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
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Given $\sin \alpha=\frac{5}{7}$, $\cos (\alpha+\beta)=\frac{11}{14}$, and $\alpha, \beta$ are acute angles, find $\cos \beta$. | Solving, we get $\cos \alpha=\frac{2 \sqrt{6}}{7}, \sin (\alpha+\beta)$ $=\frac{5 \sqrt{8}}{14}$, thus
$$
\begin{aligned}
\cos \beta & =\cos [(\alpha+\beta)-\alpha]=\cos (\alpha \\
& +\beta) \cos \alpha+\sin (\alpha+\beta) \sin \alpha \\
& =\frac{1}{98}(22 \sqrt{6}+25 \sqrt{3}) .
\end{aligned}
$$
Analysis. The result ... | \frac{1}{98}(22 \sqrt{6}+25 \sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,534 |
栵11(Zhou Xueqi). $\mathrm{E}$ knows $F[f(x)]=\frac{x+1}{x+2}$, find $\mathrm{f}(\mathrm{x})$.
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。 | Analysis: As $F[y]$ varies, $f(x)$ will vary, indicating that the condition is not sufficient. For example, if the condition is $\mathbf{F}(\mathbf{y})=\frac{1}{3 \mathbf{y}}$ or if it is changed to “Given $f[f(x)]=\frac{x+1}{x+2}$,” then a specific solution can be determined. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,535 |
$2^{n}>n^{2} . \quad(n \in N)$ | Analysis: For $n=1$, it is obviously true. Suppose for $k \in N$, $2^{\mathbf{k}}>\mathrm{k}^{2}$ holds, then we have
$$
2^{k+1}=2 \cdot 2^{k}>2 k^{2}, \cdots \text {. }
$$
Is $2 k^{2}$ greater than (or equal to) $(k+1)^{2}$? Or is $k^{2} \geqslant 2 k$ +1? This is not true for all $k \in N$; reconsidering the original... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,536 |
$\begin{array}{l}\text { 9. Prove that, } \frac{a^{2}(x-b)(x \cdots c)}{(a-b)(a-c)} \\ \quad+\frac{b^{2}(x-c)(x-a)}{(b-c)(b-a)} \\ +\frac{c^{2}(x-a)(x-b)}{(c-a)(c-b)}=x^{2} .\end{array}$ | $$
\begin{array}{c}
\text { Proof: Let the left side }=f(x)=p x^{2}+a x+r \text {. } \\
\because f(a)=\frac{a^{2}(a-b)(a-c)}{(a-b)(a-c)} \\
+\frac{b^{2}(a-c)(a-a)}{(b-c)(b-a)}+\frac{c^{2}(a-a)(a-b)}{(c-a)(c-b)} \\
=a^{2},
\end{array}
$$
Similarly, $f(b)=b^{2} \quad f(c)=c^{2}$.
$$
\therefore\left\{\begin{array}{l}
p a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,538 |
To 15(Ei). Draw a chord of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passing through B $(0,-b)$. Find the maximum chord length. | Let $(\mathrm{x}, \mathrm{y})$ be a point on the ellipse, then $|\mathrm{BM}|^{2}$ $=x^{2}+(y+b)^{2}$. Substituting $x^{2}=\frac{a^{2}}{b^{2}}\left(b^{2}-y^{2}\right)$, we get: $|B M|^{2}=\left(1-\frac{a^{2}}{b^{2}}\right) y^{2}+2 b y+a^{2}$ $+b^{2}$. Since $1-\frac{a^{2}}{b^{2}}<0$, the maximum value of $|B M|^{2}$ is... | \frac{4\left(1-\frac{a^{2}}{b^{2}}\right)\left(a^{2}+b^{2}\right)-4 b^{2}}{4\left(1-\frac{a^{2}}{b^{2}}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,540 |
Example. Let the midpoint of the base $BC$ of $\triangle ABC$ be $M$, and draw a line $l$ intersecting $\mathrm{AB}$, $\mathrm{AM}$, $\mathrm{AC}$ at points $\mathrm{D}$, $\mathrm{E}$, $\mathrm{F}$ respectively, then $\left|\frac{\mathrm{BD}}{\mathrm{DA}}\right|$, $\left|\frac{\mathrm{ME}}{\mathrm{EA}}\right|$, $\left|... | As shown in Figure 4, establish a coordinate system. Let the coordinates of A, B, and C be $A(b, c)$, $B(-a, 0)$, and $C(a, 0)$. The equation of line l is $mx + ny + l = 0$. Then,
$$
\begin{array}{l}
\frac{\mathrm{BD}}{\mathrm{DA}}=-\frac{-\mathrm{ma}+1}{\mathrm{mb}+\mathrm{nc}+1} \\
\frac{\mathrm{ME}}{\mathrm{EA}}=-\f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,544 |
Example 2. Given that $\mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$, $\mathrm{E}$ is a point on $AB$, and $CE$ intersects $AD$ at $F$.
Find the maximum area of $\triangle EDF$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result d... | Slightly solve: Taking D as the origin and BC as the X-axis to establish a coordinate system, $\frac{\mathrm{AE}}{\mathrm{EB}}=\lambda$, write out the coordinates of $\mathrm{E}$ and $\mathrm{F}$, and get the area of $\triangle \mathrm{EOF}$
$$
S(\lambda)=-\frac{b c \lambda}{2(1+\lambda)(1+2 \lambda)} .
$$
Using the d... | \lambda=\frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,546 |
Example 1. Given circles
$$
\begin{array}{l}
C_{1}:(x-a)^{2}+y^{2} \\
=a^{2} \text { and } C_{2}:(x- \\
2 a)^{2}+y^{2}=4 a^{2}
\end{array}
$$
$(a>0)$. Find the locus of the centers of circles that are externally tangent to $C_{1}$ and internally tangent to $C_{2}$. | Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the center of the moving circle (positioning element)
(Figure 6), and the points of tangency be $\mathrm{K}$ and $\mathrm{H}$. Then, $\mathrm{PH} = \mathrm{PK}$. But $\mathrm{PH} = \sqrt{(x-a)^{2}+y^{2}} - a$, $\mathrm{PK} = 2a - \sqrt{(x-2a)^{2}+y^{2}}$. Substituting these i... | 8 x^{2} + 9 y^{2} - 24 a x = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,547 |
9. Prove that $\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-c)(b-a)}$ $+\frac{a-b}{(c-a)(c-b)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}$. | Proof
$$
\begin{aligned}
& \frac{b-c}{(a-b)(a-c)}=\frac{(a-c)+(b-a)}{(a-b)(a-c)} \\
= & \frac{1}{a-b}+\frac{1}{c-a} .
\end{aligned}
$$
Similarly,
$$
\begin{array}{l}
(b-c) \frac{c-a}{(b-a)}=\frac{1}{b-c}+\frac{1}{a-b} . \\
(c-a) \frac{a-b}{(c-b)}=\frac{1}{c-a}+\frac{1}{b-c} .
\end{array}
$$
Adding the two equations ab... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,549 |
Example 1. Given angles $\alpha, \beta, \gamma$ satisfy $0 \leqslant \alpha<\beta<\gamma<2 \pi$, and
$$
\left\{\begin{array}{l}
\sin \alpha+\sin \beta+\sin \gamma=0, \\
\cos \alpha+\cos \beta+\cos \gamma=0 .
\end{array}\right.
$$
Prove: $\alpha, \beta, \gamma$ form an arithmetic sequence. | To prove: Take three points $\mathrm{A}(\cos \alpha, \sin \alpha)$, $B(\cos \beta, \sin \beta)$, $C(\cos \gamma, \sin \gamma)$. Clearly, they all lie on the unit circle
$$
\left\{\begin{array}{l}
x=\cos \theta, \\
y=\sin \theta
\end{array} \quad(0 \leqslant \theta<2 \pi)\right.
$$
Therefore, the circumcenter of $\tria... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,550 |
Example 2. Find the length of the minor arc of the circle $\mathrm{O}: \mathrm{x}^{2}+\mathrm{y}^{2}=9$ intercepted by $\odot \mathrm{O}_{1}$ : $(x-3)^{2}+y^{2}=27$. | Solve: Convert the equations of the two circles into parametric equations (one with parameter $\theta$, the other with $\varphi$), and eliminate $x$ and $y$, then
$$
\left\{\begin{array}{l}
\sqrt{3} \cos \varphi=\cos \theta-1 \\
\sqrt{3} \sin \varphi=\sin \theta .
\end{array}\right.
$$
Eliminating $\varphi$ yields $\c... | 2\pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,551 |
Example 2. Find the volume of the solid of revolution formed by rotating the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ around the $x$-axis. | If we take $s_{1}=s_{2}=0, s_{0}=\pi a b, h=2 b$ we get $V=\frac{2 b}{6}(0+4 \pi a b+0)=\frac{4}{3} \pi a b^{2}$. Or if we take $s_{1}=s_{2}=0, s_{0}=\pi b^{2}, h=2 a$ we get $\boldsymbol{V}=\frac{2 a}{6}\left(0+4 \pi b^{2}+0\right)=\frac{4}{3} \pi a b^{2}$. (Verification using definite integral is completely correct) | \frac{4}{3} \pi a b^{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,553 |
Example 4. Find the volume of the ellipsoid enclosed by the ellipsoid surface $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$. | Given $s_{1}=s_{2}=0, s_{0}=\pi a b, h=2 c$,
$$
\therefore V=\frac{2 c}{6}(0+4 \pi a b+0)=\frac{4}{3} \pi a b c \text {. }
$$
In particular, when $c=b$, we get the
volume of an ellipsoid of revolution (Example 2)
When $a=b=c=R$, we get
$$
V_{\text {sphere }}=\frac{4}{3} \pi R^{3} \text { . }
$$
Using the formula for ... | \frac{4}{3} \pi a b c | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,555 |
The volume of the cone formed by rotating the right-angled triangle around one of its legs is $800 \pi \mathrm{cm}^{3}$, and the volume of the cone formed by rotating around the other leg is $1920 \pi$ $\mathrm{cm}^{3}$. What is the length of the hypotenuse of this triangle? (in cm) | The volume of a cone with a base radius of $\mathrm{r}$ and height $\mathrm{h}$ is $\frac{\pi}{3} \mathrm{hr}$. Let $\mathrm{a}, \mathrm{b}$ represent the two legs of a right triangle, then we have
$$
\frac{\pi}{3} \mathrm{ba}^{2}=800 \pi \text { and } \frac{\pi}{3} \mathrm{ab}^{2}=1920 \pi \text {. }
$$
By comparing ... | 26 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,557 |
Question 3. If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are positive integers, satisfying $\mathrm{c}=$ $(a+b i)^{3}-107 i$, find $c$. (where $i^{2}=-1$) | $$
\text { Solve } \begin{array}{l}
c=(a+b i)^{3}-107 i \text { can be written as } \\
c=a\left(a^{2}-3 b^{2}\right)+i\left[b \left(3 a^{2}\right.\right. \\
\left.\left.-b^{2}\right)-107\right] .
\end{array}
$$
Since $c$ is a real number, the imaginary part must be 0, i.e.,
$$
\text { b }\left(3 a^{2}-b^{2}\right)-10... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,558 |
Question 4. No figure, divide each side of a single square into $\mathrm{n}$ equal parts, then connect each vertex to the nearest division point of the opposite vertex, thus creating a smaller square inside the square. If the area of the smaller square (shaded in the figure) is exactly $\frac{1}{1985}$, find the value ... | Solve as shown in the figure. $\mathrm{DM}=\frac{\mathrm{n}-1}{\mathrm{n}}-$
$$
\begin{array}{l}
\therefore A M= \\
\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} \\
=\frac{1}{\cos \angle D E H=\cos \angle D A M=\frac{D A}{A M}} \\
\sqrt{1+\left(1-\frac{1}{n}\right)^{2}}
\end{array}
$$
The area of the small square in the... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,559 |
5. Select an integer sequence $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \cdots$, such that:
$$
a_{n}=a_{n-1}-a_{n-2} \text { (where } n \geqslant 3 \text { ) }
$$
If the sum of the first 1492 terms is 1985, and the sum of the first 1935 terms is 1492, what is the sum of the first 2001 terms? | Solve We calculate the first eight terms of this sequence, $a_{1}$,
$$
\begin{array}{l}
a_{2}, a_{3}=a_{2}-a_{1}, a_{4}=a_{3}-a_{2}=a_{2}-a_{1} \\
-a_{2}=-a_{1}, a_{5}=a_{4}-a_{3}=-a_{1}-a_{2}+a_{1} \\
=-a_{2}, a_{6}=a_{5}-a_{4}=a_{1}-a_{2}, a_{7}=a_{0}-a_{5} \\
=\left(a_{1}-a_{2}\right)-\left(-a_{2}\right)=a_{1}, a_{8... | 986 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,561 |
B6. As shown in the figure, $\triangle \mathrm{ABC}$ is divided into six smaller triangles by three lines passing through its three vertices and an interior point. The areas of four of the smaller triangles are marked in the figure. Find the area of $\triangle \mathrm{ABC}$.
untranslated part:
The areas of four o... | If the heights of two triangles are the same, then the ratio of their areas is equal to the ratio of their corresponding base lengths. Therefore, from the figure, we get
$$
\frac{40}{30}=\frac{40+y+84}{30+35+x}, \quad \frac{35}{x}=\frac{35+30+40}{x+84+y},
$$
$\frac{84}{y}=\frac{84+x+35}{y+40+30}$. Solving the first two... | 315 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,562 |
Question 7. Let $a, b, c, d$ be positive integers, satisfying $a^{8}=b^{4}, c^{3}=d^{2}$, and $c-a=19$. Find $d-b$.
The text above is translated into English, preserving the original text's line breaks and format. | Since the prime factorization of a positive integer is unique, and because 4 and 5 are coprime, 2 and 3 are coprime, it follows that there exist two positive integers \( m \) and \( n \) such that
$$
a=m^{4}, b=m^{6}, c=n^{2}, d=n^{3}.
$$
Thus, \( 19=c-a=n^{2}-m^{4} \):
$$
=\left(n-m^{2}\right)\left(n+m^{2}\right) \te... | 757 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,563 |
Question 9. In a certain circle, parallel chords of lengths $2, 3, 4$ correspond to central angles $\alpha, \beta, \alpha+\beta$. Here $\alpha+\beta<\pi$. If $\cos \alpha$ (which is a positive rational number) is expressed as a reduced (simplest) fraction, what is the sum of the numerator and the denominator? | Solution: Because in the same circle, equal chords subtend equal central angles, the parallelism of these chords is irrelevant. We can select points A, B, C such that AB = 2, BC = 3. Since \( A \widehat{C} = \alpha + \beta \), it follows that \( AC = 4 \). From \( \angle AC3 = \frac{\alpha}{2} \), using the cosine rule... | 49 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,565 |
How many of the first 1000 positive integers can be expressed as:
$$
\lfloor 2 x\rfloor+\lfloor 4 x\rfloor+\lfloor 6 x\rfloor+\lfloor 8 x\rfloor \text { of }
$$
form?
where $\mathbf{x}$ is some real number, and $\mathrm{Lz}\rfloor$ denotes the greatest integer not exceeding $z$. | $$
\begin{array}{l}
\text { Let } f(x)=\lfloor 2 x\rfloor+\lfloor 4 x\rfloor \\
+\lfloor\lfloor 6 x\rfloor+\lfloor 8 x\rfloor .
\end{array}
$$
It can be seen that if $n$ is an integer, then from (1) we have:
$$
f(x+n)=f(x)+20 n .
$$
If some integer $k$ can be expressed as $f\left(x_{0}\right)$, where $x_{0}$ is some ... | 600 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,566 |
Question 12. Let the four vertices of a regular tetrahedron be A, B, C, and D, with each edge length being 1 meter. There is a small insect that starts from point A and moves forward according to the following rules: at each vertex, it chooses one of the three edges connected to that vertex with equal probability and c... | For $\mathrm{n}=0,1,2, \cdots$, let $\mathrm{a}_{\mathrm{n}}$ denote the probability that the bug returns to point $\mathrm{A}$ after walking $\mathrm{n}$ meters. Then we have $\mathrm{a}_{\mathrm{n}+1} = \frac{1}{3} (1 - a_{n})$ (1), because the necessary and sufficient condition for the bug to reach point $\mathrm{A}... | 182 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,568 |
Question 13. The general term of the sequence $101, 104, 109, 116, \cdots$ is $a_{n}=100+n^{2}$. Here $n=1,2,3$, $\cdots$. For each $n$, let $d_{n}$ denote the greatest common divisor of $a_{n}$ and $a_{n+1}$. Find the maximum value of $\mathrm{d}_{\mathrm{n}}$, where $n$ takes all positive integers. | We can prove more generally: If $a$ is a positive integer, and $d_{n}$ is the greatest common divisor of $a+n^{2}$ and $a+(n+1)^{2}$, then when $n=2a$, $d_{n}$ reaches its maximum value of $4a+1$. From this, the answer is $4(100)+1 = 401$.
To prove the above proposition, first note that if $d_{n}$ divides both $a+(n+1... | 401 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,569 |
In a certain competition, each player plays exactly one game with every other player. The winner of each game gets 1 point, the loser gets 0 points. If it's a tie, each gets $\frac{1}{2}$ point. After the competition, it is found that each player's score is exactly half from the games played against the ten lowest-scor... | Let's assume there are $n$ participants in the competition, and we will find two quantities related to $n$: one is the total points of the scores, and the other is the total points of the "losers" (let's call this sum). We need to note that if $k$ participants compete, the total number of matches should be $\frac{k(k-1... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,570 |
Question 15. Three $12 \mathrm{~cm} \times 12 \mathrm{~cm}$ squares are each divided into two pieces, $A$ and $B$, by a line connecting the midpoints of two adjacent sides, as shown in the first figure. The six pieces are then attached to the outside of a regular hexagon, as shown in the second figure. The pieces are t... | Fold into a polyhedron as shown in the first figure, where the vertices marked with $P, Q, R, S$ have three right angles. From this, we can recognize that this polyhedron is part of a cube. We can imagine obtaining a cube with an edge length of $12 \mathrm{~cm}$, as shown in the second figure, making a section (through... | 864 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,572 |
Given $a x^{3}=b y^{3}=c z^{3}$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Prove $\sqrt[3]{a x^{2}+b y^{2}+c z^{2}}=\sqrt[3]{a} +\sqrt[3]{b}+\sqrt[3]{c}$ | Proof: Let $a x^{3}=b y^{8}=c z^{2}=k$,
then $a x^{2}=\frac{k}{x}, b y^{2}=\frac{k}{y}, c z^{2}=\frac{k}{z}$.
$$
\begin{array}{l}
\sqrt[3]{a x^{2}+b y^{2}+c z^{2}} \\
=\sqrt[8]{k\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}=\sqrt[3]{k} \text {. } \\
\end{array}
$$
Let $x=\sqrt[3]{\frac{k}{a}}, y=\sqrt[8]{\frac{k}{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,574 |
11. Given that $x, y, z$ are three distinct real numbers, and $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}$. Prove: $x^{2} y^{2} z^{2}=1$. | Prove that from $x+\frac{1}{y}=z+\frac{1}{x}$
$$
\begin{array}{l}
x-z=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{x y}, \\
x y=\frac{x-y}{z-x}
\end{array}
$$
Using the substitution method, we can derive
$$
\begin{array}{c}
z x=\frac{z-x}{y-z} \\
y z=\frac{y-z}{x-y}
\end{array}
$$
Multiplying the three equations together yield... | x^{2} y^{2} z^{2}=1 | Algebra | proof | Yes | Yes | cn_contest | false | 702,575 |
Example 1. Let $x+y=1, x^{2}+y^{2}=2$, find the value of $x^{7}+y^{7}$. (Japanese University Entrance Exam Question, 1979) | From formula (1), we can get,
$$
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2} \text {. }
$$
From formula (2), we can get,
$$
x y=-\frac{1}{2}\left(1^{2}-2\right)=-\frac{1}{2} \text { . }
$$
Thus, \( x^{7}+y^{7}=\left(x^{3}+y^{3}\right)\left(x^{4}+y^{4}\right)-x^{4} y^{3} \)
$$
\begin{array}{l}
-y^{4} x^{3} ... | \frac{71}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,576 |
Example 2. Let $p \neq 0$, and the quadratic equation with real coefficients $z^{2}-2 p z+q=0$ has imaginary roots $z_{1}$ and $z_{2}$. The corresponding points of $z_{1}$ and $z_{2}$ in the complex plane are $Z_{1}$ and $Z_{2}$. Find the length of the major axis of the ellipse with foci at $Z_{1}$ and $Z_{2}$ and pass... | From formula (1) combined with the concept of the modulus of a complex number and the arithmetic root, we get
$$
\begin{aligned}
\left|z_{1}-z_{2}\right| & =\sqrt{\left|z_{1}-z_{2}\right|^{2}} \\
& =\sqrt{\left|z_{1}^{2}+z_{2}^{2}-2 z_{1} z_{1}\right|} \\
& =\sqrt{\left|\left(z_{1}+z_{2}\right)^{2}-4 z_{1} z_{2}\right|... | 2 \sqrt{q} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,577 |
Example 1, in the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, find the equation of the line passing through $M(2,1)$ and bisecting at point I. | According to Theorem 3,
$$
\begin{array}{c}
\because k_{A B} \cdot k_{O M}=-\frac{12}{16}, \\
k_{O M}=\frac{1}{2}, \\
\therefore k_{A B}=-\frac{12}{16} \cdot 2=-\frac{3}{2} .
\end{array}
$$
Therefore, the equation of $A B$ is: $y-1=-\frac{3}{2}(x-2)$. | y-1=-\frac{3}{2}(x-2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,579 |
Example 2. A chord with slope $t$ is drawn through the endpoint $(-a, 0)$ of the real axis of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Find the equation of the line passing through the other endpoint of the chord and the endpoint $(a, 0)$ of the real axis. | According to Theorem 6, the slope $k$ of the required line and $t$ should have the relationship $k \cdot t=\frac{b^{2}}{a^{2}}$, therefore $k=\frac{b^{2}}{a^{2} t}$. Hence, the required line is: $y=\frac{b^{2}}{a^{2} \dot{t}}(x-a)$. | y=\frac{b^{2}}{a^{2} t}(x-a) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,580 |
Example 3. $P$ is any point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $AB$ is any diameter, and a tangent line $l$ is drawn through $P$. Prove that the product of the slope of the tangent line $l$ and the slope of the line $OP$ is equal to the product of the slopes of the lines $PA$ and $PB$. That is,... | From Theorem 1, we know:
$$
k_{l} \cdot k_{O P}=-\frac{b^{2}}{a^{2}}
$$
From Theorem 3, we know:
$$
k_{A P} \cdot k_{P B}=-\frac{b^{2}}{a^{2}} \text {. }
$$
From (1) and (2), we get $k_{l} \cdot k_{O P}=k_{A P} \cdot k_{P B}$. Therefore, it can be deduced that: $k_{0 P}: k_{A P}=k_{P B}: k_{i}$.
Similarly, for a hyp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,581 |
Example 4. Find the locus of the midpoint of the chord passing through the imaginary vertex $B(0,-b)$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, | Let $M (x, y)$ be the midpoint of the chord that meets the condition, then we have
$$
k_{B \boldsymbol{\mu}}=\frac{y+b}{x}, k_{\circ}=\frac{y}{x} .(x \neq 0)
$$
By $k_{B M} \cdot k_{o_{\mathrm{M}}}=\frac{b^{2}}{a^{2}}$, i.e., $\frac{y+b}{x} \cdot \frac{y}{x}$
$$
=\frac{b^{2}}{a^{2}},
$$
we get $\quad b^{2} x^{2}-a^{2... | \frac{\left(y+\frac{b}{2}\right)^{2}}{\frac{b^{2}}{4}}-\frac{x^{2}}{\frac{a^{2}}{4}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,582 |
Example 2. For any non-empty finite set, all its subsets can be arranged in a sequence such that, except for the first one, each subset can be obtained from the previous one by adding or removing one element. | Let the number of elements in a set be $n$. We will apply induction on $n$. When $n=0$, this is the empty set $\phi$, which has only one subset - the empty set $\phi$ itself. The conclusion is obviously true. Assume the conclusion holds for $n=k$. When $n=k+1$, then $A \backslash\left\{a_{k+1}\right\}=\left\{a_{1}, a_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,587 |
Example 3. Solve the equation $\sin x + \cos x + \sin x \cos x = 1$ | Let $\sin x + \cos x = y$.
From formula (4), we have
$$
\sin x \cos x = \frac{1}{2}(y^2 - 1)
$$
Thus, the original equation becomes $y + \frac{1}{2}(y^2 - 1) = 1$,
(2) $y^2 + 2y - 3 = 0$.
Solving this, we get: $y = 1, y = -3$ (discard).
Therefore, $\sin x + \cos x = 1$,
which means $\sqrt{2 \sin}(45^\circ + x) = 1$.
... | x = n\pi - \frac{\pi}{4} + (-1)^n \frac{\pi}{4}, \, n \in \mathbb{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,588 |
Example 3. Prove: For any $n$ complex numbers, we have
$$
\begin{array}{l}
r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}\right. \\
\left.+i \sin \theta_{2}\right) \cdots r_{n}\left(\cos \theta_{n}+i \sin \theta_{n}\right) \\
= r_{1} r_{2} \cdots r_{n}\left[\cos \left(\theta_{1}+\... | Prove: When $n=1$, the equation obviously holds. When $n=2$, we have
\[ r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \]
\[ = r_{1} r_{2}\left[\cos \theta_{1} \cos \theta_{2} - \sin \theta_{1} \sin \theta_{2} + i \sin \theta_{1} \cos \theta_{2} + i \cos \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,589 |
Example 4. Prove: For all natural numbers $\boldsymbol{n}$, we have
$$
2^{n}+2>n^{2} \text {. }
$$ | Proof: Since $2^{1}+2=4, 1^{2}=1; 2^{2}+2=6, 2^{2}=4; 2^{8}+2=10, 3^{2}=9, 2^{4}+2=18, 4^{2}=16$, the inequality holds for $n=1,2,3,4$. Also, when $n=5$, since $2^{5}=32, 5^{2}=25$, we have $2^{5}+2>2^{5}>5^{2}$, so the inequality also holds. Assuming that for $n=k \geqslant 5$, it has been proven that $2^{k}+2>2^{k}>k... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,590 |
Example 6. Prove that for any natural number $n \geqslant 6$, any square can be divided into $n$ squares. | First, a square can be divided into $6$, $7$, or $8$ smaller squares in the manner shown in the figure, meaning the proposition holds for $n=6,7,8$. Secondly, considering that each square can easily be divided into 4 smaller squares, any square can thus be further divided into $n=\boldsymbol{k}+3$ smaller squares. That... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,592 |
Example 8. Prove that the arithmetic mean of any $n$ positive numbers is not less than their geometric mean.
Preserve the original text's line breaks and format, and output the translation result directly. | We prove the proposition by induction on $n$. When $n=1$, it is clear that $A_{1}=G_{1}$, so the proposition holds. When $n=2$, since $\left(\sqrt{c_{1}}-\sqrt{c_{2}}\right)^{2} \geqslant 0$, i.e., $c_{1}+c_{2}-2 \sqrt{c_{1} c_{2}} \geqslant 0$, it follows that $A_{2}=\frac{c_{1}+c_{2}}{2} \geqslant \sqrt{c_{1} c_{2}}=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,594 |
Example 10. If the sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfies $a_{0}=\frac{1}{2}$ and $a_{k+1}=a_{k}+\frac{1}{n} a_{k}^{2}$ (for $k=0,1,2$, $\cdots$). Here $n$ is a given natural number, prove that
$$
1-\frac{1}{n}<a_{n}<1 .
$$ | Given that $a_{1}=a_{0}+\frac{1}{n} a_{0}^{2}=\frac{1}{2}+\frac{1}{4 n}$
$$
=\frac{2 n+1}{4 n} \text {, }
$$
Therefore, we have $\frac{n+1}{2 n+1}\frac{n+1}{2 n-k+2} \\
+\frac{(n+1)^{2}}{n(2 n-k+2)^{2}}=\frac{n+1}{2 n-k+1} \\
-\frac{n+1}{(2 n-k+2)(2 n-k+1)} \\
+\frac{(n+1)^{2}}{n(2 n-k+2)^{2}}=\frac{n+1}{2 n-(k+1)+2} ... | 1-\frac{1}{n}<a_{n}<1 | Inequalities | proof | Yes | Yes | cn_contest | false | 702,596 |
Example 2. In space, there are 10 points, 4 of which lie on the same plane, and no other set of 4 points are coplanar; find the number of circular cones (not necessarily right circular cones) with one of the points as the vertex and a circle passing through 3 other points as the base. | Analysis: From the problem, we can infer that among the 10 points in space, no 3 points are collinear. Otherwise, each of the remaining 7 points would be coplanar with the 3 collinear points, which contradicts the problem statement. Now, let's denote the plane where 4 points are coplanar as \( a \). We need to consider... | 836 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,598 |
Example 4. Simplify: $\frac{1-a^{2}}{(1+a x)^{2}-(a+x)^{2}}$,
(Shanghai Mathematics Competition Question) | $$
\begin{array}{l}
=\frac{(1+a)(1-a)}{[1+(a+x)+a x][1-(a+x)+a x]} \\
=\frac{(1+a)(1-a)}{(1+a)(1+x)(1-a)(1-x)} \\
=\frac{1}{(1+x)} \frac{1}{(1-x)} \text {. } \\
\end{array}
$$
The above text translated into English, keeping the original text's line breaks and format, is as follows:
$$
\begin{array}{l}
=\frac{(1+a)(1-... | \frac{1}{(1+x)(1-x)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,599 |
Example 5. A parade team whose number of people is a multiple of 5 and no less than 1000, if lined up in rows of 4, is short of 3 people; if lined up in rows of 3, is short of 2 people; if lined up in rows of 2, is short of 1 person. Find the minimum number of people in this parade team. | Analysis: Given that the total number of people in the parade is a multiple of 5, let the total number of people be $5n$. "When arranged in rows of 4, there are 3 people short," which can be understood as "1 person more" from the opposite perspective. Similarly, when arranged in rows of 3 or 2, it can also be understoo... | 1045 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,602 |
Example 1.
Given: $\triangle A B C$,
$P$ is any point
on the external
bisector of $\angle A$. Prove:
$$
\begin{array}{l}
A B+A C \\
<P B+P C .
\end{array}
$$ | Since $A P$ is the external angle bisector, extend $B A$ to $D$ such that $A D=A C$, and connect $P D \cdot$ Thus,
$$
\triangle A P C \cong \triangle A P D,
$$
hence $P D=P C$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,603 |
Example 2. Given:
$\triangle A B C$, through any point $P$ on $B C$
, draw a line parallel to the median $A$
$M$ and intersect the
sides $A B, A C$ or
their extensions at $D, E$.
Prove: $P D+P E=$ constant. | Extend the median $A M$ to $K$ such that $A M = M K$, and connect $K C, K B$.
Let the extension of $D P$ intersect $K C$ (or $K B$) at $F$. At this point, $A B K C$ is a parallelogram, and $A K F D$ is also a parallelogram.
Since $A M = M K$, then $P E = P F$,
Thus, $P D + P E = P D + P F = A K$,
which means $P D + P... | 2AM | Geometry | proof | Yes | Yes | cn_contest | false | 702,604 |
Example 3. Given:
$$
\begin{array}{l}
\triangle A B C, \\
\angle B=2 \angle C \\
A D \perp B C
\end{array}
$$
$D$ is the foot of the perpendicular, $M$ is the midpoint of $B C$. Prove: $A B=2 M D$. | Take the midpoint $G$ of $A B$, and connect $G D, G M$. Then $G D$ is the median of the hypotenuse of $\triangle A B D$, and $G M$ is the midline of $\triangle A B C$.
Prove that $\triangle D G M$ is an isosceles triangle, hence $G D=M D$.
Also, $A B=2 G D$,
so $A B=2 M D$. | A B=2 M D | Geometry | proof | Yes | Yes | cn_contest | false | 702,605 |
Example 4. Given: As shown in the figure, $E$ is the midpoint of $DC$.
Prove: $\triangle EAB = \frac{1}{2}(\triangle DAB + \triangle CAB)$. | Draw the altitudes $D D^{\prime}, B_{E^{\prime}}, C C^{\prime}$. Then the problem is reduced to proving
$$
E^{\prime}=\frac{1}{2}\left(D D^{\prime}+C C^{\prime}\right) .
$$
To prove this, it suffices to show that $E E^{\prime}$ is the midline of the trapezoid $D D^{\prime} C^{\prime} C$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,606 |
Example 5. Given:
$$
\begin{array}{l}
\triangle A B C, D B \\
=E C ;
\end{array}
$$
Extend $D E$ to intersect the extension of $B C$ at $F$.
Prove: $\frac{A B}{A C}=\frac{F E}{F D}$. | Draw $E \boldsymbol{K} / / A B$, then $\frac{F}{F} \boldsymbol{E}=\frac{E K}{D B}$.
From $D B=E C$, we get $\frac{F E}{F D}=\frac{E K}{E C}$.
Comparing with what we need to prove, it is sufficient to show that $\frac{E K}{E C}=\frac{A B}{A C}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,607 |
Example 6. Given:
$P$ is any point inside rectangle $A B$ $C D$.
Prove: $P A^{2}+P C^{2}=P B^{2}+P D^{2}$. | Connect the diagonals $A C$ and $D B$, intersecting at point $O$. Connect $P O$.
$P O$ is the median of $\triangle P A C$, by the median theorem we have
$$
P A^{2}+P C^{2}=2\left(A O^{2}+P O^{2}\right) \text {. }
$$
Similarly, $P B^{2}+P D^{2}=2\left(B O^{2}+P O^{2}\right)$.
Since $A O=B O$, the proof is complete. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,608 |
Example 7. Given: Trapezoid $A B C D, A D / / B C$. $A B>C D$.
Prove: $D B>A C$.
| Draw $A F / /$
$D B, A E / / D C$.
Draw $A G \perp B C$.
From $A B>C D$ we get $A B>A E$.
According to the theorem “from a point to a straight line, draw a perpendicular and two oblique lines, the foot of the longer oblique line is farther from the foot of the perpendicular” we get $B G>G E$. Adding $F B=E C$ we get $F... | D B>A C | Geometry | proof | Yes | Yes | cn_contest | false | 702,609 |
Example 5. Solve the equation $5 x^{2}+x-x \sqrt{5 x^{2}-1}$ $-2=0$. (National Mathematics Competition Problem) | $$
\begin{array}{l}
\left(\sqrt{5 x^{2}-1}\right)^{2}-[1+(x-1)] \sqrt{5 x^{2}-1} \\
+(x-1)=0 \\
\left(\sqrt{5} \bar{x}^{2}-1-1\right)\left(\sqrt{5 x^{2}-1}-x+1\right)=0
\end{array}
$$
The original equation can be transformed as follows:
$$
\begin{array}{l}
\left(\sqrt{5 x^{2}-1}\right)^{2}-[1+(x-1)] \sqrt{5 x^{2}-1} \... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,610 |
Example 8. Given:
(-) In $O$, $AB$ is a diameter, point $C$ is on the circumference. Connect $AC, CB$.
Through another point $P$ on the circumference, draw a line perpendicular to $AB$ intersecting the extensions of $AB, AC, BC$ at $D, E, F$.
Prove: $DP^2 = DE \cdot DF$. | Connect $P A$ and $P B$.
Since $\angle A P B=90^{\circ}$, we have $D P^{2}=A D \cdot D B$.
To prove $A D \cdot D B=D E \cdot D F$,
we only need to prove $\triangle A D E \sim \triangle F D B$, and the problem is solved. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,611 |
Example 9. Given:
$\bigcirc O$ and an external point $P$,
$P A, P B$ are tangents to $\odot$
$O$ at A, B,
draw any secant line $P C D$ through $P$.
$M$ is the midpoint of chord $C D$, the extension of $B M$ intersects $\odot O$ at $E$, and connect $E A$.
Prove: $A E \parallel C D$. | Connect $O M, O B, O P, O A$. Then $O M \perp P D, O B \perp P B$. Therefore, $O, M, B, P$ are concyclic.
$\angle P M B=\angle P O B$.
Also, $\angle A O P=\angle P O B$,
so $\angle E=\angle P O B, \angle E=\angle P M B$,
thus $A E / / C D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,612 |
By1. The relationship between the empty set $\phi$ and the set $\{0\}$ is
(A) $\phi=\{0\} ; \quad(B) \phi \subset\{0\}$;
(C) $\phi \supset\{0\}$; (D) $\phi \subseteq\{0\}$. | This is a question about the concept of sets. This question examines the concepts that the set with no elements, the empty set, the set of the number 0 is a non-empty set, and the empty set is a subset of any set and a proper subset of any non-empty set. The correct answer is $(B)$. | B | Other | MCQ | Yes | Yes | cn_contest | false | 702,614 |
Example 2. The angle $\alpha$ formed by a straight line and a plane must be
(A) acute;
(B) a first quadrant angle;
(C) $0^{\circ} \leqslant \alpha \leqslant 90^{\circ}$,
(D) $0^{\circ}<\alpha \leqslant 90^{\circ}$. | This question is knowledge-intensive, as the angle formed by a line and a plane should include four cases: oblique intersection, perpendicular intersection, parallel, and coincident. If not considered comprehensively, it's easy to answer $(A),(D)$. However, the first quadrant angle should be $2 k \pi<\alpha<2 k \pi+\fr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,615 |
Example 3. If the three sides of a triangle are $a, b, c$, and the three interior angles are $A, B, C$, and they satisfy $a \cos A + b \cos B = c \cos C$, then this triangle must be
(A) an equilateral triangle; (B) a right triangle with $a$ as the hypotenuse; (C) a right triangle with $b$ as the hypotenuse; (D) none of... | Since the edge-angle relationship given in the problem is symmetric with respect to $a, b$, and the answer is unique, options $(B), (C)$ should be excluded simultaneously. Also, when $a=b=c$, the given equation obviously does not hold, so option $(A)$ should also be excluded. Therefore, the answer is $(D)$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,616 |
Example 4. The positional relationship between the line $y=k x+1$ and the circle $x^{2}+y^{2}=\sqrt{2}$ is
(A) Separate:
(L) Tangent;
(C) Intersect;
(D) Determined by the value of $k$. | If we consider using the discriminant method, it would be quite troublesome. Considering the intercepts of the line and the circle on the $y$-axis are 1 and $\sqrt[4]{2}$ respectively, and $\sqrt[4]{2}>1$, we can quickly determine that the answer is $(C)$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,617 |
Example 2. $\sin 2 \alpha=-\frac{24}{25}$ and $\frac{3}{4} \pi<\alpha<\pi$, then the value of tg $\boldsymbol{\alpha}$ is
(A) $\frac{4}{3}$ or $\frac{3}{4}$,
(B) $-\frac{4}{3}$ or $-\frac{3}{4}$,
(C) $-\frac{4}{3}$,
(D) $-\frac{3}{4}$. | Using the universal formula to verify, $-\frac{4}{3}$ and $-\frac{3}{4}$ are both possible, which would lead to mistakenly choosing $B$. However, an implicit condition is overlooked: due to the monotonicity of $\tan \alpha$, when $\frac{3}{4} \pi < \alpha < \pi$, $-1 < \tan \alpha < 0$. Therefore, $-\frac{4}{3}$ cannot... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,619 |
5. Choose $(x, y)$ in the coordinate plane so that its distances to the $x$-axis, $y$-axis, and the line $x+y=2$ are all equal, then the value of $x$ is
(A) $\sqrt{2}+2$,
(B) $2-\sqrt{2}$,
(C) $\sqrt{2}$,
(D) Not unique. | The above $(A),(B),(C)$ all fit when substituted into the problem, but according to the geometric characteristics, the incenter and the three excenters of the triangle formed by the $x$-axis, $y$-axis, and $x+y=2$ all fit the problem, so (D) should be selected. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,623 |
Example 3. The population of a city compared to the previous year is: in 81 it increased by 1%, in 82 it increased by 1% again, in 83 it decreased by 1%, and in 84 it decreased by 1% again. By the beginning of 85, compared to 80, it is
(A) increased by 1%;
(B) decreased by 1%;
(C) neither increased nor decreased;
(D) d... | Let's assume the population in 1980 is 1, then at the beginning of 1985 it is
$$
\begin{array}{ll}
(1+1 \%) (1+1 \%) & (1-1 \%) (1-1 \%) \\
=\left[1-(1 \%)^{2}\right]^{2}<1,
\end{array} \text { so (A), (C), }
$$
(B) should be eliminated. And $\left[1-(1 \%)^{2}\right]^{2} \neq 1-1 \%$, so (B) should be eliminated. Ther... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,629 |
If $x+\frac{1}{x}=2 \cos \theta$, then for $n \in N$, $x^{*}+\frac{1}{x^{n}}$ equals
(A) $2 \cos n$
(B) $2^{n} \cos \theta_{3}$
$(C) 20 \cos _{3} \theta_{3}$
(D) $2 \cos \theta$;
(E) $2^{n} \cos n \theta$ | Let $n=2$, then $x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2=2 \cos 2 \theta$, so the answer is $(A)$. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,631 |
Example 7. The parabola $y=a x^{2}+b x+c （ a \neq 0 ）$ intersects the $x$-axis at points $A$ and $B$. Prove that the equation of the circle with $A B$ as its diameter is $a x^{2}+b x+c+a y^{2}=0$. | Prove that the radius of the circle $R=\frac{AB}{2}=\frac{\sqrt{\bar{\Delta}}}{2|a|}$.
$\because A$ and $B$ are symmetric with respect to the parabola, and $A$ and $B$ are the endpoints of the diameter,
$\therefore$ the coordinates of the center of the circle are $\left(-\frac{b}{2a}, 0\right)$.
Thus, the equation of t... | a x^{2}+b x+c+a y^{2}=0 | Algebra | proof | Yes | Yes | cn_contest | false | 702,632 |
Example 6. The smallest positive angle $x$ that satisfies the equation $\sin \left(2 x+45^{\circ}\right)=\cos \left(30^{\circ}-x\right)$ is
(A) $60^{\circ}$;
(B) $45^{\circ}$;
(C) $30^{\circ}$;
(D) $25^{\circ}$; (E) $15^{\circ}$. | Solve by substituting $x=15^{\circ}$, $\sin 75^{\circ}=\cos 15^{\circ}$ is correct, so the answer is (E). | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,633 |
Example 1. Given $(a+1)(b+1)=2$, then the value of $\operatorname{arctg} a+\operatorname{arctg} b$ is
(A) $\frac{\pi}{3}$;
(B) $\frac{\pi}{4}$;
(C) $\frac{\pi}{6}$,
(D) $\frac{\pi}{4}$ or $-3 \pi$. | Take $a=0, b=1$, then we have $\operatorname{arctg} a+\operatorname{arctg} b$ $=\frac{\pi}{4}$,
(B) and
(D) both fit. Then take $a=-3$,
$b=-2$, then we have arctg $x+\operatorname{arctg} \dot{b}=-\frac{3 \pi}{4}$,
only (D) fits, so the answer is (D). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,634 |
Example 2. Given that $a, b$ are two unequal positive numbers, among the following three algebraic expressions:
甲. $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)$,
乙. $\left(\sqrt{a b}+\frac{1}{\sqrt{a b}}\right)^{2}$,
丙. $\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^{2}$,
the one with the largest value is
(A) Expr... | If we take $a=1, b=2$, then the value of expression 甲 is the largest. If we then take $a=1, b=\frac{1}{3}$, the value of expression 甲 is still the largest. If we conclude that the answer is (A) based on this, it would be incorrect. In fact, if we take $a=3, b=2$, then the value of expression 丙 is the largest, so the co... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,635 |
Example 1. Find the value of $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$. | \begin{array}{l} \text { Original expression }=\sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \\ \cdot \sin \left(60^{\circ}+20^{\circ}\right) \sin 60^{\circ} \\ =\frac{1}{4} \sin ^{2} 60^{\circ}=\frac{3}{16} .\end{array} | \frac{3}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,636 |
Example 1, find the values of $\cos 18^{\circ}$ and $\operatorname{tg} 18^{\circ}$. | Solve $\theta=18^{\circ}, \operatorname{ctg} 5 \theta=\operatorname{ctg} 90^{\circ}=0$.
Given $5 \operatorname{tg}^{4} 18^{\circ}-10 \operatorname{tg}^{2} 18^{\circ}+1=0$.
Since $\operatorname{tg} 18^{\circ}<1$, hence
$\operatorname{tg} 18^{\circ}=\frac{1}{5} \sqrt{25-10 \sqrt{5}}$.
$$
\begin{aligned}
\cos 18^{\circ}=... | \cos 18^{\circ} = \frac{1}{4} \sqrt{10+2 \sqrt{5}}, \operatorname{tg} 18^{\circ} = \frac{1}{5} \sqrt{25-10 \sqrt{5}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,639 |
Example 2. Prove that $\sum_{k=1}^{n} \operatorname{ctg}^{2}\left(\frac{k \pi}{2 n+1}\right)=\frac{n(2 n-1)}{3}$. | ```
\begin{array}{l}
\text { +1) } \theta=\sin ^{2 n+1} \theta\left[C \frac{1}{2 n+1} \operatorname{ctg}^{2 n} \theta-C_{2 n+1}^{3}\right. \\
\text { - } \operatorname{ctg}^{2 n-2} \theta+C_{2 n+1}^{5} \operatorname{ctg}^{2 n-4} \theta-\cdots+(-1)^{n} \\
\text { - } \left.C_{2 n+1}^{2 n-1} \operatorname{ctg}^{2} \theta... | \frac{n(2 n-1)}{3} | Algebra | proof | Yes | Yes | cn_contest | false | 702,640 |
Example 3. If $x^{2}+y^{2} \leqslant a^{2}(a>0)$, prove:
$$
\begin{array}{c}
u_{3}=\left|x^{3}+3 x^{2} y-3 x y^{2}-y^{3}\right| \\
\leqslant \sqrt{2} a^{3} .
\end{array}
$$ | Let $x=r \cos c, y=r \sin a$, then $r^{2} \leqslant a^{2}$, $|r| \leqslant a$. Then,
$u_{3}=|r^{3}\left(\cos ^{3} \alpha+3 \cos ^{2} \alpha \sin \alpha - 3 \cos \alpha \sin ^{2} \alpha-\sin ^{3} \alpha\right)|$ $=|r|^{3} \mid\left(\cos ^{3} \alpha-3 \cos \alpha \sin ^{2} \alpha\right) + \left(3 \cos ^{2} \alpha \sin \a... | \sqrt{2} a^{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,641 |
Example 8. Calculate: $\operatorname{tg} 5^{\circ}+\operatorname{ctg} 5^{\circ}-2 \sec 80^{\circ}$. (79 National College Entrance Examination Supplementary Question) | $$
\begin{aligned}
\text { Original expression } & =\frac{2}{\sin \left(2 \times 5^{\circ}\right)}-\frac{2}{\cos 80^{\circ}} \\
& =\frac{2}{\sin 10^{\circ}}-\frac{2}{\sin 10^{\circ}}=0 .
\end{aligned}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,643 |
Example 5. The curve $5 x^{5}+5 x^{4} y-50 x^{3} y^{2}-10 x^{2} y^{3}+25 x y^{4}+y^{5}=0$ intersects the unit circle at ten points $A_{1}$, $A_{2} \cdots, A_{10}$ (arranged counterclockwise along the circumference), then $A_{1} A_{2} \cdots A_{10}$ forms a regular decagon. | Substitute $x=\cos \varphi, y=\sin \varphi(0 \leqslant \varphi<2 \pi)$ into the curve equation, and apply (1), (2) (for $n=5$ case), we get
$$
\begin{array}{l}
5 \cos 5 \varphi+\sin 5 \varphi=0, \quad \sin \left(5 \varphi^{\circ}+\alpha\right) \\
=0, \quad(\alpha=\operatorname{arctg} 5) \\
\varphi_{\mathrm{k}}=\frac{k ... | A_{1} A_{2} \cdots A_{10} \text{ forms a regular decagon} | Geometry | proof | Yes | Yes | cn_contest | false | 702,644 |
Example 1. Find the value of $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}$ $-\cos \frac{4 \pi}{7}+\cos \frac{5 \pi}{7}-\cos \frac{6 \pi}{7}$. | Solve: In (1), let $\alpha=0, n=6$, and transform to get
$$
1+\sum_{k=1}^{0} \cos k \beta=\frac{\cos 3 \beta \sin \frac{7}{2} \beta}{\sin \frac{\beta}{2}} \text {. }
$$
Let $\beta=\pi-\frac{\pi}{7}, \cos k \beta=\cos k \left( \pi -\frac{\pi}{7}\right)=(-1)^{k} \cos \frac{k \pi}{7}$. Therefore,
$$
\begin{array}{l}
\sum... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,645 |
Example 1. Let $a, b, c$ be the sides of $\triangle ABC$. Prove that $a^{2}, b^{2}, c^{2}$ form an arithmetic sequence if and only if $\operatorname{ctg} A, \operatorname{ctg} B, \operatorname{ctg} C$ form an arithmetic sequence. | $\begin{array}{l}\text { Proof: Since ctg } A+\operatorname{ctg} C=\frac{1}{4 \Delta}\left(b^{2}+c^{2}\right. \\ \left.-a^{2}+a^{2}+b^{2}-c^{2}\right)=\frac{b^{2}}{2 \Delta} \\ 2 \operatorname{ctg} B=\frac{c^{2}+a^{2}-b^{2}}{2 \Delta}, \\ \begin{array}{c}\therefore a^{2}+c^{2}=2 b^{2} \longleftrightarrow \operatorname{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,647 |
Example 1. In $\triangle A B C$, if $a \cos ^{2} \frac{C}{2}$ $+c \cos ^{2} \frac{A}{2}=\frac{3}{2} b$, then $a, b, c$ form an arithmetic sequence. | Prove that from the condition we get $\alpha \frac{1+\cos C}{2}+c \frac{1+\cos A}{2}$
$$
\begin{array}{l}
=\frac{3}{2} b . \\
\text { i.e. } \begin{array}{l}
a+c+(a \cos C+c \cos A)=3 b, \\
a+c=2 b .
\end{array}
\end{array}
$$ | a+c=2b | Geometry | proof | Yes | Yes | cn_contest | false | 702,649 |
$$
\begin{array}{l}
a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C\right. \\
\left.-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=0 .
\end{array}
$$
Example 2. In $\triangle A B C$, prove: | Prove $\begin{aligned} \text { Left } & =\left(a^{2} \cos ^{2} B-b^{2} \cos ^{2} A\right) \\ + & \left(b^{2} \cos ^{2} C-c^{2} \cos ^{2} A\right)+\left(c^{2} \cos ^{2} A\right. \\ & \left.-a^{2} \cos ^{2} C\right)=c(a \cos B-b \cos A) \\ & +a(b \cos C-c \cos B)+b(c \cos A \\ & -a \cos C)=0\end{aligned}$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,650 |
Example 3. $A+B+C=180^{\circ}$. Prove: $\cos A+\cos B+\cos C>1$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Prove } \because b+c=(a \cos C+c \cos A) \\
+(a \cos B+b \cos A) \\
=(b+c) \cos A+a(\cos B+\cos C), \\
\therefore \quad \frac{\cos B+\cos C}{1-\cos A}=\frac{b+c}{a}>1, \\
\end{array}
$$
which is to be proved.
| null | Inequalities | proof | Yes | Yes | cn_contest | false | 702,651 |
\{䟝4. Given $a=\frac{b+c}{\cos B+\cos C}$, determine the shape of $\triangle A B C$.
| \begin{array}{l}\text { Solution From the given } a \cos B+a \cos C=(a \cos C \\ +c \cos A)+(a \cos B+b \cos A), \\ (b+c) \cos A=0, \\ A=90^{\circ} . \triangle A B C \text { is a right triangle. }\end{array} | \triangle ABC \text{ is a right triangle.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,652 |
Property 1. $\frac{1}{2}(a+c)-\frac{1}{2} \sqrt{b^{2}+(c-a)^{2}}$ $\leqslant f(\theta) \leqslant \frac{1}{2}(a+c)+\frac{1}{2} \sqrt{b^{2}+(c-a)^{2}}$. | Prove that if $b \neq 0$, take $\operatorname{tg} \varphi=\frac{c-a}{b}$, then
$$
\begin{aligned}
f(\theta) & =\frac{1}{2}(c+a)+\frac{1}{2}{\sqrt{b^{2}+(c-a)}}^{2} \\
& \cdot \sin (2 \theta+\varphi) .
\end{aligned}
$$
The property is true.
$$
\begin{array}{r}
\text { If } b=0, f(\theta)=a \sin ^{2} \theta+\cos ^{2} \t... | proof | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,653 |
Example 9. Simplify $1-\frac{1}{4} \sin ^{2} 2 \alpha-\sin ^{2} \beta-\cos ^{4} \alpha$ into a product form of trigonometric functions. (84 College Entrance Examination for Liberal Arts)
(84年高考文科试题)
Note: The last line is kept in Chinese as it is a reference to the source of the problem and does not need to be trans... | $$
\begin{aligned}
\text { Original expression }= & \left(1-\sin ^{2} \beta\right)-\frac{1}{4} \sin ^{2} 2 \alpha \\
& -\cos ^{4} \alpha \\
= & \cos ^{2} \beta-\sin ^{2} \alpha \cos ^{2} \alpha \\
& \left.\quad-\cos ^{4} \alpha\right) \\
= & \cos ^{2} \beta-\cos ^{2} \alpha\left(\sin ^{2} \alpha\right. \\
& \quad+\cos ... | \sin (\alpha+\beta) \sin (\alpha-\beta) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,654 |
Property 2. For the equation $f(\theta)=d$, let $\Delta=b^{2}-4(a-d)(c-d)$, then $\Delta \geqslant 0 \Longleftrightarrow \quad f(\theta)=d$ has real roots. | $$
\begin{array}{c}
\text { Prove that the equation can be transformed into } \\
{\left[b^{2}+(c-a)^{2}\right] \sin ^{2} 2 x-2 b(2 d-a} \\
-c) \sin 2 x+(2 d-a-c)^{2}-(c-a)^{2}=0
\end{array}
$$
It is known that. | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,655 |
Example 2. If $x^{2}+y^{2} \leqslant 1$, prove that $\mid y^{2}+2 x y$ $-x^{2} \mid \leqslant \sqrt{2}$. | $\begin{array}{c}\text { Proof: Let } x=\lambda \cos \alpha, y=\lambda \sin \alpha, \text { then } \\ \lambda^{2}=x^{2}+y^{2} \leqslant 1, \text { and } \\ \left|y^{2}+2 x y-x^{2}\right|=\mid \lambda^{2}\left(\cos ^{2} \alpha\right. \\ +2 \sin \alpha \cos \alpha-\sin ^{2} \alpha \mid \\ =\lambda^{2} \mid \cos 2 \alpha+... | \sqrt{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,657 |
(1) For the arithmetic sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=3 n-2$, find the formula for the sum of the first $n$ terms.
(2) For the arithmetic sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms given by the formula $S_{n}=5 n^{2}+3 n$, find its first 3 terms, and determ... | Solving (1) yields $S_{n}=\frac{3}{2} n^{2}-\frac{1}{2} n$. The general term formula of this arithmetic sequence is a linear function of $n$, and the sum of the first $n$ terms is a quadratic function of $n$. Solving (2) yields $a_{n}=S_{n}-S_{n-1}$ $=\left(5 n^{2}+3 n\right)-\left[5(n-1)^{2}+3(n-1)\right]$ $=10 n-2 \q... | S_{n}=\frac{3}{2} n^{2}-\frac{1}{2} n, \quad a_{n}=10 n-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,659 |
Example 1. Let $a, b, c$ be positive integers, and $\sqrt[n]{a} + \sqrt[n]{b} = \sqrt[n]{c}$. Prove that:
$$
a + b \geqslant \frac{c}{2^{n-1}}.
$$ | Prove (I) formula: $a+b=(\sqrt[n]{a})^{n}$
$$
+(\sqrt[n]{b})^{n} \geqslant \frac{(\sqrt[n]{c})^{n}}{2^{n-1}}=\frac{c}{2^{n-1}}
$$ | a + b \geqslant \frac{c}{2^{n-1}} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,660 |
Example 4. A cubic water tank (Figure 6), with an edge length of $1 \mathrm{dm}$, has small holes $P, Q, R$ on edges $A A_{1}, B B_{1}, A_{1} L^{*}$, respectively, and $A_{1} P=A_{1} R=\frac{1}{3} \mathrm{dm}, B_{1} \dot{Q}=\frac{1}{4} \mathrm{dm}$. How much water can the tank hold at most? (Pipes can be placed arbitra... | Slightly explained: Let points $P, Q, N$ intersect $B_{1} C_{1}$ at S. $\because$ plane $A D D_{1} A_{1} / /$ plane $B C C_{1} B_{1}, \quad \therefore P R$ $/ / Q S$. The maximum volume that can be filled is the difference between the volume of the cube and the volume of the frustum of a triangular pyramid $B_{1} Q S-A... | \frac{827}{864} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,662 |
Example 5, Given a cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1 inscribed in sphere $O$. Find the surface area of sphere $O$. | Slightly explained: Draw the section through the diagonal plane of the cube (Figure 7), where the diagonal $A C_{1}=\sqrt{3}$ is the diameter of the sphere. Therefore, the surface area of the sphere $S=4 \pi\left(\frac{\sqrt{3}}{2}\right)^{2}$ $=3 \pi$. | 3 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,663 |
Example 6 - To make a cone with the same volume as a cube with an upper edge length of $a$, what dimensions should the cone have for its total surface area to be the smallest? What is the central angle of the sector in the lateral surface development of the cone at this time? | Slightly solved: Let the radius of the base of the cone be $r$, and the apex angle of the cross-section be $2 \theta$. According to the problem, $\frac{1}{3} \pi r^{3} \operatorname{ctg} \theta=a^{3}$, then $r^{3}=\frac{3 a^{3}}{\pi} \operatorname{tg} \theta$.
$\therefore S_{\text {cone total }}=\pi r^{2}+\frac{\pi r^{... | \frac{2 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,664 |
Example 1. Given the sequence $\left\{a_{\mathrm{n}}\right\}$, where $a_{1}=1$. It satisfies the relation $a_{\mathrm{n}}=a_{\mathrm{n}-1}+2 n(n \geqslant 2, n \in N)$, find $a_{n}$. | Solve: Since $a_{n}-a_{n-1}=2n$, and the sequence with the general term $2n$ is an arithmetic sequence, the sequence $\left\{a_{\mathrm{n}}\right\}$ is a second-order arithmetic sequence, whose general formula is $a_{n}=A n^{2}+B n+C$, and it is only necessary to determine the coefficients $A, B, C$.
Given, we have $a... | a_{\mathrm{n}}=n^{2}+n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,666 |
Example 3. Given the sequence $\left\{a_{\mathbf{n}}\right\}$, where $a_{1}=1$, and it satisfies $a_{n+1}=1+2 a_{n}$. Find the general term $a_{n}$. | $$
\begin{array}{c}
\text { Sol } \because a_{n+1}=1+2 a_{n}, \\
\therefore a_{n}=1+2 a_{n-1} \text {. Subtracting the two equations gives } \\
a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right), \text { and } a_{1}=1, \\
a_{2}=1+2 \times 1=3.
\end{array}
$$
From the relation $a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right)$, we can... | 2^{n}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,668 |
Example 4. In the sequence $\left\{a_{\mathrm{n}}\right\}$, $a_{2}=1, a_{3}=2$, and it satisfies: $a_{\mathrm{r}}=(n-1)\left(a_{\mathrm{n}-1}+a_{n-2}\right)(a n \geqslant 4)$, try to find the general term formula. | Let $a_{\mathrm{n}}=n!b_{\mathrm{n}}(n \geqslant 4)$, then we have $n!b_{\mathrm{n}}=(n-1)\left[(n-1)!b_{\mathrm{n}-1}+\right.$ $(n-2)!\cdot b_{n-2}$. Therefore, $n b_{n}=(n-1) b_{n-1} + b_{n-2}$. We get $b_{n}-b_{n-1}=-\frac{1}{n}\left(b_{n-1}-b_{n-2}\right)$ $(n \geq 4)$. By recursion, we have:
$$
\begin{array}{l}
b_... | a_{\mathrm{n}}=n!\sum_{\mathrm{k}=2}^{\mathrm{n}} \frac{(-1)^{\mathrm{k}}}{k!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,669 |
Example 5. Let the sequence of positive numbers $\left\{a_{\mathrm{n}}\right\}$ have the sum of the first $n$ terms $S_{\mathrm{n}}$ satisfying: $\quad S_{\mathrm{n}}=\frac{1}{2}\left(a_{\mathrm{n}}+\frac{1}{a_{\mathrm{n}}}\right)$, find $a_{\mathrm{n}}$. | $$
\begin{array}{l}
\text { Sol } \because S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right), \\
\therefore S_{n-1}=\frac{1}{2}\left(a_{n-1}+\frac{1}{a_{n-1}}\right) .
\end{array}
$$
Subtracting the two equations gives:
$$
a_{n}=\frac{1}{2}\left[\left(a_{n}-a_{n-1}\right)+\frac{1}{a_{n}}-\frac{1}{a_{n-1}}\right]
$$
... | a_{n}=\sqrt{n}-\sqrt{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,670 |
Example 3. Let $a>0, x+y+z:=a$, and $x^{2}+y^{2}+z^{2}=\frac{a^{2}}{2}$. Prove:
$x, y, z$ are all non-negative numbers and do not exceed $\frac{2}{3} a$. | Proof: From the given conditions, it is evident that $x, y, z$ cannot all be negative numbers. Thus, we can divide into the following three cases:
(1) $x, y, z$ are all positive numbers, then
$$
x+y=a-z>0 \text{, }
$$
(2) Two of $x, y, z$ are positive, without loss of generality, assume $x>0, y>0$, then
$$
x+y=a-z>0
$$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,672 |
Given the sequence $\left\{a_{\mathrm{n}}\right\}$ satisfies the recurrence relation $a_{\mathrm{a}}=8 a_{n-1}-6 a_{\mathrm{n}-2}$, and $a_{1}=1, a_{2}=5$, find its general term formula. | The sequence $\left\{a_{\mathrm{n}}\right\}$ satisfies a homogeneous linear recurrence relation with constant coefficients. Let $a_{\mathrm{n}}=A_{1} \cdot q_{1}^{\mathrm{n}}+A_{2} \cdot q_{2}^{\mathrm{n}}$, where $q_{1}, q_{2}$ are the roots of the characteristic equation $x^{2}-5 x+6=0$. Solving, we get $q_{1}=x_{1}=... | a_{n}=3^{n}-2^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,675 |
Example 1. Given $S_{\mathrm{n}}=n^{s}-n$, find its general term $a_{\mathrm{n}}$ and prove that it can be expressed as the sum of the first $n$ terms of an arithmetic sequence.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given $\because S_{\mathrm{n}}=n^{3}-n=n\left(n^{2}-1\right)$, from the above proof, we know that $a_{1}$ satisfies the general formula,
$$
\begin{aligned}
\therefore a_{\mathrm{n}} & =S_{\mathrm{n}}-S_{\mathrm{n}-1} \\
& =n^{3}-n-(n-1)^{3}+n-1 \\
& =3 n(n-1) .
\end{aligned}
$$
Also, notice that
$$
\sum_{i=1}^{n} 6(i-... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,676 |
Example 2. Given $a_{1}=7, a_{\mathrm{n}}=a_{\mathrm{n}-1}+3 n+1$, find $a_{n}$.
untranslated text:
例2. 已知 $a_{1}=7, a_{\mathrm{n}}=a_{\mathrm{n}-1}+3 n+1$,求 $a_{n}$.
translated text:
Example 2. Given $a_{1}=7, a_{\mathrm{n}}=a_{\mathrm{n}-1}+3 n+1$, find $a_{n}$. | Solve $k=1, a_{\mathrm{k}}=a_{1}=7, b=3, c=1$.
From equation (5), we have
$$
\begin{aligned}
a_{\mathrm{n}} & =7+\left(\frac{1+1+n}{2} \times 3+1\right)(n-1) \\
& =\frac{3}{2} n^{2} + \frac{5}{2} n + 2 .
\end{aligned}
$$
Three, given $a_{\mathrm{k}}$, and $a_{\mathrm{n}}=b a_{\mathrm{n}-1}+c$, find $a_{\mathrm{n}}$ (w... | a_{\mathrm{n}}=\frac{3}{2} n^{2} + \frac{5}{2} n + 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,677 |
Example 4. If it is known that $a_{1}=3, a_{n}=3^{4 \mathrm{n}^{2}+1} a_{\mathrm{n}-1}$, find $a_{\mathrm{A}}$.
| Given the conditions, we know $k=1, a_{\mathbf{k}}=a_{1}=3, b=3$.
Substituting into formula (18), we get
$$
\begin{aligned}
a_{\mathrm{n}} & =3 \cdot 3^{\sum_{\mathrm{i}=2}^{\mathrm{n}}\left(4 n^{2}+1\right)} \\
& =3 \cdot 3^{4 \cdot \frac{n}{6}(2 n+1)(n+1)+n-5} \\
& =3^{\frac{1}{3}\left(4 n^{3}+4 n^{2}+5 n-12\right) .... | 3^{\frac{1}{3}\left(4 n^{3}+4 n^{2}+5 n-12\right)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,679 |
Example 2. Given $a_{\mathrm{n}}=3 n-1$, find $S_{\mathrm{n}}$.
| Solve the system
\[
\left\{\begin{array}{l}
2 a=3, \\
-a+b=-1
\end{array}\right.
\]
to get
\[
\left\{\begin{array}{l}
a=\frac{3}{2}, \\
b=\frac{1}{2} .
\end{array}\right.
\]
Therefore,
\[
S_{n}=\frac{3}{2} n^{2}+\frac{1}{2} n.
\] | S_{n}=\frac{3}{2} n^{2}+\frac{1}{2} n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,681 |
Example 3. Given $S_{\mathrm{n}}=a n^{3}+b n^{2}+c n(a \neq 1)$, find $a_{n}$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Since $S_{0}=0$, we have
$$
\begin{array}{l}
a_{\mathrm{n}}=S_{\mathrm{n}}-S_{\mathrm{n}-1} \\
=\left(a n^{3}+b n^{2}+c n\right)-\left[a \cdot(n-1)^{3}\right. \\
\left.\quad+b(n-1)^{2}+c(n-1)\right) \\
=3 a n^{2}+(-3 a+2 b) n+(a-b+c) .
\end{array}
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,682 |
Example 4. Let $a_{\mathrm{n}}=n^{2}+2 n-3$ : Find $S_{\mathrm{s}}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve
$$
\left\{\begin{array}{l}
3 b=1 \\
-3 a+2 b=2 \\
a-b+c=-3
\end{array}\right.
$$
to get
$$
\left\{\begin{array}{l}
a=\frac{1}{3} \\
b=\frac{3}{2} \\
c=-\frac{11}{6}
\end{array}\right.
$$
Thus,
$$
S_{n}=\frac{1}{3} n^{3}+\frac{3}{2} n^{2}-\frac{11}{6} n .
$$
In general, if
$$
a_{\mathrm{n}}=b_{1} n^{\mathrm{p}}... | S_{n}=\frac{1}{3} n^{3}+\frac{3}{2} n^{2}-\frac{11}{6} n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,684 |
Example 5. Let $a_{n}=5 n^{4}+2 n^{3}+4 n^{2}+3 n-3$, find $S_{\mathrm{n}}$. | $$
\begin{array}{l}
\text { Let } S_{n}=c_{1} n^{5}+c_{2} n^{4}+c_{3} n^{3}+c_{4} n^{2} \\
\text { + } c_{5} n \text {. } \\
\end{array}
$$
Then
$$
\left\{\begin{array}{c}
C_{5}^{1} c_{1}=5, \\
-C_{5}^{2} c_{1}+C{ }_{4}^{1} c_{2}=2, \\
C_{6}^{3} c_{1}-C_{4}^{2} c_{2}+C_{8}^{1} c_{3}=4, \\
-C_{5}^{4} c_{1}+C_{4}^{3} c_... | S_{n}=n^{5}+3 n^{4}+4 n^{3}+4 n^{2}-n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,685 |
Example 1. As shown in Figure 1, in the right triangle $\triangle OAB$, $|\overrightarrow{OA}|$ $=|\overrightarrow{AB}|$, let $z_{A}=a+bi$, find $z_{B}$. | Solve $\begin{aligned} z_{\overrightarrow{\mathrm{OC}}} & =z_{A}(\cos \theta+i \sin \theta) \\ & =\left(a+b_{i}\right)\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \\ & =-b+a i, \\ \therefore z_{\mathrm{B}}= & z_{\mathrm{A}}+z \overrightarrow{\mathrm{OC}} \\ = & (a-b)+(a+b) i .\end{aligned}$ | (a-b)+(a+b)i | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,687 |
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