problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Three, it is known that there exists a positive integer $n$, such that the number $11 \cdots 11$ can be divided by $n$ 1987s. Prove that the numbers
$$
\begin{array}{l}
p=\underbrace{11 \cdots 1199 \cdots 9988 \cdots 8877 \cdots 77}_{n \uparrow} \underbrace{9 \uparrow}_{n \uparrow} \\
q=11 \cdots 1199 \cdots 9988 \cdot... | $$
\begin{array}{l}
\text { III. Summoning } p=\underbrace{11 \cdots 11}_{n \text { digits }}\left(10^{3 n}+9\right. \\
\left.\times 10^{2 n}+8 \times 10^{n}+7\right) . \\
\because \underbrace{11 \cdots 11}_{n \text { digits }} \text { is divisible by } 1987, \\
\end{array}
$$
$\therefore \quad p$ is divisible by 1987.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,239 |
3. Find an integer-coefficient polynomial for which $a=\sqrt[3]{2}+\sqrt[3]{3}$ is a root.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
3 \alpha^{3}=(\sqrt[3]{2}+\sqrt[3]{3})^{3} \\
=5+3 \sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3}) \\
=5+3 \sqrt[3]{6} a \\
\therefore\left(a^{3}-5\right)^{3}=162 a^{3} \\
\therefore a^{9}-15 a^{8}-87 a^{3}-125=0 .
\end{array}
$$
Therefore, the required polynomial can be
$$
p(x)=x^{9}-15 x^{8}-87 x^{3}-125 \t... | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 703,242 |
4. Each cell of an infinite grid paper is colored with one of $n$ colors $(n \geqslant 2)$. Prove that it is possible to find four cells of the same color, the centers of which are the vertices of some rectangle, the sides of which are parallel to the grid lines of the paper. | 4. From the grid paper, separate out a horizontal strip containing $n+1$ squares. Each vertical column in the strip contains $n+1$ squares, each of which is colored with no more than $n$ colors, so there must be at least two squares of the same color in each column. The number of columns in the strip is infinite, while... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,243 |
3. Four settlements are located at the vertices of a square with a side length of $10 \mathrm{~km}$. Is it possible to connect these settlements with a road network such that the total length of the roads is less than $28 \mathrm{~km}$, and each settlement is connected to every other settlement? | 3. If straight roads connect $A C$ and $B D$, as shown in Figure 5, the road network clearly meets the second requirement, however
$$
A C+B D=2 \cdot 10 \cdot \sqrt{2}>28
$$
does not satisfy the first requirement.
Suppose $E$ and $F$ are the midpoints of the opposite sides $B C$ and $A D$ of the square $A B C D$, and ... | 10(\sqrt{3}+1)<28 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,247 |
Example 4. Take a point $M$ inside the rectangle $A B C D$, prove that there exists a quadrilateral, the lengths of its sides are equal to $A M, B M, C M$ and $D M$, and its diagonals are perpendicular to each other and their lengths are equal to $A B$ and $B C$.
untranslated text remains the same for the part that i... | $$
\begin{array}{l}
\text { Prove } M \xrightarrow{T(\vec{A} \vec{B})} M^{\prime}, \\
A \xrightarrow{T(\vec{A} \vec{B})} B, \\
D \xrightarrow{T(\vec{A} \vec{B})} C, \\
\end{array}
$$
then
$$
\begin{array}{l}
B M^{\prime}=A M, \quad C M=D M, \\
M M^{\prime}=A B, \quad H M N^{\prime} \perp B C .
\end{array}
$$
$\therefo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,249 |
5. There are 9 points distributed in the space outside a sphere. Prove that there is a point on the sphere's surface from which no more than 3 of the 9 points can be seen.
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。
5. There are 9 points distributed in the space outside a sphere. Prove that there is a point on the sphere's su... | 5. Let $\alpha$ be any plane passing through the center of a sphere, and let $l$ be a line passing through the center of the sphere and perpendicular to the plane $\alpha$. It intersects the sphere at the endpoints of a diameter, as shown in Figure 9. The plane $\alpha$ divides the space into two half-spaces. Another p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,250 |
4. Three circles with the same radius $R$ have a common point. Prove that if these circles intersect each other in three other points, then the radius of the circle passing through these three points is also $R$.
| 4. [Method 1] As shown in Figure 13, let $a = \angle A O_{1} D = \angle A O_{3} D$, $\beta = \angle D O_{1} B = \angle D O_{2} B$, $\gamma = \angle D O_{2} C = \angle C O_{3} D$. It is easy to prove that $\angle B D C = 180^{\circ} - \frac{\beta + \gamma}{2}$, $\angle B A C = \frac{\beta + \gamma}{2}$. Therefore, $\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,254 |
5. As shown in Figure 12, line $l$ is the boundary of the forest. The rabbit and the wolf are located at points $A$ and $B$ on the perpendicular line $AC$ to line $l$ $(AB = BC = a)$.
They run at fixed speeds, with the rabbit's speed being twice that of the wolf: if the wolf arrives at a point earlier than or at the sa... | 5. Draw a Cartesian coordinate system as shown in Figure 14. Suppose the initial positions of the rabbit and the wolf are $A(0,2a)$ and $B(0,a)$, respectively, and their speeds are $2v$ and $v$. If the rabbit runs along a straight line to point $M(x, y)$, then at time $t_{1}=\frac{A M}{2 v}$, it can reach point $M$. Th... | C D > \frac{2a}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,255 |
Example 5. Seven lines intersect each other pairwise, and among the angles formed, at least one is less than $26^{\circ}$. | Proof: Select any point $P$ on the plane, and translate the seven known lines so that they all pass through point $P$, becoming seven lines intersecting at $P$. These lines divide the circle centered at $P$ into 14 adjacent angles, which we can denote as $a_{1}, \alpha_{2}, \cdots, \alpha_{14}$. Each of these 14 angles... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,256 |
Example 6. Given a regular hexagon $A B C D E F$, points $M$ and $K$ are the midpoints of sides $C D$ and $D E$ respectively, and $L$ is the intersection of segments $A M$ and $B K$. Prove that the area of triangle $A B L$ is equal to the area of quadrilateral $M D K L$, and find the angle between lines $A M$ and $B K$... | $E^{R\left(1,60^{\circ}\right)} \rightarrow I$,
$g \xrightarrow{R\left(O, 60^{\circ}\right)} \rightarrow C$.
$K$ is the midpoint of $E D$, $M$ is the midpoint of $C D$.
$$
\begin{array}{l}
\therefore K \xrightarrow{P\left(O, 60^{\circ}\right)} \rightarrow M, \\
C \xrightarrow{P\left(O, 60^{\circ}\right)} \rightarrow B,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,257 |
Example 7. In the square $A_{1} A_{2} A_{3} A_{4}$, take a point P, draw lines from $A_{3}$ to $A_{4} P$ and from $A_{4}$ to $A_{1} P$. Prove: the four lines drawn (including their extensions) intersect at one point. | Prove: Take the center $O$ of $A_{1} A_{2} A_{3} A_{4}$. Under $P \xrightarrow{\left(O,-90^{\circ}\right)} \rightarrow$, $l_{1}$ becomes $A_{2} P, l_{2}$ becomes $A_{3} P, l_{3}$ becomes $A_{4} P, l_{4}$ becomes $A_{1} P$. Since $A_{1} P$, $A_{2} P, A_{3} P, A_{4} P$ intersect at point $P$, when under $R\left(O, 180^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,258 |
Theorem 3 For three distinct and non-collinear rotation centers
$\mathrm{A}, \mathrm{B}, \mathrm{C}$, perform three successive rotations $\mathrm{R}(\mathrm{A}, \alpha), \mathrm{R}$ (B
$\beta), R(C, \gamma)$. If $\alpha+\beta+\gamma=2 \pi$, and
$R(A, \alpha) R(B, \beta) R(C, \gamma)=I$, then
$$
\angle \mathrm{CAB}=\fra... | Given $\because \alpha+\beta \neq 2 \pi$, then by
Theorem 2 we have
$$
\begin{array}{l}
R(A, \alpha) R(B, \beta) \\
=R(O, \alpha+\beta) .
\end{array}
$$
Assume point $\mathrm{O}$ does not coincide with point $\mathrm{C}$,
since $\alpha+\beta+\gamma=2 \pi$, then by Theorem 2, $R(O, \alpha+\beta) R(C, \gamma)$ is a tran... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,261 |
$\begin{array}{c}\text { For example, solve the equation } \\ 4 x^{2}-40 [ x ] \\ +51=0 .\end{array}$ | Let $x=[x]+r, \quad-$ $0 \leqslant r<1$.
Then we have $4 r^{2}+8 r\lceil x\rfloor$ $+4[x]^{2}-40[x]+51=0$,
Clearly, $[x] \geqslant \frac{51}{40}$, and since $r \geqslant 0$, we have
$$
0 \leqslant \frac{-2[x]+\sqrt{40[x]-51}}{2}<1 .
$$
Solving this inequality, noting that $[x]$ is an integer, we get
$$
[x]=2,6,7,8 \t... | x_{1}=\frac{\sqrt{29}}{2}, x_{2}=3 \sqrt{21}, x_{3}=\sqrt{\frac{229}{2}}, x_{4}=\sqrt{269} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,263 |
Example 4. Prove that the equation
$$
[x]+[2 x]+[4 x]+[8 x]+[16 x]
$$
$+[32 x]=12345$ has no solution in integers. | Assume the equation has a real solution $x$,
Let $x=[x]+r, 0 \leqslant r<1$.
Thus, $n x=n[x]+n r$,
$$
[n x]=n[x]+[n r] .
$$
Therefore, the original equation can be transformed into
$$
\begin{array}{l}
63[x]+[r]+[2 r]+[4 r]+[8 r] \\
+[16 r]+[32 r]=12345 .
\end{array}
$$
Since $12345=63 \times 195+60$,
Thus, $[r]+[2 r... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,264 |
Example 5. Let $x$ be a positive real number, $n \in N$, prove
$$
\begin{aligned}
& {[n x] \geqslant \frac{[x]}{1}+\frac{[2 x]}{2}+\frac{[3 x]}{3}+\cdots } \\
+ & \frac{[n x]}{n} .
\end{aligned}
$$ | Proof: Let $x_{n}=\frac{[x]}{1}+\frac{[2 x]}{2}+\frac{[3 x]}{3}+\cdots$
$$
+\frac{[n x]}{n} \text {. }
$$
Thus, the problem reduces to proving
$$
[n x] \geqslant x_{n} .
$$
We will use mathematical induction on $n$.
For $n=1$, $[x]=x_{1}$, the inequality holds;
Assume the inequality holds for $k \leqslant n-1$, i.e.,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,265 |
Example 6. Solve the inequality
$$
[x]\{x\}<x-1 \text {. }
$$ | Let $x=[x]+\{x\}$, then
$$
\begin{array}{l}
{[x]\{x\}0,
$$
[x]>1 \text {. }
$$
Therefore, $x \geqslant 2$. | x \geqslant 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 703,266 |
Example 7. Let $a, b, m$ be integers, if $a$ and $m$ are coprime, find the value of the sum
$$
\begin{array}{l}
\left\{\frac{b}{m}\right\}+\left\{\frac{a+b}{m}\right\}+\left\{\frac{2 a+b}{m}\right\}+\cdots \\
+\left\{\frac{(m-1)}{m} \frac{a+b}{m}\right\} \text{. }
\end{array}
$$ | We first prove that $\left\{\begin{array}{c}a x+b \\ m\end{array}\right\}$ when $x=0$,
$1,2, \cdots, m-1$, they are all distinct.
Proof: If there exist $x_{1}, x_{2} \in\{0$, $1,2, \cdots, m-1\}$ such that
$$
\left\{\frac{a x_{1}+b}{m}\right\}=\left\{\frac{a x_{2}+b}{m}\right\},
$$
then $\frac{a x_{1}+b}{m}-\frac{a x... | \frac{m-1}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,267 |
$$
\begin{array}{l}
{\left[\frac{1^{2}}{1980}\right],\left[\frac{2^{2}}{1980}\right],\left[\frac{3^{2}}{1980}\right], \cdots,} \\
{\left[\frac{1900^{2}}{1980}\right] \text { How many different numbers are there in the sequence? }}
\end{array}
$$ | Solution: First, note that when $\alpha-\beta>1$, the value of $[a] 1_{j} [\beta]$ is definitely different. Therefore, we solve the inequality
$$
\frac{(k+1)^{2}}{1980}-\frac{k^{2}}{1980}>1,
$$
which simplifies to
$$
\begin{array}{l}
2 k+1>1980, \\
k>989 .
\end{array}
$$
Thus, starting from the 990th term, these 1980... | 1486 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,268 |
Example 9. Let $S=1+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}+\cdots$
$$
\begin{array}{l}
+\frac{1}{\sqrt{2 k-1}}+\cdots \frac{1}{\sqrt{2 k+1}} \\
+\cdots+\frac{1}{\sqrt{(2 n+1)^{2}}} .
\end{array}
$$
Prove: $[S]=2 n$. | $$
\begin{array}{l}
\text { Prove that when } k \geqslant 2, \text { still } \\
(\sqrt{k+2}-\sqrt{k})^{2}>0, \\
(\sqrt{k}-\sqrt{k}-2)^{2}>0. \\
\end{array}
$$
For example,
$$
\begin{array}{l}
\text { ( } k=2 \text { ) } \\
\end{array}
$$
$$
\text { Let } k=3,5,7, \cdots,(2 n+1)^{2} \text {, then }
$$
Add the inequali... | [S]=2 n | Algebra | proof | Yes | Yes | cn_contest | false | 703,269 |
Example 10. Prove the Hermite's identity
$$
\begin{array}{l}
{[x]+\left[x+\frac{1}{n}\right]+\cdots+\left[x+\frac{n-1}{n}\right]} \\
=[n x] . \quad(n \in N)
\end{array}
$$ | Obviously, for a determined $x$ and $n$, we can choose such a $k$ that
$$
[x]+\frac{k-1}{n} \leqslant x<[x]+\frac{k}{n} .
$$
From this, we can derive
$$
\begin{array}{l}
{[x]+k-\frac{1}{n} n-k<x+\frac{n-k}{n}} \\
<[x]+\frac{k+n-k}{n} . \\
\end{array}
$$
Thus, for the first $n-h+1$ terms on the left side of the equati... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,270 |
Example 15. For $\triangle \mathrm{ABC}$, construct equilateral triangles on the three sides outward, $\mathrm{ABC}^{\prime}, \mathrm{BCA}^{\prime}, \mathrm{CAB}^{\prime}$. Their circumcenters are $\mathrm{O}_{1}$, $\mathrm{O}_{2}$, $\mathrm{O}_{3}$ respectively. Prove that $\triangle \mathrm{O}_{1} \mathrm{O}_{2} \mat... | $$
\begin{array}{l}
\angle \mathrm{AO}_{1} \mathrm{~B}=\angle \mathrm{BO}_{2} \mathrm{C}=\angle \mathrm{CO}=\mathrm{A}=120^{\circ}, \\
\mathrm{AO}_{1}=3 \mathrm{O}_{1}, 13 \mathrm{O}_{2}=\mathrm{CO}_{2}, \mathrm{CO}=\mathrm{AO}_{3} . \\
\mathrm{A} / \mathrm{R}\left(\mathrm{O}_{2}, 120^{\circ}\right)=\mathrm{C}\left(\ma... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,272 |
Example 2. (IMO12-2)
Given natural numbers $a, b, n, \Pi a>1, b>1$, $n>1, A_{n-1}$ and $A_{n}$ are numbers in the base-$a$ numeral system, $B_{n-1}$ and $B_{n}$ are numbers in the base-$b$ numeral system. $A_{n-1}, A_{n}$, $B_{n-1}$, and $B_{n}$ are in the following forms:
$$
A_{n-1}=x_{n-1} x_{n-2} \cdots x_{0}, A_{n}... | $$
\begin{array}{l}
\text{This problem involves number bases. We know that in decimal, we have} \\
1234=1 \times 10^{3}+2 \times 10^{2}+3 \times 10+4 . \\
\text{In base } k, \text{ we have} \\
1234=1 \times k^{3}+2 \times k^{2}+3 \times k+4 . \\
\text{Thus, in this problem, we have} \\
A_{n}=x_{n} a^{n-1}+x_{n-1} a^{n-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,273 |
Example 3. (IMO6-2)
Let $a, b, c$ be the lengths of the sides of a triangle. Prove that
\[
\begin{array}{l}
a^{2}(b+c-a)+b^{2}(c+a-b) \\
+c^{2}(a+b-c) \leqslant 3 a b c .
\end{array}
\] | Proof 1 To prove the inequality symmetric in $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c>0$. Thus,
$$
\begin{array}{l}
3 a b c-a^{2}(b+c-a)-b^{2}(c+a-b) \\
-c^{2}(a+b-c) \\
\quad=a(a-b)(a-c)+b(b-c)(b-a) \\
\quad+c(c-a)(c-b) \\
\geqslant a(a-b)(a-c)+b(b-c)(b-a) \\
\geqslant a(a-b)(a-c)+a(b-c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,274 |
Example 4. Let $a, b, A, B$ be known real numbers. If for any real number $x, f(x)=1-a \cos x-b \sin x$ $-A \cos 2 x-B \sin 2 x>0$. Prove: $a^{2}+b^{2} \leqslant 2$; $A^{2}+B^{2} \leqslant 1$. | Prove that for $\varphi$ and $\theta$, we can choose 0 and $\varphi$:
$$
\begin{aligned}
f(x)= & 1-\sqrt{a^{2}+b^{2}} \cos (x+\theta) \\
& -\sqrt{A^{2}+B^{2}} \cos 2(x+\varphi) .
\end{aligned}
$$
By taking $x=-\theta+\frac{\pi}{4}$ and $-\theta-\frac{\pi}{4}$, we get
$$
\begin{array}{l}
f\left(-\theta+\frac{\pi}{4}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,275 |
Example 6. (IMO17-1)
Let $x_{1}, y_{i}(i=1,2, \cdots, n)$ be real numbers, and $x_{1}>x_{2}>\cdots \geq x_{n}, y_{1}>y_{2} \geq \cdots \geq y_{n}, z_{1}, z_{2}$,
$$
\sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} \leqslant \sum_{i=1}^{n}\left(x_{i}-z_{i}\right)^{2} .
$$ | Prove that $\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} z_{i}^{2}$, so the original inequality is equivalent to
$\sum_{i=1}^{n} x_{i} y_{i}($ ordered sum $) \geqslant \sum_{i=1}^{n} x_{i} z_{i}$ (disordered sum $)$. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,277 |
Example 7. (IMO11-6)
Prove that for all real numbers $x_{1}$, $x_{2}$, $y_{1}$, $y_{2}$, and $z_{1}$, $z_{2}$ satisfying $x_{1}>0, x_{2}>0$, $x_{1} y_{1}-z_{1}^{2}>0, x_{2} y_{2}-z_{2}^{2}>0$, the inequality
$$
\begin{array}{l}
\frac{8}{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}}- \\
... | Let $a=x_{1} y_{1}-z_{1}^{2}>0, b=x_{2} y_{2}-z_{2}^{2}>0$, then $x_{1} y_{1}=a+z_{1}^{2}, x_{2} y_{2}=b+z_{2}^{2}$. Therefore,
$$
\begin{array}{l}
\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2} \\
=a+b+x_{1} y_{2}+x_{2} y_{1}-2 z_{1} z_{2} \\
=a+b+\frac{x_{1}}{x_{2}} x_{2} y_{2}+\frac{x_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,278 |
Example 8. Let there be an infinite sequence $\left\{x_{i}\right\}: x_{0}=1$, $x_{i+1} \leqslant x_{i}(i=0,1,2, \cdots)$.
a) Prove that for every such sequence, there is an $n \geqslant 1$, such that
$$
\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geqslant 3.999 .
$$
b) Find such a ... | a) We first use mathematical induction to prove the following conclusion: for any sequence $1=x_{0} \geqslant x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant \cdots \geqslant 0$, there exists a positive sequence $c_{n}(n \geqslant 1)$, such that
$$
\begin{array}{l}
\frac{x_{i}^{2}}{x_{i+1}}+\frac{x_{i+... | 3.999 | Inequalities | proof | Yes | Yes | cn_contest | false | 703,279 |
Example 1. Solve the equation $|\operatorname{tg} x+\operatorname{ctg} x|=\frac{4}{\sqrt{3}}$. | $$
\left.\pm \frac{5 \pi}{8}, k \in Z\right\} \text {. }
$$
Interpretation 1: The original equation is transformed into $\sin \angle x= \pm \frac{\sqrt{ } 3}{2}$.
Solving, we get $x=\frac{\sqrt{2}}{2} \pm \frac{\pi}{6} \cdot(k \in Z)$
Solution 2. After simplification, we get
$$
3 \operatorname{tg}^{4} x-10 \operatorn... | x=k \pi \text{ or } x=\frac{k \pi}{4}+\frac{\pi}{8}, k \in Z | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,281 |
Example 1. Given that $x_{1}$ and $x_{2}$ are the two roots of the equation $a x^{2}+b x+c=0$ (where $a \neq 0$). Prove: $S_{n}=x_{1}^{n}+x_{2}^{n}$
$$
=-\frac{b S_{n-1}+c S_{n-2}}{a} \text {. }
$$ | $$
\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a} . \\
\therefore s_{n}= x_{1}^{n}+x_{2}^{n}=\left(x_{1}+x_{2}\right) S_{n-1} \\
-x: x_{2} S_{n-2} \\
= \frac{b}{a} S_{n-1} - \frac{c}{a} S_{n-2} \\
=-\frac{b S_{n-1}+c S_{n-2} .}{a} .
\end{array}
$$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,282 |
Example 2. Given $\sin \alpha+\cos \alpha=a$.
(1) Find the value of $\sin ^{5} \alpha+\cos ^{5} \alpha$;
(2) If $a=1$, find the value of $\sin ^{n} \alpha+\cos ^{n} \alpha$. | Let $f(n)=\sin ^{n} \alpha+\cos ^{n} \alpha$, then (1) $=a$.
Given $f(2)=\sin ^{2} \alpha+\cos ^{2} \alpha=1$,
$f(3)=a f(2)-\frac{a^{2}-1}{2} \cdot a$
$=\frac{-a^{3}+3 a}{2}$,
$f(4)=a f(3)-\frac{a^{2}-1}{2} f(2)$
$=-a^{4}+2 a^{2}+1$
2
$\therefore f(5)=\sin ^{5} \alpha+\cos ^{5} \alpha$
$=a f(4)-\frac{a^{2}-1}{2} f(3)$
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,284 |
Example 3. Solve the equation: $\sqrt[5]{171-x}+\sqrt[5]{104+x}=5$.
| Let $u=\sqrt[5]{171-x}, v=\sqrt[5]{104+x}$, then
$$
\left\{\begin{array}{l}
u+v=5 \\
u^{5}+v^{5}=275
\end{array}\right.
$$
From (2) we get $u^{5}+v^{5}=(u+n)\left(u^{2}+v^{4}\right)$
$$
\begin{array}{l}
-u v\left(u^{5}+v^{3}\right) \\
=\cdots=(u+v)^{5}-5 u v(u+v)^{3} \\
-1_{2}^{2} v^{2}(u+v) \\
=5^{5}-5^{4} u v-5^{2} ... | x_1=139, x_2=-72 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,285 |
Example 4. Let $\alpha, \beta$ be the roots of the equation $x^{2}-4 x+1=0$. Prove that $\alpha^{n}+\beta^{n}$ is always an integer.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof (Mathematical Induction)
(1) When $n=1$, from $\alpha \alpha+\beta=4$, we know that the conclusion holds for $n=1$.
(2) Assume that the conclusion holds for $n \leqslant k-1(k \geqslant 2)$, then
$$
\begin{aligned}
\alpha^{k} & +\beta^{k}=(\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
& -\alpha \beta\left... | null | Algebra | proof | Yes | Yes | cn_contest | false | 703,286 |
Example 5. Given $x+\frac{1}{x}=2 \cos 0$. Prove:
$$
x^{n}+\frac{1}{x^{n}}=2 \cos n 0 \text {. }
$$ | Prove (1) When $n=1$, the conclusion is obviously true.
For $n \geqslant 2$, we have $x^{n}+\frac{1}{x^{n}}$
$=\left(x+\frac{1}{x}\right)\left(x^{n-1}+\frac{1}{x^{n}-1}\right)$
$-\left(x^{n-2}+\frac{1}{x^{\frac{1}{n}-2}}\right)=2 \cos \theta\left(x^{n-1}\right.$
$\left.+\frac{1}{x^{n-1}}\right)-\left(x^{n-2}+\frac{1}{x... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,287 |
Example 6. In the sequence $\left\{a_{n}\right\}$, for any natural number $n(n \geqslant 2)$, we have $a_{n}=3 a_{n-1}-2 a_{n-2}$, and $a_{0}=2, a_{1}=3$, find the general term formula of this sequence. | Solve for $\begin{array}{l} a_{0}=2=2^{0}+1, \\ a_{1}=3=2^{1}+1, \\ a_{2}=3 a_{1}-2 a_{0}=2^{2}+1, \\ a_{3}=3 a_{2}-2 a_{1}=9=2^{3}+1, \cdots, \\ \text { and } 2^{n}+1=(2+1) a_{n-1}-(2 \cdot 1) a_{n-2} \\ =3 a_{n-1}-2 a_{n-2}(n \geqslant 2),\end{array}$ | a_n = 2^n + 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,288 |
Example 7. Prove that $(3+\sqrt{5})^{n}+(3-\sqrt{5})^{n}$ is divisible by $2^{n}$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format, feel free to let me know! | Proof
（1）When $n=1$, the conclusion holds.
（2）Assume that when $n \leqslant k-1 （ k \geqslant 2 ）$, the conclusion holds, i.e., $\left[(3+\sqrt{5})^{k-1}+(3-\sqrt{5})^{k-1}\right]$ and $\left[(3+\sqrt{5})^{k-2}+(3-\sqrt{5})^{k-2}\right]$ can be divided by $2^{k-1}, 2^{k-2}$ respectively.
$$
\begin{array}{l}
\therefore(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,289 |
Given: $I=C, A=Q^{-}$. Find $\bar{A}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve the following problems.
(1) $\bar{A}=Q^{+}$;
(2) $Z=Q^{+} \cup \bar{R}$;
(3) $\bar{A}=Q^{+} \cup \bar{R} \backslash\{0\}$. The complex plane is divided into $\bar{Q}, Q^{+}, Q^{-}, \{0\}$, so $\bar{A}=\bar{Q} \cup Q^{+} \cup\{0\}$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,292 |
Proof: The intersection lines intersect at one point, or are parallel.
Given: $\alpha \cap \beta=a, a \cap \gamma=b, \beta \cap \gamma=c$.
To prove: $a \cap \beta \cap \gamma=A$, or $a / / b / / c$. | Analysis: Because the positional relationship of three lines is relatively complex, it is difficult to approach the problem by merely considering the positional relationship of the three intersection lines. We know that the positional relationship between two lines only has three possibilities: parallel, intersecting, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,293 |
Example 17. Once upon a time, there was a wealthy and adventurous young man who, in his great-grandfather's legacy, found a piece of parchment. On it was recorded the location of a treasure, which read as follows:
“Sail to latitude $\times \times$, longitude $\times \times$, and you will find a deserted island. On the ... | $$
\begin{array}{l}
\quad \mathrm{R}\left(\mathrm{B}, 90^{\circ}\right) \\
\therefore\left.\mathrm{R}, \mathrm{A}, 90^{\circ}\right) \mathrm{R}\left(\mathrm{E}, 180^{\circ}\right) \mathrm{R}\left(\mathrm{B}, 90^{\circ}\right) \\
= \mathrm{I}, \\
\text { Also } 90^{\circ}+180^{\circ}+90^{\circ}=360^{\circ}, \\
\therefo... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,294 |
Learn, there are 2 teachers who can teach mathematics, and there are 4 teachers who can teach both English and Japanese. Now, 3 mathematics teachers and 3 Japanese teachers are being dispatched to teach outside the school during the holiday. How many ways are there to select them?
保留源文本的换行和格式，直接输出翻译结果。 | Let the 3 teachers competent in English teaching be set $A$, 2 teachers competent in Japanese teaching be set $B$, and 4 teachers competent in both English and Japanese teaching be set $C$.
Method 1: Classify set $A$
(1) Select 3 English teachers (choose 3 Japanese teachers from $B$ and $C$), total $C_{3}^{3} \cdot C_... | 216 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,295 |
Example 1. Find the value of $\sin \frac{3 \pi}{10}-\sin \frac{\pi}{10}$. | Solution: Let $z=\cos \frac{\pi}{10}+i \sin \frac{\pi}{10}$.
By De Moivre's Theorem, we get, $z^{10}=-1, z^{5}=i$. | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,296 |
Example 4. Solve the trigonometric equation: $\sin 3 x=\sin 2 x$. | Given $z=\cos x+i \sin x$.
$$
\frac{z^{3}-1}{2 i z^{3}}=\frac{z^{2}-1}{2 i z}
$$
Expanding and rearranging, we get $z^{\theta}-z^{4}+z^{2}-1=0$.
Solving the equation, we have
$$
\begin{array}{l}
\left(z^{2}-1\right)\left(z^{4}+1\right)=0 . \\
\text { When } z^{2}-1=0, \\
z^{2}=1, \text { i.e., } \cos 2 x+i \sin 2 x=1 ... | x=k \pi, k \in \mathbb{Z} \cup x=\frac{1}{4}(2 k-1) \pi, k \in \mathbb{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,299 |
Given: $a^{2}, b^{2}$ and $1 c^{2}$ are squares of consecutive integers. If $a^{2}$ and $c^{2}$ are known, without directly calculating the square roots of $a^{2}$ or $c^{2}$, determine the value of $b^{2}$.
Without using the calculation of the square roots of $a^{2}$ or $c^{2}$, find the value of $b^{2}$. | [Method 1] Given: $b^{2}=\frac{a^{2}+c^{2}}{2}-1$,
think of $c=a+2$. Then,
$$
\begin{array}{l}
4^{2}+\frac{(a+2)^{2}}{2}-1=\frac{a^{2}+a^{2}+4 a+4}{2}-1 \\
=\frac{2 a^{2}+4 a+4}{2}-1=a^{2}+2 a+1=(a+1)^{2}
\end{array}
$$
By definition, this is $b^{2}$.
[Method 2] Prove: $b^{2}=\left(\frac{c^{2}-a^{2}}{4}\right)^{2}$, t... | b^{2} = (a+1)^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,300 |
Example $1 . a$ For what value of $a$,
$$
(a+1) x^{2}+(a-3) x+(a-5)=0
$$ | Given: Taking $n=1$, $m=3$, then
$$
\begin{array}{c}
(a-3)^{2}-4(a+1)(a-5)=9(a+1)^{2}, \\
a=1 \text { or } -\frac{5}{3} .
\end{array}
$$ | a=1 \text { or } -\frac{5}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,301 |
Example 2. The ratio of the two roots of the equation $a x^{2}+b x+c=0$ is $2: 3$, prove that $6 b^{2}=25 a_{c}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The notation $a_{c}$ in the original problem seems to be ... | Solution: Let the two roots be $x_{1}, x_{2}$, then $x_{1}=2 x_{2} / 3$. We have
$$
\frac{2}{3} b^{2}-\left(\begin{array}{l}
2 \\
3
\end{array}+1\right)^{2} a c=0,
$$
which is $6 b^{2}=25 a c$. | null | Algebra | proof | Yes | Yes | cn_contest | false | 703,302 |
$$
\begin{array}{l}
4(m-1)^{2} x^{2}+4(m-1)(m+3) x \\
+(m+1)(m+5)=0
\end{array}
$$
Always has one root that is 1 more than three times the other. | To prove that the conclusion holds, it is sufficient to show that when $m \neq 1$,
$$
\begin{array}{l}
3[4(m-1)(m+3)]^{2}-4^{2} \cdot 4(m-1)^{2} \\
\cdot(m+1)(m+5) \\
=4(m-1)^{2}\left[4(m-1)^{2}\right. \\
-2 \cdot 4(m-1)(m+3)],
\end{array}
$$
which simplifies to $3(m+3)^{2}-4(m+1)(m+5)$ $=(m-1)^{2}-2(m-1)(m+3)$ being ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,303 |
Example 18. Given $\triangle A_{1} A_{2} A_{3}$ and a point $P_{0}$ in the plane, define $A_{B}=A_{s-3}, S \geqslant 4$, and select the point sequence $P_{0}, P_{1}, P_{2}, \cdots$, such that $P_{k+1}$ is the position reached by rotating $P_{k}$ $120^{\circ}$ clockwise around the center $A_{k+1}$. $k=0,1,2, \cdots$. If... | $$
\begin{array}{l}
\text { - }[\underbrace{\mathrm{R}\left(\mathrm{A}_{1}, 120^{\circ}\right) \mathrm{R}\left(\mathrm{A}_{2}, 120^{\circ}\right) \mathrm{R}\left(\mathrm{A}_{3}, 120^{\circ}\right.})] \\
=\mathrm{I} \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { but } \mathrm{R}\left(\mathrm{A}_{1}, 120^{\circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,305 |
Example 1. If $p, q \in R^{+}, p^{3}+q^{3}=2$. Prove that $p+q \leqslant 2$. | $$
\begin{array}{l}
(p+q)^{3}=p^{3}+q^{3}+3\left(p^{2} q+p q^{2}\right) \\
\leqslant p^{3}+q^{3}+3\left(p^{3}+q^{3}\right)=8, \\
\therefore p+q \leqslant 2 .
\end{array}
$$
Prove: By the theorem,
$$
\begin{array}{l}
(p+q)^{3}=p^{3}+q^{3}+3\left(p^{2} q+p q^{2}\right) \\
\leqslant p^{3}+q^{3}+3\left(p^{3}+q^{3}\right)=... | p+q \leqslant 2 | Inequalities | proof | Yes | Yes | cn_contest | false | 703,307 |
For example, a number in the form of $42 \cdots$ multiplied by 2, with 42 moved to the end, find this number. | Given the theorem $x=42, k=2, c=2, \omega=\frac{x}{10^{2}-2}$ $=\frac{42}{98}=0.42857^{\circ} \mathrm{i}$, therefore, the numbers that meet the condition are $428571,428571428571, \cdots$ | 428571,428571428571, \cdots | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,310 |
Example 1. Find a particular solution to $308 x+211 y=1$.
The above text is translated into English, preserving the original text's line breaks and format. Directly output the translation result. | Solution:
$2\left|\begin{array}{cc}308 & 211 \\ 211 & 194 \\ \hline 97 & 17 \\ 85 & 12 \\ 12 & -5 \\ 10 & 4 \\ -2 & 1\end{array}\right| 2$
\begin{tabular}{c|c|c|}
& 308 & 211 \\
\hline-1 & 0 & 1 \\
-2 & 1 & -1 \\
-5 & -2 & 3 \\
-1 & 11 & -16 \\
-2 & -13 & 19 \\
-2 & 37 & -54
\end{tabular}
$$
-87 \mid 127 r_{\Delta}-1
... | x_{0}=-87, y_{0}=127 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,311 |
Example 1. Factorize:
$$
2 x^{2}+x y-3 y^{2}+3 x z+7 y z-2 z^{2} \text {. }
$$ | Let $f(x, y, z)$ denote the original expression, then
$$
\begin{array}{l}
f(0, y, z)=(-y+2 z)(3 y-z), \\
f(x, 0, z)=(x+2 z)(2 x-z), \\
f(x, y, 0)=(x-y)(2 x+3 y) . \\
\therefore f(x, y, z) \\
\quad=(x-y+2 z)(2 x+3 y-
\end{array}
$$
Considering $z$ as 1, we can factorize $6 x^{2}+7 x y-3 y^{2}+3 x+10 y-3$. Let $x_{1}=x,... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,312 |
Example 2. Determine the reducibility of $f(x, y, z)=5 x^{2}+9 x y-2 y^{2}$ $-14 x z+7 y z-3 z^{2}$ over the field $R$.
| Solution:
$$
\begin{array}{l}
f(0, y, z)=(-2 y+z)(y-3 z) \\
f(x, 0, z)=(5 x+z)(x-3 z), \\
f(x, y, 0)=(5 x-y)(x+2 y) .
\end{array}
$$
It is known that the original expression is irreducible over $R$.
(Author Yangzhu: Shanghai Shicaiming Normal)
| not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,313 |
Example. Prove:
$$
\begin{array}{l}
D=\begin{array}{llll}
a & a & a & a \\
a & a & a & b \\
a & a & c & b \\
a & d & c & b
\end{array} \\
=a(a-b)(a-c)(a-d) .
\end{array}
$$ | Proof: Since $a=0, a=b, a=c$, or $a=d$ all result in $D=0$, it follows that $c, b-a, a-0, a-d$ are factors of $D$. Both sides of the equation are polynomials of the same degree, so $D=k a \cdot (a-b)(a-c)(a-d)$. Let $a=1, b=c=d=0$, then $1=k \cdot 1$ which means $k=1$. Proof completed.
（Author: Education Bureau of Ping... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,314 |
Example 1. Find the inverse of the function $y=\frac{x+1}{x-2}(x+2)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve for the range: $y=\begin{array}{l}x+1 \\ x-2\end{array}-1+\frac{3}{x-2}$,
$$
\because \frac{3}{x-2} \neq 0, \quad \therefore y=1+0=1 \text {. }
$$
Solve for the inverse: $x=\frac{3}{y-1}+2$.
F. Check: $y=\frac{3}{x-1}+2, x \in R, x=1$ is the required inverse function. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,318 |
Example 2. Find the inverse function of $y=3+\sqrt{x-1} \quad(x \geq 1)$. | Solve for the range: $\because x=1, \therefore y-3=\sqrt{x}-1$ $\geqslant 0$, i.e., $y \geqslant 3$.
Inverse solution: From $y-3=\sqrt{x-1}$, we get $x=(y-3)^{2}+1$.
Interchange: $y=(x-3)^{2}+1, x \geqslant 3$ is the required inverse function. | y=(x-3)^{2}+1, x \geqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,319 |
Example 3. Find the inverse of the function $y=\frac{1}{2} \log _{2} \frac{1+x}{1-x}$ | Solve for domain, range: From $\frac{1+x}{1-x}>0$, we get $-1<x<1$.
Inverse solution: From $y=\frac{1}{2} \log _{2} \frac{1+x}{1-\frac{x}{x}}$, we get
$$
\begin{array}{l}
2^{2 y}=\begin{array}{c}
1+x \\
1-x
\end{array}, \\
\therefore \quad x=\frac{2^{y}-2^{-y}}{2^{y}+2^{-y}} \text {. } \\
\end{array}
$$
Interchange: $... | y=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,320 |
Example 4. Find the inverse function of $y=\pi-\arcsin (3-x)$, $(2 \leqslant x \leqslant 4)$. | Solve $\because-1 \leqslant 3-x \leqslant 1$
$$
\therefore \quad-\frac{\pi}{2} \leqslant-\arcsin (3-x) \leqslant \frac{\pi}{2} \text {, }
$$
$\therefore$ The range is $\frac{\pi}{2} \leqslant y=\pi-\arcsin (3-x)$
$$
\leqslant \frac{3 \pi}{2}
$$
Also, $y-\pi=-\arcsin (3-x)$.
From (1) and (2), we solve $x=3+\sin (y-\pi)... | y=3-\sin x, x \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,321 |
Example 5. Find the inverse function of $y=2^{x^{2}-2 x}, x \in(-\infty, 1]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{c}
\text { Sol } \quad \text { Let } t=x^{2}-2 x=(x-1)^{2}-1 . \\
\because \quad x \in(-\infty, 1], \therefore t \geqslant-1, \\
\therefore \quad y=2^{t} \geqslant 2^{-1}=\frac{1}{2} .
\end{array}
$$
That is, the domain is $x \leqslant 1$,
the range is $y \geqslant \frac{1}{2}$.
From $y=2^{x^{2}-2 x}... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,322 |
Example 6. Find the inverse function of the function $y=\sin x, x \in[-\pi$, $-\frac{\pi}{2}$]. | Given $\because \quad x \in\left[-\pi,-\frac{\pi}{2}\right]$,
$$
\therefore \quad y \in[-1,0] \text {. }
$$
According to the definition of the inverse sine, if $A \in\left(-\frac{\pi}{2}\right.$, $\left.\frac{\pi}{2}\right]$, and $\sin A=N, \quad|N| \leqslant 1$, then $\arcsin N=A$. Therefore, the problem must be tran... | y=-\pi-\arcsin x, x \in[-1,0] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,323 |
Example 1. Given $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$.
Prove. $\sin 2 A+\sin 2 B+\sin 2 C=0$. | Note that here we do not say that $A$, $B$, and $C$ are the interior angles of a triangle, so the constant transformations in a triangle are invalid. However, due to the familiar formula $\sin^2 a + \cos^2 a = 1$, we think of three points
$$
D(\cos A, \sin A), E(\cos B, \sin B),
$$
$F(\cos C, \sin C)$ lying on the same... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,325 |
Example 2. Prove that for $0<a<1,0<b<1$, we have
$$
\begin{array}{l}
\quad \therefore \sqrt{a^{2}+b^{2}}+\sqrt{(1-a)^{2}+b^{2}} \\
+\sqrt{a^{2}+(1-b)^{2}}+\sqrt{(1-a)^{2}+(1-b)^{2}} \\
\geqslant 2 \sqrt{2} .
\end{array}
$$ | The left side is the sum of four square roots. If we try to simplify it by squaring, it will obviously cause significant trouble. However, noticing the structure of the roots, we find that they can be regarded as the modulus of a certain complex number. Thus, we construct the following four complex numbers:
$$
\begin{a... | 2 \sqrt{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 703,326 |
Example 11. Find the number of consecutive zeros at the end of 1987!.
untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated text:
Example 11. Find the number of consecutive zeros at the end of 1987!.
Note: The note at the end is not part of the original text and is provided for context. | To solve this problem, it is equivalent to finding the exponent of $1 \mathrm{C}$ in 1987!, which is the same as finding the exponent of 5 in 1987!, since
$$
\begin{array}{l}
{\left[\frac{1987}{5}\right]+\left[\frac{1987}{5^{2}}\right]+\left[\begin{array}{c}
1987 \\
5^{3}
\end{array}\right]} \\
+\left[\frac{1987}{5^{4}... | 494 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,327 |
Example 3. Let $x, y$ satisfy $3 x^{2}+2 y^{2}=6 x$, find the maximum value of $x^{2}+y^{2}$:
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | If not analyzed, from the known conditions:
$$
\begin{array}{l}
x^{2}+y^{2}=-\frac{1}{2} x^{2}+3 x \\
=-\frac{1}{2}(x-3)^{2}+\frac{9}{2},
\end{array}
$$
it would be incorrect to say that when $x=3$, $x^{2}+y^{2}$ achieves its maximum value $\frac{9}{2}$. This is because from $y^{2}=-\frac{3}{2} x^{2}+3 x \geqslant 0$,... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,328 |
Example 3. Prove: $\lim _{n \rightarrow \infty}[n(\sqrt[n]{z-1})]$ $=\ln r+i \varphi+2 k \pi i$.
where $z=r(\cos \varphi+i \sin \varphi), k=0, \pm 1$, $\pm 2, \cdots$. | Prove: Since $n(\sqrt[n]{\bar{z}}-1)=n\left[r^{\frac{1}{n}}\left(\cos \frac{\varphi+2 k \pi}{n}+i \sin \frac{\varphi+2 k \pi}{n}\right)-1\right]$ for any fixed $k$ value (integer), then $k$ does not change with $n$. Therefore,
\[
\lim _{n \rightarrow \infty} n r^{\frac{1}{i}} \sin \frac{\varphi+2 k \pi}{n}=\lim _{1 / ... | \ln r+i(\varphi+2 k \pi) | Calculus | proof | Yes | Yes | cn_contest | false | 703,329 |
Example 1. Let $n$ be a natural number, prove that $1 +\frac{1}{2!}+\frac{1}{3!}$
$$
+\cdots+\frac{1}{n!}<2 .
$$ | Prove: First, prove that when $n \geqslant 3$ groups, $\frac{1}{k!}<\frac{1}{2^{k-1}}$.
$$
\begin{array}{l}
=\frac{1}{2^{i-1}}, \\
\end{array}
$$
(Replace 3, 4, 5, ..., k with 2)
$$
\begin{array}{l}
\therefore \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots+\frac{1}{n!} \\
<\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,330 |
Example 2. Prove: $2 \leqslant\left(1+\frac{1}{n}\right)<3$.
| Prove: Since $\left(1+\frac{1}{n}\right)^{n}=1+C_{n}^{1} \frac{1}{n}$
$$
+C_{n}^{2} n^{2}+\cdots+C_{n}^{n} \frac{1}{n^{n}} \text {. }
$$
Therefore, to show $2 \leqslant 1+C_{n}^{1} \frac{1}{n}+C_{n}^{2} \frac{1}{n^{2}}$ $+\cdots+C_{n}^{n} \frac{1}{n^{n}}<3$, we need to consider the general term $C_{n}^{k} \frac{1}{n^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,331 |
Example 4. Prove: $S_{n}=\frac{1}{n+1}+\frac{1}{n+2}$
$$
+\frac{1}{n+3}+\cdots+\frac{1}{2 n}>\frac{13}{24} \cdot(n>1)
$$ | Proof: (1) When $n=2$, the problem is valid. (Proof omitted)
(2) Assume that when $n=k$ (where $k>1$), the proposition is valid, i.e., $S_{k}=\frac{1}{k+1}+\frac{1}{k}+\frac{1}{2}+\cdots+\frac{1}{2 k}>\frac{13}{24}$.
Then when $n=k+1$,
\[
\begin{aligned}
S_{k+1} & =\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+\cdots+\frac{1}{2(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,333 |
2. Find the maximum distance between two points, one on the surface of a sphere centered at $(-2$, $-10,5)$ with a radius of 19, and the other on the surface of a sphere centered at $(12,8,-16)$ with a radius of 87. | 2. Let $O$ and $O_{1}$ be the centers of two spheres, and $P, P_{1}$ be the intersection points of the extended line segment $O_{1}$ with the two spherical surfaces, such that $O$ is inside $P O_{1}$ and $O_{1}$ is inside $O P_{1}$. Clearly, the maximum distance between these two points is $P P_{1}=P O+O O_{1}+O_{1} P_... | 137 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,336 |
3. A natural number greater than 1, if it is exactly equal to the product of its distinct proper divisors (factors excluding 1 and itself), then it is called "good". Find the sum of the first ten "good" natural numbers. | 3. Let $k$ be a positive integer, and let $1, d_{1}, d_{2}, \cdots$, $d_{n_{-1}}, d_{n}, k$ be all its divisors, arranged in increasing order $\Rightarrow 1 \cdot k=d_{1} \cdot d_{n}=d_{2} \cdot d_{n_{-1}}=\cdots$. If $k$ is "good," then by definition, these products are also equal to $d_{1} \cdot d_{2} \cdots \cdots d... | 182 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,337 |
Example 12. Let
$$
\begin{array}{l}
\boldsymbol{f}(x)=\left[\begin{array}{c}
x \\
12 \frac{1}{2}
\end{array}\right] \cdot\left[\frac{-12 \frac{1}{2}}{x}\right] \text {, } \\
x \in(0,90) . \\
\end{array}
$$
Find the range of $f(x)$. | When $x<12 \frac{1}{2}$, $0<\frac{x}{12}<1$, then $f(x)=0$.
When $x \geqslant 12 \frac{1}{2}$, $-1 \leqslant \frac{-12 \frac{1}{2}}{x}<0$, then
$$
\begin{array}{l}
{\left[\begin{array}{c}
-12 \frac{1}{2} \\
x
\end{array}\right]=-1} \\
f(x)=-\left[\frac{x}{12 \frac{1}{2}}\right] . \\
\end{array}
$$
By $12 \frac{1}{2} \... | \{0,-1,-2,-3,-4,-5,-6,-7\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,338 |
4. Find the area of the region enclosed by the graph of the equation $|x-60|+|y|=\left|\frac{x}{4}\right|$.
untranslated text remains the same as requested. | 4. First, the graph of this equation is symmetric about the $x$-axis, so we only need to find the area enclosed by
$$
\left\{\begin{array}{l}
\left.y=\left|\frac{x}{4}\right|-1 x-60 \right\rvert\,, \\
y \geqslant 0
\end{array}\right.
$$
The region enclosed by the difference of $0 \cdots \times 1$ and the $x$-axis in t... | 480 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,339 |
$5 . x 、 y$ are integers that satisfy the equation $y^{2}+3 x^{2} y^{2}=30 x^{2} +517$, find the value of $3 x^{2} y^{2}$ . | 5. The original equation transforms to $\left(y^{2}-10\right)\left(3 x^{2}+1\right)$ $=3 \times 13^{2}>y^{2}-10=1 ， 3 ， 13 ， 39 ， 169$ or $507=>y^{2}=11,13,23,42,179$ or 517. Since $y$ is an integer $>y^{2}=49 \rightarrow y^{2}-10=39 \Rightarrow 3 x^{2}+1=13$ $\therefore 3 x^{2}=12>3 x^{2} y^{2}=12 \times 49=588$. | 588 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,340 |
6. As shown in the figure below, rectangle $A B C D$ is divided into 4 equal-area parts by 5 line segments. Given $X Y=Y B+B C$ $+C Z=Z W=W D+D A+A X, P Q / / A B$. If $B C=19 \mathrm{~cm}, P Q=87 \mathrm{~cm}$, find the length of $A B$ (in cm). | 6.18 The trapezoids $X Y Q P$ and $Z W P Q$ have equal areas, $H A Y-\| Z$. Both are equal to $\frac{B C}{2}$. Also, $X Y$ is $\frac{1}{4}$ of the perimeter of rectangle $A B C D$, so
$$
\begin{array}{l}
X Y=\frac{A B+B C}{2} \text { by } \frac{(P Q+X Y)}{2} \times \frac{B C}{2} \\
=\frac{A B \cdot B C}{4} \Rightarrow ... | 193 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,341 |
8. Find the largest positive integer $n$, such that the inequality $\frac{8}{15}<$ $-\frac{n}{n+k}<\frac{7}{13}$ holds for exactly one integer $k$. | 8 . Transform the original inequality to $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$, which is equivalent to $\frac{7}{8}>\frac{k}{n}>\frac{6}{7} \Leftrightarrow 49 n>56 k>48 n$. Therefore, the problem is converted to finding the largest open interval ( $48 n$, $49 n$ ) that contains only one multiple of 56. Since the in... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 703,342 |
10. Pay for an escalator moving upwards. A walks down from its top to its bottom, totaling 150 steps, B walks up from its bottom to its top, totaling 75 steps. Assuming A's speed (number of steps walked per unit time) is 3 times B's speed, how many steps of the escalator are visible at any given moment? (Assume this nu... | 10. Let $v_{1}, v_{2}, v$ represent the speeds (number of steps walked per unit time) of $A, B$, and the automatic escalator, respectively, and let $t_{1}, t_{2}, t$ represent the time taken by $A, B$, and the automatic escalator, respectively. From the problem, we have: $v_{1}=3 v_{2}, v_{1} t_{1}=150, v_{2} t_{2}=75$... | 120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,344 |
11. Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers. | i. Find the maximum positive integer $k$ such that $3^{11}=(n+1)+(n+2)+\cdots+(n+k)$ holds (where $n$ is a non-negative integer). From the right side, we have $=\frac{k(k+2 n+1)}{2}$ $->K \cdot(k+2 n+1)=2 \cdot 3^{11}$. To make the smaller factor $k$ as large as possible, $n$ must be non-negative $\Rightarrow k=2 \cdot... | 486 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,345 |
12. The square root of a number without $m$ is $\cdots$ numbers of the form $n+r$, where $n$ is a positive integer, and $r$ is a real number less than $\frac{1}{1000}$. If $m$ is the smallest positive integer satisfying the above condition, find the value of $n$ when $m$ is the smallest positive integer. | 12. From the problem: $\sqrt[3]{m}=n+r, 0\frac{1000}{3}
\end{array}
$
$\rightarrow n^{2} \approx \frac{1000}{3}$. Since $18^{2}<\frac{1000}{3}<19^{2}$, we can guess $n=18$ or $n=19$. Upon verification, when $n=18$, the inequality does not hold, but when $n=19$, the inequality is satisfied. Therefore, $n=19$ is the smal... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,346 |
13. For a known sequence of distinct real numbers $r_{1}, r_{2}, r_{3}, \cdots, r_{\text {n}}$, a single operation involves comparing the second term with the first term, and swapping them if and only if the second term is smaller; then comparing the third term with the new second term, and swapping them if and only if... | 13. Notice that the operation defined in the problem, when applied to any sequence $r_{1}, r_{2}, \cdots, r_{k}$ once, will always result in the last number being the largest in the sequence. Therefore, for the initial sequence $r_{1}, r_{2}, \cdots, r_{20}, \cdots, r_{30}, r_{31}, \cdots, r_{40}$, $r_{20}$ can be move... | 931 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,347 |
Example 1. Draw the graph of $[x][y]=1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | From the given information, we have
$$
\begin{array}{c}
\left\{\begin{array}{l}
x=1+r_{1}, \\
y=1+r_{2}
\end{array}\right. \\
\text { domain }\left\{\begin{array}{l}
x=-1+r_{1}, \\
y=-1+r_{2} .
\end{array}\right. \\
\left(0 \leqslant r_{1}, \quad r_{2}<1\right)
\end{array}
$$
Therefore, the figure consists of two squa... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,348 |
Example 22. ABCDEF is a regular hexagon, K is the midpoint of diagonal BD, and $\mathrm{M}$ is the midpoint of side $\mathrm{EF}$.
Prove: $\triangle A M K$ is an equilateral triangle. | $$
\begin{array}{l}
\text{Let O be the center of the hexagon, BODC be a rhombus, and the diagonals bisect each other, so K is the midpoint of OC.} \\
O \xrightarrow{\mathrm{R}\left(\mathrm{A}, 60^{\circ}\right)} \rightarrow F, C \xrightarrow{\mathrm{R}\left(\mathrm{A}, 60^{\circ}\right)} \rightarrow \mathrm{E}, \\
\the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,349 |
Example 23. A billiard table is in the shape of a regular hexagon ABCDEF. A ball is struck from point P on AB, hits a certain point Q on BC, and then successively strikes the sides CD, DE, EF, FA, and finally hits a certain point V on AB. Let $\angle \mathrm{BPQ} = \theta$. Find the range of values for $\theta$.
Point... | Solve the following figure, after a series of reflections, transform the broken line $P Q R-$ STJ $\mathrm{Y}$ into a straight line segment $\mathrm{PV}$. Then determine
$$
\angle \mathrm{PB}^{\prime} \mathrm{M}<\angle \mathrm{PV}^{\prime} \mathrm{M}=0<\angle \mathrm{PA}^{\prime} \mathrm{M} \text {. }
$$
Without loss ... | \operatorname{arctg} \frac{3 \sqrt{3}}{10}<\theta<\operatorname{arctg} \frac{3 \sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,350 |
Example 10. (IMO2-3)
Given the side lengths $a, b, c$ and the area $S$ of a triangle, prove that:
$$
a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S .
$$
and find the condition for equality.
This inequality is known as the Weitzenböck inequality, and it has many proofs. | Proof 1 Let $C$ be the angle opposite to side $c$. By the cosine theorem and the area formula of a triangle, we get
$$
\begin{array}{l}
\quad a^{2}+b^{2}+c^{2}-4 \sqrt{3} S \\
=a^{2}+b^{2}+a^{2}+b^{2}-2 a b \cos C \\
-2 \sqrt{3} a b \sin C \\
=2\left[a^{2}+b^{2}-2 a b \sin \left(C+30^{\circ}\right)\right] \\
\geqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,352 |
Example 12. (IMO24-6)
Let $a, b, c$ be the lengths of the sides of a triangle. Prove that:
$$
\begin{array}{l}
b^{2} c(b-c)+c^{2} a(c-a) \\
+a^{2} b(a-b) \geqslant 0 .
\end{array}
$$
and determine when equality holds. | Let $c=y+z, i=z+x, c=x+y$ $\left(x, y, z \in R^{+}\right)$, substitute into the original equation, expand and simplify to get
$$
\begin{array}{l}
x y^{3}+y z^{3}+z x^{3}-x^{2} y z-x y^{2} z \\
-x y z^{2} \geqslant 0,
\end{array}
$$
which is $x y z\left(\frac{y^{2}}{z}+\frac{z^{2}}{x}+\frac{x^{2}}{y}-x-y-z\right)$
$$
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,354 |
Example 13. (IMO9-2)
In a tetrahedron, exactly one edge is greater than 1. Prove that its volume $V \leqslant \frac{1}{8}$.
| Prove: As shown in the figure, in the tetrahedron $ABCD$, $AB > 1$, and the other edges are no more than 1. Draw the height $AH$ of the tetrahedron, with $H$ as the foot of the perpendicular, and then draw $AF \perp CD$, $BE \perp CD$, and connect $FH$. Let $CD = x (\leq 1)$, then one of $CF$ and $DF$ must be $\geq \fr... | \frac{1}{8} | Geometry | proof | Yes | Yes | cn_contest | false | 703,355 |
Example 14. (IMO7-1)
Given the system of equations
$$
\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=0, \\
a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=0, \\
a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=0
\end{array}\right.
$$
the coefficients satisfy the following conditions: a) $a_{11}, a_{22}, a_{33}$ are posit... | Obviously, $x_{1}=x_{2}=x_{3}=0$ satisfies the equation. Below we prove that this is the only solution. Without loss of generality, assume there is a set of roots $x_{1}$, $x_{2}$, $x_{3}$ satisfying: $\left|x_{1}\right| \geqslant\left|x_{2}\right| \geqslant\left|x_{3}\right|$ (otherwise, by changing the indices, this ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,356 |
Example 15. (IMO21-6)
Find all real numbers $a$ such that there exist non-negative real numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$ satisfying the following relations:
$$
\sum_{\mathrm{k}=1}^{\mathrm{s}} k x_{k}=a, \sum_{\mathrm{k}=1}^{5} k^{3} x_{k}=a^{2}
$$
and $\quad \sum_{k=1}^{5} k^{5} x_{k}=a^{3}$. | Proof: Let there be non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, then we have
$$
\begin{aligned}
\mathfrak{a}^{4} & =\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2} \\
& =\left[\sum_{k=1}^{5}\left(k^{\frac{1}{2}} \sqrt{x_{k}}\right)\left(k^{5} \sqrt{x_{k}}\right)\right)^{2} \\
& \leqslant\left(\sum_{k=1}^... | a=0,1,4,9,16,25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,357 |
Theorem 2 Let $x_{1}$ be positive integers, $\sum_{i=1} x_{i}=q n+r$, then the function $f:=s t(t \geqslant 2)$ attains its minimum value $(n-r) q^{t}+r(q+1)^{\iota}$ if and only if $\left\{x_{1}, x_{2}\right.$, $\left.\cdots, x_{n}\right\}=\left\{\begin{array}{c}q, \cdots, q, q+1, \cdots, q+1 \\ n-r \text { times }\e... | Proof points: Just note that:
1. If $x_{i}<q, x_{j}=q+a(a \geqslant 1)$, an adjustment can be made:
$$
x_{i}^{\prime}=x_{i}+a, x_{j}^{\prime}=q,
$$
then $s^{\prime}{ }_{t}-s_{t}=\left[\left(x_{i}+a\right)^{t}+q^{t}\right]$
$$
\begin{aligned}
& -\left[x_{i}^{t}+(q+a)^{t}\right] \\
= & \sum_{m=1}^{t-1} C_{i}^{m} x_{i}^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,359 |
Example 24. $\triangle \mathrm{ABC} 4, \mathrm{AB}=\mathrm{AC}, \mathrm{O}$ is a point inside the shape,
$$
\begin{array}{c}
\angle \mathrm{A}=80^{\circ}, \angle \mathrm{OBC}=10^{\circ}, \angle \mathrm{OCB}=20^{\circ} . \\
\text { Find } \angle \mathrm{CAO}=\text { ? }
\end{array}
$$ | Solve $\angle \mathrm{ACO}=30^{\circ}$,
$$
O \xrightarrow{\mathrm{S}(\mathrm{AC})} \mathrm{P} \text {, }
$$
then $\triangle \mathrm{CPO}$ is an equilateral triangle.
$$
\begin{array}{l}
\angle \mathrm{OAC}=\angle \mathrm{PAC} \text {. } \\
\angle \mathrm{BOC}=150^{\circ}, \angle \mathrm{BOP}=150^{\circ} .
\end{array}
... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,361 |
Example 1. $P$ is any point on the ellipse, and $Q$ is the midpoint of the line segment connecting point $P$ and focus $F$. Try to prove: the locus of point $P$ is an ellipse. | Proof 1 As shown in Figure 1, let the equation of the ellipse be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, Q\left(x_{1}, y_{1}\right),
$$
with the focus $F(c, 0), P(x, y)$.
According to the problem, $x=\frac{x_{1}+c}{2}, y=\frac{y_{1}}{2}$. Therefore, $x_{1}=2x-c, y_{1}=2y$.
Since $Q$ is on the ellipse, we have $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,363 |
Example 1. Some equilateral triangles with side lengths of $1,3,5,7, \cdots$ are placed sequentially on a straight line with their base vertices touching. Prove: Their vertices lie on a parabola, and the distances from each vertex to the focus of this parabola are all integers. | Take the base of all triangles as the $x$-axis, and take the $y$-axis through the vertex of the first triangle. Then the coordinates of the vertices of these triangles are
$$
\begin{array}{l}
\left(0, \pm \frac{1}{2} \sqrt{3}\right),\left(2, \pm \frac{3}{2} \sqrt{3}\right), \\
\left(6, \pm \frac{5}{2} \sqrt{3}\right), ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,365 |
Example 2. $P$ is a moving point on the fixed circle $(x-a)^{2}+y^{2}=r^{2}$, and $O$ is the origin. An equilateral triangle $O P Q(O P Q$ are arranged in a counterclockwise direction) is constructed with $O P$ as one side. Find the equation of the trajectory of point $Q$. | Let $P\left(x_{0}, y_{0}\right), Q(x, y)$, the complex number corresponding to vector $\overrightarrow{O Q}$ is $x+y i$, and the complex number corresponding to vector $\overrightarrow{O P}$ is $x_{0}+y_{0} i$. By the knowledge of complex number rotation, we have
$$
x+y i=\left(x_{0}+y_{0} i\right)\left(\cos \frac{\pi}... | \left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2} a\right)^{2}=r^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,366 |
Example 3. $M\left(x_{0}, y_{0}\right)$ is a fixed point in the first quadrant, and two circles passing through point $M$ are tangent to both coordinate axes, with their radii being $r_{1}, r_{2}$. Prove that $r_{1} \cdot r_{2}=x_{0}^{2}+y_{0}^{2}$. | Proof: Let the centers of the two circles be $O_{1}, O_{2}$, and the radii be $r_{1}, r_{2}$.
$\because$ Circle $O_{1}$ and circle $O_{2}$ are tangent to both coordinate axes,
$\therefore O_{1}\left(r_{1}, r_{1}\right), O_{2}\left(r_{2}, r_{2}\right)$, and the equations of the two circles are
$\left(x-r_{1}\right)^{2}+... | r_{1} \cdot r_{2}=x_{0}^{2}+y_{0}^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 703,367 |
Example 4. Prove: The circle with the chord passing through the focus of the parabola as its diameter is tangent to the directrix of the parabola. | Proof: Let chord $AB$ pass through focus $F$, with the center of the circle being $C$. Draw perpendiculars from $A, B, C$ to the directrix $l$, and let the feet of these perpendiculars be $M, N, D$.
By the property of the midline of a trapezoid,
$$
|CD|=\frac{1}{2}(|AM|+|BN|),
$$
By the definition of a parabola, $|AM... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,368 |
Example 5. Find the maximum area of a trapezoid inscribed in an ellipse with the major axis as one base. | Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $P(a \cos \theta, b \sin \theta)$.
By symmetry, $P^{\prime}(-a \cos \theta, b \sin \theta)$.
$$
\begin{array}{l}
\therefore \sqrt{3 \cos ^{3} \frac{\theta}{2} \cdot \sin ^{2} \frac{\theta}{2}} \leqslant\left(\frac{3}{4}\right)^{2} \text {. ... | \frac{3 \sqrt{3} a b}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,369 |
1. Solve the equations:
$$
\text { (1) } \begin{aligned}
& (a+b)(a x+b)(a-b x) \\
= & \left(a^{2} x-b^{2}\right)(a+b x) ;
\end{aligned}
$$
(2) $x^{\frac{1}{3}}+(2 x-3)^{\frac{1}{3}}=\{12(x-1)\}^{\frac{1}{3}}$. | (1) It is obvious that $x=1$ is a solution to the equation. Simplify this equation to
$$
\begin{array}{l}
\quad(a+b)\left\{a b+\left(a^{2}-b^{2}\right) x-a b x^{2}\right\} \\
=a^{3} x-a b^{2}+a^{2} b x^{2}-b^{3} x . \\
\quad \text { From the equation }\left[a^{2} b+a b(a+b)\right) x^{2} \\
+\left[a^{3}-i^{2}-(a+i)\left... | x=1, x=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,370 |
$$
\begin{array}{c}
\text { 2. Let }(y-z)^{2}+(z-x)^{2}+(x-y)^{2}= \\
(y+z-2 x)^{2}+(z+x-2 y)^{2}+(x+y-2 z)^{2}
\end{array}
$$
and $x, y, z$ are real numbers.
Prove: $x=y=z$. | Prove that from $(y+z-2 x)^{2}-(y-z)^{2}$
$$
\begin{array}{l}
=(2 y-2 x)(2 z-2 x)=4(x-y)(x-z) \text {, we know that } \\
(x-y) \cdot(x-z)+(y-z)(y-x)+(z-x)
\end{array}
$$
- $(z-y)=0$.
Let $y-z=a, z-x=b, x-y=c$, then we have
$$
b c+c a+a b=0,
$$
and $a+b+c=0$.
$$
\text { Therefore, }(a+b+c)^{2}-2(b c+c a+a b)=0 \text {,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,371 |
4. Find two numbers, such that the product of their squares is 5500, and the product of their difference and the square of the latter is 352. | Let $x, y$ represent the two numbers we are looking for, then
$$
\begin{array}{l}
(x+y)\left(x^{2}+y^{2}\right)=5500, \\
(x-y)\left(x^{2}-y^{2}\right)=352 .
\end{array}
$$
Therefore, $\frac{(x+y)\left(x^{2}+y^{2}\right)}{(x-y)\left(x^{2}-y^{2}\right)}=\frac{5500}{352}$,
which simplifies to $\frac{x^{2}+y^{2}}{(x-y)^{2... | x=13, y=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,374 |
5. Solve the equation:
$$
\begin{array}{c}
\left\{\left(x^{2}+x+1\right)^{3}-\left(x^{2}+1\right)^{3}-x^{3}\right\} \\
\cdot\left\{\left(x^{2}-x+1\right)^{3}-\left(x^{2}+1\right)^{3}+x^{3}\right\} \\
=3\left\{\left(x^{4}+x^{2}+1\right)^{3}-\left(x^{4}+1\right)^{3}-x^{8}\right\}=0 .
\end{array}
$$ | From the identities $(a+\dot{b})^{3}-a^{3}-b^{3}=3 a b \cdot(a+c)$ and $(a-b)^{3}-a^{3}-b^{3}=-3 a b(a-b)$, we can obtain:
$$
\begin{aligned}
& \left(x^{2}+x+1\right)^{3}-\left(x^{2}+1\right)^{3}-x^{3} \\
= & 3 x\left(x^{2}+1\right)\left(x^{2}+x+1\right), \\
& \left(x^{2}-x+1\right)^{3}-\left(x^{2}+1\right)^{3}+x^{3} \... | x=0, x^{2}+x+1=0, x^{2}-x+1=0, \left(x^{2}+1\right)^{2}=x^{4}+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,375 |
6. If the equations $x^{2}+a x+b=0$ and $x y+l(x+y)+m=0$ are used to eliminate $x$, resulting in a quadratic equation in $y$ whose roots are the same as those of the original quadratic equation in $x$, prove:
$$
a=2 l, b=m \text { or } b+m=a l \text {. }
$$ | Prove that from the second equation, we have $y(x+l)$
$=-(l x+m)$, substituting into the first equation
$$
\begin{array}{l}
(l x+m)^{2}-a(x+l)(l x+m)+b(x+l)^{2} \\
=0,
\end{array}
$$
we get
$$
\begin{array}{l}
\quad\left(l^{2}-a l+b\right) x^{2}+\left(2 l m-a l^{2}-a m\right. \\
+2 b l) x+\left(m^{2}-a l m+b l^{2}\rig... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,376 |
7. Let $\left(\beta\right.$ be a root of the quadratic equation $x^{2}+p x+q=0, x^{2 n}$ $+p^{n} x^{n}+q^{n}=0$, where $n$ is an even number. Prove that: $\frac{\alpha}{\beta}, \frac{\beta}{\alpha}$ are roots of the equation $x^{n}+1+(x+1)^{n}=0$. | Proof Given $\alpha+\beta=-p, \alpha \beta=q$, and $x^{2 n}+p^{n} \alpha^{n}+q^{n}=0$ and $\beta^{2 n}+p^{n} \beta^{n}+q^{n}=0$, therefore $\alpha^{2 n}-\beta^{2 n}+p^{n}\left(\alpha^{n}-\beta^{n}\right)=0$ or $a^{n}+\beta^{n}+p^{n}=0$.
Thus, $\alpha^{n}+\beta^{n}+(\alpha+\beta)^{n}=0$.
Since $n$ is even, it is easy t... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,377 |
9. Prove that the equation $(y+z-8 x)^{\frac{1}{3}}+(z+x-8 y)^{\frac{1}{3}}+(x+y-8 z)^{\frac{1}{3}}=0$ has the same solutions as the equation $x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}=0$. | Prove that if $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0$, then $a+b+c=3 a^{\frac{1}{3}} b^{\frac{1}{3}} c^{\frac{1}{3}}$. Therefore, from the given equation, we have:
$$
3\{(y+z-8 x)(z+x-8 y)(x+y
$$
$-8 z)\}^{\frac{1}{3}}=-6(x+y+z)$. Cubing both sides, we get
$$
\begin{array}{l}
(y+z-8 x)(z+x-8 y)(x+y-8 z) \\
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,379 |
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