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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class | __index_level_0__ int64 0 742k |
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30. For any set $S$, let $|S|$ denote the number of elements in $S$, and let $n(S)$ denote the number of subsets of $S$ (including the empty set and $S$ itself). If sets $A, B, C$ satisfy:
$$
n(A)+n(B)+n(C)=n(A \cup B \cup C),
$$
and $|A|=|B|=100$, then the minimum possible value of $|A \cap B \cap C|$ is ( ).
(A) 96.... | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 705,044 |
3. The decimal representation of the natural number $A$ is $\overline{a_{n} a_{n-1} \cdots a_{0}}$. Let $f(A)=2^{\mathrm{n}} a_{0}+2^{\mathrm{n}-1} a_{1}+\cdots+2 a_{\mathrm{n}-1}+a_{\mathrm{n}}$, and denote $A_{1}=f(A), A_{1+1}=f\left(A_{1}\right) . \quad(i=1,2$, $\cdots, n$ ) Prove:
i) There must be a $k \in \mathbb{... | Proof (i) When $n=0$. For any $k$, we have $A_{\mathrm{s}}=A$.
$$
\begin{array}{l}
\text { When } n=1 \text {, } \\
A-f(A)-10 a_{1}+a_{0}-2 a_{0}-a_{1} \\
=9 a_{1}-a_{0} \geqslant 9 \times 1-9=0 . \\
\text { When } n \geqslant 2 \text {, } \\
A-f(A) \\
\geqslant 10^{\mathrm{n}}-\left(2^{\mathrm{n}}+2^{\mathrm{n}+1}+\cd... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false | 705,048 |
4. Given $\triangle A B C, \angle C=90^{\circ}$. Prove that for any $n$ points inside $\triangle A B C$, it is always possible to label them as $P_{1}$, $P_{2}, \cdots, P_{\mathrm{n}}$, such that
$$
P_{1} P_{2}^{2}+P_{2} P_{3}^{2}+\cdots+P_{\mathrm{n}-1} P_{\mathrm{n}}^{2} \leqslant A B^{2} .
$$
Note: The proposition ... | Prove the strengthened proposition using mathematical induction.
When $n=1$, $\because \angle A P_{1} B \geqslant 90^{\circ}$,
$\therefore A P_{1}^{2}+P_{1} B^{2} \leqslant A B^{2}$, the proposition holds.
Assume the proposition holds for $n<k$, we will prove that the conclusion also holds for $n=k$.
Draw a perpendicul... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,049 |
5. Square $A B C D$ has a side length of $1, A B, A D$ each have a point $P, Q$. If the perimeter of $\triangle A P Q$ is 2, find the degree measure of $\angle P C Q$.
---
Note: The translation preserves the original text's formatting and structure. | Let $\angle B C P=\alpha, \angle D C Q=\beta$.
$$
\begin{aligned}
\because B P & =\tan \alpha, Q D=\tan \beta, \\
A P & =1-\tan \alpha, A Q=1-\tan \beta . \\
\therefore P Q & =2-A P-A Q=\tan \alpha+\tan \beta .
\end{aligned}
$$
From $A P^{2}+A Q^{2}=P Q^{2}$, we know
$$
(1-\tan \alpha)^{2}+(1-\tan \beta)^{2}=(\tan \al... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,050 |
6. Given tetrahedron $ABCD$, points $E, F, G$ are on edges $AB, AC, AD$ respectively. Denote the area of $\triangle XYZ$ as $S_{\triangle XYZ}$ and the perimeter as $p_{\triangle XYZ}$. Prove:
i) $S_{\triangle EFG} \leqslant \max \left\{S_{\triangle ABC}, S_{\triangle ABD}, S_{\triangle ACD}, S_{\triangle BCD}\right\}$... | Proof: Let the plane containing $\triangle E F G$ be $\pi$. Without loss of generality, assume the distance from point $B$ to $\pi$ is small. Then, when $B$ does not coincide with $E$, $G^{\prime}$ lies on $A D$. Thus, we have:
$$
E F\left\|B F^{\prime}, E G\right\| B G^{\prime}, F G \| F^{\prime} G^{\prime} \text {. }... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,051 |
8. Label $4k$ points on a circle arbitrarily with $1, 2, \cdots, 4k$. Prove:
i) It is possible to connect these $4k$ points with $2k$ non-intersecting chords such that the difference between the labels at the ends of each chord does not exceed $3k-1$,
ii) For any natural number $k$, the number $3k-1$ in i) cannot be re... | Proof $\mathrm{i})$ Divide the numbers $1,2, \cdots,-k$ into
A, Braid.
$$
\begin{array}{l}
A=\{1,2, \cdots, k, 3 k+1,3 k+2, \cdots, 4 k\} ; \\
B=\{k+1, k+2, \cdots, 3 k\} .
\end{array}
$$
Then the absolute value of the difference between any number in $A$ and any number in $B$ is no greater than $3 k-1$. Therefore, it... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,053 |
In the May 1989 issue of this journal, there was a problem from the Zhejiang Province High School Mathematics Competition as follows:
Arrange the positive rational numbers: $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \cdots$, then t... | The original answer is $(1+2+\cdots+3937)+1949$ $=7753902$.
In fact, the original sequence is formed by segments of fractions with the sum of the numerator and denominator being 2, 3, 4, ․ respectively. The fraction 1989/1949 has a sum of numerator and denominator of 3938, making it the 1949th number in the 3937th seg... | 7749965 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,054 |
Let $m, n, p$ be positive real numbers, and $m^{2}+n^{2}-p^{2}=0$, find the minimum value of $\frac{p}{m+n}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Given positive real numbers $m, n$, $p$ satisfy $m^{2}+n^{2}=p^{2}$, then $m+n=\sqrt{(m+n)^{2}}>\sqrt{m^{2}+n^{2}}=p$, so the segments of length $m, n$, $p$ can form a right triangle $\triangle A B C: A C=m$, $B C=n, A B=p$.
That is, $\frac{m+n}{p}=\cos A+\sin A \leqslant \sqrt{2}$, $\frac{p}{m+n} \geqslant \frac{\sqr... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,055 |
On a plane, there is a convex quadrilateral $A B C D$.
(1) If there exists a point $P$ on the plane such that the areas of $\triangle A B P, \triangle B C P, \triangle C D P, \triangle D A P$ are all equal, what condition should the quadrilateral $A B C D$ satisfy?
(2) How many such points $P$ can there be on the plane... | . Solving (i) First consider the case where $P$ is inside the quadrilateral.
If points $A, P, C$ are collinear, and points $B, P, D$ are also collinear, then quadrilateral $ABCD$ is a parallelogram, and $P$ is the intersection of the two diagonals.
If points $A, P, C$ are not collinear, by the fact that $\triangle PAB... | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 705,058 |
Let $I=[0,1], G=\{(x, y) \mid x \in I$, $y \in I\}$, find all mappings $f$ from $G$ to $I$ such that for any $x, y, z \in I$, the following conditions hold:
(1) $f[f(x, y), z]=f(x, f(y, z)]$,
(2) $f(x, 1)=x, f(1, y)=y$;
(3) $f(z x, z y)=z^{\mathrm{k}} f(x, y)$, where $k$ is a positive number independent of $x, y, z$. | Solve: From (3) and (2) we have
$$
\begin{array}{l}
f(x, y)=y^{k} f\left(\frac{x}{y}, 1\right)=y^{k-1} x, \\
x \leqslant y_{3} \\
f(x, y)=x^{k} f\left(1, \frac{y}{x}\right)=x^{k-1} y, \\
x \geqslant y .
\end{array}
$$
Take $0 < x < y$.
\end{array}\right. \\
f_{2}(x, y)=x y^{\prime} \text{.} \\
\end{array}
$$
It is ea... | f(x, y) = \begin{cases} y^{k-1} x, & x \leq y \\ x^{k-1} y, & x \geq y \end{cases} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,059 |
Example 3. Divide each side of a triangle into $n$ equal parts, and connect the points of division with line segments parallel to each side. How many parallelograms can be formed in the triangle? | Solve: First, consider parallelograms whose sides are not parallel to side $BC$. Extend the sides of these parallelograms, and they will intersect with side $BC$. Therefore, these parallelograms correspond to four points or three points on side $BC$ (one vertex is on $BC$). Extend sides $AB$ and $AC$ by $\frac{1}{n}$, ... | 3 C_{n+2}^{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,060 |
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?
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。
---
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are... | We use 10 points to represent 10 birds.
(1) Among the 10 points, there must be 5 points that are concyclic. If not, then any 5 points among the 10 points are not concyclic. The 10 known points can form $C_{10}^{5}=252$ groups of 5 points. According to the given, each group of 5 points has four points that are concyclic... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,061 |
Find all positive integer solutions $(x, y, z, n)$ to the following equation:
$$
x^{20+1}-y^{2 n+1}=x y z+2^{2 n+1}
$$
where $n \geqslant 2, \quad z \leqslant 5 \cdot 2^{20}$. | Solution: Clearly, $x-y>0$ and $x$ and $y$ have the same parity, let $x-y=2$.
$$
\begin{array}{l}
\text { When } y=1, x=3 \text {, let } \\
z=3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) .
\end{array}
$$
If $x \leqslant 5 \cdot 2^{2 n}$, then
$$
\begin{aligned}
3^{2 n} & \leqslant 5 \cdot 2^{2 n}+\frac{1}{3}\left(1+2^{... | (3,1,70,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,062 |
1. (Taiyuan City) The four vertices of a regular tetrahedron $ABCD$ lie on the surface of a sphere with radius $R$, then the length of $AB$ is ().
(A) $\sqrt{2} R$.
( B) $\frac{2 \sqrt{2}}{3} R$.
(C) $\frac{2 \sqrt{ }}{3} R$.
(D) $-\frac{3}{4}-\frac{6}{4} R$. | 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 | 705,065 |
2. (Taiyuan Rong) 9:990 divided by 11 leaves a remainder of ( ).
(A) 1 .
(B) 3 .
(C) 5 .
(D) 7 . | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,066 |
3. (Taiyuan awkward) If point $A(a, 0)$ is outside the circle $O: x^{2} + y^{2} = 1$, points $P, Q \in$ circle $O$, point $M, P$ are symmetric with respect to point $A$, and $O P \perp O Q$, $a > 0$, then the maximum length of $M Q$ must be ( ).
(A) $\sqrt{4} a^{2} + 4 a + 2$.
(B) $\sqrt{4 a^{2}} - 4 a + 2$.
(C) $\sqrt... | A
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 | 705,067 |
5. (Suzhou) There are 32 chess pieces on the board, and two people, A and B, take turns to take some of the pieces. The rule is: A goes first, and each person must take at least 1 piece and at most 6 pieces per turn, and the one who takes the last piece wins. To ensure victory, the number of pieces A should take first ... | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 705,069 |
Example 4. If from the numbers $1,2, \cdots, 141$, we sequentially select $a_{1}, a_{2}, a_{2}$, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, then how many different ways of selection are there that meet the above requirements? | Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1, 2, \cdots, 10\}$. Suppose $\left\{a_{1}, a_{2}, a_{3}\right\}$ is a subset of $S$ and satisfies $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$, i.e., the following mapping is established:
$$
\begin{array}{c}
\left(a_{1}, a_{2}, a_{3}\right) \rightarrow... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,071 |
8. (Wuhu City) A point inside a convex quadrilateral is called a "good point" if the lines connecting it to the four vertices divide the quadrilateral into four triangles of equal area.
Proposition A: There exists a "good point" inside a convex quadrilateral.
Proposition B: One diagonal of the convex quadrilateral bise... | 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 | 705,073 |
10. (Fuzhou News) Among the following groups of numbers, which one cannot be the four altitudes of a certain tetrahedron ( ).
(A) $1, \frac{3 \sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{3 \sqrt{2}}{2}$.
(B) $4, \frac{25 \sqrt{3}}{3}, \frac{25 \sqrt{3}}{3}, \frac{25 \sqrt{3}}{3}$.
(C) $\sqrt{5}, \frac{\sqrt{30}}{2}, \frac{\... | B
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 | 705,075 |
11. (Sichuan Province) If $\alpha$ is an acute angle,
$$
\begin{array}{l}
P=\log _{11 \alpha \alpha}\left(\sin 50^{\circ}+\cos 50^{\circ}\right), \\
Q=\log _{41 \Delta \alpha}\left(\sin 65^{\circ}+\cos 65^{\circ}\right),
\end{array}
$$
then the relationship between $P$ and $Q$ is ( ).
(A) $P<Q$.
(B) $P>Q$.
(C) $P=Q$.
... | A
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 | 705,076 |
12. (Sichuan Province) If the curve $x^{2} \cos \alpha-y^{2} \sin \alpha$ $+2=0$ is an ellipse, then the center of the circle $(x+\cos \alpha)^{2}+(y$ $+\sin \alpha)^{2}=1$ is in ( , ).
(A) The first quadrant.
(B) The second quadrant.
(C) The third quadrant.
(D) The fourth quadrant. | $\mathrm{D}$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 705,077 |
13. (Sichuan Province) Let the function $f(x)$ satisfy $f(1+x)=f(1-x)$ for all real numbers. If the equation $f(x)=0$ has exactly two distinct real roots, then the absolute value of the difference between these two roots is ( ).
(A) 1.
(B) 2.
(C) 4.
(D) Uncertain. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,078 |
15. (First Calm "Hope朴" · Grade 1) $A B C D$ $-A_{1} B_{1} C_{1} D_{1}$ is a unit cube. Two ants, one black and one white, start from point $A$ and crawl along the edges. Each time they complete an edge, it is called "completing a segment." The white ant's route is $A A_{1} \rightarrow$ $A_{1} D_{1} \rightarrow \cdots$... | B
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 | 705,080 |
16. (First "Hope Cup" - Grade 11) $f$ is a one-to-one mapping from $R \rightarrow R$, and the function $y=f(x)$ is strictly increasing. The solution set of the equation $x=f(x)$ is $P$, and the solution set of the equation $x=f[f(x)]$ is $Q$. Then ( ).
(A) $P \subset Q$.
(B) $P=Q$.
(C) $Q \subset P$.
(D) $P \neq Q$, an... | 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 | 705,081 |
Example 5. On a $6 \times 6$ chessboard, 11 dominoes of size $1 \times 2$ are placed, each domino exactly covering two squares. Prove: no matter how these 11 dominoes are placed, it is always possible to place one more domino. | Proof Consider the following set:
$X=\{$ the set of the bottom $5 \times 6$ grid $\}$,
$Y=\{$ the set of the top $5 \times 6$ grid $\}$.
If after placing these 11 dominoes, no more dominoes can be placed, then it must be true that:
1) The space above an empty cell must correspond to a domino (otherwise, if the space ab... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,082 |
17. (1st "Hope Cup" - Grade 11) A point $(x, y)$ is called a rational point if both coordinates $x$ and $y$ are rational numbers. For a circle $\dot{1}$ with radius $r$ and center $(a, b)$, if $a \in Q$, $b \in \bar{Q}$, the number of rational points on this circle ( ).
(A) is at most one.
(B) is at most two.
(C) is at... | B
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 | 705,083 |
18. (First "Hope Cup" - Senior High School Grade 2) The set $P$ of integers has the following properties:
(1) $P$ contains positive and negative numbers;
(2) $P$ contains odd and even numbers;
(3) $-1 \notin P$; (4) If $x, y \in P$, then $x+y \in P$. For this set $P$, it can be concluded that ( ).
(A) $0 \in P, 2 \noti... | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,084 |
1. (Suzhou City) If for any pair of real numbers $x, y$ that are not both zero, we have $f(x, y)=5 x^{2}+8 x y+3 y^{2}+\lambda\left(x^{2}+y^{2}\right)>0$, then the range of values for $\lambda$ is $\qquad$ | 1. $(\sqrt{17}-4,+\infty)$. | (\sqrt{17}-4,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 705,085 |
2. (Suzhou City) The roots of the equation $x^{2}-a x+b=0$ are $\alpha, \beta$ and the roots of $3 x^{2}-b x+c=0$ are $\gamma, \delta$. Let $M$ $=\{\alpha, \beta, \gamma, \delta\}, S=\{x \mid x=u+v, u \neq v, u$, $v \in M\}, P=\{x \mid x=u v, u \neq v, u, v \in M\}$. It is known that $S=\{5,7,8,9,10,12\}, P=\{6,10,14$,... | \[
\begin{aligned} 2, a & =7, b=10, \\ c & =21 .\end{aligned}
\] | a=7, b=10, c=21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,086 |
5. (Fuzhou City) The equation
$$
\begin{aligned}
& (6 x+5) \cdot\left(1+\sqrt{(6 x+5)^{2}+4}\right) \\
+ & x\left(1+\sqrt{\left.x^{2}+4\right)}=0\right. \text { has the solution }
\end{aligned}
$$ | 5. $-\frac{5}{7}$ | -\frac{5}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,089 |
7. (Sichuan Province) $\operatorname{tg} 20^{\circ} \cdot \operatorname{tg} 40^{\circ} \cdot \operatorname{tg} 60^{\circ}$ $+\operatorname{ctg} 70^{\circ}+\operatorname{tg} 40^{\circ}$ value is | 7.
$$
\sqrt{3} .
$$ | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,091 |
8. (Sichuan Province) Given a regular tetrahedron inscribed in a sphere with a radius of 3, and the side length of the base is $2 \sqrt{3}$, then the angle between the side edge of the tetrahedron and the base is $\qquad$ . | 8. $\operatorname{arctg} \frac{3+\sqrt{5}}{2}$ or $\operatorname{arctg} \frac{3-\sqrt{5}}{2}$. | \operatorname{arctg} \frac{3+\sqrt{5}}{2} \text{ or } \operatorname{arctg} \frac{3-\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,092 |
4 complex ones, their product is exactly a square number.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided translation is a direct interpretation of the given Chinese text. However, the phrase "斯4个繁" is som... | From the observation, if $(a, b)$ is a pair of numbers among the 48 given numbers, for example, $a=2^{7} \cdot 3^{8} \cdot 5^{9} \cdot 11, b=2^{8} \cdot 3^{7} \cdot 5^{l} \cdot 11 \cdot 13$, then the product $a b=\left(2^{4} \cdot 3^{7} \cdot 5^{*} \cdot 11\right)^{2} \cdot 3 \cdot 13$. Since $\left(2^{8} \cdot 3^{7} \... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,093 |
9. (Sichuan Province) If the function $y=a^{2}-2 a \cos x$ $-\sin ^{2} x$ has a minimum value of 1 (where $a$ is a real number), then the maximum value of the function $y$ is $\qquad$. | $9.5+4 \sqrt{2}$. | 5+4 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,094 |
10. (Shanghai · Senior 3) Let the parabola $y=x^{2}+m x+2$ intersect the line segment with endpoints $(0,1),(2,3)$ at two distinct points, then the range of values for $m$ is $\qquad$ | 10. $-\frac{3}{2} \leqslant m < -1$ | -\frac{3}{2} \leqslant m < -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,095 |
11. (Shanghai - Senior High School) Given that one side of a rectangle is $m$ cm, three mutually perpendicular lines divide this rectangle into 4 smaller rectangles, where the area of 3 of the smaller rectangles is no less than $m$ square cm, and the area of the 4th smaller rectangle is no less than $2m$ square cm, the... | 11.
$$
3+2 \sqrt{2} \text {. }
$$ | 3+2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,096 |
16. (First "Hope Cup" - Grade 11) In a regular tetrahedron wooden block $ABCD$ with edge length 2, there is a point $P$ on edge $AB$ ($AP<1$). A cross-section perpendicular to edge $AB$ is to be sawed through point $P$. When the sawing stops at a certain position, it is measured that the saw line $PM=1$ on face $ABD$, ... | 16.1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,101 |
Example 7. Generate $2^{n-1}$ distinct sequences, all of length $n$, composed of 0s and 1s. It is known that for any 3 of these sequences, there exists a number $p$ such that the $p$-th term in all three sequences is 1. Prove: there exists a number $k$ such that the $k$-th term in all $2^{n-1}$ sequences is 1, and $k$ ... | Prove that for a number sequence $x$ of length $n$, establish a mapping of 1.
Then, the new sequence $z$ produced by the corresponding term product of sequence $x$ and its complement $\bar{x}$ is called the product of sequence $x$ and $\bar{x}$, denoted as $x \bar{x}=z$.
First, prove that among the known $2^{n-1}$ se... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,104 |
1. (Shanghai $\cdot$ Senior High School) If positive integers $p, q, r$ make the quadratic equation $p x^{2}-{ }_{y} x+r=0$ have two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.
| Let $\alpha, \beta$ be the two roots of the quadratic equation $p x^{2}-q x+\gamma=0$, $\alpha \neq \beta, 0<\alpha, \beta<1$. Then
$$
\alpha \beta(1-\alpha)(1-\beta) = \frac{r}{p} \left[1 - \frac{q}{p} + \frac{r}{p}\right] = \frac{r}{p^{2}}(p-q+r).
$$
Therefore, $p^{2} > 16 r(p-q+r)$.
Since the quadratic term coeffic... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,105 |
2. (First State "Killing Hope Cup" - Grade 10) Let the function $f(n)=k$, where $n$ is a natural number, and $k$ is the $n$-th digit after the decimal point of the irrational number $\pi=3.1415926535 \cdots$. It is also defined that $f(0)=3$.
$$
\text { Let } F(n)=\frac{f(f(f(f \cdots(f(n)) \cdots)))}{10 \uparrow \bar{... | It can be easily proven that for non-negative integers $n$, $F(n)=1$ always holds.
Below $f(1990)+f(5)+f(13)$ and $f(1990) f(3)$
- $f(25)$ are all non-negative integers, so $F(f(1990)+f(5)$
$$
+f(13) \equiv 1 \equiv F[(1990) f(3) f(25)] .
$$ | 1 | Number Theory | proof | Yes | Yes | cn_contest | false | 705,106 |
3. (First "Hope Cup" - 䯩․) The sum of the first $n$ terms of the sequence $\left\{\operatorname{arcctg} 2 n^{2}\right\}$ is $S_{\mathrm{u}}$. Prove: for all $n \in N$, we have $S_{0}<\frac{\pi}{4}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translat... | Prove that for any $-k \in N, \quad k \geqslant 2$.
$$
\begin{array}{l}
\operatorname{arcctg} 2 k^{2}=\operatorname{arcctg} \frac{k^{2}+k+k^{2}-k}{k^{2}-\left(k^{2}-1\right)} \\
=\operatorname{arcctg}\left(\frac{k+1}{k}\right)\left(\frac{k}{k-1}\right)+1 \\
\left(\frac{k}{k-1}\right)-\left(\frac{k+1}{k}\right) \\
=\ope... | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,107 |
4. (Xiuzhou City) Each of several lines divides the square $ABCD$ into two quadrilaterals with an area ratio of $k$. Prove: Among these nine lines, at least three pass through the same point. | Proof: If $k=1$, then every line bisects the square, so every line must pass through the center of the square, and the proposition is obvious.
If $k \neq 1$, according to the problem,
no line can intersect two adjacent
sides of the square. Suppose line $l$
intersects $A B$ and $C D$ at $M$ and $N$
(as shown in the figu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,108 |
5. (Dongtai City) In the tetrahedron $S-ABC$, $SA$ is perpendicular to the base $ABC$, and the lateral faces $SAB$ and $SBC$ form a right dihedral angle. If $\angle BSC = 45^{\circ}, SB = a$, find the volume of the circumscribed sphere of this tetrahedron. | Let the plane of $\triangle A B C$ be $\gamma$, the plane of $\angle S A B$ be $\alpha$, and the plane of $\triangle S B C$ be $\rho$.
$\because S A \perp \gamma, \therefore S A \perp B C$.
Also, $\alpha \perp P$, draw $A D \perp S B$ in $\alpha$ through $A$, then $A D \perp B$, so $A D \perp B C$, and $S A \cap D A=\L... | \frac{\sqrt{2}}{3} \pi a^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,109 |
6. (She Di Province) Let $S$ be the sum of the lengths of two chords on the unit circle that are $d$ (a real number) apart. Try to find the range of values for $S$.
保留源文本的换行和格式，直接输出翻译结果如下：
6. (She Di Province) Let $S$ be the sum of the lengths of two chords on the unit circle that are $d$ (a real number) apart. Try t... | Solve: First, when $d>2$,
it is meaningless;
$$
\begin{array}{l}
\text{When } d=2, S=0, \\
\because \text{When } 10-d<2, \text{ set }
\end{array}
$$
Prevent chord distance from the center $r_{1}, r_{2}$
$$
\begin{array}{l}
\left(\left|r_{1} \pm r_{2}\right|=d\right), \text{ then } \\
S=2 \sqrt{1-r_{1}^{2}}+2 \sqrt{1-r... | 2 \sqrt{2 d-d^{2}} < S < 2 \sqrt{4-d^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,110 |
7. (Shaanxi Province) In the square $ABCD$, from vertex $A$, draw two rays such that the angle between them is $45^{\circ}$, intersecting $BC$ and $CD$ at points $E$ and $F$ respectively, and intersecting $BD$ at points $P$ and $O$. Prove:
$$
S_{\triangle T K F}=2 S_{\triangle A P K}
$$ | Proof of who soothes EO.
$\because, E A O=\angle E B O=45^{\circ}, A, B$ two points are on the same side of line $E Q$.
$\therefore A, B, E, Q$ are concyclic.
Therefore, $\angle A Q E=180^{\circ}-\angle A B E=90^{\circ}$,
$\triangle A E Q=\angle E A Q=45^{\circ}$,
thus $A E=\sqrt{2} A Q$. Similarly, $A F=\sqrt{2} A P$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,111 |
8. (Shanghai · Senior High School 3) In the spatial quadrilateral $ABCD$, $AD \perp AB, BC \perp AB, AD, BC$ form a $60^{\circ}$ angle, $AD=BC=a, C_{1}, C_{2}, D_{1}, D_{2}$ are points on the extensions of $BC, AD$ respectively, and $CC_{1}=C_{1}C_{2}=DD_{1}=D_{1}D_{2}=a$. Also, the lengths of $CD, C_{1}D_{1}, C_{2}D_{... | Let $AB = x, CD = m, C_2D_2 = n$, then $C_1D_1 = \sqrt{mn}$. Draw $AE \perp BC$, then $\angle DAE = 60^\circ$ or $120^\circ$.
If $\angle DAE = 60^\circ$, then $DE = a$.
Since $AD \perp AB, AB \perp AE$. Hence $AB \perp$ plane $ADE$, thus $CE \perp$ plane $ADE$. Therefore, $CE \perp DE$, and $\triangle CDE$ is a right t... | \frac{\sqrt{14}}{2}a \text{ or } \frac{\sqrt{42}}{2}a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,112 |
9. (Shanghai City • Grade 12) In an isosceles triangle $ABC$ with base angles $\alpha$ and leg length $a$, every inscribed triangle in $\triangle ABC$ has a largest side (Note: the inscribed triangle referred to here is one that has at least one vertex on each side of $\triangle ABC$). Find the minimum value of these l... | Solve in three steps:
1. For any inscribed triangle $D E F$ ($D$ on the base $B C$), if $D E \neq D F$, then an inscribed isosceles triangle (with the vertex on $B C$) can be found, whose longest side is no greater than the longest side of the original triangle. There are two cases:
i) If the perpendicular bisector of ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,113 |
10. (First "Hope Cup" - Grade 10) $ABCD$ is a square, $M$ is a point on $AB$, $N$ is a point on $BC$, and $AM = BN$. Connect $DM$ and $DN$ to intersect the diagonal $AC$ at points $P$ and $Q$, respectively, and cut off $\triangle MNB$. Prove:
(1) By folding along $DM$ and $DN$, $DA$ and $DC$ can coincide to form a side... | Prove (1) From $A M=B N$, then $C N$ $=B M$, fold along $D M$, $D N$ so that $D A$ coincides with $D C$. Since $M N>B N=A M$, $M N>M B=C N$. And $M N-M B+B N=A M+C N$, so $A M$, $M N, Q N$ form a triangle. The faces $D A M, D M N$, $D N C$ form the sides of a triangular pyramid.
(2) After forming the above triangular p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,114 |
Example 1. (14th All-Russian Mathematics Competition) Let
$$
\frac{1}{1+\frac{1}{1+\frac{1}{1+}}+\frac{1}{1}}=-\frac{m}{n},
$$
where $m$ and $n$ are coprime natural numbers, and the left side of the equation has 1988 fraction lines. Calculate $m^{2}+m n-n^{2}$. | Let the value of the complex fraction with $k$ layers of fraction lines be $\frac{m_{k}}{n_{k}},\left(m_{k}, n_{k}^{*}\right)=1$, then
$$
\frac{m_{\mathrm{k}+1}}{n_{\mathrm{k}+1}}=\frac{1}{1+\frac{m_{\mathrm{k}}}{n_{\mathrm{k}}}}=\frac{n_{\mathrm{k}}}{m_{\triangle}+n_{\mathrm{k}}},
$$
i.e., $m_{\mathrm{k}+1}=n_{\mathr... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,115 |
12. (First "Xipeng Cup" - Grade 11) The vertices $Q, R$, and $S$ of the square $P Q R S$ slide along the sides $A B, B C, C A$ of an equilateral triangle $A B D$ with side length 2, respectively. Find the locus of point $P$. | Consider a general position (as shown in the figure).
Let $\angle C R S=\theta$,
then $\angle P R C=45^{\circ}$
$+\theta$, the distance from $P$ to $B C$
is $|P R| \sin \left(45^{\circ}+\theta\right)=\sqrt{2}|R S|$
$\cdot \sin \left(45^{\circ}+\theta\right)$. Let
$|R S|=x$, in $\triangle R S C, \triangle Q R S$,
we ha... | 3-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,117 |
13. (Zhejiang Summer Camp) Let the positive integer $k$ make the set
$$
X=\left\{3^{31}, 3^{31}+1, \cdots, 3^{31}+k\right\}
$$
decomposable into three disjoint subsets $A, B, C$ of the same sum. Prove that $k \equiv 1 \pmod{3}$. Find a sequence of such $k$ values. | Proof From the problem, it is easy to know
$$
3 \mid\left[3^{31}+\left(3^{81}+1\right)+\cdots+\left(3^{81}+k\right)\right] \text {, }
$$
which means $3 \left\lvert\,\left[(k+1) 3^{31}+\frac{k(k+1)}{2}\right)\right.$.
Since $k(k+1)$, it follows that $3 \mid k$ or $3 \mid k+1$, i.e., $k+1(\bmod 3)$.
$k=6 m-1 \quad(m \in... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,118 |
16. (1: Sea concern・
Let the numbers on the tiles be: 1, 2,
$3, \cdots, n^{2}$. Arrange them
in an $n$ by $n$ matrix:
The first row is $1,2, \cdots$,
$n$, the second row is $n+1$,
$n+2, \cdots, 2 n$, and so on. Now, starting from the top-left corner to the bottom-right corner, number these tiles sequentially (as shown... | $$
\begin{array}{l}
\text { (1) Step } 3 \leqslant i+j \leqslant n+1 \text {, the number is } \\
k(i, j)=[1+2+\cdots+(i+j-2)]+j \\
-\frac{1}{2}(i+j-1)(i+j-2)+j
\end{array}
$$
Example 1: The number is $(i-1) n+i$. If
$$
\begin{array}{l}
\frac{1}{2}(i+j-1)(i+j-2)+j \\
=(i-1) n+j
\end{array}
$$
then $(i+j-1)(i+j-2)=2 n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,121 |
17. (Zhejiang Summer Camp) Suppose there are 1990 distinct points on a plane. Can a regular triangle be drawn such that 995 points are inside it and the remaining 995 points are outside? | Establish a Cartesian coordinate system $x \circ y$, such that the $y$-axis is not parallel to the line connecting any two of the 1990 known points. Then the x-coordinates of the 1990 known points are all different. Arrange them in order of their x-coordinates, and denote them as $A_{1}\left(x_{1}, y_{1}\right)$, where... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,122 |
18. (Wuhu City) Prove: In any convex quadrilateral, there always exists a diagonal and a side, such that the two circles constructed with them as diameters cover the quadrilateral. | Prove that in a convex quadrilateral $ABCD$, there is at least one angle not less than $90^{\circ}$, let it be $\angle A$.
Take the diagonal $BC$ opposite to $A$ as the diameter to draw $\odot O_{1}$.
If $\angle D \geqslant 90^{\circ}$, then $\odot O$ already covers the entire quadrilateral.
If $\angle D < 90^{\circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,123 |
19. (Wuhu City)
In a square plate $ABCD$ with a side length of 8 units, there is an equilateral triangular colored plate $OEF$ (vertex $O$ is painted red, $E, F$ are white) placed as shown (E coincides with D, and $EF$ coincides with $AD$). Now, the triangular colored plate rotates to the left with $F$ as the pivot un... | The "left advance" rule for the triangular color board is:
?.
(1) Every 3 left advances, the red vertex is inside the disk;
(2) Every 16 left advances, the triangular board returns to its original starting position, but the red vertex moves to the point F on the AD side;
(3) Every 48 left advances, the triangular board... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,124 |
20. (Suzhou Technique) There are $3 n(n>1)$ points equally dividing a circle. From these $3 n$ points, any $n+2$ points are taken. Prove: Among the taken points, there always exist two points such that the arc connecting these two points on the circle is greater than $\frac{1}{3}$ of the circumference of the circle. | Prove that among $n+2$ points, taking any point $A$ as a division point, dividing the circle into three parts, the other two divisions are $B, C$ (Figure 1).
$1^{\circ}$ If $\widehat{B} C$ contains one of the taken points, let one of these points be $D$, then $\overparen{A B D}$ and $\overparen{A C D}$ are each greater... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,125 |
2. (Bulgaria) Study the polynomial determined by the equation $a_{0}+a_{1} x$
$$
+a_{2} x^{2}+\cdots+a_{2 \mathrm{n}} x^{2 \mathrm{n}}=\left(x+2 x^{2}+\cdots
$$
$\left.+n x^{\mathrm{n}}\right)^{2}$. Prove that,
$$
\begin{array}{l}
a_{n}+1+a_{n}+2+\cdots+a_{20} \\
=\frac{n(n+1)\left(5 n^{2}+5 n+2\right)}{24} .
\end{arra... | Prove the coefficient $a_{\mathrm{k}}$ is calculated first:
$$
\begin{array}{l}
=1 \cdot(k-1)+2 \cdot(k-2)+\cdots+(k-1) \cdot 1 \\
=k(1+2+\cdots+k)-\left(1^{2}+2^{2}+\cdots+k^{2}\right) \\
=k \cdot \frac{k(k+1)}{2}-\frac{k(k+1)(2 k+1)}{6} \\
=\frac{(k-1) k(k+1)}{6}=C_{k+1}^{3} .
\end{array}
$$
From this, we get $a_{0}... | \frac{n(n+1)\left(5 n^{2}+5 n+2\right)}{24} | Algebra | proof | Yes | Yes | cn_contest | false | 705,129 |
3. (Yugoslavia) Let $\triangle A_{1} B_{1} C_{1}$ be the orthic triangle of the non-equilateral acute triangle $A B C$, and let $A_{2}, B_{2}, C_{2}$ be the points where the incircle of $\triangle A_{1} B_{1} C_{1}$ touches its sides. Prove that the Euler line of $\triangle A_{2} B_{2} C_{2}$ coincides with the Euler l... | Prove that, as shown in the figure, $\triangle A_{1} B_{1} C_{1}$ is the pedal triangle of $\triangle A B C$, and $A A_{1}$ is the altitude on side $B C$. It is easy to prove that $A A_{1}$ is the angle bisector of $\angle B_{1} A_{1} C_{1}$. Since points $B_{2}, C_{2}$ are two tangent points, then $A_{1} B_{2}=A_{1} C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,130 |
4. (Lokni Army) Let $A$ be a finite set, $N=\{1, 2, 3, \cdots\}$. If there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j)$. Try to find the minimum number of elements in $A$.
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。 | Consider in $N$ taking $i=1,3,6,8$. The absolute value of the difference between any two of these numbers is a prime number, so their images $f(1), f(3), f(6), f(8)$ are the same, meaning that the set $A$ must have at least 4 elements.
Below, we prove that the minimum number of elements in set $A$ is 4. For this, we c... | 4 | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 705,131 |
1. Given $x y+x+y=71, x^{2} y+x y^{2}$
$=830 . x, y$ are positive integers, find $x^{2}+y^{2}$. | 1. Answer: 146. (Hint: Let $a=x+y$, $b=x y$ ) | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,132 |
3. Expand $(1+0.2)^{1000}$ using the binomial theorem, i.e. let $C_{1000}^{0}(0.2)^{0}+C_{1000}^{1}(0.2)^{1}+\cdots+$ $C_{1000}(0.2)^{1000}=A_{0}+A_{1}+\cdots+A_{1000}$. Let $A_{h}=C_{1000}^{k}(0.2)^{h}, k=0,1,2, \cdots$, 1000 . Find the value of $k$ for which $A_{k}$ is the largest. | 3. For $1 \leqslant k \leqslant 1000$,
$$
\begin{array}{l}
\frac{A_{\mathrm{s}}}{A_{\mathrm{k}-1}}=\frac{\frac{1000!}{k!(1000-k)!}(0.2)^{k}}{\frac{1000!}{(k-1)!(1001-k)!}(0.2)^{k}} \\
=\frac{1001-k}{k}(0.2) \text {. } \\
\end{array}
$$
When and only when $1001-k>5k$, this ratio is greater than 1. That is, when $k \leq... | 166 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,134 |
4. The number of real solutions to $\frac{1}{5} \log _{2} x=\sin (5 \pi x)$ is? | Consider only those $x$ that satisfy $\left|\frac{1}{5} \log _{2} x\right| \leqslant 1$. Solving this, we get $\frac{1}{32} \leqslant x \leqslant 32$.
First, consider $\frac{1}{32} \leqslant x<1$, in this case $-1 \leqslant \frac{1}{5} \log _{2} x <0$. When $\frac{1}{5} \leqslant x \leqslant \frac{2}{5}, \frac{3}{5} \... | 159 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 705,135 |
5. For a given rational number, represent it as a reduced fraction. Then find the product of its numerator and denominator. How many rational numbers between 0 and 1 have a product of their numerator and denominator equal to 20! ? | 5. 20! has 8 prime factors: $2, 3, 5, 7, 11, 13, 17, 19$. A prime factor appearing in the denominator cannot appear in the numerator (and vice versa). For each prime factor, it can be in the numerator or in the denominator, so there are $2^8 = 256$ choices. However, not all of them are less than 1. In fact, the numbers... | 128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,136 |
6. Find the value of $r$ such that
$$
\begin{array}{l}
{\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots} \\
+\left[r+\frac{91}{100}\right]=546 .
\end{array}
$$
Find $[100 r]$. (Where $[x]$ denotes the greatest integer not greater than $x$) | 6. The left side of the equation has 73 terms, each being $[r]$ or $[r+1]$. This is because $\frac{19}{100}, \frac{20}{100}, \cdots, \frac{91}{100}$ are all less than 1, and $73 \times 7 < 546 < 73 \times 8$. To make the sum 546, $[r]$ must be 7.
Assume $\left[r+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m),[r+$ ... | 743 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,138 |
7. If $A$ is the sum of the absolute values of all roots of the equation
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\cdots \sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}
$$
Find $A^{2}$. | 7. The fraction to the right of the equation can be simplified to $\frac{a x+b}{c x+d}$ $(a, b, c, d$ are real numbers), then the equation is a quadratic equation, with at most two roots. H $x=\sqrt{19}+\frac{91}{x}$ has roots that are also $x=\frac{\sqrt{19}-\sqrt{38}}{2}$, which are the only two roots of the original... | 383 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,139 |
8. How many real numbers $a$ are there such that $x^{2}+a x+6 a$ $=0$ has only integer solutions?
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 but is provided to clarify that the format and structure of the original text have been maintained in the translation. | 8. Let $x^{2}+a x+6 a=$ O have integer solutions $m, n$ $(m \leqslant n)$. Then we have
$$
a=-(m+n), \quad 6 a=m n .
$$
Since $a$ must be an integer, we also have
$$
-6(m+n)=m n \text { . }
$$
H $(m+6)(n+6)=36$. Since $36=36 \times 1=18 \times 2=12 \times 3=6 \times 6$, the solutions satisfying $m \leqslant n$ are
$$... | 10 | Other | math-word-problem | Yes | Yes | cn_contest | false | 705,140 |
9. If $\sec x+\tan x=\frac{22}{7}, \csc x+\cot x$ $=\frac{m}{n}$, where $\frac{m}{n}$ is a reduced fraction, find $m+n$.
| $$
\begin{array}{l}
\text { 9. } \because \sec ^{2} x-\tan ^{2} x=1 \text {, let } p=\frac{22}{7}, \\
\therefore \sec x-\tan x=\frac{1}{p}, \sec x+\tan x=p .
\end{array}
$$
Adding these two equations, we get $2 \sec x=p+\frac{1}{p}$. Subtracting them, we get
$$
\begin{aligned}
2 \tan x & =p-\frac{1}{p} . \\
& \therefo... | 44 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,141 |
10. For the string of three letters $\hat{\mathrm{i}}$ "aaa" and "bbb", they are transmitted through a circuit, with each string being sent one letter at a time. Due to issues with the line, each of the 6 letters has a $\frac{1}{3}$ chance of being transmitted incorrectly (an $a$ is received as a $b$, or a $b$ is recei... | $x_{1}=a, y_{1}=b, i=1,2,3$.
Let $S_{1} \propto S_{2}$ represent that $S_{1}$ is lexicographically before $S_{2}$, and $P\left(S_{1} \propto S_{2}\right)$ represent the probability that $S_{1}$ is lexicographically before $S_{2}$. Due to the independence of the correct transmission of letters, we have
$$
\begin{array}... | 532 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,142 |
11. As shown in the figure, 12 congruent disks are placed on the circumference of a circle $C$ with radius 1, such that they cover $C$. No two disks overlap, and adjacent disks are tangent to each other. The sum of the areas of these disks can be written as
$$
\pi(a-b \sqrt{c}) \text {, where }
$$
$a, b, c$ are positiv... | 11. By the people's knowledge, 6-certainly passes through the tangency points of each circle, and each tangent line passes through the center $O$ of circle $C$ (as shown in the figure).
$P$ is the center of a circle,
$$
\begin{array}{l}
\angle P B O=90^{\circ}, \\
\angle B O A=\frac{1}{12} 2 \pi=\frac{\pi}{6}, \\
\angl... | 135 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,143 |
12. Rhombus $P Q R S$ is inscribed in rectangle $A B C D$, such that $P, Q, R, S$ are interior points on $\overline{A B}, \overline{B C}, \overline{C D}, \overline{D A}$. Given that $P B=15, B Q=20, P R=30, Q S$ $=40$. If the reduced fraction $\frac{m}{n}$ is the perimeter of rectangle $A B C D$, find $m+n$. | 12. As shown in the figure, let $x, y$ represent the lengths of $Q C, R C$ respectively. By symmetry, the lengths of $SA, PA$ are also $x, y$.
The diagonals of the rectangle intersect at the center H of the rectangle, and they are perpendicular to each other, thus forming eight right triangles. Six of these right trian... | 677 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,144 |
13. In a drawer, there are red and blue socks, no more than 1991 in total. If two socks are drawn without replacement, the probability that they are the same color is $\frac{1}{2}$. How many red socks can there be at most in this case? | 13. Let the number of red and black balls be $x$ and $y$ respectively. Given that the probability of randomly picking two balls of different colors is $\frac{1}{2}$, we have $\frac{x y}{C_{x+y}^{2}}=\frac{1}{2}$.
Therefore, $(x+y)(x+y-1)=4 x y$,
Thus, $(x-y)^{2}=x+y$.
Hence, the total number of balls is a perfect squa... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,145 |
15. For a positive integer $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value. Where $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ are positive integers, and their sum is 17. There is a unique value of $n$ for which $S_{n}$ is an integer, find $n$. | 15. Consider each $l_{i}=\sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the hypotenuse of a right-angled triangle, with the two legs being $2 k-1$ and $a_{b}$. When these right-angled triangles are placed together to form a ladder, let $A, B$ be the starting and ending points, respectively.
The distance from $A$ to $B$ is
$$
\begin{... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,147 |
Example 1. Let $x$ be a cube root of 1 different from 1, find the value of $x^{\text {D}}$ $+x^{2}$. $(n \in N)$ | Solve: From $x^{3}=1$, i.e., $x^{3}-1=0$, which is also $(x-1)\left(x^{2}+x+1\right)=0$.
Since $x \neq 1$, it follows that $x^{2}+x=-1$.
When $n=3k$,
$$
x^{\mathrm{n}}+x^{2 n}=\left(x^{3}\right)^{k}+\left(x^{3}\right)^{2 k}=2 \text {; }
$$
When $n=3 k+1$,
$$
\begin{array}{l}
x^{\mathrm{n}}+x^{2 \mathrm{n}}=\left(x^{3}... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,152 |
Example 3. Prove that the equation: $x^{3}+y^{3}=1987$ has no positive integer solutions.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The note is not part of the translation but is provided to clarify the instructi... | Proof: Let $x=3k+r(k \in N, r=0,1,2)$,
then $x^{3}=(3k+r)^{3}$
$=9(3k^{2}+3kr+kr^{2})+r^{3}$.
It can be seen that when $x^{3}$ is divided by 9, the remainder can be 0, or 1, or 8.
Similarly, when $y^{3}$ is divided by 9, the remainder can be 0, or 1, or 8.
Therefore, when $x^{3}+y^{3}$ is divided by 9, the remainder ca... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,154 |
Example 4. $p$ is a prime number not equal to 3, and $2p+1$ is a prime number. Prove: $4p+1$ is not a prime number. | Prove that if a prime $p$ is not divisible by 3, then $p$ can only be expressed in one of the two forms: $3k+1, 3k+2 (k \in \mathbb{N})$.
If $p=3k+1$, then $2p+1=3(2k+1)$ is a composite number, which contradicts the premise, so this is not possible.
If $p=3k+2$, then $4p+1=12k-3=3(4k-1)$ is a composite number. Theref... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,155 |
Example 5. Prove that there are infinitely many prime numbers (i.e., prime numbers) $p \in$ $\{3k+2\}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that if $p \in\{3 k+2\}$ is a prime, there are only $p_{1}$, $p_{2}$, $\cdots, p_{\mathrm{u}}$, a total of $n$ primes (we can assume $p_{1}>3, i=1, 2$, $\cdots, n$).
Let $a=3 p_{1} p_{2} \cdots p_{0}+2$ (by assumption, $a$ is not a prime).
$\because a \in\{3 k+2\}, \quad \therefore a$ is not a perfect square. Th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,156 |
Example 6. (USAMO problem) Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3,
\end{array}\right.
$$
Find all real or complex roots. | Let $x, y, z$ be the roots of the cubic equation $r^{3}-a r^{2}+b r -c=0$. Then, by (1) and (2), we know
$$
x y+y z+z x=3 \text {, }
$$
Therefore, the cubic equation is
$$
r^{3}-3 r^{2}+3 r-c=0 \text {. }
$$
Let $c=1+m^{3}$ (where $m$ is a complex number), then $(r-1)^{3} = m^{3}$. Thus, $r=1+m, 1+\omega m, 1+\omega^... | x=y=z=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,157 |
Example 8. Find the prime number $p$ such that $8 p^{2}+1$ is also a prime number.
The text above is translated into English, preserving the original text's line breaks and format. | When $p=2$, $8 p^{2}+1=33$ is not a prime number.
When $p=3$, $8 p^{2}+1=73$ is a prime number.
For $p>3$, since $p$ is a prime, $p \in\{3 k+1\}$ or $p \in\{3 k+2\}$.
(i) If $p \in\{3 k+1\}$, then $p^{2} \in\{3 k+1\}$. $8 \in\{3 k+2\}$.
Thus $8 p^{2} \in\{3 k+2\}$, so $8 p^{2}+1$ $\in\{3 k\}$ is a composite number.
(i... | p=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,159 |
Example. Find all arithmetic sequences of three prime numbers with a common difference of $\varepsilon$.
Find all arithmetic sequences of three prime numbers with a common difference of $\varepsilon$. | Solution: Clearly, these three prime numbers must be odd. If the first term is $2n+1$, then the three terms of this sequence are $2n+1, 2n+9, 2n+17$. Consider $n$ modulo 3:
If $n \equiv 0(\bmod 3)$, i.e., $n=3k$, then $2n+9=6k+9$, which is not a prime number;
If $n \equiv 1(\bmod 3)$, i.e., $n=3k+1$, then $2n+1=6k+3$... | 3, 11, 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,161 |
Example 10. (IMO6-1(1)) Find all positive integers $n$ such that $2^{n}-1$ is divisible by 7. | ```
When $n=3 k$, $2^{n}-1=2^{3 k}-1=8^{k}-1$ \\
$=(7+1)^{k}-1 \equiv 0(\bmod 7)$; \\
When $n=3 k+1$, $2^{n}-1=2^{3 k+1}-1$ \\
$=2(7+1)^{b}-1 \equiv 1(\bmod 7)$; \\
When $n=3 k+2$, $2^{n}-1=2^{3 k+2}-1$ \\
$=4(7+1)^{k}-1 \equiv 3(\bmod 7)$. \\
Thus, only when $n$ is a multiple of 3, $2^{n}-1$ is divisible by 7.
``` | n = 3k | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,162 |
Example 11. (Soviet Mathematical Competition Problem) Prove: there exist infinitely many natural numbers $n$ such that $n^{2} \neq x^{2}+p$, where $p$ is a prime number, and $x$ is an integer. | Prove that taking $n=3k+2$, where $k$ is a natural number.
From the equation $n^{2}=x^{2}+p$, we get $p=n^{2}-x^{2}$ $=(n-x)(n+x)$. Since $p$ is a prime number, and $n>x$, then $n-x=1, n+x=p$. Therefore, $p=2n-1$ $=3(2k+1)$, which is impossible.
Thus, numbers of the form $3k+2$ (there are infinitely many such numbers)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,163 |
Example 14. (22nd All-Russian School Mathematics Olympiad for 9th Graders) Prove that in the sequence with the general term formula
$$
a_{n}=1+2^{2}+3^{3}+\cdots+n^{n}
$$
there are infinitely many special numbers. | Prove that for any positive integer $n, n^{n}+(n+1)^{n+1} \div (n+2)^{n / 2} \pm (n+3)^{n+3}$ is even. Because among $n$ to $n+3$ there must be two odd numbers and two even numbers, so the sum must have two odd terms and two even terms, hence the sum is even.
Since $a_{1}=1$ is odd, all terms $a_{4m+1}$ (where $m$ is ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,164 |
Example 1. Let $n$ be an integer, calculate the following expression:
$$
\begin{array}{l}
{\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{2^{2}}\right]+\left[\frac{n+2^{2}}{2^{3}}\right]} \\
+\cdots .
\end{array}
$$
where the symbol $[x]$ denotes the greatest integer not exceeding $x$.
(IMO-10)
Analysis: In the expressio... | Let $n=\left(a_{m} a_{n-1} \cdots a_{1} a_{0}\right)_{2} ， a_{m}=1$. $a_{1}=0$ or $1(1=0,1,2, \cdots, m-1)$. F is, for $k=1,2, \cdots, m$ we have
$$
\begin{aligned}
& \frac{n+2^{\mathrm{k}-1}}{2^{\mathrm{k}}} \\
= & \frac{\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{1} a_{0}\right)_{2}+\left(100^{\mathrm{k}-1 \uparr... | n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,165 |
Example 2. Prove that the number of coefficients in the expansion of $(1+x)^{\mathrm{n}}$ is $2^{\mathrm{s}} n$. Here, $s(n)$ is the sum of the digits in the binary representation of $n$.
To prove this conclusion, we introduce a concept related to binary numbers. Let $n=\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{... | Let $n=\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{1} a_{0}\right)_{2}$. Consider the $(k+1)$-th coefficient $C_{\mathrm{n}}^{\mathrm{k}}=\frac{n!}{k!(n-k)!}$ of $(1+x)^{k}$. By (*), the power of 2 in $n!$ is $n-s(n)$. Therefore, the power of 2 in $C_{\mathrm{n}}^{\mathrm{k}}$ is
$$
\begin{array}{l}
(n-s(n))-(k-s(k... | 2^{s(n)} | Number Theory | proof | Yes | Yes | cn_contest | false | 705,166 |
Example 4. Let the side length of the square $ABCD$ be 1. Try to find the largest and the smallest inscribed equilateral triangles within it, and calculate the areas of these two triangles. (1978, National Competition)
Translate the above text into English, please retain the original text's line breaks and format, and... | By symmetry, we only need to discuss the variation of the inscribed equilateral triangle $GEF$ as shown in the figure, with one vertex on $AN$ and the other two vertices on $AD$ and $BC$, respectively. Taking the midpoint $K$ of $GF$, and drawing $KA, KB, KE$, we have $KE \perp GF$, so $G, A, E, K$ and $E, B, F, K$ are... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,171 |
Given that $A M$ is the median of $\triangle A B C$ on side $B C$, and a line is drawn intersecting $A B, A C, A M$ at points $P, Q, N$ respectively. Prove that $\frac{A B}{A P}, \frac{A M}{A N}, \frac{A C}{A Q}$ form an arithmetic sequence. (1978, Liaoning Province High School Competition) | Prove that from the viewpoint of $A$, by applying the sine rule to $\triangle APN$ and $\triangle AQB$, we get
$$
\begin{array}{l}
\sin (\alpha+\beta) \\
A N \\
=\frac{\sin \beta}{A P}+\frac{\sin \alpha}{A Q}, \\
\frac{\sin (\alpha+\beta)}{A M}=\frac{\sin \beta}{A B}+\frac{\sin \alpha}{A C} .
\end{array}
$$
In $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,172 |
Example 4. In the "equal" shaped $A B C D$, $A B=A D$, $B C=C D$, through the intersection point $O$ of $A C$ and $B D$, draw any two lines, intersecting $A D$ at $E$, intersecting $B C$ at $F$, intersecting $A B$ at $G$, intersecting $C D$ at $H, G F$ and $E H$ intersect $B D$ at $I, J$ respectively. Prove: $O I=O J$.... | Prove as shown in the figure,
It is easy to prove $A C \perp B D$.
Let $\angle E O J=\alpha$,
$\angle G O I=\beta$, then
$\angle A O G=90^{\circ}-\beta$
(because $\angle A O B=90^{\circ}$ ).
Let $O I=i, O F=f, O G=g, O J=j$,
$O E=e, O H=h, O A=a, O B=b, O C$ $=c, O D=d$, then with $O$ as the viewpoint, applying the Ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,173 |
Proposition 1 Let the area of $\triangle A B C$ be $\triangle$, the lengths of the three sides be $a, b, c$, and the corresponding median lengths be $m_{a}, m_{b}, m_{c}$. If there exists a relationship
$$
f(a, b, c, \Delta) \geqslant 0,
$$
then it must be true that
$$
f\left(m_{\mathrm{a}}, m_{\mathrm{b}}, m_{\mathrm... | Line $m_{\mathrm{a}}, m_{\mathrm{h}}, m_{\mathrm{e}}$ can serve as the three sides of a triangle similar to a given triangle. Let this triangle be $\triangle A_{1} B_{1} C_{1}$, then it is easy to prove: the area of $\triangle A_{1} B_{1} C_{1}$ is $\frac{3}{4}$ of the area of $\triangle A B C$. Applying (1) to $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,174 |
Example 1. Let $m, n$ be natural numbers. Prove:
$$
\frac{(1+\sqrt{m})^{2}-(1-\sqrt{m})^{2}}{\sqrt{m}}
$$
is an integer. | Proof: Let $(1+\sqrt{m})^{0}=a+b \sqrt{m}(a, b$ be natural numbers), then $(1-\sqrt{m})^{\mathrm{n}}=a-b \sqrt{ } \bar{m}$.
$$
\text { Hence } \begin{aligned}
& \frac{(1+\sqrt{m})^{n}-(1-\sqrt{m})^{n}}{\sqrt{m}} \\
= & \frac{a+b \sqrt{m}-(a-b \sqrt{m})}{\sqrt{m}} \\
= & \frac{2 b \sqrt{m}}{\sqrt{m}}=2 b
\end{aligned}
$... | 2b | Algebra | proof | Yes | Yes | cn_contest | false | 705,177 |
Theorem: For $n$ equally spaced points $A_{1}, A_{2}, \cdots, A_{\mathrm{n}}(n \geqslant 2)$ on the circumference of a circle with radius $R$, we have $\left|A_{1} A_{2}\right| \cdot\left|A_{1} A_{3}\right| \cdot \cdots \cdot\left|A_{1} A_{n}\right|=n R^{n-1} \cdot$ | Prove that the $n$ roots of $X^{n}-R^{n}=0$ are denoted as $R$, $R \omega_{1}, \cdots, R \omega_{\mathrm{n}-1}$, and $S_{1}, \cdots, S_{\mathrm{n}-1}$ represent the elementary symmetric polynomials of $\omega_{1}$, $\cdots, \omega_{n-1}$:
$$
\begin{array}{l}
S_{1}=\omega_{1}+\cdots \\
+\omega_{\mathrm{n}-1}, S_{2}=\ome... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,179 |
Example 5. On two parallel lines $AB$ and $CD$, take fixed points $M$ and $N$ respectively. On line $AB$, take a fixed-length segment $ME=a$. On the perpendicular segment $MN \perp$, take a point $K$, connect $EK$ and extend it to intersect $CD$ at $F$. If the distance between the two parallel lines is $d$, at what pos... | Given the figure, let the length of $MN$ be $l$, the angle between $MN$ and the parallel lines be $\alpha$, and $MK = x$. Then,
$$
\begin{array}{l}
NK = l - x \\
\text{Since } \Delta E M K
\end{array}
$$
$\rightarrow \triangle F N K$,
thus $\frac{a}{F N} = -\frac{x}{l - x}$,
hence $F N = \frac{a(l - x)}{x}$.
$$
\begin{... | (\sqrt{2} - 1) a d | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,182 |
2. In the Cartesian coordinate system, a 100-sided polygon $P$ is given, satisfying:
i) The coordinates of the vertices of $P$ are all integers;
ii) The sides of $P$ are all parallel to the coordinate axes;
iii) The side lengths of $P$ are all odd.
Prove that the area of $P$ is odd. | Proof: First, we provide a lemma.
Lemma: Given an $n$-sided polygon $P$ in the complex plane, with vertices (complex numbers) $z_{1}, z_{2}, \cdots, z_{n}$. Then the oriented area of $P$ is
$$
\begin{array}{l}
S=\frac{1}{2} I_{m}\left(z_{1} \bar{z}_{2}+z_{2} \bar{z}_{3}+\cdots+z_{n-1} \bar{z}_{n}\right. \\
\left.+z_{n... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,188 |
5. Try to find all positive integers $n$ such that the equation
$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$
has positive integer solutions. | Given $x \geq y \geq z$, then
$$
\begin{array}{l}
3 x^{3}=x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2} . \\
\therefore \quad 3 x=n y^{2} z^{3} . \\
\because \quad y, \quad z \in \mathbb{N}, \\
\therefore \quad 1+y^{3} z^{3} \geq y^{3}+z^{3} \\
\quad=n x^{2} y^{2} z^{2}-x^{3} \\
\quad=x^{2}\left(n y^{2} z^{2}-x\right) \geqslan... | n=1 \text{ or } 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,191 |
6. Let there be $2 n(n>1)$ points in space, where no three points are collinear, and they are connected by $n^{2}+1$ line segments. Prove that these line segments must form two triangles with a common side. | Prove that when $n=2$, $n^{2}+1=5$, i.e., there are five line segments connecting four points, which of course form two triangles with a common side. It is evident that the proposition holds when $n=2$.
Assume the proposition holds when $n=k$. When $n=k+1$, let $AB$ be any line segment, and let the number of line segm... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,192 |
Example 6. Draw any line through the centroid of a triangle, dividing the triangle into two parts. Prove that the difference in the areas of these two parts is no greater than $\frac{1}{9}$ of the area of the entire triangle. (1978, Anhui Party Committee) | Prove that by trisecting each side of the triangles and drawing lines parallel to the other sides through each division point, these lines divide $\triangle A B C$ into 9 smaller triangles of equal area. The internal intersection point $G$ is exactly the centroid of this triangle. Drawing any line through $G$ divides $... | \frac{1}{9} | Geometry | proof | Yes | Yes | cn_contest | false | 705,193 |
Example 7. In an equilateral $\triangle ABC$ with side length $a$, moving points $P$, $Q$, $R$ start simultaneously from $A$, $B$, $C$ and move along $AB$, $BC$, $CA$ in a uniform linear motion in the counterclockwise direction. It is known that point $P$ takes 1 second to travel from $A$ to $B$, point $Q$ takes 2 seco... | Let the moving point start to move for $t$ seconds, and $S \triangle \mathrm{P} Q \mathrm{R}$ is minimized, then $V_{\mathrm{P}}=a, V_{Q}=\frac{1}{2} a, V_{\mathrm{B}}=\frac{1}{3} a, A P=a t$,
$$
\begin{array}{l}
P B=a(1-t), B Q=\frac{a t}{2}, Q C=a\left(1-\frac{t}{2}\right), \\
C R=\frac{a t}{3}, R A=a\left(1-\frac{t}... | \frac{23 \sqrt{3}}{576} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,196 |
Example 8. As
many, let from a fixed point $P$
draw two tangents $P A, P B$,
and on the minor arc $\overparen{A B}$
1: take any point
$C$, draw a tangent through $C$ intersecting $P A, P B$ at $Q, R$. When does the point $C$ move to a position on the minor arc $\widehat{A B}$ such that the area of $\triangle P Q R$ is ... | Given that $S_{\mathrm{PAOB}}$ is a constant, when $S_{\mathrm{AOBRQ}}$ is minimized, $S \triangle \mathrm{PQR}$ is maximized.
Connect $O R, O Q$, it is easy to prove that $S_{\triangle O B R Q}=2 S_{\triangle O R Q}$.
Also, $\angle R O Q=\frac{1}{2} \angle B O A=$ constant,
with the vertex angle being constant,
the b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,197 |
Example 9. Fold a regular hexagonal paper sheet $A B C D$ with side length $a$, so that $B$ falls on $A D$. At what position on $A D$ should $B$ fall to minimize the area of the folded part? Find this minimum area. | From the given, we know that $M N$ is the fold line, and the trapezoid $B^{\prime} C^{\prime} N M$ is the folded part. Therefore, $B^{\prime}, B$ are symmetric about $M N$, so $B B^{\prime} \perp M N$. From $\mathrm{Rt} \triangle M P B$:
$$
\begin{array}{l}
\text { In } \triangle B^{\prime} A B \text {. Let } A B^{\pri... | \frac{3 a^{2}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,198 |
Example 1. In a square with a side length of 1000 units, there is a non-self-intersecting broken line. Prove: there always exists a straight line, parallel to one side of the square, that intersects the broken line at least 500 times. | Let $l_{\mathrm{i}}$ denote the length of the $i$-th segment of the broken line, and let its projections on the adjacent sides of the square be $a_{1}$ and $b_{1}$. Then we have:
$$
l_{1} \leqslant a_{1}+b_{1}, \quad i=1,2, \cdots, n .
$$
Thus, $1000=l_{1}+l_{2}+\cdots+l_{0}$
$$
\leqslant\left(a_{1}+\cdots+a_{\mathrm{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,200 |
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