problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Example 4. In a militia squad marching, the main and deputy machine gunners must be adjacent, and they must not stand at the head or the tail of the line. How many ways can the squad be arranged? | Solution: Still using the insertion method: The other 8 people can be arranged arbitrarily, with $8!$ arrangements. The main and deputy machine gunners can only be inserted into the 7 intervals formed by these 8 people, and the 2 people have a sequence, so there are $7 \times 2!$ ways to insert. Therefore, the total nu... | 564480 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,537 |
Example 3. As shown in Figure 3, let $\mathrm{P}$ be any point on the bisector of $\angle A O B$, extend $\mathrm{AP}$ to intersect $\mathrm{BO}$ at $\mathrm{D}$, and extend $\mathrm{BP}$ to intersect $\mathrm{A} 0$ at C.
Prove: $\frac{1}{\mathrm{AO}}+\frac{1}{\mathrm{DO}}=$
$$
\frac{1}{\mathrm{EO}}+\frac{1}{\mathrm{CO... | Proof $\because S_{\triangle \triangle P O}+S_{\triangle D P O}=S_{\triangle \triangle O D}$, then
$$
\begin{array}{l}
\frac{1}{2}(A O+D O) \cdot O P \sin \frac{\angle A O B}{2} \\
=\frac{1}{2} \cdot \mathrm{AO} \cdot \mathrm{OD} \cdot \sin \angle \mathrm{AOB} \text {, } \\
\therefore 2 A O \cdot O D \cos \angle \frac{... | \frac{1}{\mathrm{AO}}+\frac{1}{\mathrm{DO}}=\frac{1}{\mathrm{EO}}+\frac{1}{\mathrm{CO}} | Geometry | proof | Yes | Yes | cn_contest | false | 701,540 |
Example 4. As shown in Figure 4, $D$ and $E$ are two points on side $BC$ of $\triangle ABC$, and $\angle BAD = \angle CAE$.
Prove: $\frac{BD \cdot BE}{CD \cdot CE} = \frac{AB^2}{AC^2}$. | \begin{array}{l}\text { Prove } \quad-\frac{S}{S \triangle \cdots D}=\frac{B D}{C D}=\frac{A B \cdot A \bar{D} \sin \alpha}{A C \cdot A D \sin (A-\alpha)} \\ =\frac{A E \sin \alpha}{A C \sin (A-\alpha)} . \\ \text { Similarly, } \frac{B E}{C E}=\frac{A B \sin (A-\alpha)}{A C \sin \alpha} . \\ \text { Therefore, } \frac... | \frac{BD \cdot BE}{CD \cdot CE} = \frac{AB^2}{AC^2} | Geometry | proof | Yes | Yes | cn_contest | false | 701,541 |
Example 1. Discuss the differentiability of the function $\max (x,-x)$ on $(-\infty,+\infty)$. | Let $\mathrm{x}=-\mathrm{x}$, then $\mathrm{x}=0$. Thus, the functions $\mathrm{x}$ and $-\mathrm{x}$ are equal at the point 0 but not equal at other points. By Theorem 1, the function $\max (x,-x)$ is differentiable on $(-\infty,+\infty)-\{0\}$.
Also, $\lim _{x \rightarrow 0} \frac{f(x)-g(x)}{x-0}=\lim _{x \rightarrow... | not found | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,542 |
Example 6. As shown in Figure 6, given that $\odot \mathbf{0}$ is the circumcircle of $\triangle A B C$, with radius $R$, $\mathrm{AO}$, $\mathrm{BO}$, $\mathrm{CO}$ intersect the opposite sides at L, M, N respectively.
Prove: $\frac{1}{\mathrm{AL}}+\frac{1}{\mathrm{BM}}+$
$$
\frac{1}{C N}=\frac{2}{R} \text {. }
$$ | Prove $\frac{S_{\triangle \cup C C}}{S_{\triangle \triangle B C}}=\frac{A L-R}{A L}=1-\frac{R}{A L}$.
Similarly, $\frac{S_{\triangle \triangle O C}}{S_{\triangle \triangle B C}}=1-\frac{R}{B M}, \frac{S_{\triangle \triangle O B}}{S_{\triangle \triangle B C}}=1-\frac{R}{C N}$.
From this, we can get
$3-\frac{R}{A L}-\fra... | \frac{1}{\mathrm{AL}}+\frac{1}{\mathrm{BM}}+\frac{1}{\mathrm{CN}}=\frac{2}{\mathrm{R}} | Geometry | proof | Yes | Yes | cn_contest | false | 701,544 |
Example 8. As shown in Figure 8, if $\mathrm{AE}$ is the diameter of the circumcircle of $\triangle \mathrm{ABC}$, and $A E$ and $B C$ intersect at $D$, prove that $\frac{A D}{D E}=\operatorname{tg} B \cdot \operatorname{tg} C$. | ```
Connect $B E$ and $C E$, draw $A F \perp B C, E G \perp B C$,
\[
\begin{array}{l}
=\frac{A B \cdot A C}{B E \cdot E C}=-\frac{2 R \sin C \cdot 2 R \sin B}{2 R \cos C \cdot 2 R \cos B} \\
=\operatorname{tg} B \cdot \operatorname{tg} C \text {, } \\
\end{array}
\]
``` | \frac{AD}{DE} = \operatorname{tg} B \cdot \operatorname{tg} C | Geometry | proof | Yes | Yes | cn_contest | false | 701,546 |
Example 9. As shown in Figure 9, given that $\mathrm{AC}$ and $\mathrm{BD}$ are two diagonals of the cyclic quadrilateral $\mathrm{ABCD}$.
Prove: $\frac{e}{f}=\frac{a d+b c}{a b+c d}$. | $$
\begin{array}{c}
\text { Prove } \because S_{A B c D}= \\
a d \sin B A D+b c \sin B A D \\
=a b \sin A B C+c d \sin A B C, \\
\therefore-\frac{d}{a b+c} d=\frac{\sin A B C}{\sin B A D} \\
=\frac{2 R \sin A B C}{2 R \sin B A D}=\frac{A C}{B D}=\frac{e}{f} .
\end{array}
$$
(Where $\mathbf{R}$ is the radius of the know... | \frac{e}{f}=\frac{a d+b c}{a b+c d} | Geometry | proof | Yes | Yes | cn_contest | false | 701,547 |
Example 1. The edge length of the cube $\mathrm{ABCD}-\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$ is $\mathrm{a}$, find the distance between $\mathrm{A}, \mathrm{C}$ and $\mathrm{BD}$. | Solve as shown in Figure 1, on line segment $\mathrm{A}_{1} \mathrm{C}$, take any point $\mathrm{M}$, draw $M N \perp$ the base $\mathrm{AC}$, the foot of the perpendicular falls on the diagonal $A C$ of the base square $\mathrm{ABCD}$, the intersection of $A \subset$ and $\mathrm{BD}$ is denoted as $\mathrm{P}$, conne... | \frac{a\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,548 |
Example 2. The radius of the base of a right circular cone is $\mathbf{R}$ (Figure 2), $\mathrm{AG}$ is the bisector of the base angle of the axial section $S A B$, and $B D$ is a chord of the base circle, forming a $30^{\circ}$ angle with $\mathrm{AB}$. Find the distance between $\mathrm{AC}$ and $\mathrm{BD}$. | Take any point M on AC, draw MN perpendicular to the base, with the foot N on the diameter AB of the base circle, and set MN = x. We have
$$
\begin{array}{c}
\Delta N = MN \operatorname{ctg} 30^{\circ} = \sqrt{3} x, \\
N B = 2 R - \sqrt{3} x .
\end{array}
$$
In the base plane, draw NP perpendicular to BD,
we have $\ma... | \frac{2 \sqrt{7}}{7} \mathrm{R} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,549 |
Example 1. As shown in Figure 1, the edge length of the cube is $\mathrm{a}$. (1) Find the angle $\theta$ between $\mathrm{AC}^{\prime}$ and $\mathbf{D B}^{\prime}$, (2) Draw a perpendicular line $\mathbf{A E}$ from $\mathbf{A}$ to the diagonal $\mathbf{A}^{\prime} \mathrm{C}$, prove that $\frac{\mathrm{AE}^{2}}{\mathr... | (1) As long as we guide the students to observe that $\mathrm{AC}^{\prime}$ and $\mathrm{DB}^{\prime}$ lie on the diagonal plane $\mathrm{DAB}^{\prime} \mathrm{C}^{\prime}$, the problem clearly becomes one of finding the angle between the two diagonals of the rectangle $\mathrm{SAR}^{\prime \prime} \mathrm{C}^{\prime}$... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,551 |
Example 2. $\mathrm{AB}$ and $\mathrm{CD}$ are skew lines, $\mathrm{CA}=\mathrm{CB}$, $\mathbf{D A}=\mathbf{D B}$.
Prove: $\mathrm{AB} \perp \mathrm{CD}$. | To prove that two skew lines are perpendicular, i.e., $AB \perp CD$, we can first let one of them (such as $AB$) be perpendicular to the plane $P$ through $CD$. From $CA = CB, DA = DB$, we get that $\triangle CAB$ and $\triangle DAB$ are both isosceles triangles with $AB$ as the common base. If we can find a point $K$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,552 |
Example 3. As shown in Figure 5, two squares $\mathrm{ABCD}$ and $\mathrm{ABEF}$, with side length $\mathrm{a}$, lie in planes that are perpendicular to each other. Points $\mathrm{M}$ and $\mathrm{N}$ are on the diagonals $\mathrm{AC}$ and $\mathrm{BF}$, respectively, and $\mathrm{AM}=F N=b$. Prove that $M N \parallel... | To prove that $MN \parallel$ plane $\mathrm{CBE}$, it is sufficient to prove that the plane containing $\mathrm{MN}$ is parallel to plane $\mathrm{CBE}$.
Given that $\mathrm{ABCD}$ is a square, hence $\mathrm{CB} \perp \mathrm{AB}$.
Draw $MG \perp AB$ at $G$, then $MG \parallel CB$.
Connect $GN$. Since $\triangle AGM ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,554 |
Example 4. As shown in Figure 6, the dihedral angle N-PQ-M is a $120^{\circ}$ dihedral angle. From a point $\mathrm{A}$ inside it, perpendicular lines $A B$ and $A C$ are drawn to the planes $\mathrm{M}$ and $\mathrm{N}$, respectively. If $A B=3, A C=1$, find the distance from point $A$ to the edge $P Q$. | Analyzing this problem, if we do not approach it by constructing an auxiliary plane, but instead draw $AD \perp PQ$ through $\mathbf{A}$, and assume that $AC$, $AD$, and $AB$ are not in the same plane, it would be incorrect, and the problem would be unsolvable.
Let's look at the conditions given in the problem. Since ... | \frac{2 \sqrt{21}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,555 |
Example 5. Given a right circular cone with base radius R and height H. Find an inscribed right circular cone of maximum volume (the vertex of the inscribed cone is on the base of the given cone, and the circumference of its base is on the lateral surface of the given cone).
The translation is provided as requeste... | $$
\begin{array}{l}
\because \mathbf{R} \backslash \triangle A F G \sim R \mathrm{t} \triangle A B D, \\
\therefore-\frac{\mathrm{r}}{\mathbf{R}}=\frac{\mathrm{H}-\mathrm{h}}{\mathrm{H}}, \quad \mathrm{h}=\mathrm{H}\left(1-\frac{\mathrm{r}}{\mathbf{R}}\right),
\end{array}
$$
Therefore, the volume of the inscribed cone... | r=\frac{2}{3} R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,556 |
Example 6. As shown in Figure 8, from a point M on the surface of a sphere with radius $\mathbf{R}$, three chords MA, MB, MC are drawn, each perpendicular to the other two. Then $M A^{2}+M B^{2}+M C^{2}$ is a constant. | Let M be any point on the sphere, MA and MB be two perpendicular chords. Through MA and MB, construct a section $\boldsymbol{\alpha}$ that intersects the sphere to form a small circle $0^{\prime} \triangle \simeq B$.
$\because \angle \mathrm{AMB}=90^{\circ}$,
$\therefore \mathrm{AMB}$ is a rectangle. The diameter of th... | 4 \mathbf{R}^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 701,557 |
Five, let $ABCD$ be a square, $M$ is the midpoint of side $AB$, $MN \perp DM$, $BN$ bisects the exterior angle of $\angle ABC$, prove by analytic method: $|MD|=|MN|$.
---
The translation is provided as requested, maintaining the original formatting and line breaks. | Prove: Draw $\mathbf{NE} \perp \mathrm{AB}$ extended at $\mathrm{E}$. In right $\triangle \mathrm{AMD}$ and right $\triangle \mathrm{ENM}$,
$$
\begin{array}{c}
\because \angle \mathrm{AMD} + \angle \mathrm{ADM} \\
= 90^{\circ}, \\
\angle \mathrm{AMD} + \angle \mathrm{NME} = 90^{\circ}, \\
\therefore \angle \mathrm{ADM}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,558 |
"Prove that the three products $(1-\mathrm{a}) \mathrm{b}$, $(1-\mathrm{b}) \mathrm{c}$, $(1-\mathrm{c}) \mathrm{a}$, formed by positive numbers $\mathrm{a}<1$, $\mathrm{b}<1$, $\mathrm{c}<1$, cannot all be greater than $\frac{1}{4}$. " (61st Hungarian Mathematical Competition Problem) | Proof 1. $\because 00$,
$$
\therefore(1-a) a \leqslant\left[\frac{(1-a)+a}{2}\right]^{2}=\frac{1}{4} \text {. }
$$
Similarly, $(1-b) b \leqslant \frac{1}{4},(1-c) c \leqslant \frac{1}{4}$. Therefore,
$$
[(1-a) b][(1-b) c][(1-c) a]
$$
$=(1-a) a(1-b) b(1-c) c \leqslant\binom{ 1}{4}^{3}$. Hence, the products $(1-a) b,(1-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,559 |
Example 3. Given three different positive numbers, the sum of their reciprocals is
1. Try to find these three numbers. | Let the three numbers be $\mathrm{a}>\mathrm{b}>\mathrm{c}>1$, then $-\frac{1}{a}+\frac{1}{b}+\frac{1}{\mathrm{c}}$ $<\frac{3}{c}$, i.e., $c<3$. Therefore, $c=2, \frac{1}{a}+\frac{1}{b}=\frac{1}{2} . \because \frac{1}{a}+\frac{1}{b}$ $<\frac{2}{b}, \therefore b<4$, hence $\mathrm{b}=3$. From this, we get $\mathrm{a}=6$... | a=6, b=3, c=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,562 |
Example 4. Solve the system of equations:
$$
\left\{\begin{array}{l}
\left|a_{1}-a_{2}\right| x_{2}+\left|a_{1}-a_{3}\right| x_{3}+\left|a_{1}-a_{4}\right| x_{4}=1, \\
\left|a_{2}-a_{1}\right| x_{1}+\left|a_{2}-a_{3}\right| x_{3}+\left|a_{2}-a_{4}\right| x_{4}=1, \\
\left|a_{3}-a_{1}\right| x_{1}+\left|a_{3}-a_{2}\righ... | Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>a_{4}$, then the original system of equations becomes:
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right) x_{2}+\left(a_{1}-a_{3}\right) x_{3}+\left(a_{1}-a_{4}\right) x_{4}=1, \\
\left(a_{2}-a_{1}\right) x_{1}+\left(a_{2}-a_{3}\right) x_{3}+\left(a_{2}-a_{4}\right... | x_{1}=x_{4}=\frac{1}{a_{1}-a_{4}}, x_{2}=x_{3}=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,563 |
Example 3. Discuss the differentiability of the function $\max (\sin x, \cos x)$ on $(-\infty, +\infty)$. | Solve the equation $\sin x - \cos x = 0 \Longleftrightarrow x = k\pi + \frac{\pi}{4} (k$ is an integer). Therefore, when $x \neq k\pi + \frac{\pi}{4}$, $\sin x \neq \cos x$. By Theorem 1, the function $\max (\sin x, \cos x)$ is differentiable. Because
$$
\begin{array}{l}
\lim _{x \rightarrow k\pi + \frac{\pi}{4}} \frac... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,564 |
$$
\begin{array}{l}
\text { 2. Let } f(x)=x-p^{!}+\left|x-15^{\prime}+x-p-15\right|, \text { where } \\
\text { } 0<p<15 \text {. }
\end{array}
$$
Find: For $x$ in the interval $p \leqslant x \leqslant 15$, what is the minimum value of $f(x)$? | \begin{array}{l}\text { Solve } \because p \leqslant x \leqslant 15 \text{ and } p>0, \therefore p+15 \geqslant x . \\ \therefore f(x)=x-p+15-x+p+15-x=30-x, \\ \text { Also } \because x \leqslant 15, \\ \therefore f(x) \geqslant 30-15=15 . \\ \therefore \text { the minimum value of } f(x) \text { is } 15 .\end{array} | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,568 |
Three, find the product of the real roots of the equation $x^{2}+18 x+30=2 \sqrt{x^{2}+18 x+45}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{ll}
\text { Let } t=\sqrt{x^{2}+18 x+45}, \\
\therefore & t^{2}-2 t-15=0, \\
& t_{1}=5, t_{2}=-3(\because t>0, \text { discard }) . \\
\therefore & \sqrt{x^{2}+18 x+45}=5, \\
\therefore & x^{2}+18 x+20=0 . \\
\text { Also } \because \Delta=18^{2}-4 \cdot 20>0,
\end{array}
$$
$\therefore$ The equation h... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,569 |
Five, let the sum of the squares of two complex numbers x, $\mathrm{y}$ be 7, and their sum of cubes be 10. The largest real value that $x+y$ can take is what?
---
Translation:
Five, let the sum of the squares of two complex numbers x, $\mathrm{y}$ be 7, and their sum of cubes be 10. The largest real value that $x+y$... | $$
\begin{array}{l}
\because x^{2}+y^{2}=7, \\
\therefore x^{3}+y^{3}=10=(x+y)\left(x^{2}+y^{2}-x y\right) \\
=(x+y)\left[7-\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2}\right] \\
=(x+y)\left[7-\frac{\left.(x+y)^{2}-7\right] .}{2}\right] .
\end{array}
$$
Let $x+y=t$ , then
$$
\begin{array}{c}
10=\frac{21 t-t^{3}}{2}, \... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,571 |
Seven, the 25 knights of King Arthur sat at their round table, and three knights (any of whom were chosen with equal probability) were selected to slay the dragon. Let $\mathrm{p}$ be the probability that at least two of the three selected knights are seated next to each other. If $\mathrm{p}$ is written as a reduced f... | The number of ways to choose 3 people from 25 is
$$
C_{25}^{3}=\frac{25 \cdot 24 \cdot 23}{3 \cdot 2 \cdot 1}=25 \cdot 23 \cdot 4 \text { (ways). }
$$
The number of ways to choose 3 people from 25 such that at least 2 are adjacent can be calculated as follows: treat the adjacent two people as one, then there are 25 wa... | 57 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,573 |
$\begin{array}{l} \text { Eight, the largest two-digit number factor of the number } \mathrm{n}=\binom{200}{100} \text { is } \\ \text { what? } \quad \text { [Note: }\binom{200}{100} \text { is } \mathrm{C}_{2}^{1} 00 \% \text { ] }\end{array}$ | Solve $\mathrm{n}=\mathrm{C}_{200}^{+} \times 0=\frac{200 \cdot 199 \cdot 198 \cdots 102 \cdot 101}{100!}$
$$
=2^{50} \cdot \frac{199 \cdot 197 \cdot 195 \cdots 103 \cdot 101}{50!} .
$$
This number is an integer, and the numerator $199 \cdot 197 \cdot 195 \cdots 103 \cdot 101$
is the product of three-digit odd numbers... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,574 |
Theorem 4 If the functions $\mathrm{f}(\mathrm{x})$ and $g(\mathrm{x})$ are right (left) differentiable at $x_{0}$, and $f\left(x_{0}\right)=g\left(x_{0}\right)$, and the limit $\lim _{x \rightarrow x_{0}+} \frac{f(x)-g(x)}{x-x_{0}}$ $\left(\lim _{x \rightarrow x_{0}-} \frac{f(x)-g(x)}{x-x_{0}}\right)$ exists, then the... | Prove the right derivative case only. Similarly to Theorem 2, by $\max (f(x), g(x))=\frac{1}{2}(f(x)+g(x))+$ $\frac{1}{2}|f(x)-g(x)|$, it suffices to prove that the function $|f(x)-g(x)|$ is right-differentiable at point $x_{0}$ (because if two functions are right-differentiable at a point, then their sum is also right... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,575 |
Nine, find the minimum value of $f(x)=\frac{9 x^{2} s \sin ^{2} x+4}{x \sin x}(0<x<\pi)$. | Let $\mathbf{x} \sin \mathrm{x}=\mathrm{t}$, then $\mathrm{t}$ is a real number.
Thus $\quad \mathrm{f}(\mathrm{x})=\frac{9 \mathrm{t}^{2}+4}{\mathrm{t}} \geq \frac{2 \cdot 3 \cdot 2}{t}=12$.
$\therefore \mathrm{f}(\mathrm{x})$'s minimum value is 12. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,576 |
Ten, $1447, 1005, 1231$ have many things in common: they are all four-digit numbers, their highest digit is 1, and they each have exactly two identical digits. How many such numbers are there? | For the same digit being 1, since the highest digit is 1, the other 1 has three positions to choose from, and the remaining positions have $\mathrm{P}_{8}^{2}$ arrangements, i.e., $3 \cdot P_{0}^{*}=3 \times 72=216$ (ways).
For the same digit not being 1, there are 9 choices for the same digit, 8 choices for the other... | 432 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,577 |
Twelve, as shown in Figure 5, the length of the diameter AB of a circle is a two-digit integer (in decimal). By reversing the two digits, we get the length of the chord CD, which is perpendicular to the diameter AB. The distance from the intersection point H to the center O of the circle is a positive rational number. ... | Let $\mathrm{AB}=10 \mathrm{x}+\mathrm{y}$, where $\mathrm{x}, \mathrm{y}=0,1,2, \cdots, 9$, and $x \neq 0$.
Then $C D=10 y+x$.
$$
\begin{aligned}
\mathrm{OH}^{2} & =\left(\frac{10 \mathrm{x}+\mathrm{y}}{2}\right)^{2}-\left(\frac{10 \mathrm{y}+\mathrm{x}}{2}\right)^{2} \\
& =\frac{9}{4} \cdot 11\left(\mathrm{x}^{2}-\ma... | 65 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,579 |
Thirteen, for $\{1,2,, 3 \cdots, n\}$ and each of its non-empty subsets, we define the alternating sum as follows: arrange the numbers in the subset in descending order, then alternately add and subtract the numbers starting from the largest (for example, the alternating sum of $\{1,2,4,6,9\}$ is $9-6+4-2+1=6$, and the... | Solution: Obviously, the number of non-empty subsets of $\{1,2,3,4,5,6,7\}$ is $2^{7}-1=127$. Each element appears 64 times in these subsets. According to the problem's requirements, the numbers $1,2,3,4,5,6$ each appear 32 times in odd positions and 32 times in even positions, so the alternating sum of these numbers i... | 448 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,580 |
Fourteen, in Figure 6, the radii of the two circles are 8 and 6, the distance between the two centers is 12, a line passing through one of the intersection points of the two circles intercepts equal chords $\mathrm{QP}$ and $\mathrm{PR}$ on the two circles, find the square of the length of $QP$. | Solve as shown in Figure 7, construct $\mathbf{O}_{1} M \perp Q \mathbf{Q}$ at $M, \mathbf{O}_{2} \mathbf{N} \perp$ $Q R$ at. Given $Q P=$
$\mathrm{PR}$, then $\mathrm{QM}=\mathrm{NR}=\frac{1}{2} \mathrm{Q} P$.
Let $Q \mathrm{M}=\mathrm{x}$, then
$$
\begin{array}{c}
\mathrm{O}_{1} \mathrm{M}=\sqrt{ } \overline{\mathrm{... | 130 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,581 |
Fifteen, as shown in Figure 8, if two intersecting chords of a circle, with $\mathrm{B}$ on the smaller arc $\mathrm{AD}$, and the circle's radius is $5, \mathrm{BC} = 6$, $\mathrm{AD}$ is bisected by $\mathrm{BC}$. Also, from $\mathrm{A}$, the chord $\mathrm{AD}$ is the only one that can be bisected by $\mathrm{BC}$. ... | Solve as shown in Figure 9, let $\mathrm{AD}$ and $\mathrm{BC}$ intersect at $P$, connect $O P$, then $O P \perp A D$. Connect $\mathrm{OA}, \mathrm{O} 3, \mathrm{O} C$, draw $\mathrm{OE} \perp . \mathrm{BC}$ at $E$, and let $P A=P D=y, P E=$ $x$, then $B P=3-x$. Also, from the given conditions, $\mathrm{OE}=4$.
By the... | 175 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,582 |
Question 11. Let $\mathrm{R}$ be the set of all real numbers, and for the function
$$
f(x)=x^{2}+a x+b(a, b \in R),
$$
define the sets
$$
\begin{array}{l}
A=\{\mathbf{x} \mid \mathbf{x}=f(\mathbf{x}), \mathbf{x} \in \mathbf{R}\}, \\
B=\{\mathbf{x} \mid \mathbf{x}=\mathrm{f}(\mathrm{f}(\mathbf{x})), \mathbf{x} \in \mat... | Solve (1) From the given conditions, the function
$$
f(x)=x^{2}-x-2 .
$$
The equation $x=f(x)$ becomes $x^{2}-2 x-2=0$.
Its solution set is $\mathrm{A}$, so
$$
A=\{1-\sqrt{3},1+\sqrt{3}\} .
$$
Similarly, the equation $x=f(f(x))$ is
$$
x=\left(x^{2}-x-2\right)^{2}-\left(x^{2}-x-2\right)-2,
$$
Simplifying, we get $\le... | A \cup B=\{1-\sqrt{3}, 1+\sqrt{3},-\sqrt{2}, \sqrt{2}\}, A \cap B=\{1-\sqrt{3}, 1+\sqrt{3}\}, B=\{-\sqrt{3}, \sqrt{3},-1,3\}, A \cap B=\{\alpha\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,583 |
Question 12. Let $a, b, c$ be the sides opposite to the angles $A, B, C$ of $\triangle ABC$, prove that the equation
$$
x^{2}+2 a b x \sin C+a b c^{2} \sin A \sin B=0
$$
has two equal real roots. | Proof: Let the heights on sides $\mathrm{a}$ and $\mathrm{b}$ of $\triangle \mathrm{A} 3 \mathrm{C}$ be $\mathrm{h}_{\mathrm{A}}$ and $\mathrm{h}$, respectively. Then,
$$
\begin{array}{l}
\mathrm{h}_{\mathrm{a}}=c \sin \mathrm{B}=\mathrm{b} \sin \mathrm{C}, \\
\mathrm{h}_{\mathrm{b}}=c \sin \mathrm{A}=\mathrm{a} \sin \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,584 |
Question 13. Try to use parametric equations to prove the tangent line theorem of a circle in plane geometry. | Given the circle with center $\mathrm{O}$ as the origin, establish a rectangular coordinate system xoy (as shown in the figure).
Let the radius of the circle be $R$,
then the equation of the circle is:
$$
x^{2}+y^{2}=R^{2} \text{ (1) }
$$
Let $P \left(x_{0}, y_{0}\right)$ be a point outside the circle,
and the param... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,585 |
1. $\left|x^{2}-9\right|<\frac{1}{2} x^{2}+1$. | There are two intersection points: one is the intersection of $y=\frac{1}{2} x^{2}+1$ and $y=x^{2}-9$, and the other is the intersection of $y=\frac{1}{2} x^{2}+1$ and $y=9-x^{2}$.
$$
\begin{array}{r}
v=x^{2}-\theta=\frac{1}{2} x^{2}+1, \\
\frac{1}{2} x^{2}=10, x^{2}=20, \\
x= \pm 2 \sqrt{5} ;
\end{array}
$$
and
$$
\b... | \left\{x \in R:-2 \sqrt{5}<x<-\frac{4}{3} \sqrt{3} \text{ or } \frac{4}{3} \sqrt{3}<x<2 \sqrt{5}\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,588 |
2. $\left|\frac{x-2}{x-1}\right|<6\left|\frac{x+1}{x+2}\right|, x \neq 1, -2$. | The first step to solve this problem is to transform the equation into
$$
\begin{array}{c}
|(x-2)(x+2)|<5|(x+1)(x-1)|, \\
x \neq 1,-2
\end{array}
$$
or $\left|x^{2}-4\right|<5\left|x^{2}-1\right|, x \neq 1,-2$. We need to find the $x$-coordinates of the intersection points of $y=4-x^{2}$ and $y=5\left(x^{2}-1\right)$,... | \left\{x \in R: x<-2 \text { or }-2<x<-\sqrt{\frac{6}{2}} \text{ or } -\frac{1}{2}<x<\frac{1}{2} \text{ or } \sqrt{\frac{6}{2}}<x\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,589 |
Example 3. Let $P(n)$ denote the following equation
$$
1+3+\cdots+(2 n-1)=n^{2} .
$$
We wish to prove that $P(n)$ is true for all natural numbers. | Prove (1) When $n=1$, $1=1^{2}$. So $P(1)$ is true;
(2) ${ }^{L_{6}}$ Assume $P(n-1)$ is true, that is, we have the equation
$$
1+3+\cdots+(2 n-3)=(n-1)^{2} \text {. }
$$
Adding $2 n-1$ to both sides of this equation, we get
$$
\begin{array}{l}
1+3+\cdots+(2 n-3)+(2 n-1)= \\
(n-1)^{2}+2 n-1=n^{2},
\end{array}
$$
whic... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,590 |
Example 5. Given $a \sin \theta + b \cos \theta = c, b \sin \theta + a c \cos \theta$ $= \mathrm{d}$, prove: $\left(\frac{c+d}{a+b}\right)^{2}+\left(\frac{c-d}{a-b}\right)^{2}=2$. | Proof Given by the Master
$$
\begin{array}{l}
a \sin \theta+b \cos \theta=c, \\
b \sin \theta+a \cos \theta=d, \\
\begin{array}{l}
\text { (1) }+ \text { (2) } \quad a(\sin \theta+\cos \theta)+b(\sin \theta+\cos \theta) \\
=c+d,
\end{array} \\
\begin{aligned}
\therefore & \sin \theta+\cos \theta=\frac{c+d}{a+b} .
\end... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,592 |
Example 7. If $a+b+c>0, ab+bc+ca>0$, $abc>0$, prove that $a>0, b>0, c>0$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: Let $a+b+c=p, ab+bc+ca=q, abc=r$. By Vieta's formulas, $a, b, c$ are the three real roots of the equation
$$
x^{3}-p x^{2}+q x-r=0
$$
Let $f(x)=x^{3}-p x^{2}+q x-r$.
If $x_{1}<x_{2}<x_{3}$ are the roots of $f(x)$, and $a>0, b>0, c>0$.
The translation preserves the original text's line breaks and formatting. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,593 |
Example 8. Find the sum: $\cos \alpha+\cos 3 \alpha+\cos 5 \alpha+\cdots$ $+\cos (2 n-1) \alpha$. | Solve the original expression $\begin{array}{l}=\frac{1}{2 \sin \alpha}[2 \sin \alpha \cos \alpha+2 \sin \alpha \cos 3 \alpha \\ +2 \sin \alpha \cos 5 \alpha+\cdots+2 \sin \alpha \cos (2 n-1) \alpha] \\ =\frac{1}{2 \sin \alpha}[\sin 2 \alpha \div(\sin 4 \alpha-\sin 2 \alpha) \\ +(\sin 6 \alpha-\sin 4 \alpha)+\cdots+\si... | \frac{\sin 2 n \alpha}{2 \sin \alpha} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,594 |
Example 1. Let $z$ be a complex number, solve the equation
$$
\frac{1}{2}(z-1)=\frac{\sqrt{3}}{2}(1+z) \text { i. }
$$ | Solving by eliminating $\frac{1}{2}$ from both sides, we get
$$
z-1=\sqrt{3} i+\sqrt{3} i \cdot z .
$$
Rearranging and combining terms, we get
$$
(1 -\sqrt{3} i) z=1+\sqrt{3} i \text {. }
$$
This leads to $z=\frac{1+\sqrt{3}}{1-\sqrt{3}} \frac{i}{i}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$.
Or, let $z=a+b i,(a, b \in R)$
F... | z=-\frac{1}{2}+\frac{\sqrt{3}}{2} i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,595 |
Example 2. There are three balls, A, B, and C. The volume of ball A is $\frac{1}{2}$ the sum of the volumes of B and C, and the volume of ball B is $\frac{1}{3}$ the sum of the volumes of A and C. What fraction of the sum of the volumes of A and B is the volume of C? | Let the volumes of spheres $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$ be $\mathrm{V}_{1}$, $\mathrm{V}_{2}$, and $\mathrm{V}_{3}$, respectively. Suppose the volume of sphere $\mathrm{C}$ is $-\frac{1}{x}$ times the sum of the volumes of spheres $\mathrm{A}$ and $\mathrm{B}$. Then,
$$
\left\{\begin{array}{l}
\mathrm{V... | \frac{5}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,596 |
Example 2. Given: $\mathbf{P}$ is any point inside $\triangle \mathrm{ABC}$, extending $\mathrm{AP}$, $B P$, and $C P$ intersect the opposite sides at D, E, F (as shown in Figure 2). Prove: $\frac{\mathrm{PD}}{\mathrm{AD}}+\frac{\mathrm{PE}}{\mathrm{BE}}+\frac{\mathrm{PF}}{\mathrm{CF}}=1$. | $$
\begin{array}{l}
\frac{\mathrm{PF}}{\mathrm{CF}}=\frac{\mathrm{S}_{\triangle M A B}}{\mathrm{~S}_{\triangle \Delta B}} . \text { (3) } \\
\text { From }(1)+(2)+(3) \text {, we get: } \\
\frac{P D}{A D}+\frac{P E}{B E}+\frac{P F}{C F}=\frac{S_{\triangle r B C}+S_{\triangle P C A}+S}{S_{\triangle A B C}} \text {, } \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,598 |
Example 3. Given: In $\triangle \mathrm{ABC}$, $\mathrm{AD}$ is the angle bisector of $\angle \mathrm{BAC}$. Prove: $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$. | Prove as shown in Figure 3, draw $\mathrm{AH} \perp \mathrm{B}, \mathrm{H}$ is the foot of the perpendicular.
$$
\begin{aligned}
\because \quad S_{\triangle \triangle B D} & =\frac{1}{2} \mathrm{BD} \cdot \mathrm{AH}, \\
S_{\triangle \triangle C D} & =\frac{1}{2} \mathrm{DC} \cdot \mathrm{AH}, \\
\therefore \quad \frac... | \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}} | Geometry | proof | Yes | Yes | cn_contest | false | 701,599 |
Example 4. Draw perpendiculars from any point inside an equilateral triangle to its three sides, and prove that the sum of these three perpendiculars is equal to the height of this equilateral triangle. Given: In Fig. 4, $\mathbf{P}$ is any point inside the regular $\triangle \mathrm{ABC}$, $P D \perp B C$, $P E \perp ... | ```
\begin{array}{l}
\because S_{\triangle B P C}+S_{\triangle C P A}+S_{\triangle \triangle P B}=S \triangle \triangle B C, \\
\text { and } S_{\triangle \triangle P C}=\frac{1}{2} B C \cdot P D, \\
S_{\triangle C P A}=\frac{1}{2} C A \cdot P E, \\
S_{\triangle A P B}=\frac{1}{2} A B \cdot P F, \\
S_{\triangle A B C}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,600 |
Example 5. Given: $\odot 0$ is the circumcircle of the equilateral $\triangle \mathrm{ABC}, \mathbf{P}$ is any point on $\widehat{\mathrm{B} C}$. PA intersects $\mathrm{BC}$ at D (as shown in Figure 5). Prove: $\frac{1}{P D}=\frac{1}{P C}+\frac{1}{P B}$. | Proof $\because \triangle \mathrm{ABC}$ is an equilateral triangle,
$$
\begin{array}{l}
\therefore \quad \angle \mathrm{BPD}=\angle \mathrm{CPD}=60^{\circ}, \\
\angle \mathrm{BPC}=120^{\circ} . \\
\because \quad \mathrm{S}_{\triangle \mathrm{BPC}}=\mathrm{S} \triangle \mathrm{BPD}+\mathrm{S}_{\triangle \mathrm{CPD}},
\... | \frac{1}{P D}=\frac{1}{P C}+\frac{1}{P B} | Geometry | proof | Yes | Yes | cn_contest | false | 701,601 |
Example 1. Let $y=f(x)$ be a cubic function of $x$. When $x=x_{0}$, $y=f(x)$ has an extremum $M=f\left(x_{0}\right)$, then $y=f(x)$ $=a\left(x-x_{0}\right)^{2}(x-m)+M$, where $a, m$ are two undetermined constants. | Proof: Let $y=f(x)=a x^{3}+b, x^{2}+c x+d$.
$\because f\left(x_{0}\right)=M$,
$\therefore x_{0}$ is a root of the equation $f(x) - M=0$.
According to the factor theorem, $x-x_{0}$ is a factor of $f(x) - M$, let $f(x)-M=a\left(x-x_{0}\right) f_{1}(x)$, i.e., $f(x) = a\left(x-x_{0}\right) f_{1}(x)+M$, where $f_{1}(x)$ is... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,602 |
Example 6. Given: H is the orthocenter of acute $\triangle \mathrm{ABC}$.
Prove: $\frac{\mathrm{BC}}{\mathrm{AH}}+\frac{\mathrm{CA}}{\mathrm{BH}}+\frac{\mathrm{AB}}{\mathrm{CH}}=\frac{\mathrm{BC} \cdot \mathrm{CA} \cdot \mathrm{AB}}{\mathrm{AH} \cdot \mathrm{BH} \cdot \mathrm{CH}}$. | Proof As shown in Figure 6, construct the circumcircle $\odot \mathrm{O}$ of $\triangle \mathrm{ABC}$, and let its radius be $R$; extend $\mathrm{AH}$ to intersect $\odot$ O at $\mathrm{D}$, and connect $B D, C D$.
$$
\begin{array}{l}
\angle \mathrm{BCD}=\angle \mathrm{BAD}=\angle \mathrm{BCH}, \\
\because \mathrm{BC}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,603 |
Example 7. Given: As shown in Figure $7, \mathrm{AT}$ is the tangent of the circumcircle of $\triangle \mathrm{ABC}$, and $\mathrm{T}$ lies on the extension of $\mathrm{BC}$.
Prove: $\frac{A B^{2}}{\mathrm{~A}} \mathrm{C}^{2}=\frac{B T}{C T}$.
保留源文本的换行和格式，翻译结果如下：
Example 7. Given: As shown in Figure $7, \mathrm{AT}$ ... | Proof Given the known conditions, we have:
$\triangle T A B \sim \triangle T C A$,
$$
\begin{array}{l}
\therefore \frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{S}_{\triangle T A B}}{\mathrm{~S}_{\triangle T C A}} . \\
\text { Also } \because \frac{\mathrm{S}_{\triangle T A B}}{\mathrm{~S}_{\triangle T C A}}=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,604 |
Example 1. Let the integer part of $\frac{1}{3-\sqrt{7}}$ be $a$, and the decimal part be $b$, find the value of $a^{2}+(1+\sqrt{7}) a b$. | $$
\begin{aligned}
\because \frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{(3-\sqrt{7})(3+\sqrt{7})} \\
\quad=\frac{3+\sqrt{7}}{2},
\end{aligned}
$$
and $2<\sqrt{7}<3,0<\frac{\sqrt{7}-1}{2}<1$,
$$
\begin{array}{l}
\therefore \quad \frac{1}{3-\sqrt{7}}=\frac{3+1+\sqrt{7}-1}{2} \\
=2+\frac{\sqrt{7}-1}{2} .
\end{array}
$$
Ther... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,605 |
Solve the equation
$$
\sqrt{x^{2}+5 x-2}-\sqrt{x^{2}+5 x-5}=1 . \text { (1) }
$$ | Solve: Multiply both sides of (1) by the conjugate radical
$\sqrt{x^{2}+5 x-2}+\sqrt{x^{2}+5 x-5}$, we get
$$
\checkmark \sqrt{x^{2}+5 x-2}+\sqrt{x^{2}+5 x-5}=3 \text {. }
$$
$$
\text { By } \frac{(1)+(2)}{2} \text { we get } \sqrt{x^{2}+5 x-2}=2 \text {, }
$$
i.e., $x^{2}+5 x-6=0$.
Solving it, we get $x_{1}=1, x_{2}=... | x_{1}=1, x_{2}=-6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,609 |
Column 6, when $\lim _{x \rightarrow-\infty}\left(4 x-\sqrt{a x^{2}-b x+c}=3\right.$,
untranslated part:
when $\lim _{x \rightarrow-\infty}\left(4 x-\sqrt{a x^{2}-b x+c}\right)=3$, | $$
\text { Let } \begin{aligned}
& \lim _{x \rightarrow \infty}\left(4 x-\sqrt{a x^{2}-b x+c}\right. \\
= & \lim _{x \rightarrow \infty}\left(4 x-\sqrt{a x^{2}-b x+c} \cdot \frac{4 x+\sqrt{a x^{2}-b x+c}}{4 x+\sqrt{a x^{2}-b x+c}}\right. \\
= & \frac{(16-a) x+b}{4+\sqrt{a}}
\end{aligned}
$$
Given the condition $\frac{... | a=16, b=24 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,610 |
Example 7. Given $y=5 \sqrt{3 \bar{x}-1}+2 \sqrt{1-3 \mathbf{x}}+\frac{1}{2}$, find the value of $\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}}$, | Solving, we get $x=\frac{1}{3}$,
From this, $y=\frac{1}{2}$,
$$
\begin{array}{l}
\text { Hence } \frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}} \\
=\frac{\sqrt{y}(\sqrt{x}+\sqrt{y})}{x-y}+\frac{\sqrt{y}(\sqrt{x}-\sqrt{y})}{x-y} \\
=\frac{2 \sqrt{x y}}{x-y}=-2 \sqrt{x-} . \\
\end{array}
$$ | -2 \sqrt{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,611 |
Example 8. Given $1 x=(3+2 \sqrt{2})^{-1}$, $y=(3-2 \sqrt{2})^{-1}$, find the value of $(x+1)^{-1}+(y+1)^{-1}$. | $$
\begin{array}{c}
\text { Sol } \because x=(3+2 \sqrt{2})^{-1} \\
=\frac{3-2 \sqrt{2}}{(3+2 \sqrt{2})(3-2 \sqrt{2})}=3-2 \sqrt{2},
\end{array}
$$
$\therefore x, y$ are conjugate radicals, thus we have
$$
\begin{array}{l}
y=x^{-1}=3+2 \sqrt{2} . \\
\therefore(x+1)^{-1}+(y+1)^{-1}=(4-2 \sqrt{2})^{-1} \\
\quad+(4+2 \sqr... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,612 |
Example 2. The graph of the function $\mathrm{y}=\mathrm{f}(\mathrm{x})$ passes through points $\mathrm{A}(-1$, $-5)$ and $B(-2,3)$, and when $x=1$, $y$ has an extremum, which is 3. Find the expression for $\mathrm{f}(\mathrm{x})$. | Solve $\because \mathrm{x}=1$ when, $\mathrm{y}$ has an extremum, the extremum is 3, so we can set $y=f(x)=a(x-1)^{2}(x-m)+3$. $\because y=f(x)$'s graph passes through A $(-1,-5)$, B $(-2,3)$,
$$
\therefore\left\{\begin{array}{l}
4 a(-1-m)+3=-5, \\
9 a(-2-m)+3=3 .
\end{array}\right.
$$
From (1) and (2), we get $a=-2, ... | f(x) = -2 x^{3} + 6 x - 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,613 |
Example 9. If one root of the equation $x^{2}-(\operatorname{tg} \theta+2 \operatorname{tg} \theta) x+1=0$ is $2+\sqrt{3}$, and $: g 0 \dot{v}+\mathrm{tt} \theta \theta$ is a rational number, find the value of $\sin 2 \theta$.
Translate the above text into English, please retain the original text's line breaks and ... | Solving: Since the irrational roots of a rational equation always appear in pairs, the other root of the equation is $2-\sqrt{3}$.
From Vieta's formulas, we have $\operatorname{tg} \theta+\operatorname{ctg} \theta=4$,
$$
\begin{array}{l}
\text { i.e., } \frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}=4... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,614 |
Example 10. Simplify: $\sqrt{(-45)^{2}}+\frac{1}{\sqrt{12-\sqrt{140}}}$
$$
-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}} .
$$ | $$
\begin{array}{l}
-\frac{1}{\sqrt{8+\frac{v^{\prime} 64-60}{2}}}-\sqrt{\frac{8-\sqrt{64}-60}{2}} \\
2 \\
\sqrt{\frac{10+\sqrt{100-84}}{2}}+\sqrt{\frac{10-\sqrt{100-84}}{2}} \\
=3 \sqrt{5} \text {. } \\
\end{array}
$$
This problem uses the following theorem:
If $a>0, b>0$, and $a^{2}>b$, then
$$
\sqrt{a \pm \sqrt{b}}... | 3 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,615 |
Example 11. When $x \geqslant 1, x \neq 2$, simplify
$$
y=(x+2 \sqrt{x-1})^{-\frac{1}{2}}+(x-2 \sqrt{x-1})^{-\frac{1}{2}}
$$ | Solve $\because y^{2}=(x+2 \sqrt{x-1})-1$
$$
\begin{array}{l}
+2(x+2 \sqrt{x-1})^{-\frac{1}{2}}(x-2 \sqrt{x-1})-\frac{1}{2} \\
+(x-2 \sqrt{x-1})^{-1} \\
=\frac{2}{x^{2}-4 x+4}+\frac{2}{|x-2|} .
\end{array}
$$
(1) When $1 \leqslant x < 2$, hence $\mathrm{y}=\sqrt{\frac{4}{(\mathrm{x}-2)^{2}}}=\frac{2}{-(\mathrm{x}-2)}$
... | \frac{2}{2-x} \text{ for } 1 \leqslant x < 2 \text{ and } \frac{2 \sqrt{x-1}}{x-2} \text{ for } x > 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,616 |
Example 12. Compare $\sqrt{7-4 \sqrt{3}}$ with $\sqrt{5-2 \sqrt{6}}$. | Solve $\because \sqrt{7-4 \sqrt{3}}=\sqrt{(2-\sqrt{3})^{2}}$
$$
\begin{array}{l}
=2-\sqrt{3}=\frac{1}{2+\sqrt{3}}, \\
\sqrt{5-2 \sqrt{6}}=\sqrt{(\sqrt{3}-\sqrt{2})^{2}} \\
=\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{2}+\sqrt{3}} \text {. } \\
\end{array}
$$
And $2+\sqrt{3}>\sqrt{2}+\sqrt{3}$,
$$
\begin{array}{l}
\therefore \fra... | \sqrt{7-4 \sqrt{3}} < \sqrt{5-2 \sqrt{6}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,617 |
Example 13. Find the domain of the function $y=\frac{1}{\sqrt{x+1}-\sqrt{1-x}}$. | $$
\begin{array}{l}
\text { Sol } \because y=\frac{1}{\sqrt{x+1}-\sqrt{1-x}} \\
=\frac{\sqrt{x+1+\sqrt{1-x}}}{(x+1)-(1-x)}=\frac{\sqrt{x+1}+\sqrt{1-x}}{2 x}, \\
\therefore\left\{\begin{array} { l }
{ x + 1 \geqslant 0 , } \\
{ 1 - x \geqslant 0 , } \\
{ x \neq 0 }
\end{array} \quad \left\{\begin{array}{l}
x \geqslant-... | -1 \leqslant x < 0 \text{ and } 0 < x \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,618 |
Example 14. Determine the monotonicity of the function $y=x-\sqrt{x^{2}-1}(x \geqslant 1)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Sol } y=x-\sqrt{x^{2}-1}=\frac{\left(x-\sqrt{x^{2}-1}\right)\left(x+\sqrt{x^{2}-1}\right)}{x+\sqrt{x^{2}-1}} \\
=\frac{1}{x+\sqrt{x^{2}-1}} .
\end{array}
$$
As \( x \) increases, \( \sqrt{x^{2}-1} \) also increases, hence \( x+\sqrt{x^{2}-1} \) increases accordingly, which indicates that \(... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,619 |
Example 15. The function $\mathrm{f}(\mathrm{x})=\log _{2}\left(\sqrt{\mathrm{x}^{2}+1}+\mathrm{x}\right)$ is an odd function, or an even function? | \[
\begin{array}{l}
\text { Solution } \because \sqrt{x^{2}+1}-x=\frac{1}{\sqrt{x^{2}+1}+x}, \\
\text { and } f(-x)=\log _{a}\left[\sqrt{(-x)^{2}+1}+(-x)\right] \\
\quad=\log _{4}\left(\sqrt{x^{2}+1}-x\right) \\
=\log _{2} \frac{1}{\sqrt{x^{2}+1}+x}=\log _{2}\left(\sqrt{x^{2}+1}+x\right)^{-1} \\
=-\log _{a}\left(\sqrt{... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,620 |
Example 16. Solve the equation: $(\sqrt{5+2 \sqrt{6}})^{x}$
$$
+(\sqrt{5-2 \sqrt{6}})^{x}=10 .
$$ | Solve $\because y^{\prime} 5-2 \sqrt{6}=\frac{1}{\gamma^{i} 5+2 v^{\prime} 6}$,
$\therefore$ the equation can be rewritten as $(\sqrt{5+2 \sqrt{6}})^{x}$ $+\frac{1}{(\sqrt{5+2 \sqrt{6}})^{x}}=10$.
Let $\mathrm{y}=(\sqrt{5+2 \sqrt{6}})^{\mathrm{x}}$, then $\mathrm{y}+\frac{1}{y}=10$,
which simplifies to $y^{2}-10 y+1=0$... | x_1=2, x_2=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,621 |
Example 17. Solve the equation:
$$
\left(x-\frac{1}{x}\right)^{\frac{1}{2}}+\left(1-\frac{1}{x}\right)^{\frac{1}{2}}=x .
$$ | Multiply both sides of (1) by the conjugate radical $\left(x-\frac{1}{x}\right)^{\frac{1}{2}}-\left(1-\frac{1}{x}\right)^{\frac{1}{2}}$, and simplify to get
$$
\left(x-\frac{1}{x}\right)^{\frac{1}{2}}-\left(1-\frac{1}{x}\right)^{\frac{1}{2}}=1-\frac{1}{x} \text {. }
$$
By (1) + (2), we get
$2\left(x-\frac{1}{x}\right)... | x=\frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,622 |
Example 3. For the cubic function $y=f(x)$, when $x=1$, $y$ has a local minimum of -3; when $x=-3$, $y$ has a local maximum of 29. Find the expression for $f(x)$ and the center of symmetry of the graph of $y=f(x)$. | Solve $\because$ when $x=1$, $y$ has a minimum value of -3,
$\therefore$ we can set $y=a(x-1)^{2}(x-m)-3 .(a \neq 0)$
$\because$ when $\mathbf{x}=-3$, $\mathrm{y}$ has a maximum value of 29,
$\therefore$ we can set $y=a(x+3)^{2}(x-n)+29$.
Thus, $a(x-1)^{2}(x-m)-3$
$$
=a(x+3)^{2}(x-n)+29 \text { is an identity. }
$$
... | y=x^{3}+3 x^{2}-9 x+2, \text{ center of symmetry } (-1,13) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,624 |
Example 19. Prove that $\cos \left(\frac{1}{2} \arcsin x\right)$
$$
=\frac{|x|}{\sqrt{2\left(1-\sqrt{1-x^{2}}\right.}} \quad(x \neq 0) .
$$ | Let $\arcsin = a, \quad\left(-\frac{\pi}{2} \leqslant \alpha < 0\right.$,
$$
0 < \alpha \leqslant \frac{\pi}{2} \text { ) }
$$
Therefore, $\cos \left(\frac{1}{2} \arcsin I\right) = \sqrt{\frac{1 + \sqrt{1} - x^{2}}{2}}$
$$
\begin{array}{l}
= \frac{\sqrt{x^{2}}}{\sqrt{2\left(1 - \sqrt{\left.1 - x^{2}\right)}\right.}} =... | \frac{|x|}{\sqrt{2\left(1 - \sqrt{1 - x^{2}}\right)}} | Algebra | proof | Yes | Yes | cn_contest | false | 701,625 |
Example 21. Divide a line segment of length $10 \mathrm{~cm}$ into two parts, use one part as the side length of an equilateral triangle to construct an equilateral triangle, and use the other part as the side length of a square to construct a square. Find the minimum value of the sum of their areas.
Divide a line seg... | $(10-\mathrm{x}) \mathrm{cm}$. For the
$\mathrm{S}_{\text {equilateral triangle }}=\frac{1}{2} \cdot \frac{\sqrt{3}}{2}x^{2}=\frac{\sqrt{3}}{4} x^{2}$;
$\mathrm{S}_{\text {square }}=(10-\mathrm{x})^{2}$.
Let their combined area be $\mathrm{sm}^{2}$, then
$$
\begin{array}{l}
\mathrm{S}=\frac{\sqrt{3}}{4} \mathrm{x}^{2}+... | \frac{100(4 \sqrt{3}-3)}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,627 |
Example 22. Using the lower base of a frustum as the base, and the center of the upper base as the vertex, construct a cone. If the lateral surface of this cone divides the volume of the frustum into two equal parts, find the ratio of the radii of the upper and lower bases of the frustum. | Solve the external figure. Let the radii of the upper and lower bases of the frustum be $\mathrm{r}$ and $\mathrm{R}$, and the height be $\mathrm{h}$, then
$$
\begin{array}{l}
\frac{1}{3} \pi h\left(r^{2}+r R+R^{2}\right) \\
=2 \cdot \frac{1}{3} \pi \hbar^{2} h .
\end{array}
$$
Simplifying and rearranging, we get $R^{... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,628 |
Example 1. When $\alpha \in\left(0, \frac{\pi}{2}\right)$, prove: $\sin \alpha<\alpha<\tan \alpha$. | For the unit circle, as shown in Figure 3, let the central angle \(\angle \mathrm{AOB}\) be an acute angle, with its radian measure being \(\alpha\). Since
\(\mathrm{S}_{\Delta 0 \Delta \pi} < \mathrm{S}_{\text{sector } 0 \Delta \pi}\)
\( < S_{\triangle O A C}\), and from geometry, we know:
\(S_{\Delta 011} = \frac{1}{... | \sin \alpha < \alpha < \tan \alpha | Inequalities | proof | Yes | Yes | cn_contest | false | 701,629 |
Example 4. Solve the equation $\sin x + \sin 2x + \sin 3x = 0$. | Solution 1 $\sin x+\sin 2 x+\sin 3 x$
$$
\begin{array}{l}
=(\sin x+\sin 3 x)+\sin 2 x \\
=2 \sin 2 x \cdot \cos x+\sin 2 x=\sin 2 x(2 \cos x+1) .
\end{array}
$$
Thus, the original equation becomes
$$
\sin 2 x \cdot(2 \cos x+1)=0 \text {. }
$$
Solving it, we get: $\sin 2 x=0, \cos x=-\frac{1}{2}$.
$\therefore$ The sol... | x=\frac{n \pi}{2}, x=2 n \pi \pm \frac{2 \pi}{3}, n \in \mathbb{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,632 |
Example 6. Find the maximum value of the area of a triangle inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad(a>b>0)$. | Solve: Using the parametric equations of the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1 \quad(\mathrm{a}>\mathrm{b}>0)$:
$$
\left\{\begin{array}{l}
x=a \cos \theta, \\
y=b \sin \theta .
\end{array} \text { ( } \theta\right. \text { is the parameter) }
$$
As shown in Figure ... | \frac{3}{8} \sqrt{3} \mathrm{ab} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,634 |
1. In $\triangle \mathrm{ABC}$, $\angle \mathrm{A}$ is twice $\angle \mathrm{B}$, given sides $\mathrm{b}$ and $c$, find side $a$. | Point $D$ and divide it into two parts with the ratio $b: c$. Secondly, since $\triangle A C D$ and $\triangle A B C$ are similar, we have,
$$
\frac{A C}{C D}=\frac{B C}{A C} \text { or written as } \frac{b}{\frac{a b}{b+c}}=\frac{a}{b} \text {. }
$$
Therefore, $\mathrm{a}=\sqrt{b^{2}+b c}$. | a=\sqrt{b^2+bc} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,636 |
2. In a right-angled triangle, the legs are equal to $b$ and $c$, respectively. Find the length of the angle bisector of the right angle.
The above text has been translated into English, preserving the original text's line breaks and format. | 2. Let $\mathrm{AD}$ be the external angle bisector of the right angle $\mathrm{A}$ in $\triangle \mathrm{ABC}$, and $\mathrm{DE} \parallel \mathrm{AC}$ (Figure 2). Given that $\angle \mathrm{DAE}=\frac{\pi}{4}$, then $\mathrm{AE}=\mathrm{DE}=\frac{x}{\sqrt{2}}$, where the solution $x=\mathrm{AD}$ is the required lengt... | \frac{bc\sqrt{2}}{b+c} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,637 |
3. From point $\mathbf{P}$ on the hypotenuse $\mathrm{AC}$ of the right triangle $\mathrm{ABC}$, draw a perpendicular line $\mathrm{PM}$ to $\mathrm{AC}$. If it divides the triangle into two figures of equal area, and $A P=a, P C=b$, find the area of $\triangle A B C$.
Translate the above text into English, please kee... | 3. If the foot of the perpendicular $\mathrm{M}$ is on the right-angle side $\mathrm{AB}$, and $M$ does not coincide with $B$ (Figure 3), according to the problem, we have $\frac{1}{2} \mathrm{a} \cdot \mathbf{P M} > \frac{1}{2} \mathrm{~b} \cdot \mathbf{P M}$, thus $a > b$.
Conversely, it is easy to see that when con... | S = \frac{\sqrt{2}}{2} a \sqrt{b^{2} + 2 ab - a^{2}} \text{ or } S = \frac{\sqrt{2}}{2} b \sqrt{a^{2} + 2 ab - b^{2}} \text{ or } S = a^{2} | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 701,638 |
4. Given the two sides $\mathrm{a}$ and $\mathrm{b}$ of a triangle, and it is known that the medians on these two sides intersect at a right angle, find the third side of the triangle. Under what conditions does this triangle exist? | $\mathrm{BC}=\mathrm{a}$. Now find $\mathrm{AB}$, denoted by $\mathrm{c}$.
Let $O D=x, O E=y$. Then, in $\triangle A O E$, $\triangle A O B$, and $\triangle B O D$, we can obtain:
$$
4 x^{2}+y^{2}=\frac{b^{2}}{4}, 4 x^{2}+4 y^{2}=c^{2}, 4 x^{2}+16 y^{2}=a^{2} \text {. }
$$
Eliminating $\mathbf{x}$ and $y$, we get
$$
\... | c^2 = \frac{a^2 + b^2}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,639 |
5. The two sides enclosing a vertex of a triangle are $\mathrm{a}$ and $\mathrm{b}(\mathrm{a}<\mathrm{b})$, and the angle at this vertex is trisected by straight lines. The ratio of the segments of these lines inside the triangle is $\mathrm{m}: \mathrm{n}(\mathrm{m}<\mathrm{n})$. Find the lengths of these segments. | 5. $\angle \mathrm{ACD}=$
$$
\begin{array}{l}
\angle \mathrm{DCE}=\angle \mathrm{ECB}=\alpha, \text { given } \\
\mathrm{HCE}=\mathrm{x}, \mathrm{CD}=\mathrm{v} \text { (Figure 5). }
\end{array}
$$
The area of $\triangle \mathrm{ABC}$ can be expressed in three different ways:
$$
\begin{array}{l}
S_{A C D}+S_{D C D}=\fr... | x=\frac{\left(n^{2}-m^{2}\right) a b}{n(b m-a n)}, y=\frac{\left(n^{2}-m^{2}\right) a b}{m(b m-a n)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,640 |
6. In $\triangle \mathrm{ABC}$, $\angle \mathrm{A}=\arccos \frac{7}{8}, \mathrm{BC}=\mathrm{a}$, the altitude from vertex $\mathrm{A}$ is equal to the sum of the other two altitudes. Find the area of $\triangle \mathrm{ABC}$. | 6. Let $\mathrm{AB}=\mathrm{x}, \mathrm{AC}=\mathrm{y}$ (Figure 6).
At this time, the area of $\triangle \mathrm{ABC}$ is
$$
S=\frac{1}{2} x y \sin B A C=x y \frac{\sqrt{15}}{16}
$$
The altitudes of the triangle can be expressed as:
$$
\mathrm{h}_{\mathrm{a}}=\frac{2 \mathrm{~S}}{\mathrm{a}}, \mathrm{h}_{\mathrm{x}}=... | S=a^{2} \sqrt{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,641 |
7. A line is parallel to side $\mathrm{BC}$ of the known $\triangle \mathrm{ABC}$, intersects the other two sides at $D$ and $E$, and makes the area of $\triangle BDE$ equal to the constant $\mathrm{k}^{2}$. What is the relationship between $\mathrm{k}^{2}$ and the area of $\triangle \mathrm{ABC}$ for the problem to be... | 7. Let S denote the area of $\triangle ABC$ (Fig. 7), and set $\frac{AD}{AB}=x$. Then the area of $\triangle ADE$ is equal to $x^{2} S$, and the area of $\triangle ABE$ is equal to $\mathrm{J} \mathbf{x S}$ (this conclusion is not difficult to obtain - poem difficulty). According to the conditions of the problem, we ca... | S \geqslant 4 k^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,642 |
9. In $\triangle \mathrm{ABC}$, side $\mathrm{BC}=\mathrm{a}, \mathrm{AB}=\mathrm{c}$, altitudes are $\mathbf{A D}$ and $\mathbf{C E}$. If $\frac{\mathrm{DE}}{\mathrm{AC}}=k$, find $\mathrm{A} \mathbf{C}$. | 9. Let's first consider: $\triangle A B C \sim \triangle D \bar{D} E$, and if $\angle B$ is acute, the ratio of areas is $\cos B$; if $\angle B$ is obtuse, the similarity ratio is $-\cos B$. We only examine the case where $\angle B$ is acute (as shown in Figure 9), and the case where $\angle B$ is obtuse is completely ... | A C=\sqrt{a^{2}+c^{2} \pm 2 a c k} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,644 |
10. In isosceles triangle $\mathrm{ABC}$ with base $\mathrm{AC}$, draw altitudes $\mathrm{AA}^{\prime}, \mathrm{BB}^{\prime}$, and $\mathrm{CC}^{\prime}$. If $\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}$, find the ratio of the areas of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and ... | 10. $\triangle \mathrm{ABC} \cos \triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}$ (Fig. 10), and the cosine ratio is $\cos \mathrm{C}$ (refer to Question 9). Thus, $\cos \mathrm{C}=\frac{1}{\sqrt{3}}$.
Let $A B=B C=a, \angle A B C=\beta$. Therefore, $A^{\prime} B^{\prime}=\frac{a}{\sqrt{3}}$. Since $\tria... | \frac{2}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,645 |
Example 5. Find the number of distinct real roots of the cubic equation $x^{3}-3 x^{2}-a=0$, where $\mathrm{a}$ is a constant. | Let $y=f(x)=x^{3}-3 x^{2}-a$, then $f^{\prime}(x)=3 x^{2}-6 x$.
$$
\begin{array}{l}
\text { By } f^{\prime}(x)=0 \text { we get } x=0,2 . \\
y_{\text {max }}=f(0)=0 . \\
y_{\text {min }}=f(2)=-4 .
\end{array}
$$
Therefore, (i) when $-4 < a < 0$ or $a > 0$ or $a < -4$
the original equation has one real root (and two c... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,646 |
11. In $\triangle \mathrm{ABC}$, construct altitudes $\mathrm{AA}_{1}, \mathrm{BB}_{1}, \mathrm{CC}_{1}$, and connect the feet of the perpendiculars. If the angles of $\triangle A B C$ are known, try to calculate the ratio of the areas of $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$. | 11. First, assume that $\triangle \mathrm{ABC}$ is an acute triangle (Figure 11).
For: $S_{\triangle B C}-S_{\Delta_{1}} B_{1} c_{1}=S \boldsymbol{B}_{1} \wedge c_{1}+S_{\mathbf{c}} \boldsymbol{\Delta}{ }_{1}+S_{1} \mathrm{cs}$
Since $S, \wedge c_{1}=\frac{1}{2} A B_{1} \cdot A C_{1} \sin A=\frac{1}{2} A B \cos A$
$$
... | \frac{S_{A_{1}B_{1}C_{1}}}{S_{\triangle ABC}}=\cos^2 A+\cos^2 B+\cos^2 C-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,647 |
12. In $\triangle \mathrm{ABC}$, a line parallel to $BC$ is drawn through the intersection of the angle bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$. $M$ and $N$ are the points where this line intersects sides $\mathrm{AB}$ and $\mathrm{AC}$, respectively. Try to find the relationship between the segments $... | 12. 1) Let $BO$ and $CO$ be the angle bisectors of $\triangle ABC$ (Figure 13). It is easy to see that $\triangle BOM$ and $\triangle CON$ are isosceles, hence $\mathrm{MN}=\mathrm{BM}+\mathrm{CN}$.
2) The relation $MN=BM$
$+\mathrm{CN}$ also holds for the external angle bisectors.
3) If one is an internal
angle bisect... | MN=|\mathrm{CN}-\mathrm{BM}| | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,648 |
13. Points $M$ and $N$ are on sides $A B$ and $B C$ of $\triangle A B C$, respectively, and $\frac{A M}{B M}=m, \frac{C N}{B N}=n$. Line $M N$ intersects the median $B D$ of the triangle at point $O$. Find $\frac{\mathrm{DO}}{\mathrm{BO}}$. | 13. For definiteness, let $m>n$. When $m<n$, the positions of vertices $\mathrm{A}$ and $\mathrm{C}$ need to be exchanged in the following discussion.
The lines $\mathrm{CE}$ and $\mathrm{DF}$ through points $\mathrm{C}$ and $\mathrm{D}$ are parallel to $\mathrm{MN}$ (Figure 15). At this time, $\frac{E M}{B M}=\frac{C... | \frac{1}{2}(m+n) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,649 |
14. In an equilateral triangle $\mathrm{ABC}$, take any point $\mathbf{P}$ inside it, and draw perpendiculars $\mathrm{PD}, \mathrm{PE}$, and $\mathrm{PF}$ from this point to $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ respectively, with $\mathrm{D}$, $\mathrm{E}$, and $\mathrm{F}$ being the feet of the perpendiculars. Cal... | 14. Through point $\mathbf{P}$, draw three lines parallel to the three sides of a triangle (Figure 16). The three triangles formed (the shaded parts in the figure) are also equilateral triangles, and the perimeter of each of these triangles equals the side $A B=a$ of $\triangle A B C$. The sum of their heights equals t... | \frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,650 |
15. Find the ratio of the area of $\triangle A B C$ to the area of the triangle formed by its medians.
untranslated text remains in its original format and line breaks are preserved. | 15. Let point $O$ be the intersection of the medians of $\triangle ABC$ (Figure 17). On the extension of median $\mathrm{BE}$, construct $\mathrm{ED}=0 \mathrm{E}$. From the properties of medians, the sides of $\triangle \mathrm{CDO}$ are $\frac{2}{3}$ of the sides of the triangle formed by the medians. If $S_{1}$ repr... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,651 |
16. In a triangle with sides $a, b, c$, there is a semicircle with its diameter on side $c$ and tangent to sides $a$ and $b$. Find the radius of this semicircle. | And $S_{c O A}=\frac{1}{2} b_{\text {r }}$. Adding these two quantities and expressing the area of $\triangle A B C$ according to Heron's formula, we get:
$$
\begin{array}{l}
r=\frac{2}{a+b} \sqrt{p(p-a)(p-b)(p-c)} \text {. } \\
\text { where } p=\frac{1}{2}(a+b+c) .
\end{array}
$$ | r=\frac{2}{a+b} \sqrt{p(p-a)(p-b)(p-c)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,652 |
20. In isosceles triangle $\mathrm{ABC}$, there is a point $\mathrm{D}$ on the base $\mathrm{AB}$. It is known that $A D=a, D B=b(a<b) . \triangle A C D$ and $\triangle B C D$ have incircles that touch line $C D$ at points $M$ and $N$, respectively. Find $M N$. | 20. As shown in Figure 22, $\mathrm{MN}=\mathrm{DN}-\mathrm{DM}$.
Let $P$ be the semiperimeter of $\triangle B C D$. Since $D N=D F, C N=C_{E}$, and $B F=B E$, we have $P=D N + B C$, which means $D Y=P-B C$.
From this, we get $D Y=P-AC$ (A.1), and since $\mathbf{A C}=$
$B C$, we obtain $M N=P-p$.
Furthermore, because ... | \frac{b-a}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,656 |
Example 3. The cubic equation $x^{3}-a x^{2}-a^{2} x+2 a^{2}+a=0$ has three distinct real roots. Find the range of positive real values of $\mathrm{a}$. | Let $y=f(x)=x^{3}-a x^{2}-a^{2} x+2 a^{2}+a$,
$$
\begin{array}{l}
\left.f^{\prime} x\right)=3 x^{2}-2 a x-a^{2}=(3 x+a)(x-a). \\
\text { By } f^{\prime}(x)=0 \text { we get } x_{1}=-\frac{a}{3}, x_{2}=a, \\
\because a>0, \therefore x_{1}<0, x_{2}>0. \\
\text { When } x \in(-\infty, -\frac{a}{3}), f^{\prime}(x)>0, \text... | a>1+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,657 |
21. Construct a new circumferential ring outside the given rectangle so that it has a known area of $\mathrm{m}^{2}$. For the problem to have a solution, what value should m take?
In the given text, the phrase "m个S取何值" is a bit ambiguous. A more precise translation might be "what value should m take for the area S?" H... | 21. Let $\mathrm{a}, \mathrm{b}$ represent the sides of the known rectangle, and let $\phi$ represent the angle between the sides of the circumscribing rectangle and the known rectangle (Figure 23). In this case, the sides of the circumscribing rectangle will be:
$$
\mathrm{a} \cos \phi+\mathrm{b} \sin \phi \text { and... | \sqrt{a L} \leqslant \mathrm{~m} \leqslant \frac{a+b}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,658 |
Find the distance from the point $(5,1)$ to the line $y=2x+3$.
. | Solve: Translating the coordinate axes so that $(5,1)$ becomes the center of the new coordinate system, the translation formula is $\left\{\begin{array}{l}x=x^{\prime}+5, \\ y=y^{\prime}+1\end{array}\right.$ The corresponding line equation will be $y^{\prime}+1=2\left(x^{\prime}+5\right)+3$. After rearrangement, it bec... | \frac{12}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,662 |
Question 1. The sum of $\mathrm{n}$ positive numbers is 1983. When is their product maximized, and what is the maximum value? | Since the geometric mean of $n$ positive numbers is not greater than their arithmetic mean, when the product of $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ is 1983,
$$
\begin{array}{l}
a_{1} a_{2} \cdots a_{n} \leqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n} \\
=\left(\frac{1983}{n}\right)^{n}
\end{a... | \left(\frac{1983}{n}\right)^n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,663 |
Question 2. The sum of several positive numbers is 1983. When is their product the largest, and what is the maximum value?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | From the previous problem, if the sum of $n$ positive numbers is 1983, then their product reaches its maximum value when these $n$ numbers are equal, and the maximum value $M_{2}$ is $\left(\frac{1983}{n}\right)^{n}$. However, unlike in problem 1, in this problem, $n$ is not a predetermined number; we need to determine... | \left(\frac{1983}{730}\right)^{730} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,664 |
Five, if $\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}-\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}=2 \tan \alpha$, find the quadrant in which the angle $\alpha$ lies. | Five, Brief Solution: Simplify the given conditions to $\sqrt{\frac{(1+\sin \alpha)^{2}}{\cos ^{2} \alpha}}$
$$
\begin{array}{l}
-\sqrt{\frac{(1-\sin \alpha)^{2}}{\cos ^{2} \alpha}}=2 \operatorname{tg} \alpha, \frac{1+\sin \alpha}{|\operatorname{cos} \alpha|}-\frac{1-\sin \alpha}{|\cos \alpha|} \\
=2 \operatorname{tg} ... | \alpha \in I | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,666 |
1. Through four points not lying on the same straight line, the maximum number of planes that can be determined is _, and the minimum number of planes that can be determined is _. | 1. 4, 1. | 4, 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,669 |
2 . In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $P$ and $Q$ are the midpoints of $A_{1} B_{1}$ and $B_{1} C_{1}$, respectively, $B_{:} B=B C=3, A B=4$. Then (1) the angle between $A_{2} C_{1}$ and $\mathrm{AB}_{1}$ is $\qquad$ , (2) the distance between $\mathrm{PQ}$ and $\mathrm{CD}$ is $\qquad$ . | $$
\text { 2. (1) } \arccos \frac{16}{25}
$$
(2) 3 . | \arccos \frac{16}{25}, 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,670 |
Theorem 2 When the last digit of $\mathrm{N}_{\mathrm{r}}$, $\mathrm{a}_{1}$, is only taken from $1,3,7,9$, then the last digit of $N_{\mathbf{r}}{ }^{4} \mathbf{k}$ is 1. | Prove that $N_{x}^{4} \mathbf{k}=10 \mathrm{M}+\mathrm{a}_{1}^{4 \mathbf{k}}$, where $\mathrm{M}$ is a positive integer. It is not difficult to verify: when $\mathrm{a}_{1}$ takes the values $1,3,7,9$, the last digit of $\mathrm{a}_{1}^{4} \mathrm{k}$ is 1.
Therefore, the last digit of $\mathrm{N}_{\mathrm{r}}^{4}$ is ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,676 |
6. Two lines perpendicular to the same plane are parallel; | 6 . Correct;
The above text has been translated into English, retaining the original text's line breaks and formatting. | Correct | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,679 |
8. If a straight line is perpendicular to two intersecting lines in a plane, then this line is perpendicular to this plane. | 8. Correct.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,681 |
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