end{array}\right.$
When combined with the line $y-x+m$, we have $\sqrt{5} \sin \theta=-\sqrt{20} \cos \theta+m$, or $5 \sin (\theta+\phi)=m .(\phi=\operatorname{arctg} 2)$
Since the parameter (e... | m = \pm 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,812 |
Example 2. A triangle with equal angle bisectors is an isosceles triangle. | Proof As shown in Figure 2, let $\mathrm{BC}=\mathrm{a}$,
$$
\begin{array}{l}
B D=C F=t, A B=c, \\
\mathrm{AC}=\mathrm{b}, \angle \mathrm{ABD}=\angle \mathrm{CBD}=\alpha, \\
\angle \mathrm{ACE}=\angle \mathrm{BCE}=\beta \\
\because S_{\triangle \triangle \because C}=S_{\triangle \triangle \Delta D}+S_{\triangle D \beca... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,813 |
Example 3. As shown in Figure 3, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}$, a circle is drawn with $BC$ as the diameter, and $P$ is any point on the circle (excluding B and C). Prove: $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{1}{4}$. | Proof As shown in Figure 3, $\because$ BC is the diameter of $\odot$ O,
$$
\begin{array}{l}
\therefore \angle \mathrm{BPC}=90^{\circ}, \\
\because \mathrm{AB}=\mathrm{BC}=\mathrm{CD}, \\
\therefore S_{\triangle r \triangle B}=S_{\triangle r M e}=S S_{\triangle P C D}=S. \\
\end{array}
$$
Thus, $\frac{1}{2} \mathrm{PA}... | \operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{1}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 701,814 |
Example 6. As shown in Figure 7, given that O is any point inside $\triangle \mathrm{ABC}$, extend $\mathrm{AO}, \mathrm{BO}$, and $\mathrm{CO}$ to intersect $\mathrm{BC}, \mathrm{CA}$, and $\mathrm{AB}$ at points D, E, and F, respectively. Prove:
(1) $\frac{\mathrm{BD}}{\mathrm{BC}} \cdot \frac{\mathrm{CE}}{\mathrm{EA... | $$
\therefore \frac{B D}{D C}=\frac{S_{\triangle A B D}-S_{\triangle B O D}-S_{\triangle C O D}}{S_{\triangle C A D}}=\frac{S_{\triangle A B Q}}{S_{\triangle A C O}} \text {. }
$$
Similarly, $\frac{C E}{E A}=\frac{S_{\triangle B O C}}{S_{\triangle B O A}}, \frac{A F}{F B}=\frac{S_{\triangle A O C}}{S_{\triangle B O A}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,817 |
Example 7. As shown in Figure 8, given two circles $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ intersect at points $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{P}$ is any point on line $\mathrm{AB}$ outside the circles. From point $\mathrm{P}$, draw secants to the two circles, intersecting $\odot \mathrm{O}_{1}$ at $\... | Proof $\because D F \cdot C F$ is the product of two adjacent sides of $\triangle C D F$,
$$
\therefore \mathrm{DF} \cdot \mathrm{CF}=\frac{2 \mathrm{~S} \Delta \mathrm{CrD}}{\sin \angle \mathrm{CFD}},
$$
Similarly, $\mathrm{DE} \cdot \mathrm{CE}=\frac{2 \mathrm{~S} \triangle \mathrm{CED}}{\sin \angle C E D}$.
$$
\beg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,818 |
Example 8. Given $\mathrm{P}$ is any point on the minor arc $\widehat{\mathrm{BC}}$ of the circumcircle of equilateral $\triangle \mathrm{AOBC}$. Prove: $A P^{2}$ $=A B^{2}+P B \cdot P C$ | $$
\begin{array}{l}
=S \Delta \Delta r \Delta+S \Delta \Delta r c \\
=\frac{1}{2} \mathrm{BP} \cdot \mathrm{APsin} 60^{\circ} \frac{1}{2} \mathrm{AP} \cdot \mathrm{PCsin} 60^{\circ} \\
=\frac{1}{2} \mathrm{AP}(\mathrm{BP}+\mathrm{PC}) \sin 60^{\circ}=\frac{1}{2} \mathrm{AP}^{2} \sin 60^{\circ} \text { (here we use } \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,819 |
Example 9. As shown in Figure 10, in $\triangle \mathrm{ABC}$, if $BC > AB$, and $AD$, $CE$ are the altitudes to sides $BC$ and $AB$ respectively, prove that: $\mathrm{BC} + \mathrm{AD} \geqslant \mathrm{AB} + \mathrm{CE}$. | $$
\begin{array}{l}
\frac{1}{2} \mathrm{AD} \cdot \mathrm{BC}=\frac{1}{2} \mathrm{AB} \cdot \mathrm{CE}, \\
\therefore \mathrm{AD} \cdot \mathrm{BC}=\mathrm{AB} \cdot \mathrm{CE} . \\
\because \mathrm{BC}>\mathrm{AB}, \therefore \mathrm{AD}<\mathrm{CE} . \\
\text { Also } \because \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,820 |
Example 4. Prove that the intersection points of the ellipse $\frac{x^{2}}{20}+\frac{y^{2}}{5}=1$ and the hyperbola $\frac{x^{2}}{12}-\frac{y^{2}}{3}=1$ are the vertices of a rectangle. (Problem 19, Page 143) | Prove that substituting the parametric equations of the ellipse $\left\{\begin{array}{l}x=\sqrt{20} \cos \theta, \\ y=\sqrt{5} \sin \theta\end{array}\right.$ into the curve equation yields
$$
\frac{5}{3} \cos ^{2} \theta-\frac{5}{3} \sin ^{2} \theta=1
$$
or $\cos 2 \theta=\frac{3}{5}$.
Thus, $\cos \theta= \pm \sqrt{\f... | (4,1), (-4,1), (-4,-1), (4,-1) | Geometry | proof | Yes | Yes | cn_contest | false | 701,823 |
Example 12. In $\triangle \mathrm{ABC}$, if $\mathrm{AB}=\mathrm{c}, \mathrm{AC}=\mathrm{b}$, $BC=a, c^{2}=a^{2}+b^{2}$, prove: $\angle C=90^{\circ}$. | Prove $\because S=\frac{1}{2} absinC$,
$$
\begin{array}{l}
S=\sqrt{p(p-a)(p-b)(p-c)}, \\
p=\frac{1}{2}(a+b+c),
\end{array}
$$
(8) Substituting into (2) and after calculation, we get,
$$
S=\frac{1}{2} b \sqrt{a^{2}+b^{2}-b^{2}}=\frac{1}{2} a b \text {, }
$$
Comparing with (1) we have $\operatorname{sinC}=1$.
$$
\begin{... | \angle C=90^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 701,824 |
Example 13. As shown in Figure 13, ABCD is a square, CE//BD, $\mathrm{BE}=\mathrm{BD}, \mathrm{BE}$ intersects $\mathrm{CD}$ at $F$. Prove that $\mathrm{DF}=\mathrm{DE}$. | Proof: Let $\mathrm{AB}=1$,
$$
\begin{array}{l}
\text { then } \mathrm{BD}=\mathrm{BE}=\sqrt{2} \text {. } \\
\because B D / / C E \text {, } \\
\therefore S_{\triangle D X D}=S_{\triangle D D C} \\
=\frac{J}{2} \text {. } \\
\end{array}
$$
Let $\angle D E E-1$,
then $\frac{1}{2}(\sqrt{2})^{2} \sin x=\frac{1}{2}$, thu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,825 |
Example 1. Solve the equation
$$
\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)=0 \text {. }
$$ | Solve $-\cos \left(x+\frac{\pi}{6}\right)=\sin \left(x+\frac{\pi}{6}-\frac{\pi}{2}\right)$
$$
=\sin \left(x-\frac{\pi}{3}\right),
$$
The original equation can be transformed into
$$
\begin{array}{l}
\sin \left(x+\frac{\pi}{6}\right)=\sin \left(x-\frac{\pi}{3}\right) . \\
\therefore x+\frac{\pi}{6}=n \pi+(-1) \cdot\lef... | x=k\pi+\frac{7}{12}\pi, k \in Z | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,826 |
Theorem 2. The equation $\cos f(x)=\cos \varphi(x)$ is equivalent to the equation $f(x)=2 n \pi \pm \varphi(x) \quad(n \in Z)$. | To prove that due to $\operatorname{cost}$ being an even function, the solutions of the latter are obviously the solutions of the former.
Suppose $x_{0}$ is a solution to the equation $\cos f(x)=\cos \varphi(x)$, then $\cos f\left(x_{0}\right)=\cos \varphi\left(x_{0}\right)$,
$\therefore f\left(x_{0}\right)=2 n \pi \pm... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,827 |
Example 2. Prove: The equations $\sin ^{2} x=\sin ^{2} \alpha, \cos ^{2} x$ $=\cos ^{2} \alpha, \operatorname{tg}^{2} x=\operatorname{tg}^{2} \alpha$ have the same solutions. | Prove that the equation $\sin ^{2} x=\sin ^{2} \alpha$ is equivalent to $\sin x$
$$
\begin{array}{l}
=\sin ( \pm \alpha), \\
\quad \therefore \quad x=n \pi+(-1)^{2}( \pm \alpha)=n \pi \pm \alpha .(n \in Z)
\end{array}
$$
The equation $\cos ^{2} x=\cos ^{2} \alpha$ is equivalent to the equations $\cos x=\cos \alpha$ an... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,829 |
Example 1. Given $\left\{\begin{array}{l}a \sin A+b \sin B+c \sin C=0, \\ a \cos A+b \cos B+c \cos C=0 \text {. }\end{array}\right.$
where $A, B, C$ are not equal to each other, $\sin C \neq 0, \cos C \neq 0$,
prove that $\quad \frac{\sin (B-C)}{a}=\frac{\sin (C-A)}{b}=\frac{\sin (A-B)}{c}$. | Prove that (1) can be organized into a system of equations about $a$ and $b$:
$$
\left\{\begin{array}{l}
\frac{\sin A}{\sin C} \cdot a + \frac{\sin B}{\sin C} \cdot b = \cdots c, \\
\frac{\cos A}{\cos C} \cdot \frac{\cos B}{\cos C} \cdot b = -c.
\end{array}\right.
$$
From (2), we know that $A\left(\frac{\sin A}{\sin C... | \frac{\sin (B-C)}{a} = \frac{\sin (C-A)}{b} = \frac{\sin (A-B)}{c} | Algebra | proof | Yes | Yes | cn_contest | false | 701,830 |
Example 2 If $A+B+C=\pi$, prove that,
$\operatorname{si}^{2} A$ ctg $B+\sin ^{2} B \operatorname{ctg} C+\sin ^{2} C \operatorname{ctg} A-$ $\sin ^{2} A \operatorname{cig} C-\sin ^{2} B \operatorname{ctg} A-\sin ^{2} C \operatorname{ctg} B=0$.
Note: There seems to be a typo in the original problem. The function "si" sh... | $$
\begin{array}{l}
=\left|\begin{array}{ccc}
1 & \sin ^{2} A & \operatorname{ctg} A \\
1 & \sin ^{2} B & \operatorname{ctg} B \\
1 & \sin ^{2} C & \operatorname{ctg} C
\end{array}\right| \\
=\left\lvert\, \begin{array}{ccc}
1 & \sin ^{2} A & \operatorname{ctg} \mathbf{A} \\
0 & \sin ^{2} B-\sin ^{2} A & \operatorname{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,831 |
Example 1. Let $\mathrm{x}$ be a non-negative real number, then
$$
\sqrt{x+3}-\sqrt{x+2}<\sqrt{x+1}-\sqrt{x} .
$$ | Prove that if $\sqrt{x+3}-\sqrt{x+2}<\sqrt{x+1}-\sqrt{x}$ holds, then
$$
\sqrt{x+3}+\sqrt{x}<\sqrt{x+1}+\sqrt{x+2}
$$
$\because$ each radical is positive, $\therefore$ squaring both sides of (1), we get
$$
\begin{array}{l}
x+3+2 \sqrt{x+3} \sqrt{x}+x<x+1+2 \sqrt{x+1} \\
-\sqrt{x+2}+x+2
\end{array}
$$
$\because x$ is a ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,834 |
If a tangent line $\mathrm{PT}$ is drawn through $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ of a conic section, and $P Q, T, Q$ are the points of tangency, then the equation of the line through $T, Q$ is:
$$
y-y_{0}=-\frac{2 A x_{0}+B y_{0}+D}{B x_{0}+2 C y_{0}+E}\left(\mathbf{x}-x_{0}\right) .
$$ | Proof: Let the coordinates of $T$ and $Q$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. From a known proof, the equation of the tangent line through $T$ is
$$
\begin{array}{l}
A x_{1} x + B \frac{x_{1} y + y_{1} x}{2} + C y_{1} y + D \frac{x + x_{1}}{2} \\
+ E \frac{y + y_{1}}{2} + F = 0... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,835 |
Then $\left|\frac{a+b}{1+a b}\right|<1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in a mathematical form which is universal and does not need translation, the content remains ... | Proof Assuming the conclusion of the proposition holds,
$$
\begin{array}{l}
\Leftrightarrow\left\{\begin{array} { l }
{ | \mathrm { a } | > 1 , } \\
{ | \mathrm { b } | > 1 }
\end{array} \text { or } \left\{\begin{array}{l}
|\mathrm{a}|<1, \\
|\mathrm{~b}|<1
\end{array}\right.\right. \\
\end{array}
$$
(assumption) and... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,845 |
$$
\text { 5. } f(x)=e^{2}+5 \quad(x \geqslant 0) \text {, where } e=2.718 \cdots \text {, }
$$
then $\mathrm{f}^{-1}(\mathbf{x})=$ $\qquad$ , the domain of $\mathrm{f}^{-1}(\mathrm{x})$ is $\qquad$ | $5 . f^{-1}(x)=\ln (x-5),\{x \mid 6 \leqslant x<+\infty\}$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5 . f^{-1}(x)=\ln (x-5),\{x \mid 6 \leqslant x<+\infty\}$ | f^{-1}(x)=\ln (x-5),\{x \mid 6 \leqslant x<+\infty\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,851 |
8. If $\mathrm{P}(\sin \alpha, \cos \alpha)$ is a point on the terminal side of angle $\beta$, then $\sin \beta=$ $\qquad$ , $\operatorname{tg} \beta=$ $\qquad$ | $8 . \cos \alpha, \operatorname{ctg} \alpha$.
$8 . \cos \alpha, \cot \alpha$. | \cos \alpha, \cot \alpha | Other | math-word-problem | Yes | Yes | cn_contest | false | 701,854 |
Example 1. A newly built city has purchased a large number of public buses to solve the city's public transportation problems. They plan to set up bus stops at 1983 different locations and connect them by opening several bus routes. Their wishes are:
(i) to open as many routes as possible,
(ii) every two routes should ... | Let a total of $k$ bus routes be established, and let the set of 1983 bus stops be denoted as $\mathrm{A}$. The sets of bus stops served by each route are respectively denoted as
$$
A_{1}, \hat{H}_{2}, \cdots, \hat{A}_{K},
$$
It can be known from the problem statement that these subsets have the following properties:
... | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,856 |
Example 3. Given $\odot \mathrm{O}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-4 \mathrm{y}+1$ $=0, \odot \mathrm{O}_{2}: \mathrm{x}^{2}+\mathrm{y}^{2}-10 \mathrm{x}-4 \mathrm{y}-71=0$. If a moving circle is externally tangent to $\odot \mathrm{O}_{1}$ and internally tangent to $\odot \mathrm{O}_{2}$, find the equa... | $$
\begin{array}{c}
\text { Solving } \odot \mathrm{O}_{1}: \quad(\mathrm{x} \\
+1)^{2}+(\mathrm{y}-2)^{2}=4, \\
\text { radius } \mathrm{R}_{1}=2 ; \\
\odot \mathrm{O}_{2}: \\
(\mathrm{x}-5)^{2}+(\mathrm{y}-2)^{2} \\
=100, \text { radius } \mathrm{R}_{2}=10 .
\end{array}
$$
Let the center of the moving circle be $P(x... | \frac{(x-2)^{2}}{36}+\frac{(y-2)^{2}}{27}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,858 |
Example 4. A moving circle is externally tangent to the circle $x^{2}+y^{2}=2 a x$ $(a>0)$, and is also tangent to the y-axis. Find the equation of the locus of the center of the moving circle. | Solve As shown in Figure 5, given the center of the circle $\mathrm{O}_{1}(\mathrm{a}, 0)$, with radius $a$, the equation is: $(x-a)^{2}+y^{2}=a^{2}$. Let the center of the moving circle be $\mathrm{O}_{2}(\mathrm{x}, \mathrm{y})$. Introduce a fixed line $\mathrm{x}=-\mathrm{a}$, then, $\left|\mathrm{O}_{1} \mathrm{O}_... | y=\left\{\begin{array}{ll} \pm 2 \sqrt{\mathrm{ax}}, & (x \geq 0) ; \\ 0, & (x<0) . \end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,859 |
Example 6. Through point A $(a, 0)$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, a chord $\mathrm{AB}$ is drawn. On $\mathrm{AB}$ (or its extension), take a point $\mathrm{M}$ such that $\frac{\mathrm{AM}}{\mathrm{MB}}=\lambda$. When $\mathrm{B}$ moves on the ellipse, find the equation of the locus of po... | Let the coordinates of the moving point $\mathrm{M}$ be $\left(x_{M}, y_{M}\right)$, and the coordinates of point $B$ be $\left(x_{B}, y_{B}\right)$. We have
$$
\left\{\begin{array}{l}
x_{M} = \frac{a + \lambda x_{B}}{1 + \lambda}, \\
y_{M} = \frac{0 + \lambda y_{B}}{1 + \lambda}
\end{array} \Rightarrow \left\{\begin{a... | \frac{(2 x - a)^{2}}{a^{2}} + \frac{2 y^{2}}{b^{2}} = 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,861 |
Example 7. The base of a triangle is of fixed length and position, and the height on this side is of fixed length. Find the equation of the locus of its orthocenter.
untranslated text:
例7. 三角形的某底边长一定, 位置不变, 且这边上的高是定长, 求它的垂心的轨迹方程。
translated text:
Example 7. The base of a triangle is of fixed length and position, a... | Solve: Establish a rectangular coordinate system with the line containing the base $\mathrm{AB}$ of $\triangle \mathrm{ABC}$ as the $x$-axis and the midpoint of $\mathrm{AB}$ as the origin.
As shown in Figure 7, let the orthocenter $\mathrm{H}$ be $(\mathrm{x}, \mathrm{y}), |\mathrm{AB}|=2 \mathrm{a}, |C D|=h$, then $A... | y=-\frac{1}{h}\left(x^{2}-a\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,862 |
Example 8. As shown in Figure 8, a ray passing through the origin intersects the circle $x^{2}+(y-1)^{2}=1$ at point $Q$. On the ray $O Q$, find a point $P$ such that the distance from $P$ to the line $y=2$ equals $|P Q|$. When this ray rotates around the origin, find the equation of the trajectory of point $P$. | Let the equation of the ray be
$$
\left\{\begin{array}{l}
x=t \cos \theta, \\
y=t \sin \theta .
\end{array}(0 \leqslant \theta < 2\pi, \quad t \geqslant 0)\right.
$$
When the ray intersects the line $y=2$, we have
$$
t \sin \theta=2, \quad \therefore t_{1}=\frac{2}{\sin \theta} .
$$
When the ray intersects the circle $... | x^{2}+y^{2}=4 \text{ (upper half)} \text{ and } x=0(y>0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,863 |
Example 9. As shown in Figure $9, \mathrm{AB}$ is the diameter of a fixed circle $\mathrm{O}$, and the moving chord $M N \perp A B$. Find the equation of the locus of the intersection point $P$ of the lines $A N$ and $M B$. | Solve: Take the line through B and A as the $x$-axis and the circle center $O$ as the origin to establish a Cartesian coordinate system. The moving point $\mathrm{P}$ is the intersection of line $\mathrm{AN}$ and $\mathrm{BM}$, with its coordinates set as $(x, y)$. Also, let the fixed circle radius be $\mathrm{R}$. The... | x^{2}-y^{2}=R^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,864 |
Example 10. Draw any line L through the origin, intersecting the circles $x^{2}+y^{2}-2 a x=0$ and $x^{2}+y^{2}-3 a x=0$ at points $A$ and $B$. Take a point $P$ on line $L$ such that $|AP|: |BP| = 1: 2$. Find the equation of the locus of point $P$. | Tears 10
Solution Let the polar equations of the two circles be: $\rho=2 a \cos \theta, p=3 a \cos \theta$ (as shown in Figure 10).
The polar coordinates of points on the same ray L are: $\mathrm{A}\left(\rho_{1}, \theta\right)$, B $\left(\rho_{2}, \theta\right), \mathrm{P}(\rho, \theta)$.
By the given condition $|\m... | x^{2}+y^{2}-a x=0 \text{ or } x^{2}+y^{2}-\frac{7}{3} a x=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,865 |
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