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
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ on the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by... | 5.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
\begin{aligned}
w+z & =2 \bar{z}+(1+\mathrm{i}) z, \text { i.e., } w=2 \bar{z}+\mathrm{i} z . \\
\therefore|w|^{2} & =(2 \bar{z}+\mathrm{i} z)(2 z-\mathrm{i} \bar{z}) \\
& =4+1+2 \mathrm{i}\left(z^{2}-\bar{z}^{2}\ri... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,028 |
Three. (Full marks 20 points) Through the height of a triangular pyramid with all lateral faces being equilateral triangles, a plane is drawn, intersecting the planes of the three lateral faces at three lines. These three lines form angles $\alpha, \beta, \gamma$ with the base. Prove:
$$
\operatorname{ctg}^{2} \alpha+\... | Three, as shown in Figure 2, from the conditions
we know that $D-ABC$ is a regular tetrahedron. Let $H$ be the projection of $D$ onto the base $\triangle ABC$, then $H$ is the center of $\triangle ABC$. Let $M, N, P$ be the points where the plane through the line $HD$ intersects the lines $BC, CA, AB$ respectively. Fro... | \operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta+\operatorname{ctg}^{2} \gamma \geqslant \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,030 |
Four, (Full marks 20 points) Given the quadratic function $f(x)=$ $a x^{2}+b x+c(a \neq 0)$. If the equation $f(x)=x$ has no real roots, prove: the equation $f(f(x))=x$ also has no real roots. | Given the function $f(x)=a x^{2}+b x+c(a \neq 0)$, substituting it into the equation $f(x)=x$, and rearranging, we get
$$
a x^{2}+(b-1) x+c=0(a \neq 0) .
$$
Since this equation has no real roots, its discriminant is negative, i.e.,
$$
\Delta_{1}=(b-1)^{2}-4 a c<0 \text {. }
$$
Furthermore, from $f(f(x))=a(f(x))^{2}+b... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,031 |
Five. (Full marks 20 points) There is a quantity $W$, after "modeling" the relationship is given by
$$
W=\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1-t^{2}}}\right),
$$
where $a, b, c, u, t$ are all positive, $u<1, t<1$, and satisfy $a t+b u=c, a^{2}+2 b c u=b^{2}+c^{2}$. Please design a method to find... | Given $a^{2}=b^{2}+c^{2}-2 b c u$. All $a, b, c, u$ are positive, and $ub^{2}+c^{2}-2 b c=(b-c)^{2}$.
From this, we get $|b-c|<a<b+c$.
Therefore, the positive numbers $a, b, c$ can be the lengths of the three sides of a triangle. Let the vertices opposite these sides be $A, B, C$, respectively. By transforming the con... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,032 |
One. (Full marks 50 points) $\odot O_{1}$ is externally tangent to $\odot O_{2}$ at point $P$, and $QR$ is a common tangent of the two circles, where $Q$ and $R$ are the points of tangency on $\odot O_{1}$ and $\odot O_{2}$, respectively. The line through $Q$ perpendicular to $QO_{2}$ intersects the line through $R$ pe... | $$
\begin{array}{l}
\angle I Q O_{2}=\angle I N O_{2} \\
=90^{\circ}. \text{ Thus, } I, Q, \\
\end{array}
$$
N, and $O_{2}$ are concyclic.
$$
\begin{array}{c}
\therefore \angle Q I N= \\
\angle Q O_{2} O_{1}. \\
\because \angle I Q M+ \\
\angle M Q O_{2}=\angle I Q O_{2} \\
=90^{\circ}=\angle R Q O_{1}= \\
\angle M Q O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,033 |
II. (Full marks 50 points) Let $AB$ and $A'B'$ be chords of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the circle $x^{2}+y^{2}=a^{2}$, respectively, with the endpoints $A$ and $A'$, $B$ and $B'$ having the same x-coordinates and the same sign for their y-coordinates. Prove that when $AB$ passes ... | Let the coordinates of $A$ and $B$ be $A\left(x_{A}, y_{A}\right)$ and $B\left(x_{B}, y_{B}\right)$, respectively, then
$$
\frac{x_{A}^{2}}{a^{2}}+\frac{y_{A}^{2}}{b^{2}}=1, \frac{x_{B}^{2}}{a^{2}}+\frac{y_{B}^{2}}{b^{2}}=1 .
$$
Transforming the equations, we get
$$
x_{A}^{2}+\left(\frac{a}{b} y_{A}\right)^{2}=a^{2} \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,034 |
Example 1: Prove that $\sum_{k=1}^{50} \mathrm{C}_{1948+k}^{k}=\mathrm{C}_{1999}^{1949}-1$.
Analysis: Observing the characteristics of the equation, it is easy to think of using the properties of binomial coefficients $\mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{n-k}$ and $\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k}=\mathrm{C}_{n+... | Prove: $\begin{aligned} & \sum_{k=1}^{50} \mathrm{C}_{1948+k}^{k}=\sum_{k=1}^{50} \mathrm{C}_{1948+k}^{1948} \\ & =\mathrm{C}_{1999}^{1949}+\sum_{k=1}^{50} \mathrm{C}_{1948+k}^{1948}-1 \\ & =\mathrm{C}_{1999}^{1949}-1 .\end{aligned}$ | \mathrm{C}_{1999}^{1949}-1 | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,035 |
Three, (Full marks 50 points) Divide a rectangle of length $m$ and width $n$ into $m \times n$ small squares. A particle passes through one diagonal of each small square continuously without repetition or omission. Is this possible? If not, please explain the reason; if it is possible, please provide a route:
| Three, It can be done.
In two cases:
(1) If at least one of $m$ and $n$ is odd, let's assume $n$ is odd. As shown in Figure 5, the particle starts from $A_{0}$ and travels to $A_{1,1}; A_{0,2}; A_{1,3}; \cdots; A_{0, n-1}; A_{1, n}$, which gives a route for the particle when $\bar{n}=1$. If $m>1$, the particle can cont... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,036 |
79. (1) Prove that 1998 cannot be expressed as the sum of any number of consecutive odd numbers;
(2) If the numbers from 1 to 1997 are connected using “+” and “-” signs, then no matter how many “-” signs are used, the result cannot be 1998;
(3) If the numbers from 1 to 1999 are connected using “+” and “-” signs, then w... | Solution: (1) If it can be expressed, let $n_{0}$ be the first odd number, then
$$
\begin{array}{l}
1998= n_{0}+\left(n_{0}+2\right)+\cdots+\left(n_{0}+2 k\right) \\
=\left(n_{0}+2 k\right)+\left(n_{0}+2 k-2\right)+\cdots \\
+\left(n_{0}+2\right)+n_{0} . \\
\therefore 2 \times 1998=(k+1)\left(2 n_{0}+2 k\right) .
\end... | 586 | Number Theory | proof | Yes | Yes | cn_contest | false | 711,037 |
Given $CD$ is the altitude on the hypotenuse $AB$ of the right triangle $\triangle ABC$, $O, O_{1}, O_{2}, O_{3}$ are the incenter of $\triangle ABC, \triangle ACD, \triangle CBD$, and $\triangle O_{1}O_{2}D$, respectively. Connect $O_{3}O_{1}$ and $O_{3}O_{2}$ and extend them to intersect $AC$ and $BC$ at points $E$ a... | Proof: As shown in the figure, connect \(EO\), \(FO\), \(CO_1\), \(OO_1\), and \(OB\) (where \(O_2\) is on \(OB\)), and connect \(O_2O_1\) intersecting \(AC\) at \(M\). According to the problem, it is easy to prove that
\[
\triangle CO_1D \sim \triangle BO_2D.
\]
\[
\therefore \frac{O_1D}{O_2D} = \frac{CD}{BD}.
\]
It i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,038 |
79. If four circles are internally tangent to a fifth circle, the length of the common external tangent between the first and second circles is denoted as $l_{12}$, and the lengths of the common external tangents between other pairs of the first four circles are denoted similarly, with the first four circles arranged i... | Proof: As shown in the figure, let the first four circles be $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}$, and the fifth circle be $\odot O$. The first four circles are internally tangent to $\odot O$ at points $A, B, C, D$, respectively. It is easy to see that points $A, O_{1}, O$ are collinear. Similarly, poi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,039 |
In $\triangle A B C$, prove that:
$$
\cos A(\sin B+\sin C) \geqslant-\frac{2 \sqrt{6}}{9} \text {, }
$$
and determine the necessary and sufficient conditions for equality. | Proof: Without loss of generality, let $\angle A$ be an obtuse angle (otherwise, the original inequality obviously holds), then $\cos A<0$.
The original inequality is equivalent to
$2 \cos A \sin \frac{B+C}{2} \cos \frac{B-C}{2} \geqslant-\frac{2 \sqrt{6}}{9}$
$\Leftrightarrow \cos A \cos \frac{A}{2} \cos \frac{B-C}{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,040 |
Example 1 As shown in Figure 1, $D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ respectively, $DE \parallel BC$, $BE$ intersects $CD$ at $O$, and $AO$ intersects $BC$ at $M$. Prove: $BM = MC$.
(1978, National Junior High School Mathematics Competition) | Prove: Draw a line through $B$ parallel to $DC$ intersecting the extension of $AM$ at $P$, and connect $PC$. Then,
$$
\frac{AO}{AP}=\frac{AD}{AB}=\frac{AE}{AC}.
$$
It follows that $PC \parallel BE$.
Thus, quadrilateral $BPCO$ is a parallelogram. Therefore, $BC$ and $OP$ bisect each other.
Hence, $BM = MC$.
This proble... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,041 |
Example 2 As shown in Figure $2, A D$ and $C E$ are altitudes of $\triangle A B C$. Take a point $F$ on $A B$ such that $A F=A D$. Draw a line through $F$ parallel to $B C$ intersecting $A C$ at $G$. Prove: $F G=C E$.
untranslated text remains the same as the original, only the content has been translated into Englis... | Prove: Draw a line through $F$ parallel to $AC$ intersecting $BC$ at $H$, connect $HE$, $ED$, and $DF$. It is known that quadrilateral $FHCG$ is a parallelogram, so $FG=HC$.
It is easy to see that points $A$, $E$, $D$, and $C$ are concyclic, thus $\angle BED = \angle BCA = \angle BHF$. Therefore, points $F$, $H$, $D$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,042 |
Example 4 The sequences $a_{0}, a_{1}, \cdots$ and $b_{0}, b_{1} \cdots$ are defined as follows:
$$
\begin{array}{c}
a_{0}=\frac{\sqrt{2}}{2}, a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}, n=0, \\
1,2, \cdots, b_{0}=1, b_{n+1}=\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}, n=0,1,
\end{array}
$$
$2, \cdots$. Prove:
For e... | Given: $a_{0}=\sin \frac{\pi}{2^{2}}$, let $a_{n}=\sin \frac{\pi}{2^{n+2}}$, then
$$
a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\cos \frac{\pi}{2^{n+2}}}=\sin \frac{\pi}{2^{n+3}} \text {. }
$$
Using mathematical induction, we can prove: $b_{n}=\tan \frac{\pi}{2^{n+2}}$.
Since for $x \in\left(0, \frac{\pi}{2}\right)$, $\sin x ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,043 |
List $5 x_{n+1}=\frac{2 x}{1-x_{n}^{2}}, x_{1}=\sqrt{3}$. Find $x_{n}$. | Let $y_{n}=\operatorname{arc} g i_{n}$. Then $x_{n}=\operatorname{tg} y_{n}$, substituting into the original equation we get $\operatorname{tg} y_{n+1}=\operatorname{tg} 2 y_{n}$. We can take $y_{n+1}=2 y_{n}$, then $x_{n}=\operatorname{tg}\left[\frac{2^{n-1} \pi}{3}\right]$. | x_{n}=\operatorname{tg}\left[\frac{2^{n-1} \pi}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,044 |
Example 6 Given $x_{n+1}=\frac{x_{n}}{\left(1+c x_{n}^{k}\right)^{\frac{1}{k}}}$, where $k \in$ $\mathbf{N}, c \in \mathbf{R}, c \neq 0$, and $x_{n}>0$. Find $x_{n}$. | Let $y_{n}=\frac{1}{x_{n}^{k}}$, then $x_{n}=\left(\frac{1}{y_{n}}\right)^{\frac{1}{k}}$. The original expression becomes $y_{n+1}=y_{n}+c$. Therefore, $y_{n}=y_{1}+(n-1) c$. Thus, we have
$$
x_{n}=\frac{x_{1}}{\sqrt[k]{1+(n-1) c x_{1}^{k}}} .
$$ | x_{n}=\frac{x_{1}}{\sqrt[k]{1+(n-1) c x_{1}^{k}}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,045 |
Example $7 a_{1}=1, a_{2}=2, a_{n+2}=\frac{2 a_{n} a_{n+1}}{a_{n}+a_{n+1}}$. Find $a_{n}$. | Let $b_{n}=\frac{1}{a_{n}}$, then we have $b_{n+2}=\frac{1}{2} b_{n+1}+\frac{1}{2} b_{n}$. This can be rewritten as $b_{n+2}-b_{n+1}=-\frac{1}{2}\left(b_{n+1}-b_{n}\right)=\cdots=\left(-\frac{1}{2}\right)^{n}\left(b_{2}-b_{1}\right)$.
Thus, we have $\quad b_{k}-b_{k-1}=\left(-\frac{1}{2}\right)^{k-1}$.
Let $k=1,2, \cdo... | a_{n}=\frac{3}{2+\left(-\frac{1}{2}\right)^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,046 |
Example 9 Given $a_{n}-7 a_{n-1}+10 a_{n-2}=3^{n}, a_{0}$ $=0, a_{1}=1$. Find $a_{n}$. | Solution: The characteristic equation is $x^{2}-7 x+10=0$, with roots $x_{1}=5, x_{2}=2$. Therefore, the homogeneous solution is
$$
a_{n}^{(h)}=A_{1} \cdot 2^{n}+A_{2} \cdot 5^{n} \text {. }
$$
Assume the particular solution is $a_{n}^{(p)}=p \cdot 3^{n}$. Substituting into the original recurrence relation gives $p=-\... | a_{n}=\frac{8}{3} \cdot 2^{n}+\frac{11}{6} \cdot 5^{n}-\frac{9}{2} \cdot 3^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,048 |
Example 11 Let $1<x_{1}<2$, for $n=1,2,3 \cdots$ define
$$
x_{n+1}=1+x_{n}-\frac{1}{2} x_{n}^{2} .
$$
Prove that for $n \geqslant 3$ we have $\left|x_{n}-\sqrt{2}\right|<2^{-n}$.
(17th Canadian Mathematical Olympiad) | $$
\begin{array}{l}
=\frac{3}{2}-\frac{1}{2}\left(x_{2}-1\right)^{2} \\
=\frac{3}{2}-\frac{1}{2}\left(x_{1}-\frac{1}{2} x_{1}^{2}\right)^{2} .
\end{array}
$$
Since on the interval $(1,2)$, the quadratic function $-\frac{1}{2} x^{2}+x$ decreases from $\frac{1}{2}$ to 0, we have $\frac{3}{2}-\frac{1}{8}<x_{3}<\frac{3}{2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,050 |
For example, let $12 a_{0}$ be a positive real number, consider the sequence defined by
$$
a_{n+1}=\frac{a_{n}^{2}-1}{n+1}, \quad n \geqslant 0
$$
Prove that there exists a real number $a>0$, such that
(i) for all real numbers $a_{0} \geqslant a$, the sequence $\left\{a_{n}\right\} \rightarrow +\infty$;
(ii) for all r... | Proof: (i) To find $a$, first take $n=0$. From $a_{1} = a_{0}^{2} - 1 > a_{0}$, we get $a_{0} > \frac{\sqrt{5} + 1}{2}$. Let's try $a=2$, then when $a_{0} \geqslant 2$, we have $a_{1} \geqslant 2^{2} - 1 = 3$, and $a_{2} \geqslant \frac{3^{2} - 1}{2} = 4$. Therefore, we can conjecture that $a_{n} \geqslant n + 2$, and ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,051 |
1. If $x_{n+1}=\frac{n+2}{n} x_{n}+\frac{1}{n}, x_{1}>0$, find $x_{n}$. | $$
\begin{array}{l}
\quad x_{n+1}+\frac{1}{2}=\frac{n+2}{n}\left(x_{n}+\frac{1}{2}\right) \Rightarrow x_{n}=\frac{(n+1)!}{(n-1)!} \times \\
\left.\frac{1}{4}-\frac{1}{2}(n \geqslant 2) .\right)
\end{array}
$$
(Tip: I derived the fixed point $\alpha=-\frac{1}{2}$. We have
$$
\begin{array}{l}
\quad x_{n+1}+\frac{1}{2}=\... | x_{n}=\frac{(n+1)!}{(n-1)!} \times \frac{1}{4}-\frac{1}{2} \text{ (for } n \geqslant 2\text{)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,052 |
Example 3 Let $P$ be a point inside $\square ABCD$, $\angle BAP = \angle BCP$. Prove: $\angle PBC = \angle PDC$.
(1979, Qingdao City Junior High School Mathematics Competition) | Prove: As shown in Figure 3, construct a square $APQD$ with $AD$ and $AP$ as sides, connect $QC$, and it is known that quadrilateral $BCQP$ is a parallelogram.
It is easy to see that $\angle CDQ = \angle BAP = \angle BCP = \angle CPQ$, i.e., $\angle CDQ = \angle CPQ$. Therefore, points $P, C, Q, D$ are concyclic, whic... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,053 |
2 . If $x_{n+1}=\frac{3 x_{n}+4}{2 x_{n}+1}, x_{1}=1$, find $x_{n}$.
| (Solution: Fixed points $\alpha=-1,2, \frac{x_{n-1}+1}{x_{n+1}-2}$ $=-2(-5)^{n}$. Solving for $x_{n+1}=\frac{4(-5)^{n}+1}{2(-5)^{n}+1}$, i.e., $\left.x_{n}=\frac{4(-5)^{n} \cdot 1-1}{2(-5)^{n} \cdot 1+1}.\right)$ | x_{n}=\frac{4(-5)^{n}-1}{2(-5)^{n}+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,054 |
6. $a_{n} \in \mathbf{R}^{+}$ and $a_{1}=1, a_{2}=10, a_{n}^{2} a_{n-2}=10 a_{n-1}^{3}$, $n=3,4, \cdots$. Find $a_{n}$. 1987, China Mathematical Olympiad Training Team Selection Test) | (Solution: Taking logarithms on both sides, we get $\lg a_{n}=\frac{3}{2} \lg a_{n-1}$ $-\frac{1}{2} \lg a_{n-2}+\frac{1}{2}$
(1). From the characteristic equation: $x^{2}=\frac{3}{2} x-$ $\frac{1}{2}$, we get $x_{1}=1, x_{2}=\frac{1}{2}$. The particular solution is $d_{n}=n d$. Substituting into (1), we get $(n+2) d=\... | a_{n}=10^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,058 |
Example 1 Let $x_{1}, x_{2}, \cdots, x_{9}$ all be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is $\qquad$
$(1992$, National Junior High School Mathematics League) | Solution: From $x_{1}+x_{2}+\cdots+x_{9}=220$,
we know $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>110$,
or $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
From (1), then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant$ $27, x_{8} \geqslant 28, x_{9} \geqslant 29$. We get
$$
\begin{aligned}
\left(x_{1}+x_{2}+x_{3}+x_{4}+... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,062 |
Example 2 Given $x^{2}+y^{2}=a^{2}, a^{2}<1$. Try to find: $د=\sqrt{1-x^{2}}+\sqrt{1-y^{2}}$ the maximum and minimum values. (1993, Shandong Province Junior High School Mathematics Competition) | $$
\begin{aligned}
\text { Solution: } s & =\sqrt{1-x^{2}}+\sqrt{1-y^{2}} \\
& =\sqrt{\left(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\right)^{2}} \\
& =\sqrt{2-\left(x^{2}+y^{2}\right)+2 \sqrt{1-\left(x^{2}+y^{2}\right)+x^{2} y^{2}}} . \\
\because x^{2} & +y^{2}=a^{2}, \\
\therefore s & =\sqrt{2-a^{2}+2 \sqrt{1-a^{2}+a^{2} x^{2}-x... | s_{\text {max }}=\sqrt{4-2 a^{2}}, \quad s_{\text {min }}=1+\sqrt{1-a^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,063 |
Example 4 As shown in Figure 4, $P$ is a point on the midline $DE$ of $\triangle ABC$, $BP$ intersects $AC$ at $N$, and $CP$ intersects $AB$ at $M$. Prove: $\frac{AN}{NC} + \frac{AM}{MB} = 1$.
(1985, Qiqihar and Daqing Junior High School Mathematics Competition) | Prove: Draw a line through $A$ parallel to $BC$ intersecting lines $BN$ and $CM$ at $G$ and $H$ respectively. Connect $GC$ and $HB$. It is easy to see that $HG \parallel DE \parallel BC$.
Since $D$ is the midpoint of $AB$, it is known that $P$ is the midpoint of $BG$ and $CH$. Therefore, quadrilateral $DUGH$ is a para... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,064 |
Example 3 Given $p^{3}+q^{3}=2$, where $p, q$ are real numbers. Then the maximum value of $p+q$ is $\qquad$
(1987, Jiangsu Province Junior High School Mathematics Competition) | Solution: Let $s=p+q$.
From $p^{3}+q^{3}=2$ we get
$(p+q)\left(p^{2}+q^{2}-p q\right)=2$,
$(p+q)^{2}-3 p q=\frac{2}{s}$,
Thus, $p q=\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)$.
From (1) and (2), $p$ and $q$ are the two real roots of the equation
$$
x^{2}-s x+\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)=0
$$
We know $\Delt... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,065 |
Example 4 If $x \neq 0$, find the maximum value of
$$
\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}
$$
(1992, National Junior High School Mathematics League) | Solution: Original expression
$$
\begin{array}{l}
= \pm\left(\frac{\sqrt{1+x^{2}+x^{4}}}{|x|}-\frac{\sqrt{1+x^{4}}}{|x|}\right) \\
= \pm\left(\sqrt{\frac{1}{x^{2}}+1+x^{2}}-\sqrt{\frac{1}{x^{2}}+x^{2}}\right) \\
= \pm \frac{1}{\sqrt{\left(x-\frac{1}{x}\right)^{2}+3}+\sqrt{\left(x-\frac{1}{x}\right)^{2}}+2} .
\end{array... | \sqrt{3}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,066 |
For example, $5 x, y, z$ are real numbers, and satisfy $x+y+z=0, xyz=2$. Find the minimum value of $|x|+|y|+|z|$.
(1990, Beijing Junior High School Mathematics Competition). | Solution: From the problem, we know that among $x, y, z$, there are 2 negative numbers and 1 positive number. Without loss of generality, let $x>0, y<0, z<0$. When $x>0, -y>0, -z>0$, we have
$(-y)(-z) \leqslant \left[\frac{(-y)+(-z)}{2}\right]^{2}=\frac{x^{2}}{4}$,
which means $2 x=\frac{4}{(-y)(-z)} \geqslant \frac{4}... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,067 |
Example 6 Given three non-negative real numbers $a, b, c$ satisfying the conditions: $3a+2b+c=5, 2a+b-3c=1$. Let $s=$ $3a+b-7c$ have a maximum value of $M$ and a minimum value of $m$. Find $\mathrm{Mm}$.
(1989, Shanghai Junior High School Mathematics Competition) | Solution: From the given conditions, we have $a=7c-3, b=7-$ $11c$, substituting into $s=3a+b-7c$, we get $s=3c-2$.
Since $a, b, c$ are non-negative real numbers,
$$
\therefore\left\{\begin{array}{l}
a = 7c - 3 \geqslant 0, \\
b = 7 - 11c \geqslant 0
\end{array} \Rightarrow \left\{\begin{array}{l}
c \geqslant \frac{3}{7... | \frac{5}{77} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,068 |
Example 7 Given that $a$ is a real number, and makes the quadratic equation in $x$, $x^{2}+a^{2} x+a=0$, have real roots. Find the maximum value that the root $x$ of the equation can take.
(1994, Beijing Junior High School Mathematics Competition). | Solution: Consider $a$ as the unknown. When $x \neq 0$, the condition for the equation $x a^{2} + a + x^{2} = 0$ to have real roots $a$ is
$$
\Delta = 1 - 4 \cdot x^{3} \geqslant 0 \Rightarrow x \leqslant \frac{3}{2} \text{.}
$$
Substituting $x = \frac{\sqrt[3]{2}}{2}$ into the original equation and completing the squ... | \frac{\sqrt[3]{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,069 |
Example 8 If $a<b<c$, find the minimum value of the function
$$
y=|x-a|+|x-b|+|x-c|
$$
(1985, Shanghai Mathematics Competition) | Solution: In the right figure, $a$, $b$, and $c$ are three numbers on the $x$-axis. The problem is reduced to finding a point on the number line such that the sum of its distances to points $a$, $b$, and $c$ is minimized. It is easy to see that when the point $x=b$, the sum of its distances to points $a$, $b$, and $c$ ... | c-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,070 |
4. $A B C D E$ is a convex pentagon, $A D$ is a diagonal. It is known that $\angle E A D>\angle A D C, \angle E D A>\angle D A B$. Prove: $A E +$ $E D > A B + B C + C D$. | (Let the intersection of $AB$ and $DC$ be $F$. Construct $\square AFDG$ with $AF$ and $FD$ as adjacent sides, and it is easy to see that $G$ is inside $\triangle ADE$. Let the extension of $AG$ intersect $FD$ at $M$, then we know
$$
\begin{array}{l}
AE + ED = AE + EM + MD > AM + MD \\
= AG + GM + MD > AG + GD = FD + AF... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,071 |
5. In quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. $M$ and $N$ are the midpoints of $AD$ and $BC$, respectively, and $MN$ intersects $AC$ at $E$ and $BD$ at $F$. Prove that $OE: AC = OF: BD$. | (Tip: Take a point $G$ on the extension of $D N$, connect $G A, G B, G C$. It is easy to know that quadrilateral $D B G C$ is a parallelogram, thus $\triangle C A G$ $\backsim \triangle O E F$.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,072 |
6. In quadrilateral $ABCD$, $O$ is the intersection of $AC$ and $BD$, $AO > CO$, $BO = DO$. Prove: $\angle BAD < \angle BCD$. | (Tip: Take a point $E$ on $AO$ such that $OE = OC$, then we know
$$
\begin{array}{l}
\angle B A D = \angle B A O + \angle O A D \\
= \angle B E O - \angle A B E + \angle O E D - \angle A D E \\
\angle \angle B E O + \angle O E D = \angle B E D = \angle B C D .)
\end{array}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,073 |
7. In $\triangle A B C$, the medians $A D$ and $B E$ intersect at $G$. A circle passing through $A$ is tangent to $B E$ at $G$, and intersects the extension of $C G$ at $F$. Prove that $A G^{2}=G F \cdot G C$.
untranslated text remains the same as requested. | (Tip: Take a point $H$ on the extension of $GD$ such that $DH = DG$, and connect $HC$, $HB$. It is easy to see that quadrilateral $BHCG$ is a parallelogram. Connect $FA$, then we have
$$
\angle AFG = \angle AGE = \angle BGH = \angle CHG.
$$
It follows that points $A$, $F$, $H$, and $C$ are concyclic, thus
$$
AG^2 = AG... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,074 |
Example 5 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Find $B C+D E$.
(1994, Beijing Junior High School Mathematics Competition) | Solution: As shown in Figure 5, let $F A$ and $C B$ intersect at $P$, and $F E$ and $C D$ intersect at $Q$. It is easy to see that quadrilateral $P C Q F$ is a parallelogram, and $\triangle A P B$ and $\triangle Q E D$ are both equilateral triangles. Therefore,
$$
\begin{aligned}
& F A + A B \\
= & F P = Q C \\
= & D E... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,075 |
8. In $\triangle A B C$, $\angle A=90^{\circ}$, $D$ is a point inside $\triangle A B C$, $B D=A B=A C$, $\angle A B D=30^{\circ}$. Prove: $A D=D C$. | ( Hint: Construct a regular $\triangle A C E$ outside $\triangle A B C$ with $A C$ as one side, and connect $D E$. It is known that quadrilateral $A B D E$ is a parallelogram, and $D E$ is the perpendicular bisector of $A C$. ) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,076 |
Example 1 Given that $\angle A, \angle B, \angle C$ are the interior angles of $\triangle ABC$. Find
$$
W=\operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+\operatorname{ctg} \frac{A}{2} \operatorname{ctg} \frac{B}{2} \operatorname{ctg} \frac{C}{2}
$$
the minimum value. | Solution: Given that $\angle A, \angle B, \angle C$ are the interior angles of a triangle, we know that $\operatorname{tg} \frac{A}{2}, \operatorname{tg} \frac{B}{2}, \operatorname{tg} \frac{C}{2}$ are all greater than 0. For simplicity, let $x, y, z$ represent $\operatorname{tg} \frac{A}{2}, \operatorname{tg} \frac{B}... | \frac{28}{9} \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,077 |
Example 2 Let $x, y, z, w$ be real numbers, not all zero. Find the maximum value of $P=\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}}$. | Solution: Introducing undetermined constants $\alpha, \beta, \gamma > 0$, we have
$$
\begin{array}{l}
\alpha^{2} x^{2}+y^{2} \geqslant 2 \alpha x y, \beta^{2} y^{2}+z^{2} \geqslant 2 \beta y z, \\
\gamma^{2} z^{2}+w^{2} \geqslant 2 \gamma z w .
\end{array}
$$
Thus, $\frac{\alpha}{2} x^{2}+\frac{y^{2}}{2 \alpha} \geqsl... | \frac{\sqrt{2}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,078 |
Example 3 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd Hope Cup for High School Grade 2) | Solution: Completing the square, we get
$$
u=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2 \text {. }
$$
Consider points $P_{1}\left(x, \frac{9}{x}\right)$, $P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,079 |
Example 4 If $3 x-y-1=0$, find
$$
z=\left|\sqrt{x^{2}+y^{2}-8 x-2 y+17}-\sqrt{x^{2}+y^{2}-8 y+16}\right|
$$
the maximum value. | Solution: The symmetric point of point $B$ with respect to the line $3 x-y-1=0$ is $B_{1}(3,3)$. Then, by the knowledge of plane geometry, we have
$$
z_{\text {max }}\left|A B_{1}\right|=\sqrt{5} .
$$ | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,080 |
Example 5 If $2 x^{2}+3 x y+2 y^{2}=1$, find the minimum value of $k=x+y+x y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Example 5 If $2 x^{2}+3 x y+2 y^{2}=1$, find the minimum value of $k=x+y+x y$. | Solution: From $2 x^{2}+3 x y+2 y^{2}=1$, we get $2(x+y)^{2}=1+x y$, which means $2(x+y)^{2}=k-(x+y)+1$. Rearranging terms, we have
$$
2(x+y)^{2}+(x+y)-(1+k)=0.
$$
Considering this equation as a quadratic equation in terms of $x+y$, and since $x+y$ is a real number, we have
$$
\Delta=1+8(1+k) \geqslant 0.
$$
Solving ... | -\frac{9}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,081 |
Example 6 If $x, y$ are real numbers, find the minimum value of $S=5 x^{2}-4 x y$ $+y^{2}-10 x+6 y+5$. | Solution: From the given, we have
$$
\begin{array}{l}
5 x^{2}-(10+4 y) x+y^{2}+6 y+5-S=0 . \\
\Delta_{x}=-4\left(y^{2}+10 y-5 S\right) \geqslant 0 . \\
\text { Hence, } 5 S \geqslant y^{2}+10 y=(y+5)^{2}-25 \\
\quad \geqslant-25 .
\end{array}
$$
Thus, $5 S \geqslant y^{2}+10 y=(y+5)^{2}-25$
$\therefore S \geqslant-5$,... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,082 |
Example 7 Given that $a, b, c, d$ are positive real numbers. Find
\[
\begin{aligned}
S= & \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b} \\
& +\frac{d}{a+b+c}
\end{aligned}
\]
the minimum value. | Solution: Let $b+c+d=A, c+d+a=B, d+$ $a+b=C, a+b+c=D$.
Adding the above four equations, we get
$$
a+b+c+d=\frac{1}{3}(A+B+C+D) .
$$
Subtracting the four equations from equation (1) respectively, we get
$$
\begin{array}{c}
a=\frac{1}{3}(B+C+D-2 A), \\
b=\frac{1}{3}(C+D+A-2 B), \\
c=\frac{1}{3}(D+A+B-2 C), \\
d=\frac{1}... | \frac{4}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,083 |
Example 9 Given $x^{2}-y^{2}=16$. Find the minimum and maximum values of the function
$$
f(x, y)=\frac{1}{x^{2}}+\frac{y}{8 x}+1
$$ | Solution: $x^{2}-y^{2}=16$ is a hyperbola with equal axes centered at the origin, and its parametric equations are
$$
x=4 \sec \theta, y=4 \tan \theta
$$
$(\theta$ is the parameter, and $\left.0 \neq \pm \frac{\pi}{2}\right)$.
Substituting them, we get
$$
\begin{array}{l}
f(x, y)=\frac{1}{x^{2}}+\frac{y}{8 x}+1 \\
=\fr... | \frac{9}{8} \text{ and } \frac{7}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,085 |
Example 6 The lengths of the two diagonals of a quadrilateral are $m$ and $n$, and the angle between them is $a$. Prove: the area of the quadrilateral is
$$
S=\frac{1}{2} m n \sin \alpha .
$$ | Proof: As shown in Figure 6, in quadrilateral $ABCD$, $AC = m$, $BD = n$, and the angle between $AC$ and $BD$ is $\alpha$. Draw lines through $B$ and $D$ parallel to $AC$, and draw lines through $A$ and $C$ parallel to $BD$ to form parallelogram $PQMN$. We have
$S_{\text{PQMN}} = mn \sin \alpha$.
It is easy to see that... | S_{ABCD} = \frac{1}{2} mn \sin \alpha | Geometry | proof | Yes | Yes | cn_contest | false | 711,086 |
Given $x^{2}-9 y^{2}-4 x+18 y-9=$
0. Find the extremum of the function
$$
f(x, y)=\frac{5 x^{2}+3 x y-23 x-6 y+22}{x^{2}-4 x+4}
$$ | Solving: $x^{2}-9 y^{2}-4 x+18 y-9=0$ can be transformed into $\frac{(x-2)^{2}}{4}-\frac{(y-1)^{2}}{\frac{4}{9}}=1$, its parametric equations are
$$
x=2+2 \sec \theta, y=1+\frac{2}{3} \tan \theta
$$
( $\theta$ is the parameter, $0<\theta<\frac{\pi}{2}$ ).
Substituting into $f(x, y)$ gives
$$
\begin{aligned}
f(x, y) & =... | \frac{15}{4} \text{ and } 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,087 |
Example 11 Given $-1 \leqslant 2 x+y-z \leqslant 8, 2 \leqslant x$ $-y+z \leqslant 9, -3 \leqslant x+2 y-z \leqslant 7$. Find the maximum and minimum values of the function $u=7 x+5 y-2 z$. | Solution: Let $a(2 x+y-z)+b(x-y+z)+$ $c(x+2 y-z)=7 x+5 y-2 z$. By comparing the coefficients on both sides, we get $a=1, b=2, c=3$. Multiplying the given three conditions by $1, 2, 3$ respectively, we obtain $-1 \leqslant 2 x+y-z \leqslant 8, 4 \leqslant$ $2(x-y+z) \leqslant 18, -9 \leqslant 3(x+2 y-z) \leqslant 21$. A... | -6 \leqslant u \leqslant 47 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,088 |
Example 12 Let $x, y$ be positive numbers, and $x+y=1$. Find the minimum value of the function $W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$. (Adapted from the 3rd Canadian Mathematical Competition) | Solution: Without loss of generality, let $x \geqslant y$, and set $x=\frac{1}{2}+t$, then $y=$
$$
\begin{array}{l}
\frac{1}{2}-t, 0 \leqslant t<\frac{1}{2} . \\
\therefore W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \\
=\frac{3+2 t}{1+2 t} \cdot \frac{3-2 t}{1-2 t}=\frac{9-4 t^{2}}{1-4 t^{2}} \\
\geqslant ... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,089 |
Example 13 Given that $x, y, z$ are positive numbers, and $4x + 5y + 8z = 30$. Find the minimum value of $W = 8x^2 + 15y^2 + 48z^2$. | Solution: Let $x=3+t_{1}, y=2+t_{2}, z=1+t_{3}$, substituting into the given equation yields $4 t_{1}+5 t_{2}+8 t_{3}=0$.
$$
\begin{aligned}
W= & 8 x^{2}+15 y^{2}+48 z^{2} \\
= & 180+12\left(4 t_{1}+5 t_{2}+8 t_{3}\right)+8 t_{1}^{2} \\
& +15 t_{2}^{2}+48 t_{3}^{2} \\
\geqslant & 180,
\end{aligned}
$$
(when and only wh... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,090 |
Promotion 1: Given $n$ points in a plane, where no three points are collinear, connect some points arbitrarily with line segments (these line segments are called edges). The condition to ensure that the graph contains a triangle with the given points as vertices is that the number of edges $x \geqslant \frac{2 \mathrm{... | Proof: Construct drawers: each drawer contains three distinct points, yielding a total of $\mathrm{C}_{n}^{3}$ drawers. Since the same edge will appear in $\mathrm{C}_{n-2}^{1}$ drawers, according to the Dirichlet drawer principle, when $x \cdot \mathrm{C}_{n-2}^{1} \geqslant 2 \mathrm{C}_{n}^{3}+1$, it ensures that th... | x \geqslant \frac{n(n-1)(n-2)+3}{3(n-2)} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,091 |
Given $n$ points, where no three points are collinear, and some points are arbitrarily connected by line segments (these line segments are called edges), the condition to ensure that the graph contains a complete graph of order $m(m<n)$ with the given points as vertices is that the number of edges in the graph $x \geqs... | Proof: Construction of drawers: Each drawer contains $m$ distinct points, resulting in $\mathrm{C}_{n}^{m}$ drawers. Since the same edge will appear in $\mathrm{C}_{n-2}^{m-2}$ drawers, according to the Dirichlet drawer principle, when $x \cdot \mathrm{C}_{n-2}^{m-2} \geqslant \mathrm{C}_{n}^{m}\left(\mathrm{C}_{m}^{2}... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,092 |
Proposition 1 For any given triangle, there exists a rectangle with the same perimeter and area. | Proof: Let the three sides of $\triangle ABC$ be $a, b, c$, the area be $S$, and the semi-perimeter be $p$. Then the equation
$$
t^{2}-p t+S=0
$$
if it has two positive roots, the rectangle with these two roots as length and width satisfies the proposition's requirements.
By Heron's formula, we have
$$
\begin{aligned}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,093 |
Proposition: Let the three sides of $\triangle A_{i} B_{i} C_{i}$ be $a_{i}$, $b_{i}$, and $c_{i}$, and the radii of the excircles be $r_{a_{i}}$, $r_{b_{i}}$, and $r_{c_{i}}$ ($i=1,2, \cdots, n$). Then
$$
\prod_{i=1}^{n} \frac{a_{i}}{r_{a_{i}}}+\prod_{i=1}^{n} \frac{b_{i}}{r_{b_{i}}}+\prod_{i=1}^{n} \frac{c_{i}}{r_{c_... | $$
\begin{array}{l}
\prod_{i=1}^{n} \frac{a_{i}}{r_{r_{i}}}+\prod_{i=1}^{n} \frac{b_{i}}{r_{b_{i}}}+\prod_{i=1}^{n} \frac{c_{i}}{r_{c_{i}}} \\
-\prod_{i=1}^{n} \frac{a_{i}\left(p_{i}-a_{i}\right)}{\Delta_{i}}+\prod_{i=1}^{n} \frac{b_{i}\left(\hat{p}_{i}-b_{i}\right)}{\Delta_{i}}+\prod_{i=1}^{n} \frac{c_{i}\left(p_{i}-c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,094 |
Lemma $\mathrm{C}_{m}^{n}+\mathrm{C}_{m}^{n-1}=\mathrm{C}_{m+1}^{n}(m>n, m, n$ $\left.\in \overline{\mathbf{Z}^{-}}\right)$ | ```
\begin{aligned}
\text { Left }= & \frac{m(m-1) \cdots(m-n+1)}{n!} \\
& +\frac{m(m-1) \cdots(m-n+2)}{(n-1)!} \\
= & \frac{m(m-1) \cdots(m-n+1)}{n!} \\
& +\frac{m(m-1)(m-2) \cdots(m-n+2) n}{n(n-1)!} \\
= & \frac{m(m-1) \cdots(m-n+2)[(m-n+1)+n]}{n!} \\
= & \frac{(m+1) \cdots(m-1) \cdots(m-n+2)}{n!} \\
= & \mathrm{C}_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,095 |
$\begin{array}{l} \text { Proposition } \quad 1 \times 2 \times 3 \times \cdots \times m+2 \times 3 \times 4 \times \cdots \\ \times(m+1)+\cdots+(n-m+1) \times \cdots \times n . \\ = m!\mathrm{C}_{n+1}^{n-m}=\frac{1}{m+1}(n-m+1) \\ \cdot(n-m+2) \cdots n(n+1) . \\ (n \geqslant m)\end{array}$ | $$
\begin{array}{l}
\text { Left side }=m!+\frac{(m+1)!}{1}+\frac{(m+2)!}{1 \times 2}+\cdots \\
+\frac{n!}{(n-m)!} \\
=m!\left[1+\frac{m+1}{1}+\frac{(m+1)(m+2)}{1 \times 2}\right. \\
+\frac{(m+1)(m+2)(m+3)}{1 \times 2 \times 3}+\cdots \\
\left.+\frac{n(n-1) \cdots(m+1)}{(n-m)!}\right] \\
=m!\left(1+\mathrm{C}_{m+1}^{1}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,096 |
16. Find the smallest integer \( n (n \geq 4) \), such that from any \( n \) integers, four different numbers \( a, b, c, d \) can be selected, making \( a + b - c - d \) divisible by 20. | Solution: First, we consider the different residue classes modulo 20. For a set with $k$ elements, there are $\frac{1}{2} k(k - 1)$ integer pairs. If $\frac{1}{2} k(k - 1) > 20$, i.e., $k \geq 7$, then there exist two pairs $(a, b)$ and $(c, d)$ such that $a + b \equiv c + d \pmod{20}$, and $a, b, c, d$ are all distinc... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,099 |
17. The integer sequence $a_{1}, a_{2}, a_{3}, \cdots$ is defined as follows: $a_{1}=1$, for $n \geqslant 1, a_{n+1}$ is the smallest integer greater than $a_{n}$, and for all $i, j, k \in\{1,2, \cdots, n+1\}$, it satisfies $a_{i}+a_{j}=3 a_{k}$. Find $a_{1998}$. | Solution: We first find the initial few $a_{n}$.
. $a_{3}=1$.
Since $1+2=3 \times 1$, then $a_{2} \neq 2 . a_{2}=3, a_{3}=4$.
Since $4+5=3+6=3 \times 3$, then $a_{4} \neq 5, a_{4} \neq 6 . a_{4}=$
7.
Since $1+8=3 \times 3,3+9=3 \times 4$, then $a_{5} \neq 8, a_{5} \neq$ 9. $u:=10$
Since $1+11=3 \times 4$, then $a_{6} ... | 4494 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,100 |
18. Find all positive integers $n$, such that for the positive integer $n$ sought, there exists an integer $m$, making $2^{n}-1$ a divisor of $m^{2}+9$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: If $n$ is not an integer power of 2, then it has an odd factor $l \geqslant 3$. Since $2^{l}-1$ divides $2^{n}-1$, then $2^{l}-1$ also divides $m^{2}+9$. For any $l \geqslant 3$, we have $2^{t}-1 \equiv -1 \pmod{4}$, so $2^{i}-1$ has a prime factor $p$ congruent to -1 modulo 4, i.e., $p = -1 \pmod{4}$. Assume... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,101 |
20. Prove that for each positive integer $n$, there exists a positive integer satisfying the following properties:
(1) It has $n$ digits;
(2) Each of its digits is non-zero;
(3) It is divisible by the sum of its digits. | Proof: If $n=3^{l}$, construct the simplest $n$-digit number as $11 \cdots 1$. Clearly, $l=0$ satisfies the condition. Assume that $11 \cdots 1$ is divisible by $3^{l}$. Since
$$
\begin{aligned}
\boxed{11 \cdots 1} & =\frac{1}{9}\left(10^{3^{l+1}}-1\right) \\
& =\frac{1}{9}\left(10^{3^{l}}-1\right)\left(10^{2 \times 3^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,102 |
The sum of numbers is an integer. Prove that the numbers in the columns of the matrix can be changed to $[x]$ or $\langle x\rangle$, making the sum of all numbers in each row and each column remain unchanged. Here, $[x]$ represents the greatest integer not exceeding $x$, and $\langle x \rangle$ represents the smallest ... | Proof: We first change all non-integer values $x$ to $[x]$, and denote this as “one”. Then, we restore the sum of all numbers in each column to their original values. This is achieved by changing some selected $[x]$ to $\langle x\rangle$, and denote this as “+”. Finally, we restore the sum of all numbers in each row to... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,104 |
23. Given that $n$ is an integer greater than 2. If a positive integer is 1 or can be obtained from 1 through a series of operations with the following properties:
(1) The operations are addition or multiplication;
(2) Addition and multiplication are used alternately;
(3) Each addition operation can freely choose to ad... | Proof: (1) If $n$ is even, then all odd numbers greater than $n+1$ are “unreachable”. If $n$ is odd, assume $a$ is obtained from $b$ using addition, and $b$ is obtained from $c$ using multiplication, then $b=2c$ or $nc, a=2c+2, 2c+n, nc+2$ or $nc+n$. Thus, $a \equiv 2c+2 \pmod{(n-2)}$, and adding $a^2 = -2 \pmod{(n-2)}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,105 |
24. Nine cards labeled 1 to 9 are arranged in a row at random. We can choose any adjacent cards that are in either ascending or descending order and reverse their order, which is called one operation. For example, in the sequence 916 532748, if we select 653 and reverse it to 356, the original sequence becomes 913562 7... | Proof: Let $g(\pi)$ be the minimum number of operations required to make a permutation $\pi$ of $n$ distinct numbers monotonic, and let $f(n)$ be the maximum value of $g(\pi)$ over all $n!$ permutations of $\pi$. We will prove that $f(n) \leqslant f(n-1)+2$.
For any permutation of $\{1,2, \cdots, n\}$, let the first e... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,106 |
25. Let $U=\{1,2, \cdots, n\}$, where $n \geqslant 3, S$ is a subset of $U$. If an element not in $S$ appears in a permutation of $U$ such that it is between two elements of $S$, we say $S$ is "split" by $U$. For example, 13542 can "split" $\{1,2,3\}$, but cannot "split" $\{3,4,5\}$. Prove: For any $n-2$ subsets of $U$... | Proof: We use mathematical induction on $n$.
When $n=3$, the subset family $\{i, j\}$ containing two elements can be "split" by the permutation $\langle i, k, j\rangle$, where $k$ is the third element of $U$.
Assume the conclusion holds for $n \geqslant 3$. Let $U=$ $\{1,2, \cdots, n+1\}$, and $\mathscr{F}$ be a subse... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,107 |
Example 1 Given $x_{n+1}=\frac{x_{n}\left(x_{n}^{2}+3 a^{2}\right)}{3 x_{n}^{2}+a^{2}}, x_{1}=$ $b$, where $a, b$ are real constants, $a \neq b$. Find $x_{n}$. | Solution: Let $\alpha$ replace $x_{n+1}$ and $x_{n}$, we get the fixed points $\alpha=0$, $\pm a$. Therefore,
$$
\begin{array}{l}
\frac{x_{n+1}-a}{x_{n+1}+a}=\frac{\frac{x_{n}\left(x_{n}^{2}+3 a^{2}\right)}{3 x_{n}^{2}+a^{2}}-a}{\frac{x_{n}\left(x_{n}^{2}+3 a^{2}\right)}{3 x_{n}^{2}+a}+a} \\
=\frac{\left(x_{n}-a\right)... | x_{n}=\frac{a\left[(b+a)^{3^{n-1}}+(b-a)^{3^{n-1}}\right]}{\left[(b+a)^{3^{n-1}}-(b-a)^{3^{n-1}}\right]} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,108 |
28. In a single-player card game, there are $m n$ cards that are white on one side and black on the other, played on an $m \times n$ rectangular board. Initially, the $m n-1$ small squares of the rectangular board are covered with cards white side up, and only one corner square is covered with a card black side up. In ... | Proof: Assume it is possible to remove all the cards. Each time an operation is performed, record the number of cards that have already been removed and are adjacent to the card being removed, and let the sum of these numbers be $\sigma$.
The number recorded when the first card is removed is 0. Then, for each card rem... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,110 |
Approximately 2 Giant Knowledge $a_{n+1}=\frac{q-a_{n} a_{n-1}}{2 p-a_{n}-a_{n-1}}(n=$ $1,2, \cdots)$, where $p, q \in \mathbf{R}, a_{0}=a_{1}=0$ and $p^{2} \neq$ $q$. Find $a_{n}$. | Given $\alpha=\frac{q-\alpha^{2}}{2 p-2 \alpha}$, we have
$$
\alpha^{2}-2 p \alpha+q=0 \text {. }
$$
When $p^{2} \neq q$, the roots are
$$
\begin{array}{l}
\alpha_{1}=p+\sqrt{p^{2}-q}, \alpha_{2}=p-\sqrt{p^{2}-q} . \\
\text { Also, } \quad a_{n+1}-\alpha_{1} \\
= \frac{q-a_{n} a_{n-1}-2 p a_{1}+\alpha_{1} a_{n}+\alph... | a_{n}=\frac{\alpha_{1} \cdot \alpha_{z^{n}}^{f_{n}}-\alpha_{2} \cdot \alpha_{1_{n}}^{f_{n}}}{\alpha_{z^{n}}^{f}-\alpha_{1_{n}^{n}}^{f_{n}}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,111 |
Example 3 In the positive term sequence $\left\{a_{n}\right\}$, $a_{1}=10, a_{k+1}$ $=10 \sqrt{a_{k}}$. Find the general term formula $a_{n}$.
(1991, Sichuan Province High School Mathematics Competition) | Solution: Let $x_{k}=\lg a_{k}$. Then
$$
x_{k+1}=1+\frac{1}{2} x_{k}, x_{k}=1+\frac{1}{2} x_{k-1} \text {. }
$$
From this, we get $x_{i+1}-x_{k}=\frac{1}{2}\left(x_{k}-x_{k-1}\right)$, and $x_{2}-x_{1}=\frac{1}{2}$.
Let $k$ take $1,2, \cdots, n$, the sum is
$$
\begin{array}{l}
x_{n+1}-x_{1}=1-\left(\frac{1}{2}\right)^... | a_{n}=10^{2-\left(\frac{1}{2}\right)^{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,112 |
Example: 120 farm workers cultivate 50 hectares of land, this
crop name workers required per hectare estimated output value per hectare
\begin{tabular}{lll}
vegetables & $\frac{1}{2}$ & 11000.00 yuan \\
cotton & $\frac{1}{3}$ & 7500.00 yuan \\
rice & $\frac{1}{4}$ & 6000.00 yuan
\end{tabular}
Question: How should it... | Let the land for growing vegetables, cotton, and rice be $x$ hectares, $y$ hectares, and $z$ hectares, respectively, and the expected total output value be $W$ yuan. According to the given conditions, we have
$$
\begin{array}{l}
x+y+z=50, \\
\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z=20 . \\
W=11000 x+7500 y+6000 z .
\e... | 450000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,113 |
刭 2 A real estate company owns a "defective rectangular" vacant land $A B C D E$, with side lengths and directions as shown in Figure 1. It is intended to build an apartment with a rectangular foundation running east-west on this land. Please draw the foundation of this land and calculate the maximum area of the founda... | Solution: Establish a rectangular coordinate system with line $BC$ and $AE$ as the $x$-axis and $y$-axis, respectively. The positive directions are along $BC$ and $AE$, with the unit of length being meters $(\mathrm{m})$.
The equation of line $AB$ is
$$
\frac{x}{30}+\frac{y}{20}=1 \text {. }
$$
Obviously, one vertex o... | 6017 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,114 |
5. In the border desert area, patrol vehicles travel 200 kilometers per day, and each vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their tasks, and then return along the same route to the base. To allow three of the vehicles to ... | 5. Suppose the patrol car travels to point $B$ in $x$ days, and three of the cars take $y$ days to travel from $B$ to the farthest point, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$
which simplifies to $5 x+3 y=35$.
From the problem, we know that $x>0, y>0$ and $14 \times 5-(5+2) x \leqslant 14 \times 3$, which mean... | 1800 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,115 |
6. If a merchant sells goods that cost 8 yuan each at 10 yuan each, he can sell 100 pieces per day. Now he is using the method of raising the price and reducing the purchase quantity to increase profits. Given that the price of each item is set to how much, can maximize the daily profit? And find the maximum profit. | 6. When the price is set at 14 yuan, the daily profit is maximized, with the maximum profit being 360 yuan. | 360 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,116 |
7. A company has 16 machines in inventory at locations $A$ and $B$. Now, these machines need to be transported to locations 甲 and 乙, with 15 machines required at 甲 and 13 machines required at 乙. It is known that the transportation cost from $A$ to 甲 is 500 yuan per machine, and to 乙 is 400 yuan per machine; the transpo... | 7. Send 3 units from $A$ to location 甲, 13 units to location 乙; send 12 units from $B$ to location 甲, 0 units to location 乙, to minimize the total transportation cost, the minimum cost is 10300 (yuan). | 10300 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,117 |
8. Based on market research analysis, a home appliance manufacturing company has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be at least... | 8. Every week, 30 air conditioners, 270 color TVs, and 60 refrigerators should be produced to maximize output value, with the maximum output value being 1050 thousand yuan. | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,118 |
10. Beside a circular road
There are five middle schools, No.1, No.2, No.3, No.4, and No.5, arranged in sequence, each having 15, 7, 11, 3, and 14 computers respectively. Now, to make the number of computers in each school equal, how many computers should each school transfer to its neighboring school to minimize the ... | 10. The optimal dispatch plan is: No.1 High School sends out 3 units to No.2 High School, No.2 High School does not send to No.3 High School, No.3 High School sends out 1 unit to No.4 High School; No.5 High School sends out 6 units to No.4 High School, No.1 High School sends out 2 units to No.5 High School | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,120 |
Example 2 A regular quadrilateral pyramid is inscribed in a sphere of radius $R$ and circumscribed about a sphere of radius $r$. Prove that: $\frac{R}{r} \geqslant \sqrt{2}+1$.
| Proof:
As shown in Figure 2, let
the regular quadrilateral pyramid
$S-ABCD$
have its base center as $O$, the side length of the base as
$a$, the center of the circumscribed sphere as
$\mathrm{O}_{2}$, and the center of the inscribed sphere as $O_{1}$. By symmetry, $O_{1}$ and $O_{2}$ must lie on $SO$, and $SO \perp$ th... | \sqrt{2}+1 | Geometry | proof | Yes | Yes | cn_contest | false | 711,122 |
Example 3 Prove: For any tetrahedron, the following inequality holds: $r<\frac{ab}{2(a+b)}$, where $a, b$ are the lengths of two skew lines, and $r$ is the radius of the inscribed sphere. | Prove: The radius $r$ of the inscribed sphere of a tetrahedron is $r=\frac{3 V}{S}$, where $V$ is the volume of the tetrahedron, and $S$ is its surface area. Combining Steiner's theorem $V=\frac{1}{6} a b d \sin \theta$, we know $r \leqslant \frac{1}{2} \frac{a b d}{S}$, where $d$ is the distance between the opposite e... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,123 |
Example 5: Prove that if the distances between the opposite edges of a tetrahedron are $d_{1}, d_{2}$, and $a_{3}$, then the volume $V$ of the tetrahedron is not less than $\frac{1}{3} d_{1} a_{2} d_{3}$. | Proof: As shown in Figure 3, for the tetrahedron $A-BCD$, the three pairs of opposite edges are $AB$ and $CD$, $AD$ and $BC$, $AC$ and $BD$. By drawing three pairs of parallel planes through the three pairs of opposite edges of the tetrahedron, we obtain the parallelepiped $AD_{1}BC_{1}-A^{\prime}DB^{\prime}C$. The dia... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,126 |
Example 6 Let the base of the pyramid $M-ABCD$ be a square, and $MA=MD, MA \perp AB$. If the area of $\triangle AMD$ is 1, prove that the radius $P$ of the sphere that can fit inside the pyramid is $P \leq \sqrt{2}-1$.
---
The base of the pyramid $M-ABCD$ is a square, and $MA=MD, MA \perp AB$. If the area of $\triang... | Proof: It is sufficient to prove that the maximum radius $r=\sqrt{2}-1$ for the sphere that can fit inside the pyramid.
As shown in Figure 4, since $A B \perp A D$ and $A B \perp A M$, it follows that $A B \perp$ plane $M A D$. Therefore, plane $A B C \perp$ plane $M A D$. Construct the section $M E F \perp A D$, then... | P \leq \sqrt{2}-1 | Geometry | proof | Yes | Yes | cn_contest | false | 711,127 |
Example 7 A tetrahedron has exactly one edge greater than 1. Try to prove: the volume of the tetrahedron $V \leqslant \frac{1}{8}$.
| Proof: In the tetrahedron $V-ABC$, $VA > 1$. As shown in Figure 5, draw the height $VH$, $HE \perp BC$ at $E$, $AF \perp BC$ at $F$, connect $AE$, and let $BC = x$. Then the volume of the tetrahedron is
$$
\begin{array}{l}
V = \frac{1}{3} \left( \frac{1}{2} BC \cdot AF \right) \cdot VH \\
= \frac{1}{6} x \cdot AF \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,128 |
Example 1 Let $x$ be a real number, and $f(x)=|x+1|+|x+2|+|x+3|+|x+4|+|x+5|$. Find the minimum value of $f(x)$. | Analysis: According to the geometric meaning of absolute value, draw the points $A, B, C, D, E$ corresponding to the real numbers $-1, -2, -3, -4, -5$ on the number line, as shown in Figure 1. Let $x$ correspond to the moving point $P$, then $f(x)=|PA|+|PB|+|PC|+|PD|+|PE| \geqslant |CB|+|CD|+|CA|+|CE|=2+4=6$, that is, ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,129 |
Example 2 Given $f(x)=|1-2 x|, x \in[0$, 1]. How many real solutions does the equation $f(f(f(x)))=\frac{x}{2}$ have? | Solution: To find the number of real solutions using the graphical method. First, draw the graph of $f(x) = |2x - 1|$, and then double the y-coordinates of all points while keeping the x-coordinates unchanged. Next, shift the obtained graph down by 1 unit (as shown in Figure 2), and then reflect the part of the graph b... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,130 |
Example 4 For all real numbers $x$, such that $|x-1|+|x-2|+|x-10|+|x-11| \geqslant m$ always holds. Then the maximum value of $m$ is $\qquad$ | Analysis: This type of problem is actually an extension and generalization of the function extremum problem in Example 1. It is easy to know that the function $f(x) = |x-1| + |x-2| + |x-10| + |x-11|$ has a minimum value of 18 when $x \in [2,10]$. Therefore, when $m \leqslant 18$, the inequality $f(x) \geqslant m$ alway... | 18 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,131 |
Example 6 Let the function $f_{3}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$ | Analysis: First, draw the graph of $f_{0}(x)=|x|$ and translate it downward by 1 unit to get the graph of $y=f_{0}(x)-1$, from which we obtain the graph of $f_{1}(x)=\left|f_{0}(x)-1\right|$ (as shown in Figure 7).
Next, translate the graph of $f_{1}(x)$ downward by 2 units to get the graph of $y=f_{1}(x)-2$, and refl... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,134 |
Example 7: A town has five primary schools along a circular road, sequentially named First, Second, Third, Fourth, and Fifth Primary Schools. They have 15, 7, 11, 3, and 14 computers, respectively. To make the number of computers equal in each school, some computers need to be transferred to neighboring schools: First ... | Analysis: Let $A, B, C,$
$D, E$ represent the five primary schools in a clockwise order, and let them sequentially transfer $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ computers to their neighboring schools, as shown in Figure 9. Then,
\[
\begin{array}{c}
7 + x_{1} - x_{2} = 11 + x_{2} \\
- x_{3} = 3 + x_{3} - x_{4} = 14 + x_{... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,135 |
Example 4 On a circular road, there are four middle schools arranged in sequence: $A_{1}, A_{2}, A_{3}, A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferr... | Solution: Let $A_{1}$ high school transfer $x_{1}$ color TVs to $A_{2}$ high school (if $x_{1}$ is negative, it means $A_{2}$ high school transfers $r_{1}$ color TVs to $A_{1}$ high school. The same applies below), $A_{2}$ high school transfers $x_{2}$ color TVs to $A_{3}$ high school, $A_{3}$ high school transfers $x_... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,136 |
Example 1 (Fill in the blank Question 1) Given $\frac{1}{4}(b-c)^{2}=$ $(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ . | Solution 1: From the given, we have
$$
\begin{aligned}
0 & =\frac{1}{4}[-(a-b)-(c-a)]^{2}-(a-b)(c-a) \\
& =\frac{1}{4}\left[(a-b)^{2}+(c-a)^{2}-2(a-b)(c-a)\right] \\
& =\frac{1}{4}[(a-b)-(c-a)]^{2} \\
& =\frac{1}{4}(2 a-b-c)^{2} .
\end{aligned}
$$
Thus, $\frac{b+c}{a}=2$.
Solution 2: The given condition indicates that... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,137 |
Example 2 (Fill-in-the-blank Question 5) In square $A B C D$, $N$ is the midpoint of $D C$, $M$ is a point on $A D$ different from $D$, and $\angle N M B=$ $\angle M B C$. Then $\operatorname{tg} \angle A B M$ $=$ . $\qquad$ | Solution 1: As shown in Figure 1, let $AB=1, AM=x$, then $\operatorname{tg} \angle ABM=x$. The problem is converted into finding an equation about $x$. From $MD=1-x$, $DN=\frac{1}{2}$, we get
$$
MN=\sqrt{(1-x)^{2}+\frac{1}{4}}.
$$
Draw $NE \parallel BC$, intersecting $MB$ at $F$ and $AB$ at $E$. Since $N$ is the midpo... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,138 |
1. Simplify $\frac{\sqrt{1-2 \sin 20^{\circ} \cos 20^{\circ}}}{\cos 20^{\circ}-\sqrt{1-\cos ^{2} 160^{\circ}}}$ to get ( ).
(A) $\sqrt{1-\sin 40^{\circ}}$
(B) $\frac{1}{\cos 20^{\circ}-\sin 20^{\circ}}$
(C) 1
(D) -1 | \begin{array}{l}\text {-.1.C. } \\ \begin{aligned} \text { Original expression } & =\frac{\sqrt{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}-2 \sin 20^{\circ} \cos 20^{\circ}}}{\cos 20^{\circ}-\sqrt{1-\cos ^{2} 20^{\circ}}} \\ & =\frac{\cos 20^{\circ}-\sin 20^{\circ}}{\cos 20^{\circ}-\sin 20^{\circ}}=1 .\end{aligned}\end{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,141 |
2. Let $P_{1} P_{2}$ be a chord of the parabola $x^{2}=y$. If the equation of the perpendicular bisector of $P_{1} P_{2}$ is $y=-x+3$, then the equation of the line containing the chord $P_{1} P_{2}$ is ( ).
(A) $y=x+3$
(B) $y=x-3$
(C) $y=x+2$
(D) Cannot be determined
Translate the above text into English, please reta... | 2.C.
Let $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$, obviously $k_{P_{1} P_{2}}=1$. Then the equation of the line $P_{1} P_{2}$ is $y=x+b$.
From $\left\{\begin{array}{l}x^{2}=y, \\ y=x+b\end{array}\right.$ we have $x^{2}-x-b=0$.
Thus, $x_{1}+x_{2}=1$.
Then the midpoint of $P_{1} P_{2}$ is $P\left... | C | Number Theory | proof | Yes | Yes | cn_contest | false | 711,142 |
3. The tetrahedron S-ABC is empty, with three pairs of edges being equal, sequentially $\sqrt{34}, \sqrt{41}, 5$. Then the volume of the tetrahedron is ( ).
(A) 20
(B) $10 \sqrt{7}$
(C) $20 \sqrt{3}$
(D) 30 | 3. A.
As shown in Figure 1, complete the rectangular prism, and let its length, width, and height be $x$, $y$, and $z$ respectively. Then we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=5^{2}, \quad A \\
y^{2}+z^{2}=(\sqrt{41})^{2}, \\
z^{2}+x^{2}=(\sqrt{34})^{2} .
\end{array}\right.
$$
Solving, we get $x=3, y=4, z=5$... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,143 |
4. Given the complex number $z$ satisfies $z^{3}=27$. Then the value of $z^{5}+3 z^{4}+2242$ is ( ).
(A) 2728
(B) 1999
(C) 2247
(D) 2728 or 1999 | 4. D.
From $z^{3}=27$ we get $(z-3)\left(z^{2}+3 z+9\right)=0$.
When $z=3$, the original expression $=3^{5}+3 \times 3^{4}+2242=2728$.
When $z \neq 3$, i.e., $z$ is a complex number, then $z^{2}+3 z+9=0$.
$$
\begin{aligned}
\text { The original expression } & =z^{3}\left(z^{2}+3 z+9\right)+2242-9 z^{3} \\
& =2242-9 \t... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,144 |
5. Let $S_{n}$ be the sum of the first $n$ terms of the arithmetic sequence $\left\{a_{n}\right\}$, $S_{9}=18$, $a_{n-4}=30(n>9)$, $S_{n}=336$. Then the value of $n$ is ( ).
(A) 16
(B) 21
(C) 9
(D) 8 | 5.B.
From $S_{9}=18$, we have $\frac{9\left(a_{1}+a_{9}\right)}{2}=18$, which means $a_{1}+a_{9}=4$, hence $2 a_{5}=4, a_{5}=2$.
$$
\text { Also } \begin{aligned}
S_{n} & =\frac{n\left(a_{1}+a_{n}\right)}{2}=\frac{n}{2}\left(a_{5}+a_{n-4}\right) \\
& =\frac{n}{2}(2+30)=336,
\end{aligned}
$$
thus $n=21$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,145 |
6. Let $0<x<\pi$, then the minimum value of the function $y=\frac{2-\cos x}{\sin x}$ is ( ).
(A) 3
(B) $\sqrt{3}$
(C) 2
(D) $2-\sqrt{3}$ | 6. B.
$$
\begin{aligned}
y & =\frac{1}{\sin x}+\frac{1-\cos x}{\sin x}=\frac{1-\tan^2 \frac{x}{2}}{2 \tan \frac{x}{2}}+\tan \frac{x}{2} \\
& =\frac{1}{2 \tan \frac{x}{2}}+\frac{3}{2} \tan \frac{x}{2} . \\
\because 0 & < x < \pi, \quad \tan \frac{x}{2} > 0,
\end{aligned}
$$
i.e., $y \geqslant 2 \sqrt{\frac{1}{2 \tan \f... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,146 |
For example, ship $B$ is located $45^{\circ}$ north of ship $A$, and the distance between the two ships is $10 \sqrt{2} \mathrm{~km}$. If ship $A$ sails west and ship $B$ sails south at the same time, with ship $B$'s speed being twice that of ship $A$, what is the closest distance between ships $A$ and $B$ in kilometer... | Solution: As shown in Figure 3, let after $t$ hours, ships $A$ and $B$ have sailed to positions $A_{1}$ and $B_{1}$, respectively, and let $A A_{1}=x$. Thus, $B B_{1}=2 x$.
Given $A B=10 \sqrt{2}$, we have $A C=B C=10$. $\therefore A_{1} C=|10-x|, B_{1} C=|10-2 x|$.
Therefore, $A_{1} B_{1}=\sqrt{|10-x|^{2}+|10-2 x|^{2... | 2 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,147 |
1. Given $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$. Then
$$
\frac{1+\sin \alpha+\cos \alpha}{1+\sin \alpha-\cos \alpha}=
$$
$\qquad$ | II. $1 . \frac{3}{4}$ or 0.
From $\sin \frac{\alpha}{2}-2 \cos \frac{\alpha}{2}=1$, we have
$\operatorname{tg} \frac{\alpha}{2}-2=\sqrt{1+\operatorname{tg}^{2} \frac{\alpha}{2}}$.
Thus, $\operatorname{tg} \frac{a}{2}=\frac{3}{4}$.
Also, $\operatorname{tg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=\frac{\sin \a... | \frac{3}{4} \text{ or } 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,148 |
2. The range of the function $y=\frac{1}{(x-1)(2 x-1)}$ is
保留了源文本的换行和格式，这是翻译结果。 | 2. $\{y \mid y \leqslant-8$ or $y>0\}$.
$$
y=\frac{1}{2 x^{2}-3 x+1}=\frac{1}{2\left(x-\frac{3}{4}\right)^{2}-\frac{1}{8}} \text {, }
$$
i.e., $16\left(x-\frac{3}{4}\right)^{2} y-y=8$.
$$
\therefore\left(x-\frac{3}{4}\right)^{2}=\frac{8+y}{16 y} \geqslant 0 \text {, }
$$
i.e., $y(y+8) \geqslant 0$, and $y \neq 0$.
Th... | \{y \mid y \leqslant-8 \text{ or } y>0\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,149 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.