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Example 2. Let $P$ be a moving point on the parabola $y=x^{2}$, construct an equilateral triangle $O P Q (O, P, Q$ in counterclockwise order $)$, find the equation of the locus of its center.
untranslated text:
例2. 设 $P$ 为挞物线 $y=x^{2}$ 上的动点,作正三角形 $O P Q(O 、 P 、 Q$ 按逆时针方向 $)$,求其中心的轨迹方程.
translated text:
Example 2. L... | Let $P$ correspond to the complex number $z_{\mathrm{P}}=t+t^{2} i$, then
$$
\begin{aligned}
z_{Q} & =\left(t+t^{2} i\right)\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) \\
= & \left(\frac{t}{2}-\frac{\sqrt{3}}{2} t^{2}\right) \\
& +\left(\sqrt{3} \frac{3}{2} t+\frac{t^{2}}{2}\right) i .
\end{aligned}
$$
If $\t... | 3 \sqrt{3} x^{2}+6 x y+\sqrt{3} y^{2}+2 x -2 \sqrt{3} y=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,688 |
Please practice yourself: Let $z_{\mathrm{A}}=1, z_{B}=2+i$ $\triangle A B C$ be an equilateral triangle, $C$ is in the first quadrant, find $z$.
| (Answer $\frac{1}{2}(3-\sqrt{3})+\frac{1}{2}(1+\sqrt{3}) i$.
) | \frac{1}{2}(3-\sqrt{3})+\frac{1}{2}(1+\sqrt{3}) i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,689 |
Example?. Prove:
$$
\sin ^{5} \theta=\frac{1}{16}(\sin 5 \theta-5 \sin 3 \theta+10 \sin \theta) .
$$ | $\begin{array}{l}\text { Let } z=\cos \theta+i \sin \theta, \\ \sin \theta=\frac{1}{2 i}\left(z-\frac{1}{z}\right), \\ \sin n \theta=1 \\ \sin ^{5} \theta=\left[\frac{1}{2 i}\left(z-\frac{1}{z}\right)\right]^{5} \\ =\frac{1}{32 i}\left(z^{5}-5 z^{3}+10 z-\frac{10}{z}+\frac{5}{z^{3}}\right. \\ \left.-\frac{1}{z^{5}}\rig... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,692 |
$4. Prove \arcsin \frac{1}{\sqrt{10}}+\arccos \frac{5}{\sqrt{26}}$ $+\operatorname{arctg} \frac{1}{7}+\operatorname{arcctg} 8=\frac{\pi}{4}$ | Prove $\arcsin \frac{1}{10}=\operatorname{arctg} \frac{1}{3}=\arg (3+i)$
$$
\begin{array}{l}
\arccos \frac{5}{\sqrt{26}}=\operatorname{arctg} \frac{1}{5} \\
=\arg (5+i), \\
\operatorname{arctg} \frac{1}{7}=\arg (7+i), \\
\operatorname{arcctg} 8=\arg (8+i) . \\
\therefore \quad \text { the left side of the original equa... | \frac{\pi}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 702,695 |
Example 5. Find the value of $\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}$. | Let $z=\cos \frac{\pi}{7}+i \sin \frac{\pi}{7}$. Then
$$
\begin{array}{l}
\left(\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}\right) \\
+i\left(\sin \frac{\pi}{7}+\sin \frac{3 \pi}{7}+\sin \frac{5 \pi}{7}\right) \\
=z+z^{3}+z^{5}=\frac{z^{7}-z}{z^{2}-1}=\frac{1}{1-z} \\
=\frac{1}{1-\cos \frac{\pi}{7}+i \... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,696 |
Example 1. Prove:
$$
\sin (k \pi+\alpha)=(-1)^{\mathbf{k}} \sin \alpha \cdot(k \in Z)
$$ | \begin{array}{l}\text { Proof: Since } e^{\mathrm{k}}{ }_{\pi}{ }^{1}=\cos k \pi+i \sin k \pi \\ =\cos k \pi=(-1)^{\mathbf{k}}, \quad(k \in \mathbb{Z}) \\ \therefore \sin (k \pi+\alpha) \\ \left.\left.=\frac{1}{2 i}\left(e^{k \pi+\alpha i}\right)^{1}-e^{-(k \pi+\alpha i)}\right)^{1}\right) \\ =(-1)^{k} \frac{1}{2 i}\le... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,697 |
Example 3. Express $\sin ^{8} \alpha \cos ^{3} \alpha$ as a sum. | Let $z=\cos \alpha+i \sin \alpha$. Then,
$$
\begin{array}{l}
\sin ^{8} \alpha \cos ^{3} \alpha=\left[\frac{1}{2 i}\left(z-\frac{1}{z}\right)\right]^{5} \\
=-\frac{1}{256 i} \quad\left(z-\frac{1}{z}\right)^{2}\left(z^{2}-\frac{1}{z^{2}}\right)^{8} \\
=-\frac{1}{256 i}\left[\left(z^{8}-\frac{1}{z^{8}}\right)-2\left(z^{6... | \frac{1}{128} \sin 8 \alpha+\frac{1}{64} \sin 6 \alpha+\frac{1}{64} \sin 4 \alpha | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,699 |
Example 4. Convert $\cos \alpha+\cos 3 \alpha+\cos 5 \alpha$ $+\cos 7 \alpha$ into a product. | Let $z=\cos \alpha+i \sin \alpha$. Then
$$
\begin{aligned}
\text { Original expression }= & \frac{1}{2}\left(z+z^{-1}+z^{3}+z^{-3}+z^{8}\right. \\
& \left.+z^{-5}+z^{7}+z^{-7}\right) \\
= & \frac{1}{2}\left(z+z^{-1}\right)\left(z^{2}+z^{-2}\right) \\
& \cdot\left(z^{4}+z^{-4}\right) \\
= & \frac{1}{2} \cdot 2 \cos \alp... | 4 \cos \alpha \cos 2 \alpha \cos 4 \alpha | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,700 |
Example 2. Find the maximum and minimum values of the function $y=\sin \left(x-30^{\circ}\right) \cos x$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | \begin{array}{l}\text { Solve } y=\sin \left[(x-10)-15^{2}\right] \\ \text { - } \cos \left[\left(x-15^{\circ}\right)+15^{\circ}\right] \\ =\sin \left(x-15^{\circ}\right) \cos \left(x-15^{\circ}\right) \\ -\sin 15^{\circ} \cos 15^{\circ} \\ =\frac{1}{2} \sin \left(2 x-30^{\circ}\right)-\frac{1}{4} \text {. } \\ \theref... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,702 |
Example 3. In $\triangle A B C$, the three interior angles satisfy $\lg \sin A-\lg \cos B-\lg \sin C=\lg 2$. Prove that $\triangle A B C$ is an isosceles triangle. | Given $\operatorname{Ig} \frac{\sin A}{\cos B \sin C}=\lg 2$. Therefore $\sin$
$$
\begin{array}{l}
=2 \cos B \sin C=2 \sin \left(\frac{C+B}{2}+\frac{C-B}{2}\right) \\
\cdot \cos \left(\frac{C+B}{2}-\frac{C-B}{2}\right)=\sin (C+B) \\
+\sin (C-B)=\sin A+\sin (C-B) . \\
\therefore \sin (C-B)=0 \text{. But } C, B \text{ ar... | C=B | Geometry | proof | Yes | Yes | cn_contest | false | 702,703 |
Example 1. If $\frac{x^{2}}{4}+y^{2} \leqslant 1$, find the extremum of $z=x-y$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Example 1. If $\frac{x^{2}}{4}+y^{2} \leqslant 1$, find the extremum of $z=x-y$. | Let $x=2 \lambda \sin \theta, y=\lambda \cos \theta$,
$$
\begin{aligned}
\frac{x^{2}}{4}+y^{2} & =\lambda^{2} \leqslant \lambda, \text { then } \\
|z| & =|2 \lambda \sin \theta-\lambda \cos \theta| \leqslant \sqrt{4 \lambda^{2}+\lambda^{2}} \\
& \leqslant \sqrt{5 .}
\end{aligned}
$$
The equality holds when $\lambda=1,... | z_{\text {max }}=\sqrt{5}, \quad z_{\text {min }}=-\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,704 |
Example 1. Simplify $\sin x\left(1+\operatorname{tg} 2 \operatorname{tg} \frac{x}{2}\right)$. | Let $\operatorname{tg} \frac{x}{2}=t$, then
$$
\begin{aligned}
& \text { Original expression }=\frac{2 t}{1+t^{2}}\left(1+\frac{2 t \cdot t}{1-t^{2}}\right)=\frac{2 t}{1-t^{2}} \\
= & \operatorname{tg} x .
\end{aligned}
$$ | \operatorname{tg} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,707 |
Example 3. Given $2 \sin x + 3 \cos x = 2$, find the values of $\sin x$ and $\cos x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $\operatorname{tg} \frac{x}{2}=t$. Then
$$
\frac{2 \cdot 2 t}{1+t^{2}}+\frac{3\left(1-t^{2}\right)}{1+t^{2}}=2 \text {. }
$$
$t_{1}=1, t_{2}=-\frac{1}{5}$. Then $\sin x=1$ or $-\frac{5}{13}$. Accordingly, $\cos x=0$ or $\frac{12}{13}$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,709 |
Example 4, $\sin x + \cos x > 1$, find the range of values for $x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $\operatorname{tg} \frac{x}{2}=\mathrm{t}$, substitute into the inequality to solve for $0<t<1$, thus $2 k \pi<x<2 k \pi+\frac{\pi}{2}, k \in Z$.
| 2 k \pi < x < 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 702,710 |
Example 5. Given the equation $x^{2}-(\tan \theta+\cot \theta) x$ $+1=0-$ with roots $2+\sqrt{3}$, find the value of $\sin 2 \theta$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve the other root of the equation is $\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$. Let $\operatorname{tg} \theta=t$, then $t+\frac{1}{t}=4$,
$$
\begin{aligned}
t & =2 \pm \sqrt{3}, \sin 2 \theta=\frac{2 t}{1+t^{2}} \\
& =\frac{2(2 \pm \sqrt{3})}{1+(2 \pm \sqrt{3})^{2}}=\frac{1}{2} .
\end{aligned}
$$
(2) Prove that $\frac{\sin ... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,711 |
Example 1. Find the value of $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$ | Let $\cos \frac{\pi}{7}=x$, then the original expression $=2 x^{2}-1+3 x$
$$
-4 x^{3}-x=4 x^{3}+2 x^{2}-1 \text {. }
$$
Notice: $\cos \frac{4 \pi}{7}=-\cos \frac{3 \pi}{7}$,
i.e., 2
$$
\begin{array}{l}
\left(2 x^{2}-1\right)^{2}-1=3 x-4 x^{3}, \\
\therefore(x+1)\left(8 x^{3}-4 x+1\right)=0 .
\end{array}
$$
But $x \ne... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,712 |
Question 2. If $x$ and $y$ are two acute angles, prove that $\cos (x+y)=\cos x \cos y-\sin x \sin y$. | Prove in Figure 2, by drawing a perpendicular line $BF$ from point $B$ to $DE$, intersecting $DE$ at $F$.
By the chord division theorem,
$$
B E^{2}=1^{2}+1^{2}-2 \cos \left[180^{\circ}-(x+y)\right]
$$
Also, $B E^{2}=B F^{2}+F E^{2}$, which means
$B E^{2}=(\cos x+\cos y)^{2}$
$$
+(\sin y-\sin x)^{2}
$$
From equations ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,714 |
Question 3. If $x$
is an obtuse angle,
is an acute angle, prove
that: $\sin (x-y)$
$=\sin x \cos y$
$-\cos x \sin y$. | Prove that on a straight line $l$, take any point $A$, with point $A$ as the vertex and $l$ as one side, construct angles $x$ and $y$. Cut $A B = A E$, and let $A B = A E = 1$. Draw $B C \perp C D, E D \perp C D$, and connect $B E$ (Figure 3). Then
$$
\begin{array}{c}
D E = \sin y, A D = \cos y, \\
B C = \sin \left(180... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,715 |
Same question 5. Cauchy-Schwarz inequality: If $a$, $b$, $c$, and $d$ are four arbitrary real numbers, then
$$
\begin{array}{l}
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
\geqslant(b c+a d)^{2} .
\end{array}
$$ | Prove that in Figure 5, we have
$$
A B=\sqrt{a^{2}+b^{2}}, A E=\sqrt{c^{2}+d^{2}}.
$$
Meanwhile, the area ${ }_{O D D}=\frac{1}{2}(b+d)(a+c)$.
For $\triangle_{B A B}=\frac{1}{2} \Delta E \cdot A E \sin B A E$.
The area of the triangle is influenced by the sine of the angle $\angle B A E$ between the two sides, and the... | \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \geqslant(b c+a d)^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,718 |
If $a$ and $b$ are positive real numbers, then
$$
\frac{a^{2}+b^{2}}{2} \geqslant\left(\frac{a+b}{2}\right)^{2} .
$$ | $$
\begin{array}{c}
\angle B C A \cong \triangle A D E, \angle B A E=90^{\circ} \mathbb{Z} A B=A E . \\
\therefore B E=\sqrt{2 c}=\sqrt{2} \sqrt{a^{2}+b^{2}} . \\
\text { Also } \because B E \geqslant C D, \text { we have } \\
\sqrt{2} \sqrt{a^{2}+b^{2}} \geqslant a+b .
\end{array}
$$
Dividing both sides by 2, we get
... | \frac{a^{2}+b^{2}}{2} \geqslant\left(\frac{a+b}{2}\right)^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,719 |
3. Find the value of $\sin ^{2} 10^{\circ}+\cos ^{2} 40^{\circ}+\sin 10^{\circ} \cos 40^{\circ}$. | Let $\cos 40^{\circ}=x$, then
$$
\begin{array}{l}
-\frac{1}{2}=\cos 120^{\circ}=4 x^{3}-3 x, \\
f(2)=4 x^{3}-3 x+\frac{1}{2}=0, \\
\text { Original expression } 2=\cos ^{2} 80^{\circ}+\cos ^{2} 40^{\circ}+\cos 80^{\circ} \\
\cdot \cos 40^{\circ}=\left(2 x^{2}-1\right)^{2}+x^{2}+\left(2 x^{2}-1\right) x \\
= x f(x)+\fra... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,720 |
3. The larger solution of the quadratic equation
$$
x(x-5)=6
$$
is a solution of the equation
$$
x^{2}+(a-3) z+8=0
$$
Find the value of $a$ and the solutions of equation (2). | $3, a=-4$; (2) solutions $1, 6$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. If you intended to translate a different text, please provide the c... | a=-4; (2) solutions 1, 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,724 |
5. There are 20 cards labeled with digits $1-20$. Now, the card labeled “1” is recorded as $1 \mathrm{~cm}^{2}$, the card labeled “2” is recorded as $2 \mathrm{~cm}^{2}, \cdots$ to represent the area of the card.
(1) When one card is randomly drawn from these 20 cards, try to find: the probability that the area of the ... | 5. (1) $\frac{1}{5} ;$ (2) 22 kinds. | \frac{1}{5} ; 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,726 |
Example 1. In the cube $A C$ (Figure 1), prove that $A C$ $\perp D_{1} B$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | A brief proof: The simpler proof method is to use the three perpendiculars theorem. Connect $B D, B D$ as the projection of $D_{1} B$ on the base $A B C D$. $\because A C \perp B D, \therefore A C \perp D_{1} B$.
From Figure 1, we can also deduce: $A C$ is perpendicular to the line connecting any point on $D_{1}$ and ... | null | Geometry | proof | Yes | Yes | cn_contest | false | 702,727 |
6、The table below shows the score distribution of 50 students in Ye's class:
\begin{tabular}{c|c|c|c|c|c|c|c|c}
\hline Score & 0 & 1 & 2 & 3 & 4 & $\mathbf{5}$ & 6 & Total \\
\hline \begin{tabular}{c}
Number \\
of \\
Students
\end{tabular} & 4 & 4 & $\mathbf{5}$ & $A$ & 15 & $B$ & $\mathbf{2}$ & 50 \\
\hline
\end{tabu... | 6. ( 1 ) $A$ is 18 people, $B$ is 2 people; （2）19 people or more. | A = 18, B = 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,728 |
7. In the figure below, $ABCD$ is a trapezoid, $AD // BC$, $\frac{BC}{AD}=5$, and the area of $\triangle OAD$ is $S$.
(1) Express the area of $\triangle OBC$ in terms of $S$;
(2) How many times the area of trapezoid $ABCD$ is the area of $\triangle OAD$? | 7. (1) $25 S$; (2) 36 times. | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,729 |
8. A cylindrical container with a base diameter of $16 \mathrm{~cm}$ contains 2 spheres of the same size. The height from the bottom of the container to the top of the upper sphere is $18 \mathrm{~cm}$.
（1）Find the radius of the spheres;
（2）If a third sphere is added, find the height from the bottom of the container to... | 8. ( 1 ) $5 \text{ cm} ;$ (2) $26 \text{ cm}$. | 5 \text{ cm}; 26 \text{ cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,730 |
Example 2. In the cube $A C_{1}$, the edge length is $a$, find:
(1) The distance from point $A_{1}$ to the plane $B C_{1} D$; (2) The distance between the line $A B$ and $A_{1} C$; (3) The distance between the plane $A B_{1} D_{1}$ and the plane $B C_{1} D$.
保留源文本的换行和格式，直接输出翻译结果。 | "Body only law". It is easy to prove that $A_{1} C \perp$ plane $B C_{1} D$, let the foot of the perpendicular be $S, A_{1} S$ is the desired length. $A_{1} C=\sqrt{3} a$, $C S=\frac{1}{\sqrt{3}} a$, so $A_{1} S=A_{1} C-C S$ $=\sqrt{3} a-\frac{1}{\sqrt{3}} a=\frac{2 \sqrt{3}}{3} a$.
(2) (Figure 3) This is to find the d... | \frac{2 \sqrt{3}}{3} a, \frac{\sqrt{3}}{3} a, \frac{\sqrt{3}}{3} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,731 |
Example 3. Let the sequence be $u_{1}=2, u_{n+1}=u_{n}+3 n$ +2, find its general term.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Let $v_{n}=v_{n+1} \cdots u_{n}=3 n+2, v_{n}$ be an arithmetic sequence, find the sum:
$$
\sum_{n=1}^{n} v_{k}=3 \cdot \frac{n(n+1)}{2}+2 n \text {. On the other hand, }
$$
$$
\begin{array}{c}
\sum_{k=1}^{n} v_{k}=\sum_{k=1}^{n}\left(u_{n+1}-u_{n}\right) \\
=\left(u_{n+1}-u_{n}\right)+\left(u_{n}-u_{n-1}\right)... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,736 |
Example 4. Let the sequence be $u_{1}=2, u_{n+1}=2 u_{n}+3^{n}$, find its general term.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Solution: Let } u_{n}=v_{n}+n \cdot 3^{n} \text {, substituting into the original equation, we get } \\
v_{n+1}+\alpha \cdot 3^{n+1}=2 \quad\left(v_{n}+\alpha \cdot 3^{n}\right) \\
+3^{n} .
\end{array}
$$
By comparing the coefficients of $3^{n}$, we find that when $\alpha=1$, the equation t... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,737 |
Example 5. Let the sequence be $u_{1}=2, u_{n+1}=2 u_{n} +3 \cdot 2^{n}$, find its general term. | In this example, since the base of the additional term and the coefficient of $x_{n}$ are both 2, the transformation used in Example 4 is effective. At this point, we can let
$$
\begin{array}{l}
u_{n}=v_{n}+a \cdot a \cdot 2^{n}, \text { thus we have } \\
v_{n+1}+a \cdot(n+1) \cdot 2^{n+1} \\
=2\left(v_{n}+a n \cdot 2^... | u_{n}=(3 n-1) \cdot 2^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,738 |
Example 8. Let the sequence be $u_{1}=2$, $u_{n+1}=\frac{n+1}{n^{2}} u_{n}^{2}$, find its general term. | Solve: First, rewrite the sequence as $\frac{u_{n+1}}{n+1} = \left(\frac{u_{n}}{n}\right)^{2}$, then let $v_{n} = \log _{2} \frac{u_{n}}{n}$ to get the geometric sequence $v_{1} = 1, v_{n+1} = 2 v_{n}$, thus
$$
\begin{aligned}
v_{n} = 2^{n-1}, & u_{n} = n \cdot 2^{2^{n-1}} \cdot \\
& \equiv 、 \mathbf{u}_{n+1} = \frac{a... | u_{n} = n \cdot 2^{2^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,741 |
Example 10. Let the sequence be $u_{1}=3$, $s_{n+1}=\frac{2 u_{n}+3}{u_{n}+4}$, find its general term. | Let $u_{n}=v_{n}+\alpha$, then the original equation becomes
$$
v_{n+1}=\frac{(2-\alpha) v_{n}-\left(\alpha^{2}+2 \alpha-3\right)}{v_{n}+\alpha+4}
$$
When $a^{2}+2 a-3=0$, it reduces to the case of Example 9. For this, take $a=1$, and let $v_{1}=2$,
$$
v_{n+1}=\frac{v_{n}}{v_{n}+5} .
$$
Let $\mu_{n}=\frac{1}{v_{n}}$,... | u_{n}=\frac{3 \cdot 5^{n-1}+3}{3 \cdot 5^{n-1}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,743 |
拊3. Let the circumradius of $\triangle A B C$ be $R$. Prove: $a=2 R \sin A$.
| Proof: As shown in the figure, from the circumcenter $O$ of $\triangle ABC$, draw $OD \perp BC$. Connect $OB, OC$.
Then $OB=R, \angle BOD = \frac{1}{2} \angle BOC = \angle A$,
$$
\begin{array}{l}
OD = \frac{1}{2} a . \\
\therefore \frac{BD}{OB} = \sin \angle BOD .
\end{array}
$$
$$
\text{Thus } \frac{a}{2R} = \sin A \... | a = 2R \sin A | Geometry | proof | Yes | Yes | cn_contest | false | 702,745 |
The second root, find the locus of point $(a, b)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve according to Vieta's theorem
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sec \theta+\csc \theta=a, \\
\sec \theta \cdot \csc \theta=b
\end{array}\right. \\
\therefore\left\{\begin{array}{l}
\sin \theta \cdot \cos \theta=\frac{1}{b}, \\
\sin \theta+\cos \theta=\frac{a}{b}
\end{array}\right.
\end{array}
$$
Elimin... | (b+1)^{2}-a^{2}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,750 |
Example 5. Find the extremum of $y=\frac{\sqrt{3} x+1}{\sqrt{x^{2}+1}}+2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve using trigonometric substitution. Let $x=\operatorname{tg} \theta$, $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$,
we get $y=\frac{\sqrt{3} \operatorname{tg} \theta+1}{\sec \theta}+2=\sqrt{3} \sin \theta+\cos \theta+2=2 \sin \left(\theta+\frac{\pi}{6}\right)+2$, $-\frac{\pi}{2}+\frac{\pi}{6}<\theta<\frac{\pi}{2}+\frac{\... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,751 |
Example 6. For all real numbers, prove
$$
|\cos x|+|\cos 2 x| \geqslant \frac{1}{\sqrt{2}} .
$$ | Solve $y=|\cos x|+\left|2 \cos ^{2} x-1\right|$, considering the range constraints of $|\cos x|$ and $\cos ^{2} x$, we use the transformation $|\cos \boldsymbol{x}|=\boldsymbol{t}, 0 \leqslant t \leqslant 1$. This allows us to conveniently remove the absolute value symbols. When $0 \leqslant t \leqslant \frac{\sqrt{2}}... | \frac{1}{\sqrt{2}} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,752 |
Example 1. Given the circle $C: x^{2}+y^{2}=1$ and point $A(2, 0)$, $B$ is a moving point on the circle. Construct an isosceles right triangle $\triangle A B P$ (with $A, B, P$ in clockwise order) using $A B$ as one leg. Try to find the equation of the trajectory of point $P$. | Place circle $C$ and point $\boldsymbol{A}$ on the corresponding complex plane, then the complex number equation of circle $C$ is $|z|=1$. Let $B$ and $P$ correspond to the complex numbers $z^{\prime}$ and $z$, respectively. Clearly, $\left|z^{\prime}\right|=1$.
Vectors $\overrightarrow{B A}$ and $\overrightarrow{B P}... | |z-2i|=\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,753 |
Example 2. Given the circle $x^{2}+y^{2}=4$ and a point $A(2, 0)$ on it, the moving chord $BC$ of this circle always satisfies the condition $\angle BAC = \frac{\pi}{3}$. Try to find the equation of the locus of the centroid of $\triangle ABC$. | Solve: Place the circle and $A$ on the corresponding complex plane, and let point $B$ correspond to the complex number $z^{\prime}$.
$$
\because \text{ the complex equation of the circle is } |z|=2
$$
$$
\therefore |z^{\prime}|=2 \text{, }
$$
$$
\begin{aligned}
& \because |B O| \\
= & |C O|,
\end{aligned}
$$
$$
\begin{... | \left(x-\frac{2}{3}\right)^{2}+y^{2}=\left(\frac{2}{3}\right)^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,754 |
Example 3. Let points $A$ and $B$ slide on the $x$-axis and $y$-axis respectively, and $|A B|=a$. Construct an isosceles triangle with $A B$ as the base. Find the equation of the locus of point $C$.
Translate the text into English, please retain the original text's line breaks and format, and output the translation re... | Rhodium $A, B, C$ correspond to the complex numbers $x^{\prime}$, $y^{\prime} i, x+i y\left(x^{\prime}, y^{\prime}, x, y \in R\right)$, respectively. Therefore, $\overrightarrow{B A}$ and $\overrightarrow{B C}$ correspond to the complex numbers $x^{\prime}-y^{\prime} i, x+\left(y-y^{\prime}\right) i$
$$
\begin{array}{l... | (x-y)^{2}+y^{2}=a^{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,755 |
Example 2. Given: $0<a<1,0<b<1$. Prove:
$$
\begin{array}{l}
+\sqrt{a^{2}+(1-b)^{2}}+\sqrt{(1-a)^{2}+(1-b)^{2}} \\
\geqslant 2 \sqrt{2} . \\
\end{array}
$$ | Prove that, as shown in the figure, on two adjacent sides of the square $ABCD$, segments $AE=a$ and $AF=b$ are taken respectively. Perpendiculars are drawn from these points to the sides, dividing the square into four rectangles, and $EG$ and $FH$ intersect at point $P$. Then $PA=\sqrt{a^{2}+b^{2}}$,
$$
\begin{aligned}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,756 |
Example 1. On the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, points $A\left(x_{1}, y_{1}\right), B\left(4, \frac{\mathfrak{G}}{5}\right), C\left(x_{2}, y_{2}\right)$ and the focus $\boldsymbol{F}(4,0)$ form an arithmetic sequence in terms of their distances. Prove that $x_{1}+x_{2}=8$. | Proof: $a=5, b=3, c=4$.
By the focal radius formula
$$
\begin{array}{l}
\left|A F_{2}\right|=5-\frac{4}{5} x_{1}, \\
\left|C F_{2}\right|=5-\frac{4}{5} x_{2} .
\end{array}
$$
$\because$ The focal radii form an arithmetic sequence,
$$
\begin{array}{l}
\therefore\left|A F_{2}\right|+\left|C F_{2}\right|=2\left|B F_{2}\ri... | x_{1}+x_{2}=8 | Geometry | proof | Yes | Yes | cn_contest | false | 702,758 |
Example 2. Prove: The distance from any point on an equilateral hyperbola to the center is the mean proportional between its distances to the two foci. | Proof: $\because$ the eccentricity of an equilateral hyperbola is $\sqrt{2}$,
$$
\begin{array}{l}
\therefore\left|P F_{2}\right|=|e x-a| \\
=|\sqrt{2} x-a| . \\
\left|P F_{1}\right|=|e x+a| \\
=|\sqrt{2} x+a| \\
\therefore r_{1} \cdot r_{2}=|\sqrt{2} x-a| \\
\cdot|\sqrt{2} x+a| \\
=\left|2 x^{2}-a^{2}\right| \\
=\left... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,759 |
Example 3. Prove that the circle with the
focal chord of a parabola
as its diameter is tangent to the
directrix of the parabola.
Proof: Let the equation of the parabola be $y^{2}=2 p x$. The coordinates of the endpoints of the focal chord are $P_{1}\left(x_{1}, y_{1}\right)$ and $P_{2}\left(x_{2}, y_{2}\right)$. | Analysis: To prove that the distance from the midpoint of the focal chord $P_{1} P_{2}$ to the directrix is equal to $\frac{1}{2}\left|P_{1} P_{2}\right|$, we need to show that the x-coordinate of the midpoint $M$ of $P_{1} P_{2}$ is
$$
\frac{1}{2}\left(x_{1}+x_{2}\right) \text {. }
$$
Then, $|M D|=\frac{1}{2}\left(x_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,760 |
$\begin{array}{c}\text { Example 2. Find the sum } S_{n}=\frac{1}{1 \cdot 5}+\frac{1}{5 \cdot 9}+\cdots \\ +\frac{1}{(4 n-3)(4 n+1)} .\end{array}$ | $$
\text{Solve } \begin{aligned}
& S_{n}=\frac{1}{4}\left[\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)\right. \\
& \left.+\cdots+\left(\frac{1}{4 n-3}-\frac{1}{4 n+1}\right)\right] \\
& =\frac{n}{4 n+1} .
\end{aligned}
$$
In general, if $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is an arithmetic sequence... | \frac{n}{4 n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,762 |
$\begin{array}{l}\text { Example 4. Find } S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\text {. } \\ +\frac{1}{\sqrt{1002001}} \text { the integer part of } S. \\\end{array}$ | Solve the inequality:
$$
\begin{array}{l}
2(\sqrt{k+1}-\sqrt{k})<\frac{1}{\sqrt{k}} \\
<2(\sqrt{k}-\sqrt{k-1}) \\
\end{array}
$$
By setting \( k=1,2, \cdots, 1002001 \), we get:
$$
\begin{array}{l}
2(\sqrt{2}-\sqrt{1})<\frac{1}{\sqrt{1}} \leqslant 1, \\
2(\sqrt{3}-\sqrt{2})<\frac{1}{\sqrt{2}} \\
<2(\sqrt{2}-\sqrt{1}),... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,764 |
Example 1. Find the sum of the arithmetic sequence $a_{1}, a_{1}+d, \cdots$, $a_{1}+(n-1) d$. | $$
\left.\begin{array}{l}
a_{1}+d+d+\cdots+d \\
a_{1}+d+d+\cdots+d \\
a_{1}+d+d+\cdots+d \\
\cdots \cdots \cdots \cdots \cdots \cdots+\cdots \\
a_{1}+\underbrace{d+d+\cdots+d}_{n-1}
\end{array}\right\}
$$
Except for the 1st column, the number of $d$ in the upper right triangle is the same as in the lower left triangle... | S_{n}=n a_{1}+\frac{(n-1) n}{2} d | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,765 |
Example 2. Find the sum $S_{n}=1+(1+2)+(1+2+3)$ $+\cdots+(1+2+3+\cdots+n)$. | Solve by forming a triangular array and then completing it to a square array:
$$
\begin{array}{c}
1+2+3+4+\cdots+n \\
1+2+3+4+\cdots+n \\
1+2+3+4+\cdots+n \\
1+2+3+4+\cdots+n \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
1+2+3+4+\cdots+n
\end{array}
$$
Except for the elements on the diagonal, th... | \frac{n(n+1)(n+2)}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,766 |
Example 1. Prove:
$$
\begin{array}{l}
1+2 C_{n}^{1}+4 C_{n}^{2} \\
+\cdots+2{ }_{n}^{n}=3^{n}
\end{array}
$$ | Prove that
$$
x^{n}+C_{n}^{1} a x^{n-1}
$$
$$
\begin{array}{l}
+C_{n}^{2} a^{2} x^{n-2}+\ldots \\
+C_{n}^{n-1} a^{n-1} x+C_{n}^{n} a^{n} \\
=(x+a)^{n},
\end{array}
$$
Let $x=1, a=2$, then we get
$$
\begin{array}{l}
1+C_{n}^{1} \cdot 2+C_{n}^{2} \cdot 2^{2}+\cdots+C_{n}^{n-1} \\
\cdot 2^{n-1}+C_{n}^{n} \cdot 2^{n}=(1+2... | 3^{n} | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,767 |
Example 1. Prove: $\sqrt{1-\sqrt{1-\sqrt{1-\ddots}}}$
$$
=\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}
$$ | Let the left side $=k_{1}$, then $\sqrt{1-k_{1}}=k_{1}$, the root is $k_{1}=\frac{-1+\sqrt{5}}{2}$. Let the right side $=k_{2}$, then $\frac{1}{1+k_{2}}=k_{2}$. Solving yields $k_{2}=\frac{-1+\sqrt{5}}{2}$, hence $k_{1}=k_{2}$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,768 |
Example 2. Let $x$ be a real number, prove that the value of the fraction $\frac{x^{2}+34 x-71}{x^{2}+2 x-7}$ cannot be between 5 and 9. | Proof: Let the original equation be $y$, then $(y-1) x^{2}+(2 y-34) x-7 y+71=0$. $x$ is a real number, then $y^{2}-14 y+45 \geqslant 0$. Therefore, $y \leqslant 5$ or $y \geqslant 9$.
(Author's affiliation: Nanyang Health School, Henan) | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,769 |
Example 1. Factorize:
$$
\left(x^{2}-x+15\right)\left(x^{2}-x-5\right)+51
$$ | \begin{aligned} \text { Sol } & \text { Let } y=\frac{x^{2}-x+15+x^{2}-x-5}{2} \\ = & x^{2}-x+5 . \text { Then } \\ \text { Original expression } & =(y+10)(y-10)+51=y^{2}-49 \\ & =(y-7)(y+7) \\ & =(x-2)(x+1)\left(x^{2}-x+12\right) .\end{aligned} | (x-2)(x+1)\left(x^{2}-x+12\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,770 |
Example 2. Find the real solutions of $\left\{\begin{array}{l}x^{4}+y^{4}=272, \\ x-y=2\end{array}\right.$. | Let $z=\frac{x+y}{2}$, combining with $x-y=2$, the first equation of the system becomes $\left(z^{2}-9\right)\left(z^{2}+15\right)=0$, $z= \pm 3$. Therefore, we get two sets of solutions:
$$
\left\{\begin{array} { l }
{ x _ { 1 } = - 2 , } \\
{ y _ { 1 } = - 4 , }
\end{array} \quad \left\{\begin{array}{l}
x_{2}=4 \\
y... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,771 |
Example 4. Real numbers $a, b, c, d, e$ satisfy $a+b+c+d$ $+e=8, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$. Find the maximum value of $e$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve: $a+b+c+d=8-e$. Let $a=\frac{8-e}{4}+a, b=\frac{8-e}{4}+\beta, c=\frac{8-e}{4}+\gamma, d=\frac{8-e}{4}+\delta$, then $a+\beta+\gamma+\delta=0$, and
$$
\begin{array}{l}
\quad 16-e^{2}=a^{2}+b^{2}+c^{2}+d^{2}=\frac{(8-e)^{2}}{4} \\
\quad+a^{2}+\beta^{2}+\gamma^{2}+\delta^{2} \geqslant \frac{(8-e)^{2}}{4} . \\
\ther... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,773 |
Example. Factorization: $x^{7}+2 x^{5}+x+2$. | $$
\begin{array}{l}
x^{7}+2 x^{5}+x+2=\left(x^{7}+x^{5}+1\right) \\
+\left(x^{5}+x+1\right)=\left(x^{2}+x+1\right) \\
\left(x^{5}-x^{4}+2 x^{3}-x^{2}-x+2\right)
\end{array}
$$
(Author's affiliation: Dao County No.1 High School, Hunan)
| (x^{2}+x+1)(x^{5}-x^{4}+2 x^{3}-x^{2}-x+2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,774 |
Example 2. Solve the equation $\sqrt{x^{2}-x-2}+\sqrt{x^{2}-3 x+5}$ $=3$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve the equation $\left(x^{2}-x-2\right)-\left(x^{2}-3 x+5\right)$ $=2 x-7$ by dividing both sides by the original equation, then adding it to the original equation, and squaring to organize:
$$
8 x^{2}-11 x-19=0 \text {. }
$$
Solving yields: $x_{1}=-1, x_{2}=2 \frac{3}{8}$ (extraneous root). (Author's affiliation: ... | x_{1}=-1, x_{2}=2 \frac{3}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,776 |
The sum of the binomial expansion coefficients of three terms equals 22, and the middle term of this expansion equals 540000, find the value of $x$.
untranslated part:
的㑑。
Note: The last part "的㑑。" seems to be a typographical error or a non-standard phrase, and it is not translated as it does not have a clear mean... | From the given, we have
$$
C_{n}^{n-2}+C_{n}^{n-1}+C_{n}^{n}=22 .
$$
That is, $\frac{n(n-1)}{2}+n+1=22$.
Simplifying and rearranging, we get: $n^{2}+n-42=0$.
Solving for $n$, we get $n=6, n=-7$ (discard $n=-7$).
Therefore, the middle term is the 4th term.
By the general term formula, the middle term is
$$
T_{3+1}=C_{0... | x=10 \text{ or } x=\frac{1}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,778 |
3. Given point $P_{1}(3, a)$ is symmetric to point $P_{2}$ $(b,-2)$ with respect to the $x$-axis, find the values of $a, b$ and the coordinates of the point $P_{1}$ symmetric to the $y$-axis and the origin. | (Ans: $a=2, b=3$;
$$
(-3,2) ;(-3,-2)) \text {. }
$$ | a=2, b=3; (-3,2); (-3,-2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,780 |
10. (1) Given the parabola $y=-2 x^{2}+8 x-8$. Without moving the vertex, reverse the direction of the opening, and try to find the equation of the resulting parabola. | (筥: $y=2 x^{2}-8 x+8$ )
(Basket: $y=2 x^{2}-8 x+8$ ) | y=2 x^{2}-8 x+8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,782 |
Example 3. If the coefficients of the $x^{n}$ term in the expansions of $(a x+1)^{2 n}$ and $(x+a)^{2 n+1}$ are equal $(a \neq 0)$, prove that $\frac{1}{\boldsymbol{a}}$ must be a root of the equation $n^{2}(n+1) x^{2}+(2 n+1)^{2} x^{2}$ $-(2 n+1)^{2}=0$. | Prove that in $(a x+1)^{2 n}$, the term containing $x^{n}$ is the $(n+1)$-th term, which is $T_{n+1}=C_{2 n}^{n}(a x)^{n}$. And in $(x+a)^{2 n+1}$, the term containing $x^{n}$ is the $(n+2)$-th term, which is $T_{n+2}=C_{2 n+1}^{n+1} x^{n} \cdot a^{n+1}$.
From the problem, we have: $a^{n} C_{2 n}^{n}=a^{n+1} C_{2 n+1}^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,789 |
25. Given in $\triangle A B C$ that $a$, $b$, and $c$ are the three sides of the triangle, and $a: b: c=20: 29: 21$.
Find: (1) the cosine of the largest angle of $\triangle A B C$;
(2) the sine of the angle opposite side $a$;
(3) the tangent of the angle opposite side $c$. | ( Answer $\cos B=0$ )
( Answer $\sin A=\frac{20}{29}$ )
( Answer $\operatorname{tg} C=\frac{21}{20}$ ) | \cos B=0, \sin A=\frac{20}{29}, \operatorname{tg} C=\frac{21}{20} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,792 |
26. $a, b, c$ are the sides opposite to the internal angles $A$, $B$, $C$ of $\triangle ABC$, respectively. Given $a=2, b=\sqrt{2}$, and $\frac{\sin C}{\sin B}=\frac{\sqrt{6}+\sqrt{2}}{2}$.
Try to find: (1) the side $c$ and the supplementary angle of angle $A$;
(2) Calculate:
i)
$$
\begin{array}{l}
\sin \left(180^{\cir... | (Given: $\left.=\sqrt{3}+1,135^{\circ}\right)$ (Answer: $\frac{1}{2}$ )
(Answer: $2-\sqrt{2}$ ) | \sqrt{3}+1, 135^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,793 |
30. For which values of $b$, the system of inequalities
$$
\left\{\begin{array}{l}
0.5(2 x+5)<2(x-2)+5 \\
2(b x-1)<3
\end{array}\right.
$$
has no solution? | (Answer: $b \in\left[\frac{5}{3},+\infty\right)$ ) | b \in\left[\frac{5}{3},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,796 |
31. For which values of $n$, is the function $y=n x^{2}$
$+(n-x) x+n-1$ negative for all $x \in R$? | (Answer: $n \in\left(-\infty, \frac{2}{3}\right)$ ) | n \in\left(-\infty, \frac{2}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,797 |
32. For what $m$, the inequality $\frac{x^{2}-m x-2}{x^{2}-3 x+4}$ $>-1$ holds for all $x$? | (Answer: $m \in(-7,1)$ ) | m \in(-7,1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,798 |
35. Find the general term formula (involving complex numbers or trigonometric functions) for the following sequences:
$$
\text { (1) } 1,-1,1,-1, \cdots,
$$
(2) $1,0,1,0, \cdots$,
(3) $1, i,-1,-i, \cdots$,
(4) $1,0,0,1,0,0,1, \cdots$
(5) $1,0,0,0,1,0,0,0,1 \cdots$. | (Ans: $(-1)^{n-1}=\cos (n-1) \pi$ )
(2) $\left.\frac{1+(-1)^{n-1}}{2}=\sin ^{2} \frac{n \pi}{2}\right)$
(Ans: $i^{\mathrm{n}-1}$ )
(Ans $\frac{1+\omega^{n-1}+\omega^{2(n-1)}}{3}$,
$$
\begin{array}{c}
\left.\omega=\frac{-1+\sqrt{3} i}{2}\right) \\
\left(\text { Ans } \frac{1+i^{n-1}+(-1)^{n-1}+(-i)^{n-1}}{4}\right)
\end... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,799 |
Given 4. $m, n$ are positive integers, and in the polynomial $f(x)=(1+x)^{m}+(1+x)^{n}$, the coefficient of $x$ is 19.
1) Try to find the minimum value of the coefficient of $x^{2}$ in $f(x)$;
2) For the $m, n$ that make the coefficient of $x^{2}$ in $f(x)$ the smallest, find the term containing $x^{7}$ at this time. | Solve for the coefficient of $x$ in $f(x)=(1+x)^{m}+(1+x)^{n}$:
$$
\begin{array}{r}
C_{m}^{1}+C_{n}^{1}=m+n=19 . \\
\therefore \quad m=19-n .(1 \leqslant n \leqslant 18)
\end{array}
$$
1) The coefficient of $x^{2}$ in $f(x)$ is:
$$
\begin{aligned}
C_{m}^{2} & +C_{n}^{2}=\frac{1}{2}[m(m-1)+n(n \\
& -1)] \\
& =\frac{1}{2... | 156 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,800 |
Example 5. Prove that the sum of the quotients obtained by dividing each coefficient of the expansion of $x(1+2)^{n}$ by the power of $x$ in that term equals
$$
\frac{2^{n+1}-1}{n+1}
$$ | Prove that the $(k+1)$-th term of the expansion of $x(1+x)^{n}$ is:
$$
T_{\mathrm{k}+1}=C_{n}^{b} x^{k+1} \quad(k=0,1,2, \cdots, n,)
$$
Therefore, the required sum is
$$
\begin{aligned}
S=1+ & \frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\cdots+\frac{1}{n} C_{n}^{n-1} \\
& +\frac{1}{n+1} C_{n}^{n}
\end{aligned}
$$
If w... | \frac{2^{n+1}-1}{n+1} | Algebra | proof | Yes | Yes | cn_contest | false | 702,801 |
Example 2. Let the x-coordinates $x_{1}, x_{2}, x_{3}$ of three points $A, B, C$ on an ellipse form an arithmetic sequence (due to 1), and $F$ is a focus of the ellipse. Prove that $|A F|,|B F|,|C F|$ also form an arithmetic sequence. | Proof: Let the directrix corresponding to the focus $F$ be $l: x=m$. Draw perpendiculars from $A, B, C$ to the directrix $l$, and let the feet of these perpendiculars be $A_{1}, B_{1}, C_{1}$, respectively. Then, by the definition of an ellipse, we have
$$
\begin{aligned}
\quad \frac{|A F|}{\left|A A_{1}\right|}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,803 |
Example 3. Prove: If $AB$ is a chord passing through the focus of the parabola $y^{2}=2px$, then the circle with $AB$ as its diameter must be tangent to the directrix of the parabola. | Proof: Let $M$ be the midpoint of $AB$, and the distances from $A, M, B$ to the directrix $x=-\frac{p}{2}$ are $d_{1}, d_{2}, d$ respectively.
Then $d_{1}=|AF|, d_{3}=|BF|$,
$$
\therefore d_{1}+d_{3}=|AF|+|BF|=|AB|.
$$
By the midpoint theorem of the parabola, we get
$$
l_{2}=\frac{d_{1}+d_{3}}{2}=\frac{1}{2}|AB|.
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,804 |
Example 2. Two rectangular boards are joined as shown in Figure 2. Given $A E=6, C D=2, E D=1$, if it is folded along $E F$ to form a $120^{\circ}$ dihedral angle, prove: $A C \perp D F$, | Prove that for the "circumscribed" rectangular prism $AC$ (Fig. 3) formed by folding, $EG = AE \cos 60^{\circ} = 3$. Connect $CG$ intersecting $FD$ at $H, \operatorname{tg} \angle CGD = \frac{CD}{DG} = \frac{1}{2}, \operatorname{tg} \angle CDF = \frac{FC}{CD} = \frac{1}{2}$, thus $\angle CGD = \angle CDF, \triangle CHD... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,805 |
Example 3. The radius of the top base of the frustum is 1, the radius of the bottom base and the height are $2, AC$ is the diagonal drawn along the axis, $BD$ is a chord of the bottom circle, $\angle ABD=\frac{\pi}{4}$ (Figure 4), find the distance between the skew lines $AC, BD$.
Translate the above text into Englis... | Solve for cutting $A D$, then $A D \perp B D$. In the axial section, extend $A C$ to intersect the perpendicular line $B F$ of $A B$ at $F$, and then construct a rectangular prism $A F$ with $A D, B D, B F$ as length, width, and height respectively (Figure 5). From Figure 4, we can find $F B=2 \frac{2}{3}$. In Figure 5... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,806 |
Example. The edge length of the cube $A C^{\prime}$ is $a$, find the distance between $B D$ and $A B^{\prime}$. | Solution One: As shown in Figure 1, where $O E \perp A O^{\prime}$, thus $O E \perp$ plane $A B^{\prime} D^{\prime}$. But $B D / A$ plane $A B^{\prime} D^{\prime}$, so the length of $O E$ is the required distance. (Calculation omitted)
Solution Two: As shown in Figure 2, let $E$ be a point on $B D$, with $B E = x$. Dr... | \frac{\sqrt{3}}{3} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,808 |
Example 4. Prove: The four intersection points of $y^{2}=2\left(x+\frac{7}{2}\right)$ and $x^{2}=2\left(y+\frac{7}{2}\right)$ are shared.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that by adding the known two equations and simplifying, we get:
$$
(x-1)^{2}+(y-1)^{2}=4^{2} \text { . }
$$
This is a circle passing through all the intersection points of the known two curves, so the four intersection points of the known two curves are concyclic. | null | Algebra | proof | Yes | Yes | cn_contest | false | 702,815 |
. The sum of the greatest integer less than or equal to $x$ and the smallest integer greater than or equal to $x$ is 5, then the solution set of $x$ is
(A) $\{5 / 2\}$; (B) $\{x \mid 2 \leqslant x \leqslant 3\}$;
(C) $\{x \mid 2 \leqslant x<3\}$; (D) $\{x \mid 2<x \leqslant$
$3\}$; (E) $\{x \mid 2<x<3\}$. | 7
$\mathrm{E}$
7. Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$, then $x-1<\lfloor x\rfloor \leqslant x, \cdots$ (1)
Let $\lceil x\rceil$ denote the smallest integer greater than or equal to $x$, then $x \leqslant\lceil x\rceil<x+1 \cdots$ (2).
It is also known that $\lfloor x\rfloor+\lc... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,816 |
8.19 In 1980, the population of the United States was $226,504,825$, its area was $3,615,122$ square miles, and each square mile is (5280) ${ }^{2}$ square feet. Which of the following numbers is the best approximation of the average square footage per person?
(A) $5,0 \div 0$;
(B) 10,003 ;
(C) 50,000 ;
(D) 100,000 ; (... | 8
$\mathrm{E}$
8. With about 230 million people in an area of 4 million square miles, there are approximately 60 people per square mile. Since 1 square mile is roughly $(5000 \text{ ft })^{2}=25$ million square feet, this results in each person having about 25 million square feet divided by 60, which is closest to $(E)... | E | Other | MCQ | Yes | Yes | cn_contest | false | 702,817 |
9 . The product $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \cdots(1$ $\left.-\frac{1}{9^{2}}\right)\left(1-\frac{1}{16^{2}}\right)$ equals
(A) $\frac{5}{12}$;
(B) $\frac{1}{2}$;
(C) $\frac{11}{26}$;
(D) $\frac{2}{3}$;
(E) $\frac{7}{10}$. | 9
C
9. Factorize each term in the given expression as a difference of squares, then group the terms according to their signs:
$$
\begin{array}{l}
{\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\right.} \\
\left.\cdots\left(1-\frac{1}{10}\right)\right]\left[\left(1+\frac{1}{2}\right... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,818 |
10. Consider the 12C permutations of $A H S M E$ as ordinary five-letter words and arrange them in dictionary order. The last letter of the 86th word is
(A) $A$; (B) $H$; (C) $S$; (D) $M$; (E) $E$. | $\frac{10}{\mathrm{E}}$
10. The first $4!=24$ words start with the letter $A$, the next 24 words start with $E$, and the following 24 words start with $H$. Therefore, the 86th word starts with $M$. And it is the $86-72=14$th word of such: $\quad / M$ starts with the first 6 words starting with $M A$, the next few words... | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 702,819 |
12. John ( $J o h n$ ) scored 93 points in this $A H S M E$. If the old scoring method were still in effect, he would have scored only 84 points with the same answers. How many questions did he leave unanswered? (The new scoring rules are explained on the cover; under the old scoring method, the base score was 30 point... | 12
$\mathrm{B}$
12. Let John answer $c$ questions correctly, $w$ questions incorrectly, and leave $u$ questions unanswered. According to the old scoring method, he scores 84 points, i.e., $30+4 c-w=84 \cdots$ (1). According to the new scoring method, he scores 93 points, i.e., $5 c+2 u=93 \cdots$ (2). Also, $c+w+u=30 \... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,821 |
16. In $\triangle A B C$
$$
A B=8, B C=7 \text {, }
$$
$C A=6$. Also, as shown in the figure, extend side $B C$ to point $P$, such that $\triangle P A B$ is similar to $\triangle P C A$. The length of $P C$ is
$$
\text { (A) } 7 \text {; (B) } 8 \text {; (C) } 9 \text {; (D) } 10 \text {; (E) } 11 .
$$ | $$
\frac{16}{\mathrm{C}}
$$
16. From the similarity of the two triangles, we get $\frac{P A}{P B}=\frac{P C}{P A}=\frac{C A}{A B}$. Therefore, $\frac{P A}{P C+7}=\frac{P C}{P A}=\frac{6}{8}$, which leads to two equations $6(P C+7)=8 P A$ and $6 P A=8 P C$, from which we can find $P C=9$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,825 |
Example 5. A line segment connecting any two points on a parabola is called a chord of the parabola. Prove that the midpoints of parallel chords of a parabola lie on a straight line. | Proof: Let the equation of the parabola be $y^{2}=2 p x(p)$ 0 ), the slope of its parallel chords is $k$, and the equation is $y=k x+b$. From the system of equations $\left\{\begin{array}{l}y^{2}=2 p x, \\ y=k x+b\end{array}\right.$, eliminating $x$ yields $y^{2}-\frac{2 p}{k} y+\frac{2 p b}{\hbar}=0$.
The two roots o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,826 |
17. In a drawer in a dark room, there are 160 red socks, 80 green socks, 60 blue socks, and 40 black socks. A young man picks one sock from the drawer each time, but he cannot see the color of the sock he picks. How many socks must he pick to ensure that he has at least 10 pairs of socks?
(A) 21 ; (B) 23; (C) 24; (D) 3... | $\left|\frac{17}{\mathrm{~B}}\right|$
17. In any selection, at most one sock of a certain color will not form a pair, which happens if and only if an odd number of socks of that color are taken. Therefore, taking 24 socks is sufficient: because at most 4 of them will not form pairs, and the rest will have at least 20 p... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 702,827 |
18. A plane intersects a right circular cylinder of radius 1 to form an ellipse. If the major axis is 50% longer than the minor axis, the length of the major axis is
(A) 1 ; (B) $\frac{3}{2}$;
(C) 2 ;
(D) $\frac{9}{4}$;
(E) 3 . | 18
E
18. The minor axis of the ellipse is parallel to the diameter of the circular base of the cylinder, so the length of the minor axis is 2. At this time, the length of the major axis is
$$
2+0.5 \times 2=3
$$ | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,828 |
19. A park is in the shape of a regular hexagon with each side 2 kilometers long. Alice starts from a corner and walks 5 kilometers along the park's perimeter. How far is she from her starting point?
(A) $\sqrt{13}$;
(B) $\sqrt{14}$;
(C) $\sqrt{15}$
(D) $\sqrt{16}$;
(E) $\sqrt{17}$ | 19
A
19. As shown in the figure, let's assume Alice starts from point $A$ and reaches point $B$. By the cosine rule, we get $A C=2 \sqrt{3}$. In $\triangle A B C$, $\angle A C B=90^{\circ}$. By the Pythagorean theorem, we have
$$
(A B)^{2}=13, A B=\sqrt{13} .
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,829 |
20. Let two positive numbers $x$ and $y$ be inversely proportional. If $x$ increases by $P \%$, then $y$ decreases by
(A) $p \%$;
(B) $\frac{p}{1+p} \%$;
(C) $\frac{100}{p} \%$
(D) $\frac{p}{100+p} \%$;
(E) $\frac{100 p}{100+p} \%$ | 26. When $x$ and $y$ are inversely proportional, it means that if $x$ is multiplied by $k$, then $y$ should be divided by $k$. Let $x^{\prime}$ and $y^{\prime}$ be the new values of $x$ and $y$ after $x$ increases by $p \%$, then
$$
\begin{aligned}
x^{\prime} & =\left(1+\frac{p}{100}\right) x, \\
y^{\prime} & =\frac{y}... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,830 |
21. In the mechanism shown in the figure, $\theta$ is measured in radians, $C$ is the center of the circle, $BCD$ and $ACE$ are straight line segments, and $AB$ is the tangent to the circle at point $A$. Given $0<\theta<\frac{\pi}{2}$, the necessary condition for the areas of the two shaded regions to be equal is
(A) $... | $\frac{21}{\mathrm{~B}}$
21. The area of the shaded sector is $\frac{\theta}{2}(A C)^{2}$, which must equal half the area of $\triangle A B C$, so it is true if and only if
$\frac{\theta}{2}(A C)^{2}=\frac{1}{4}(A C) \quad(A B)$, that is,
$2 \theta=\frac{A B}{A C}=\tan \theta$,
when the two shaded regions have the sam... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,831 |
22. From $\{1,2,3, \cdots, 10\}$, six different integers are randomly selected. Among these selections, the probability that the second smallest number is 3 is
(A) $\frac{1}{60}$;
(B) $\frac{1}{6}$;
(C) $\frac{1}{3}$;
(D) $\frac{1}{2}$;
(E) none of the above. | $\frac{22}{C}$
22. To choose six different integers from $\{1,2, \cdots, 10\}$, there are $C_{10}^{\mathrm{B}}=210$ different ways. However, if the second smallest number is 3, then one number must be chosen from $\{1,2\}$, and four numbers must be chosen from $\{4,5, \cdots, 10\}$. This results in $C_{2}^{1} C_{7}^{4}... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 702,832 |
$$
\begin{aligned}
\text { 23. Let } N=69^{5}+5 \cdot 69^{4}+10 \cdot 69^{3}+10 \cdot 69^{2} \\
+5 \cdot 69+1
\end{aligned}
$$
How many positive integers are divisors of $N$?
(A) 3; (B) 5 ; (C) 69; () 125; (D) 216 .
| $\frac{23}{\mathrm{E}}$
23. By the binomial theorem, $N=(69+1)^{5}$ $=(2,5,7)^{5}=2^{5}, 5^{5}, 7^{5},$ the number of factors of $N$ is $(5+1)(5+1)(5+1)=6^{3}=216$. | 216 | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,833 |
24. Let $p(x)=x^{2}+b x+c$, where $b$ and $c$ are integers. If $p(x)$ is a factor of both $x^{4}+6 x^{2}+25$ and $3 x^{4}+4 x^{2}+28 x+5$, then $p(1)$ is
(A) 0 ; (B) 1 ; (C) 2 ; (D) 4 ; (E) 8 . | $\frac{24}{\mathrm{D}}$
24. Since $p(x)$ is a factor of $x^{4}+6 x^{2}+25$ and $3 x^{4}+4 x^{2}+28 x+5$, it is also a factor of $3\left(x^{4}+6 x^{2}+25\right)-\left(3 x^{4}+4 x^{2}+28 x+5\right)$. The latter equals $14 x^{2}-28 x+70$, i.e., $14\left(x^{2}-2 x+5\right)$. Therefore, $p(x)=x^{2}-2 x+5$, $p(1)=4$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,834 |
26. In the coordinate plane, construct a right triangle such that its two legs are parallel to the $x$ and $y$ axes, and such that the midline of one of its legs lies on one of the lines $y=3x+1$ and $y=mx+2$. How many constants $m$ allow for the existence of such a triangle?
(A) 0 ; (B) 1 ; (C) 2 ; (D) 3 ; (E) More th... | 26
C
26. It is easy to prove using analytic geometry that: "In a right-angled triangle with its two legs parallel to the coordinate axes, the slope of the median on one leg is necessarily 4 times the slope of the median on the other leg." As shown in the figure, a triangle of this kind is drawn, where
$a, b, c, d$ are... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,836 |
Example 6. Given the line $\frac{x}{n}+\frac{y}{n}=1$ is a tangent to the ellipse $b^{2} x^{2}+ a^{2} y^{2}=a^{2} b^{2}$. Prove: $a^{2}+b^{2}=n^{2}$. | Prove that the equation of the tangent line to an ellipse is
$y=l x \pm \sqrt{a^{2} k^{2}+b^{2}}$.
$\because k=1, \quad \therefore y=x \pm \sqrt{a^{2}+b^{2}}$.
$=-n$. Therefore, $a^{2}+b^{2}=n^{2}$.
Comparing with the line $x-y=n$ gives $\pm \sqrt{a^{2}+b^{2}}$
(Author's affiliation: Taiyuan Normal University, Shanxi) | a^{2}+b^{2}=n^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 702,837 |
27. As shown in the figure, $AB$ is the diameter of a circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$, with $\angle AED = a$. The ratio of the area of $\triangle CDE$ to the area of $\triangle ABE$ is
(A) $\cos a$; (B) $\sin a$; (C) $\cos^2 a$;
(D) $\sin^2 a$;
(E) $1-\sin a$. | $\frac{27}{C}$
27. In the figure, $A B / / D C, A \overarc{D}=\overarc{C B}, \triangle C D E$ is similar to $\triangle A B E$, thus
$$
\frac{\text { area of } \triangle C D E}{\text { area of } \triangle A B E}=\left(\frac{D E}{A E}\right)^{2}.
$$
Connecting $A D$, since $A B$ is a diameter, $\angle A D B=30^{\circ}$.... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,838 |
29. In a scalene triangle $ABC$, the lengths of two altitudes are 4 and 12. If the length of the third altitude is also an integer, what is the maximum it can be?
(A) 4 ;
(B) 5 ;
(C) 6 ;
(D) 7 ;
(E) None of the above. | $\left\lvert\, \frac{29}{\mathrm{~B}}\right.$
29. Suppose the length of the base on $a$ is $4$, the height on $b$ is $12$, and the height on $c$ is the unknown $h$. Let $K$ be the area of $\triangle A B C$, then $4 a=12 b=h c=2 K$. By the triangle inequality $c<a+b$, in other words,
$$
\frac{2 K}{h}\frac{2 K}{4}-\frac{... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,840 |
30. Solve the system of equations
$$
\begin{array}{l}
2 y=x+\frac{17}{x}, 2 z=y+\frac{17}{y} \quad 2 w=z+\frac{17}{z}, \\
2 x=w+\frac{17}{w}
\end{array}
$$
The number of real solutions $(x, y, z, w)$ is
(A) 1 ; (B) 2 ; (C) 4 ; (D) 8 ; (E) 16 . | 30
B
30. First, either $x>1$, or $x<0$, consider each equation in sequence, we can deduce that $y \geqslant \sqrt{17}$, $z \geqslant \sqrt{17}$, $w \geqslant \sqrt{17}$, and $x \geqslant \sqrt{17}$. Assume $x>\sqrt{17}$, then
$$
y-\sqrt{17}=\frac{x^{2}+17}{2 x}-\sqrt{17}=\left(\frac{x-\sqrt{17}}{2 x}\right)
$$
$\cdot(x... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,841 |
1. Find the sum of all roots of the following equation:
$$
\sqrt[4]{x}=\frac{12}{7-\sqrt[4]{x}}
$$ | 1. Let $y=\sqrt[4]{x}$, then the original equation can be transformed into $y^{2}-7 y+12=0$, whose roots are 3 and 4, so the solution is $s^{4}+4^{4}$ $=337$. | 337 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,842 |
$$
\begin{array}{l}
(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6} \\
\quad-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7}) \\
\cdot(-\sqrt{5}+\sqrt{6}+\sqrt{7}) .
\end{array}
$$ | $$
\begin{array}{l}
=x^{2}-y^{2} \text {, we have: } \\
(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6} \\
-\sqrt{7})=(\sqrt{5}+\sqrt{6})^{2}-(\sqrt{7})^{2} \\
=11+2 \sqrt{30}-7=4+2 \sqrt{30}, \\
(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6} \\
+\sqrt{7})=(\sqrt{7})^{2}-(\sqrt{5}-\sqrt{6})^{2} \\
=7-(5-2 \sqrt{30})=-4... | 104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,843 |
$\begin{array}{l}\text { 3. If } \operatorname{tg} x+\operatorname{tg} y=25, \text { and } \\ \quad \operatorname{ctg} x+\operatorname{ctg} y=30, \\ \text { find } \quad \operatorname{tg}(x+y) .\end{array}$ | $\begin{array}{l}\text { 3. From ctg } x+\operatorname{ctg} y=30 \text { we get } \frac{1}{\operatorname{tg} x}+\frac{1}{\operatorname{tg} y} \\ =30 \Rightarrow \operatorname{tg} x+\operatorname{tg} y=30 \operatorname{tg} x \cdot \operatorname{tg} y . \\ \text { That is } \operatorname{tg} x \cdot \operatorname{tg} y=\... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,844 |
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the following system of equations
$$
\cdot\left\{\begin{array}{c}
2 x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=6, \\
x_{1}+2 x_{2}+x_{3}+x_{4}+x_{5}=12, \\
x_{1}+x_{2}+2 x_{3}+x_{4}+x_{5}=24, \\
x_{1}+x_{2}+x_{3}+2 x_{4}+x_{5}=48, \\
x_{1}+x_{2}+x_{3}+x_{4}+2 x_{5}=96
\end{array}\ri... | 4. Adding up the 5 equations, and then dividing both sides by 6, we get:
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=31 .
$$
By subtracting (1) from the 4th and 5th equations respectively, we obtain
$$
x_{4}=17, x_{5}=65 \text {, }
$$
Therefore, $3 x_{4}+2 x_{5}=51+130=181$. | 181 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,845 |
5 . Find the largest positive integer $n$, such that $n^{3}+100$ can be divided by $n+10$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
5 . Find the largest positive integer $n$, such that $n^{3}+100$ can be divided by $n+1... | 5 . By the division algorithm we get $\left.n^{3}+10\right)=(n+10)$
$$
\text { - }\left(n^{2}-10 n+100\right)-900 .
$$
If $n+10$ divides $n^{3}+100$, it must also divide 900. By the maximality of $n$, we have $n+10=903$, so $n=890$. | 890 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,846 |
6. The page numbers of a book are from 1 to $n$. When these page numbers were added up, one page number was mistakenly added one extra time. As a result, the incorrect sum obtained was 1986. What is the page number that was added one extra time? | 6. Let $k$ be the page number that was added one extra time, then $0<k<n+1$, so $1+2+\cdots+n+k$ is between $1+2+\cdots+n$ and $1+2+\cdots+n+(n+1)$, thus
$$
\frac{n(n+1)}{2}<1988<\frac{(n+1)(n+2)}{2},
$$
which means $n(n+1)<3972<(n+1)(n+2)$. By trial (since $n$ is slightly greater than 60), we get
$$
\begin{array}{l}
... | 33 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,847 |
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