problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 4. Calculate $\int_{1}^{n}+1 \ln [x] d x$, where [x] represents the greatest integer part of $x$.
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 { Solution When } n \leqslant x<n+1 \text {, we have }[x]=n . \\ \int_{1}^{n+1} \ln [x] d x=\sum_{i=1}^{n} \int_{i}^{i+1} \ln i d x \\ =\sum_{i=1}^{n} \ln i=\ln (n!) .\end{array} | null | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,992 |
Example 1. If $x_{1}>0, x_{2}>0$, and $x_{1}+x_{2}=$ $k$, then $x_{1} x_{2} \leqslant \frac{k^{2}}{4}$. | $$
\begin{array}{l}
\text { Proof: Let } x_{1}=\frac{k}{2}+t, x_{2}=\frac{k}{2}-t \text {. } \\
\left(0 \leqslant t<\frac{1}{2}\right)
\end{array}
$$
Then $\quad x_{1} x_{2}=\left(\frac{k}{2}+t\right)\left(\frac{k}{2}-t\right)$
$$
=\frac{k^{2}}{4}-t^{2} \leqslant \frac{k^{2}}{4} \text {. }
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,993 |
Example 2. If $x_{1}>0, x_{2}>0$. And $x_{1} x_{2}=k$, then $\mathrm{x}_{1}+\mathrm{x}_{2} \geqslant 2 \sqrt{\mathrm{k}}$. | Let $y=x_{1}+x_{2}=x_{1}+\frac{k}{x_{1}}$, then $\mathrm{x}_{1}^{2}-\mathrm{yx} \mathrm{x}_{1}+\mathrm{k}=0$.
Since $\mathrm{x} \in \mathrm{R}$, we have $\triangle=\mathrm{y}^{2}-4 \mathrm{k} \geqslant 0$. Solving this, we get $\mathrm{y} \leqslant$ $-2 \sqrt{\mathrm{k}}$ (discard), or $\mathrm{y} \geqslant 2 \sqrt{\m... | x_{1}+x_{2} \geqslant 2 \sqrt{k} | Inequalities | proof | Yes | Yes | cn_contest | false | 701,994 |
Example 4. If $A, B, C$ are the three interior angles of a triangle, prove that: $\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \leqslant \frac{1}{8}$. | $$
\begin{array}{l}
\leqslant \frac{a}{2 \sqrt{bc}}. \\
\end{array}
$$
Similarly, we can prove
$$
\begin{array}{l}
\sin \frac{B}{2} \leqslant \frac{b}{2 \sqrt{b c}}. \\
\sin \frac{C}{2} \leqslant \frac{c}{2 \sqrt{a b}}.
\end{array}
$$
Multiplying the three inequalities, we get
$$
\sin \frac{A}{2} \cdot \sin \frac{B}{... | \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \leqslant \frac{1}{8} | Inequalities | proof | Yes | Yes | cn_contest | false | 701,995 |
Example 3. If $x+y+z=k$, then $x^{2}+y^{2}+z^{2}$
$$
\geqslant \frac{\mathrm{k}^{2}}{3} \text {. }
$$ | $\begin{array}{l}\text { Prove: Let } x=\frac{k}{3}-t, y=\frac{k}{3}-2 t, z=\frac{k}{3} \\ +3 t(t \in R) \text {, then } x^{2}+y^{2}+z^{2}=\left(\frac{k}{3}-t\right)^{2} \\ +\left(\frac{k}{3}-2 t\right)^{2}+\left(\frac{k}{3}+3 t\right)^{2}=\frac{k^{2}}{3}+14 t^{2} \\ \geqslant \frac{k^{2}}{3}\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,996 |
Lemma: If four positive numbers $\mathrm{a}, \mathrm{~b}, \mathrm{c}, \mathrm{~d}$ satisfy $\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}$, and $a+d=b+c$, then $a d<b c$.
Proof: Let $\mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}=\mathrm{k}, \mathrm{a}=\frac{\mathrm{k}}{2}-\mathrm{t}$, $d=\frac{k}{2}+t, b=\frac{k}{2}-t^{\p... | Proof: Let the $n$ positive numbers be $x_{1}, x_{2}, \cdots, x_{n}$.
(1) When $n=2$, it has been proven (see Example 1).
(2) Suppose when $n=m$, $x_{1} x_{2} \cdots x_{m} \leqslant\left(\frac{k}{m}\right)^{m}$
holds.
Then when $n=m+1$, among these $m+1$ numbers, there must be a number not greater than $\frac{k}{m+1}$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,997 |
For example, if $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, find the extremum of $\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}$, and the corresponding values of $\alpha$ and $\beta$. | $$
\begin{array}{l}
\geqslant 2 \sqrt{\sin ^{2} \alpha} \cos ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta \\
=\frac{8}{\sin 2 \alpha \sin 2 \beta} \geqslant 8 \text {, it is easy to list the above formula to get the minimum value } \\
\end{array}
$$
The condition for 8 is $\sin 2 \alpha=\sin 2 \beta=1$, at this time $\a... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,998 |
Example 1. In the cube $A B C D$ $-\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{\mathrm{i}} \mathrm{D}_{1}$, find the distance between $A B_{1}$ and $\mathrm{A}_{1} \mathrm{C}_{1}$. | Solve As shown in Figure 2, the dihedral angle $A-C_{1} A_{1}-B_{1}$ has a plane angle of $90^{\circ}, \mathrm{A}$ is at a distance of $\mathrm{a}$ from $\mathrm{A}_{1} \mathrm{C}_{1}, \mathrm{B}_{1}$ is at a distance of $\frac{\sqrt{2}}{2} \mathrm{a}$ from $\mathrm{A}_{1} \mathrm{C}_{1}$, and by formula (2), the dista... | \frac{\sqrt{3}}{3} \mathrm{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,999 |
Example 2. In the quadrilateral pyramid $\mathrm{S}-\mathrm{ABCD}$, $\mathrm{ABCD}$ is a rectangle, $\mathrm{AB}=\sqrt{2} \mathrm{a}, \mathrm{BC}=\mathrm{a}, \mathrm{AC} \cap \mathrm{BD}=$ $\mathrm{O}, \mathrm{SO} \perp$ plane $\mathrm{ABCD}$, and $\mathrm{SO}=\frac{\mathrm{a}}{2}$, find the distance between $\mathrm{A... | As shown in Figure 3, the dihedral angle $\mathrm{S}-\mathrm{AC}-\mathrm{B}$ has a plane angle of $90^{\circ}$, the distance from $\mathrm{S}$ to $\mathrm{AC}$ is $\mathrm{SO} = \frac{\mathrm{a}}{2}$, and the distance from $\mathrm{B}$ to $\mathrm{AC}$ is
From formula (2), the distance $\mathrm{h}$ between the skew li... | \frac{\sqrt{22}}{11} \mathrm{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,000 |
Example 3. For a cylinder with a base radius of $\mathrm{R}$, the radii of the two bases $\mathrm{OM}$ and $\mathrm{O}_{1} \mathrm{~N}_{1}$ form a $60^{\circ}$ angle. Find the distance between the skew lines $\mathrm{MN}_{1}$ and the cylinder axis $\mathrm{OO}_{1}$. | From Fig. 4, it is easy to know that the dihedral angle $\mathrm{M}-\mathrm{OO}_{1}-\mathrm{N}_{1}$ is $60^{\circ}, \mathrm{M}$ and $\mathrm{~N}_{1}$ are both at a distance of $R$ from $\mathrm{OO}_{1}$. Using formula (3), the distance $\mathrm{h}$ between the skew lines $\mathrm{MN}_{1}$ and $\mathrm{OO}_{1}$ is $\mat... | \frac{\sqrt{3}}{2} \mathrm{R} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,001 |
Example 1. In $\triangle A B C$, if $\sin ^{8} A+\cos ^{8} A$ $=1$, then $\triangle A B C$ is a right triangle. | $$
\begin{array}{c}
\text { Prove } \because \sin ^{8} A+\cos ^{8} A=1, \\
\therefore \frac{1}{2^{7}}[\cos 8 A-8 \cos 6 A+28 \cos 4 A \\
-56 \cos 2 A+35]+\frac{1}{2^{7}}[\cos 8 A+8 \cos 6 A \\
+28 \cos 4 A+56 \cos 2 A+35]=1, \\
2 \cos 8 A+56 \cos 4 A+70=128, \\
\cos 8 A+28 \cos 4 A+35-64=0, \\
2 \cos ^{2} 4 A-1+28 \cos... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,003 |
Example 5. The base of the prism $\mathrm{ABCD}-\mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime} \mathrm{D}^{\prime}$ is an isosceles trapezoid, $\mathrm{AB} / / \mathrm{CD}, \angle \mathrm{DAB}=60^{\circ}$, $\mathrm{CD}=5$, the leg length $\mathrm{BC}=3$, and the two non-parallel sides are perpendicular to ... | Solve as shown in Figure 6, let the intersection of line $B^{\prime} \mathrm{C}^{\prime}$ and line $\mathrm{A}^{\prime} \mathrm{D}^{\prime}=\mathrm{E}^{\prime}$, the intersection of line $\mathrm{BC}$ and line $\mathrm{AD}=\mathrm{E}$, connect $\mathrm{E}^{\prime} \mathrm{E}$.
Plane $\mathrm{B}^{\prime} \mathrm{C}^{\pr... | \frac{15}{14} \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,004 |
Example 6. The radius of the base of a regular circular cone is $R$, the base angle $A$ of the axial section $\mathrm{SAB}$ is bisected by $\mathrm{AC}$, and $\mathrm{BD}$ is a chord of the base circle, forming a $30^{\circ}$ angle with $\mathrm{AB}$. Find the distance between the generatrix $\mathrm{AC}$ and the chord... | Given Figure 7, construct
$\mathrm{SO} \perp$ plane $\mathrm{ABD}$, then $\mathrm{O}$ is the center of the base circle.
In the base $\mathrm{ABD}$, construct $\mathrm{OE} \perp \mathrm{BD}$ at $\mathrm{E}$, and connect $\mathrm{SE}$, then $\mathrm{SE} \perp \mathrm{BD}$.
$$
\begin{aligned}
\because O E & =\frac{1}{2} R... | \frac{2 \sqrt{7}}{7} R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,005 |
Example 1. Let CEDF be an inscribed rectangle of a known circle. Draw the tangent line to the circle through D, which intersects the extension of CE at point A and the extension of CF at point B. Prove: $\frac{\mathrm{BF}}{\mathrm{AE}}=\frac{\mathrm{BC}^{3}}{\mathrm{AC}^{3}}$. | Prove as shown in Figure 1, let the right-angle vertex $\mathrm{C}$ be the pole, the altitude $\mathrm{CD}$ be the polar axis, point $\mathrm{D}(1,0), \mathrm{F}(\cos \alpha, \alpha)$, then $\mathrm{B}(\sec \alpha, \alpha), \mathrm{E}\left(\sin \alpha, \alpha-\frac{\pi}{2}\right)$, A $\left(\csc \alpha, \alpha-\frac{\p... | \frac{\mathrm{BF}}{\mathrm{AE}}=\frac{\mathrm{BC}^{3}}{\mathrm{AC}^{3}} | Geometry | proof | Yes | Yes | cn_contest | false | 702,006 |
Example 2. As shown in Figure 2, if $\mathrm{P}$ is any point on the arc $\overparen{\mathrm{AC}}$ outside the equilateral $\triangle \mathrm{ABC}$, prove that
$$
\frac{1}{\mathrm{PD}}=\frac{1}{\mathrm{PA}}+\frac{1}{\mathrm{PC}} \text {. }
$$ | Proof: Let the moving point $\mathrm{P}$ be the pole, the chord $\mathrm{PC}$ be the polar axis, and the circle equation be $\rho=\cos \left(\theta-\theta_{0}\right) .(2 \mathrm{R}=1)$. By setting $\theta=0$ and $\frac{2}{3} \pi$, we get $\rho_{\mathrm{C}}=\mathrm{PC}=\cos \theta_{0}, \rho_{\mathrm{A}}=\mathrm{PA}=\cos... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,007 |
Example 3. Given that $A$ and $B$ are common points of two circles (A and B may coincide), draw a line $\mathrm{CAD}$ and $\mathrm{EBF}$ through the common points, ending on the two circles. Prove: CE//DF. (There are twenty-three cases in this problem, only three are illustrated here)
---
The translation maintains th... | Proof: In 23 cases, we assume the common point $\mathrm{A}$ as the pole and the common chord $\mathrm{AB}$ as the polar axis. The polar angles of points F, D, E are $\alpha, \beta, \gamma$, respectively, and the conjugate of point $\mathrm{C}$ is $\beta$ (D, A, C are collinear). Then, by the collinearity of F, B, E, th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,008 |
In circle $\mathrm{O}$, diameter $\mathrm{AOB}$ and radius $\mathrm{OC}$ are perpendicular to each other, $\odot \mathrm{O}^{\prime}$ is tangent to $\mathrm{OB}$ and $\mathrm{OC}$ at $\mathrm{D}$ and $\mathrm{E}$, and is internally tangent to semicircle $\mathrm{O}$ at $\mathrm{F}$. Prove that $\mathrm{A}$, $\mathrm{E}... | Proof As shown in the figure, with $\mathrm{O}$ as the pole and the extension line of $\mathrm{OA}$ in the opposite direction as the polar axis, establish a polar coordinate system. Let the radius of the semicircle $\mathrm{O}$ be $\mathrm{R}$, and the radius of $\odot \mathrm{O}^{\prime}$ be $\mathrm{r}$. Then,
$\math... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,009 |
Example 1. As shown in the figure, $\mathrm{OP}$ is a radius of $\odot 0$. $\odot \mathrm{O}^{\prime}$, which has $\mathrm{OP}$ as its diameter, intersects the chord $\mathrm{PB}$ of $\odot \mathrm{O}$ at $\mathrm{C}$. Prove: $\mathrm{C}$ is the midpoint of $\mathrm{PB}$ | Prove that with P as the pole and the ray of the tangent line passing through P as the polar axis, establish a polar coordinate system.
Let the radius of $\odot \mathrm{O}$ be $R$, then the equation of $\odot \mathrm{O}$ is $\rho=2 R \sin \theta$. The equation of $\odot \mathrm{O}^{\prime}$ is $\rho=R \sin \theta$, $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,010 |
Example 2. Given $\mathrm{A}_{0}, \mathrm{~A}_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ are five equally spaced points on the unit circle.
Prove: $\left(\mathrm{A}_{0} \mathrm{~A}\right.$,
$$
\left.\mathrm{A}_{0} \mathrm{~A}_{2}\right)^{2}=5
$$ | Prove that $A_{0} A_{1}$, $A_{0}$
$A_{2}$ are chords of a circle with a common endpoint at $A_{0}$. Taking $A_{0}$ as the pole and the diameter $A_{0} A$ as the polar axis, we establish a polar coordinate system. According to the problem, $\angle A_{1} A_{0} A_{2} = 36^{\circ}$, $\angle A_{2} A_{0} X = 18^{\circ}$, and... | 5 | Geometry | proof | Yes | Yes | cn_contest | false | 702,011 |
Example 3. Given that $\mathrm{AP}$ is the altitude from $\mathrm{A}$ to side $\mathrm{BC}$ of $\triangle \mathrm{ABC}$, and a circle is drawn with $\mathrm{AP}$ as its diameter, intersecting $A B$ and $A C$ at $D$ and $E$ respectively. Prove that B, D, $\mathrm{E}$, and $\mathrm{C}$ are concyclic. | Proof: With A as the pole and the ray of the diameter AP as the polar axis, establish a polar coordinate system. Let $AP = 2R$, $\angle PAB = \alpha$, $\angle PAC = \beta$. Then the equation of the circle is $\rho = 2R \cos \theta$ (1). The equation of the line BC is $\rho = \frac{2R}{\cos \theta}$ (2). By substituting... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,012 |
Example 4. Given $\mathrm{F}$ is any point on the bisector of $\angle \mathrm{P}$, and any two lines $\mathrm{AD}, \mathrm{BC}$ through $\mathrm{F}$ intersect the two sides of $\angle \mathrm{P}$ at $A, D, B, C$. Prove:
$$
\frac{1}{\mathrm{PA}}+\frac{1}{\mathrm{PD}}
= \frac{1}{P B}+\frac{1}{P C} \text{. }
$$ | Proof As shown in the figure, with $\mathrm{P}$ as the pole and the ray $\mathrm{PF}$ as the polar axis, we establish a polar coordinate system. Let $\mathrm{PA}=\mathrm{a}, \mathrm{PB}=\mathrm{b}, \mathrm{PF}=\mathrm{e}, \angle \mathrm{BAX}=\angle \mathrm{DAX}=\alpha$, then $\mathrm{A}(\mathrm{a}, \alpha), \mathrm{B}(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,013 |
Example 2. Find the extremum of the function $y=\sin ^{10} \theta+\cos ^{10} \theta$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
y=\frac{1}{2^{7}}[5 \cos 8 \theta+60 \cos 4 \theta+63] \\
=\frac{1}{2^{7}}\left[10 \cos ^{2} 4 \theta-5+60 \cos 4 \theta+63\right] \\
=\frac{1}{2^{6}}\left[5 \cos ^{2} 4 \theta+30 \cos 4 \theta+26\right] \\
=\frac{1}{2^{\theta}}\left[5(\cos 4 \theta+3)^{2}-19\right] . \\
\because \cos 4 \theta \neq-... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,014 |
Example 5. Given an isosceles right $\triangle \mathrm{ABC}$ with hypotenuse $\mathrm{BC}$ and any point $\mathrm{D}$ on $\mathrm{BC}$. Prove: $\mathrm{BD}^{2}+\mathrm{CD}^{2}=2 \mathrm{AD}$ | Take $\mathrm{A}$ as the pole and $A C$ as the polar axis to establish a polar coordinate system. Let $\mathrm{AC}=\mathrm{AB}=\mathrm{a}$, and $\angle \mathrm{DAC}=\alpha$, then $\mathrm{C}(\mathrm{a}, 0), \mathrm{B}\left(\mathrm{a}, \frac{\pi}{2}\right)$.
The equation of the line $B C$ is $P=\frac{a^{2} \sin \frac{\p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,015 |
Example 2 - (83 College Entrance Examination
Mathematics for Science Students, Question 7)
As shown in the figure, it is known that the length of the major axis of the ellipse
is $\left|A_{1} A_{2}\right|=6$, and the focal distance
$\left|\mathbf{F}_{\mathbf{1}} \mathbf{F}_{\mathbf{2}}\right|=4 \sqrt{\mathbf{2}}$. A l... | Take the focus $\mathrm{F}_{1}$ as the pole, and the ray starting from $\mathrm{F}_{1}$ passing through $F_{3}$ as the polar axis to establish a polar coordinate system.
Given the ellipse's semi-major axis $\mathrm{a}=3$, semi-focal distance $\mathrm{c}=2 \sqrt{2}$, semi-minor axis $\mathrm{b}=1$, eccentricity $\mathr... | \alpha=\frac{\pi}{6} \text{ or } \alpha=\frac{5 \pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,017 |
Example. Determine whether $\left(5, \frac{7}{4} \pi\right) 、\left(5,-\frac{9}{4} \pi\right)$ 、 $\left(-5,-\frac{13}{4} \pi\right) 、\left(-5,-\frac{\pi}{4}\right)$ represent the same point. | Solve: Let $\frac{7 \pi}{4}+2 \mathrm{k} \pi=-\frac{9}{4} \pi$, we get $\mathrm{k}=-2$, which is an integer, so $\left(5, \frac{7}{4} \pi\right)$ and ( $\left.5,-\frac{9}{4} \pi\right)$ represent the same point.
$$
\text { Let } \frac{7}{4} \pi+(2 \mathrm{k}+1) \pi=-\frac{13}{4} \pi \text {, we get } \mathrm{k}=-3 \tex... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,018 |
Example 2. Determine whether the equations
$$
\rho=\pi+\frac{1}{2} \theta
$$
and the equation
$$
\rho=\frac{3}{2} \pi-\frac{1}{2} \theta
$$
represent the same curve. | In the solution (1), by substituting $\theta$ with $\theta+2 \mathrm{k} \pi$ and keeping $\rho$ unchanged, we get
$$
\rho=\pi+\frac{1}{2}(\theta+2 \mathrm{k} \pi),
$$
which simplifies to
$$
\rho=(\mathrm{k}+1) \pi+\frac{1}{2} \theta \text {. }
$$
No matter what integer value $\mathrm{k}$ takes, it does not satisfy th... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,019 |
Example 3. Determine whether the point $\mathrm{P}(1, \pi)$ lies on the hyperbola
$$
\rho=\frac{2}{1-3 \cos \theta} \text { . }
$$ | The general expression for the coordinates of point $\mathrm{P}$ is
$$
(1, \pi+2 \mathrm{k} \pi)
$$
or $(-1,2 \mathrm{k} \pi)$.
Substituting (1) into the curve equation, we get
$$
1=\frac{2}{1-3(-1)}=\frac{1}{2},
$$
which is a contradiction, so $\mathrm{k}$ does not exist.
Substituting (2) into the curve equation, we... | -1=\frac{2}{1-3 \cdot 1}=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,020 |
Example 1. An isosceles trapezoid is inscribed in a semicircle with a radius of 1, where the lower base is the diameter of the semicircle. (1) Try to find the functional relationship between its perimeter \( \mathrm{y} \) and the leg length \( \mathrm{x} \), and write down the domain; (2) When the leg length is how muc... | Solve the projection orbit easily given by
$$
y=-x^{2}+2 x+4 \text {. }
$$
When finding the domain of this polynomial function, the answers are varied: some think it is all real numbers, some believe it is positive real numbers, some think that $x$ must be greater than a certain value but cannot increase indefinitely,... | y=5 \text{ when } x=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,021 |
Example 2. The vertex angle and base of a triangle - given, when is the area maximum? What is the maximum area? | As shown in Figure 2, for $\triangle BAC$ with a fixed base $BC = a$ and vertex angle $A = a$, the trajectory of vertex $A$ is two circular arcs with $BC$ as the chord and the angle subtended by the arc as $a$. When the moving point travels from $B$ to $C$ along the arc, due to the symmetry, when the moving point reach... | S = \frac{a^2}{4} \operatorname{ctg} \frac{a}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,022 |
Example 3. When is the area of a triangle maximized if two of its sides are of fixed length? | As shown in Figure 3, for triangle $\mathrm{ABC}$ with $\mathrm{BC}=\mathrm{a}$ as the base and one side $\mathrm{AB}$ equal to a fixed length $\mathrm{c}$, the locus of vertex $\mathrm{A}$ is a circle with center at $B$ and radius $c$. Clearly, when the moving point reaches a position perpendicular to $B C$, i.e., $\m... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,023 |
Solve the system of inequalities for $\mathbf{x}$, and discuss their solutions:
$$
\left\{\begin{array}{l}
\mathrm{m}(\mathbf{x}-1)>\mathbf{x}-2 . \\
3(\mathrm{~m}+1) \mathrm{x}>3 \mathrm{~m} \mathbf{x}+5 .
\end{array} \text { ( } \mathrm{m}\right. \text { is a real parameter) }
$$ | $$
\begin{array}{l}
\text { Solve } \begin{array}{l}
m(x-1)>x-2 . \\
3(m+1) x>3 m x+5
\end{array} \\
\Leftrightarrow\left\{\begin{array}{l}
m x-m>x-2 \\
3 m x+3 x>3 m x+5
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{r}
(m-1) x>m-2 . \\
x>\frac{5}{3} .
\end{array}\right.
\end{array}
$$
1. If $m=1$.
(I) $\Le... | x>\frac{5}{3}, \text { if } m \geqslant 1 \text { or } -\frac{1}{2}<m<1. \text{ No solution, if } m \leqslant-\frac{1}{2}. | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,024 |
Question 27. Given the quadratic equation in $\mathbf{x}$: $x^{2}+2 p \mathbf{x}-6 \mathbf{x}+2 p^{2}-5 p-3=0$ has real roots, find the maximum and minimum values of the product of the roots. | Let the product of the two roots be $y$, by Vieta's formulas,
$$
\mathrm{y}=2 \mathrm{p}^{2}-5 \mathrm{p}-3=2\left(\mathrm{p}-\frac{5}{4}\right)^{2}-\frac{49}{8} \text {. }
$$
Given that the original equation has real roots,
$$
\begin{array}{l}
\therefore \quad(2 p-6)^{2}-4\left(2 p^{2}-5 p-3\right) \geqslant 0, \\
p... | y_{\text{max}}=49, \; y_{\text{min}}=-\frac{49}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,026 |
Question 30. Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a.
(1) Find the distance between the edge $\mathrm{A}_{1} \mathrm{~A}$ and the diagonal $\mathrm{BD}_{1}$,
(2) Find the distance between the diagonals $A_{1} B$ and $B_{1} C$ on adjacent faces:
(3) Let $P=\{\mathbf{x} \mid \mathbf{x}$ is ... | (1) As shown in Figure 1, take point E as the midpoint of AA1, and point F as the midpoint of BD1. Connect FA, FA1, FE, EB, and ED1. Since
$\triangle D1A1E \cong \triangle BAE \Rightarrow D1E = BE$,
$\Rightarrow \triangle BED1$ is an isosceles triangle
$\Rightarrow EF \perp BD1$;
$FA = FA1 \Rightarrow \triangle FAA... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,029 |
Example 1. Draw the graph of the function $\mathrm{y}=$ $|x|+\frac{1}{2}$. | According to the definition of absolute value, the given function can be expressed as:
(1) When $x \geqslant 0$, draw the graph of $y=x+\frac{1}{2}$, which is the ray $\mathrm{AC}$ in Figure 1.
(2) When $x<0$, draw the graph of $y=-x+\frac{1}{2}$, which is the ray $\mathrm{AB}$ in Figure 1;
(3) The broken line BAC in F... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,030 |
Example 2. Draw the graph of the function $\mathrm{y}=\frac{1}{4} \mathrm{x}^{2}-|\mathrm{x}|-3$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 2. Draw the graph of the function $\mathrm{y}=\frac{1}{4} \mathrm{x}^{2}-|\m... | Solve: Since $x^{2}=\left|x^{2}\right|=|x|^{2}$, the function $y=\frac{1}{4} x^{2}-|x|-3$ is of the type $y=f(|x|)$.
(1) Draw the graph of $y=\frac{1}{4} x^{2}-x-3$ when $x \geqslant 0$. This is the part of the parabola opening upwards to the right of the y-axis. From
$$
\frac{1}{4} x^{3}-x-3=3
$$
we can see that the ... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,031 |
Example 2. Prove that the polynomial $x^{2}-3 x y+y^{2}+x$ $-4 y-1$ cannot be factored over the field of rational numbers, but can be factored over the field of real numbers. | To prove that because $\sqrt{b^{2}-4ac}=\sqrt{(-3)^{2}-4 \cdot 1 \cdot 1}$ $=\sqrt{5}$ is not a rational number, the original polynomial cannot be factored over the rational numbers.
Since $b^{2}-4ac=5>0, d^{2}-4a\hat{a}i=5>0$, $\Delta'=[(-3)-2 \cdot(-4)]'-5 \cdot 5=0$, the original polynomial can be factored over the... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,032 |
Example 1. Find the equation of the chord of contact $\mathrm{AB}$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ from an external point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$. | Let the points of tangency be $A \left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and the equation of the tangent line PA be:
$$
\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1
$$
Since $P \in PA$, we have $\frac{x_{1} x_{0}}{a^{2}}+\frac{y_{1} y_{0}}{b^{2}}=1$.
Similarly, $\frac{x_{2} x_{0}}{a^{2}}+\frac{y_{... | \frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,033 |
Example 2. Find the length of the chord of contact $\mathrm{AB}$ of the point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ with respect to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ $=1$. | Solve: Combine the equation of AB $\mathrm{b}^{2} x_{0} \mathrm{x}+\mathrm{a}^{2} y_{0} \mathrm{y}=\mathrm{a}^{2} \mathrm{~b}^{2}$
with the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, eliminate $y$
and rearrange to get
$$
\begin{array}{l}
\left(a^{2} y_{0}^{2}+b^{2} x_{0}^{2}\right) x^{2}-2 a^{2} b^{2} x \circ x \\
... | \frac{2 \sqrt{\left(a^{2} y_{0}^{2}+b^{2} x_{0}^{2}-a^{2} b^{2}\right)\left(b^{4} x_{0}^{2}+a^{4} y_{0}^{2}\right)}}{a^{2} y_{0}^{2}+b^{2} x_{0}^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,034 |
Example 3. Prove: $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ is such that the chord of contact $A B$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is bisected by $O P$ | Let M $(\bar{x}, \bar{y})$ be the midpoint of AB.
From Example 2, we know:
$$
\begin{array}{c}
\bar{x}=\frac{a^{2} b^{2} x_{0}}{a^{2} y_{0}^{2}+b^{2} x_{0}^{2}}, \\
\bar{y}=\frac{a^{2} b^{2} y_{0}}{a^{2} y_{0}^{2}+b^{2} x_{0}^{2}} .
\end{array}
$$
Thus, $\frac{\bar{y}}{\bar{x}}=\frac{y_{0}}{x_{0}}$, which means $k_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,035 |
Example 4. P $\left(x_{0}, y_{0}\right)$ is the point of tangency of the chord $A B$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ $=1$ passing through the focus $F$, then $P F \perp A B$. | \begin{array}{l}\text { Prove } A B \text { is } \\ \text { the equation } b^{2} x_{0} x \\ +a^{2} y_{0} y=a^{2} b^{2} . \\ \because F \in A B, \\ \therefore b^{2} x_{0} c-a^{2} b^{2} \\ \therefore \quad x_{0}=\frac{d^{2}}{c} . \\ k_{A B}=-\frac{b^{2} x_{0}}{a^{2} y_{0}}=-\frac{b^{2}}{c y_{0}}, \\ \quad k_{P F}=\frac{y... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,036 |
Example 5. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, the chord of contact of the tangents from the point $P\left(x_{0}, y_{0}\right)$ is $AB$. A line $l$ through $P$ intersects the chord of contact at $Q$ and the ellipse at $R$ and $S$. Prove that $\frac{PR}{RQ}=-\frac{PS}{SQ}$. | Given $Q(x, y)$, the first equation is:
$$
\left\{\begin{array}{l}
x=\frac{x_{0}+\lambda \bar{x}}{1+\lambda}, \\
y=\frac{y_{0}+\lambda \bar{y}}{1+\lambda} .
\end{array}\right.
$$
( $\lambda$ is a parameter, $\lambda \neq-1$ )
Substitute into $\lambda \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and rearrange into a quad... | \frac{\mathrm{PR}}{\mathrm{RQ}}=-\frac{\mathrm{PS}}{\mathrm{SQ}} | Geometry | proof | Yes | Yes | cn_contest | false | 702,037 |
Example 6. The chord $\mathrm{AB}$ of the ellipse $\frac{x^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ passes through a fixed point $\mathrm{M}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$, find the locus of the intersection point $\mathrm{P}$ of the two tangents through $\mathrm{A}$ and $\mathrm{B}$. | Let $\mathrm{P}(\overline{\mathrm{x}}, \overline{\mathrm{y}})$, then the equation of $\mathrm{AB}$ is
$$
\frac{\bar{x} x}{a^{2}}+\frac{\bar{y} y}{b^{2}}=1
$$
$\because M \in A B, \quad \therefore \frac{\bar{x} x_{0}}{a^{2}}+\frac{\bar{y} y_{0}}{b^{2}}=1$.
$\therefore$ The locus of point $\mathrm{P}$ is the line: $\frac... | \frac{x_{0} x}{\mathrm{a}^{2}}+\frac{\mathrm{y}_{0} \mathrm{y}}{\mathrm{b}^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,038 |
Example 7. The chord $\mathrm{AB}$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ has a constant slope $k$. Find the coordinates of the intersection point $P$ of the two tangents passing through $A$ and $E$.
Translate the above text into English, please keep the original text's line breaks and format, and ... | Let $\mathrm{P}(\overline{\mathrm{x}}, \overline{\mathrm{y}})$, then the equation of $\mathrm{AB}$ is
$$
\begin{array}{l}
\frac{x x}{a^{2}}+\frac{y y}{b^{2}}=1 \\
\therefore \quad-\frac{b^{2} x}{a^{2} y}=k .
\end{array}
$$
Therefore, the trajectory of point $P$ is the part of the line $y=-\frac{b^{2}}{a^{2} k} x$ outs... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,039 |
Example. Draw two tangent lines to the ellipse $4 x^{2}+y^{2}-16 x$ $-4 y+16=0$ from the origin $O$. Any secant line through $O$ intersects the ellipse and the line connecting the two tangent points at points $P, Q, R$ respectively. Prove that $|O P|, |O R|, |O Q|$ form a harmonic sequence (i.e., their reciprocals form... | Proof: Let the parameter equation of any secant line passing through $\mathrm{O}$ be
$$
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=t \sin \alpha .
\end{array}\right.
$$
It is known that the chord of the tangents drawn from point $\mathrm{O}$ to the ellipse has the equation
$$
4 \mathrm{x} + y - 8 = 0 . \quad (2)
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,040 |
Example 1. Find the coordinates of the intersection point of the two tangent lines to the ellipse at the points of intersection with the line $x+4 y-2=0$ and the ellipse $\frac{x^{2}}{2}$ $+\mathrm{y}^{2}=1$. | Let $\mathrm{P}^{\prime}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be the point sought. Then the chord of contact of the circle with respect to $P^{\prime}$ is given by $\frac{x^{\prime} x}{2}+y^{\prime} y-1=0$, which should coincide with the line equation $\mathrm{x}+4 \mathrm{y}-2=0$. Therefore, we have $... | (1,2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,041 |
For example, plot the graph of the function $\mathrm{y}=\sin |\mathrm{x}|$. | (1) Draw the graph of the function $y=\sin x$ for $x \geqslant 0$, as the curve to the right of the $y$-axis in Figure 5,
(2) Using the $y$-axis as the axis of symmetry, draw the symmetric image of the curve to the right of the $y$-axis;
(3) Combining the two curves, we get the graph of $y=\sin |x|$, as shown in Figure... | not found | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,042 |
Example 2. The tangent line at point $\mathrm{P}_{0}\left(x_{0}, y_{0}\right)$ on the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ intersects the hyperbola $b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2}$ at points $P_{1}$ and $P_{2}$. Find the equation of the locus of the intersection point $\mathrm{P}^{\prime}$ of the tangents... | From the problem, we know that the tangent line of the ellipse at $\mathrm{P}_{0}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ is exactly the chord of the tangent points of the hyperbola about the moving point $\mathrm{P}^{\prime}$. Thus, the equation
$$
b^{2} x_{0} x+a^{2} y_{0} y
=a^{2} b^{2}
$$
should be consistent... | b^{2} x^{2}+a^{2} y^{2} = a^{2} b^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,043 |
Example 3. Find the locus of the intersection of the two tangents to the hyperbola $\frac{x^{2}}{2}-y^{2}$ $=1$ at the points of intersection with the line passing through (1) $\mathrm{P}_{1}^{\prime}\left(2, \frac{1}{2}\right)$, (2) $P_{2}^{\prime}(-2,3)$. | (1) Let $\mathrm{A}=\frac{1}{2}, \mathrm{C}=-1, \mathrm{D}=0$,
$$
\mathrm{E}=0, \mathrm{~F}=-1, \mathrm{x}^{\prime}=2, \mathrm{y}^{\prime}=\frac{1}{2} \text { into }
$$
we get $x-\frac{1}{2} y-1=$
0, which simplifies to $2 x-y-2$
$$
=0\left(1_{1}\right) \text {. }
$$
Since point $P_{1}$ is
inside the
hyperbola, the
r... | 2x-y-2=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,044 |
Example 1. Try to calculate the sum of the distances from point $\mathrm{P}$ on the circumference of the circumscribed circle of a regular n-sided polygon to each vertex, and find its maximum and minimum values. | Solve as shown in Figure 2.
For a regular n-gon inscribed in a circle,
the radius of the circle is $R$, and $P$ is
a point on the arc $\overparen{A_{1} A_{n}}$, and
let $\angle \mathrm{POA}_{\mathrm{n}}=\alpha$,
by the Law of Sines, we have
$$
\begin{array}{l}
\mathrm{PA}_{1}=2 \mathrm{R} \cdot \sin \frac{\theta-\alpha... | s_{\text{max}}=2 R \cdot \csc \frac{\pi}{2 n}, \quad s_{\text{min}}=2 R \cdot \cot \frac{\pi}{2 n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,046 |
Example 2. Given that the radius of the circumscribed circle of a regular dodecagon is $a$, find the sum of all the diagonals (i.e., all the lines connecting non-adjacent vertices) of the regular dodecagon. | Let the sum of distances from one vertex to all other vertices be $\mathrm{s}_{1}$. According to the conclusion from the previous problem:
$$
\mathrm{s}_{1}=2 \mathrm{a} \cdot \operatorname{ctg} \frac{\pi}{2} \times \frac{\pi}{12}=2 \mathrm{a} \cdot \operatorname{ctg} \frac{\pi}{24}.
$$
$\mathrm{s}_{1}$ includes two ed... | 6(4+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}) \mathbf{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,047 |
Example 1. In a triangle, the median on one side is less than half the sum of the other two sides. | 〔As shown in Figure 1, prove:
$$
\left.A D=\frac{1}{2}(A B+A C) .\right]
$$
Extend $\mathrm{AD}$ to E, such that $\mathrm{DE}=\mathrm{AD}$. Then, from $\triangle \mathrm{ADC} \cong \triangle \mathrm{EDB}$, we get $\mathrm{BE}=\mathrm{AC}$. From $\mathrm{AE}<\mathrm{AB}+\mathrm{BE}$, the proof is easily obtained. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,048 |
Example 2. In $\triangle \mathrm{ABC}$, $\mathrm{AB}>\mathrm{AC}$, and $\mathrm{AD}$ is the angle bisector. Prove: $\mathrm{BD}>\mathrm{DC}$. | As shown in Figure 2, we intercept $\mathrm{AE}=\mathrm{AC}$ on $\mathrm{AB}$. It is easy to prove that $\mathrm{ED}=\mathrm{DC}$. Then, we prove that $\angle \mathrm{BED}>\angle \mathrm{EBD}$ (using $\angle 1$ and $\angle 2$), which leads to $\mathrm{BD}>\mathrm{ED}=\mathrm{D}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,049 |
Example 3. Secondary Proof: In a triangle, the altitude on the larger side is less than the altitude on the smaller side. (As shown in Figure 3, prove: $\mathrm{CF}<\mathrm{BE}$.) | We intercept $\mathrm{AC}^{\prime}=\mathrm{AC}$ on AB, and construct $\mathrm{C}^{\prime} \mathrm{F}^{\prime} \perp \mathrm{AC}, \mathrm{C}^{\prime} \mathrm{D} \perp \mathrm{BE}$. It is easy to prove that $\mathrm{C}^{\prime} \mathrm{F}^{\prime}=\mathrm{CF}$. Therefore, $\mathrm{BE}>\mathrm{DE}=\mathrm{C}^{\prime} \mat... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,050 |
Example 4. The product of the hypotenuse of a right-angled triangle and the altitude to it is greater than the sum of the legs. | Proof 1: On $\mathrm{AB}$, take $\mathrm{BE}=\mathrm{BC}$, construct $\mathrm{EF} \perp \mathrm{A} \%$ EGł.P, C, C river foot,
$$
\begin{array}{l}
\text { then } \mathrm{CD}=\mathrm{EG}=\mathrm{CF} \text{. Also, } \\
\mathrm{AE}= \mathrm{AB}-\mathrm{BE}, \\
\mathrm{AF}= \mathrm{AC}-\mathrm{CF}, \\
\mathrm{AE}> \mathrm... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,051 |
Example 4. Draw the graph of $y=|x-1|$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed for the example itself. However, if the task is to translate ... | (1) Draw the graph of the function $\mathrm{y}=\mathrm{x}-1$, which is the line $\mathrm{C}^{\prime} \mathrm{AB}$ in Figure 8 (1);
(2) Using the $x$-axis as the axis of symmetry, draw the symmetric figure of $\mathrm{C}^{\prime} \mathrm{A}$, which is CA, as shown in Figure 8 (2);
(3) The broken line CAB in Figure 8 (2)... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,053 |
Example 6. $\angle \mathrm{A}$ is the largest angle in $\triangle \mathrm{ABC}$, and $\mathrm{E}$, $\mathrm{F}$ are points on $\mathrm{AB}$, $\mathrm{AC}$ respectively, then $\mathrm{EF}<\mathrm{BC}$ | $$
\begin{array}{c}
\angle \mathrm{A}>\angle \mathrm{B} \Rightarrow \mathrm{BC}>\mathrm{AC}, \\
\angle \mathrm{A}>\angle \mathrm{C} \Rightarrow \mathrm{BC}>\mathrm{AB} . \text { Connect } \mathrm{EC}, \\
\angle 1>\angle \mathrm{A}>\angle \mathrm{B} \Rightarrow \mathrm{BC}>\mathrm{EC} \\
\angle 2>\angle \mathrm{A}>\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,054 |
Example 7. $D$ is a point inside $\triangle ABC$, and $DB - DC = AB - AC$, then $\angle BAD > \angle CAD$.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
Example 7. $D$ is a point inside $\triangle ABC$, and $DB - DC$ $=AB - AC$, then $\angle BAD$... | Prove: As shown in Figure 7, construct $\mathrm{AE}$
$$
=\mathrm{AC}, \mathrm{DF}=\mathrm{DC} \text {. }
$$
It is easy to see that $\mathrm{BE}=\mathrm{BF}$.
By the complementary relationship, we get $\angle 1=\angle \mathrm{AEF}$.
Thus, $\angle 1>\angle 2 \Rightarrow \mathrm{DE}>\mathrm{DF}=\mathrm{DC}$. Q.E.D. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,055 |
Example 8. From a point $\mathrm{A}$ outside $\odot \mathrm{O}$, draw tangents that touch the circle at points $\mathrm{B}$ and $\mathrm{C}$. On the arc $\overparen{\mathrm{BC}}$ enclosed by $\angle \mathrm{BAC}$ and not exceeding $90^{\circ}$, choose any point $\mathrm{P}$. The tangent through $\mathrm{P}$ is intercep... | $$
\text { Prove } \begin{array}{l}
\mathrm{AM}+\mathrm{AN}>\mathrm{MN}, \\
\mathrm{AM}+\mathrm{MN}+\mathrm{AN}=\mathrm{AB}+\mathrm{AC} \\
\Rightarrow \mathrm{AB}+\mathrm{AC}>2 \mathrm{MN}, \\
\\
\widehat{\mathrm{BC}} \leqslant 91)^{\circ} \Rightarrow \angle \mathrm{MAN} \geqslant 90^{\circ} \Rightarrow \mathrm{MN} \te... | \frac{1}{3}(\mathrm{AB}+\mathrm{AC})<\mathrm{MN}<\frac{1}{2}(\mathrm{AB}+\mathrm{AC}) | Geometry | proof | Yes | Yes | cn_contest | false | 702,056 |
Example 1. In square $\mathrm{ABCD}$, $\mathrm{M}$ is on $\mathrm{CD}$, and extending $\mathrm{AM}$ intersects $\mathrm{BC}$ at $\mathrm{N}$ (Figure 1).
Prove: $\mathrm{AM}+\mathrm{AN} > 2 \mathrm{AC}$. | Prove that for a square with side length $a$, $\angle \mathrm{CAN}=\alpha$, then
$$
\begin{array}{l}
\angle \mathrm{N}=\angle \mathrm{DAN} \\
=45^{\circ}-\alpha, \\
\mathrm{AC}=\sqrt{2} \mathrm{a}, \\
\mathrm{AM}=\frac{\mathrm{a}}{\cos \left(45^{\circ}-\alpha\right)}.
\end{array}
$$
In $\triangle \mathrm{ACN}$, $\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,057 |
Example 2. In right $\triangle \mathrm{ABC}$, $\angle \mathrm{C}=90^{\circ}$, $\mathrm{CD} \perp \mathrm{AB}$. Prove:
$$
\mathrm{AC}+\mathrm{BC}<\mathrm{AB}+
$$
CD. | Proof: Let $\mathrm{BC}=$
$\mathrm{a}$, then $\mathrm{AC}=\operatorname{atg} \mathrm{B}$,
A $\mathrm{B}=$ asec $B$, $\mathrm{CD}=$ $a \sin B$.
$\mathrm{AC}+\mathrm{BC}-(\mathrm{AB}+\mathrm{CD})$
$=\operatorname{atg} B+a-(a \sec B+a \sin B)$
$=\frac{\sin B+\cos B-1-\sin B \cos B}{\cos B} \cdot a$
$=\frac{(\cos B-1)(1-\s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,058 |
3. In $\triangle \mathrm{ABC}$, $\mathrm{AC}>\mathrm{AB}$, $\angle \mathrm{A} \neq 90^{\circ}$, $\mathrm{BD} \perp \mathrm{AC}$ at $\mathrm{D}$, $\mathrm{CE} \perp \mathrm{AB}$ at $\mathrm{E}$. Prove: $\mathrm{AC}+\mathrm{BD}>\mathrm{AB}+\mathrm{CE}$ | Prove that in $\triangle \mathrm{ABC}$, let $\mathrm{BC}=\mathrm{a}$, then
$$
\begin{array}{l}
\mathrm{AC}=\frac{a \sin \mathrm{ABC}}{\sin \mathrm{A}}, \\
\mathrm{AB}=\frac{a \sin \mathrm{ACB}}{\sin \mathrm{A}} .
\end{array}
$$
In the right triangles $\triangle E C D$ and $\triangle C B E$, $B \bar{D}=a \sin \mathrm{A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,059 |
Example 4. Given in quadrilateral $\mathrm{ABCD}$, $\mathrm{AB}=\mathrm{AC}$, $\mathrm{BD}>\mathrm{DC}$, $\angle \mathrm{BAC} = \angle \mathrm{BDC}$. Prove:
$$
\mathrm{AB}>\frac{1}{2}(\mathrm{BD}+\mathrm{DC})
$$ | Proof: Since $\angle \mathrm{BAC}=\angle \mathrm{BDC}$, it follows that $\mathrm{ABCD}$ is cyclic. Let the radius of the circle be $R$, $\angle \mathrm{CAD}=\alpha$, and $\angle \mathrm{ABD}=\beta$. According to geometric theorems, it is easy to see that $\angle \mathrm{ACB}=\alpha+\beta$.
Thus, $\mathrm{AB}=2 \mathrm{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,060 |
Dream 5. In $\triangle \mathrm{ABC}$, $\mathrm{AB}=\mathrm{AC}$, and $\mathrm{P}$ is any point on the circumcircle $\overparen{\mathrm{AC}}$. Prove: $\mathrm{AB}+\mathrm{AC}>\mathrm{PB}+\mathrm{PC}$ | $$
\begin{array}{l}
\text{Given the circumradius } R, \angle ACB = \alpha, \angle ACP = \beta, \text{ then } \angle PBC = \alpha - \beta. \\
\because AB = AC = 2R \sin \alpha, \\
PB = 2R \sin (\alpha + \beta), \\
PC = 2R \sin (\alpha - \beta), \\
\therefore PB + PC = 2R[\sin (\alpha + \beta) + \sin (\alpha - \beta)] = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,061 |
Example 6. In $\triangle \mathrm{ABC}$, $\mathrm{AB}=\mathrm{AC}$, and $\mathrm{P}$ is a point inside this triangle such that $\angle \mathrm{APB}>\angle \mathrm{APC}$. Prove that $\mathrm{PB}<\mathrm{PC}$. | Proof: Let $\mathrm{PA}=\mathrm{a}$,
$$
\begin{array}{l}
\mathrm{PB}=\mathrm{b}, \mathrm{PC}=\mathrm{c}, \\
\angle \mathrm{APB}=\alpha, \angle \mathrm{APC}=\boldsymbol{\beta} .
\end{array}
$$
By the cosine rule: $\cos \alpha=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{AB}^{2}}{2 \mathrm{ab}}$,
$$
\begin{array}{l}
\cos... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,062 |
Example 7. Given in $\triangle \mathrm{ABC}$, $\mathrm{C}=90^{\circ}$, and the altitude $C D=h$. Prove:
$$
(\sqrt{2}-1) \mathrm{h} \leqslant \mathrm{r}<\frac{1}{2} \mathrm{~h} .
$$ | Let $\mathrm{Ar}_{2}=\mathrm{c}$,
$$
\begin{array}{l}
\because \frac{1}{2} c h=r s, \quad\left(\text { where } s=\frac{a+b+c}{2}\right) \\
\therefore \frac{r}{h}=\frac{c}{2 s}=\frac{c}{c+a+b} \\
=\frac{c}{c+c \sin A+c \sin B} \\
=\frac{1}{1+\sin A+\sin B}
\end{array}
$$
$$
\begin{array}{l}
=\frac{1}{1+\sqrt{2} \cos \fr... | (\sqrt{2}-1) \mathrm{h} \leqslant \mathrm{r}<\frac{1}{2} \mathrm{~h} | Geometry | proof | Yes | Yes | cn_contest | false | 702,063 |
Example 5. Draw the graph of the function $y=\left|x^{2}-6 x+5\right|$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 5. Draw the graph of the function $y=\left|x^{2}-6 x+5\right|$. | Solve (1) Draw the graph of the function $y=x^{2}-6 x+5$. The graph is a parabola with the vertex at $(3,-4)$, intersecting the $x$-axis at points $(1,0)$ and $(5,0)$, and opening upwards, as shown by the parabola $\mathrm{ABC}^{\prime} \mathrm{DE}$ in Figure 9;
(2) Using the $x$-axis as the axis of symmetry, draw the ... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,064 |
Example 1. The moving circle $\mathrm{M}$ is externally tangent to the fixed circle $\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}=0$, and also tangent to the $\mathrm{y}$-axis. Find the equation of the locus of the center $\mathrm{M}$. | Analysis: As shown in Figure 1, the moving circle $\mathrm{M}(\mathrm{x}, \mathrm{y})$ is externally tangent to the fixed circle $\mathrm{O}_{1}(2,0)$ at $\mathrm{A}$, and is tangent to the $\mathrm{y}$-axis at B. Therefore, the condition for the motion of $\mathrm{M}$ must satisfy
$$
\left|\mathrm{O}_{1} \mathrm{M}\ri... | y^2=8x | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,065 |
Example 2. In
$\triangle \mathrm{ABC}$, it is
known that the coordinates of
$\mathrm{B}$ and $\mathrm{C}$ are
(0, 0) and (a,
0), respectively. When $\mathrm{A}$ moves
on the line $l$:
$x+2 y-8=0$, find the equation of the locus of the centroid of $\triangle A B C$. | Analysis: As shown in Figure 2, $P(x, y)$ is the centroid of $\triangle ABC$. Clearly, the movement of $P$ is restricted by $A$. If $M$ is the midpoint of $BC \left(\frac{a}{2}, 0\right)$, and $A$ is $\left(x_{1}, y_{1}\right)$, then $|AP|: |PM| = 2: 1$. By the section formula, we get: $x_{1}=3x-a, y_{1}=3y$. Using the... | 3x+6y-a-8=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,066 |
Example 3. Point $\mathrm{P}$ is a moving point on a fixed line segment $\mathrm{AB}$, where $\mathrm{AB}=\mathrm{a}$. On the same side of $\mathrm{AB}$, equilateral triangles $\mathrm{APM}$ and $\mathrm{BPN}$ are constructed with $\mathrm{AP}$ and $\mathrm{PB}$ as sides, respectively. Find the locus of the midpoint $\... | Analysis: As shown in Figure 3, with point $\mathrm{A}$ as the origin and line $\mathrm{AB}$ as the $\mathrm{x}$-axis, we establish a coordinate system. Clearly, the position of $\mathrm{Q}$ depends on the position of the moving point $\mathrm{P}$ on $\mathrm{AB}$, i.e., the distance from point $\mathrm{P}$ to the orig... | y=\frac{\sqrt{3}}{4}a, \quad\left(\frac{a}{4}<x<\frac{3}{4}a\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,067 |
Example $4 . \triangle \mathrm{ABC}$ has three sides $\mathrm{a}>\mathrm{b}>\mathrm{c}$ forming an arithmetic sequence, and $b=|A C|=2$, find the locus of vertex $\mathrm{B}$.
---
The translation maintains the original text's line breaks and format. | Analysis: As shown in Figure 4, establish a coordinate system with the line of side $\mathrm{AC}$ as the $\mathrm{x}$-axis and the perpendicular bisector of $\mathrm{AC}$ as the $\mathrm{y}$-axis. From the given information, we have $a+c=2b=4$, which means the sum of the distances from the moving point $\mathrm{B}$ to ... | \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,068 |
Example 5. Let $\mathrm{M}$ be the midpoint of any chord $\mathrm{AB}$ parallel to the $\mathrm{x}$-axis (not on the $\mathrm{x}$-axis) of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=1$, and let $\mathrm{C}$ be the intersection point of the known circle with the positive direction of the $x$-axis. Find the locus of the i... | Analysis: As shown in Figure 5, to find the equations of the two lines OB and MC, we need to set \(\angle BOC = \theta\), obtaining the coordinates of points B and M as \((\cos \theta, \sin \theta)\) and \((0, \sin \theta)\), respectively. Point C has coordinates \((1,0)\). Therefore, the equation of line OB is \(y = x... | y^2 = -2x + 1 (|y| < 1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,069 |
Example 1. A function of the natural number $\mathrm{n}$ satisfies the following conditions:
(1) $f(n)=f(n-1)+a^{n}(n \geqslant 2$, $a$ is a non-zero constant $)$;
(2) $f(1)=18$.
Try to find: (1) the analytical expression of $\mathrm{f}(\mathrm{n})$;
(2) when $a=1$, what is the value of $f(1983)$? | From the given recurrence relation, we get
$$
\begin{array}{l}
f(2)=f(1)+a^{2}, \\
f(3)=f(2)+a^{3}, \\
f(4)=f(3)+a^{4}, \\
\cdots \cdots \\
f(n-1)=f(n-2)+a^{n-1}, \\
f(n)=f(n-1)+a^{n} .
\end{array}
$$
By adding both sides of the above equations, we get
$$
\begin{array}{l}
f(n)=f(1)+a^{2}+a^{3}+a^{4}+\cdots+a^{n}, \tex... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,071 |
Example 2. In the set of non-negative integers $\overline{\mathbf{Z}}-1$: Define the function $\mathrm{f}(\mathrm{n})$:
$$
\left\{\begin{array}{l}
f(0)=2, f(1)=3, \\
f(k+1)=3 f(k)-2 f(k-1) . \text { where, } k \geqslant 1 .
\end{array}\right.
$$
Try to find the formula for $f(n)$ expressed in terms of $n$. | Given $f(0)=2, f(1)=3$, by the recursive formula: $f(k+1)=3 f(k)-2 f(k-1)$, we get
$$
\begin{array}{l}
f(2)=3 f(1)-2 f(0)=5, \\
f(3)=3 f(2)-2 f(1)=9, \\
f(4)=3 f(3)-2 f(2)=17,
\end{array}
$$
Since $2=2^{4}+1,3=2^{1}+1,5=2^{2}+1,9=2^{3}+1,17=2^{4}+1$, by incomplete induction, we have
$$
f(n)=2^{n}+1.
$$
Next, we prove... | f(n)=2^{n}+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,072 |
Example 3. Let the function $\mathrm{f}(\mathrm{x})$ have the properties:
i) It is defined for all $x$ except $x=0$ and $x=1$;
ii) $f(x)+f\left(\frac{x-1}{x}\right)=1+x$.
Try to find the analytical expression of the function $\mathrm{f}(\mathrm{x})$. | Given, $\mathrm{f}(\mathrm{x})+\mathrm{f}\left(-\frac{\mathrm{x}-1}{\mathrm{x}}\right)=1+\mathrm{x}$. (1)
Substitute $\frac{\mathrm{x}-1}{\mathrm{x}}$ and $\frac{1}{1-\mathrm{x}}$ for $\mathrm{x}$ in (1), we get
$$
\begin{array}{l}
f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-x}\right)=\frac{2 x-1}{x} . \\
f\left(\fra... | f(x)=\frac{x^{3}-x^{2}-1}{2 x(x-1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,073 |
Example 4. The cubic function $\mathrm{f}(\mathrm{x})$ passes through points $\mathrm{A}(1,4), \mathrm{B}(3,8)$, and has an extremum of 0 at $x=2$, find $f(x)$. | Let $f(x)=a x^{3}+b x^{2}+c x+d$, then $f^{\prime}(x)=3 a x^{2}+2 b x+c$.
Since the function passes through $A(1,4)$ and $B(3,8)$, and $f(2)=0$, $f^{\prime}(2)=0$,
we can obtain the system of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b+c+d=4, \\
27 a+9 b+3 c+d=8, \\
12 a+4 b+c=0, \\
8 a+4 b+2 c-d=0, \tex... | f(x)=2 x^{3}-6 x^{2}+8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,074 |
Example 6. Draw the graph of the function $y=|\cos x|$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the request is to translate the given text, here is the translation:
Example 6. Draw the graph of the fu... | (1) Draw the graph of the function $y=\cos x$;
(2) Reflect the graph of $y=\cos x$ across the $x$-axis;
(3) In Figure 10, the curve above the $x$-axis (including the points where the curve intersects the $x$-axis) is the graph of the function $y=|\cos x|$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,075 |
Example 5. Try to find the function $f(x)$ defined on the set of natural numbers, such that $f(x+y)=f(x)+f(y)+xy, f(1)=1$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Translation:
Example 5. Try to find the function $f(x)$ defined on the set of natural numbers, such that $f(x+y)=f(x)+f(y)+xy, f(1)=1$... | Let $\mathrm{y}=1$, then we have
$$
f(x+1)=f(x)+x+1,
$$
i.e., $f(x+1)-f(x)=x+1$.
In the above equation, let $\mathrm{x}=1,2,3, \cdots, \mathrm{n}$ respectively, substitute them in, and add both sides of the equations respectively, considering $\mathrm{f}(1)=1$, we get $f(n)=\frac{1}{2} n(n+1)$.
This is the required fu... | f(n)=\frac{1}{2} n(n+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,076 |
Example 6. $f\left(\frac{2}{x}+1\right)=\lg x$, find $f(x)$. | Let $\frac{2}{x}+1=y$, then $x=\frac{2}{y-1}$, so we have $f(y)=\lg \left(\frac{2}{y-1}\right)=\lg 2-\lg (y-1)$, therefore $f(x)=\lg 2-\lg (x-1)$. | f(x)=\lg 2-\lg (x-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,077 |
Example 8. If $\mathrm{P}(\mathrm{x})$ represents an $\mathrm{n}$-degree polynomial, and it is known that $\mathrm{P}(\mathrm{k})$ $=\frac{k}{1+k}(k=0,1,2,3, \cdots, n)$, try to find $\mathrm{P}(\mathrm{n}+\mathrm{1})$. | Construct a polynomial of degree $\mathrm{n}+1$:
$Q(x)=(x+1) P(x)-x$, according to the given conditions we have
$Q(\mathrm{k})=0 \quad(\mathrm{k}=0,1,2, \ldots, \mathrm{n})$,
and $Q(-1)=1$, hence we can obtain
$$
Q(x)=\frac{(-1)^{n+1}}{(x+1)!} x(x-1)(x-2) \cdots(x-n) \text {. }
$$
Thus,
$$
P(x)=\frac{(-1)^{n+1} \frac{... | P(n+1) = \left\{\begin{array}{cc} \frac{n}{n+2} ; & (n \text{ is even) } \\ 1 . & \text{ ( } n \text{ is odd) } \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,079 |
Example 1. Write a general term formula for the following sequence:
(1) $1,-4,9,-16,25, \cdots$, | $\left(a_{n}=(-1)^{n+1} n^{2}\right)$ | a_{n}=(-1)^{n+1} n^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,081 |
Example 2. Write a general term formula for the following sequence
(1) $\frac{1}{2}, \frac{5}{6}, \frac{11}{12}, \frac{19}{20}, \frac{29}{30}, \cdots$; | $\left(a_{n}=1-\frac{1}{n(n+1)}\right)$ | a_{n}=1-\frac{1}{n(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,084 |
Example 7. Draw the graph of the function $\mathrm{y}=\left|\log _{2}\right| \mathrm{x}||$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Example 7. Draw the graph of the function $\mathrm{y}=\left|\log _{2}\right| \mathrm{x... | Solve (1) Draw the graph of $y=\log _{2} x$ when $x \geqslant 0$, as shown in Figure 11(1);
(2) Using the y-axis as the axis of symmetry, draw the symmetric curve of the curve in (1) above. The curve in Figure 11(2) is the graph of $\mathrm{y}=\log _{2}|\mathrm{x}|$;
(3) Using the x-axis as the axis of symmetry, draw t... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,086 |
Example 5. Given $a_{1}=3, a_{n+1}=a_{n}+2 n+3$ $(n \geqslant 1)$, find the general term $a_{n}$ of the sequence. | Solution 1
$$
\begin{array}{l}
\because a_{1}=3, \quad a_{n+1}=a_{n}+2 n+3, \\
\therefore \mathrm{a}_{2}=8=2 \times(2+2) \text {, } \\
\mathrm{a}_{3}=15=3 \times(3+2) \text {, } \\
a_{4}=24=4 \times(4+2) \text {, } \\
\text {... } \\
a_{n}=\left\{\begin{array}{ll}
3, & (n=1) \\
n & (n+2) .(n>1)
\end{array}\right. \\
\e... | a_{n}=n(n+2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,091 |
Example 6. Given $a_{1}=1, a_{n}=3 a_{n-1}+2(n \geqslant 2)$, find the general term $\mathrm{a}_{\mathrm{n}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Solution } \because a_{1}=1, a_{n}=3 a_{n-1}+2, \quad(n \geqslant 2) \\
\therefore a_{2}=3 a_{1}+2=3+2, \\
a_{3}=3 a_{2}+2=3^{2}+2 \times 3+2 \\
a_{4}=3 a_{3}+2=3^{3}+2 \times 3^{2}+2 \times 3+2 \\
\cdots \cdots \\
a_{n}=3^{n-1}+2 \times 3^{n-2}+2 \times 3^{n-3}+\cdots \\
+2 \times 3+2 \\
\q... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,092 |
Example 7. Given: $a_{1}=1, a_{n+1}=3+\frac{1}{2} a_{n}$ $(n \geqslant 1)$, find the general term of the sequence $a_{n}$. | Solution 1 $\because a_{1}=1, a_{n+1}=3+\frac{1}{2} a_{n}$, $(n \geqslant 1)$ then
$$
\begin{array}{l}
a_{1}=1, a_{2}=\frac{1}{2}+3, \\
a_{3}=\frac{1}{2^{\frac{2}{2}}}+3 \times \frac{1}{2}+3, \\
a_{4}=\frac{1}{2^{3}}+3 \times \frac{1}{2^{2}}+3 \times \frac{1}{2}+3, \\
\cdots \cdots \\
a_{n}=\frac{1}{2^{n-1}}+3 \times ... | 6-\frac{5}{2^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,093 |
Example 1. Prove:
(1) $2^{n}-C_{n}^{1} 2^{n-1}+C_{n}^{2} 2^{n-2}+\cdots$ $+(-1)^{n-1} C_{n}^{n-1} 2+(-1)^{n}=1 ;$
$$
\begin{array}{l}
\text { (2) } 3^{n}-C_{n}^{1} 3^{n-1}+C_{n}^{2} 3^{n-2}+\cdots \\
+(-1)^{n-1} C_{n}^{n-1} 3+(-1)^{n}=2^{n} .
\end{array}
$$ | For these two questions, just set $a=2, b=-1$ and $a=3, b=-1$ in the formula to obtain the two identities to be proven.
2. Comparison method. The following two comparison methods are commonly used:
(1) Appropriately select a known identity, then compare the coefficients of the same power of $\mathrm{x}$ on both sides, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,095 |
Example 2. Prove: (1) $\left(C_{n}^{0}\right)^{2}+\left(C_{n}^{1}\right)^{2}+\left(C_{n}^{2}\right)^{2}+\cdots$
$$
+\left(C_{n}^{n}\right)^{2}=\frac{(2 n)!}{n!n!},
$$
$$
\begin{array}{l}
\text { (2) }\left(C_{n}^{0}\right)^{2}-\left(C_{n}^{1}\right)^{2}+\left(C_{n}^{2}\right)^{2}-\cdots \\
+(-1)^{n}\left(C_{n}^{n}\righ... | (1) Choose $(1+x)^{n}(1+x)^{n}=(1+x)^{2 n}$, comparing the coefficients of $x^{n}$ on both sides can prove it. Similarly, comparing the coefficients of $X^{n}$ in the identity $(1+x)^{n}(1-x)^{n}=\left(1-x^{2}\right)^{n}$ can prove the identity in question (2).
Of course, the constant term can also be compared. For q... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,096 |
Example 8. Draw the graph of the function $\mathrm{y}=|2-| \mathrm{x}||$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 8. Draw the graph of the function $\mathrm{y}=|2-| \mathrm{x}||$. | Solution 1
(1) Draw the graph of $y=2-x$ when $x \geqslant 0$, as shown by the ray ABC in Figure 12(i).
(2) Using the $y$-axis as the axis of symmetry, draw the symmetric image of the ray ABC, which is the ray $AB^{\prime}C^{\prime}$ in Figure 12(2).
The broken line $C^{\prime}B^{\prime}ABC$ in Figure 12(2) is the gr... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,097 |
Example. Prove: (1) $C_{n}^{0}-C_{n}^{2}+C_{n}^{4}-C_{n}^{6}+\ldots$
$$
=2^{\frac{n}{2}} \cos \frac{n \pi}{4} \text {; }
$$
(2) $C_{n}^{1}-C_{n}^{3}+C_{n}^{5}-C_{n}^{7}+\cdots$
$$
= 2^{\frac{n}{2}} \sin \frac{n \pi}{4} .
$$ | The proof method is to choose $(1+\mathrm{i})^{n}=\left[\sqrt{2}\left(\cos \frac{\pi}{4}\right.\right.$ $\left.\left.+i \sin \frac{\pi}{4}\right)\right]^{n}=2^{\frac{n}{2}}\left(\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right)$, compare the real part coefficients on both sides to get the first question, and compare t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,098 |
Example 4. Prove $C_{n}^{1}+2 C_{n}^{2}+3 C_{n}^{3}+\cdots+n C_{n}^{n}=n \cdot 2^{n-1}$ | $$
\begin{aligned}
\because \text { Left side } & =n+n(n-1) \\
& +\frac{n(n-1)(n-2)}{2 \cdot 1}+\cdots \\
& +\frac{n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1}{(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1} \\
= & n\left[1+(n-1)+\frac{(n-1)(n-2)}{2 \cdot 1}\right. \\
& \left.+\cdots+\frac{(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1}{(n-1)(n-2)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,099 |
Example 5. Prove: $C_{m}^{m}+C_{m+1}^{m}+C_{m+2}^{m}+\cdots+C_{m+n}^{m}$ $=\mathrm{C}_{\mathrm{m}+\mathrm{n}+1}^{\mathrm{m}+1}$. | Proof: Left side $=C_{m+1}^{m+1}+C_{m+2}^{\mathrm{m}+1}-C_{m+1}^{\mathrm{m}+1}+C_{m+3}^{\mathrm{m}+1}$
$$
\begin{array}{l}
-C_{m+2}^{m+1}+\cdots+C_{m+n+1}^{m+1}-C_{m+n}^{m} \\
= C_{m+n+1}^{m+1}=\text { Right side. }
\end{array}
$$
$\therefore$ The original equation holds.
Here, we used $\mathrm{C}_{\mathrm{n}+1}^{m}=\m... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,100 |
Example 6. Prove: $1+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\cdots+\frac{1}{n+1} C_{n}^{n}=$ $\frac{1}{n+1}\left(2^{n+1}-1\right)$. | Prove $\because C_{n}^{k}=\frac{n}{k} C_{n-1}^{k-1}$,
$$
\begin{aligned}
\therefore & C_{n+1}^{1}+C_{n+1}^{2}+C_{n+1}^{3}+\cdots+C_{n+1}^{n+1} \\
= & \frac{n+1}{1} C_{2}^{3}+\frac{n+1}{2} C_{n}^{1}+\frac{n+1}{3} C_{n}^{2} \\
& +\cdots+\frac{n+1}{n+1} C_{n}^{n} \\
= & (n+1)\left(C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{... | \frac{1}{n+1}\left(2^{n+1}-1\right) | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,101 |
$\begin{array}{c}\text { Example 7. Prove: } C_{n}^{2}+2 C_{n}^{3}+3 C_{n}^{4}+\cdots+(n-1) C_{n}^{n} \\ =1+(n-2) 2^{n-1} .\end{array}$ | Prove that from $C_{n}^{0}+C_{n}^{1}+C_{n}^{2} \cdots +C_{n}^{n}=2^{n}$,
$$
C_{n}^{1}+2 C_{n}^{2}+3 C_{n}^{3}+\cdots+n C_{n}^{n}=n \cdot 2^{n-1}
$$
(2) - (1) gives:
$$
\begin{aligned}
C_{n}^{2} & +2 C_{n}^{3}+3 C_{n}^{4}+\cdots+(n-1) C_{n}^{n} \\
& =1+(n-2) 2^{n-1}
\end{aligned}
$$
Using the same method, we can prove ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,102 |
Example 1. Prove that the sequence $\left\{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\right.$ $\left.\frac{1}{n^{2}}\right\}$ converges using the Cauchy criterion. | Prove $\forall \varepsilon>0, \exists N=\left[\frac{1}{\varepsilon}\right]+1>0$, such that for $m, n>N$ (assuming $m<n$), we have $\left|a_{m}-a_{n}\right|$
$$
\begin{array}{l}
=\left|\frac{1}{(m+1)^{2}}+\cdots+\frac{1}{n^{2}}\right|=\frac{1}{(m+1)^{2}} \\
+\frac{1}{(m+2)^{2}}+\cdots+\frac{1}{n^{2}}<\left(\frac{1}{m}-\... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 702,104 |
峢?？江㺺： $\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!}<2$.
Jiang Muxi: $\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!}<2$. | $$
\begin{array}{l}
\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!} \\
<\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4^{2}}+\cdots+\frac{1}{n^{2}}=\frac{1}{1}+\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3} \\
+\frac{1}{4^{2}}+\cdots+\frac{1}{n^{2}}<1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\fra... | 2-\frac{1}{n}<2 | Inequalities | proof | Yes | Yes | cn_contest | false | 702,105 |
Example 3. Find $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}+\frac{1}{n^{2}}\right)^{n}$. | \begin{array}{l}\text { Sol } \because \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=e, \\ \text { and } \quad \frac{1}{n^{2}}<\frac{1}{n-1}-\frac{1}{n} \Longrightarrow 1+\frac{1}{n}+\frac{1}{n^{2}}<1+\frac{1}{n-1} \\ \Rightarrow \quad\left(1+\frac{1}{n}+\frac{1}{n^{2}}\right)^{n}<\left(1+\frac{1}{n-1}\rig... | e | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,106 |
Example 9. Draw the graph of the function
$$
y=|x-1|+|x-2|
$$ | According to the definition of absolute value, $\mathrm{x}_{1}=1$ and $\mathrm{x}_{2}=2$ divide the domain of the given function into three intervals. On these intervals, the given function is respectively expressed as:
$$
y=\left\{\begin{array}{ll}
3-2 x, & x<1, \\
1, & 1 \leqslant x<2, \\
2 x-3, & x \geqslant 2 .
\en... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,108 |
$2, \mathrm{n}$ is the smallest integer with the following property: it is a multiple of 15, and each of its digits is 0 or 8. Find
$$
\frac{\mathrm{n}}{15}
$$ | Solve $n$ should be a common multiple of 5 and 3. The last digit of a multiple of 5 can only be 0 or 5, and since $n$ consists only of the digits 0 and 8, the last digit of $n$ must be 0. As a multiple of 3, the number of 8s in $n$ must be a multiple of 3. Given the requirement for $n$ to be the smallest, we should hav... | 592 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,109 |
3. As shown in the figure, $P$ is a point inside $\triangle ABC$. Lines are drawn through $P$ parallel to the sides of $\triangle ABC$.
The smaller triangles
$t_{1}, t_{2}$, and $t_{3}$
have areas of 4, 9,
and 49, respectively. Find the area of $\triangle ABC$. | Given the figure, let $\mathrm{MP}=\mathrm{p}, \mathrm{PN}=\mathrm{q}, \mathrm{RT}=\mathrm{r}$. Since triangles $t_{1}, t_{2}, t_{3}$ are all similar to $\triangle A B C$, the ratio of their areas is equal to the square of the ratio of corresponding sides.
Let $\mathrm{AB}=c, \triangle \mathrm{ABC}$ have an area of $\m... | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,110 |
4. Let S be a table consisting of positive integers - the table can contain the same numbers - and it contains the number 68. The mean (arithmetic average) of the numbers in S is 56. However, if 68 is removed, the mean of the remaining numbers drops to 55. What is the largest number that could appear in S? | Let $\mathrm{n}$ denote the number of positive integers in $\mathrm{S}$, then the sum of these $\mathrm{n}$ numbers is $56 n$. After removing 68, the sum of the remaining $n-1$ numbers is $55(n-1)$, thus we get $56 n-55(n-1)$ $=68$, solving for $\mathrm{n}$ gives $\mathrm{n}=13$. After removing 68, the sum of the remai... | 649 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,111 |
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