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 |
|---|---|---|---|---|---|---|---|---|---|
6. $\frac{a^{\frac{1}{6}}}{a^{\frac{1}{4}}+a^{\frac{1}{2}}}+\frac{a^{\frac{2}{7}}}{a^{\frac{7}{7}}+a^{\frac{1}{2}}}+\frac{a^{\frac{3}{7}}}{a^{\frac{1}{8}}+a^{\frac{1}{2}}}+\cdots+\frac{a^{\frac{8}{4}}}{a^{\frac{1}{8}}+a^{\frac{1}{2}}}$ (where $a>0, a \neq 1$ ) is equal to ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 6. D
$$
\frac{a^{\frac{1}{8}}}{a^{\frac{1}{8}}+a^{\frac{1}{2}}}+\frac{a^{\frac{8}{8}}}{a^{\frac{8}{8}}+a^{\frac{1}{2}}}=\frac{2 a+a^{\frac{11}{16}}+a^{\frac{25}{18}}}{2 a+a^{\frac{1}{18}}+a^{\frac{25}{18}}}=1 .
$$
Similarly, the sum of the 2nd and 7th terms, the sum of the 3rd and 6th terms, and the sum of the 4th and... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,706 |
7. Given that $x$ is an acute angle, and $2 \sin ^{2} x-7|\sin x|+3<0$. Then the range of values for $x$ is ( ).
(A) $0^{\circ}<x<30^{\circ}$
(B) $30^{\circ}<x<45^{\circ}$
(C) $30^{\circ}<x<90^{\circ}$
(D) $60^{\circ}<x<90^{\circ}$ | 7. C
Let $\sin x=t$, then $2|t|^{2}-7|t|+3<0$. Solving this, we get $\frac{1}{2}<|t|<3$. Therefore, $-3<t<-\frac{1}{2}$ or $\frac{1}{2}<t<3$. Since 0 $<\sin x<1$, it follows that $\frac{1}{2}<\sin x<1$. Hence, $30^{\circ}<x<90^{\circ}$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 706,707 |
8. In $\triangle A B C$, it is known that $A B=17, A C=10, D$ is a point on $B C$, $A D=12$, and both $B D$ and $D C$ are integers. Then the number of all possible values of $B C$ is ( ).
(A) 21
(B) 4
(C) 2
(D) 0 | 8. D
Let $B D=x, D C=y$. In $\triangle A B D$ and $\triangle A D C$, by the cosine rule we get $\frac{144+x^{2}-289}{x}=-\frac{144+y^{2}-100}{y}$. Simplifying, we have $(x y+44)(x+y)=7 \cdot 3^{3} \cdot y$.
From $A B+B C>B C$ and $A B-B C<B C$, we get $7<B C<27$, i.e., $7<x+y<27$. Thus, there are only four possibilit... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,708 |
1. Given the function $y=y_{1}+y_{2}$, where $y_{2}$ is a direct proportion function of $x$, and $y_{2}$ is an inverse proportion function of $x$, and when $x=2$, $y=8$, and when $x=4$, $y=13$. Then the function expression of $y$ with respect to $x$ is $\qquad$ . | =. 1. $y=3 x+\frac{4}{x}$.
Let $y_{1}=k_{1} x, y_{2}=\frac{k_{2}}{x}$. From the given $8=2 k_{1}+\frac{k_{2}}{2}$ and $13=4 k_{1}+\frac{k_{2}}{4}$, we solve to get $k_{1}=3, k_{2}=4$. | y=3 x+\frac{4}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,709 |
3. If the inequality $|x-2|+|x-1| \geqslant a$ holds for all real numbers $x$, then the maximum value of $\boldsymbol{a}$ is $\qquad$ . | 3. 1.
Make $y=|x-2|+|x-1|$ $=\left\{\begin{array}{l}3-2 x, \quad x<1 \\ 1, \quad 1 \leqslant x \leqslant 2 \\ 2 x-3, \quad x>2\end{array}\right.$ Obviously, $|x-2|+|x-1| \geqslant 1$. Therefore, the maximum value of $a$ is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,711 |
Let $A, B$ be any two points on plane $\alpha$, and their images under $H\left(O_{1}, k_{1}\right)$ are $A^{\prime}$, $B^{\prime}$; the images of $A^{\prime}$, $B^{\prime}$ under $H\left(O_{2}, k_{2}\right)$ are $A^{\prime \prime}$, $B^{\prime \prime}$. Since $\overrightarrow{A^{\prime} B^{\prime}}=k_{1} \cdot \overrig... | Below, we prove that $O$ lies on the line $\mathrm{O}_{1} \mathrm{O}_{2}$.
\[
\overrightarrow{\sigma_{1}} \overrightarrow{O_{1} A^{\prime}}=k_{1} \overrightarrow{O_{1} A} \cdot O_{2} A^{n}=k_{4} \cdot U_{2} A^{\prime},
\]
so
\[
\overrightarrow{C_{2} A^{\prime \prime}}-k_{2}\left(\overrightarrow{O_{2} O_{1}}+\overrighta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,712 |
Example 1. Three congruent
circles have a common point $Q$, and
are all inside a given triangle,
each circle is tangent to two sides of the triangle. Prove: The incenter $I$, circumcenter $O$ and point $Q$
are collinear. (IMO22-5) | To prove that three points are collinear, if we can prove that two of them are a pair of corresponding points with the third point as the homothetic center, it is sufficient. As shown in the figure, it is easy to know that $\triangle O_{1} O_{2} O_{3} \sim \triangle A B C$, and $A O_{1}, B O_{2}, C O_{3}$ are the angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,713 |
Example 2. On the sides $AB, BC, CA$ of $\triangle ABC$, take points $P, Q, S$ respectively. Prove that the triangle formed by the circumcenters of $\triangle APS, \triangle BQP, \triangle CSQ$ is similar to $\triangle ABC$. (B - Polaslov's High School | Let $O_{1}$, $O_{2}, O_{3}$ be the circumcenters of $\triangle A P S$, $\triangle B Q P$, $\triangle C S Q$. After constructing the hexagon $\mathrm{O}_{1} \mathrm{PO}_{2} \mathrm{QO}_{3} \mathrm{~S}$, by the properties of circumcenters, we have
$$
\begin{array}{l}
\angle P O_{1} S=2 \angle A, \\
\angle Q O_{2} P=2 \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,714 |
7. Let the height of the cone be 1, and the radius of the base be $\sqrt{3}$, then the maximum value of the area of the section passing through the vertex of the cone is ( ).
(A) $\sqrt{3}$.
(B) 2
(C) $2 \sqrt{3}$
(D) 3 | 7. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,716 |
8. Skew lines $a, b$ form an $80^{\circ}$ angle, $P$ is any point, then the number of lines passing through $P$ and forming $50^{\circ}$ angles with both $a$ and $b$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 8. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,717 |
9. Draw a line through point $P(1,2)$, such that it is equidistant from $A(2,3)$ and $B(4, -5)$. Then, the equation of the line is ().
(A) $4 x+y-6=0$
(B) $x+4 y-6=0$
(C) $3 x+2 y-7=0$ or $4 x+y-6=0$
(D) $2 x+3 y-7=0$ or $x+4 y-6=0$ | 9. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,718 |
14. The domain of the function $y=\sqrt{9-x^{2}} \cdot \lg (1-\sqrt{2} \cos x)$ is $\qquad$ . | $\begin{array}{l}\text { 14. }[-3, \\ \left.\left.-\frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, 3\right)\right]\end{array}$
The translation is:
$\begin{array}{l}\text { 14. }[-3, \\ \left.\left.-\frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, 3\right)\right]\end{array}$
Note: The text is already in a mathematical not... | [-3, -\frac{\pi}{4}) \cup (\frac{\pi}{4}, 3] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,723 |
15. In a dihedral angle $\alpha-l-\beta$ of $60^{\circ}$, a moving point $A \in \alpha$, a moving point $B \in \beta, A A_{1} \perp \beta$ at point $A_{1}$, and $\left|A A_{1}\right|=a,|A B|=\sqrt{2} a$. Then, the maximum distance from point $B$ to the plane $\alpha$ is $\qquad$. | 15. $\frac{(\sqrt{3}+1) a}{2}$ | \frac{(\sqrt{3}+1) a}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,724 |
Example 3. $A D, \ddot{B} \dot{E}, C F$ are the three medians of $\triangle A B C$, and $P$ is an arbitrary point. Prove that among $\triangle P A D, \triangle P B E, \triangle P C F$, one of the areas is equal to the sum of the other two areas. (26th Moscow Mathematical Olympiad) | Let $G$ be the centroid of $\triangle ABC$, and the line $PG$ intersects $AB$ and $BC$. Perpendiculars are drawn from $A, C, D, E, F$ to this line, with the feet of the perpendiculars being $A^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}, F^{\prime}$, respectively.
It is easy to prove that $A A^{\prime}=2 D D^{\prime}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,725 |
17. If the line $a x+2 y+m=0$ is perpendicular to the line $2 x-y$ $+2=0$, and the chord length intercepted by the circle $x^{2}+y^{2}=1$ is 1, then $m=$ $\qquad$ . | 17. $\pm \frac{\sqrt{15}}{2}$ | \pm \frac{\sqrt{15}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,727 |
19. Symmetric about the $y$-axis, passing through the focus of the parabola $y^{2}=4 x$, and divided into two arcs with a length ratio of $1: 2$ by the line $y=x$, the equation of the circle is $\qquad$ | $\begin{array}{l}\text { 19. } x^{2}+(y-1)^{2}=2 \text { or } x^{2} \\ +(y+1)^{2}=2\end{array}$ | x^{2}+(y-1)^{2}=2 \text{ or } x^{2}+(y+1)^{2}=2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,729 |
20. Point $P$ moves on the parabola $y^{2}=a x$, point $Q$ is symmetric to $P$ about $(1,1)$. Then the equation of the trajectory of point $Q$ is $\qquad$. | 20. $(y-2)^{2}=a(2-x)$ | (y-2)^{2}=a(2-x) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,730 |
23. Given a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with each edge length being $a$, a plane $M$ is constructed through $A B_{1}$ and is parallel to $B C_{1}$.
(1) Find the dihedral angle $\alpha$ ($\alpha < 90^{\circ}$) formed between plane $M$ and plane $A B C$,
(2) Find the distance from point $A_{1}$ to ... | 23. $\alpha=\operatorname{arctg} 2$.
The distance from $A_{1}$ to the plane $M$ is $\frac{\sqrt{5}}{5} a$. | \alpha=\operatorname{arctg} 2, \frac{\sqrt{5}}{5} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,733 |
24. If the inequality $\cos ^{2} x+2 p \sin x-2 p-2<0$ holds for any real number $x$, find the range of $p$.
| 24. The range of values for $p$ is $(1-\sqrt{2},+\infty)$. | (1-\sqrt{2},+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,734 |
25. In the ellipse $C: \frac{\left(x-x_{0}\right)^{2}}{a^{2}}+\frac{\left(y-y_{0}\right)^{2}}{b^{2}}=1(a>b$ $>0)$, $F_{1}$ is the left focus, $O^{\prime}$ is the center, $A, \mathscr{B}$ are its right vertex and upper vertex, $P$ is a point on $\bigcirc$, $P F_{\text {、 }}$ is perpendicular to the x-axis, and $O^{\prim... | 25. 11) First, flatten $C$, making $\bigcirc^{\prime}$ coincide with the center. The equation of $C$ is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text {. }
$$
Let $P\left(x^{\prime}, y^{\prime}\right)$,
where $x^{\prime}=-c$.
$$
\begin{array}{l}
\because P O / / B . A, \\
\therefore \frac{y^{\prime}}{x^{\prime}}=-... | 3 \sqrt{2}-3 \leqslant a \leqslant 3 \sqrt{2}+3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,735 |
Example 4. If the squares of the three sides of a triangle form an arithmetic sequence, then the triangle is similar to the new triangle formed by its three medians. The converse is also true. (self-compiled) | Let $\triangle ABC$ be denoted as $\triangle$, and the new triangle formed by the three medians $AD$, $BE$, $CF$ be denoted as $\triangle'$. $G$ is the centroid. Connect $DE$ to $H$ such that $EH = DE$, then connect $HC$, $HF$. Thus, $\triangle'$ is $\triangle HCF$.
(1) $a^{2}, b^{2}, c^{2}$ form an arithmetic sequenc... | a^{2}+c^{2}=2 b^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 706,736 |
One. (35 points) The lines $l_{1}$ and $l_{2}$ are tangent to a circle at points $A$ and $B$, respectively. On lines $l_{1}$ and $l_{2}$, take 1993 points $A_{1}, A_{2}, \cdots, A_{1993}$ and $B_{1}, B_{2}, \cdots, B_{1993}$, such that $A A_{i}=(i+1) B B_{i} (i=1, 2, \cdots, 1993)$. If the extension of $A_{i} B_{i}$ in... | For a fixed $i$,
consider the point $A_{i}$ on $l_{1}$ and the point $B_{i}$ on $l_{2}$. At this time, $A A_{i} = (i+1) B B_{i}$ and the extension of $A_{i} B_{i}$ intersects the extension of $A B$ at point $M_{i}$ (as shown in the figure). Draw a line through point $B_{i}$ parallel to $A A_{i}$, intersecting $A B$ at ... | \frac{1}{1994} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,737 |
II. (35 points) Let $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$ be real numbers, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1, y_{1}^{2}+y_{i}^{2}+\cdots+y_{n}^{2}=1$. Prove: There exist not all zero numbers $a_{1}$, $a_{2}, \cdots, a_{n}$, taking values in $\{-1,0,1\}$, such that
$$
\left|a_{1} x_{1... | By the Cauchy inequality, we have
$$
\begin{array}{l}
\left(\left|x_{1}\right| \cdot\left|y_{1}\right|+\cdots+\left|x_{n}\right| \cdot\left|y_{n}\right|\right)^{2} \\
\leqslant\left(\left|x_{1}\right|^{2}+\cdots+\left|x_{n}\right|^{2}\right)\left(\left|y_{1}\right|^{2}+\cdots+\left|y_{n}\right|^{2}\right)=1 .
\end{arra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,738 |
Three. (35 points) Let the sequence $\left\{U_{n}\right\}$ satisfy: $U_{\mathrm{b}}=1, U_{1}=3$. $U_{n+2}=4 U_{n+1}-U_{n}$. For non-negative integers $n$, determine the last two digits (i.e., the units and tens digits) of $2 U_{n}^{3}-U_{n}$.
Translate the above text into English, preserving the original text's line b... | For $n=0,1,2 \cdots$, the first step is to divide $U_{n}$ by 4 to get a sequence with a period of 4: $1,3,3,1,1,3,3,1, \cdots$,
that is, when $n=4 k, 4 k+3$, it leaves a remainder of 1 when divided by 4;
when $n=4 k+1,4 k+2$, $U_{n}$ leaves a remainder of 3 when divided by 4.
The second step is to divide $U_{n}$ by 25 ... | 01 \text{ or } 51 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,739 |
1. Every natural number $n$ greater than 2 can be expressed as the sum of several distinct natural numbers. Let the maximum number of such distinct natural numbers be $A(n)$, find $A(n)$ (expressed in terms of $n$). | 1. Let $n=a_{1}+a_{2}+\cdots+a_{A(n)}, a_{1}<a_{2}<\cdots<a_{A(n)}$ be all natural numbers.
Obviously, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{a} \geqslant A_{G}$. Therefore,
$$
n \geqslant 1-1-2+\cdots+A(n):=\frac{1}{2} A(n)(A(n)+1) \text {. }
$$
Thus, the natural number $A(n) \leqslant\left[\frac{\sqrt{8 n... | A(n)=\left[\frac{\sqrt{8 n+1}-1}{2}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,740 |
2. For a finite set of points $M$ in the plane, it has the property: for any two points $A, B$ in $M$, there must exist a third point $C \in M$ such that $\triangle A B C$ is an equilateral triangle. Find the maximum value of $|M|$. | 2. Let $A, B \in M$ and $AB$ be the longest, by the given condition there exists $C \in M$, such that $\triangle ABC$ is an equilateral triangle. Clearly, the point set $M$ is entirely within the union of the three segments $CAB$, $ABC$, $BCA$ with $A$, $B$, $C$ as centers and $AB$ as the radius, as shown in the figure... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,741 |
(ii) $a, b$ are both square numbers;
(iii) Writing $b$ after $a$ produces a square number.
(For example, $a=16, b=81, 1681=41^{2}$) | 3. Let $a=\left(5 \times 10^{n-1}-1\right)^{2}, b=\left(10^{n}-1\right)^{2}$, then $a, b$ are both $2n$-digit numbers, and
$$
\begin{array}{l}
\left(5 \times 10^{n-1}-1\right)^{2} \times 10^{2 n}+\left(10^{n}-1\right)^{2} \\
=\left(5 \times 10^{2 n-1}\right)^{2}-\left(10^{n}-1\right) \times 10^{2 n}+\left(10^{n}-1\righ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,742 |
4. For $\triangle ABC$ with side lengths $AB=c, BC=a, CA=b$. Extend $AB$ to $A''$ such that $BA''=a$, and extend in the opposite direction to $B'$ such that $AB'=b$. Similarly, obtain $A', C', B'', C''$ as shown in the figure. Prove: $\frac{S_{A' B^* B^{\top} C C^* A^*}}{S_{ABC}} \geqslant 13$.
---
Note: The symbols ... | 4. Obviously, $\triangle A B^{\prime} C^{\prime \prime} \cong \triangle C^{\prime} B A^{\prime \prime} \cong \triangle B^{\prime \prime} C A^{\prime} \cong \triangle A B C$.
$$
\frac{S_{\triangle A A^{*} A}}{S_{\triangle A B C}}=\frac{(a+b)(a+c)}{b c}=1+\frac{a(a+b+c)}{b c},
$$
Therefore, $\frac{S_{\triangle A A^{-} A... | 13 | Geometry | proof | Yes | Yes | cn_contest | false | 706,743 |
1. The vertices of a regular nonagon are each colored either red or green. Any three vertices determine a triangle; if the three vertices are the same color, it is called a green (red) triangle. Prove that there must exist two monochromatic triangles of the same color that are congruent. | 1. The circumference of the circumscribed circle of a regular nonagon is divided into nine segments by the vertices, let each arc length be 1. Without loss of generality, assume the number of red points $\geqslant 5$. For any three red points, consider the three segments of arcs they divide: the possible combinations o... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,744 |
2. Given a real number $a$ satisfies: there is one and only one square, whose four vertices all lie on the curve $y=x^{3}+a x$. Try to find the side length of the square.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: ... | 2. Since the curve $y=x^{3}+a x$ is symmetric about the origin $(0,0)$, the center of the square must be the origin (otherwise, reflecting the square about the origin would yield another square with vertices on the curve).
Let one vertex of the square be $A(m, n)$, then $B(n,-m)$, $C(-m,-n), D(-n, m)$ are the other th... | \sqrt[4]{72} | Number Theory | proof | Yes | Yes | cn_contest | false | 706,745 |
3. In $\triangle A B C$, the angle bisector of $\angle A$ intersects the perpendicular bisector of side $A B$ at $A^{\prime}$, the angle bisector of $\angle B$ intersects the perpendicular bisector of side $B C$ at $B^{\prime}$, and the angle bisector of $\angle C$ intersects the perpendicular bisector of side $C A$ at... | 3. (1) When $A^{\prime}$ coincides with $B^{\prime}$, the incenter and circumcenter of $\triangle A B C$ coincide, making it an equilateral triangle.
(2) Rotate $\triangle A^{\prime} A C^{\prime}$ around $A^{\prime}$ so that $A$ coincides with $B$. Let $C^{\prime}$ rotate to $K$, then $B K$
$$
\begin{array}{l}
=A C^{\p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,746 |
4. Does there exist a natural number $n$, such that the first four digits of $n!$ are 1993? | 4. Existence. Let $m=1000100000$. When $k<99999$, if $(m+k)!=\overline{a b c d} \cdots$, then $(m+k+1)!=(m+k)!\times (m+k+1)=\overline{a b c d} \times 10001 \cdots=\overline{a b c x \cdots}$, where $x=d$ or $d+1$. Therefore, if $n!=\overline{a b c d \cdots}$, then the first four digits of $(m+1)!,(m+2)!, \cdots,(m+9999... | 1993 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,748 |
1. On the side $BC$ of the equilateral triangle $ABC$, take a point $D$. Draw a line through point $C$ parallel to $AL'$, intersecting the line $AB$ at point $E$. Prove: $\frac{CE}{CD} \geqslant 2 \sqrt{3}$. | 1. We have $\frac{C E}{C D}=\frac{C E}{A D} \cdot \frac{A D}{C D}=\frac{B C}{B D} \cdot \frac{A D}{C D}$ $=B C \cdot \frac{A D}{B D \cdot C D}$. Given that $A D$ is not shorter than the height from $B C$, we have $A D \geqslant \frac{\sqrt{3}}{2} B C$; also, since $B D + C D = B C$, let's denote $B D$ as $x B C$, where... | 2 \sqrt{3} | Geometry | proof | Yes | Yes | cn_contest | false | 706,749 |
2. Arbitrarily choose a natural number as $a_{0}$, and then arbitrarily select $a_{1}$ $\in\left\{a_{0}+54, a_{0}+77\right\}$, and so on, that is, after selecting $a_{k}$, select $a_{k+1} \in\left\{a_{k}+54, a_{k}+77\right\}$. Prove: In the obtained sequence $a_{0}$. $a_{1}, a_{2}, \cdots$, there must be a term $a_{n}$... | 2. Since 77 and 100 are coprime, if we let
$k \cdot 77 \equiv b_{k}(\bmod 100), 0 \leqslant b_{k} \leqslant 99$,
then $B=\left\{b_{k} \mid k=1,2, \cdots, 100\right\}$ is a complete residue system modulo 100, i.e., $B=\{0,1, \cdots, 99\}$. Now, arrange $b_{1}, b_{2}, \cdots, b_{100}$ in a clockwise direction on a circle... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,750 |
3. Now we have $n k$ stones, which are divided into $n$ piles in some way. The following operation is allowed: move some stones from the other piles into one of them, the number of stones moved should be equal to the original number of stones in that pile. For which natural numbers $k$, is it always possible to make th... | 3. Only when $k$ is a power of 2 can the remaining piles have an equal number of blocks. Suppose $2k$ blocks are divided into two piles, one with 1 block and the other with $2k-1$ blocks, denoted as $(1, 2k-1)$. Assuming it can be transformed into $(k, k)$ (or into $(0, 2k)$, which is discussed similarly) after a finit... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,751 |
4. Prove: If the natural number $A$ is not a perfect square, then there exists a natural number $n$, such that $A=\left[n+\sqrt{n}+\frac{1}{2}\right]$. | 4. Let $n=A-[\sqrt{A}]$, we will prove that $\left[n+\sqrt{n}+\frac{1}{2}\right]=A$. Let $\left[\sqrt{A}\right]=x$, then we have $x^{2}<A<(x+1)^{2}$, which implies $x^{2}+1 \leqslant A \leqslant x^{2}+2 x$, equivalent to $x^{2}-x+1 \leqslant A-x = n \leqslant x^{2}+x$, thus we have
$$
\sqrt{x^{2}-x+1}+\frac{1}{2} \leqs... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,752 |
5. Rotate a regular $n$-sided polygon $M$ around its center by an angle of $\frac{\pi}{n}$ to get $M^{\prime}$. How many convex $n$-sided polygons can the figure $M \cup M^{\prime}$ be divided into at minimum? | 5. The minimum number of convex polygons is $n+1$. A division of the figure $M \cup M^{\prime}$ into $n+1$ convex polygons clearly exists; we only need to prove that it cannot be divided into fewer convex polygons. For this, we examine all the "outer" vertices of the figure $M \cup M^{\prime}$ and number the convex pol... | n+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,753 |
6. The sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=1, F_{2}=2$, and for any $n \in N$, $F_{n+2}=F_{n+1}+F_{n}$. Prove: For any $n \in N$, $\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}}$. | 6. Let $F_{0}=\bar{F}_{2}-F_{1}=1$, then
$$
F_{k+1}=F_{k}+F_{k-1}, k=1,2, \cdots, n \text {, }
$$
which means
$$
1=\frac{F_{k}}{F_{k+1}}+\frac{F_{k-1}}{F_{k+1}}, k=1,2, \cdots, n .
$$
Thus, $n=\sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}}$, hence
$$
\begin{array}{l}
1=\frac{1}{n} \sum_{k... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,754 |
7. Color every point in the plane with one of 1992 colors, and ensure that points of each color exist. Given any triangle \( T \). Prove: there exists a triangle in the plane that is congruent to \( T \), such that on each of its two sides, there are points of the same color (excluding the vertices). | 7. Place triangle $T$ on the plane, and denote its circumcenter as $O$. The triangle obtained by rotating $T$ counterclockwise around point $O$ by $\alpha$ degrees is denoted as $T(\alpha)$, clearly $T(\alpha) \cong T$. It is easy to see that as long as the rotation angle $a_{0}>0$ is not too large, it can make the fir... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,755 |
8. There are 128 ones written on the blackboard. In each step, you can erase any two numbers $a$ and $b$ on the blackboard, and write the number $a \cdot b + 1$. After doing this 127 times, only one number remains. Denote the maximum possible value of this remaining number as $A$. Find the last digit of $A$.
| 8. Let's prove that by operating on the smallest two numbers on the board at each step, we can achieve the maximum final number. For convenience, we use $a * b$ to denote the operation of removing $a$ and $b$, and immediately adding $a b + 1$. Suppose at some step we do not operate on the smallest two numbers $x$ and $... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,756 |
9. Given $n$ segments. It is known that any $n-1$ segments can form an $n-1$-sided polygon. Prove: It is possible to form a triangle using any 3 of these segments.
保留源文本的换行和格式，直接输出翻译结果。 | 9. We arrange the given $n$ segments in length as $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. If no 3 segments can form a triangle, then we have $a_{3} \geqslant a_{1}+a_{2}, a_{6} \geqslant a_{2}+a_{3}, \cdots, a_{n} \geqslant a_{n-2} +a_{n-1}$, thus $a_{n} \geqslant a_{n-2}+a_{n-3}+\cdots+a_{3}+2 a_{2}+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,757 |
Example 6. $H$ is the centroid of $\triangle ABC$, and $D, E, F$ are the midpoints of $BC, CA, AB$ respectively. A circle $\odot H$ centered at $H$ intersects the lines $EF, FD, DE$ at $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$.
Prove: $A A_{1}=A A_{2}=B B_{1}=B B_{2}=C C_{1}=C C_{2}$. | To prove that $A A_{1}=B B_{1}=C C_{1}$ is sufficient. Let $B C = a, C A = b, A B = c$, the circumradius of $\triangle A B C$ be $R$, and the radius of $\odot H$ be $r$.
Connect $H A_{1}, A H$ intersecting $E F$ at $M$.
$$
\begin{aligned}
A A_{1}^{2} & =A M^{2}+A_{1} M^{2}=A M^{2}+r^{2}-M H^{2} \\
& =r^{2}+\left(A M^{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,758 |
10. Find all such triples of natural numbers: the product of any two numbers plus 1 is divisible by twice the third number.
Find all such triples of natural numbers: the product of any two numbers plus 1 is divisible by twice the third number. | 10. There is only one such 3-tuple $(1,1,1)$. Hint: If the 3-tuple of natural numbers $(a, b, c)$ satisfies the condition, then $a, b, c$ should be coprime and all odd. Since $a b+b c+c a+1$ should be divisible by each of $a, b, c$, it should therefore be divisible by their product $a b c$. | (1,1,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,759 |
13. If several lines are distributed in the same plane, and no two of them are parallel, and no three of them are concurrent, then these lines are said to be in "general position".
It is known that for any 4 lines in general position, the circumcircles of the 4 triangles formed by each set of 3 lines all pass through ... | 13. It is possible. Let $l_{1}$ and $l_{2}$ intersect at a point $O$. Take any point $P$ outside these two lines. Construct 43 distinct circles, all passing through points $O$ and $P$, and all intersecting $l_{1}$ and $l_{2}$. Let the $i$-th circle intersect $l_{1}$ at point $A_{i}$ and $l_{2}$ at point $B_{i}$, where ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,761 |
14. Paddle Country has 100 cities, which are connected by roads. It is known that if any city $A$ and all the roads leading to and from $A$ are removed, it is still possible to travel from any city to any other city (excluding $A$). Prove: The country can be divided into two main countries, each with 50 cities, such th... | 14. We use induction to prove that 100 cities can be divided into two countries $A$ and $B$, such that $A$ has $k$ cities and $B$ has $100-k$ cities.
When $k=1$, it is obvious by the problem statement. To complete the induction proof, we only need to assume that 100 cities have been divided into the two countries as re... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,762 |
15. In an $m \times n$ grid, fill each small square with a natural number, such that the numbers in each column are strictly increasing from top to bottom, and the numbers in each row are non-strictly increasing from left to right. We denote by $L\left(a_{1}, a_{2}, \cdots, a_{4}\right)$ the number of all different val... | 15. It is sufficient to prove that \( L\left(a_{1}, \cdots, a_{p-1}, a_{p}, a_{p+1}, \cdots, a_{k}\right) = L\left(a_{1}, \cdots, a_{p-1}, a_{p+1}, a_{p}, a_{p+2}, \cdots, a_{k}\right) \), i.e., the value of \( L \) remains unchanged when two adjacent variables are swapped. We will examine "cell groups" \( T \) in an \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,763 |
16. There are some unit circles distributed on a plane. Is it possible to mark some points on the plane so that there is exactly one marked point inside each unit circle? | 16. Assuming that what is stated can be realized, draw a unit circle with each marked point as the center, so that the center of each original unit circle is precisely covered (and lies within) one of the newly drawn unit circles. Thus, the original problem can be transformed into: Given several points on a plane, is i... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,764 |
Example 1. Solve the equation
$$
\sqrt{x^{2}+4 x+8}+\sqrt{x^{2}-8 x+20}=10 \text {. }
$$ | Analysis: Removing the square root from this equation is not easy, but if we let $y^{2}=4$, the original equation becomes $\sqrt{(x+2)^{2}+y^{2}}+\sqrt{(x-4)^{2}+y^{2}}=10$, from which we know that the trajectory of $(x, y)$ is an ellipse $\frac{(x-1)^{2}}{25}+\frac{y^{2}}{16}=1$. Substituting $y^{2}=4$ into it, we fin... | x=1 \pm \frac{5 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,765 |
Example 2. Let positive numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a^{2}+a b+\frac{b^{2}}{3}=25, \\ \frac{b^{2}}{3}+c^{2}=9, \\ c^{2}+a c+a^{2}=16 .\end{array}\right.$ Find the value of $a b+2 b c+3 c c$. | Analysis: The conditional equation resembles the cosine theorem, while the value equation is similar to the sum of areas. Transform it in this direction:
\[ a^{2}+\left(\frac{b}{\sqrt{3}}\right)^{2}- 2 a\left(\frac{b}{\sqrt{3}}\right) \cos 150^{\circ}=5^{2}, \left(\frac{b}{\sqrt{3}}\right)^{2}+c^{2}=3^{2}, c^{2}+a^{2}... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,766 |
Example 3. Find the minimum value of the function $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u^{2}}\right.$ $\left.-\frac{9}{v}\right)^{2}$. (1983 Putnam Competition) | Analysis: The problem is equivalent to finding the shortest distance (squared) between points on two curves. A sketch reveals that $[f(u, v)]_{\min }=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,767 |
Example 1. Find the range of $y=3 \sin x+4 \cos x-2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 1. Find the range of $y=3 \sin x+4 \cos x-2$. | $3 \sin x+4 \cos x-2-y=0$, since $\sin ^{2} x+\cos ^{2} x=$ 1, it is known that the line $3 v+4 u-2-y=0$ intersects the circle $u^{2}+v^{2}=1$ at a common point $(\sin x, \cos x)$, the distance from the center of the circle $(0,0)$ to the line $\frac{|-2-y|}{\sqrt{3^{2}+4^{2}}}$ $\leqslant 1, \therefore-1 \leqslant y \... | -1 \leqslant y \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,768 |
Example 8. Given $\odot O$ is the circumcircle of $\triangle ABC$, and $\odot Q$ is tangent to $AB$, $AC$ at $E$, $F$ respectively, and is internally tangent to $\odot O$. Prove: the midpoint $P$ of $EF$ is the incenter of $\triangle ABC$. (B. Polastov, "High School Mathematics Olympiad") | In the 20th IMO, a problem provided by the United States is actually a special case of Example 8, but it adds the condition $AB = AC$. How can we prove it when $AB \neq AC$?
As shown in the figure, it is obvious that the midpoint $P$ of $EF$, the center $Q$, and the midpoint $K$ of $\widehat{BC}$ are all on the angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,769 |
Example 2. Find the extremum of $f(\alpha)=\cos \alpha+\cos \frac{\pi}{12}-\cos \left(\alpha+\frac{\pi}{12}\right)$. | Using an example similar to solving, the minimum value is $\cos \frac{\pi}{12}-$ $\sqrt{2-2 \cos \frac{\pi}{12}}$, the maximum value is $\cos \frac{\pi}{12}+\sqrt{2-2 \cos \frac{\pi}{12}}$. | \cos \frac{\pi}{12} \pm \sqrt{2-2 \cos \frac{\pi}{12}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,770 |
Example 3. Find the maximum and minimum values of $y=\sqrt{x+3}+\sqrt{1-x}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\because \quad \sqrt{x+3}+\sqrt{1-x}-y=0 \text {, }
$$
there is $(\sqrt{x+3})^{2}+(\sqrt{1-x})^{2}=2^{2}$,
$\therefore$ the line $v+u-y=0$ intersects with the quarter circle $v^{2}+u^{2}=2^{2}(v$ $\geqslant 0, u \geqslant 0$ ) at the common point $(\sqrt{x+3}, \sqrt{1-x})$.
As shown in the figure, the line $v+u-y=0... | y_{\text {min }}=2, y_{\max }=2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,771 |
*1. A set $M=\{19,93,-1,0,25, -78,-94,1,17,-2\}$ consists of ten elements. All non-empty subsets of $M$ are denoted as $M_{i}(i=1,2, \cdots, 1023)$, and the product of all elements in each non-empty subset is denoted as $m_{i}(i=1,2, \cdots, 1023)$. Then $\sum_{i=1}^{1023} m_{i}=(\quad)$.
(A) 0
(B) 1
(C) -1
(D) None of... | 1. C. Let the ten elements be $a_{1}, a_{2}, \cdots, a_{10}$. Then
$$
\begin{aligned}
\sum_{i=1}^{1023} m_{i}= & a_{1}+a_{2}+\cdots+a_{10}+a_{1} a_{2}+a_{1} a_{3}+\cdots \\
& +a_{8} a_{10}+a_{1} a_{2} a_{3}+\cdots+a_{9} a_{10} a_{1}+\cdots \\
& +a_{1} a_{2} \cdots a_{10} \\
= & \prod_{i=1}^{10}\left(1+a_{i}\right)-1
\e... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,772 |
$A^{2 .} \triangle ABC$ 's three interior angles $A, B, C$ form an arithmetic sequence, and the heights $h_{a}, h_{b}, h_{c}$ on the three sides $a, b, c$ also form an arithmetic sequence. Then $\triangle ABC$ is ( ).
(A) Isosceles but not equilateral triangle
(B) Equilateral triangle
(C) Right triangle
(D) Obtuse non-... | 2. B. From $A, B, C$ forming an arithmetic sequence, we know that $B=60^{\circ}$. Let the area of $\triangle A B C$ be $S$. Then, from $2 h_{b}=h_{a}+h_{c}$, we get $\frac{2 S}{b}=\frac{S}{a}+\frac{S}{c}$. Therefore, $2 a c=a b+b c$.
By the Law of Sines, and noting $A=60^{\circ}-\theta, C=60^{\circ}+\theta, 0^{\circ} ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,773 |
3. For all real numbers $x$, the inequality $x^{4}+(a-1) x^{2}+1 \geqslant 0$ always holds. Then the range of values for $a$ is ().
(A) $a \geqslant-1$
(B) $a \geqslant 0$
(C) $a \leqslant 3$
(D) $a \leqslant 1$ | 3. A. When $x=0$, $x^{4}+(a-1) x^{2}+1=1 \geqslant 0$ always holds. When $x \neq 0$, the original inequality is equivalent to
$$
x^{2}+\frac{1}{x^{2}} \geqslant 1-a .
$$
Since the minimum value of $x^{2}+\frac{1}{x^{2}}$ is 2, we get $1-a \leqslant 2$, which means $a \geqslant -1$. | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 706,774 |
4. If four points in space $A, B, C, D$ satisfy $A B=C D=8, A C$ $=B D=10, A D=B C=13$, then the number of such tetrahedra $A B C D$ is ( ).
(A) 0
(B) 1
(C) 2
(D) More than 2 | 4. A. Construct $\triangle A B C$, such that $A B=8$, $B C=13$, $A C=10$. Using $B C$ as one side, continue to construct $\triangle B D C$. Clearly, when $\triangle A B C$ and $\triangle B C D$ are in the same plane, $A D$ has its maximum value, at which point quadrilateral $A B C D$ is a parallelogram, thus we have
$$... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,775 |
1. The range of the function $y=\sqrt{3 x+6}+\sqrt{8-x}$ is | 1. $y \in[\sqrt{10}, 2\sqrt{10}]$.
The original function can be transformed into
$$
\begin{array}{l}
y=\sqrt{3} \sqrt{x+2}+ \\
\sqrt{8-x}.
\end{array}
$$
Let point $P(u, v)=P(\sqrt{x+2}, \sqrt{8-x})$, then
$$
u^{2}+v^{2}=10(u, v \geq 0),
$$
which means $P$ lies on the circle $u^{2}+v^{2}=10(u, v \geqslant 0)$. Also, ... | \sqrt{10} \leqslant y \leqslant 2 \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,778 |
2. Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$, with foci $F_{1}, F_{2}$, and $P$ as any point on the ellipse (but $P$ is not on the $x$-axis), the incenter of $\triangle P F_{1} F_{2}$ is $I$. A line parallel to the $x$-axis through $I$ intersects $P F_{1}, P F_{2}$ at $A$,
$B$. Then $\frac{S_{\triangle P A ... | 2. $\frac{9}{16}$. Given, $a=3, b^{2}=8, c=1$, then
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=2 a=6,\left|F_{1} F_{2}\right|=2 c=2 \text {. }
$$
The perimeter of $\triangle P F_{1} F_{2}$ is 8.
Let the inradius of $\triangle P F_{1} F_{2}$ be $r$, and point $P(x, y)$.
By $\triangle P A B \backsim \triangle P F_{1} F... | \frac{9}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,779 |
Example 9. In a right-angled triangle, prove:
$r+r_{d}+r_{b}+r_{s}=2 p$.
where $r, r_{0}, r_{b}$ are the radii of the incircle and the excircles opposite to $a, b, c$ respectively, and $p$ denotes the semiperimeter. (Hangzhou University "High School Mathematics Competition Problems") | Consider a right-angled triangle $ABC$ with $c$ as the hypotenuse. Let's prove the following property:
$$
p(p-c)=(p-a)(p-b) \text {. }
$$
$\because p(p-c)=\frac{1}{2}(a+b+c) \cdot \frac{1}{2}(a+b-c)$
$=\frac{1}{4}\left[(a+b)^{2}-c^{2}\right]$
$=\frac{1}{2} a b$;
$(p-a)(p-b)$
$=\frac{1}{2}(-a+b+c) \cdot \frac{1}{2}(a-b+... | 2p | Geometry | proof | Yes | Yes | cn_contest | false | 706,780 |
* 3. $A, B, C$ are the three interior angles of $\triangle A B C$, and $\operatorname{ctg} \frac{A}{2}$ $+\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}-2(\operatorname{ctg} A+\operatorname{ctg} B+\operatorname{ctg} C) \geqslant T$. Then $T_{\max }=$ $\qquad$ | 3. $\sqrt{3}$. From $\operatorname{ctg} \frac{A}{2}-2 \operatorname{ctg} A=\operatorname{tg} \frac{A}{2}$,
we have $\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{C}{2} \geqslant T$.
$$
\begin{array}{l}
\text { (tg } \left.\frac{A}{2}+\operatorname{tg} \frac{B}{2}+\operatorname{tg}... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,781 |
* 4. Real numbers $a, b, c$ satisfy $a+b-2 c=-3, a^{2}+b^{2}+2 c^{2}$ $+8 c=5$. Then the minimum value of $a b$ is $\qquad$. | 4. $\frac{25}{16}$. From the given conditions, we have $a+b=2 c-3, a b=c^{2}-2 c+2$.
Therefore, $a$ and $b$ are the real roots of the equation
$$
t^{2}-(2 c-3) t+\left(c^{2}-2 c+2\right)=0
$$
Thus,
$$
\begin{array}{c}
\Delta=(2 c-3)^{2}-4\left(c^{2}-2 c+2\right) \geqslant 0, \\
c \leqslant \frac{1}{4} .
\end{array}
$... | \frac{25}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,782 |
*5. In a football championship, each team is required to play a match against all other teams. Each match, the winning team gets 2 points, a draw results in 1 point for each team, and the losing team gets 0 points. It is known that one team has the highest score, but it has won fewer matches than any other team. If the... | 5. $n=6$. We call the team $A$ with the highest score the winning team. Suppose team $A$ wins $n$ games and draws $m$ games, then the total score of team $A$ is $2n + m$ points.
From the given conditions, each of the other teams must win at least $n+1$ games, i.e., score no less than $2(n+1)$ points. Therefore,
$2n + ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,783 |
* 6 . If $2^{10}+2^{13}+2^{m}$ is a perfect square, then the natural number $m=$ $\qquad$ | 6. $m=14$. Let $k^{2}=2^{10}+2^{13}+2^{m}$, then
$$
\begin{aligned}
2^{m} & =k^{2}-\left(2^{10}+2^{13}\right)=k^{2}-\left(2^{5} \cdot 3\right)^{2} \\
& =(k+96)(k-96) .
\end{aligned}
$$
Therefore, there exist two non-negative integers $s$ and $t$, such that
$$
\left\{\begin{array}{l}
k-96=2^{s}, \\
k+96=2^{m-t}=2^{t}, ... | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,784 |
Three. (This question is worth 20 points) If the distance from a vertex of the base of a regular tetrahedron to the centroid of the opposite face is 4, find the maximum volume of this regular tetrahedron.
| Three, as shown in the figure, for the regular triangular pyramid $P-ABC$, $G$ is the centroid of the side face $\triangle PBC$, and $AG=4$.
Extend $PG$ to intersect $BC$ at $D$. Clearly, $D$ is the midpoint of $BC$.
Draw $PO \perp$ the base $ABC$ at $O$. Then $O$ lies on $AD$, and $O$ is the centroid of $\triangle A... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,785 |
Four. (This question is worth 20 points) A point moves at a speed of 2 meters/second on the $x$-axis, and at a speed of 1 meter/second in other parts of the plane. Try to find the boundary line of the region that the point can reach within 1 second from the origin.
| By symmetry, we only need to consider the boundary points $(a, b)$ in the first quadrant. Clearly, $0 \leqslant a \leqslant 2$.
$b$ is the maximum ordinate of the point that can be reached on the line $x=a$ within 1 second.
Let the last point on the $x$-axis of the most time-saving route from $(0,0)$ to $(a, b)$ be $(... | |b|=\left\{\begin{array}{ll}
\sqrt{1-a^{2}}, & |a| \leqslant \frac{1}{2} \\
\frac{1}{\sqrt{3}}|2-| a||, & \frac{1}{2} \leqslant|a| \leqslant 2
\end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,786 |
Five. (This question is worth 20 points) Given that $x$ is an imaginary number, and $x+\frac{1}{x}$ is a real root of the equation $y^{2}-a y+a+1=0$. Find the range of real values for $a$.
---
Translate the above text into English, preserving the original text's line breaks and format, and output the translation dire... | Let $x+\frac{1}{x}=y$, then
$$
x^{2}-y x+1=0 \text {, }
$$
Since $y$ is a real number and $x$ is an imaginary number, i.e., equation (1) has imaginary roots, then
$$
\begin{array}{l}
\Delta=y^{2}-40, \\
f(-2)=4+2 a+a+1>0 .
\end{array}\right.
$$
Solving, we get $-\frac{5}{3}<a<2-2 \sqrt{2}$ or $a>5$ or $a \leqslant-\f... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,787 |
*One. (This question is worth 35 points) In $\triangle A B C$, $M$ is any point on side $A B$, $M E \perp A B$ at $E$, $M F \perp A C$ at $F$, $A N \perp$ $E F$ intersects $B C$ at $N$.
Prove: $A M \cdot A N+\sqrt{B M \cdot B N \cdot C M \cdot C N}=$ $A B \cdot A C$. | Construct the circumcircle of $\triangle A B C$, extend $A M, A N$ to intersect the circumcircle at $M^{\prime}, N^{\prime}$, and connect $B M^{\prime}, C N^{\prime}$.
Since $M E \perp A B, M F \perp A C$, then $A, E, M, F$ are concyclic.
Thus, $\angle 1=\angle 2$.
Also, $\angle 2=\angle 3$, so $\angle 1=\angle 3$.
Fu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,788 |
*II, (Full score for this question: 35 points) Use $n$ numbers (repetition allowed) to form a sequence of length $N$, and $N \geqslant 2^{n}$. Prove: It is possible to find several consecutive terms in this sequence, whose product is a perfect square. | Let $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$ form a sequence of length $N$: $b_{1}, b_{2}, \cdots, b_{N}$.
Here $b_{i} \in\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, i=1,2, \cdots, N$.
Establish the mapping
$$
B=\left\{b_{1}, b_{2}, \cdots, b_{N}\right\} \rightarrow V=\left\{v_{1}, v_{2}, \cdots, v_{n}\right\} .
$$
w... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,789 |
*Three. (This question is worth 35 points) In space, there are 100 points, where no four points lie on the same plane, and no three points lie on the same line. Each point is connected to 33 other points with red lines, to another 33 points with yellow lines, and to the last 33 points with blue lines. Prove: There will... | If two lines extending from a point have different colors, we say that these two lines of different colors form a bichromatic angle.
From a point, 33 red lines and 33 yellow lines form $33^{2}$ bichromatic angles, 33 red lines and 33 blue lines form $33^{2}$ bichromatic angles, and 33 yellow lines and 33 blue lines fo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,790 |
Example 10. $M$ is any point on the side $AB$ of $\triangle ABC$. $r_{1}, r_{2}$, $r$ are the radii of the incircles of $\triangle AMC, \triangle BMC, \triangle ABC$, and $q_{1}, q_{2}, q$ are the radii of the excircles of the above triangles that lie inside $\angle ACB$. Prove: $\frac{r_{1}}{q_{1}} \cdot \frac{r_{2}}{... | For any $\triangle A^{\prime} B^{\prime} C^{\prime}$, by the Law of Sines, we have
$$
\begin{array}{l}
O D=O A^{\prime} \cdot \sin \frac{A^{\prime}}{2} \\
=A^{\prime} B^{\prime} \cdot \frac{\sin \frac{B^{\prime}}{2}}{\sin \angle A^{\prime} O B^{\prime}} \\
\text { - } \sin \frac{A^{\prime}}{2} \\
=A^{\prime} B^{\prime}... | \frac{r_{1}}{q_{1}} \cdot \frac{r_{2}}{q_{2}}=\frac{r}{q} | Geometry | proof | Yes | Yes | cn_contest | false | 706,791 |
Example 11. In a convex hexagon $A B C D E F$ inscribed in a circle, $A B=$ $B C, C D=D E, E F=F A$. Prove:
(1) The three diagonals $A D, B E, C F$ intersect at one point;
(2) $A B+B C+C D+D E+E F+F A \geqslant A K+B E$ $+C F$ (1991, National Education Commission Mathematics Experimental Class Admission Test) | Analyzing the connection $A C, C E$, $E A$, from the given information, we can prove that $A D, C F$, $E B$ are the three angle bisectors of $\triangle A C E$, and $I$ is the incenter of $\triangle A C E$. Therefore, we have $I D=C D=D E$,
$$
\begin{array}{l}
I F=E F=F A, \\
I B=A B=B C .
\end{array}
$$
Considering $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,792 |
Example 12. The circumcenter of $\triangle ABC$ is $O, AB=AC, D$ is the midpoint of $AB$, $E$ is the centroid of $\triangle ACD$. Prove that $OE \perp CD$. (Canadian Mathematical Olympiad Training Problem) | Let $A M$ be both the altitude and the median. Taking the midpoint $F$ of $A C$, $E$ must lie on $D F$ and $D E: E F=2: 1$. Let $C D$ intersect $A M$ at $G$, $G$ must be the centroid of $\triangle A B C$. Connect $G E, M F$, and let $M F$ intersect $D C$ at $K$. It is easy to prove:
$$
\begin{array}{l}
\text { Prove: }... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,793 |
Example 4. In quadrilateral $ABCD$, with the ratio of a pair of opposite sides $AB$ : $CD$, internally divide the other pair of opposite sides $AD, BC$ at $E, F$, extend $BA, CD$ to meet the extension of $FE$ at $G, Q$ respectively. Prove: $\angle BGF$ $=\angle FQC$. | Analyzing the two angles to be proven, they are in two triangles, and the given conditions involve the ratio of internal division of two line segments that are not on the same line, making the conditions somewhat scattered. It is necessary to draw auxiliary lines to concentrate the conditions. Let's connect $B D$, then... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,794 |
Example 13. In $\triangle ABC$, $\angle C=30^{\circ}$, $O$ is the circumcenter, $I$ is the incenter, point $D$ on side $AC$ and point $E$ on side $BC$ are such that $AD=BE=AB$. Prove: $OI \perp DE, OI=DE$. (1988, China Mathematical Olympiad) | Given the auxiliary lines as shown in the figure, construct the angle bisector of $\angle D A O$ intersecting $B C$ at $K$.
It is easy to prove $\triangle A I D \cong \triangle A I R$
$$
\begin{array}{l}
\cong \triangle E I B, \angle A I D=\angle A I B \\
=\angle E I B .
\end{array}
$$
Using the angle bisector formula... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,795 |
2. Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$.
Find the minimum value of $\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+$ $\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}$. (1982, China Mathematical Competition) | Let $S=\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots$ $+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}$. Given $\sum_{i=1}^{n} a_{i}=1$, we have $S=\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}$.
Let $\lambda<0$. From the condition $\sum_{i=1}^{n} a_{i}=1$, we get $\lambda\left(\sum_{i=1}^{n} a_{i}-1... | \frac{n}{2 n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,799 |
Column 3. Let $a, b, c, d$ be positive real numbers satisfying $a b+b c+c d+d a=1$. Prove that: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$.
(31st IMO Preliminary Selection) | Proof Let $\lambda>0$, from the condition we have
$\lambda(a b+b c+c d+d a-1)=0$.
Let $S=\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c}$, then
$$
\begin{aligned}
S=S+ & \lambda(a b+b c+c d+d a-1) \\
= & S+\frac{\lambda}{3}((a b+a c+a d)+(b a+b c+b d) \\
& +(c a+c b+c d)+(d a+d b+d c)]-\... | \frac{1}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 706,800 |
Example 2. Determine all real solutions of the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3
\end{array}\right.
$$
(2nd United States of America Mathematical Olympiad) | Let the plane $\pi: x+y+z=3$, the sphere $O: x^{2}+y^{2}+z^{2}=3$, and $d=\frac{|-3|}{\sqrt{3}}=\sqrt{3}=R$. By Theorem 2 (2), the plane $\pi$ is tangent to the sphere $O$, and the coordinates of the point of tangency are the solutions to the system of equations $\left\{\begin{array}{l}x+y+z=3, \\ x^{2}+y^{2}+z^{2}=3\e... | x=1, y=1, z=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,802 |
Example 4. Let $a, b, c, d \in R^{+}, a+b+c+d=1$. Prove that
$$
\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}+\sqrt{4 d+1}<6 \text {. }
$$
(1980, Leningrad Mathematical Competition) | Let $\sqrt{4 a+1}=x, \sqrt{4 b+1}=y, \sqrt{4 c+1}=z, \sqrt{4 d+1}=w, \sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}+$ $\sqrt{4 d+1}=S$. Then we have $x^{2}+y^{2}+z^{2}+w^{2}=8$. The plane $\pi: x+y$ $+z+(\omega-S)=0$, the sphere $O: x^{2}+y^{2}+z^{2}=8-w^{2}$.
By Theorem 2 (2), (3) we have
$\frac{|w-S|}{\sqrt{3}} \leqslant \sq... | \sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}+\sqrt{4 d+1} \leqslant 4 \sqrt{2} < 6 | Inequalities | proof | Yes | Yes | cn_contest | false | 706,804 |
Example 2. Let $H$ be the orthocenter of $\triangle ABC$, $L, M, N$ be the midpoints of sides $BC, CA, AB$ respectively, $D, E, F$ be the feet of the altitudes from $A, B, C$ respectively, and $P, Q, R$ be the midpoints of $HA, HB, HC$ respectively. Prove that $L$, $M, N, D, E, F, P, Q, R$ are concyclic. | Since $P, Q, R$ are the midpoints of $H A, H B, H C$ respectively, a homothety centered at $H$ with a ratio of 2 transforms $\odot_{P Q R}$ into $\odot_{A B C}$. Therefore, to prove that $L, M, N, D$, $E, F$ lie on $\odot_{P Q R}$, it suffices to prove that these points, under the aforementioned homothety, when $H L$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,805 |
Example 6. Let $P$ be a point inside an acute $\triangle ABC$, and the feet of the perpendiculars from $P$ to the sides $BC, CA, AB$ are $D, E, F$ respectively. Find (and prove) the point that minimizes $PD^2 + PE^2 + PF^2$ (IMO-28).
Translate the above text into English, please retain the original text's line breaks ... | Let the area of $\triangle ABC$ be $S$, $AB=c$, $BC=a$, $CA=b$, $PD=x$, $PE=y$, $PF=z$.
From $S_{\triangle PBC} + S_{\triangle PCA} + S_{\triangle PAB} = S$, we have $ax + by + cz = 2S$. The plane $\pi: ax + by + cz = 2S$. The sphere $O: x^2 + y^2 + z^2 = R^2$. By Theorem 2 (2), (3), we know $\frac{|-2S|}{\sqrt{a^2 + ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,807 |
Example 7. Let $a, b$ be real numbers, and the equation $x^{4}+a x^{3}+b x^{2}$ $+a x+1=0$ has at least one real solution. Determine the minimum possible value of $a^{2}+b^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $x_{0}$ be a real root of the equation, obviously $x_{0} \neq 0$. Therefore, we have $x_{0}^{2}+a x_{0}+b+\frac{a}{x_{0}}+\frac{1}{x_{0}^{2}}=0$.
The line $l:\left(x_{0}+\frac{1}{x_{0}}\right) a+b+\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)=0$,
The circle $O: a^{2}+b^{2}=R^{2}$.
By Theorem 1 (2), (3), we know
$$
\fr... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,808 |
Example 8. Given that $a, b, c, d, e$ are real numbers satisfying $a+b+c+d+e$ $=8, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$. Determine the maximum value of $e$. (7th American High School Mathematics Examination) | Let the plane $\pi: a+b+c+(d+e-8)=0$, and the sphere $O: a^{2} +b^{2}+c^{2}=16-d^{2}-e^{2}$. From Theorem 2 (2), (3), we have
$$
\frac{|d+e-8|}{\sqrt{3}} \leqslant \sqrt{16-d^{2}-e^{2}},
$$
which simplifies to $2 d^{2}+(e-8) d+2 e^{2}-8 e+8 \leqslant 0$, a quadratic inequality in $d$. Therefore,
$$
\begin{array}{l}
\q... | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,809 |
Example 9. Find the range of $y=\frac{4 \sin x+5 \cos x}{\sin x+\cos x+3}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $X=\cos x, Y=\sin x$. Then we have $y=\frac{4 Y+5 X}{Y+X+3}$, which means $(5-y) X+(4-y) Y-3 y=0$,
the line $l:(5-y) X+(4-y) Y-3 y=0$, and the circle $O: x^{2}+y^{2}=1$. By Theorem 1 (2), (3) we know
$$
\begin{array}{l}
\frac{|-3 y|}{\sqrt{(5-y)^{2}+(4-y)^{2}}} \leqslant 1 \text {. That is, } 7 y^{2}+18 y-41 \leqsl... | \left[\frac{-9-4 \sqrt{23}}{7}, \frac{-9+4 \sqrt{23}}{7}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,810 |
1. Segment Equation and Proof
Example 1. As shown in the figure, $A B$ //
$$
\begin{array}{c}
C D / / E F \text {. If } A B=x, \\
C D=y, E F=z, \text { then } \\
\frac{1}{x}+\frac{1}{y}=\frac{1}{z} .
\end{array}
$$ | Proof From the given,
$$
\begin{array}{l}
\frac{z}{x}=\frac{C F}{C A}, \frac{z}{y}=\frac{F A}{C A}, \\
\therefore \frac{z}{x}+\frac{z}{y}=\frac{C F+F A}{C A}=1 .
\end{array}
$$
That is, $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,811 |
Example 2. As shown in the figure, in the isosceles right $\triangle ABC$, $AD$ is the median on the right side $BC$, $BE \perp AD$, and intersects $AC$ at $E$, $EF \perp BC$. If $AB = BC = a$, then $EF$ equals ( ).
(A) $\frac{1}{3} a$
(B) $\frac{1}{2}-2$
(C) $\frac{2}{3} a$
(D) $\frac{2}{5} a$ O100, Xi'an Junior High ... | Draw $CG$ perpendicular to $BC$ intersecting the extension of $BE$ at $G$.
There is
$\triangle A B D \cong \triangle B C G$.
So $C G=\frac{a}{2}$.
That is $\frac{1}{a}+\frac{1}{\frac{a}{2}}=\frac{1}{E F}$. Thus $E F=\frac{1}{3} a$. Therefore, the answer is (A). | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,812 |
Example 3. As shown in the figure, there are three adjacent equilateral triangles $\triangle ABC$, $\triangle ADE$, $\triangle AFG$ on the same side of line $l$, and points $G$, $A$, $B$ are all on line $l$. Let the side lengths of these three triangles be $a$, $b$, $c$ respectively. Connect $GD$ to intersect $AE$ at $... | Prove: Draw $N M / / A C, L H / / A E \approx l$ at points $M, N$. By the equalities, we have
$$
\frac{1}{H L}=-\frac{1}{A N}+\frac{1}{H C} \cdot \frac{1}{M N}=\frac{1}{G F}+\frac{1}{A D} \text {. }
$$
And $H i=A L, M N=A N$,
$$
\therefore \frac{1}{A L}=\frac{1}{A N}+\frac{1}{a}, \frac{1}{A N}=\frac{1}{b}+\frac{1}{c}=... | AL=\frac{abc}{ab+bc+ca} | Geometry | proof | Yes | Yes | cn_contest | false | 706,813 |
3. Given the equation $x^{2}+(a-2) x+a+1=0$ with two complex roots $x_{1}, x_{2}$, and the point $\left(x_{1}, x_{2}\right)$ lies on the circle $x^{2}+y^{2}=4$. Then the value of $a$ $=$ $\qquad$ | $3.3-\sqrt{11}$ | 3-\sqrt{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,817 |
5. Given $\cos ^{2}(\alpha+\beta)-\cos ^{2}(\alpha-\beta)=x, \cos ^{2} \alpha$
$\cdot \cos ^{2} \beta=y, y \neq 0$. Then express $\operatorname{tg} \alpha \operatorname{ctg} \beta=$ $\qquad$ | 5. $-\frac{x}{4 y}$ | -\frac{x}{4 y} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,819 |
Example 5. In quadrilateral $ABCD$, the extensions of opposite sides $AD, BC$ meet at $E$, and the extensions of $AB, CD$ meet at $F$. $O$ is the intersection of the diagonals. A line through $O$ parallel to $AB$ intersects $EF$ at $Q$. Prove: $OG=GQ$. | Let $Q G O$ intersect $A D$ at $R$, and intersect $E B$ at $P$. Perform a homothety transformation:
Then $\frac{R Q}{A F}=\frac{R P}{A B}, \frac{R O}{A B}=\frac{R G}{A F}$,
$$
\frac{O P}{A B}=\frac{O G}{A F} \text {. }
$$
From $\frac{G Q}{A F}=\frac{R Q-R G}{A F}=\frac{R Q}{A F}-\frac{R G}{A F}=\frac{R P}{A B}-\frac{... | GQ=OG | Geometry | proof | Yes | Yes | cn_contest | false | 706,827 |
13. Let point $A_{0}$ be at the origin of the Cartesian coordinate system, and $A_{1}, A_{2}, \cdots$, $A_{n}, \cdots$ be sequentially on the positive half of the $x$-axis, with $\left|A_{n-1} A_{n}\right|=2 n-1$, $n=1,2, \cdots$. Construct equilateral triangles $A_{n-1} A_{n} B_{n}$ above the $x$-axis with $A_{n-1} A_... | 13. $y^{2}=3\left(x-\frac{1}{4}\right)$ | y^{2}=3\left(x-\frac{1}{4}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,828 |
15. Given that the area of the inscribed sphere of a cone is the arithmetic mean of the area of the base and the lateral surface area of the cone. Then the angle between the slant height of the cone and the base is $\qquad$ . | 15. $\arccos \frac{1}{3}$ | \arccos \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,830 |
16. Through the center $O$ of the equilateral $\triangle A B C$, draw a line intersecting $A B$ and $A C$ at points $D$ and $E$ respectively. If $D O=3, C E=2$, then the side length of the equilateral triangle is $\qquad$ . | 16. $\frac{18 \sqrt{7}}{7}$ | \frac{18 \sqrt{7}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,831 |
20. Let $n \in N, n \geqslant 3$, and let $f(n)$ denote the smallest natural number that is not a divisor of $n$, for example $f(12)=5$. If $f(n) \geqslant 3$, then $f(f(n))$ and so on can be defined. If $\underbrace{f(f(\cdots f}_{k \uparrow f}(n) \cdots))=2$, then $k$ is called the length of $n$. For all $n \in N, n ... | 20. $\{1,2,3\}$ | \{1,2,3\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,835 |
Example 6. A line segment is drawn from $P(4,3)$ to any point $Q$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and point $M$ internally divides $P Q$ in the ratio $2: 1$. Find the equation of the locus of $M$, and determine where $M$ is located when the area of the triangle formed by $M$ and the two foci $F_{1},... | Given the conditions, the homothetic center is $P(4,3)$, and the homothetic ratio is $\frac{\lambda}{1+\lambda}=\frac{2}{3}$. By property 5 of homothetic transformations, the relevant data of curve $C'$ are the corresponding data of $C$ multiplied by the homothetic ratio. Therefore, the desired trajectory is an ellipse... | N\left(\frac{4}{3}, \frac{7}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,838 |
3.8 Arrange 8 people in a row, where A and B are two of them, the number of different arrangements where there are exactly three people between A and B is $\qquad$ (express the answer as a numerical value).
Translate the above text into English, please keep the original text's line breaks and format, and output the tr... | 3. 5760 | 5760 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,839 |
II. (25 points) Given 14 points within a circle of radius 1. Prove that there must be two points whose distance is less than 0.72. | Given a circle with radius $r=0.35$, draw a smaller concentric circle with the same center. Divide the annular region into 12 equal parts, resulting in 13 regions: one small circle with radius $r$, and 12 congruent curved trapezoids (as shown in the figure).
There are 14 points within the known circle, so at least two... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,842 |
Four, (30 points) Let natural numbers $a, b, c, d$ satisfy $\frac{a}{b}+\frac{c}{d}<1$ and $a+c=20$. Find the maximum value of $\frac{a}{b}+\frac{c}{d}$.
| Given $a, c$, and without loss of generality, assume $a \leqslant c$. Let $b=a+p, d=c+q$, where $p, q \in \mathbb{N}$. We have
$$
\begin{array}{l}
1>\frac{a}{b}+\frac{c}{d}=\frac{a}{a+p}+\frac{c}{c+q} \\
=\frac{a c+a q+a c+c p}{a c+a q+p q+c p} .
\end{array}
$$
This implies $a c < a c$. Therefore, $p q \geqslant a c+1... | \frac{1385}{1386} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,844 |
1. Given $\operatorname{tg} \frac{x}{2}=\sqrt{2}-\sqrt{3}$. Then the value of $\csc x-|\operatorname{ctg} x|$ is ( ).
(A) $\sqrt{2}-\sqrt{3}$
(B) $\sqrt{3}-\sqrt{2}$
(C) $\sqrt{2}+\sqrt{3}$
(D) $-\sqrt{3}-\sqrt{2}$ | 1. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,845 |
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