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Example 5. The hypotenuse of a right-angled triangle is a fixed length c. Prove that its perimeter is maximized when it is an isosceles triangle.
保留源文本的换行和格式，直接输出翻译结果。
Example 5. The hypotenuse of a right-angled triangle is a fixed length c. Prove that its perimeter is maximized when it is an isosceles triangle. | Prove that, as shown in (*), from (1),
the perimeter $\mathrm{l}=\mathrm{c}+\mathrm{a}+\mathrm{b} \leqslant \mathrm{c}+2 \sqrt{\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{2}}$
$$
=c+2 \sqrt{\frac{\mathrm{c}^{2}}{2}}=(\sqrt{2}+1) \mathrm{c}
$$
The equality in the above inequality holds if and only if
$$
\mathrm{a}=\mathrm{b}, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,248 |
Example 6. Given the perimeter of a rectangle is $2 \mathrm{p}$, and the radius of the circumscribed circle is $\mathrm{R}$.
Prove: $1<\frac{p}{2} R$ $\leqslant \sqrt{2}$ | Prove that, as shown in the figure, according to the problem, $\mathrm{a}+\mathrm{b}=\mathrm{p}$.
From (1) in (*), $\frac{\mathrm{p}}{2 \mathrm{R}}=\frac{\mathrm{a}+\mathrm{b}}{2 \mathrm{R}}$
$$
\leqslant \frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{2}}=\frac{1}{\mathrm{R}} \sqrt{\frac{4 \mathrm{R}^{... | 1<\frac{\mathrm{p}}{2 \mathrm{R}} \leqslant \sqrt{2} | Geometry | proof | Yes | Yes | cn_contest | false | 702,249 |
Example 7. Given: $\alpha \in\left(0, \frac{\pi}{2}\right)$. Prove:
$$
\left(1+\frac{1}{\sin \alpha}\right)\left(1+\frac{1}{\cos \alpha}\right) \geqslant 3+2 \sqrt{2} .
$$ | Proof: First, prove $\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha} \geqslant 2 \sqrt{2}$.
Given $\alpha \in\left(0, \frac{\pi}{2}\right)$, we know: $\sin \alpha>0$, $\cos \alpha>0, \sin 2 \alpha>0$.
$$
\begin{array}{l}
\therefore \frac{1}{\sin \alpha}+\frac{1}{\cos \alpha} \geqslant 2 \sqrt{\frac{1}{\sin \alpha} \cos \al... | 3+2 \sqrt{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,250 |
Example 1. Given: Rt $\triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}$.
Prove: $\sqrt{1-\sin \mathrm{A}}+\sqrt{1-\sin \mathrm{B}}$
$$
+\sqrt{\frac{3}{2}-\sin A-\sin B}>2-\frac{\sqrt{3}}{2} .
$$ | $$
\begin{array}{l}
\text { Given } \mathrm{A}=90^{\circ}-\mathrm{B}, \sin \mathrm{A}=\cos B, \cos \mathrm{A}=\sin \mathrm{B}, \text { according to inequality (2), } \\
2 \sqrt{2}, \\
\sqrt{\sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}} \\
+\sqrt{1-2 \sin \mathrm{A}+\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}} \\
+\sqrt{... | 2-\frac{\sqrt{2}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,256 |
Example 5. Draw the graph of $\mathrm{y}=\log _{2}(x+2)+3$.
Make the graph of $\mathrm{y}=\log _{2}(x+2)+3$. | Solve: Write the original equation as $y-3=\log _{2}(x+2)$. In the XOY coordinate system, draw the graph of $\mathrm{y}=\log _{2} \mathrm{x}$, then translate this graph to $(-2,3)$, which gives the graph of $\mathrm{y}-3=\log _{2}(x+2)$, or $\mathrm{y}=\log _{2}(\mathrm{x}+2) +3$.
Since moving the graph to draw a new ... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,257 |
Example 2 - Given: $\triangle \mathrm{ABC}$ is an acute triangle, $a, b, c$ are the three sides, and $R$ is the circumradius. Prove:
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}+c^{2}+\sqrt{(a-2 R)^{2}+b^{2}+c^{2}}} \\
+\sqrt{a^{2}+(b-2 R)^{2}+c^{2}} \\
+\sqrt{a^{2}+b^{2}+(c-2 R)^{2}} \\
+\sqrt{(a-2 R)^{2}+(b-2 R)^{2}+c^{2}}... | Proof: By the Law of Sines: $\mathrm{a}=2 \mathrm{R} \sin \mathrm{A}$, $b=2 R \sin B, c=2 R \sin C$, the left side of the inequality
$$
\begin{array}{l}
=2 R \sqrt{\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}+\sin ^{2} \mathrm{C}} \\
+2 R \sqrt{(1-\sin \mathrm{A})^{2}+\sin ^{2} \mathrm{~B}+\sin ^{2} \mathrm{C}} \\
+2 R ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,258 |
Theorem 2 Let the sequence $\left\{a_{n}\right\}$ satisfy $A a_{k+2} + \mathrm{B}a_{\mathrm{k}+1} + \mathrm{C}a_{\mathrm{k}}=0$ (where $\mathrm{AC} \neq 0, \mathrm{~A}, \mathrm{~B}, \mathrm{C}$ are constants, $\mathrm{k}=1,2,3, \cdots). \mathrm{x}_{1}, \mathrm{x}_{2}$ are the roots (real or complex) of the equation $\m... | Proof: From Theorem 1, the sequence $\left\{a_{n+1}-x_{1} a_{n}\right\}$ is a geometric sequence with common ratio $x_{2}$,
$$
\therefore a_{n+1}-x_{1} a_{n}=\left(a_{2}-x_{1} a_{1}\right) x_{2}^{n-1}.
$$
Similarly, $a_{n+1}-x_{2} a_{n}=\left(a_{2}-x_{2} a_{1}\right) x_{1}^{n-1}$. (2)
(i) When $x_{1} \neq x_{2}$, from... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,260 |
Example 1. Let the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ satisfy $\mathrm{a}_{1}=\mathrm{a}_{2}=1$, $a_{k+2}=a_{k+1}+a_{k} \cdot(k=1,2,3, \cdots)$ Find $a_{n}$. | Solving the equation $\mathrm{x}^{2}-\mathrm{x}-1=0$, we get $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. According to Theorem 2, the general term is $\mathrm{a}_{\mathrm{n}}=\mathrm{A}_{0}\left(\frac{1+\sqrt{5}}{2}\right)^{\mathrm{n}-1}$
$$
\begin{array}{l}
+\mathrm{B}_{0}\left(\frac{1-\sqrt{5}}{2}\right)... | a_{n}=-\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,261 |
Example 2. The sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ satisfies the conditions $\mathrm{a}_{1}=1$,
$$
\begin{array}{c}
\mathrm{a}_{2}=-\frac{3}{2}, 4 \mathrm{a}_{\mathrm{k}+2}-12 \mathrm{a}_{\mathrm{k}+1}+9 \mathrm{a}_{\mathrm{k}}=0 . \\
(\mathrm{k}=1,2,3, \cdots) \text { Find the general term } \mathrm{a}_{... | Solving $4 x^{2}-12 x+9=0$, we get $x_{1}=x_{2}=\frac{3}{2}$. By Theorem 2, let $a_{n}=\left(A_{0}+n B_{0}\right)$
- $\left(\frac{3}{2}\right)^{n-1}$, from $a_{1}=1, a_{2}=-\frac{1}{2}$ we get the system of equations
$$
\left\{\begin{array}{l}
\mathrm{A}_{0}+\mathrm{B}_{0}=1, \\
\frac{3}{2} \mathrm{~A}_{0}+3 \mathrm{~B... | a_{n}=(3-2 n)\left(\frac{3}{2}\right)^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,262 |
Example 1. In a plane, there are $\mathrm{n}$ lines $(\mathrm{n} \in \mathrm{N})$, among which no two are parallel, and no three intersect at the same point. Question: How many regions does these $\mathrm{n}$ lines divide the plane into? | Let's assume that $k$ lines divide the plane into $f(k)$ regions. Then, $(k+1)$ lines will divide the plane into $f(k+1)$ regions.
The $(k+1)$-th line $l_{k+1}$ intersects with the original $k$ lines, creating $k$ intersection points. These $k$ points divide the line $l_{k+1}$ into $(k+1)$ segments, and each segment d... | \frac{1}{2}\left(n^{2} + n + 2\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,264 |
Example 6. Prove that the locus of the foci of the curve system $\mathrm{y}^{2}-4 \mathrm{x}+4 \mathrm{my}$ $+4 \mathrm{~m}^{2}+4 \mathrm{~m}=0$ is a straight line, and find its equation. | The original equation can be written as $(y-2 m)^{2}=4 (x$ $-\mathrm{m}) .(\mathrm{m}, 2 \mathrm{~m})$ is the translation increment of $\mathrm{y}^{2}=4 \mathrm{x}$. The focus of $y^{2}=4 x$ is $(1,0)$,
so the translation trajectory of $(1,0)$ is $(1+\mathrm{m}$, $2 \mathrm{~m})$, i.e.,
$$
\left\{\begin{array}{l}
x=1+... | 2x-y=2 | Algebra | proof | Yes | Yes | cn_contest | false | 702,268 |
Example 2. Given n spheres, each pair of which intersects in a circle. How many regions do these n spheres divide the space into?
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。
Note: The last sentence is a repetition of the instruction and should not be included in the translated content. Here is the final version:
Example 2. ... | Consider the following problem first:
On a sphere, there are $\mathrm{n}$ circles, and every two circles have two common points. Suppose such $\mathrm{k}$ circles divide the sphere into $\mathrm{f}(\mathrm{k})$ regions, then $(\mathrm{k}+1)$ such circles divide the sphere into $\mathrm{f}(\mathrm{k}+1)$ regions. The $(... | \frac{1}{3} n\left(n^{2}-3 n+8\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,269 |
The article in the fifth issue of our journal in 1983, titled "A Simple Method for Compiling Radical Equations," states that the equation
$$
\sqrt{5 \mathrm{x}-1}+\sqrt{2 \mathrm{x}}=3 \mathrm{x}-1
$$
"will produce a quartic equation after squaring twice, which may be quite troublesome to solve." In fact, this equation... | Solve the equation $3 x-1=(5 x-1)-2 x$
$=(\sqrt{5 x-1}+\sqrt{2 x})$
- $(\sqrt{5 x-1}-\sqrt{2 x})$
and $\sqrt{5 x-1}+\sqrt{2 x}>0$, from (1) we can get
$$
\sqrt{5 x-1}-\sqrt{2 x}=1 \text {. }
$$
This indicates $\sqrt{5 x-1} \geqslant 1$, i.e., $x \geqslant \frac{2}{5}$.
$$
\begin{array}{l}
\text { From (1)+(2) we get ... | x=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,270 |
Example. Solve the equations
(a) $\sqrt{3 \mathrm{x}+1}=\sqrt{2 \mathrm{x}-1}+1$;
(b) $2 \sqrt{x-1}=\sqrt{x+4}+1$. | Solve (a) The original equation is
$$
\sqrt{3 x+1}-\sqrt{2 x-1}=1 \text {. }
$$
Multiplying both sides by $\sqrt{3 x+1}+\sqrt{2 x-1}$ gives
$$
\begin{array}{l}
\sqrt{3 x+1}+\sqrt{2 x-1}=x+2 . \\
\text { (1) }+(2): 2 \sqrt{3 x+1}=x+3 . \\
\left\{\begin{array}{l}
x^{2}-6 x+5=0, \\
x>-\frac{1}{3} .
\end{array}\right.
\en... | x=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,271 |
Let $\mathrm{ABCD}$ be a square, $\mathrm{M}$ is the midpoint of $\mathrm{AB}$, $\mathrm{MN} \perp \mathrm{DM}$, $\mathrm{BN}$ bisects the exterior angle of $\angle \mathrm{ABC}$, prove by analytic method that $|\mathrm{MD}|=|\mathrm{MN}|$. (High School Self-Examination Question 5, Issue 2, 1983, Intermediate Mathemati... | I proved it as follows: As shown in Figure 1, connect DB and DN. It is easy to know that $\angle \mathrm{DBN}=45^{\circ}+45^{\circ}=$ $90^{\circ}$. Since $M N \perp D M$, points $D$, $M$, $B$, and $N$ are concyclic, so $\angle 1=\angle 2=45^{\circ}$. Therefore, $\triangle D N N$ is an isosceles right triangle, which me... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,272 |
In the article "Four Methods for Finding a Class of Linear Equations" in Issue 5, 1983 of *Intermediate Mathematics*, an example is given: "A line is drawn through a point $(1, -2)$ inside the conic section $C: 14x^2 + 24xy + 21y^2 - 4x + 18y - 139 = 0$, such that the chord intercepted by the line is bisected by point ... | Solution 5: Let the equation of the required chord be
$$
\begin{array}{l}
\left\{\begin{array}{l}
x=1+t \cos \alpha, \\
y=-2+t \sin \alpha
\end{array} \text { ( } t\right. \text { is a parameter), substitute into the } \\
\text { quadratic curve equation } 14(1+\mathrm{t} \cos \omega)^{\circ}+21(1+\mathrm{t} \cos \alph... | 4x+7y+10=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,273 |
Example 1. There are several different books, which are initially numbered as No. 1, No. 2, $\cdots$, up to No. $\mathrm{n}$. Later, they are renumbered again. This time, the numbering is not entirely the same, so each book has two numbers: the new number and the old number. If the two numbers of each book are differen... | The 1st book can be placed in any pile at will. Suppose that the first $\mathrm{k}-1$ books have been successfully sorted into piles, we now consider the placement of the $\mathrm{k}$-th book: the old number in this book may match the new number of one of the $\mathrm{k}-1$ books already sorted, in which case this book... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,274 |
Example 1. A cube toy made of uniform material, with the numbers $1,2,3,4,5$, 6 marked on each face. If this toy is tossed once, what is the probability that the face up is an odd number? (Textbook P92 Exercise 6 Question 4) | Because the cube toy is uniform, the possibility of each face landing up is equal when it is tossed, with a total of six basic events $(\mathrm{n}=6)$. “The face that lands up shows an odd number” includes 3 basic events $(m=3)$, so $\mathrm{P}(\mathrm{A}) = \frac{3}{6} = \frac{1}{2}$.
2. Tabulation Method
List the bas... | \frac{1}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,276 |
Example 2. Toss a coin three times, what is the probability of getting "2 heads and 1 tail"? (Exercise 6, Question 5) | Make a tree diagram:
First time Second time Third time
There are 8 equally likely basic events, and the event $\{2$ heads, 1 tail $\}$ includes 3 basic events, so the required probability is $\frac{3}{8}$. | \frac{3}{8} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,277 |
Example 3. A pocket contains 7 white balls and 3 black balls of the same size. If two balls are drawn at random, what is the probability of getting one white ball and one black ball? (Exercise 6, Question 6) | From 7 white balls and 3 black balls, any 2 balls can be drawn with $\mathrm{C}_{10}^{2}$ equally possible outcomes, among which the outcomes of getting one white and one black ball are $\mathrm{C}_{7}^{1} \cdot \mathrm{C}_{3}^{1}$, so the required probability is
$$
\frac{\mathrm{C}_{7}^{1} \cdot \mathrm{C}_{3}^{1}}{\m... | \frac{7}{15} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,278 |
Example 2. Solve the equation $\sin ^{2} x=\cos ^{2} x$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | \begin{aligned} \text { Sol } & \because \sin ^{2} x=\cos ^{2} x, \\ & \therefore \sin x=\cos x, \\ & \therefore x=n \pi+\frac{\pi}{4} \cdot(n \in \mathbb{Z})\end{aligned} | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,280 |
Example 3. Given: $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{d}}{\mathrm{a}}$, find the value of $\frac{a+b+c+d}{b+a+c-d}$. | Solution 1: By the ratio theorem, we have:
$$
\begin{array}{c}
\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a} \\
=\frac{a+b+c+d}{a+b+c+d}=1, \text { so } a=b=c=d, \\
\therefore \frac{a+b+c+d}{a+b+c-d}=\frac{4 d}{2 d}=2 .
\end{array}
$$
Solution 2: Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, then
$a=bk, b... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,281 |
Example 4. What real number is $m$ when the two roots of the equation $x^{2}-2 m x$ $+\mathrm{m}^{2}-1=0$ are both greater than -2 and less than 4? | Solution 1: Since $\Delta=(-2 m)^{2}-4\left(m^{2}-1\right)$ $=4>0$,
$\therefore$ the equation has two distinct real roots.
Let the two roots be $x_{1}$ and $x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=2 m \\
x_{1} x_{2}=m^{2}-1
\end{array}\right.
$$
$\because$ both roots of the equation are greater than -2 and ... | \sqrt{5}<\mathrm{m}<4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,282 |
Example 6. Father's age is 48 years old, son's age is 20 years old. How many years later will the father's age be 3 times the son's age? | Solution: Let $\mathrm{x}$ years later, the father's age is 3 times the son's age.
We have $48+x=3(20+x)$. Solving for $\mathrm{x}$, we get $\mathrm{x}=-6$.
This means that 6 years ago, the father's age was 3 times the son's age. Conventionally, there is no such expression as “-6 years later.” However, according to the... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,284 |
Theorem 2 Let $\mathrm{P}_{2}$ be the intersection of the trisectors of $\angle \mathrm{BAC}$ with $\mathrm{EC}$ in $\triangle \mathrm{ABC}$, then
$$
\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BP}_{1} \cdot \mathrm{BP}_{2}}{\mathrm{CP}_{1} \cdot \mathrm{CP}_{2}} .
$$
Can the above result be generalized to a... | Let $P_{1}$,
$P_{2}, \cdots, P_{n_{-1}}$
be the points of intersection of the $\mathrm{n}$ angle bisectors of
$\angle \mathrm{BAC}$ with $\mathrm{BC}$ (as shown in the figure),
Following the above method, we have
Multiplying the above $(n-1)$ equations, we get
$$
\frac{\mathrm{AB}^{\mathrm{n}-1}}{\mathrm{AC}^{\mathr... | \frac{\mathrm{AB}^{\mathrm{n}-1}}{\mathrm{AC}^{\mathrm{n}-1}}=\frac{\mathrm{BP}_{1} \cdot \mathrm{BP}_{2} \cdots \mathrm{BP}_{\mathrm{n}-1}}{\mathrm{CP}_{1} \cdot \mathrm{CP}_{2} \cdots \mathrm{CP}_{\mathrm{n}-1}} | Geometry | proof | Yes | Yes | cn_contest | false | 702,287 |
Example 2. Given the sequence $2,4,9,17,28,42$, $\cdots$, find its general term.
Analysis: If the method of undetermined coefficients is used, a system of six linear equations needs to be solved; if the method of observation and induction is used, it is also quite difficult. If the method of successive differences is ... | The difference sequence $\left\{\mathrm{b}_{\mathrm{A}}\right\}$ of the sequence $\left\{\mathrm{a}_{\mathrm{A}}\right\}$ is: $2,5,8,11,14, \cdots$. It is an arithmetic sequence with the first term $2$ and common difference $3$. Therefore,
$$
\begin{array}{l}
b_{n}=2+(n-1) \times 3=3 n-1 \text {. } \\
\text { Hence } a... | \frac{1}{2}\left(3 n^{2}-5 n+6\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,288 |
Example 3. Prove: The equation of the tangent line at point $P\left(x_{0}, y_{0}\right)$ on the circle $x^{2}+y^{2}+D x+E y+F$ $=0$ is
$$
x_{0} x
+y_{0} y+D \cdot \frac{x_{0}+x}{2}+E \cdot \frac{y_{0}+y}{2}+F=0
$$ | Prove that the equation of the point circle $\mathrm{P}$ is
$$
\left(\mathrm{x}-\mathrm{x}_{0}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{0}\right)^{2}=0 .
$$
From I, the required tangent equation is
$$
\begin{array}{c}
\left(x^{2}+y^{2}+D x+E y+F\right) \\
-\left[\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}\right]=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,289 |
Example 4. Find the circle intersecting the given circle $x^{2}+y^{2}-7 y+10$ $=0$ such that the common chord is parallel to the given line $2 \mathbf{x}$ $-3 y-1=0$, and passes through the points $(-2,3)$ and $(1,4)$. | The equation of the circle passing through the points $(-2,3),(1,4)$ is
$$
\begin{aligned}
& (x+2)(x-1)+(y-3)(y-4) \\
+\lambda & {[(x+2)-(y-3) \cdot 3]=0 }
\end{aligned}
$$
which simplifies to \(x^{2}+y^{2}+(\lambda+1) x-(3 \lambda+7) y+11 \lambda\)
$$
+10=0 \text {. }
$$
Since (·) intersects with the known circle, s... | x^{2}+y^{2}+2 x-10 y+21=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,290 |
Example 6. There are n circles, each pair of which intersects at two points, and no three circles intersect at the same point. Prove that these $\mathrm{n}$ circles divide the plane into $\mathrm{n}^{2}-n+2$ parts. | Proof: Let $n$ circles divide the plane into $f(n)$ regions. When adding the $(n+1)$-th circle, it is divided into $2n$ arcs by the original $n$ circles. Each arc divides its region into two parts, thus adding $2n$ regions in total. Therefore, there are $f(n+1) = f(n) + 2n$ regions. Clearly, $f(1) = 2$. By taking $n = ... | n^2 - n + 2 | Geometry | proof | Yes | Yes | cn_contest | false | 702,292 |
Example 1. Given A, B, C, D are four points on the same plane, if $A B \perp C D, A C \perp B D$, then $A D \perp B C$. | Proof: As shown in the figure, extend $\mathrm{CD}$ to intersect $\mathrm{AB}$ at $\mathrm{P}$, and extend $\mathrm{BD}$ to intersect $\mathrm{AC}$ at $\mathrm{Q}$. Then, $\mathrm{D}$ is the orthocenter of $\triangle \mathrm{ABC}$. Therefore, extending $A D$ to intersect $B C$ at $R$, $A R$ is the altitude on $\mathrm{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,294 |
Example 3. Given points A, B and line l in the same plane, try to find a point P on l such that
$$
P A+P B \text { is minimal. }
$$ | Slightly explained: Suppose A and B are on the same side of l, construct the symmetric point of A with respect to l as $\mathrm{A}^{\prime}$, connect $\mathrm{A}^{\prime} \mathrm{B}$ intersecting $\mathrm{l}$ at $\mathrm{P}$. Then $P$ is the desired point. | P | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,295 |
Example 4. The sum of the squares of the four sides of a quadrilateral is not less than the sum of the squares of the two diagonals. (This can be proven using the cosine theorem) | If the four vertices are not in the same plane, it can be generalized to: the sum of the squares of the four sides of a spatial quadrilateral is not less than the sum of the squares of the two diagonals. In fact, as shown in the figure, take the midpoint M of BD, and in triangle ABD, by the median formula
$$
\begin{arr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,296 |
(2) $\left(\frac{1-\mathrm{i}^{\wedge}}{1+\mathrm{i}^{1}}\right)^{\circ}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
(2) $\left(\frac{1-\mathrm{i}^{\wedge}}{1+\mathrm{i}^{1}}\right)^{\circ}$. | Correct solution: $\left(\frac{1-i}{1+i}\right)^{5}$
$$
\begin{array}{l}
=\left[\frac{(1-i)^{2}}{(1+i)(1-i)}\right]^{5}=\left(\frac{-2 i}{2}\right)^{5} \\
=(-i)^{5}=-i .
\end{array}
$$ | -i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,299 |
Example 3. Solve the equation: $\operatorname{ctg}^{2} \frac{x}{2}=1+\sec x$. | Correct solution: Upon inspection, $x=2 n \pi+\pi(n \in Z)$ satisfies the original equation, so the solutions to the original equation are $\mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}, x=2 n \pi+\pi,(n \in Z)$ | x=2n\pi \pm \frac{\pi}{3}, x=2n\pi+\pi, (n \in Z) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,301 |
Example 4. A tangent line is drawn through the point $\mathrm{P}(1,-1)$ to the parabola $\mathrm{y}^{2}-2 \mathrm{x}$ $-2 y+3=0$. Find the equation of the tangent line. | Correct answer: When the slope exists, the tangent line is $x+4 y+3=0$, and when the slope does not exist, the tangent line is $x=1$.
From this example, we can see that when using the point-slope form, it is necessary to consider the condition for the point-slope equation to hold, which is that the slope must exist. I... | x+4 y+3=0 \text{ and } x=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,302 |
$\begin{array}{l}\text { Given } \frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y} \\ =k \text {. Find the value of } k \text { . }\end{array}$ | Correct answer: When $x+y+z \neq 0$, we get $k=\frac{1}{2}$; when $x+y+z=0$, by substituting $y+z=-x$ into $\frac{x}{y+z}=k$, we get $k=-1$. | k=\frac{1}{2} \text{ or } k=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,303 |
Example 6. Given that $a$, $b$, $c$, $d$ are real numbers. Find the necessary condition for the equation $x^{2}+(a+b i) x+c+d i=0$ to have real roots.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let the real root of the equation be $\mathrm{x}_{0}$, then
$$
\mathrm{x}_{0}^{2}+(\mathrm{a}+\mathrm{bi}) \mathrm{x}_{0}+(\mathrm{c}+\mathrm{di})=0 \text {. }
$$
Using the condition for equality of complex numbers, we get
$$
\left\{\begin{array}{l}
\mathrm{x}_{0}^{2}+\mathrm{a} \mathrm{x}_{0}+\mathrm{c}=0, \\
\mathrm... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,304 |
Example 7. Let $f_{1}(x)=x^{2}-2, x=2 \cos \theta$, and $f_{m}(x)=f_{1}\left[f_{m-1}(x)\right](m=2,3, \cdots)$, then $f_{n}(x)=2 \cos 2^{n} \theta, \quad(n \in N)$ | $$
\begin{array}{l}
\text { That is } f_{k+1}(x)=f_{1}\left[f_{k}(x)\right]=f_{1} \quad\left(2 \cos 2^{k} \theta\right) \\
\quad=\left(2 \cos 2^{k} \theta\right)^{2}-2=2\left[2 \cos ^{2} 2^{k} \theta-1\right] \\
\quad=2 \cos 2^{k+1} \theta \text {, that is, when } n=k+1 \text {, the proposition also }
\end{array}
$$
h... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,305 |
Example 1. Given fixed points $\mathrm{A}$ and $\mathrm{B}$ and a fixed line $\mathrm{L}$, determine a point on line $\mathrm{L}$ such that the sum of its distances to points $\mathrm{A}$ and $\mathrm{B}$ is minimized. | (1) As shown in Figure 1, if points $\mathrm{A}$ and $\mathrm{B}$ are on opposite sides of $L$, then connect $A B$ which intersects $L$ at point $P_{1}$. Point $P_{1}$ is the solution.
(2) As shown in Figure 2, if points $\mathrm{A}$ and $\mathrm{B}$ are on the same side of $L$, then construct the symmetric point $\mat... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,306 |
Example 2. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of $\triangle \mathrm{ABC}$, determine points $P$ and $Q$ such that $B Q+Q P+P C$ is minimized. | As shown in Figure 3, construct the symmetric point $\mathrm{B}^{\prime}$ of point $\mathrm{B}$ with respect to $\mathrm{AC}$, and the symmetric point $\mathrm{C}^{\prime}$ of point $\mathrm{C}$ with respect to $\mathrm{AB}$. Connect $\mathrm{B}^{\prime} \mathrm{C}^{\prime}$, which intersects $\mathrm{AB}$ and $\mathrm... | B'Q+QP+PC' = B'C' | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,307 |
Example. Let $x^{2}+y^{2}-6 x-4 y+11=0$. Try to find the minimum and maximum values of $\mu=\sqrt{x^{2}+(y+1)^{2}}$.
| Solve as shown in Figure 4, the equation $x^{2}+y^{2}-6x-6y+i1=0$ can be rewritten as $(x-3)^{2}+(y-2)^{2}=2$. It represents a circle with center C (3, 2) and radius $\sqrt{2}$. Thus, the problem is to find a point $(x, y)$ on $\mathrm{QC}$, such that the distance to the fixed point $\mathrm{A}(0,-1)$ is maximized or m... | 4 \sqrt{2}, 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,309 |
Example 4. Given point $\mathrm{A}$ as a fixed point on a plane, and $\mathrm{F}$ as the focus of a parabola, try to find a point $\mathrm{P}$ on the parabola such that $|P A|+|P F|$ is minimized. | Solve as shown in the figure
5, discuss in two cases: (1)
If $\mathrm{AF}$ intersects the parabola
at $P_{1}$, then
$\mathrm{P}_{1} \mathrm{~A} + \mathrm{P}_{1} \mathrm{~F}$
$=\mathrm{AF}$, i.e., $\mathrm{P}_{1}$ is
the solution. (2) If $\mathrm{AF}$
does not intersect the parabola,
then draw $\mathrm{AB} \perp \mathrm... | P_1 \text{ or } P_2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,310 |
Example 5. Find a point P on a fixed line L such that $\mathrm{PA}^{2} + PB^{2}$ is minimized (A and B are two fixed points). | Solve: As shown in Figure 6, take the midpoint C of AB, and let P be the point on L that satisfies the condition, then $PA^2 + PB^2 = 2(PC^2 + AC^2)$, so PC must be the shortest. Therefore, draw CP perpendicular to L through C, and the foot of the perpendicular P is the desired point. | P | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,311 |
Example 6. In $\triangle \mathrm{ABC}$, it is known that $\mathrm{AB}>\mathrm{AC}, \mathrm{P}$ is any point on $\mathrm{BC}$, the symmetric point of $\mathrm{P}$ with respect to $\mathrm{AB}$ is $\mathrm{E}$, and the symmetric point of $\mathrm{P}$ with respect to $\mathrm{AC}$ is $\mathrm{F}$. When is the area of $\tr... | From the given conditions,
(As shown in Figure 7),
\[
\begin{array}{l}
\angle \mathrm{EAB} \\
=\angle \mathrm{PAB}, \\
\angle \mathrm{FAC}
\end{array}
\]
\[
\begin{array}{l}
=\angle \mathrm{PAC}, \text{ hence } \\
\angle \mathrm{EAP}=2 \angle \mathrm{PAB}, \angle \mathrm{FAP}=2 \angle \mathrm{PAC}. \\
\therefore \angle... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,312 |
Example 1. As shown in the figure, points P, R, and Q are located on the edges $\mathrm{A}_{1} \mathrm{D}_{1}$, $\mathrm{C}_{1} \mathrm{C}$, and $\mathrm{BC}$ of the cube $\mathrm{ABCD}-\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$, respectively. Find the trace of the plane passing through these three p... | Drawing method:
(1) Connect $QR$ and extend it to intersect $B_{1}C_{1}$ at point $H$;
(2) $\because A_{1}D_{1}$ and $B_{1}C_{1}$ are in the same plane,
$\therefore$ connect $PH$ to intersect $D_{1}C_{1}$ at point $K$,
(3) $\because K$ and $R$ are on the side face $D_{1}C$,
$\therefore$ connect $KR_{3}$
(4) Extend
$KP$... | EFQRKP | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,317 |
Example 3. Using a car crane with a height of 1.5 meters and a boom length of 15 meters, to lift a cylindrical oil tank with a diameter of 6 meters and a height of 2 meters. To find the maximum height at which the car crane can lift the oil tank, it is necessary to first determine the functional relationship between th... | Solve as shown in the figure, according to the algebraic sum of line segments:
$$
\begin{array}{l}
\mathrm{h}=\mathrm{AB}+1.5 \text {. } \\
\text { but } \mathrm{AB}=\mathrm{AD} \\
-\mathrm{BC}-\mathrm{CD},
\end{array}
$$
and $\mathrm{AD}=\mathrm{ED} \sin \varphi$
$$
=15 \sin \varphi \text {, }
$$
$$
\begin{aligned}
\... | h=15 \sin \varphi-3 \operatorname{tg} \varphi-0.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,319 |
Example 3. If $\mathrm{n}$ is a positive integer, then the value of $\frac{1}{8}\left[1-(-1)^{n}\right]\left(n^{2}-1\right)$ is
(A) definitely odd;
(B) definitely even;
(C) an integer but not necessarily even;
(D) not necessarily an integer. | For the sake of simplicity, let $\left[1-(-1)^{n}\right]\left(n^{2}-1\right)$ $=S$. It is easy to see that, regardless of any positive integer $n$, $8 \mid S$, so $\frac{1}{8} S$ is an integer, thus eliminating (D); it is also evident that when $n$ is odd, $\frac{1}{8} S \neq 0$, so (A) cannot be eliminated; furthermor... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,322 |
Example 4. The necessary and sufficient condition for $\arccos (-x)$ to be greater than $\arccos x$ is:
(A) $x \in(0,1)$;
(B) $x \in(-1,0)$;
(C) $x \in[0,1]$;
(D) $x \in\left[0, \frac{\pi}{2}\right]$. | Among the four answers, the domain of $\arccos x$, $|x| \leqslant 1$, directly eliminates (D); then by $\arccos (-x) > \arccos x$, using the monotonicity of the arccosine function, we can deduce that $-x < 0$, thus eliminating (B). When $x=0$, we have $\arccos (-x) = \arccos x$, which eliminates (C); therefore, (A) is ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,323 |
Side 5. If $\theta$ is an angle in the second quadrant; and satisfies $\cos \frac{\theta}{3}-\sin \frac{\theta}{2}=\sqrt{1}-\sin \theta$, then $\frac{\theta}{2}$
(A) is an angle in the first quadrant;
(B) is an angle in the third quadrant;
(C) may be an angle in the first quadrant; or may be an angle in the third quadr... | Among the four answers, since $\theta$ is in the second quadrant, it is known that $\frac{\theta}{2}$ cannot be in the second quadrant, so (D) is eliminated; also, because $\cos \frac{\theta}{2} - \sin \frac{\theta}{2} = \sqrt{1 - \sin \theta} \geqslant 0$, it follows that $\cos \frac{\theta}{2} \geqslant \sin \frac{\t... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,324 |
Example 1. As shown in the figure, $\mathrm{ABCD}$ and $\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$ are both squares, and $\mathrm{A}_{1} 、 \mathrm{~B}_{1} 、 \mathrm{C}_{1} 、 \mathrm{D}_{1}$ divide $\mathrm{AB}$, $\mathrm{BC}$, $\mathrm{CD}$, and $\mathrm{DA}$ in the ratio $\mathrm{m}: \mathrm{n}$, re... | $\mathrm{AB}=1, \frac{\mathrm{AA}_{1}}{\mathrm{~A}_{1} \mathrm{~B}}=\frac{\mathrm{m}}{\mathrm{n}}$, so $\mathrm{AA}_{1}=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}$, $A_{1} B=\frac{n}{m+n}$.
Thus $\mathrm{A}_{1} \mathrm{~B}_{1}=\sqrt{\mathrm{BB}_{1}{ }^{2}+\mathrm{A}_{1} \mathrm{~B}^{2}}$
$$
=\sqrt{\mathrm{AA}_{1}{ }^{2}+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,325 |
Example 2. Prove: $2+\sin \alpha+\cos \alpha$
$$
\geqslant \frac{2}{2-\sin \alpha-\cos \alpha}
$$ | $\begin{array}{l}\text { Prove (using the ratio comparison method) } \\ \because 2+\sin \alpha+\cos \alpha>0, \\ \frac{2}{2-\sin \alpha-\cos \alpha}>0, \text { then } \\ \frac{2+\sin \alpha+\cos \alpha}{2}=\frac{4-(\sin \alpha+\cos \alpha)^{2}}{2} \\ =\frac{3-\sin 2 \alpha}{2} \geqslant 1 . \\ \therefore 2+\sin \alpha+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,326 |
Example 3. Prove: $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \cdots \times \frac{99}{100}<\frac{1}{10}$. | $$
\begin{array}{l}
\text { Prove (using the method of partial comparison) } \\
\because \frac{1}{2}<\frac{2}{3}, \frac{3}{4}<\frac{4}{5}, \frac{5}{6}<\frac{6}{7}, \cdots, \\
\frac{99}{100}<\frac{100}{101},
\end{array}
$$
Multiplying all the inequalities, we get: $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \tim... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,327 |
Example 4. On a plane, there are $\mathrm{n}$ lines, among which no two are parallel, and no three or more pass through the same point. How many regions does the plane get divided into by these $\mathrm{n}$ lines? | Solve by first exploring the pattern using $\mathrm{n}=1,2,3,4$, etc.
Let the number of regions into which the plane is divided by $\mathrm{n}$ lines be $\mathrm{P}_{\mathrm{n}}$. From the diagram, we can see that
$$
\begin{array}{l}
\mathrm{P}_{1}=2, \\
\mathrm{P}_{2}=4=2+2=\mathrm{P}_{1}+2, \\
\mathrm{P}_{3}=7=4+3=\... | \frac{n(n+1)}{2} + 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,328 |
Example 5. Given a square $\mathrm{ABCD}$ with side length $\mathrm{a}$ inscribed in $\odot \mathrm{O}$, extend $\mathrm{DC}$ to $\mathrm{Z}$ such that $\mathrm{CZ}=\frac{1}{2} \mathrm{a}$, and take $\mathrm{BF}=\frac{1}{3} \mathrm{a}$ on $\mathrm{BC}$. The extension of $\mathrm{AF}$ intersects $\mathrm{BZ}$ at $\mathr... | Proof: Let the parameters be
$\alpha=\angle \mathrm{BAG}$,
$\beta=\angle \mathrm{CBZ}$, then
$\operatorname{tg} \alpha=\frac{1}{3}, \quad \operatorname{tg} \beta$ $=\frac{1}{2}$.
$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,329 |
Example 6. When $a$ varies, find the range of solutions for the inequality $|x+a| + |x| < 2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve by transforming the original inequality:
$$
\begin{array}{c}
|\mathrm{x}+\mathrm{a}|<2-|\mathrm{x}| . \\
\text { Let } \mathrm{y}_{1}=|\mathrm{x}+\mathrm{a}|, \\
\mathrm{y}_{2}=2-|\mathrm{x}| .
\end{array}
$$
(1) The graph of (1) is a set of parallel rays with endpoints at $(\cdots a, 0)$ and slopes $\mathrm{k}= ... | -\frac{a+2}{2}<x<\frac{2-a}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,331 |
Example 1. Find a point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ such that the product of its distances to the two foci is 16. | Let the point to be found be $\mathrm{P}(\mathrm{x}, \mathrm{y}), \mathrm{F}_{1}$, $\mathrm{F}_{2}$ be the left and right foci of the ellipse.
Since $=5, \mathrm{~b}=3, \mathrm{c}=4, \mathrm{e}=\frac{4}{5}$,
it follows from formula (1) that
$$
\begin{array}{l}
\left|P F_{1}\right|=a+e x=5+\frac{4}{5} x, \\
\left|P F_{2... | \left(\frac{15}{4}, \frac{3 \sqrt{7}}{4}\right),\left(\frac{15}{4},-\frac{3 \sqrt{7}}{4}\right),\left(-\frac{15}{4}, \frac{3 \sqrt{7}}{4}\right),\left(-\frac{15}{4},-\frac{3 \sqrt{7}}{4}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,332 |
Example 2. Optical property of an ellipse: The tangent of an ellipse externally bisects the angle between the two focal radii at the point of tangency. | Proof: Let $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ be any point on the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}} + \frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$, and $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ be the two foci of the ellipse.
This problem only requires proving that the normal at $\mathrm{P}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,333 |
Example 1. Find $\mathrm{S}_{\mathrm{n}}=1 \cdot 2 \cdot 4+2 \cdot 3 \cdot 5+\cdots$ $+n(n+1)(n+3)$. | \begin{array}{l}\text { Solve } u_{k}=k(k+1)(k+3) \\ =k(k+1)(k+2+1)=k(k+1)(k+2) \\ +k(k+1), \text { which meets the condition of using ( } *) \text { formula. } \\ \text { First, find } \\ S_{n}^{1}=\sum_{k=1}^{n} k(k+1)(k+2), \text { where } \\ a=0, b=1, r=3. \\ \text { Then, } S_{n}^{1}=\frac{(n+3)n(n+1)(n+2)}{4 \cdo... | \frac{n(n+1)(n+2)(3 n+13)}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,334 |
$\begin{array}{l}\text { Example } 3 . u_{k}=\frac{1}{k(k+1)(k+2)(k+3)} \\ \text { Find } \sum_{k=1}^{n} u_{k} .\end{array}$ | \begin{array}{l} \text { Sol } u_{k}=\frac{1}{3}\left[\frac{1}{k(k+1)(k+2)}\right. \\ -\left.\frac{1}{(k+1)(k+2)(k+3)}\right], \\ \therefore \sum_{k=1}^{n} u_{k} \\ = \frac{1}{3}\left[\frac{1}{6}-\frac{1}{(n+1)(n+2)(n+3)}\right]\end{array} | \frac{1}{3}\left[\frac{1}{6}-\frac{1}{(n+1)(n+2)(n+3)}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,336 |
$\begin{array}{l}\text { Example 5. Find } S_{n}=\frac{1}{\sin \theta \sin 3 \theta}+\frac{1}{\sin \theta \sin 5 \theta} \\ +\cdots+\frac{1}{\sin (2 n-1) \theta \sin (2 n+1) \theta} .\end{array}$ | $$
\begin{array}{l}
\text { Solve } u_{1}(\theta) \sin 2 \theta=\frac{\sin 2 \theta}{\sin \theta \sin 3 \theta} \\
=\frac{\sin (3 \theta-\theta)}{\sin \theta \sin 3 \theta} \\
=\frac{\sin 3 \theta \cos \theta-\sin \theta \cos 3 \theta}{\sin \theta \sin 3 \theta} \\
=\cot \theta-\cot 3 \theta \text {. } \\
\end{array}
$... | S_{n}^{(\theta)}=\frac{\cot \theta-\cot(2 n+1) \theta}{\sin 2 \theta} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,338 |
Example 1. Find $P_{n}=\prod_{k=1}^{n}\left(2 \cos 2^{k-1} \theta-1\right)$. | Solve $\begin{aligned} p_{k} & =2 \cos 2^{k-1} \theta-1 \\ & =\frac{\left(2 \cos 2^{k-1} \theta-1\right)\left(2 \cos 2^{\mathbf{k}-1} \theta+1\right)}{2 \cos 2^{k-1} \theta+1} \\ & =\frac{4 \cos ^{2} 2^{\mathbf{k}-1} \theta-1}{2 \cos 2^{\mathbf{k}-1} \theta+1} \\ & =\frac{2\left(\cos { }^{K} \theta+1\right)-1}{2 \cos 2... | \frac{2 \cos 2^{n} \theta+1}{2 \cos \theta+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,340 |
Example 4. Find
$$
\begin{array}{c}
\mathrm{P}_{\mathrm{n}}=\sqrt{3} \cdot \sqrt{2+\sqrt{3}} \\
\cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdots
\end{array}
$$ | $$
\cdot \sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{3}}}}
$$
$n$ nested radicals
Solve $\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$,
$$
\cos \frac{\pi}{2 \cdot 6}=\sqrt{\frac{1+\cos \frac{\pi}{6}}{2}}=\sqrt{\frac{2+\sqrt{3}}{2}}
$$
In general, when $\cos \frac{\pi}{2^{\mathbf{k}-1} \cdot 6}$
$$
=\frac{1}{2} \sqrt{2+\sqrt{2+\cdot... | P_{n}=\frac{\sqrt{3}}{2 \sin \frac{\pi}{2^{n-1} \cdot 6}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,344 |
Example 1. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $\mathrm{n}$ terms as $S_{n}=2^{n+1}-n(n+1)$, find $a_{n}$. | $$
\begin{array}{l}
\text { For } n \geqslant 2, a_{n}=S_{n}-S_{n-1}=2^{n}-2 n \text {. } \\
n=1 \text {, } a_{1}=S_{1}=2^{2}-2=2 \text {. } \\
\end{array}
$$
$\therefore$ The general term formula of the sequence is:
$$
a_{n}=\left\{\begin{array}{l}
2, n=1 ; \\
2^{n}-2 n, n \geqslant 2 .
\end{array}\right.
$$ | a_{n}=\left\{\begin{array}{l}2, n=1 ; \\2^{n}-2 n, n \geqslant 2 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,345 |
Example 2. Given the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$, the sum of the first $\mathrm{n}$ terms is $\mathrm{S}_{\mathrm{n}}$ $=1-\sin ^{2 n} \alpha$, try to find $a_{n} . \quad(\alpha \neq k \pi)$ | $$
\begin{array}{l}
n \geqslant 2 \text { when } a_{n}=S_{n}-S_{n-1} \\
=\cos ^{2} \alpha \sin ^{2}(n-1) \alpha ; \\
n=1 \text { when, } a_{1}=S_{1}=1-\sin ^{2} \alpha=\cos ^{2} \alpha \\
\quad=\cos ^{2} \alpha \sin ^{2(1-1) \alpha}
\end{array}
$$
$\therefore$ The general term formula of the sequence is:
$$
a_{n}=\cos ... | a_{n}=\cos ^{2} \alpha \sin ^{2(n-1)} \alpha | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,346 |
Example 3. The sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ is an arithmetic sequence if and only if: the sum of its first $\mathrm{n}$ terms is a quadratic function of $\mathrm{n}$ with a constant term of 0. | Proof: Sufficiency. Let $\mathrm{S}_{n}=\mathrm{pn}^{2}+\mathrm{qn}$, where $\mathrm{p}$ and $\mathrm{q}$ are constants, and $\mathrm{p} \neq 0$.
When $n \geqslant 2$, $a_{n}=S_{n}-S_{n-1}=p n^{2}+q n$
$-\left[p(n-1)^{2}+q(n-1)\right]$
$$
=2 p n-p+q \text {; }
$$
When $n=1$, $a_{1}=p+q=2 p \times 1-p+q$.
Therefore, fo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,347 |
1. Let $a, b$ be constants satisfying $a>b>0$, for the sequence $\left\{\mathrm{x}_{\mathrm{n}}\right\}$ defined by the recurrence relation
$$
\cdots x_{1}=b, x_{n}-a x_{n-1}=b^{n}(n=2,3,
$$
answer the following questions:
(1) If $y_{n}=\frac{x_{n}}{b^{n^{2}}}$, find the general term of $\left\{y_{n}\right\}$;
(2) Fin... | Solve: (1) From $x_{n}-a x_{n-1}=b^{n}$, we have $\frac{x_{n}}{b^{n}}-\frac{a}{b} \cdot \frac{x_{n-1}}{b^{n-1}}=1$,
thus $\mathrm{y}_{\mathrm{n}}-\frac{\mathrm{a}}{\mathrm{b}} \mathrm{y}_{\mathrm{n}-1}=1$.
Let $y-\frac{a}{b} y=1$, we get $y=\frac{b}{b-a}$,
hence $\frac{\mathrm{b}}{\mathrm{b}-\mathrm{a}}-\frac{\mathrm{a... | a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,348 |
4. (1) Let the function $f(x)$ have its minimum value in the interval $0 \leqslant x \leqslant 1$, then for any real number $\alpha$, prove the following inequality:
$$
\alpha^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} \frac{1}{f(x)} d x \geqslant 2 \alpha \text {; }
$$
(2) In (1), if $f(x)=\frac{2 x+1}{x^{2}+x+1}$, to der... | Solution: By the inequality of the arithmetic mean $\geqslant$ geometric mean,
$$
\begin{array}{l}
=|\alpha| \geqslant \alpha, \\
\therefore \alpha^{2} \mathrm{f}(\mathrm{x})+\frac{1}{\mathrm{f}(\mathrm{x})} \geqslant 2 \alpha. \\
\end{array}
$$
Thus, $\alpha^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} f \frac{1}{(x)} d x$... | \log 3 \geqslant \frac{2(\sqrt{7}-1)}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,351 |
Example 6. When measuring the atmospheric temperature T, it is found that T decreases with the height h in the ascending air, up to 11 kilometers, with a decrease of approximately $6^{\circ} \mathrm{C}$ for every kilometer ascended. At higher altitudes, the temperature remains almost constant. If the ground temperature... | Solve As shown in the figure, from the problem, we know:
1) From the ground to 11 kilometers above, the function relationship of T with h is
$$
T=19-6 h ;
$$
2) Above 11 kilometers, due to the temperature being almost constant,
therefore, we have
$$
\mathrm{T}=19^{\circ}-6^{\circ} \times 11=-47^{\circ} \mathrm{C} \text... | \mathrm{T}=\left\{\begin{array}{l}
19-6 \mathrm{~h}, \quad 0 \leqslant \mathrm{~h} \leqslant 11 ; \\
-47, \mathrm{~h}>11
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,352 |
Example 2. Find the minimum value of the function $y=4 x^{4}-4 x^{2}-4$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve: Completing the square for $4 x^{4}-4 x^{2}-4$ yields:
$$
\begin{aligned}
& y=\left(2 x^{2}-1\right)^{2}-5 . \\
\because \quad & \left(2 x^{2}-1\right)^{2} \geqslant 0, \text { and when } x= \pm \frac{\sqrt{2}}{2}
\end{aligned}
$$
equality holds,
$$
\therefore \text { when } x= \pm \frac{\sqrt{2}}{2} \text {, } ... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,354 |
Example 1. If $a_{1}>0, a_{2}>0$, and $a_{1}+a_{2}$ $=1$, then
$$
\left(a_{1}+\frac{1}{a_{1}}\right)^{2}+\left(a_{2}+\frac{1}{a_{2}}\right)^{2} \geqslant \frac{25}{2} .
$$ | Given that, $a_{1}, a_{2} > 0, 1 = a_{1} + a_{2}$
$\geqslant 2 \cdot \sqrt{a_{1} a_{2}}$,
$$
\begin{array}{l}
\therefore \quad a_{1} a_{2} \leqslant \frac{1}{4} \Longrightarrow \frac{1}{2_{1} a_{2}} \geqslant 4, \\
\text { hence }\left(a_{1}-\frac{1}{a_{1}}\right)^{2}+\left(a_{2}+\frac{1}{a_{2}}\right)^{2} \\
\Rightarr... | \frac{25}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,356 |
Example 2. Prove: $\frac{x^{2}+5}{\sqrt{x^{2}+4}} \geqslant 2 .(x \in R)$ According to teaching needs, we will transform it into: find the extremum of the function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}+5}{\sqrt{\mathrm{x}^{2}+4}}$. | According to $\mathrm{A} \geqslant \mathrm{G}$, it is easy to get
$$
\begin{array}{l}
f(x)=\sqrt{x^{2}+4}+\frac{1}{\sqrt{x^{2}+4}} \geqslant 2, \\
\therefore \quad f_{\text{min}}=2.
\end{array}
$$
A closer examination reveals that this solution is incorrect. In fact, the condition for equality in (1) is
$$
\sqrt{x^{2}... | \frac{5}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,357 |
Time 3. Let $\alpha, \beta, \gamma$ be the three angles of any triangle, prove the inequality $\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}<\frac{1}{4}$.
(Hungarian Mathematical Olympiad (1896-1897) Problem) | By the principle of symmetry, it is evident that this problem requires a low level of complexity. In fact, under the constraint
$$
\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=\frac{\pi}{2}
$$
we are to find the extremum of the objective function
$$
\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}... | \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2} \leqslant \frac{1}{8} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,358 |
Example 4. The three dimensions $\mathrm{x}, \mathrm{y}, \mathrm{z}$ of a rectangular prism sum to a constant $\mathrm{L}$. Find the maximum values of the volume $\mathrm{V}$ and the surface area $\mathrm{S}$. | This is to find the extremum of the objective functions
$$
\begin{array}{l}
V=x y z \\
\text { and } S=2(x y+y z+z x) \\
\end{array}
$$
under the constraint
$$
x+y+z=L
$$
Notice: (1), (2), and (3) are symmetric with respect to $x$, $y$, and $z$, and $V$ and $S$ clearly have no minimum value. Therefore, when $x=y=z=\fr... | V_{\text{max}}=\frac{L^3}{27}, \quad S_{\text{max}}=\frac{2}{9}L^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,359 |
Example 1. Let $f(x)=x^{2}+a x+b$. Prove: $|\mathrm{f}(1)|, |\mathrm{f}(2)|, |\mathrm{f}(3)|$ contain at least one number not less than $\frac{1}{2}$. | Proof 1 (Proof by Contradiction):
$$
\because f(1)-2 f(2)+f(3)=2 \text {. }
$$
$\therefore|f(1)|+2|f(2)|+|f(3)|$
$\geqslant|\mathrm{f}(1)-2 \mathrm{f}(2)+\mathrm{f}(3)|=2$. (1)
If $|\mathrm{f}(1)|<\frac{1}{2},|\mathrm{f}(2)|<\frac{1}{2}$,
$|\mathrm{f}(3)|<\frac{1}{2}$,
then $|\mathrm{f}(1)|+2|\mathrm{f}(2)|+|\mathrm{f}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,361 |
Example 2. Suppose $\mathrm{x}+\frac{1}{\mathrm{x}}=2 \cos \mathrm{a}$, prove:
$$
x^{n}+\frac{1}{x^{n}}=2 \cos n a .
$$ | Proof 1 (Mathematical Induction):
1) When $n=1$, $x+\frac{1}{x}=2 \cos a$, which is the given condition.
2) Assume that when $n=k$, $x^{k}+\frac{1}{x^{k}}=2 \cos ka$ holds,
then $\mathrm{x}^{\mathbf{k}+1}+\frac{1}{\mathrm{x}^{\mathbf{k}+1}}$
$$
\begin{aligned}
= & \left(x^{\mathbf{k}}+\frac{1}{x^{k}}\right)\left(x+\fra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,362 |
Example 3. Let $\mathrm{x}>0, \mathrm{y}>0, \mathrm{x}+\mathrm{y}=1$. Prove: $\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right) \geqslant \frac{25}{4}$. | Proof 1:
Let $y=1-x$, then
$$
\begin{array}{l}
\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right) \\
= \frac{x^{2}(1-x^{2})-2 x(1-x)+2}{x(1-x)} . \\
\because 0<x(1-x)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2} \\
\leqslant \frac{1}{4},
\end{array}
$$
$$
\begin{aligned}
\text { and } & x^{2}(1-x)^{2}-2 x(1-x)+2 \\
& =... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,363 |
Example 1. Select 4 people from 6 boys and 4 girls to participate in an extracurricular interest group. How many ways are there to select them?
(1) At least one boy and one girl participate;
(2) At most 3 boys participate. | Analysis: (1) Consider the selection of 4 students from 10 as the whole. The whole can be divided into two parts: one part is that the selected 4 students include at least one boy and one girl; the other part is that all 4 are either boys or girls, and these two parts are mutually exclusive. Therefore, the solution is:... | 195 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,365 |
Example 4. Find the minimum value that the polynomial $4 x^{2}+2 y^{2}-4 x y-4 y$
-1 can obtain.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Solve } 4 x^{2}+2 y^{2}-4 x y-4 y-1 \\
=\left(4 x^{2}-4 x y+y^{2}\right)+\left(y^{2}-4 y+4\right)-5 \\
=(2 x-y)^{2}+(y-2)^{2}-5 . \\
\because \quad(2 x-y)^{2} \geqslant 0,(y-2)^{2} \geqslant 0,
\end{array}
$$
When and only when $y=2, x=1$, this polynomial achieves its minimum value of -5. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,366 |
Example 2. If the three equations $\mathrm{x}^{2}+4 \mathrm{ax}-4 \mathrm{a}+3=0$, $x^{2}+(a-1) x+a^{2}=0, x^{2}+2 a x-2 a=0$, have at least one real solution, try to find the range of the real number $\mathrm{a}$.
Translate the above text into English, please retain the original text's line breaks and format, and out... | Solve $\left\{\begin{array}{l}(4 a)^{2}-4(-4 a+3)<0, \\ (a-1)^{2}-4 a^{2}<0, \\ 4 a^{2}-4(-2 a)<0 .\end{array}\right.$
to get $-\frac{3}{2}<\mathrm{a}<-1$. When $\mathrm{a}$ satisfies the above inequality,
all three equations have no real solutions, so when at least one of the three equations has a real solution, the r... | \left(-\infty, -\frac{3}{2}\right] \cup [-1, +\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,367 |
Example 3. On 99 cards, each is marked with a non-negative integer, and it is known that the sum S of these 99 integers does not exceed 1583. Prove that among these 99 cards, there are at least 4 cards with the same number. | Proof: Assuming that at most 3 cards have the same number, because the notation that minimizes the sum of the numbers on 99 cards is three 0s, three 1s, ..., three 32s, thus we have
$$
\begin{array}{l}
\mathrm{S} \geqslant \mathrm{S}_{\mathrm{min}}=3 \times(0+1+2+\cdots+32) \\
=3 \times \frac{32 \times(1+32)}{2} \\
=1... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,368 |
Example 4. Given that line 1 is
$\odot$ O's tangent, the point of tangency is
A. Prove: $O A \perp 1$. | Assume that OA is not perpendicular to 1. Draw OB ⊥ 1 at B, then B is the foot of the perpendicular. ∴ OB < OA = R, thus 1 intersects with ⊙O, which contradicts the given. ∴ OA ⊥ l.
Readers have seen that the derived OB < OA also contradicts the property in a right triangle that the hypotenuse is greater than the legs... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,369 |
Example 5. Let $\mathrm{a}, \mathrm{~b} \in J, \mathrm{a}^{2}+\mathrm{b}^{2}$ can be divided by 3, prove that $a$ and $b$ must be divisible by 3. | Proof: (1) Suppose one of a and b cannot be divided by 3, without loss of generality, let $\mathrm{a}=3 \mathrm{k} \pm 1$
$$
\begin{array}{l}
(k \in \mathbb{Z}), \text { then we get } a^{2}+b^{2}=(3 k \pm 1)^{2}+b^{2} \\
=9 k^{2} \pm 6 k+1+b^{2}=[3 k(3 k \pm 2)
\end{array}
$$
$+\mathrm{b}^{2}+1$. The number in the squa... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,370 |
Example 2. Find the maximum and minimum values of the function $y=\frac{x^{2}+x-1}{x^{2}+x+1}$. | Misunderstanding: The original equation is transformed into $(y-1) x^{2}+(y-1) x+y+1=0, \because x \in R, \therefore \Delta=(y-1)^{2}-4(y-1)(y+1) \geqslant 0$, which means $-\frac{5}{3} \leqslant y \leqslant 1$, thus $y_{min}=-\frac{5}{3}, y_{max}=1$.
Analysis: In fact, when $y=1$, it implies $x^{2}+x-1=x^{2}+x+1 \Rig... | y_{\mathrm{min}}=-\frac{5}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,372 |
Example 3. Prove that $3 \arcsin x = \arcsin (3x - 4x^3)$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | Misproof: $\because \sin (3 \arcsin x)$
$$
\begin{array}{l}
=3 \sin (\arcsin x)-4 \sin ^{3}(\arcsin x) \\
=3 x-4 x^{3}, \sin \left[\arcsin \left(3 x-4 x^{3}\right)\right] \\
=3 x-4 x^{3}, \\
\therefore 3 \arcsin x=\arcsin \left(3 x-4 x^{3}\right) .
\end{array}
$$
Analysis: From $\sin \alpha=\sin \beta$, generally we c... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,373 |
Example 4. Given $\log _{18} 9=a(a \neq 2)$, $18^{b}=5$, find $\log _{38} 45$. | Analysis: $\because \log _{18} 9 \neq \log _{18} 18^{2}=2$, $\therefore \log _{18} 9 \neq 2$, so the condition $a \neq 2$ in the problem is redundant. The reason the problem setter added the condition $\mathrm{a} \neq 2$ is mainly to avoid the denominator of the result $\log _{38} 45=\frac{a+b}{2-a}$ being 0, but this ... | \frac{a+b}{2-a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,374 |
Example 5. Given the function $y=\frac{5 x^{2}-2 k x+50}{4 x^{2}-10 x+25}$ has a minimum value of 1, find the real number $\mathrm{k}$. | Misunderstanding: $\because \frac{5 x^{2}-2 kx+50}{4 x^{2}-10 x+25} \geqslant 1$, and $4 x^{2}-10 x+25=4\left(x-\frac{5}{4}\right)^{2}+\frac{75}{4}>0$, $\therefore 5 x^{2}-2 kx+50 \geqslant 4 x^{2}-10 x+25$, i.e., $x^{2}-2(k-5) x+25 \geqslant 0, \quad \therefore \Delta \leqslant 0$, i.e., $(k-5)^{2}-25 \leqslant 0$, so... | k=0 \text{ or } k=10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,375 |
Example 5. If $x^{2}-1 \leqslant 0$, find the maximum and minimum values of the function $y=x^{2}-4 x$ +5. | $\begin{array}{l} \text { Sol } \because y=(x-2)^{2}+1, \text { and }-1 \leqslant x \leqslant 1, \\ \therefore \quad \text { when } x=1, \text { } y_{\text {minimum }}=2 . \\ \text { When } x=-1, \text { } y_{\text {maximum }}=10 .\end{array}$ | y_{\text{minimum}}=2, y_{\text{maximum}}=10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,377 |
Example 7. Prove that for all $\mathrm{k} \in \mathrm{R}$, the curve $\mathrm{C}$:
$$
\begin{array}{l}
f(x, y) \equiv x^{4}+k x^{3} y-6 x^{2} y^{2}-k x y^{3}+y^{4} \\
=0 \text { always divides a circle centered at the origin into eight equal parts. }
\end{array}
$$ | Misunderstanding: $\because \mathrm{f}(\mathrm{x}, \mathrm{y})=0$ holds for all real numbers $\mathrm{k}$, $\therefore\left\{\begin{array}{l}x^{3} y-x y^{3}=0, \\ x^{4}-6 x^{2} y^{2}+y^{4}=0,\end{array}\right.$ which are the four lines $l_{1}: x=0$, $l_{2}: y=0$, $l_{3}: x+y=0$, $l_{4}: x-y=0$. These four lines $l(i=1,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,378 |
Example 1. Prove: The sum of the cosines of the angles of an acute triangle with vertices on the unit circle is less than half the perimeter of the triangle. | Analysis: Let the internal angles of the acute triangle $\triangle \mathrm{ABC}$ be $\mathrm{A}, \mathrm{B}, \mathrm{C}$, and their opposite sides be $a, b, c$ respectively. By the Law of Sines and $R=1$, we have:
$$
\begin{aligned}
a+b+c & =2 R(\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}) \\
& \Rightarrow \frac{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,379 |
Example 2. In $\triangle \mathrm{ABC}$, prove that: $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C} \geqslant \sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}$. | $$
\begin{array}{l}
\text { Prove } \because \sin 2 \mathrm{~A}+\sin 2 \mathrm{~B} \\
=2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B}) \\
=2 \sin C \cos (\mathrm{A}-\mathrm{B}) \leqslant 2 \sin \mathrm{C} \text {. } \\
\Rightarrow 2 \sin C \geqslant \sin 2 A+\sin 2 B \text {. } \\
\text { Similarly, } 2 \si... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,380 |
Example 3. Let $\triangle \mathrm{ABC}$ be an acute triangle. Prove:
\[
\begin{array}{l}
\operatorname{tg} A(\operatorname{ctg} B+\operatorname{ctg} C)+\operatorname{tg} B(\operatorname{ctg} C \\
+\operatorname{ctg} A)+\operatorname{tg} C(\operatorname{ctg} A+\operatorname{ctg} B) \geqslant 6 .
\end{array}
\] | $\begin{array}{l}\text { Prove } \because 00, \operatorname{tg} \mathrm{B}>0, \operatorname{tg} \mathrm{C}>0 \text {. } \\ \Rightarrow \operatorname{tg}^{2} A+\operatorname{tg}^{2} B \geqslant 2 \operatorname{tg} A \cdot \operatorname{tg} B \\ \Rightarrow \frac{\operatorname{tg} \mathrm{A}}{\operatorname{tg} \mathrm{B}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,381 |
Example 4. Let $\alpha, \beta, \gamma$ be the three interior angles of an acute triangle. Prove that: $\sin \alpha+\sin \beta+\sin \gamma+\tan \alpha$ $+\tan \beta+\tan \gamma>2 \pi$ | $$
\begin{array}{c}
\text { Prove } \because 04 \operatorname{tg} \frac{\alpha}{2}>4 \cdot \frac{\alpha}{2}=2 \alpha .
\end{array}
$$
$$
\because \text { in } 02 \alpha.
\text { Similarly, } \sin \beta+\operatorname{tg} \beta>2 \beta,
\sin \gamma+\operatorname{tg} \gamma>2 \gamma.
$$
$$
\begin{array}{c}
(1)+(2)+(3) \te... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,382 |
Example 5. Let $A$, $B$, and $C$ be the three interior angles of a triangle. Prove that
$$
\operatorname{tg}^{2} \frac{A}{2}+\operatorname{tg}^{2} \frac{B}{2}+\operatorname{tg}^{2} \frac{C}{2} \geqslant 1 .
$$ | Analysis: Given $\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \Rightarrow \frac{\mathrm{A}}{2}=\frac{\pi}{2}-\frac{\mathrm{B}+\mathrm{C}}{2}$
$$
\begin{array}{l}
\Rightarrow \operatorname{tg} \frac{A}{2}=\operatorname{ctg} \frac{B+C}{2}=\frac{1-\operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}}{\operatorname{tg} \fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,383 |
Example 6. Let $\triangle \mathrm{ABC}$ be an acute triangle. Prove:
\[
\begin{array}{l}
\operatorname{tg} \mathrm{A}+\operatorname{tg} \mathrm{B}+\operatorname{tg} \mathrm{C}>1 . \\
\text { Analysis: In a non- } R \operatorname{t} \triangle \text {, we have: } \\
\operatorname{tg} \mathrm{A}+\operatorname{tg} \mathrm{... | $$
\text{Proof } \begin{aligned}
& \because 0 < \frac{\pi}{2} \\
\Rightarrow & 0 < 1 .
\end{aligned}
$$
Similarly, $\operatorname{tg} B \cdot \operatorname{tg} C > 1$,
$$
\operatorname{tg} C \cdot \operatorname{tg} A > 1 \text{. }
$$
Multiplying (1), (2), and (3) we get: $\operatorname{tg}^2 A \cdot \operatorname{tg}^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,384 |
$\begin{array}{l}\text { Example 7. Prove: The three interior angles } \mathrm{A}, \mathrm{B}, \mathrm{C} \text { of a triangle satisfy: } \\ 2 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}<\sin ^{2} \mathrm{~A}(\sin \mathrm{B} \\ +\sin \mathrm{C}-\sin \mathrm{A})+\sin ^{2} \mathrm{~B}(\sin \mathrm{C}+\sin \mathrm{A}... | Analysis: Using the sine theorem, it is easy to obtain the equivalent inequality of the original problem as:
$$
\begin{aligned}
2 a b c \mathrm{c} \Rightarrow \mathrm{a}+\mathrm{b}-\mathrm{c}>0.
\text{ Similarly, } b+c-a>0.
\end{aligned}
$$
$$
c+a-b>0.
$$
Multiplying (1), (2), and (3) yields:
$$
(a+b-c)(b+c-a)(c+a-b)>0... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,385 |
Example 1. Factorize: $2 x^{3}-x^{2}-x-3$.
| Solve $\begin{aligned} & 2 x^{3}-x^{2}-x-3 \\ = & 2 x^{3}+2 x^{2}+2 x-3 x^{2} \cdots 3 x-3 \\ = & \left.2 x i x^{2}+x+1\right)-3\left(x^{2}+x+1\right) \\ = & (2 x-3)\left(x^{2}+x+1\right)\end{aligned}$ | (2 x-3)\left(x^{2}+x+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,386 |
Example 2. Factorize:
$$
x^{4}+2 x^{3}+3 x^{2}+2 x+1
$$ | Solve $\begin{aligned} & x^{4}+2 x^{3}+3 x^{2}+2 x+1 \\ = & x^{4}+2 x^{2}+1+2 x^{3}+2 x+x^{2} \\ = & \left(x^{2}+1\right)^{2}+2 x\left(x^{2}+1\right)+x^{2} \\ = & \left(x^{2}+x+1\right)^{2} .\end{aligned}$ | \left(x^{2}+x+1\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,387 |
Example 6. Find the minimum value of $y$ in the equation $x^{2}+8 x-y-11=0$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve: Treating $\mathrm{y}$ as coefficients, the discriminant of the quadratic equation in $\mathrm{x}$ is $\triangle=8^{2}-4(-y-11)$
$$
\begin{array}{l}
\quad=108+4 y . \\
\because \quad x \text { is a real number, } \\
\therefore \quad \triangle=108+4 y \geqslant 0 . \\
\therefore \quad y \geqslant-27, \\
\therefore... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,388 |
Example 3. Factorize: $x^{5}+x^{4}+1$.
| Solve $\begin{aligned} & x^{5}+x^{4}+1 \\ = & \frac{\left(x^{5}+x^{4}+1\right)(x-1)}{x-1} \\ = & \frac{x^{6}+x^{5}+x-x^{5}-x^{4}-1}{x-1} \\ = & \frac{x^{8}-1+x-x^{4}}{x-1} \\ = & \frac{\left(x^{3}+1\right)\left(x^{3}-1\right)-x\left(x^{3}-1\right)}{x-1} \\ = & \frac{\left(x^{3}-1\right)\left(x^{3}-x+1\right)}{x-1} \\ =... | (x^{2}+x+1)(x^{3}-x+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,389 |
Example 4. Factorize: $x^{10}+x^{5}+1$. | Solve: Substituting $x=\omega$ into $x^{10}+x^{5}+1$, we get $\omega^{10}+\omega^{5}+1=\omega+\omega^{2}+1=0$.
$\therefore x^{10}+x^{5}+1$ must have a factor $x^{2}+x+1$.
By synthetic division, we get $\left(x^{10}+x^{5}+1\right)$
$$
\begin{array}{c}
\div\left(x^{2}+x+1\right) \\
=x^{8}-x^{7}+x^{6}-x^{4}+x^{3}-x+1 \\
\... | x^{10}+x^{5}+1=\left(x^{2}+x+1\right)\left(x^{8}-x^{7}+x^{6}-x^{4}+x^{3}-x+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,390 |
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