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Let the sequence $\{a_n\}$ have a sum of the first $n$ terms denoted by $S_n$, with $a_1=1$, and $S_{n+1}=4a_n+2$ ($n\in\mathbb{N}^+$).
(1) If $b_n=a_{n+1}-2a_n$, find $b_n$;
(2) If $$c_{n}= \frac {1}{a_{n+1}-2a_{n}}$$, find the sum of the first 6 terms of $\{c_n\}$, denoted as $T_6$;
(3) If $$d_{n}= \frac {a_{n}... | (1) Since $a_1=1$, and $S_{n+1}=4a_n+2$ ($n\in\mathbb{N}^+$),
we have $S_{n+2}=4a_{n+1}+2$, and $S_{n+2}-S_{n+1}=4(a_{n+1}-a_n)$,
thus $a_{n+2}-2a_{n+1}=2(a_{n+1}-2a_n)$
which means $b_{n+1}=2b_n$
Therefore, $\{b_n\}$ is a geometric sequence with a common ratio of 2, and $b_1=a_2-2a_1$
Since $a_1=1$, and $a_2... | \frac {21}{32} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false | 549,289 |
Example 1. The following numbers are all approximate numbers obtained by rounding. Find their absolute error bounds and relative error bounds:
$$
\begin{array}{l}
\text { (1) } \mathrm{a}_{1} \approx 12.5, \quad(2) \mathrm{a}_{2} \approx 1.25 \times 10^{4}, \\
\text { (3) } \mathrm{a}_{3} \approx 0.0125
\end{array}
$$ | $\begin{array}{c}\text { Solution (1) } \boldsymbol{\delta}_{\mathrm{a}_{1}}=0.05, \\ \boldsymbol{\delta}^{\prime}{ }_{\mathrm{a}_{1}}=\frac{0.05}{12.5}=0.4 \% ; \\ \text { (2) } \boldsymbol{\delta}_{\mathrm{a}_{2}}=50, \\ \boldsymbol{\delta}^{\prime}{ }_{\mathrm{a}_{2}}=\frac{50}{12500}=0.4 \% ; \\ \text { (3) } \bold... | 0.4\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,224 |
Example 2. Find the absolute error bounds and relative error bounds of the following approximate numbers (obtained by rounding).
(1) $\mathrm{a}_{1} \approx 100$,
(2) $\mathrm{a}_{2} \approx 999$,
(3) $\mathrm{a}_{3} \approx 10000$,
(4) $\mathrm{a}_{4} \approx 99999$. | $\begin{array}{c}\text { Sol (1) } \delta_{\mathrm{a}_{1}}=0.5, \\ \delta_{\mathrm{a}_{1}}^{\prime}=\frac{0.5}{100}=0.5 \% \mathrm{~s}, \\ \boldsymbol{\delta}_{\mathrm{a}_{2}}=\frac{0.5}{999} \approx 0.05 \% \delta_{\mathrm{a}_{2}}=0.5, \\ \quad(3) \delta_{\mathrm{a}_{3}}=0.5, \\ \delta_{\mathrm{a}_{3}}^{\prime}=\frac{... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,225 |
Example 4 Find the sequence
... | Solve using the old method: First write out its first few terms $\frac{1}{1}, \quad \frac{1}{2}, \quad \frac{2}{3}, \quad \frac{3}{5}, \quad \frac{5}{8}, \quad \frac{8}{18}, \cdots$ then it is not difficult to observe:
(1) Sequence (1) is not monotonically arranged (but oscillates);
(2) But the two subsequences
$x_{1},... | x_{1}>x_{3}>x_{5}>\cdots>\frac{-1+\sqrt{5}}{2} | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 701,229 |
$$
\begin{array}{r}
\text { FORTY } \\
\text { TEN } \\
+\quad \text { TEN } \\
\hline \text { SIXTY }
\end{array}
$$
In the above problem, each letter represents a digit from $0,1,2,3,4,5,6,7,8,9$, and different letters represent different digits. | Solution (The column numbers are counted from right to left)
From the first column, we see $\mathbf{Y}+\mathrm{N}+\mathbf{N}=\mathrm{Y}$, so $\mathrm{N}=0$ or $N=5$. But $N \neq 5$, otherwise, from the second column, we would have $T+2 E+1$ $=T$, which is impossible. Similarly, $E=0$, or $E=5$, but since $N=0$, we have... | \begin{array}{r}
29786 \\
+\quad 850 \\
\hline 31486
\end{array} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 701,230 |
Example 4. Prove by definition that the sum of two odd functions is still an odd function. | Proof: Let $f(x)$ and $g(x)$ be odd functions, $\varphi(x)=f(x)+$
$$
\begin{aligned}
& g(x) . \\
\because & f(x) 、 g(x) \text { are odd functions, } \\
\therefore & f(-x)=-f(x), g(-x)=-g(x) . \\
\therefore \quad & \varphi(-x)=f(-x)+g(-x)=[-f(x)]+ \\
& {[-g(x)] } \\
& =-[f(x)+g(x)]=-\varphi(x) .
\end{aligned}
$$
Thus, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,233 |
Example 5. Prove that the derivative of an odd function is an even function, and the derivative of a differentiable even function is an odd function. | Proof: Let $f(x)$ be a differentiable odd function, i.e.,
$$
f(-x)=-f(x) .
$$
Differentiating both sides with respect to $x$, we have
$$
\begin{array}{l}
f^{\prime}(-x) \cdot(-1)=-f^{\prime}(x), \\
f^{\prime}(-x)=f^{\prime}(x) .
\end{array}
$$
Thus, $f^{\prime}(x)$ is an even function.
Similarly, we can prove another... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,234 |
Example 6. Prove that the following functions are neither odd nor even:
(1.) $y=x+x^{2}-x^{5}$;
(2) $y=\sin x-\cos x$. | (1) Take $x_{0}=2$, then we have
$$
f(2)=-26, f(-2)=34 \text {, }
$$
Since $f(-2) \neq-f(2)$, and $f(-2) \neq f(2)$,
$\therefore y=x+x^{2}-x^{5}$ is neither an odd nor an even function.
(2) Take $x_{0}=\frac{\pi}{4}$, then we have
$$
\begin{array}{l}
\text { F }\binom{\pi}{4}=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=0 \\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,235 |
Example 1. Calculate $8 \frac{1}{7} + 43.8 - 13.947 + 0.00375$ ( $\frac{1}{7}$ is an exact number, all others are approximate numbers) | The number with the least decimal places is 43.8, so the other numbers only need to be truncated to the hundredth place, and the result is accurate to the tenth place.
$$
\begin{array}{l}
\frac{1}{7}+43.8-13.947+0.00375 \\
\approx 0.14+43.8-13.95 \approx 30.0
\end{array}
$$ | 30.0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,236 |
Example 7. Prove that the function $\mathbf{D}(\mathrm{x})$ (see rule 2) has any rational number $\mathbf{r} \neq 0$ as a period, and has no smallest positive period. | Let $\mathrm{r} \neq 0$ be a fixed rational number, and $\mathrm{x}$ be any real number. When $x$ is a rational number, $x+r$ is still a rational number, so
$$
\mathbf{D}(\mathbf{x}+\mathbf{r})=\mathrm{D}(\mathbf{x})=1 \text {, }
$$
When $x$ is an irrational number, $x+r$ is also an irrational number, so
$$
\mathrm{D}... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,237 |
Example 8 - Prove that the following functions are not periodic:
(1) $y=x \cos x_{3}$
(2) $y=\sin x^{2}$. | (1) For any $\mathrm{T} \neq 0$,
when $T=k \pi$, take $x_{0}=0, \nu_{i} i$
$$
\begin{array}{l}
\left.\mathrm{f}\left(\mathrm{x}_{0}+\mathrm{T}\right)=k \pi \cos k \pi = 0 \cdot (-1)^k = 0\right) * \leqslant \pi, \\
\mathrm{f}\left(\mathrm{x}_{0}\right)=0,
\end{array}
$$
$\therefore f\left(x_{0}+T\right) \neq f\left(x_{... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 701,238 |
Example 3. As shown in Figure 3, from a point $Q(2,0)$ outside the circle $x^{2}+y^{2}=1$, a secant is drawn intersecting the circle at points $A$ and $B$. Find the equation of the locus of the midpoint of chord $\mathrm{AB}$. | Connect $\mathrm{OP}$,
then $O P \perp A B$, that is, $\angle \mathrm{OPQ}$ is always a right angle.
Since $OQ$ is a fixed line segment, by the locus theorem of plane geometry:
point $P$ lies on the circle with $O Q$ as its diameter.
Since $\mathrm{P}$ is the midpoint of chord $\mathrm{AB}$, point $\mathrm{P}$ must be... | (x-1)^{2}+y^{2}=1\left(x^{2}+y^{2}<1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,241 |
Example 4. For what value of $a$ does the curve $y=\sqrt{x+1}$ and the line $y=x+a$ have two intersection points? | Solve as shown in Figure 4: $\because y=\sqrt{x+1}, \therefore y \geqslant 0$ squaring both sides gives $y^{2}=x+1(y \geqslant 0)$,
$\therefore$ the curve of $y=\sqrt{x}+1$ is the part of the parabola $y^{2}=x+1$ above the x-axis (including the point on the x-axis).
Obviously, the intersection points should satisfy:
$$... | 1 \leqslant a < \frac{5}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,242 |
Example 5. Find the equation of the line containing the common chord of two intersecting circles: $x^{2}+y^{2}+D_{1} x+E_{1} y$ $+F_{1}=0, x^{2}+y^{2}+D_{2} x+E_{2} y+F_{2}=0$. | Let the common points of the two circles be \( C\left(x_{1}, y_{1}\right) \) and \( D\left(x_{2}, y_{2}\right) \). Then \( C\left(x_{1}, y_{1}\right) \) should satisfy the equations of both circles:
\[
\begin{array}{c}
x_{1}^{2} + y_{1}^{2} + D_{1} x_{1} + E_{1} y_{1} + F_{1} = 0, \\
\cdots \cdots \cdots \cdots \cdots ... | \left(D_{1} - D_{2}\right) x + \left(E_{1} - E_{2}\right) y + F_{1} - F_{2} = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,243 |
Example 6. Prove: regardless of the value of $\mathrm{k}$ $(\mathrm{k} \neq 0)$, the intersection point of the two lines
$$
a x-b y=k \text{ and } a x+b y=\frac{1}{k}
$$
always lies on a certain hyperbola. | Given that the coordinates of the intersection point of two lines are $\left(x_{0}, y_{0}\right)$, then we have
$$
\left\{\begin{array}{l}
a x_{0}-b y_{0}=k, \\
a x_{0}+b y_{0}=\frac{1}{k}
\end{array}\right.
$$
(1) $\times$ (2) gives
$$
a^{2} x_{0}^{2}-b^{2} y_{0}^{2}=1 .
$$
This equation shows that $\left(x_{0}, y_{0... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,244 |
Example 7. A moving straight line forms a triangle with the coordinate axes whose area is a constant. Find the equation of the locus of the midpoint of the line segment intercepted by the coordinate axes.
untranslated text:
例7. 一动直线与坐标轴所成的三角形的面积为定值, 求这条直线被坐标轴所截线段的中点轨迹方程。 | Let the moving line intersect the $x$-axis and $y$-axis at points $A\left(x_{0}, 0\right)$ and $B\left(0, y_{0}\right)$, respectively, and let the midpoint of $A B$ be $P(x, y)$.
Since
$$
x=\frac{x_{0}}{2}, y=\frac{y_{0}}{2}.
$$
It follows that $\mathrm{x}_{0}=2 \mathrm{x}, \mathrm{y}_{0}=2 \mathrm{y}$.
Given that the... | x y=\frac{k}{2} \text { or } x y=-\frac{k}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,245 |
Example 8. As shown in Figure 5, the endpoints of the base of $\triangle \mathrm{ABC}$ are $B(0,6)$ and $C(0,-6)$, and the product of the slopes of the other two sides is $-\frac{4}{9}$. Find the equation of the locus of vertex $\mathrm{A}$. | Let the coordinates of vertex $\mathrm{A}$ be $(x, y)$,
then $k_{A B}=\frac{y-6}{x}, k_{A C}=\frac{y+6}{x}$.
According to the problem, we have
$$
\frac{y-6}{x} \cdot \frac{y+6}{x}=-\frac{4}{9},
$$
Simplifying, we get
$$
\frac{x^{2}}{81}+\frac{y^{2}}{36}=1
$$
Is this the equation of the trajectory of vertex $\mathrm{A... | \frac{x^{2}}{81}+\frac{y^{2}}{36}=1 \quad(x \neq 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,246 |
Example 2. Calculate $7.38 \times 10^{3}+2.26 \times 10^{4}$ (all are approximate numbers). | \begin{array}{l}\text { Solve } 7.38 \times 10^{3}+2.26 \times 10^{4} \\ =0.738 \times 10^{4}+2.26 \times 10^{4} \approx 3.00 \times 10^{4} .\end{array} | 3.00 \times 10^{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,247 |
Example 10. Two vertices of an equilateral triangle are on the parabola $\mathrm{y}^{2}=2 \mathrm{px}$, and the other vertex is at the origin. Find the side length of this triangle. | To find the length of OA as shown in Figure 7, it is sufficient to determine the coordinates of point A, denoted as \(\left(x_{1}, y_{1}\right)\).
Since A is the intersection point of the parabola and the line OA, and the equation of the line OA is
\[ y = \tan 30^{\circ} \cdot x = \frac{\sqrt{3}}{3} x, \]
\(\left(x_{... | 4 \sqrt{3} p | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,249 |
Example 11. The equation of the parabola is $y^{2}=2 x$, and there is a circle with a radius of 1, whose center
moves on the $x$-axis. Ask at what position
does the circle move to
when the tangents at the intersection points
of the circle and the parabola
are perpendicular to each other? | As shown in Figure 8, let the intersection point of the circle and the parabola be \( p\left(x_{0}, y_{0}\right) \), and the center of the circle be point \( \mathrm{C} \).
From the problem, the tangent line to the parabola at point \( \mathrm{P} \) must pass through the center \( \mathrm{C} \). The equation of the ta... | \left(-\frac{\sqrt{5} - 1}{4}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,250 |
Example 1. Using the ten digits from $0-9$,
(1) How many 3-digit numbers without repeated digits can be formed?
(2) How many of these numbers are between $300-700$ and do not end in zero? | (1) Since different arrangements of three digits form different numbers (ordered), this is a permutation problem. We can arrange them in the order of hundreds, tens, and units. The first digit cannot be 0, so there are $\mathrm{P}_{\mathrm{P}}^{1}$ ways to arrange the first digit. The next two digits can be arranged fr... | 648 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,251 |
Example 1. Given $f(n)=n^{4}+n^{3}+n^{2}+n+1$, find the remainder of $f\left(2^{5}\right)$ divided by $f(2)$. | $$
\begin{aligned}
f(2) & =2^{4}+2^{3}+2^{2}+2+1 \\
& =(11111)_{2}
\end{aligned}
$$
(The subscript 2 outside the parentheses indicates a binary number, the same applies below).
$$
\begin{array}{l}
f\left(2^{5}\right)=2^{20}+2^{15}+2^{10}+2^{5}+1 \\
=(100001000010000100001)_{2} \\
=(1111100 \cdots 0)_{15 \text { ones }}... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,254 |
Example 2. Find the remainder when $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$ is divided by
81. | Solve $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$
$$
\begin{array}{l}
=\left(2^{22}-2^{20}+2^{18}-2^{16}+2^{14}-\right. \\
\left.2^{12}\right)+\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \\
=\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \\
\left(2^{12}+1\right)
\end{array}
$$
Since $9=(1001)_{2}, 81=9 \times 9$,
and $2^{1... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,255 |
Example 4. Prove: For $b \in N, 2^{2 b-1}+2^{b}+1$ cannot be divisible by 7. | Prove $A=2^{2 b-1}+2^{b}+1$
$$
\begin{aligned}
& =(\underbrace{00 \cdots 0}_{b-2} 0100 \cdots 01)_{b}, \\
7 & =(111)_{2} .
\end{aligned}
$$
When $b$ is a number of the form $3 k+2$, the remainder of $A$ divided by 7 is $(110)_{2}$;
When $b$ is a number of the form $3 k+1$, the remainder of $A$ divided by 7 is $(101)_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 701,257 |
Example 3. The circumference of a circle is measured to be 10.7 meters, find its radius. | Radius $=\frac{10.7}{2 \pi} \approx \frac{10.7}{2 \times 3.142} \approx 1.70$ (meters). | 1.70 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,258 |
Example 4. Calculate: $\frac{3.6 \times 11.74 \times 138.4}{6 \times 2437}$.
(6 is an exact number) | \begin{aligned} & \text { Sol } \frac{3.6 \times 11.74 \times 138.4}{6 \times 2437} \\ \approx & \frac{3.6 \times 11.7 \times 138}{6 \times 2440} \approx \frac{0.60 \times 11.7 \times 138}{2440} \\ \approx & \frac{7.02 \times 138}{2440} \approx \frac{969}{2440} \approx 0.40 .\end{aligned} | 0.40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,259 |
Example 6. Calculate: $17.2^{3}+82.3^{3}$ (all numbers are exact) | Solve by consulting a four-figure mathematical table, we get
$$
\begin{array}{l}
17.2^{3}+82.3^{3}=5.088 \times 10^{3}+5.574 \\
\times 10^{5} \approx 0.0509 \times 10^{5}+5.574 \times 10^{5} \\
\approx 5.625 \times 10^{5} .
\end{array}
$$ | 5.625 \times 10^{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,261 |
Example 7. Calculate:
$$
\frac{(742-108+321 \times 0.082) \times(37.2-28.7)}{517.4 \times(90.28-37.2)}
$$ | $$
\begin{array}{l}
\frac{(742-108+321 \times 0.082) \times(37.2-28.7)}{517.4 \times(90.28-37.2)} \\
\approx \frac{(634+26.3) \times 8.5}{517 \times 53.1} \approx \frac{660 \times 8.5}{517 \times 53.1} \\
\approx \frac{5610}{27500} \approx 0.20
\end{array}
$$ | 0.20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,262 |
Example 2. Study the maximum and minimum values of the function $\mathrm{y}=\frac{1}{\mathrm{x}(1-\mathrm{x})}$ on $(0,1)$.
| Given $x \in (0,1)$, therefore $x>0, 1-x>0$.
When $x \rightarrow +0$, $1-x \rightarrow 1, x(1-x) \rightarrow +0$,
When $x \rightarrow 1-0$, $1-x \rightarrow +0, x(1-x) \rightarrow +0$,
Therefore, $y(+0)=y(1-0)=+\infty$. From (v),
it is known that $y$ has only a minimum value on $(0,1)$. Also, $y' = \frac{2x-1}{x^2(1-x... | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,265 |
Example 4. In $\square \mathrm{ABCD}$, $\mathrm{E}$ is the midpoint of $\mathrm{AB}$, $\mathrm{F}$ is a point on $\mathrm{AD}$, and PF $=2 \mathrm{FA}, \mathrm{EF}$ intersects $\mathrm{AC}$ at P (Figure 4). Prove: $A P=\frac{1}{5} A C$ | Proof: (Key Points)
Draw DG//FE,
intersecting AC at G,
then $\mathrm{AP}=\frac{1}{2} \mathrm{PG}$,
then prove $\triangle A P E \backsim \triangle C C D$,
from $A E=\frac{1}{2} C D$ we get $B \cdot P=\frac{1}{2} \mathrm{GC} \cdots$ (2)
From (1) and (2) we get $\mathbf{A P}=\frac{1}{5} \mathbf{A C}$. | A P=\frac{1}{5} A C | Geometry | proof | Yes | Yes | cn_contest | false | 701,266 |
Example 1. In the equation $a x^{2}-\sqrt{2} b x+c=0$, $a, b, c$ are the three sides of an obtuse triangle, and $b$ is the longest. Prove: The equation must have two distinct real roots. | To prove that the discriminant of the given equation is $\triangle=2 b^{2}-4 a c$. How to prove $\triangle>0$?
By the cosine rule, we have $\mathbf{b}^{2}=\mathbf{a}^{2}+\mathbf{c}^{2}-2 \mathbf{a c c o s} \mathrm{B}$.
$\because B$ is an obtuse angle,
$\therefore \quad-1<\cos B<0$.
From $b^{2}=a^{2}+a^{2}-2 a c \cos B$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,268 |
Example 3. The three sides of $\triangle A B C$ are $a, b, c$. Prove:
$$
\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \leqslant \frac{1}{8} .
$$ | Prove $\because 2 \sin ^{2} \frac{A}{2}=1-\cos A$
$$
\begin{array}{l}
=1-\frac{b^{2}+c^{2}-a^{2}}{2 b c} \\
=\frac{a^{2}-(b-c)^{2}}{2 b c} .
\end{array}
$$
Since $(b-c)^{2} \geqslant 0$,
$$
\therefore 0<\sin ^{2} \frac{\mathrm{A}}{2} \leqslant \frac{\mathrm{a}^{2}}{4 \mathrm{bc}} \text {, }
$$
which means $0<\sin \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,270 |
Example 4. Prove that $\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \geqslant(a x+b y)^{2}$. | $\begin{array}{l}\text { Prove } \quad\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \\ =a^{2} x^{2}+b^{2} y^{2}+a^{2} y^{2}+b^{2} x^{2} \\ =a^{2} x^{2}+2 a x b y+b^{2} y^{2}+a^{2} y^{2}-2 a x b y+b^{2} x^{2} \\ =(a x+b y)^{2}+(a y-b x)^{2} \\ \geqslant(a x+b y)^{2} .\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,271 |
Example 5. In right $\triangle \mathrm{ABC}$, $\angle \mathrm{C}=90^{\circ}$. What is the relationship between $\mathrm{d}^{3}+\mathrm{b}^{3}$ and $\mathrm{c}^{3}$? | In the right-angled $\triangle \mathrm{ABC}$, there is $\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}$.
Multiplying both sides of the above equation by $c$ yields $a^{2} c+b^{2} c=c^{2} s$.
Expressing $c$ in terms of $a$ and $b$ ($c>a, c>b$), we get from $a^{2} c+$ $b^{2} c=c^{3}$ that
$$
a^{3}+b^{3}<c^{3} .
$$ | a^{3}+b^{3}<c^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,272 |
Example 6. In the right triangle $\mathrm{ABC}$, $\angle \mathrm{C}=90^{\circ}$. Prove: when $n \geqslant 2$, $a^{n}+b^{n} \leqslant c^{n}$. | $\begin{array}{l}\text { Proof: When } n=2, a^{2}+b^{2}=c^{2} ; \\ \text { When } n \geqslant 3, c^{n}=c^{2} \cdot c^{n-2} \\ =\left(a^{2}+b^{2}\right) c^{n-2}=c^{n-2} \cdot a^{2}+c^{n-2} \cdot b^{2} \\ >a^{n-2} \cdot a^{2}+b^{n-2} \cdot b^{2}=a^{n}+b^{n} . \\ c^{n}>a^{n}+b^{n} \text {. }\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,273 |
Example 8. If $\mathrm{n}$ is a natural number, prove that: $\sqrt{2}+\sqrt{6}+\sqrt{12}$ $+\cdots+\sqrt{n(n+1)}<n(n+1)$. | Prove that for $k>0$, from $\sqrt{\mathbf{k}(\mathbf{k}+1)}<\frac{k+(k+1)}{2}$,
we get $\sqrt{2}=\sqrt{1 \cdot 2}<\frac{1+2}{2}$,
$$
\begin{array}{l}
\sqrt{6}=\sqrt{2 \cdot 3}<\frac{2+3}{2}, \cdots, \\
\sqrt{\mathbf{n}(\mathbf{n}+1)}<\frac{\mathrm{n}+(\mathbf{n}+1)}{2} .
\end{array}
$$
Therefore, $\sqrt{2}+\sqrt{6}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,275 |
Example 3. Study the maximum and minimum values of the function $\mathrm{y}=\frac{4 \mathrm{x}^{3}-3 \mathrm{x}^{4}}{12}$ on the interval $(-1$, 2). | Solving, we know $y(-1+0)=y(-1)=-\frac{7}{12}$,
$y(2-0)=y(2)=-\frac{4}{3}$.
Also, $y^{\prime}=x^{2}-x^{3}=x^{2}(1-x)$, the critical points are $x_{1}=0, x_{2}=1$. By comparing $y(0)=0$, $y(1)=\frac{1}{12}$ with the endpoint (limit) values, we can see that $y$ has only a maximum value of $y(1)=\frac{1}{12}$ on $(-1,2)$... | y(1)=\frac{1}{12} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,276 |
2. In the following expressions, “ $\bigcirc$ ” and “ $\square$ ” represent two different operation symbols. Think about which symbols can be filled in to make both equations true at the same time:
$$
1 \bigcirc 9 \bigcirc 1 \square 9=19 ; 8 \square 3 \square 3 \bigcirc 8 \bigcirc 3=83 \text {. }
$$ | $2.1+9+1 \times 9=19$;
$$
8 \times 3 \times 3+8+3=83 \text {. }
$$ | 1+9+1 \times 9=19; 8 \times 3 \times 3+8+3=83 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 701,278 |
3. Please select eight different digits to form four two-digit numbers, and make the sum of each two-digit number and the two-digit number obtained by reversing its digits a perfect square. Try to find these four two-digit numbers.
Translate the above text into English, please keep the line breaks and format of the so... | $3.56,74,83,92$. | 3.56, 74, 83, 92 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 701,279 |
Example $1$, $, $ Given line segments $b, c$. Find the line segment $a$ such that
$$
\frac{1}{a}=\frac{1}{b}+\frac{1}{c} .
$$ | Analysis: Consider $\mathrm{a}$ as $\mathrm{f}$, $\mathrm{b}$ as $\mathrm{u}$, and $\mathrm{c}$ as $\mathrm{v}$. Using the above physical construction method, we can construct $\mathrm{a}$.
Construction:
(1) Draw a line 1, take a point 0 on 1, and on both sides of 0 on 1, mark $O D=b$ and $O E=c$;
(2) Draw perpendicula... | a=\frac{bc}{b+c} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,283 |
Example 2. Given line segments b, $\mathbf{c}$. Find: line segment $a$, such that $\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$. | Analysis: Squaring both sides of $\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$ yields
$$
\frac{1}{a}=\frac{1}{b}+\frac{1}{c}+\frac{2}{\sqrt{b c}} \text {. }
$$
Let $d=\frac{\sqrt{b c}}{2}, \frac{1}{e}=\frac{1}{b}+\frac{1}{c}$, then $d, e$ are easy to construct, from which we have $\frac{1}{\mathrm{a}}=\fr... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,284 |
Example 4. Study the maximum and minimum values of the function $y=x^{2}-6 x+4 \ln x$ on $(0,2.5)$. | Since $y(+0)=-\infty, \therefore y$ has no minimum value on $(0,2.5)$. Also, $y^{\prime}=2 x-6+\frac{4}{x}=\frac{2\left(x^{2}-3 x+2\right)}{x}$
$$
=\frac{2(x-1)(x-2)}{x},
$$
the critical points are $x_{1}=1, x_{2}=2$. Comparing $y(1)=-5, y(2)=4 \ln 2 - 2 \doteq -5.2, y(2.5-0)$
$=y(2.5)=4 \ln \frac{5}{2}-\frac{35}{4}=$... | -5 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,287 |
1. $\frac{3+\sqrt{13}}{2}$ and $\frac{-3+\sqrt{13}}{2}$ are related as ( ).
A、opposites, B、reciprocals, $O$ 、negative reciprocals, D、none of the above. | 1. (B)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,294 |
2. When $b>0$, $\sqrt{-x^{3} b}=(\quad$.
$A 、-x \sqrt{x b}, B 、 x \sqrt{-x b}, C 、-x \sqrt{-x b}$,
$D, x \sqrt{x b}$. | 2. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,295 |
3. $\left[(-2)^{4}\right]^{-\frac{3}{4}}=(\quad)$ 。
A, \frac{1}{8}, B, -8, C, -\frac{1}{8}, D, 6 | 3. (A)
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,296 |
4. If $y=m x^{1}-m$ is an inverse proportion function, then the value of $m$ is ( ).
$A$ 、not equal to zero, $B 、 1, C 、 2, D 、 m$ can take any value and it is impossible to be an inverse proportion function. | 4. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 701,297 |
Example 5. Study the function $y=x+\frac{3}{4} \ln \frac{2-x}{2+x}$ for its maximum and minimum values on $(-2,2)$.
| Solve $\because x \rightarrow-2+0$ when $2+x \rightarrow+0, 2-x \rightarrow$
4, thus $\frac{2-x}{2+x} \rightarrow+\infty, \therefore y(-2+0)=+\infty$.
Similarly, when $x \rightarrow 2-0$, $\frac{2-x}{2+x} \rightarrow+0, \therefore y(2-0)$ $=-\infty$. Therefore, $y$ has no maximum or minimum value on $(-2,2)$ (Figure 7)... | not found | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,298 |
5. The quadrilateral formed by connecting the midpoints of the four sides of an isosceles trapezoid in sequence is ( ).
$\mathrm{A}$, parallelogram, $\mathrm{B}$, rectangle, $\mathrm{C}$, rhombus, $\mathrm{D}$, square. | 5. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 701,299 |
1. $\lg x=1+2 \lg 2+\frac{\lg 3}{2}-\lg \frac{1}{2}$, find $x$ | $\begin{array}{l}\text { 1. Solve } \lg x=\lg 10+\lg 4+\lg \sqrt{3}-\lg \frac{1}{2} \\ =\lg \frac{10 \times 4 \sqrt{3}}{\frac{1}{2}} \\ =\lg 80 \sqrt{3} . \\ x=80 \sqrt{3} .\end{array}$ | 80 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,300 |
3. Solve the equation $\left(2 x^{2}-3 x+1\right)^{2}=4 x^{2}-6 x$ $+53$ | 3. Solution The original equation can be transformed into
$$
\begin{array}{l}
\left(2 x^{2}-3 x+1\right)^{2}-2\left(2 x^{2}-3 x+1\right)-3=0, \\
\left(2 x^{2}-3 x+1-3\right)\left(2 x^{2}-3 x+1+1\right)=0, \\
\left(2 x^{2}-3 x-2\right)\left(2 x^{2}-3 x+2\right)=0 .
\end{array}
$$
Thus, we have $2 x^{2}-3 x-2=0$,
(1)
o... | x_{1}=-\frac{1}{2}, x_{2}=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,302 |
4. As shown in Figure 1,
Given that the circumcircle $O$ of $\triangle \mathrm{ABC}$ has diameter $A D=6 \mathrm{~cm}, \quad \angle B=$ $\angle \mathrm{DAC}$. Find the length of $\mathrm{AC}$. | 4. Solution As shown in Figure 1, connect $\mathrm{CD}$,
$$
\begin{array}{l}
\because \mathrm{AD} \text { is the diameter of } \odot \mathrm{O}, \\
\therefore \angle \mathrm{ACD}=90^{\circ} . \\
\because \angle \mathrm{B}=\angle \mathrm{D}, \angle \mathrm{B}=\angle \mathrm{DAC}, \\
\therefore \angle \mathrm{DAC}=\angle... | 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,303 |
Four. The distance between location A and location B is 120 kilometers. A car travels from location A to location B at a speed of 30 kilometers per hour. 1. Write the function relationship between the distance \( s \) (kilometers) the car is from location B and the time \( t \) (hours) since it left location A, and det... | 4. Solution:
1) The function relationship between $\mathrm{s}$ and $\mathrm{t}$ is $s=120-30 t$.
$$
(0 \leqslant \mathrm{t} \leqslant 4)
$$
2) Draw point A $(0, 120)$ and point B $(4,0)$,
then line segment $\mathrm{AB}$ is
the required graph (Figure 5).
From the graph, we can see: When $t=2.5$, $s=45$, that is, 2.5 ho... | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,304 |
Five. As shown in Figure 2, it is known that
$ABCD$ is a cyclic quadrilateral,
\[
\begin{array}{r}
\angle D=60^{\circ}, \quad AB=3 \text{ cm}, \\
BC=2 \text{ cm}, \text{ find the length of } AC. \\
\text{ (10 points) }
\end{array}
\] | Five. Solution (as shown in Figure 2)
\[
\because A B C D \text { is a cyclic quadrilateral,}
\]
\[
\begin{aligned}
\text { and } \angle \mathrm{D} & =60^{\circ}, \\
\therefore \angle \mathrm{B} & =120^{\circ} \\
\because \mathrm{AB} & =3 \mathrm{~cm}, \mathrm{BC}=2 \mathrm{~cm}, \\
\therefore & \mathrm{AC}^{2}=3^{2}+2... | \sqrt{19} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,305 |
Seven. As shown in Figure 4, given that $\mathrm{O}^{\prime}$ is any point on $\odot \mathrm{O}$, $\odot \mathrm{O}^{\prime}$ intersects $\odot \mathrm{O}$ at $\mathrm{A}, \mathrm{E}$, the diameter $\mathrm{AB}$ of $\odot \mathrm{O}$ intersects $\odot \mathrm{O}^{\prime}$ at $\mathrm{C}$, and $\mathrm{AD}$ is the diame... | Proof 1: Connect $\mathrm{BO}^{\prime}$,
$\because \mathrm{AB}$ is the diameter of $\odot \mathrm{O}$,
$$
\begin{array}{l}
\therefore \angle \mathrm{AO}^{\prime} \mathrm{B}=90^{\circ}, \mathrm{BO}^{\prime} \perp \mathrm{AD}^{\circ} \text { . } \\
\text { Also, } \because \mathrm{AO}^{\prime}=\mathrm{DO}^{\prime},
\end{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,307 |
Eight. For what value of $\mathrm{k}$, the roots of the equation $3\left(\mathrm{x}^{2}+3 \mathrm{x}+4\right)$ $=x(2-x)(2+k)$ are the sides of a right triangle
保持了源文本的换行和格式。 | Eight, Solution: The original equation is rearranged as
$$
(\mathrm{k}+5) \mathrm{x}^{2}-(2 \mathrm{k}-5) \mathrm{x}+12=0 \text {. }
$$
Let the two roots of the equation be $x_{1}$ and $x_{2}$, then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{2 k-5}{k+5}, \\
x_{1} \cdot x_{2}=\frac{12}{k+5} .
\end{array}
$$
Since $x_{1}$ a... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,308 |
Example 6. Study the maximum and minimum values of the function $y=5-36 x+3 x^{2}+4 x^{3}$ on $(-2,+\infty)$. | Solve $\because y=x^{3}\left(\frac{5}{x^{3}}-\frac{3 b}{x^{2}}+\frac{3}{x}+4\right)$, when $x \rightarrow+\infty$, $x^{3} \rightarrow+\infty, \frac{5}{x^{3}}-\frac{36}{x^{2}}+\frac{3}{x}+4 \rightarrow 4$, $\therefore \mathrm{y}(+\infty)=+\infty$ (1), so $\mathrm{y}$ has no maximum value on $(-2,+\infty)$. $\mathrm{y}(-... | -\frac{115}{4} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,309 |
$$
\begin{array}{l}
\text { 1. Given } A=\left\{x: 6 x^{2}-x-2 \geqslant 0\right\}, \\
B=\left\{x: 7 x^{2}-5 x<0\right\}, \quad I=R .
\end{array}
$$
Then $A \cap B=$ $\qquad$ $\triangle \overline{U B}=$ $\qquad$ | $$
\begin{array}{l}
\text { 1. } A \cap B=\left\{x: \frac{2}{3} \leqslant x<\frac{5}{7}\right\}, \\
\overline{A \cup B}=\left\{x:-\frac{1}{2}<x \leqslant 0\right\}
\end{array}
$$ | A \cap B=\left\{x: \frac{2}{3} \leqslant x<\frac{5}{7}\right\}, \overline{A \cup B}=\left\{x:-\frac{1}{2}<x \leqslant 0\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 701,310 |
$2 . \operatorname{tg} \frac{\pi}{8}-\operatorname{ctg} \frac{\pi}{8}$ 的值是 $\qquad$ -
The value of $2 . \operatorname{tg} \frac{\pi}{8}-\operatorname{ctg} \frac{\pi}{8}$ is $\qquad$ - | 2. $\operatorname{tg} \frac{\pi}{8}-\operatorname{ctg} \frac{\pi}{8}=-2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
2. $\operatorname{tan} \frac{\pi}{8}-\operatorname{cot} \frac{\pi}{8}=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,311 |
3. When $k=$ $\qquad$, the principal value of the argument of the complex number $\left(2 k^{2}-3 k-2\right)+$ $\left(k^{2}+k-6\right)$ i is $270^{\circ}$. | $3 . k=-\frac{1}{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in a mathematical form which is universal and does not require translation, the output remains the same:
$3 .... | k=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,312 |
Given that $a$, $b$, and $c$ are all positive numbers, prove:
$$
\left|\begin{array}{lll}
a & b & c \\
c & a & L \\
b & c & a
\end{array}\right| \geqslant 0 \text {. }
$$
(Equality holds if and only if $a=b=c$) | $$
\because\left|\begin{array}{lll}
a & b & c \\
c & a & b \\
b & c & a
\end{array}\right|=a^{3}+b^{3}+c^{3}-3 a b c,
$$
By the basic inequality, $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3} \geqslant 3 \mathrm{abc}$ (equality holds if and only if $a=b=c$)
$$
\begin{array}{l}
\therefore a^{3}+b^{3}+c^{3}-3 a b c \geq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,316 |
$$
\text { Three. Given } f(\theta)=a \sin \theta+b \cos \theta \text {, }
$$
if $\mathrm{f}\left(\frac{\pi}{6}\right)=\sqrt{3}$, and the maximum value of $\mathrm{f}(\theta)$
is 2, find the value of $\mathrm{f}\left(\frac{\pi}{3}\right)$.
| Three. Solution
Given $f\left(\frac{\pi}{6}\right)=\sqrt{3}$, we have
$$
\operatorname{asin} \frac{\pi}{6}+b \cos \frac{\pi}{6}=\sqrt{3},
$$
which simplifies to $a+\sqrt{3} b=2 \sqrt{3}$.
Also, $\mathrm{f}(\theta)=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}} \sin (\theta+\varphi)$, with the maximum value being $\sqrt{\mathrm{... | 2 \text{ or } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,317 |
IV. From 1981 to the end of this century, a total of twenty years, our country's industrial and agricultural total output value needs to quadruple, so what is the approximate average annual growth rate? | iv. Solution Let the average annual growth rate be $x$, then according to the problem,
$$
(1+x)^{20}=4 \text{. }
$$
Taking the logarithm with base 10 on both sides,
$$
\begin{array}{l}
20 \lg (1 \div x)=2 \lg 2 . \\
\lg (1+x)=\frac{1}{10} \lg 2=0.0301 . \\
1+x \approx 1.072 . \\
\therefore x \approx 0.072=7.2\% .
\end... | 7.2\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,318 |
Six. If the sum of the absolute values of the coefficients of the first three terms in the expansion of $\left(x^{18 x}-2\right)^{n}$ is 73, and its middle term is -160000, find the value of $x$.
| Six. The first three terms of the expansion are:
$$
\begin{array}{l}
x^{n 1 g x},-2 C_{n}^{1} x^{(n-1) 1 \varepsilon x}, \\
4 C_{n}^{2} x^{(n-2) 1 \varepsilon x} .
\end{array}
$$
From the condition, $1+2 \mathrm{n}+\mathrm{C}_{9}^{2}=73$,
simplifying gives $n^{2}=36$,
$$
\therefore \mathrm{n}= \pm 6
$$
(negative value... | x_1=10, x_2=\frac{1}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,321 |
Seven. Line segment $\mathrm{AB}$ is parallel to plane $\mathrm{M}$. From $\mathrm{A}$ and $\mathrm{B}$, perpendicular lines $\mathrm{AA}_{1}$ and $\mathrm{BB}_{1}$ are drawn to this line segment, making angles with plane $\mathrm{M}$ of $\beta$ and $\alpha(\alpha>\beta)$, respectively. Given $\mathrm{AB}=\mathrm{a}, \... | Seven. Solution Let C, Y be the projections of points A, B on plane M, respectively.
We discuss in two cases.
(1) AA_1 and BB_1 are on the same side of AB,
\because AA_1 \perp AB,
BB_1 \perp AB, AB \parallel plane M,
AB \perp AC, AB \perp BD,
\therefore AB \perp plane A_1CA, AB \perp plane B_1DB,
\therefore plane A_1CA... | \frac{\sqrt{b^2 - a^2} \sin \alpha \cdot \sin \beta}{\sin (\alpha - \beta)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,322 |
Eight. $\mathrm{P}_{1}, \mathrm{P}_{2}$ are any two points on a parabola, the tangents at these two points intersect at $P, F$ is the focus of the parabola, prove that $|\mathrm{PF}|^{2}=\left|\mathrm{P}_{1} \mathrm{~F}\right| \cdot\left|\mathrm{P}_{2} \mathrm{~F}\right|$. | Eight. Proof. Let the equation of the parabola be $\mathrm{y}^{2}=2 \mathrm{px}, \mathrm{P}_{1}$, $P_{2}, P$ have coordinates respectively $\left(x_{1}, y_{1}\right),\left(x_{2}\right.$, $\left.y_{2}\right),(a, b)$.
Thus, $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ satisfy the equations
$$
\left\{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,323 |
Nine. (Bonus Question) When $n \geqslant 3$, $x_{1}, x_{2}, \cdots$, $\mathrm{x}_{\mathrm{n}} \cdots\left(\mathrm{x}_{1} \neq 0\right)$ satisfy the condition:
$$
\begin{array}{r}
\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n-1}^{2}\right)\left(x_{2}^{2}+x_{3}^{2}\right. \\
\left.+\cdots+x_{n}^{2}\right)=\left(x_{1} x_{g}+x_{2}... | (1) When $n=3$, $\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{2}^{2}+x_{2}^{2}\right)$
$$
=\left(x_{1} x_{2} i - x_{2} x_{3}\right)^{2},
$$
i.e., $x_{2}^{4} + x_{1}^{2} x_{3}^{2} - 2 x_{1} x_{2}^{2} x_{3} = 0$,
$\therefore \quad \left(x_{2}^{2} - x_{1} x_{3}\right)^{2} = 0, \Rightarrow x_{2}^{2} = x_{1} x_{3}$.
$\therefor... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,324 |
Question 1. $O$ is the centroid of the equilateral $\triangle ABC$, and $D$, $E$, and $F$ are the midpoints of $OA$, $OB$, and $OC$, respectively. The regular triangular pyramid $\mathrm{V}-\mathrm{APC}$ and the regular triangular prism $\mathrm{DEF}-\mathrm{D}^{\prime} \mathrm{E}^{\prime} \mathrm{F}^{\prime}$, both wi... | To draw a three-dimensional figure, one must first study the intersection lines of the prism and pyramid. According to the problem, it is easy to know that the lateral edges of the prism $\mathrm{DD}^{\prime}, \mathrm{EF}, \mathrm{FF}^{\prime}$ and the corresponding lateral edges of the pyramid $\mathrm{AV}, B V, C V$ ... | \frac{1}{4} \sqrt{3 \mathrm{a}^{2}+9 \mathrm{~h}^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,325 |
Question 2. $E$ is the midpoint of edge $A' D'$ of a cube $ABCD$ $\qquad$ $A'B'C'D'$ with edge length $a$. Find the lengths of the two segments into which $EB$ and $EC$ are divided by the plane $AB'D'$. | To solve as shown in Figure 3, to discuss the two segments of $\mathrm{EB}$ divided by the plane $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$, we must study the intersection point of $\mathrm{EB}$ and the plane $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$. For this, we look at the plane $\mathrm{EBD}^{\prime}$, i.e., the i... | |\mathrm{OE}| = \frac{1}{2} a, |\mathrm{OB}| = a, |\mathrm{EQ}| = \frac{3}{10} a, |QC| = \frac{4}{5} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,326 |
Example 8. Study the maximum and minimum values of the function $y=\mathrm{x}^{-\frac{\mathrm{x}^{2}}{2}}$ on $(-\infty$, $+\infty)$. | Given, $\because y(-\infty)=y(+\infty)=0^{(2)}$, and $y^{\prime}=e^{-\frac{x^{2}}{2}}\left(1-x^{2}\right)$, the critical points are $x_{1}=-1$, $\mathrm{x}_{2}=1 \cdot \mathrm{y}(-1)=-\mathrm{e}^{-\frac{1}{2}}, \mathrm{y}(1)=\mathrm{e}^{-\frac{1}{2}}$, hence we know that $y$ on $(-\infty,+\infty)$ has a maximum value $... | y(1)=e^{-\frac{1}{2}}, y(-1)=-e^{-\frac{1}{2}} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 701,327 |
Example. Solve the following problems:
1. Without using tables, find the value of $\lg ^{3} 2+\lg ^{3} 5+3 \lg 2 \cdot \lg 5$;
2. Simplify: $\frac{1-\operatorname{tg} \theta}{1+\operatorname{tg} \theta}$;
3. Solve the system of equations $\left\{\begin{array}{l}\frac{x}{y}+\frac{y}{x}=\frac{25}{12}, \\ x^{2}+y^{2}=7\en... | 1. Original expression $=\mathbf{t g}^{3} 2+\lg ^{3} 5+3 \lg 2 \cdot 1 \mathrm{~g} 5$
$$
\begin{aligned}
& (\lg 2+\operatorname{Ig} 5) \\
= & (\lg 2+\lg 5)^{2}=1,
\end{aligned}
$$
2. Original expression $=\frac{\operatorname{tg} 45^{\circ}-\operatorname{tg} \theta}{1+\operatorname{tg} 45^{\circ} \operatorname{tg} \thet... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,328 |
Example. Given the coordinates of points $A$ and $B$ are $(-2,4)$
and $(8,-2)$, find the circle with $AB$ as its diameter. | Solution 1: Let the center of the circle to be found be $O(x, y)$. Since $A(-2,4)$ and $B(8,-2)$ are the endpoints of the diameter of the circle, it is not difficult to find the coordinates of the center. Because the center is the midpoint of $A$ and $B$, we have
$$
x=\frac{-2+8}{2}=3, y=\frac{4-2}{2}=1 \text {. }
$$
... | x^{2}+y^{2}-6 x-2 y-24=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,329 |
For example, $1, \mathrm{a}, \mathrm{b}$ are skew lines, $\mathrm{a} \perp$ plane $\alpha, \mathrm{b}$ has what positional relationship with $\alpha$? | (Ans: $\mathrm{b}$ may be parallel to $\alpha$, may intersect $\alpha$ obliquely, or $b \subset \boldsymbol{\alpha}$. ) | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,330 |
Example 8. For the cube $\mathrm{ABCD}-\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$, with edge length $\mathrm{a}$, and the centers of the two bases being $\mathrm{O}, \mathrm{O}_{1}$, find the following distances:
(1) A to $\mathrm{B}_{1} \mathrm{D}_{1}$, (2) $\mathrm{A}$ to $\mathrm{A}_{1} \mathrm{C}... | (1) $\frac{\sqrt{6}}{2} \mathrm{a}$, (2)
$\frac{\sqrt{6}}{3} \mathrm{a}$,
(3) $\frac{\sqrt{2}}{2} \mathrm{a}$,
(4) $\frac{\sqrt{3}}{3} \mathrm{a}$,
(5) $\frac{\sqrt{3}}{3} \mathrm{a}$,
(6) $90^{\circ}$, (7) $60^{\circ}$, (8) $\operatorname{arctg} \sqrt{2}$, (9)
$45^{\circ},(10) \operatorname{arctg} \sqrt{2}.$ | \frac{\sqrt{6}}{2} \mathrm{a}, \frac{\sqrt{6}}{3} \mathrm{a}, \frac{\sqrt{2}}{2} \mathrm{a}, \frac{\sqrt{3}}{3} \mathrm{a}, \frac{\sqrt{3}}{3} \mathrm{a}, 90^{\circ}, 60^{\circ}, \operatorname{arctg} \ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,331 |
Example 3. Prove that the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{a^{2}}$ $-\frac{\dot{y}^{2}}{b^{2}}=1$ do not have any common tangent lines that are not parallel to the $\mathrm{y}$-axis. | Proof (Proof by Contradiction)
Assume the two have a common tangent line not parallel to the $y$-axis, $y = kx + m$, then
$$
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 . \\
y=k x+m ; \\
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1, \\
y=k x+m .
\end{array}\right.
\end{array}\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,332 |
Example 4. As shown in the figure, let the centroid, incenter, and circumcenter of $\triangle \mathrm{ABC}$ be $M, I, O$ respectively, and $M I \parallel A C$. Prove:
(1) $\operatorname{ctg} \frac{\mathrm{A}}{2}, \operatorname{ctg} \frac{B}{2}, \operatorname{ctg} \frac{\mathrm{C}}{2}$ form an arithmetic sequence;
(2) $... | To prove that $\operatorname{ctg} \frac{A}{2}$, $\operatorname{ctg} \frac{B}{2}$, $\operatorname{ctg} \frac{C}{2}$ can form an arithmetic sequence, we need to prove that $\operatorname{ctg} \frac{A}{2} + \operatorname{ctg} \frac{C}{2} = 2 \operatorname{ctg} \frac{B}{2}$. By the half-angle theorem:
$$
\begin{array}{l}
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,333 |
Example 3. Let $0<x<\frac{\pi}{2}$, prove that $\frac{2}{\pi} x<\sin x$. | Proof: Let $\phi(x)=\frac{2}{\pi} x-\sin x$, then
$$
\phi^{\prime}(x)=-\frac{2}{\pi}-\cos x .
$$
Since $0 < \cos x < 1$, $\phi^{\prime}(x)$ changes from negative to positive within $\left(0, \frac{\pi}{2}\right)$, thus $\phi(x)$ first decreases and then increases. However, as $x \rightarrow 0$, $\phi(x) \neq 0$, and a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,336 |
$$
\begin{array}{l}
\text { When } x>0 \text {, } e^{x}>1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!} ; \\
\text { When } x<0 \text {, } e^{x}<1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!} \text {. }
\end{array}
$$ | Proof: Let $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$,
$$
g(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!},
$$
then $f^{\prime}(x)=e^{x}, g^{\prime}(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}$,
$$
\begin{array}{l}
f^{\prime \prime}(x)=e^{x}, g^{\prime \prime}(a)=1+x+\frac{x^{2}}{2!} \\
f^{\prime \prime}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,339 |
Example 7. $\mathrm{n}$ is a natural number, prove that
$$
\begin{array}{l}
2(\sqrt{n+1}-1)<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} \\
\leqslant 2 \sqrt{n}-1 .
\end{array}
$$ | Proof: Let $\mathrm{y}=\frac{1}{\sqrt{x}}, S_{0}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+$
$$
\frac{1}{\sqrt{n}} \text {. }
$$
First, prove $2(\sqrt{n+1}-1)\int_{1}^{n+1} \frac{1}{\sqrt{x}} d x=2(\sqrt{n+1}-1) .
$$
Then, prove $S . \leqslant 2 \sqrt{\mathrm{n}}-1$.
$S$. is equal to the sum of the areas of $n$ re... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,340 |
Example 9. In the dihedral angle $\alpha-\mathrm{a}-\beta$ of $120^{\circ}$, $\mathrm{A} \in \alpha, \mathrm{B} \in$ $\beta$, it is known that the distances from points $A$ and $B$ to the edge $a$ are 2 and 4, respectively, and $A B=$ 10. Find the angle formed by the line $\mathrm{AB}$ and the edge $\mathrm{a}$, and th... | (Answer: The angle between $\mathrm{AB}$ and edge a is $\arcsin \frac{\sqrt{7}}{5}, AB$ and $\beta$ is $\arcsin \frac{\sqrt{3}}{10}$.)
| \arcsin \frac{\sqrt{7}}{5}, \arcsin \frac{\sqrt{3}}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,342 |
Example 9. $n$ is a natural number greater than 1, prove that
$$
\begin{array}{l}
\frac{3 n+1}{2 n+2}<\left(\frac{1}{n}\right)^{n}+\left(\frac{2}{n}\right)^{n}+\cdots \\
+\left(\frac{n}{n}\right)^{n} < \frac{2 n+1}{n+1} .
\end{array}
$$ | Proof: Let $\mathrm{y}=\mathrm{x}^{\circ}(\mathrm{x}>0)$,
$$
S_{n}=\left(\frac{1}{n}\right)^{n}+\left(\frac{2}{n_{0}}\right)^{n}+\cdots+\left(\frac{n}{n}\right)^{n} \text {. }
$$
Divide the interval $[0,1]$ into $n$ equal parts, $x_{k}$ represents the $k$-th division point,
then $x_{k}=\frac{k}{n}, x_{k+1}=\frac{k+1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 701,343 |
2. Please fill in the four boxes below with the numbers $1$, $9$, $8$, and $3$. How should you fill them to get the maximum product? How should you fill them to get the minimum product?
$\square \square \times \square \square=$ ? | \begin{aligned} \text { 2. } & \text { Maximum: } 91 \times 83=7553 \text {, } \\ \text { Minimum: } & 18 \times 39=702 .\end{aligned} | 7553 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 701,345 |
3 . Divide the nine digits 1 - 9 into two groups: $1, 9, 8, 3$ and 2,
$4, 5, 6, 7$, and fill them into the following equation to make it true:
$$
\frac{\square \square}{\square \square}=\frac{O+O+O}{O+O} \text {. }
$$ | 3. $\frac{38}{19}=\frac{4+5+7}{2+6}$. | \frac{38}{19}=\frac{4+5+7}{2+6} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 701,346 |
6. Da Guang read a book, the number of pages read on the first and second days are two two-digit numbers composed of the digits $1$, $9$, $8$, and $3$, and the number of pages read on the second day is exactly equal to the sum of these four digits more than the first day. How many pages did he read on the first and sec... | 6. On the first day, read 18 pages, on the second day, read 39 pages,
$$
39-18=1+9+8+3 \text {. }
$$ | 39-18=1+9+8+3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 701,349 |
1. If $a, b, c$ are all real numbers, then $\frac{a^{2}+b^{2}+c^{2}}{3} \geqslant$
$$
\left(\frac{a+b+c}{3}\right)^{2}
$$ | Slightly prove: $\because(a-b)^{2} \geqslant 0,(b-c)^{2} \geqslant 0$,
$$
\begin{array}{c}
(c-a)^{2} \geqslant 0, \\
\therefore \frac{2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a}{9} \geqslant 0,
\end{array}
$$
Thus, $\frac{3\left(a^{2}+b^{2}+c^{2}\right)}{9} \geqslant \frac{(a+b+c)^{2}}{9}$,
i.e., $\frac{a^{2}+b^{2}+c^{... | \frac{a^{2}+b^{2}+c^{2}}{3} \geqslant \left(\frac{a+b+c}{3}\right)^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 701,350 |
Example 1, find the tangent line equation at a point $\left(x_{0}, y_{0}\right)$ on a general quadratic curve (non-degenerate case);
$$
\mathrm{Ax}^{2}+\mathrm{Bxy}+\mathrm{Cy}^{2}+\mathrm{Dx}+\mathrm{Ey}+\mathrm{F}=0
$$ | Solve: Express $y$ as an implicit function of $x$, differentiate both sides of equation (5) with respect to $x$, we get
$$
\begin{array}{l}
2 A x + B y + B x y' + 2 C y y' + D + E y' = 0, \\
\text { i.e., } y' = \frac{-(2 A x + B y + D)}{2 C y + B x + E}, \\
\left.\therefore \quad y'\right|_{(x_0, y_0)} = -\frac{2 A x_... | A x_0 x + \frac{B}{2} (x_0 y + y_0 x) + C y_0 y + \frac{D}{2} (x + x_0) + \frac{E}{2} (y + y_0) + F = 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,353 |
1. Let $x, y$ be positive real numbers, and $x+y=1$. Prove: $\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2} \geqslant \frac{25}{2}$. | $$
\begin{array}{l}
\because \frac{\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2}}{2} \\
\geqslant\left(\frac{\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)}{2}\right)^{2} \\
=\left(\frac{1+\frac{1}{x}+\frac{1}{y}}{2}\right)^{2}
\end{array}
$$
$$
\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\... | \left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2} \geqslant \frac{25}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 701,354 |
Example 1. Prove:
$$
\sum_{i=1}^{n} \frac{(i-1)!}{(p+i)!}=\frac{1}{p}\left(\frac{1}{p!}-\frac{n!}{(n+p)!}\right)
$$
( $n, p$ are natural numbers). | Prove $\because C_{p+1}^{p+1}=\frac{(p+i)!}{(p+1)!(i-1)!}$,
$$
\therefore \frac{(i-1)!}{(p+i)!}=\frac{1}{(p+1)!C_{p+i}^{p+1}}
$$
By formula (2), we get
$$
\begin{array}{l}
\sum^{n} \frac{(i-1)!}{(p+i)!} \\
i=1 \\
=\frac{1}{(I)}\left(\frac{1}{C_{p+1}^{p+1}}+\frac{1}{C_{p+2}^{p+1}}+\cdots+\frac{1}{C_{p+n}^{p+1}}\right) ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,360 |
Example 2. Prove: $\frac{1}{1 \cdot 2 \cdots \mathrm{r}}+\frac{1}{2 \cdot 3 \cdots(\mathrm{r}+1)}$
$$
\begin{array}{c}
\quad+\frac{1}{3 \cdot 4 \cdots(r+2)}+\cdots+\frac{1}{n(n+1) \cdots(n+r-1)} \\
=\frac{1}{(r-1) \cdot(r-1)!}-\frac{1}{(r-1)(n+1) \cdots(n+r-1)},
\end{array}
$$
Here $\mathrm{r}$ is a positive integer n... | Prove that
$$
n(n+1)(n+2) \cdots(n+r-1)=A_{i+n-1}=r!C_{n+n-1}^{n} \text{, }
$$
Therefore, using formula (2), we get
$$
\begin{array}{l}
\frac{1}{1 \cdot 2 \cdots r}+\frac{1}{2 \cdot 3 \cdots(r+1)}+\frac{1}{3 \cdot 4 \cdots(r+2)}+\cdots \\
+\frac{1}{n(n+1) \cdots(n+r-1)} \\
=\frac{1}{r!C_{i}^{r}}+\frac{1}{r!C_{i+1}}+\f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 701,361 |
Example 1. Given: $\mathrm{ABCD}$ is a cyclic quadrilateral. Prove: $\mathrm{AC} \cdot \mathrm{BD}=\mathrm{AB} \cdot \mathrm{CD}+\mathrm{AD} \cdot \mathrm{BC}$. | Analysis: If a point $P$ can be found on $B D$ and makes $A G$
$$
(B P+P D)=A B \cdot C D+
$$
$AD \cdot BC$, in the equation, $AC \cdot BP=AB$
- $CD, AC \cdot PD=AD \cdot BC$,
it means the selection of $P$ should transform the equation to be proven into two basic
proportional equations.
From $AC \cdot BP = AB \cdot CD... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,362 |
$\begin{array}{l}\text { Example } 2 \text {. Given: } \mathrm{AD} \text { is the angle bisector of } \angle \mathrm{A} \text { in } \triangle \mathrm{ABC}. \\ \text { Prove: } \mathrm{AD}^{2}=\mathrm{AB} \cdot \mathrm{AC} \\ -\mathrm{BD}^{2} \cdot \mathrm{DC} .\end{array}$ | Analysis: The equation to be proven can be written as $\mathrm{AD} \cdot \mathrm{AD}=\mathrm{AB} \cdot \mathrm{AC} - \mathrm{BD} \cdot \mathrm{DC} (\cdot)$. This equation can be viewed as the difference of two ratios. If we can express $\mathrm{AD}$ as the difference of two line segments, it might be helpful for solvin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,363 |
Example 2. Given, a moving point $\mathrm{P}(\mathrm{X}, \mathrm{Y})$ draws two tangent lines to the parabola $\dot{y}^{2}=2 \mathrm{ps}(\mathrm{p}>0)$ which are perpendicular to each other, find the equation of the locus of point $\mathrm{P}$.
The above text is translated into English, preserving the original text... | $$
\left(x_{1}, y_{1}\right) 、\left(x_{2}, y_{2}\right),
$$
It is known that neither of the two tangent points is the origin (Figure 1).
From $y^{2}=2 p x$, we get $2 y y^{\prime} = 2 p, y^{\prime}=\frac{p}{y}$. Since the tangent line passes through point $P(x, y)$, from equation (4), we have
$$
\left\{\begin{array}{l... | X = -\frac{p}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,364 |
Example 4. Given: As shown in Figure 4, $\mathbf{P}$ is any point inside $\triangle \mathrm{ABC}$.
Prove: $\frac{P D}{A D}+\frac{P E}{B E}$ $+\frac{P F}{C F}=1$.
| Analysis: The equation to be proved can be transformed into
$$
\frac{\mathrm{PE}}{\mathrm{BE}}+\frac{\mathbf{P F}}{\mathrm{CF}}=1-\frac{\mathrm{PD}}{\mathrm{AD}}=\frac{\mathrm{Ar}}{\mathrm{AD}} \text {. }
$$
If we try to express AP as the sum of two segments, for example, by taking Q on AP, we have AP = AQ + QP. Thus,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,365 |
Example 3. Given: As shown in Figure 3, $D$ is a point on side $AB$ of $\triangle ABC$, $E$ is a point on the extension of $AC$, and $DE$ intersects $BC$ at $F$. Prove: $\frac{AC}{CE}=\frac{AB \cdot DF}{BD \cdot FE}$. | Analysis: We imagine a line segment $\mathrm{x}$, such that $\frac{\mathrm{AC}}{\mathrm{x}}=\frac{\mathrm{AB}}{\mathrm{BD}}$, $\frac{\mathrm{CE}}{\mathbf{x}}=\frac{\mathrm{FE}}{\mathrm{DF}}$. If the line segment $\mathrm{x}$ is found, then the entire proof is solved. To make $\mathrm{BD}$ and $\mathrm{x}$ form a triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,366 |
Example 1. Given that the following three non-parallel lines intersect at a point, find the value of $\mathbf{k}$;
$$
3 x+2 y+9 k=0,2 x+y-5=0, x+k y+2=0 .
$$ | Given that the three lines intersect at a common point,
$$
\therefore\left|\begin{array}{rrr}
3 & 2 & 9 k \\
2 & 1 & -5 \\
1 & k & 2
\end{array}\right|=0,
$$
which simplifies to $3 k^{2}+k-2=0$,
$$
\therefore k_{1}=\frac{2}{3}, k_{2}=-1 \text {. }
$$
When $k=\frac{2}{3}$, the line $3 x+2 y+9 k=0$ coincides with the l... | k=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 701,367 |
Example 2. Prove: The three medians of a triangle intersect at one point, and the distance from this point to a vertex is twice the distance from this point to the midpoint of the opposite side. | As shown in Figure 1, with D as the origin BC, the original axis is the $x$-axis, forming a rectangular coordinate system.
Thus, we have
$$
\begin{array}{l}
\mathrm{B}(0,0), \mathrm{C}(\mathrm{a}, 0), \mathrm{A}(\mathrm{c}, \mathrm{d}), \\
\mathrm{D}\left(\frac{\mathrm{a}}{2}, 0\right), \mathrm{E}\left(\frac{\mathrm{a}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,368 |
Example 3. Given that $O$ is a point inside $\triangle \mathrm{ABC}$, and the lines connecting $O$ to the three vertices of $\triangle \mathrm{ABC}$ are $\mathrm{AO}, \mathrm{BO}, \mathrm{CO}$, respectively. Their extensions intersect the opposite sides at $X, Y, Z$. Prove that $\frac{B X}{X C} \cdot \frac{C Y}{Y A} \c... | Proof As shown in Figure 2, establish a Cartesian coordinate system, where the coordinates of each point are A (h, k), B (b, 0), C (c, 0), X (0,0). Let Z divide AB in the ratio λ. Then the coordinates of Z are
$$
\left(\frac{\mathbf{h}+\lambda b}{1+\lambda}, \frac{\mathbf{k}}{1+\lambda}\right) \text {. }
$$
Let Y divi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,369 |
Find the locus equation of points whose sum of distances to the fixed points $F_{1}(-c, 0), F_{2}(c, 0)$ is $2 \mathrm{a}$ $(\mathrm{a}>c>0)$ | Let $M(x, y)$ be any point on the trajectory, and its distances to $\mathrm{F}_{1}, F_{1}$ are $r, r_{2} x$ respectively. $\mathrm{F}_{1}$ is the center with radius $r_{1}$, and $\mathrm{F}_{2}$ is the center with radius $\mathrm{r}_{2}$ of two circles (as shown in the figure). Therefore,
$$
\left\{\begin{array}{l}
(x+... | \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,370 |
Find the locus equation of points whose absolute value of the difference of distances to the fixed points $F_{1}(-c, 0), F_{2}(c, 0)$ is $2a$ $(c>a>0)$. | Let $M(x, y)$ be any point on the trajectory, we can obtain the system of equations
$$
=\left\{\begin{array}{l}
(x+c)^{2}+y^{2}=r_{1}^{2} \\
(x-c)^{2}+y^{2}=r_{2}^{2}, \\
\left|r_{1}-r_{2}\right|=2 a .
\end{array}\right.
$$
Thus, $\left\{\begin{array}{l}r_{1}+r_{2}=\frac{2 c x}{ \pm a} \\ r_{1}-r_{2}= \pm 2 a\end{arra... | \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,371 |
Find the locus equation of points that are equidistant from the fixed line $1: x=-\frac{p}{2}$ and the fixed point $F\left(\frac{p}{2}, 0\right)$. | Let $M (x, y)$ be any point on the trajectory, then it must lie on the line at a distance of $\mathrm{d}$ from 1 and on the circle centered at $\mathrm{F}$ with radius $\mathrm{d}$. Therefore,
$$
\left\{\begin{array}{l}
x=d-\frac{p}{2} \\
\left(x-\frac{p}{2}\right)^{2}+y^{2}=d^{2}
\end{array}\right.
$$
Eliminating the... | y^{2}=2 p x | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 701,372 |
Example 1. Using the sides $\mathrm{BC}$ and $\mathrm{BA}$ of $\triangle \mathrm{ABC}$ as the legs, and $B$ as the right-angle vertex, construct isosceles right triangles $\triangle P B C$ and $\triangle Q B A$ on the same side as the interior of $\triangle \mathrm{ABC}$. Try to prove: $\mathbf{P A} \mid \mathrm{QC}$ | Proof As shown in Figure 1, with
$\mathrm{B}$ as the pole and the extension of $\mathrm{BC}$
as the polar axis, establish a polar coordinate system.
Let $\mathrm{A}\left(\rho_{1}, \theta\right) 、 \mathrm{C}\left(\rho_{2}\right.$,
$$
\begin{array}{l}
\left.00^{\circ}\right), \text { then } \mathbf{P}\left(\rho_{2}, 90^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 701,373 |
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