problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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2. Let $A, B$ be two sets, and set $X$ satisfies: $A \cap X$ $=B \cap X=A \cap B$ and $A \cup B \cup X=A \cup B$. Then the relationship between $X$ and $A, B$ is ( ).
(A) $X \supset A \cap B$
(B) $X \subset A \cap B$
(C) $X=A \cap B$
(D) $X=A \cup B$ | 2. C
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 | 706,846 |
3. Given the lines $l_{1}: a x+b y+c=0, l_{2}: b x+a y+c=0$. If the angle bisector of $l_{1}$ and $l_{2}$ is $y=x$, then ( ).
(A) $a \neq 0$
(B) $c \neq 0$
(C) $a b>0$
(D) $a b \leqslant 0$ | 3. D
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 | 706,847 |
4. Given that the curves $C_{1}$, $C_{2}, C_{3}, C_{4}$ are the graphs of the logarithmic function $y=$ $\log _{\circ} x$, then the corresponding $a$ values for the curves $C_{1}$, $C_{2}, C_{3}, C_{4}$ are ( ).
(A) $3,2, \frac{1}{3}, \frac{i}{2}$
(B) $2,3 \frac{1}{3}, \frac{1}{2}$.
(C) $2,3, \frac{1}{2}, \frac{1}{3}$
... | 4. B
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 | 706,848 |
Example 7. $\triangle A B C$ and $\triangle A D E$ are two non-congruent isosceles right triangles. Now fix $\triangle A B C$, and rotate $\triangle A D E$ around point $A$ on the plane. Prove: No matter what position $\triangle A D E$ rotates to, there must be a point $M$ on segment $E C$ such that $\triangle B M D$ i... | Consider the position of $\triangle A D E$ during the rotation process as shown in the figure. Consider these two similar rotation transformations: $S\left(E, 45^{\circ}, \sqrt{2}\right)$ and $S\left(C, 45^{\circ}, \frac{\sqrt{2}}{2}\right)$. Under the first transformation, point $D$ is transformed to point $A$, and th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,849 |
5. If the function $f(x)$ satisfies for all real numbers: $f(2+x)$ $=f(2-x), f(5-x)=f(5+x)$, then the period of the function $f(x)$ is ( ).
(A) 2
(B) 5
(C) 6
(D) 7 | 5. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,850 |
6. The dihedral angle $\alpha-A B-\beta$ is $60^{\circ}, P \in \alpha, Q \in \beta, P, Q$ are not on $A B$, the angle between $P Q$ and $A B$ is $45^{\circ}$, and $P Q=$ $7 \sqrt{2}, Q$ is 3 units away from $A B$. Then the distance from $P$ to $A B$ is ( ).
(A) 5
(B) $\sqrt{37}$
(C) 8
(D) $\sqrt{79}$ | 6. As shown in the figure, construct $Q T$
$\perp A B$ at $T, Q R / /$ $A B, P R \perp Q R$ at $R$, $R S \perp A B$ at $S$, and connect $P S$. Then $P S \perp A S$.
$$
\begin{array}{l}
\therefore \angle P S R=60^{\circ} . \\
\text { Also } \angle P Q R=45^{\circ}, P Q=7 \sqrt{2}, \\
\therefore P R=\frac{\sqrt{2}}{2} P ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,851 |
2. If the equation $\lg (k x)=2 \lg (x+1)$ has only one real solution, then the range of values for $k$ is $\qquad$ . | 2. The original equation is equivalent to
$$
\left\{\begin{array}{l}
k x>0 \\
x+1>0 \\
k x=(x+1)^{2}
\end{array}\right.
$$
$y_{1}=k x(>0)$ is the part of the line above the $x$-axis, $y_{2}=$ $(x+1)^{2}$ is a parabola with its vertex at $(-1, 0)$ and opening upwards.
Thus, the problem is converted to finding the range... | k=4 \text { or } k<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,853 |
4. Given $0<x<\frac{\pi}{2}, \log _{\sin x} \cos x$ and $\log _{\cos x} \sqrt{\tan x}$ have a sum of their mantissas equal to 1, and their characteristic are both zero. Then the value of $x$ is $\qquad$ . | 4. $\arcsin \left[\frac{(\sqrt{5}-1)}{2}\right]$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,855 |
Three, (15 points) Given: $\frac{\sin ^{2} \gamma}{\sin ^{2} \alpha}=1-\frac{\tan(\alpha-\beta)}{\tan \alpha}$. Prove: $\tan^{2} \gamma=\tan \alpha \cdot \tan \beta$ | $\begin{array}{l}\text { Three, } \because \sin ^{2} \gamma=\sin ^{2} \alpha\left[1-\frac{\tan(\alpha-\beta)}{\tan \alpha}\right] \\ =\sin ^{2} \alpha \cdot \frac{\sin \alpha \cos (\alpha-\beta)-\cos \alpha \sin (\alpha-\beta)}{\sin \alpha \cos (\alpha-\beta)} \\ =\frac{\sin \alpha \sin \beta}{\cos (\alpha-\beta)}, \\ ... | \tan^{2} \gamma = \tan \alpha \cdot \tan \beta | Algebra | proof | Yes | Yes | cn_contest | false | 706,856 |
1. (Senior high school student work) Given the ellipse $C: \frac{(x-1)^{2}}{9}$ $+\frac{(y-2)^{2}}{4}=1$ on which there exist two points symmetric about the line $l_{1} y=2 x+m$, find the range of $n$.
| 1. Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$ be two points on $C$ that are symmetric with respect to $l$, and let $M(\bar{x}, \bar{y})$ be the midpoint of $A B$. Then
$$
\left\{\begin{array}{l}
4\left(x_{1}-1\right)^{2}+9\left(y_{1}-2\right)^{2}=36, \\
4\left(x_{2}-1\right)^{2}+9\left(y_{2}-2\right... | -2 < m < 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,858 |
In the figure, the vertex angle of $\triangle AOB$ varies within $\left(0, \frac{\pi}{2}\right)$, and $PQRS$ is an inscribed square in this sector (as shown in the figure). Try to find the minimum value of $OS$.
Translate the above text into English, please keep the original text's line breaks and format, and output t... | 2. Let $\angle A O B=\theta, O S=l$, then $P(l \cos \theta, 0)$, $S(l \cos \theta, l \sin \theta)$,
$$
\begin{array}{l}
R\left(\sqrt{1-l^{2} \sin ^{2} \theta}, l \sin \theta\right), Q\left(\sqrt{1-l^{2} \sin ^{2} \theta}, 0\right) . \\
\because P S=S R, \therefore l \sin \theta=\sqrt{1-l^{2} \sin ^{2} \theta}-l \cos \t... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,859 |
Six. (15 points) Let the sequence of positive numbers
$a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 5)$ satisfy:
$$
\begin{array}{l}
\frac{1}{a_{1} a_{2} a_{3}}+\frac{1}{a_{2} a_{3} a_{4}}+\cdots+\frac{1}{a_{k-2} a_{k-1} a_{k}} \\
=\frac{(k+1) a_{k-2}}{4 a_{k-1} a_{k}}(k=3,4, \cdots, n) \text { and } a_{4}=4, a_{5}=5 .
\end... | (1) For $k=3$, from the condition $\frac{1}{a_{1} a_{2} a_{3}}=\frac{4 a_{1}}{4 a_{2} a_{3}}$, we get $a_{1}=1$. For $k=4$, from $\frac{1}{a_{1} a_{2} a_{3}}+\frac{1}{a_{2} a_{3} a_{4}}=\frac{5 a_{2}}{4 a_{3} a_{4}}$, and $a_{4}=4$, we have $\frac{1}{a_{2}}+\frac{1}{4 a_{2}}=\frac{5 a_{2}}{16}$, which gives $a_{2}=4$.
... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,861 |
1. In tetrahedron $ABCD$, $AB=2\sqrt{2}$, $S_{\triangle ABC}=6$, $S_{\triangle ABD}=4$, and the dihedral angle between faces $ABC$ and $ABD$ is $\frac{\pi}{4}$. Then the volume of the tetrahedron is ( ).
(A) $2\sqrt{2}$
(B) $3\sqrt{2}$
(C) 3
(D) 4 | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,862 |
3. $A, B, C$ are three points on line $l$, and $A B=B C=5$, and $P$ is a point outside line $l$, $\angle A P B=\frac{\pi}{2}, \angle B P C=\frac{\pi}{4}$. Then the distance from $P$ to line $l$ is $\qquad$ | 3. 2
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,869 |
4. Given $\frac{1}{\cos \theta}-\frac{1}{\sin \theta}=\frac{4}{3}, 0<\theta<\frac{\pi}{2}$. Then $\theta=$ | 4. $\frac{1}{2}\left(\pi-\arcsin \frac{3}{4}\right)$ | \frac{1}{2}\left(\pi-\arcsin \frac{3}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,870 |
Example 1. Through a point $P$ on the base $BC$ of isosceles $\triangle ABC$, draw $PM \parallel CA$ intersecting $AB$ at $M$; draw $PN \parallel BA$ intersecting $AC$ at $N$. Construct the point $P'$ symmetric to $P$ with respect to $MN$. Prove: $P'$ lies on the circumcircle of $\triangle ABC$. (Hangzhou University "H... | From the given information, we have
$$
M P^{\prime}=M P=M B, N P^{\prime}=
$$
$N P=N C$, thus point $M$ is the circumcenter of $\triangle P^{\prime} B P$, and point $N$ is the circumcenter of $\triangle P^{\prime} P C$. Therefore,
$$
\begin{array}{l}
\angle B P^{\prime} P=\frac{1}{2} \angle B M P=\frac{1}{2} \angle B A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,871 |
Three. (20 points) A subset $S$ of the set of rational numbers $Q$ has the following properties:
(1) If $a \in S, b \in S$, then $a+b, a b \in S$;
(2) For each rational number $r, r \in S, -r \in S, r=0$ holds and only one of these is true.
Prove that $S$ consists of all positive rational numbers. | iii) For all positive integers $n$, from $1 \in S$ we get $2=1+1 \in S$. $3=2+1 \in S, \cdots$, hence $n=(n-1)+1 \in S$.
ii) For all positive rational numbers $\frac{n}{m}$ (where $n, m$ are coprime positive integers), $\frac{n}{m} \in S$. Otherwise, by (2) we get $-\frac{n}{m} \in S$. By (1), we can get $-2 \cdot \fra... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,872 |
One, (25 points) Let $f(z)=z^{2}+a z+b$ be a quadratic trinomial in the complex variable $z$ with complex coefficients $a, b$, and for all $z(|z|=1)$, $|f(z)|=1$. Find the values of $a$ and $b$.
保留源文本的换行和格式，直接输出翻译结果。 | Let $a=a_{1}+i a_{2}, b=b_{1}+i b_{2}, a_{1}, a_{2}, b_{1}, b_{2}$
all be real numbers. Since $|z|=1$ implies $|f(z)|=1$, let $z=1$, we get
$\left(1+a_{2}+b_{1}\right)^{2}+\left(a_{2}+b_{2}\right)^{2}=1$.
Let $z=-1$, we get $\left(1-a_{1}+b_{1}\right)^{2}+\left(-a_{2}+b_{2}\right)^{2}=1$.
(1) - (2) gives $a_{1}+a_{1} b... | a=b=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,874 |
II. (25 points) As shown in the figure, on the sides $AB$ and $CD$ of a square $ABCD$ with side length 1, points $M$ and $N$ are taken respectively. $AN$ and $DM$ intersect at $E$, and $BN$ and $CM$ intersect at $F$. Try to find the maximum area of quadrilateral $EMFN$, and indicate the positions of $M$ and $N$ when th... | II. Solution First, consider the right trapezoid XYZU, whose upper and lower bases are $m, n$, and the right leg is 1. It is obvious that $\triangle X Y O$ and $\triangle U Z O$ have equal areas. Let their area be $t$, and the areas of $\triangle X O U$ and $\triangle Y O Z$ be $p, q$, respectively. It is easy to know ... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,875 |
Three. (25 points) Let $f(n)$ denote the number of arrangements of length $n$ (such as 00101, 10100 are of length 5) composed of 0 and 1, where no two 1s are adjacent, with the convention that $f(0)=1$. Try to prove:
(1) $f(n)=f(n-1)+f(n-2), n \geqslant 2$;
(2) $f(4 k+2)$ is divisible by 3, $k \geqslant 0$. | (1) For length 1, the permutations are $\overline{1}, 1$, so $f(1)=2$. For length 2, the permutations are $00, 01, 10, 11$, i.e., $f(2) = 3$. Therefore, $f(2)=f(1)+f(0)$. When $n>2$, divide the permutations of length $n$ into two parts: those ending with 0 and those ending with 01. The number of permutations ending wit... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,876 |
1. $A \subseteq B, A \subseteq C$, and $B=\{0,1,2,3,4\}, C=\{0, 2,4,8\}$. Then the maximum number of subsets of $A$ is ( ).
(A) 1
(B) 2
(C) 4
(D) 8 | 1.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text "1.D" seems to be a label or a choice in a list, and it doesn't require translation as it is already in a form that is commonly used in both Chinese... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,877 |
2. The number of elements in the solution set of the equation $2^{x}+x^{\frac{1}{3}}=0$ is ( .).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,878 |
3. Stretch the abscissa of all points on the graph of the function $y=\sin x$ to twice its original length, keeping the ordinate unchanged, and then translate the graph $\pi$ units in the positive direction of the $x$-axis. The function corresponding to the resulting graph is ( ).
(A) $y=-\cos 2 x$
(B) $y=-\sin 2 x$
(C... | 3. C
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 | 706,879 |
4. The smallest positive period of the function $y=\sin x \cos x-\sqrt{3} \cos ^{2} x-\frac{\sqrt{3}}{2}$ is ( ).
(A) $\frac{\pi}{4}$
(B) $\frac{7}{2}$
(C) $\pi$
(D) $2 \pi$. | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,880 |
5. It is known that after 100 years, cadmium retains $95.76 \%$ of its original mass. Let the remaining mass of a mass of 1 after $x$ years be $y$, then the functional relationship between $x$ and $y$ is ( ).
(A) $y=(0.9576)^{\frac{x}{100}}$
(B) $y=(0.9576)^{100 x}$
(C) $y=\left(\frac{0.9576}{100}\right)^{x}$
(D) $y=1-... | 5. A | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,881 |
Example 1. As shown in the figure, in $\triangle A B C$, $A D \perp B C$ at $D, \angle C A B=45^{\circ}, B D=3, C D=2$. Find the area of $\triangle A B C$. | Solve: Construct $\triangle A E B$ and $\triangle A C F$ symmetric to $\triangle A B D$ and $\triangle A D C$ with respect to $A B$ and $A C$ respectively. Extend $E B$ and $F C$ to intersect at point $G$. In quadrilateral $A F G E$, according to the transformation, we have
$$
A F=A D=A E,
$$
$$
\begin{array}{l}
\angle... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,882 |
Example 2. In parallelogram $A B C D$, $E, F$ are the midpoints of $A D, C D$ respectively, and connect $B F, B E$. Prove: $S_{\triangle A B E}=S_{\triangle B C F}$ | Prove: As shown in the figure, connect $B D$, then $S_{\triangle A B D}=S_{\triangle B C D}$.
$\because E$ is the midpoint of $A D$,
$$
\begin{aligned}
\therefore S_{\triangle B A E} & =S_{\triangle B E D} \\
& =\frac{1}{4} S_{\cap A B C D} .
\end{aligned}
$$
Similarly, we can get $S_{\triangle B C F}=\frac{1}{4} S_{D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,883 |
Example 6. Prove: A quadrilateral with area $S$ and perimeter $p$ can definitely cover a circle with radius $\frac{S}{p}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that this problem is equivalent to the existence of a point inside a quadrilateral that is at a distance greater than or equal to $\frac{S}{p}$ from each side. Let this quadrilateral be $ABCD$.
Using $AB, BC, CD, DA$ as lengths and $\frac{S}{p}$ as width, construct rectangles inward towards the quadrilateral $AB... | null | Geometry | proof | Yes | Yes | cn_contest | false | 706,884 |
2. If $f(x)=a x^{2}+b x(a, b$ are non-zero real constants) has two distinct imaginary roots $x_{1}, x_{2}$, such that $f\left(x_{1}\right)=$ $f\left(x_{2}\right)=c \in R$, then the relationship between $b^{2}+4 a c$ and zero is ( ).
(A) $b^{2}+4 a c>0$
(B) $b^{2}+4 a c=0$
(C) $b^{2}+4 a c<0$
(D) Cannot be determined. | 2. C
From the given, $x_{1}, x_{2}$ are the two imaginary roots of the quadratic equation with real coefficients
$$
a x^{2}+b x-c=0
$$
with the discriminant being less than 0, i.e., $b^{2}+4 a c$ $<0$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,886 |
3. In the rectangular prism $A B C D$
$$
\begin{array}{l}
-A_{1} B_{1} C_{1} D_{1} \text {, } A B_{1} \\
=2 \sqrt{2}, A D_{1}=\sqrt{17} .
\end{array}
$$
Then the range of $A C$ is ( ).
(A) $\sqrt{17}-2 \sqrt{2}<A C$ $<5$
(B) $3<A C<\sqrt{17}+2 \sqrt{2}$
(C) $3<A C<5$
(D) $\sqrt{17}-2 \sqrt{2}<A C<\sqrt{17}+2 \sqrt{2}$ | 3. C
Obviously, $A C=B_{1} D_{1}$, and the necessary and sufficient condition for $\triangle A B_{1} D_{1}$ to be an acute triangle is
$$
\sqrt{A D_{1}^{2}-A B_{1}^{2}}<B_{1} D_{1}<\sqrt{A D_{1}^{2}+A B_{1}^{2}} \text {. }
$$
Thus, $3<A C<5$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,887 |
4. Given the coordinates of points $A\left(a, a^{2}\right), B\left(b, b^{2}\right)(a \neq b)$ satisfy
$$
\begin{array}{l}
a^{2} \sin \theta+a \cos \theta=1, \\
b^{2} \sin \theta+b \cos \theta=1 .
\end{array}
$$
Let the distance from the origin to the line $A B$ be $d$. Then the value of $d$ is suitable for ( ).
(A) $d... | 4. B
It is known from the table that $A, B$ lie on the tangent line $x \cos \theta + y \sin \theta = 1$ of the unit circle passing through the point $(\cos \theta, \sin \theta)$. According to the principle that two points determine a line, the above tangent line is the line $AB$, and it always holds that $d=1$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,888 |
5. In the geometric sequence $\left\{a_{n}\right\}$, $q$ is the common ratio $(0<|q|<1)$. $S_{n}$ is the sum of the first $n$ terms, and $S=\lim _{n \rightarrow \infty} S_{n}$. Which of the following statements is correct? ( ).
(A) $S_{n}=S\left(1-q^{n}\right)$
(B) $a_{n}=S\left(1-q^{n}\right)$
(C) $a_{n}$ is monotonic... | 5. From $S=\frac{a_{1}}{1-q}$, we know $S_{n}=\frac{a_{1}\left(1-q^{n}\right)}{1-q}=S\left(1-q^{n}\right)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,889 |
*6. If $m, n$ are non-negative integers not greater than 6, then $C_{6}^{m} x^{2}+C_{6}^{n} y^{2}=1$ represents ( ) different ellipses.
(A) $P_{7}^{2}$
(B) $P_{6}^{2}$
(C) $C_{4}^{2}$
(D) $P_{4}^{2}$. | 6. D
$C_{6}^{k}$ has four values: $1,6,15,20$. The problem is transformed into the number of permutations of choosing 2 elements from 4 different elements, so there are $P_{4}^{2}$ different ellipses. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,890 |
1. Given the set $N=\{x \mid a+1 \leqslant x<2 a-1\}$ is a subset of the set $M=\{x \mid-2 \leqslant x \leqslant 5\}$. Then the range of values for $a$ is $\qquad$ . | $$
\text { II. 1. }\{a \mid a \leqslant 3\}
$$
$N \subseteq M$ has two cases:
(1) $N=\varnothing$, i.e., $a+1 \geqslant 2 a-1$, which gives $a \leqslant 2$.
(2) $N \neq \varnothing$, then
$$
\left\{\begin{array}{l}
a+1 \geqslant-2, \\
2 a-1 \leqslant 5, \\
a+1<2 a-1
\end{array} \quad \Rightarrow 2<a \leqslant 3 .\right... | a \leqslant 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,891 |
* 2. On the three sides of $\triangle A B C$, take points $P_{1}, P_{2}$, $P_{3}, P_{4}, P_{5}, P_{6}, \cdots$, such that $P_{1}, P_{4}, P_{7}, \cdots$ are on $A C$, $P_{2}, P_{5}, P_{8}, \cdots$ are on $A B$, and $P_{3}, P_{6}, P_{9}, \cdots$ are on $B C$, and $A P_{1}=A P_{2}$, $B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_... | 2. 0
Construct the incircle of $\triangle ABC$. By the property of tangent segments, we can prove that $P_{n+6}=P_{6}$. Therefore, $P_{1994}=P_{2}$, which gives $P_{2} P_{1994}=$ 0. Now, let's handle this using complex numbers.
Set up the complex plane, and let each point's letter represent the complex number at that... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,892 |
3. The line passing through point $M(1,1)$ and the coordinate axes form a triangle with an area of 3. How many such lines exist?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 4
The line passing through point $M(1,1)$ is set as $y=k(x-1) +1$, its intersection points with the coordinate axes are $P(0,1-k), Q\left(1-\frac{1}{k}, 0\right)$. Then the area of $\triangle M P Q$ satisfies
$$
3=\frac{1}{2}|1-k|\left|1-\frac{1}{k}\right| .
$$
This yields two quadratic equations in $k$:
$k^{2}-8 ... | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,893 |
*4. If the function relationship $k=f(n)$ between natural numbers $n$ and $k$ is determined by the equation
$$
2\left(1+9+9^{2}+\cdots+9^{n-1}\right)=k(k+1)
$$
then the expression for $f(n)$ is | 4. $k=f(n)=\frac{1}{2}\left(3^{n}-1\right)$ From the given, we have
$$
k(k+1)=\frac{2\left(9^{n}-1\right)}{9-1}=\frac{3^{n}-1}{2} \cdot \frac{3^{n}+1}{2},
$$
thus $k=f(n)=\frac{1}{2}\left(3^{n}-1\right)$. | k=f(n)=\frac{1}{2}\left(3^{n}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,894 |
Example 7. Given that three rays on a number line cover the entire number line. Prove: It is possible to select two rays from them that can also cover the entire number line. | Proof: Let $l_{1}, l_{2}, l_{3}$ be three half-lines that cover the entire number line.
Assume that any two of the half-lines $l_{1}, l_{2}, l_{3}$ cannot cover the entire number line. Therefore, there must be a point $P_{1}$ not covered by $l_{2}$ and $l_{3}$. Since the three half-lines can cover the entire number li... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 706,895 |
6. There is a batch of parts, with the smallest diameter being $12 \mathrm{~mm}$ and the largest being $12.5 \mathrm{~mm}$. If $x$ parts are randomly selected, there will always be 2 parts with a diameter difference less than $0.01 \mathrm{~mm}$, then the minimum value of $x$ is | 6. 52
Divide a line segment between 12 to $12.5 \mathrm{~mm}$ into $n$ equal parts. If $n+1$ items are taken, then at least 2 items have a diameter difference that falls within the same equal part interval. From $\frac{12.5-12}{n} \leq 0.5$, we get the smallest $n=51$. | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,897 |
Three. (20 points) In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $X$ is the center of the square $A A_{1} B B_{1}$, $Y$ is the center of the square $B B_{1} C_{1} C$, and $Z$ is on the diagonal $B D$ such that $D Z=3 Z B$. Find the area of the section passing through $X, Y, Z$.
Translate the above ... | Three, first find the section. Draw the projections $X', Z'$ of $X, Z$ on the plane $B B_{1} C_{1} C$. The two parallel lines $Z Z'$ and $X X'$ determine a plane. Connect $X Z$ and $X' Z'$, intersecting at $O$, then $O$ is on the plane $B B_{1} C_{1} C$. Connect $O Y$ intersecting $B C$ at $P$ and $B_{1} C_{1}$ at $S$.... | \frac{\sqrt{2}}{2} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,898 |
Four, (20 points) If the hyperbola $y^{2}-x^{2}=1$ and $\frac{x y-x-y+1}{x^{2}-3 x+2}=k$ have a unique common point, find all possible values of $k$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation and is provided for context. The actual translation i... | $$
\text { IV. From } \begin{aligned}
k & =\frac{x y-x-y+1}{x^{2}-3 x+2} \\
& =\frac{(x-1)(y-1)}{(x-1)(x-2)}
\end{aligned}
$$
we know that when $x \neq 1$, we have
$$
y-1=k(x-2) .
$$
Substituting into the hyperbola equation, we get
$$
\begin{array}{l}
\left(1-k^{2}\right) x^{2}-2 k(1-2 k) x-4 k(k-1) \\
=0 .
\end{arra... | 1, -1, 0, \frac{4}{5}, 1+\sqrt{2}, 1-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,899 |
Five. (20 points) The function $f(x)$ defined for $x>0$ satisfies:
(1) There exists $a>1$, such that $f(a) \neq 0$.
(2) For any real number $b$, $f\left(x^{b}\right)=b f(x)$.
Prove that for $x>2$, the inequality
$$
f(x-1) f(x+1)<[f(x)]^{2}
$$
holds. | For $a>1$ such that $f(a) \neq 0$. For $x>2$, we have $f(x)=f\left(a^{\log _{a} x}\right)=f(a) \log _{a} x$.
Similarly, $f(x-1)=f(a) \log _{a}(x-1)$, $f(x+1)=f(a) \log _{a}(x+1)$.
Thus, $f(x-1) f(x+1)$
$$
\begin{array}{l}
=[f(a)]^{2} \log _{a}(x-1) \log _{a}(x+1) \\
<[f(a)]^{2}\left[\frac{\log _{a}(x-1)+\log _{a}(x+1)... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,900 |
* 1. (35 points) In $\triangle ABC$, $AD, BE, CF$ are the altitudes to sides $BC, AC, AB$ respectively. If $AE + AF = BC$, $BD + BF = AC$, and $CD + CE = AB$. Prove that $\triangle ABC$ is an equilateral triangle. | Let's assume $\angle A \geqslant \angle B \geqslant \angle C$.
First, we need to prove that $\triangle A B C$ must be an acute triangle, which only requires proving that $\angle A$ is an acute angle.
(1) If $\angle A$ is a right angle, then $A$ coincides with $E, F$, which contradicts $A E+A F = B C$.
(2) If $\angle A$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,901 |
II. (35 points) A stationery store that operates both wholesale and retail has stipulated: If a customer buys 51 pencils or more (including 51), they will be charged at the wholesale price; if they buy 50 pencils or fewer (including 50), they will be charged at the retail price. The wholesale price for 60 pencils is 1 ... | Get $x=-5+\sqrt{25+600 m}$,
$(25+600 m)$ is a perfect square.
From $40<-5+\sqrt{25+600 m} \leqslant 50$,
we get $3 \frac{1}{3}<m \leqslant 5$.
When $m=4$, $25+600 m$ is not a perfect square, discard it.
When $m=5$, $x=50$ is the solution.
Second, let the class have $x$ students, then the retail price of pencils is $\f... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,902 |
*Three, draw 9 $120^{\circ}$ sectors on the clock face, each covering 4 numbers, with no two sectors covering exactly the same set of numbers. Prove that it is always possible to find 3 sectors that together cover the entire clock face.
Provide a counterexample to show that making 8 sectors does not have the above pro... | Three, prove that 1 takes the first number covered by the constructed sector (all calculated in a clockwise direction) denoted as
$$
a_{1}, a_{2}, \cdots, a_{9} .
$$
Since the numbers covered by each sector are not all the same, the above 9 numbers are distinct. Therefore, among the 12 numbers on the clock face, there... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,903 |
Let $a_{1}, a_{2}, \cdots, a_{n}>0$,
$$
A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}, G_{n}=\sqrt[n]{a_{1} a_{2} \cdots a_{n}} .
$$
Then $A_{n} \geqslant G_{n}$,
with equality if and only if $a_{1}=a_{2}=\cdots=a_{n}$. | Proof 1 Let $a_{1}=\min \left\{a_{1}, a_{2}, \cdots, a_{k+1}\right\}$, $a_{k+1}=\max \left\{a_{1}, a_{2}, \cdots, a_{k+1}\right\}$. Then $a_{1} \leqslant G_{k+1} \leqslant a_{k+1}$, $\left(G_{k+1}-a_{1}\right)\left(G_{k+1}-a_{k+1}\right) \leqslant 0$,
$$
a_{1}+a_{k+1} \geqslant G_{k+1}+\frac{a_{1} a_{k-1}}{G_{k-1}} \te... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,904 |
Example 1. For any natural number $n$, prove the inequality $\frac{n^{2}}{3}+n>(n!)^{\frac{2}{n}}$.
(1977, Kyiv Mathematical Olympiad) | $\begin{array}{l} \text { Prove }(n!)^{\frac{2}{n}} \\ = \sqrt[n]{(n \cdot 1)[(n-1) \cdot 2] \cdots(1 \cdot n)}< \\ \sqrt[n]{\left(\frac{(n+1)+[(n-1)+2]+\cdots+(1+n)}{2 n}\right)^{2 n}} \\ = \frac{n^{2}(n+1)^{2}}{2^{2} n^{2}}=\frac{(n+1)^{2}}{4} \\ <n+\frac{n^{2}}{3} .\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,905 |
Example 2. If $S_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$, prove:
(1) $n(n+1)^{\frac{1}{n}}<n+S_{n}$;
(2) $(n-1) n^{-\frac{1}{n-1}}<n-S_{n}$. | Prove (1) $\frac{n+S_{n}}{n}$
$$
\begin{array}{l}
=\frac{2+\frac{3}{2}+\frac{4}{3}+\cdots+\frac{n+1}{n}}{n} \\
>\sqrt[n]{2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \cdots \cdot \frac{n+1}{n}}=\sqrt[n]{n+1} . \\
\end{array}
$$
Thus, we have $n(n+1)^{\frac{1}{n}} & \sqrt[n-1]{\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,906 |
Example 4. Let $k, n$ be natural numbers, $1<k \leqslant n, x_{1}, x_{2}$, $\cdots, x_{k}$ be $k$ positive numbers, and their sum equals their product.
(1) Prove: $x_{1}^{n-1}+x_{2}^{n-1}+\cdots+x_{k}^{n-1} \geqslant k n$.
(2) Determine the necessary and sufficient conditions for the equality $x_{1}^{n-1}+x_{2}^{n}+\cd... | Let $T=x_{1}+x_{2}+\cdots+x_{k}=x_{1} x_{2} \cdots x_{k}$. Show that $\frac{T}{k} \geqslant \sqrt[k]{T}$.
Thus, $T^{\frac{k}{k}} \geqslant k$,
$$
T^{\frac{1}{k}} \geqslant k^{\frac{1}{k-1}} .
$$
The equality holds if and only if all $x_{i}$ are equal.
$$
\begin{aligned}
\text { Also, } & \frac{x_{1}^{n} 1+x_{2}^{n}+\c... | x_{1}^{n-1}+x_{2}^{n-1}+\cdots+x_{k}^{n-1} \geqslant k n | Inequalities | proof | Yes | Yes | cn_contest | false | 706,908 |
Example 6. Solve the inequality
\[
\begin{array}{l}
\sqrt{\frac{\pi}{4}-\operatorname{arctg} \frac{|x|+|y|}{\pi}}+\operatorname{tg}^{2} x+1 \\
\leqslant 2|\operatorname{tg} x|(\sin x+\cos x) .
\end{array}
\] | Solve: Left side $\geqslant \operatorname{tg}^{2} x+1 \geqslant 2|\operatorname{tg} x|$.
Equality holds if and only if $\operatorname{arctg} \frac{|x|+|y|}{\pi}=\frac{\pi}{4}$, and $|\operatorname{tg} x|=1$, i.e., $|x|+|y|=\pi,|\operatorname{tg} x|=1$.
Right side $=2|\operatorname{tg} x| \sin \left(x+\frac{\pi}{4}\righ... | x=\frac{\pi}{4}, y= \pm \frac{3}{4} \pi | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,910 |
Example 7. Find the sum of all negative roots of the equation $x^{3}-\frac{3}{2} \sqrt[3]{6} x^{2}+3=0$. (Second Hope Cup Mathematics Competition) | To solve the equation with a zero root, the equation can be transformed as follows:
$$
\begin{aligned}
x+\frac{3}{x^{2}} & =\frac{3}{2} \sqrt[3]{6} . \\
x+\frac{3}{x^{2}} & =\frac{x}{2}+\frac{x}{2}+\frac{3}{x^{2}} \\
& \geqslant \sqrt[3]{\frac{3}{4}}=\frac{3}{2} \sqrt[3]{6} .
\end{aligned}
$$
From (1), we know that eq... | -\frac{\sqrt[3]{6}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,911 |
Example 3. As shown in the figure, $A D, B E, C F$ intersect at a point $P$ inside $\triangle A B C$, dividing $\triangle A B C$ into six smaller triangles, with the areas of four of these smaller triangles already given in the figure. Find the area of $\triangle A B C$. | Let the unknown areas of the two smaller triangles be $x$ and $y$, then
$$
\frac{B D}{D C}=\frac{40}{30}=\frac{84+x}{70+y},
$$
i.e., $\frac{84+x}{70+y}=\frac{4}{3}$.
Also, $\frac{A E}{E C}=\frac{70}{y}=\frac{84+x}{40+30}$.
i.e., $\frac{84+x}{70}=\frac{70}{y}$.
Dividing (1) by (2), we get
$$
\frac{70}{70+y}=\frac{4}{3... | 315 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,912 |
Example 8. There is a rectangular sheet of size $80 \times 50$. Now, we need to cut off a square of the same size from each corner and then make it into an open box. What should be the side length $y$ of the square to be cut off so that the volume of this open box is maximized? | Let the side length of the cut-out square be $x$, then the volume of the box made is
$$
V=x(80-2 x)(50-2 x) .
$$
To find the maximum value of $V$, a common approach is to try to make the product a constant, which involves converting $x$ into $4x$.
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x(80-2 x)(50-2 x) \\
& \leq... | 10 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 706,913 |
Example 1. 1000 teachers and students of a school are to visit a place 100 km away from the school. There are five cars available, each capable of carrying 50 people, with a speed of 25 km/h, while the walking speed of a person is 5 km/h. How much time is required for all teachers and students to arrive at the destinat... | Let's assume they walk $3 x$ kilometers throughout the journey. According to the problem, we have
$$
\frac{100-3 x+100-4 x}{25}=\frac{x}{5} .
$$
Thus, $\frac{200-7 x}{5}=x$, which means $3 x=50$.
The total time spent walking and riding is
$$
\frac{100-3 x}{25}+\frac{3 x}{5}=12 \text{. }
$$
Answer: The total time to r... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,914 |
Example 2. Two bus stations, $A$ and $B$, continuously send out a bus at the same intervals, with each bus traveling at the same speed. A cyclist on the road between $A$ and $B$ notices that a bus passes from behind every $a$ minutes and a bus passes from the front every $b$ minutes. How often do stations $A$ and $B$ s... | Let $A, B$ two stations send out a bus every $x$ minutes. According to the problem, we have $\left(\frac{1}{a}+\frac{1}{b}\right) x=2$.
Solving for $x$ gives $x=\frac{2 a b}{a+b}$.
Answer: $A, B$ two stations send out a bus every $\frac{2 a b}{a+b}$ minutes. | \frac{2 a b}{a+b} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,915 |
Example 3. In the figure, the large circle is a 400-meter track, and the track from $A$ to $B$ is 200 meters long. The straight-line distance is 50 meters. A father and son start running counterclockwise from point $A$ along the track for a long-distance run. The son runs the large circle, while the father runs straigh... | Let the number of laps the father and son run when they meet be $N_{1}, N_{2}\left(N_{1}, N_{2}\right.$ are positive integers), and let the distance from point $A$ to the meeting point on the left half-circle be $x$ meters $(0 \leqslant x \leqslant 200)$. According to the problem, we have
$$
\frac{250 N_{1}+x}{100 / 20... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,916 |
Problem (1988, National Training Team Selection Competition)
Let $f(x)=3x+2$, prove: there exists an integer $m$, such that $f^{(100)}(x)$ is divisible by 1988. | Analyzing, we know that
$$
f^{(100)}(x)=3^{100}(x+1)-1 \text { . }
$$
Thus, $f^{(100)}(m)=3^{100}(m+1)-1$ (*)
Since $1988=2 \times 2 \times 497$, we need to find the above factors from equation (*). It is natural to factorize equation (*). Clearly, it would be convenient if $m+1$ is a perfect square, otherwise it woul... | m=45^{20}-1 | Algebra | proof | Yes | Yes | cn_contest | false | 706,917 |
Example 1. (IMO32-1) Given $\triangle ABC$, let $I$ be its incenter, and the internal angle bisectors of $\angle A, \angle B, \angle C$ intersect the opposite sides at $A', B', C'$ respectively. Prove that:
$$
\frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A' \cdot B B' \cdot C C'} \leqslant \frac{8}{27} .
$$ | Let $\frac{A I}{A A^{\prime}}=x, \frac{B I}{B B^{\prime}}=y, \frac{C I}{C C^{\prime}}=z$. By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
A^{\prime} C=\frac{a b}{b+c}, \\
x=\frac{b}{b+A^{\prime} C}=\frac{b+c}{a+b+c}
\end{array}
$$
Similarly, $y=\frac{c+a}{a+b+c}$,
$$
z=\frac{a+b}{a+b+c}
$$
Thus, $x+y+z=2$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,918 |
In $\triangle ABC$, the angle bisector of $\angle A$ intersects the circumcircle at $A_{1}$. Similarly, define $B_{1}$ and $C_{1}$. $AA_{1}$ intersects the external angle bisectors at $D$ and $C$ at $A_{0}$. Similarly, define $B_{0}$ and $C_{0}$. Prove:
(1) $S_{A_{0} B_{0} C_{0}}=2 S_{A C_{1} B A_{1} C B_{1}} ;$
(2) $S... | Let $I$ be the incenter of $\triangle ABC$, then $I$ is the intersection point of $AA_{0}, BB_{0}, CC_{0}$,
$$
\begin{aligned}
& \angle B I A_{1}=\alpha+\beta, \\
& \angle A_{1} B I \\
& =\angle \beta+\angle A_{1} B C \\
& =\beta+a .
\end{aligned}
$$
Then $A_{1} B=A_{1} I$, it is easy to see that $B B_{1} \perp B A_{0... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,919 |
Example 3. (IMO32-5) Let $P$ be a point inside $\triangle A B C$. Prove that at least one of $\angle P A B, \angle P B C, \angle P C A$ is less than or equal to $30^{\circ}$. | $$
\begin{array}{l}
\text { Proof } \quad \because P A \sin \alpha= \\
P G=P B \sin (B-\beta), \\
P R \sin \beta \\
=P C \sin (C-\gamma), \\
P C \sin \gamma \\
=P A \sin (A-\alpha), \\
\quad \therefore \sin \alpha \cdot \sin \beta \cdot \sin \gamma \\
\quad=\sin (A-\alpha) \sin (B-\beta) \cdot \sin (C-\gamma)
\end{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,920 |
Example 4. (IMO30-4) A convex quadrilateral $ABCD$ has the property: (1) $AB = AD + BC$, (2) there is a point $P$ inside it, the distance from $P$ to $CD$ is $h$, and it satisfies $AP = h + AD$, $BP = h + BC$. Prove:
$$
\frac{1}{\sqrt{h}} \geqslant -\frac{1}{\sqrt{AD}} + \frac{1}{\sqrt{BC}}.
$$ | Prove: As shown in the figure, circles are drawn with $A, B, P$ as centers and $AD = r, EC = R, h$ as radii, respectively. A special case is the right trapezoid $ABEF$, where the distance from $P$ to $EF$ is $h' \geq h$.
$$
h + h' = HP + PG \leq 2r,
$$
Thus, at least one of $h$ and $h'$ is less than or equal to $r$, a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,921 |
Example 1. (47th Putnam Competition) Find and prove the maximum value of $f(x)=x^{3}-3 x$, where $x$ is a real number satisfying $x^{4}+36 \leqslant 13 x^{2}$.
保留源文本的换行和格式，翻译结果如下：
Example 1. (47th Putnam Competition) Find and prove the maximum value of $f(x)=x^{3}-3 x$, where $x$ is a real number satisfying $x^{4}+36... | From $x^{4}+36 \leqslant 13 x^{2}$, we can get
$$
-3 \leqslant x \leqslant-2,2 \leqslant x \leqslant 3 \text {. }
$$
Let $x_{i}3$. Then,
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right) \\
=\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}-3\right)<0, \\
f\left(x_{1}\right)<f\left(x_{2}\right) .... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,922 |
Example 2. (1979, Shanghai Competition Question) Can the sum of the areas of $n$ squares with side lengths of $1, 3, 5, 7$, $\cdots, 2 n-1$ cm be equal to $10^{6} \mathrm{~cm}^{2}$? | If possible, then
\[
\begin{aligned}
10^{6} & =1^{2}+3^{2}+\cdots+(2 n-1)^{2} \\
& =\sum_{k=1}^{2 n} k^{2}-\sum_{k=1}^{n}(2 k)^{2}=\frac{1}{3} n\left(4 n^{2}-1\right),
\end{aligned}
\]
i.e., \(4 n^{3}-n=3000000\).
It is easy to see that \(4 n^{3}-n\) is monotonically increasing. From \(4 n^{3}>3000000\), we know \(n>9... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,924 |
In the tetrahedron $S-ABC$, the lateral edges $SA, SB, SC$ are pairwise perpendicular, $M$ is the centroid of $\triangle ABC$, $D$ is the midpoint of $AB$, and $DP \parallel SC$. Prove:
(1) $DP$ intersects $SM$;
(2) The intersection point $D'$ of $DP$ and $SM$ is the center of the circumscribed sphere of $S-ABC$.
Tran... | As shown in the figure, we can set up a coordinate system with $A(a, 0,0), B(0, b, 0), C(0,0, c)(a, b, c > 0)$, then $D\left(\frac{a}{2}, \frac{b}{2}, 0\right)$, $M\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$. Since $D P / / S C$, its equation is
$$
\left\{\begin{array}{l}
x=\frac{a}{2}, \\
y=\frac{b}{2} .
\end{... | null | Geometry | proof | Yes | Yes | cn_contest | false | 706,926 |
3. Given $\triangle A B C$ with $O_{1}, O_{2}, O_{3}$ as the centers of the excircles. Prove: $\triangle O_{1} O_{2} O_{3}$ is an acute triangle. | 3. It is known that the three angle bisectors of $\triangle ABC$ are the three altitudes of $\triangle O_{1} O_{2} O_{3}$, and the incenter $O$ of $\triangle ABC$ is the orthocenter of $\triangle O_{1} O_{2} O_{3}$. Since $O$ is inside $\triangle ABC$, $O$ is also inside $\triangle O_{1} O_{2} O_{3}$. Therefore, $\tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,929 |
4. Let $P(x, y)$ be a point on $|5 x+y|+|5 x-y|=$ 20. Find the maximum and minimum values of $x^{2}-x y+y^{2}$. | 4. After simplifying and discussing the size of $|5 x|$ and $|y|$ from the equation $|5 x+y|+|5 x-y|=20$, we easily get $x= \pm 2$, $|y| \leqslant 10$ or $|x| \leqslant 2, y= \pm 10$. Therefore, the graph of the equation is a rectangle $A B C D$, where $A(2,-10), B(2,10), C$ $(-2,10), D)(-2,-10)$. By symmetry, we only ... | Q_{\text {max }}=124, Q_{\text {min }}=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,930 |
5. For what positive integer $n$ is $20^{n}+16^{n}-3^{n}-1$ divisible by 323? | 5. Let $a \mid b$ denote $a$ divides $b$. Since for polynomials $a-b \mid a^{k}-b^{k}, a+b \mid a^{2 k-1}+b^{2 k-1}, k \in N$, we have:
(1) When $2 \mid n$, $17=20-3 \mid 20^{n}-3^{n}$,
$17 \mid 255=256-1 \mid 256^{\frac{n}{2}}-1=16^{n}-1$.
Thus, $17 \mid A$;
Similarly, $19 \mid 20^n - 1$,
$19 \mid 247 \mid 256^{\frac{... | any positive even number | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,931 |
1. $p$ is an integer, prove that $x^{2}-2 x-\left(10 p^{2}+\right.$ $10 p+2)=0$ has no integer solutions. | 1. From $x(x-2)=2[5 p(p+1)+1)$ we know that $x$ is even, the left side should be a multiple of 4, but it is not. Hence, this is a contradiction. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,932 |
2. Does there exist a perfect square, the sum of whose digits is $1993 ?$ | 2. Exist. If the number is of the form $99 \cdots 99$ representing $n$ nines, then
$$
\overbrace{29 \cdots 997^{2}}^{\lambda}=\overbrace{99 \cdots 994}^{\lambda} \overbrace{00 \cdots 009}^{\lambda} .
$$
Now, if $n=220$, then it holds. (Similarly, $33 \cdots 331^{2}=\overbrace{1!\cdots!1}^{n}$
$\frac{\lambda}{3}, 55 b$... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,933 |
4. On the sides $AB, AD$ of square $ABCD$, take points $K, N$ such that $AK \cdot AN = 2 BK \cdot DN$. Segments $CK, CN$ intersect the diagonal $BD$ at $L, M$. Prove that: $\angle B L K = \angle D N C = \angle B A M$ | 4. Let $A B=u, B K$
$=b, D N=c$, then
$$
\begin{array}{l}
(a-b)(a-c)=2 b c, \\
a^{2}-b c=a(b+c) .
\end{array}
$$
and $\operatorname{tan}(\angle B C K+\angle D C N)$
$$
\begin{array}{l}
=\frac{\frac{b}{a}+\frac{c}{a}}{1-\frac{b}{a} \cdot \frac{c}{a}}=\frac{a(b+c)}{a^{2}-b c}=1, \\
\angle B C K+\angle D C N=45^{\circ},
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,936 |
5. Let necklace $A$ have 14 beads, and $B$ have 19 beads. For odd $n \geqslant 1$, use $n, n+1, n+2, \cdots, n+32$ to number these 33 beads, such that each integer is used exactly once, and the numbers on adjacent beads are coprime (here “necklace” is circular, and each bead is adjacent to 2 other beads). Prove that th... | 5. Label the pearls of $A$ in the order $n+k, n+k+1, \cdots, n+k+13$, and the pearls of $B$ in the order $n+k+14, \cdots, n+32, n, n+1, \cdots, n+k-1$, where $1 \leqslant k \leqslant$ 18. This method is feasible if and only if
$$
\begin{array}{l}
(n+k, n+k+13)=(n+32, n) \\
=(n+k-1, n+k+14)=1,
\end{array}
$$
i.e., $(n+... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,937 |
1. Do there exist integers $m, n$ satisfying $m^{2}+1954$ $=n^{2}$ ? | 1. Does not exist. Since $1954=(n+m)(n-m)$, the right side is odd or a multiple of 4, not equal to the left. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,938 |
3. In an acute triangle $\triangle ABC$, take a point $D$ on the altitude $AH$ from $A$ to $BC$, such that $D$ lies between $A$ and $H$. Draw $BD$ and $CD$ and extend them to meet $AC$ at $E$ and $AB$ at $F$ respectively. Draw $EH$ and $FH$. Prove: $\angle AHE = \angle AHF$. | 3. Taking $H$ as the origin and $HC$ as the $x$-axis to establish a Cartesian coordinate system, and let $A$ $\left(0, a^{-1}\right), B\left(-b^{-1}\right.$, $0), C\left(c^{-1}, 0\right), D(0$, $d^{-1}$ ). It is easy to see that the coordinates of $E, F$ are the solutions to the systems of equations
$$
\left\{\begin{ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,940 |
4. The positive sequence $\left\{a_{n}\right\}$ has $a_{1}=a_{2}=1, a_{3}=$ $997, a_{n+3}=\frac{1993+a_{n+2} a_{n+1}}{a_{n}}$. Prove that all $a_{n}$ are integers. | 4. Let's supplement $a_{0}=2$. From
$$
a_{n+3} a_{n}=1993+a_{n+2} a_{n+1} \text {, }
$$
by replacing $n$ with $n+1$ and subtracting, then simplifying, we get
$$
\frac{a_{n+1}+a_{n+2}}{a_{n+2}+a_{n}}=\frac{a_{n+3}}{a_{n+1}} .
$$
For this equation, let $n=0,2, \cdots, 2 m-2$, and multiply them to get
$$
\frac{a_{2 m-2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,941 |
5. The license plate numbers issued by a city consist of 6 digits (from 0 to 9), but it is required that any 2 license plates must have at least 2 different digits (for example, license plates 038471 and 030471 cannot be used simultaneously). Try to find the maximum number of different license plates that the city can ... | 5. The answer is 100000. If 100001 license plates are issued, then by the pigeonhole principle, at least 10001 numbers will have the same first digit, and similarly, at least 1001 numbers will have the same second digit, $\cdots$, at least 2 numbers will have the same fifth digit, which would violate the rule. Using th... | 100000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,942 |
1. Let $\left\{a_{k}\right\}$ be a sequence of positive numbers, such that for any $k$ we have $\left(a_{k+1}+k\right) - a_{k}=1$. Prove: all terms in the sequence are irrational numbers. | 1. Suppose a term in the sequence is $a_{k}=\frac{p}{q}$, where $p$ and $q$ are natural numbers, then it is not hard to see that the next term is $\frac{q-k p}{p}$. Naturally, the sum of the numerator and denominator of this fraction decreases. Thus, after several steps, either the numerator or the denominator will fir... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,943 |
2. In $\triangle A B C$, $A B$ $=B C$, points $D, E$ and $F$ are taken on sides $A C, A B$ and $B C$ respectively, such that $D E=D F$. At this point, $A E+F C=A C$. Prove: $\angle B A C=\angle F D E$.
| 2. Consider point $D^{\prime}$ on the base $A C$, such that $A I^{\prime} \cdots$ $F C, D^{\prime} C=A E$. Therefore, by the "side-angle-side" congruence, $\triangle A E D^{\prime} \cong \triangle F C D^{\prime}$, so $E D^{\prime}=D^{\prime} F$, which means $D^{\prime}$ coincides with $D$. Consequently, it is known tha... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,944 |
Example 1. Suppose there is a coil made of a single wire on the desktop, with a perimeter of $2 l$. We also have a circular paper piece with a diameter of $l$. It is clear:
(1) When the coil is formed into a parallelogram, we can completely cover it with the said circular paper piece;
(2) Regardless of the shape of the... | (1) Let the parallelogram be $\triangle A B C D$, with diagonals $A C$ and $B D$ intersecting at $O$. By the given condition,
$$
A B+B C+C D+D A=2 l,
$$
thus $A B+B C=l$.
Since $A C<A B+B C=l$, it follows that $O C<\frac{1}{2} l$,
and similarly, $O B<\frac{1}{2} l$.
Therefore, $A, B, C, D$ are all within the circle ce... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,945 |
3. In a $6 \times 7$ rectangular grid, the small squares in the 4th column are filled with the numbers $1,2,3,4,5,7$. Can the remaining small squares be filled with a number so that the 6 seven-digit numbers formed by the numbers in each row form an arithmetic sequence, and the 7 six-digit numbers formed by the numbers... | 3. Assuming it can be filled successfully. Since the 6 seven-digit numbers formed by the rows of digits form an arithmetic sequence, their sum is divisible by 3. Since the sum of all the numbers in the table has the same remainder when divided by 3 as this sum, the sum of all the numbers in the table is also divisible ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,946 |
4. Prove: It is impossible to divide a square into isosceles triangles with vertex angles of $10^{\circ}$. | 4. Assuming the division is successful. Then, in each of the divided triangles, there is a $10^{\circ}$ "small angle" and two $85^{\circ}$ "large angles". Therefore, among all the interior angles of the divided triangles, the "small angles" account for exactly $\frac{1}{3}$. Now, let's examine the vertices of these tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,947 |
6. 93 volleyball teams participate in a round-robin tournament. It is known that among any 19 teams, there is one team that has won against all the other 18 teams, and there is one team that has lost to all the other 18 teams. Prove that all 93 teams have different scores. | 6. Suppose there are two teams $A$ and $B$ with the same score, and $A$ has defeated $B$. Thus, it is easy to prove that there exists another team $C$, which has defeated $A$ but lost to $B$. Let's examine these 3 teams and the other 16 teams. According to the problem, among these 19 teams, there is a "last" team that ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,949 |
7. A natural number is written on the blackboard. Every second, it is increased by the sum of its even digits (i.e., the sum of the tens, hundreds, etc., digits). Prove that sooner or later the number on the blackboard will no longer change.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 7. Let the number on the blackboard be approximately an $n$-digit number, and $n$ is even. This ensures that the number will not change indefinitely.
$n=2$ is obvious.
Assume the statement holds for $n=k-2$ $(k \geqslant 4)$, we need to prove that the statement also holds for $n=k$. Suppose at some point, the number on... | null | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 706,950 |
8. Mark 4 points inside a convex quadrilateral. Prove that there exists a point on the perimeter of the quadrilateral such that the sum of its distances to the vertices of the quadrilateral is greater than the sum of its distances to the marked points. | 8. The 4 marked points may form the 4 vertices of a certain convex quadrilateral, or 3 of them may form the vertices of a triangle with the other point inside the triangle (which could be a degenerate case). For the sake of simplicity, we only discuss the first scenario (as shown in the right figure).
Let points $M, N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,951 |
10. Try to determine whether it is possible to fill each small square of a $10 \times 10$ grid with a non-zero digit, such that the 10 ten-digit numbers formed by the rows are all greater than the ten-digit number formed by the main diagonal, and this ten-digit number is also greater than the 10 ten-digit numbers form... | 10. Answer: No. It is only necessary to use the fact that "the two-digit number formed by each row is greater than the two-digit number formed by each column". Let the number in the top-left corner be $a$, the number to its right be $b$, and the number below it be $c$ (as shown in the figure below).
\begin{tabular}{|l|... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,952 |
11. Given a square $A B C D$ in the plane, find the minimum value of the ratio $\frac{O A+O C}{O B+O D}$, where $O$ is any point in the plane. | 11. First, prove $\frac{O A+O C}{O B+O D} \geqslant \frac{1}{\sqrt{2}}$. Squaring both sides of the inequality and eliminating the denominator, we get
$$
\begin{array}{l}
2\left(O A^{2}+O C^{2}+2 O A \cdot O C\right) \\
\geqslant O B^{2}+O D^{2}+2 O B \cdot O D .
\end{array}
$$
Using the well-known fact $O A^{2}+O C^{... | \frac{1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,953 |
13. Punch out 1933 points on a piece of paper, some of which are connected by lines. After that, each player must place a chess piece on a point adjacent to the previous one (two points connected by a line segment are called adjacent). The player who cannot place a chess piece loses. Prove that the first player can win... | 13. Let the first person be referred to as 甲, and the second person as 乙. Assuming that for any first move by 甲, placing a chess piece at a certain point \( A \), 乙 has a corresponding winning strategy — placing the piece at the corresponding point \( B \). We examine another scenario:
If 甲's first move is to place th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,955 |
Example 2. There are $n$ points on a plane, where any three points can be covered by a circle of radius 1, but there are always three points that cannot be covered by any circle of radius less than 1. Find the minimum radius of a circle that can cover all $n$ points. | We prove that the radius of the smallest circle covering $n$ points is 1.
Since any three points can be covered by a circle with a radius of 1, the distance between any two points is no more than 2.
Thus, a circle with a radius of 2 centered at any one of the points can cover all the points. This implies that there e... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,956 |
14. Real numbers $a_{1}, a_{2}, \cdots, a_{n}$ are all in the interval $[-1,1]$. Prove that
$$
\text { Prove: } \sum_{i=1}^{n} \frac{1}{1+a_{i} a_{i+1}} \geqslant \sum_{i=1}^{n} \frac{1}{1+a_{i}^{2}} \cdot\left(a_{n+1} \equiv a_{1}\right)
$$ | 14. For any positive numbers $x$ and $y$ not greater than 1, we have
$$
\frac{2}{1+x y} \geqslant \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}} \text {. }
$$
$$
\frac{2}{1+a_{i} a_{i+1}} \geqslant \frac{1}{1+a_{i}^{2}}+\frac{1}{1+a_{i+1}^{2}}
$$
Adding the inequalities together yields the result. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,957 |
15. Place a number on each vertex of a regular $n$-gon, with $n-1$ zeros and one 1. It is allowed to add 1 to all the numbers on the vertices of a sub-$k$-gon. Can such operations make all the numbers on the $n$ vertices equal? | 15. No. Let the center of the regular $n$-gon be denoted as $O$, and the number placed on the $i$-th vertex $A_{i}$ be denoted as $a_{i}$, and let
$$
\vec{S}=\sum_{i=1}^{n} a_{i} \overrightarrow{O A}_{i} .
$$
It is easy to see that under any valid operation, the vector $\vec{S}$ remains unchanged. Initially, we have $... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,958 |
16. There are $2 p+1$ balls in two jars. Every second, half of the balls from the jar containing an even number of balls are moved to the other jar. Let $k$ be a natural number less than $2 p+1$, and let $p$ and $2 p+1$ both be prime numbers. Prove that, sooner or later, one of the jars will contain exactly $k$ balls. | 16. Operations are performed modulo $2 p+1$. Let the first jar contain $x$ balls, and the second jar contain $y$ balls. Thus, every second, the pair $(x, y)$ transforms into the pair $\left(\frac{x}{2}, \frac{y}{2}\right)$ (considering only modulo $2 p+1$, so $y = -x$). Let $m$ be the smallest natural number such that ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,959 |
21. Prove that for any positive numbers $a_{k}, b_{k}, k=1,2, \cdots, n$, the inequality
$$
\sum_{k=1}^{\infty} \frac{a_{k} b_{k}}{a_{k}+b_{k}} \leqslant \frac{A B}{A+B},
$$
holds, where $A=\sum_{k=1}^{n} a_{k}, B=\sum_{k=1}^{n} b_{k}$. | 21. Induction on $n$. When $n=2$, the inequality
$$
\frac{a_{1} b_{1}}{a_{1}+b_{1}}+\frac{a_{2} b_{2}}{a_{2}+b_{2}} \leqslant \frac{\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)}{a_{1}+a_{2}+b_{1}+b_{2}}
$$
is equivalent to the inequality
$$
\begin{array}{l}
{\left[a_{1} b_{1}\left(a_{2}+b_{2}\right)+a_{2} b_{2}\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,963 |
23. Along the perimeter of each face of a convex polyhedron, there is a fly traveling (as many faces as there are, there are that many flies traveling). It is known that at any time their speed is no less than 1 millimeter/hour. Prove that sooner or later, at least two flies will meet. | 23. On the $i$-th face, mark a point $A_{i}(i=1,2 \cdots$, as the "center" of the face, and connect these points to form a graph denoted as $G$.
At any fixed moment $t$, label the "edges" of graph $G$ with arrows according to the following rule: If the $i$-th fly is crawling along the common edge of the $i$-th face an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,965 |
24. On a $1993 \times 1993$ grid paper, mark two small squares $A$ and $B$ on the same edge, with an odd number of small squares between them. Now cover the grid paper with $1 \times 2$ rectangles, except for one small square. Prove that the number of ways to cover the grid without covering $A$ is equal to the number o... | 24. Color the grid paper like an international chessboard, that is, alternately color the small squares black and white, and assume that the four corner small squares are all colored black. At this point, it can be considered that the two small squares $A$ and $B$ are both colored black, because if they were both white... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,966 |
Example 3. Given a convex pentagon where all interior angles are obtuse. Prove: it is possible to find two diagonals of this pentagon such that the two circular paper pieces with these diagonals as diameters can cover this pentagon.
---
The translation maintains the original text's line breaks and format. | Prove that $\odot O_{1}$ and $\odot O_{2}$ can cover the convex pentagon $A B C D E$.
Because $\angle B>90^{\circ}$, $\triangle A B C$ can be covered by $\odot O_{1}$.
Similarly, $\triangle A E D$ can be covered by $\odot O_{2}$.
In $\triangle A C D$, if $\angle A D C$ is an obtuse or right angle, then $\triangle A C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,967 |
1. If $0<a<1$, then $\sqrt{a^{2}+\frac{1}{a^{2}}-2} \div\left(1+\frac{1}{a}\right) \times$ $\frac{1}{1+a}$ can be simplified to ( ).
(A) $\frac{1-a}{1+a}$
(B) $\frac{a-1}{a+1}$
(C) $1-a^{2}$
(D) $a^{2}-1$ | $\begin{array}{l}\text { 1. } \because \sqrt{\left(a-\frac{1}{a}\right)^{2}}=\left(\frac{1}{a}-a\right)=\frac{1-a^{2}}{a}, \\ \therefore \text { original expression }=\frac{1-a^{2}}{a} \times \frac{a}{a+1} \times \frac{1}{1+a}=\frac{1-a}{1+a} .\end{array}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,968 |
3. As shown in the figure, a semicircle $O$ is inscribed in the trapezoid $ABCD$ on the base $AB$, and is tangent to the other two sides $BC$, $CD$, $DA$. If $BC=2$, $DA=3$, then the length of $AB$ ( ).
(A) is 4
(B) is 5
(C) is 6
(D) cannot be determined | 3. Connect $O C, O D$. Let the radius of the semicircle $O$ be $r$, then in $\triangle A O D$, the distances on sides $A O$ and $D A$ are both $r$, so $A O=D A$. Similarly, $B O=B C$. Therefore, $A B=B C+D A=5$.
| B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,970 |
4. When $x=\frac{1+\sqrt{1994}}{2}$, the value of the polynomial $\left(4 x^{3}-1997 x-\right.$ $1994)^{2001}$ is ( ).
(A) 1
(B) -1
(C) $2^{2001}$
(D) $-2^{200 \mathrm{i}}$ | $$
\begin{array}{l}
\text{Since } x=\frac{1+v^{\prime} 1994}{2}, \text{ then } (2, r-1)^{2}=1994, \text{ i.e., } 4 x^{2}-4 x-1993=0. \text{ Thus,} \\
\left(4 x^{3}-1997 x-1994\right)^{2001}: \\
=\left(\left(4 x^{2}-4 x-1993\right) x+\left(4 x^{2}-4 x-1993\right)\right. \\
-1)^{2001} \\
=(-1)^{2001}=-1 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,971 |
5. If the parallel lines $E F, M N$ intersect with the intersecting lines $A B, C D$ to form the figure shown, then the number of pairs of alternate interior angles is ( ).
(A) 4
(B) 8
(C) 12
(D) 16 | 5. Since each "three-line eight-angle" basic figure contains two pairs of consecutive interior angles, and from the given figure, the following 8 basic figures can be decomposed, there are a total of 16 pairs of consecutive interior angles. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,972 |
7. Let the altitudes $A D, B E, C F$ of the acute triangle $\triangle A B C$ intersect at $H$. If $B C=a, A C=b, A B=c$. Then the value of $A H \cdot A D+B H$ - $B E+C H \cdot C F$ is ( ).
(A) $\frac{1}{2}(a b+b c+c a)$
(B) $\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)$
(C) $\frac{2}{3}(a b+b c+c a)$
(D) $\frac{2}{3}\left... | 7. From the given, we know that points $H, D, C, E$ are concyclic, therefore, we have
$$
\begin{array}{l}
A D \cdot A H=A C \cdot A E \\
=A C \cdot A B \\
\quad \cdot \cos \angle B A E \\
=\frac{1}{2}\left(A C^{2}+A B^{2}-B C^{2}\right)=\frac{1}{2}\left(b^{2}+c^{2}-a^{2}\right) .
\end{array}
$$
Similarly, $B H \cdot B... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,974 |
Example 4. Using the four sides of a known quadrilateral as diameters, draw four semicircles inside the quadrilateral respectively. Prove:
(1) Any point inside the quadrilateral is covered by at least one semicircle.
(2) There is at most one point inside the quadrilateral that is covered by all four semicircles. | (1) Let $O$ be any point inside quadrilateral $ABCD$, and connect $OA, OB, OC, OD$. Since
$$
\begin{array}{c}
\angle AOB + \angle BOC \\
+ \angle COD + \angle DOA = 360^{\circ},
\end{array}
$$
then at least one of $\angle AOB, \angle BOC, \angle COD, \angle DOA$ is not less than $90^{\circ}$. Assume $\angle BOC \geqsl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,978 |
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