problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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To 9. On a plane, there is a fixed point $P$, consider all possible equilateral triangles $ABC$, where $AP=3, BP=2$. What is the maximum length of $CP$? (1961 Autumn Competition) | Solve for Ling and Hui
$$
\begin{array}{l}
\angle A P B=\alpha . \\
\angle B A P=\beta,
\end{array}
$$
Given $A B^{2}=3^{2}+2^{2}$
$$
\begin{aligned}
- & 12 \cos \alpha, \\
\cos \beta & =\frac{3^{2}+A B^{2}-2^{2}}{6 A B}, \sin \beta=\frac{2 \sin \alpha}{A B} .
\end{aligned}
$$
From this, we get
$$
\begin{aligned}
\co... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,139 |
Example 11. Let $a, b, c$ be positive real numbers.
$$
\begin{array}{l}
\text { Prove: } a^{2} b(a-b)+b \cdot c(b-c)+c^{2} a(c-a) \\
\geqslant 0 \text {, } \\
\end{array}
$$
and determine when equality holds. (IMO 1974, Problem 6) | Prove that let $a=x+y, b=y+z, c=z+x$, then the original inequality becomes
$$
\begin{array}{c}
(x+y)^{2}(y+z)(x-z)+(y+z)^{2}(z+ \\
x)(y-x)+(z+x)^{2}(x+y)(z-y) \geqslant 0.
\end{array}
$$
After simplification, we get
$$
x y^{3}+y z^{5}+z x^{3}-x y z(x+y+z) \geq 0,
$$
which is equivalent to $x y z\left(\frac{y^{2}}{z}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,141 |
Example 12. Let $P$ be a point inside $\triangle A B C$, for what position of $P$ is $P A^{2}+P B^{2}+P C^{2}$ minimized? | Let the $x$-axis, with $B C$ as the focal points and $OC$ as the unit length, establish a Cartesian coordinate system. Thus, the coordinates of points $B$ and $C$ are $(-1, 0)$ and $(1, 0)$, respectively. The coordinates of points $A$ and $P$ are $(r, q)$ and $(x, y)$, respectively. Therefore, we have
$$
\begin{array}{... | P A^{2}+P B^{2}+P C^{2} \text{ is minimized when } P \text{ is at the centroid} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,142 |
Let $\lambda_{1}, \lambda_{2}, \lambda_{3}$ be any three positive numbers, and let the side lengths of $\triangle A B C$ be $B C=a, C A=b$, $A B=c$, and the circumradius of $\triangle A B C$ be $R$. Then
$$
\begin{array}{l}
\lambda_{2} \lambda_{3} a^{2}+\lambda_{3} \lambda_{1} b^{2}+\lambda_{1} \lambda_{2} c^{2} \\
\le... | Prove that in the theorem, taking $n=3, A_{1}=A$, $A_{2}=B, A_{3}=C, m_{1}=\lambda_{1}, m_{2}=\lambda_{1}$, $m_{3}=\lambda_{3}$, and assuming $P$ is the centroid of $\triangle A B C$, then from (3) we have
$$
\begin{array}{l}
\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right) R^{2} \\
=(\bar{F} G)^{2}\left(\lambda_{1}+\l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,154 |
IMO-26-5) $\odot O$ passes through the vertices $A, C$ of $\triangle A B C$, intersects sides $A B, B C$ at $K, N$, respectively. The circumcircle of $\triangle A B C$ intersects the circumcircle of $\triangle K B N$ at two distinct points $B$ and $M$. Prove that $\angle O M B=90^{\circ}$. | Proof: First, extend $AC$, $KN$, and $BM$. These three lines must intersect at a point $P$ (this can be proven using the Double Secant Theorem). Observing the cyclic quadrilateral $ACNK$, which meets the conditions of a certain problem, we can use its conclusion to solve this problem. For this purpose, draw the angle b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,156 |
Example 3, Given five distinct points in a plane, the ratio of the maximum distance to the minimum distance among them is denoted as $\lambda$. Prove: $\lambda \geqslant 2 \sin 54^{\circ}$. (1985 National Competition Problem). | Proof: Let $A, B, C, D, E$ be five points forming a convex pentagon. By the pigeonhole principle, at least one of the angles is greater than or equal to $108^{\circ}$. Without loss of generality, assume $\angle A B C \geqslant 108^{\circ}$. Assuming $\angle B A C \geqslant \angle B C A$, we have:
$$
\begin{aligned}
\fr... | 2 \sin 54^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 704,159 |
Given $\triangle A B C$, extend the three sides by 1, 2, 3 times respectively, to get $\triangle A^{\prime} B^{\prime} C^{\prime}$. Ask how many times the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is compared to the area of $\triangle A B C$. | Theorem: Let the area of $\triangle ABC$ be $S$. Extend the sides $AB, BC, CA$ of $\triangle ABC$ such that $BB'=\lambda_{1} AB$, $CC'=\lambda_{2} BC$, $AA'=\lambda_{3} CA$, and let $\triangle A' B' C'$ be the resulting triangle with area $S'$. Then,
$$
S'=1+\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,161 |
Example 1. A circular basket is divided into two equal areas by the curve $A m B$. Prove: the length $l$ of the curve $A m B \geqslant$ the diameter $d$ of the circular disk. | Analysis: In this problem, the shape of the curve is arbitrary, which seems difficult to handle. However, considering the conclusion 1, we can try to change the curve into a polyline, that is, for any point C on the curve $A m B$, it is obvious that $l \geqslant C A+C B$. As long as we can select an appropriate point $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,164 |
Example 2. Prove: A closed curve $L$ of length $4 l$ can certainly be covered by a circle of radius $l$.
| Analysis: This problem can be divided into two steps to solve: the first is to construct the covering circle, i.e., to find the center $O$ of the circle, and the second is to prove that every point on the curve $L$ is inside or on the circumference of $\odot O$, i.e., for any point $A$ on $L$, prove that $O A \leqslant... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,165 |
Example 5. Two regular hexagons are inscribed in a circle of radius $r$, and let the area of their common part be $S$. Prove:
$2 S>\sqrt{3} r^{2}$. (26th IMO Shortlist) | Analysis: Identifying the geometric characteristics of this problem is the key to solving it. It is easy to see that $A M+1 N+N A=A B$. Then,
prove the following $S$ is
the sum of the areas of the three shaded parts,
which, due to the symmetry of the figure, is
three times the area of one of the shaded parts.
Since t... | 2 S \geqslant \sqrt{3} r^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 704,168 |
1. Prove: The set $\{1,2, \cdots, 1989\}$ can be divided into 117 mutually disjoint subsets $A_{i} (i=1,2, \cdots, 117)$, such that
（1）each $A_i$ contains 17 elements;
（2）the sum of the elements in each $A_i$ is the same. | 1. Consider a certain problem: Can the $1, 2, \cdots, m n$ positive integers be divided into $m$ groups, each containing $n$ numbers, such that the sum of the numbers in each group is equal, where $m$ is a positive integer and $n$ is an integer greater than 1? If so, how can this be done?
When $n=2$, it is sufficient ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,169 |
2. In an acute triangle $ABC$, the angle bisector of $\angle A$ intersects the circumcircle of the triangle at point $A_{1}$. Points $B_{1}, C_{1}$ are defined similarly. The line $A A_{1}$ intersects the external angle bisectors of $\angle B$ and $\angle C$ at point $A_{0}$. Points $B_{0}, C_{0}$ are defined similarly... | Let $I$ be the incenter of $\triangle ABC$, then $I$ is the intersection point of $AA_{0}, BB_{0}$, and $CC_{0}$. $\angle B_{1} A_{1}=\alpha+\beta$,
$$
\angle A_{1} B I=\beta+\angle A_{1} B C=\beta+\alpha \text {. }
$$
Thus, $A_{1} B=A_{1} I$.
It is easy to see that $B B_{1} \perp B A_{0}$, so $\angle A_{1} B A_{0}=90... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,170 |
6. The circumradius of $\triangle A B C$ is $R$, and the internal angle bisectors intersect the opposite sides at $A^{\prime}, B^{\prime}, C^{\prime}$. Prove the inequality:
$$
16 Q^{3} \geqslant 27 R^{4} P \text {. }
$$
where $Q$, $P$ are the areas of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and $\triangle A B C$... | 6. Let the internal angles of $\triangle ABC$ be $\alpha, \beta, \gamma$, then
$$
P=\frac{1}{2} R^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) .
$$
Since the internal angles of $\triangle A^{\prime} B^{\prime} C^{\prime}$ are $\frac{\beta+\gamma}{2}, \frac{\alpha+\gamma}{2}$, $\frac{\alpha}{2}+\beta$, we have
$$
\beg... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,171 |
9. A store has 10 televisions, arranged in a row. It is known that 3 of them are defective. If we conduct a random inspection of these televisions, what is the probability that all the defective ones appear in the first 5 televisions?
(A) $\frac{3}{10}$
(B) $\frac{1}{12}$.
(C) $\frac{2}{7}$.
(D) $\frac{1}{7}$.
(E) None... | 9.3 defective items placed on 10 positions is ( $\left.\begin{array}{c}10 \\ 3\end{array}\right)$, and all 3 defective items placed in the first 5 positions is ( $\left.\begin{array}{l}5 \\ 3\end{array}\right)$, so the probability that all defective items appear in the first 5 televisions is ( ( $\left.\begin{array}{l}... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,172 |
10. Let $C$ be a circle with radius $r$, centered at the point $(\sqrt{2}, \sqrt{3})$, where $r$ is a positive real number. When $x, y$ are both rational numbers, the point $(x, y)$ is called a rational point. Then the maximum number of rational points on the circle $C$ is ( ).
(A) $0_{R}$.
(B) 1 .
(C) 2 .
(D) 3 .
(E) ... | 10. Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ be two distinct rational points on circle $C$, then $x_{1}^{2}-2 \sqrt{2} x_{1}+2+y_{1}^{2}$
$$
-2 \sqrt{3} y_{1}+3=x_{2}^{2}-2 \sqrt{2} x_{2}+2+y_{2}^{2}
$$
$-2 \sqrt{3} y_{2}+3$, i.e., $\left(x_{1}^{2}-x_{2}^{2}\right)+\left(y_{1}^{2}-y_{2}^{2}\right)$ $=2... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 704,173 |
1. Let $f(x)$ be an $n$-degree polynomial, and satisfy
$$
f(k)=\frac{k}{k+1}, k=0,1,2, \cdots, n \text {. Find } f(n+1) \text {. }
$$ | 1. $k$ is $(k+1) f(i)-k=0, k=0,1$, $2, \cdots, n$; Therefore, the $n+1$ degree polynomial $g(x)=(x+1) f(x)-x$ has $n+1$ roots: $x=0$, $1,2, \cdots, n$. Thus, we can set $g(x)=c x(x-1) \cdots(x-n)$, where $c$ is a constant, i.e.,
$$
\begin{array}{c}
(x+1) f(x)-x=c x(x-1) \cdots(x-n). \\
\text{Let } x=-1, \text{ we get }... | f(n+1)=\frac{(-1)^{n+1}+n+1}{n+2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,174 |
3. Let $p_{0}=1+2^{n}+3^{n}+4^{n}$, find all positive integers $n$ such that the sum is divisible by 5. | $$
\begin{array}{l}
\text { 3. Let } 2 \equiv 2(\bmod 5), 2^{2} \equiv 4(\bmod 5) \text {, } \\
2^{3} \equiv 3(\bmod 5), 2^{4} \equiv 1(\bmod 5), \\
\text { then } 2^{\mathrm{D}} \equiv 2^{\mathrm{n}+4}(\bmod 5) \text { . } \\
\end{array}
$$
Similarly, $3^{\mathrm{n}} \equiv 3^{\mathrm{n}+4}(\bmod 5)$,
$$
4^{n} \equiv... | n \text{ is not a multiple of 4} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,176 |
5. Find all positive integer solutions $x, y, z$ that satisfy the equation:
$$
5(x y+y z+x z)=4 x y z
$$ | 5. The original equation is transformed into $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}$.
Assume $x \leqslant y \leqslant z$, then
$$
\frac{3}{x} \geqslant \frac{1}{x}+\frac{1}{y}+\frac{1}{x}=\frac{4}{5} \text {. }
$$
Therefore, $x1$.
Thus, we get $1<x<4$.
We will discuss two cases:
Case 1. When $x=2$,
$$
\frac... | (2,4,20), (2,20,4), (4,2,20), (20,2,4), (4,20,2), (20,4,2), (2,5,10), (2,10,5), (5,2,10), (10,2,5), (5,10,2), (10,5,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,178 |
Given that the two arcs have the given points as endpoints), the angle of intersection of these two arcs is at least $\left(1-\frac{2}{n}\right) \pi$.
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。
Given that the two arcs have the given points as endpoints), the angle of intersection of these two arcs is at least $\left(1-\frac{... | 7. Let $P_{1}, P_{3}$ be points in the polygon $E$. The line $P_{1} P_{2}$ intersects the boundary of $E$ at $Q_{1}, Q_{2}$. If $Q_{1}, Q_{2}$ can be transformed by two similar transformations to make $a_{1}^{\prime}, a_{1}^{\prime}$ into the arcs $\alpha_{1}, a_{2}$ connecting $P_{1}$ and $P_{3}$, and the angles at wh... | \left(1-\frac{2}{n}\right) \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,179 |
8. $R$ is a rectangle, which is the union of several rectangles $R_{i}, 1 \leqslant i \leqslant n$, satisfying
(1) The sides of $R_{i}$ are parallel to the sides of $R$,
(2) $R_{i}$ do not overlap,
(3) Each $R_{i}$ has at least one side of integer length.
Prove: $R$ has at least one side of integer length. | 8. Taking $A$ as the origin, $AB$ as the $x$-axis, and $AD$ as the $y$-axis to establish a rectangular coordinate system. We just need to prove that at least one of $B, C, D$ is a lattice point.
For every rectangle $R_{1}$, at least one side length is an integer, and the sides are parallel to the coordinate axes, so t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,180 |
9. $n$ is a non-negative integer, express $(1+4 \sqrt[3]{2} -4 \sqrt[3]{4})$ as $(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})$ " $=a_{a}+b_{\mathrm{a}} \sqrt[3]{2}+c_{\mathrm{a}} \sqrt[3]{4}$, where $a_{\mathrm{n}}, b_{\mathrm{n}}, c_{\mathrm{a}}$ are integers. Prove: if $c_{n}=0$, then $n=0$. | $$
\begin{array}{l}
\text { 9. Given } (1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{n+1} \\
=\left(a_{0}+b_{0} \sqrt[3]{2}+c_{0} \sqrt[3]{4}\right)(1+4 \sqrt[3]{2} \\
-4 \sqrt[3]{4})
\end{array}
$$
we get
$$
a_{n+1}=a_{0}-8 b_{n}+8 c_{n}
$$
Since $a_{0}=1$, all $a_{0}$ are odd.
Every non-zero integer $k$ can be expressed as $k=2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,181 |
10. $g: C \rightarrow C, \omega \in C, a \in C$, $\omega^{3}=1, \omega \neq 1$. Prove that there exists a unique function $f: C \rightarrow C$, satisfying
$$
f(z)+f(\omega z+a)=g(z), z \in C,
$$
Find $f$. | 10. By substituting $\omega z+a$ for $z$ in the functional equation, we get
$$
\begin{aligned}
& f(\omega z+a)+f\left(\omega^{2} z+\omega a+a\right) \\
= & g(\omega z+a) .
\end{aligned}
$$
Repeating this process, we obtain $\quad f(z)+f\left(\omega^{2} z+\omega a+a\right)$
$$
=g\left(\omega^{2} z+\omega a+a\right) \te... | f(z) =\frac{1}{2}\left(g(z)+g\left(\omega^{2} z+\omega a+a\right) - g(\omega z+a)\right) | Algebra | proof | Yes | Yes | cn_contest | false | 704,182 |
11. If $\sum a_{\mathrm{i}}=2$, define $a_{\mathrm{u}}$, prove $n \mid a_{5}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 11. For a sequence $s$ of length $n$ consisting of $0$s and $1$s, if for some $d \mid n$, $s$ can be divided into $d$ identical blocks, then $s$ is called cyclic. Clearly, every sequence of length $n$ can be obtained by repeating its unique, longest non-cyclic initial segment a certain number of times. Since there are ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,183 |
12. On a circular track, there are $n$ cars ready to start at 1 point. Each car runs one lap in 1 hour. Upon hearing the signal, they immediately set off in various directions. If two cars meet, they simultaneously change direction and continue at the original speed. Prove that there must be a moment when each car is b... | 12. Suppose two cars exchange their numbers when they meet, so we see each car, for example, car No. 1, moving around the circle at the same speed and direction, one lap after another. Therefore, after one hour (after several exchanges of numbers), each starting point is occupied by a car with the same number as origin... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 704,184 |
14, A bicentric quadrilateral refers to a quadrilateral that has both an incircle and a circumcircle. Prove that for such a quadrilateral, the two centers and the intersection point of the diagonals are collinear. | 14. Let quadrilateral $ABCD$ be a bicentric quadrilateral, with its circumcenter $O$ and incenter $I$, and the intersection of the diagonals $K$.
Lemma 1 For a tangential quadrilateral $ABCD$, let the points of tangency be $P, Q, R, S$, then the intersection of $PR$ and $QS$ is the intersection of the diagonals $AC$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,185 |
15. Let $a, b, c, d, m, n$ be integers, $a^{2}+b^{2}+c^{2}+d^{2}=1989, a+b+c+d$ is determined (and prove) the values of $m, n$. | 15. By Cauchy-Schwarz inequality,
\[ a + b + c + d \leqslant 2 \sqrt{1989}a^{2} + b^{2} + c^{2} + d^{2} \]
it follows that \( m^{2} = 49 \) or 81.
Assume without loss of generality that \( a \leqslant b \leqslant c \leqslant d = n^{2} \).
If \( m^{2} = 49 \), then
\[
\begin{aligned}
(49 - d)^{2} & = (a + b + c)^{2} >... | a = 12, b = 15, c = 18, d = 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,186 |
3. Let $n$ and $k$ be positive integers, and $S$ be a set of $n$ points in the plane, satisfying:
(1) No three points in $S$ are collinear.
(2) For each point $P$ in $S$, there are at least $k$ points in $S$ that are equidistant from $P$.
Prove:
$$
k<\frac{1}{2}+\sqrt{2 n} \text {. }
$$ | 3. Solution one. For any two points $A_{1}, A_{1}$ in $S$, there are at most two points of $S$ on the perpendicular bisector of $A_{1} A_{1}$ (since no three points in $S$ are collinear). Thus, on such perpendicular bisectors, there are at most $2 C_{n}^{2}$ points of $S$ (each point can be counted repeatedly).
On the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,188 |
17. In the plane, 7 points are given, connect them with some line segments so that
(1) among any three points, at least two are connected,
(2) the number of line segments is minimized.
How many line segments are there? Provide such a graph. | 17. The figure below shows that 9 line segments are sufficient.
Now prove that at least 9 line segments are needed.
If point $A$ is not an endpoint of any line segment, then due to (1), the other 6 points must connect at least $C_{6}^{2}>9$ line segments.
If point $A$ is the endpoint of only 1 line segment, then due... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,189 |
19. In an $m \times n$ rectangular table filled with natural numbers, you can add an integer $k$ to two adjacent cells simultaneously, ensuring the resulting numbers are non-negative integers (two cells sharing a common edge are called adjacent). Determine the necessary and sufficient condition so that after a finite n... | 19. In an $m \times n$ table, adjacent cells are colored with two different colors * and $\square$, and the sums of the numbers in the two types of cells are denoted as $S_{*}$ and $S_{\square}$, respectively. Let $S = S_{*} - S_{\square}$. Since $S$ remains unchanged after each operation, $S = 0$ is a necessary condit... | S = 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,191 |
24. Points $A_{1}, \cdots, A_{5}$ are on a sphere with radius 1, what is the maximum value of $\min _{1 \leqslant \mathrm{i}, \mathrm{j} \leqslant 5} A_{1} A_{\mathrm{j}}$? Determine all cases where the maximum value is achieved. | 24. We can use the size of $\angle A_{1} O A_{1}$ to replace the distance $A_{1} A_{\mathrm{j}}$, where $O$ is the center of the sphere.
There exists a set of points $A_{1}, A_{2}, \cdots, A_{5}$, such that
$$
\mathrm{mn}_{1 \leqslant \mathrm{i}, \mathrm{j}} \leqslant A_{1} O A_{\mathrm{j}} \leqslant \frac{\pi}{2} .
$$... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,193 |
25. Let $a, b$ be integers, not perfect squares. Prove: If $x^{2}-a y^{2}-b z^{2}+a b w^{2}=0$ has a non-trivial integer solution (i.e., not all zero integer solutions), then $x^{2}-a y^{2}-b z^{2}=0$ has a non-trivial integer solution. | 25. $a, b$ cannot both be negative. Without loss of generality, we can assume $a>0$ (where $a, b$ are not perfect squares, and of course, not $0$). Let $\left(x_{0}, y_{0}, z_{0}, w_{0}\right) \neq(0,0,0,0)$ be a solution to
$$
x^{2}-a y^{2}-b z^{2}+a b w^{2}=0
$$
Then,
$$
x_{0}^{2}-a y_{0}^{2}-b\left(z_{0}^{2}-a w_{0... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,194 |
32. In an acute triangle $ABC$, the distance from vertex $A$ to the circumcenter $O$ is equal to the distance from $A$ to the orthocenter $H$. Find all possible values of $\angle A$. | 32. Let $C^{\prime} C^{\prime}$
be the height. Since
$A H=R$ (the radius of the circumcircle, so
$$
\begin{array}{l}
A C^{\prime}=R \\
\cdot \sin \angle A H C^{\prime} \\
=R \sin B .
\end{array}
$$
Thus, $C C^{\prime}=R \sin B \operatorname{tg} A$.
Also, $C C^{\prime}=B C \sin B=2 R \sin A \sin B$,
comparing the two ... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,198 |
31. Let $a_{1} \geqslant a_{2} \geqslant a_{3}$ be given positive integers, and $N\left(a_{1}, a_{2}, a_{3}\right)$ be the number of solutions $\left(x_{1}, x_{2}, x_{3}\right)$ to the equation
$$
\frac{a_{1}}{x_{1}}+\frac{a_{2}}{x_{2}}+\frac{a_{3}}{x_{3}}=1
$$
where $x_{1}$, $x_{2}, x_{3}$ are positive integers. Prove... | $N_{1, \mathrm{j}}$, is the number of positive integer solutions to the equation satisfying $\frac{a}{x_{1}} \geqslant \frac{a_{\mathrm{j}}}{x_{\mathrm{j}}} \geq x_{2}$
$$
\frac{a}{x_{1}} + \frac{a}{x_{2}} + \frac{a_{3}}{x_{3}} = 1
$$
The number of solutions is $N$
$$
\mathrm{a}_{1}, \mathrm{a}_{2}, a_{3} \leqslant \s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,200 |
26. $n$ is a positive integer, $a, b$ are given real numbers, $x_{0}, x_{1}, \cdots, x_{n}$ are real variables,
$$
\sum_{i=0}^{n} x_{i}=a, \quad \sum_{i=0}^{n} x_{i}^{2}=b .
$$
Determine the range of variation for $x_{n}$. | 26. By the Cauchy inequality,
$$
\left(\sum_{1=1}^{n} x_{1}\right)^{2} \leqslant n \sum_{1=1}^{n} x_{1}^{2} \text {. }
$$
Therefore, $\left(a-x_{0}\right)^{2} \leqslant n\left(b-x_{0}^{2}\right)$,
which means $(n+1) x_{0}^{2}-2 a x_{0}+a^{2}-n b \leqslant 0$.
The discriminant of this quadratic trinomial is
$$
D=4 n(n+... | \frac{a-\sqrt{\frac{D}{4}}}{n+1} \leqslant x_{0} \leqslant \frac{a+\sqrt{\frac{D}{4}}}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,201 |
2. Calculate: $\left(\frac{1-i}{\sqrt{2}}\right)^{1080}=$ | 2. $-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i$.
untranslated as it is a mathematical expression. | -\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,203 |
4. The polar equation of a circle is $\rho=5 \cos \theta$ $-5 \sqrt{3} \sin \theta$, if the range of the polar angle is specified as $0 \leqslant \theta<2 \pi$, then the polar coordinates of its center are | 4. $\left(5, \frac{5 \pi}{3}\right)$. | \left(5, \frac{5 \pi}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,205 |
5. Prove: For any positive integer $n$, there exist $n$ consecutive positive integers, none of which are prime numbers. | 5. Let $a=(n+1)!$, then $a^{2}+k(2 \leqslant k \leqslant n+1)$ is divisible by $k$ but not by $k^{2}$ (since $a^{2}$ is divisible by $k^{2}$ and $k$ is not divisible by $k^{2}$) - if $a^{2}+k$ is a power of a prime $p^{t}$, then $k=p^{1}(t, j$ are positive integers). But $a^{2}$ is divisible by $p^{2}$, and thus by $p^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,210 |
11. If an arithmetic sequence, starting from the 1st term, the sum of any number of terms is always equal to 10 times the square of the number of terms, then the general term formula of this sequence $a_{n}=$ $\qquad$ - | 11. $10(2 n-1)$. | 10(2 n-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,213 |
12. Given the quadratic curve $x^{2}-8 x \sin ^{2} \theta+4 y$ $+16 \sin ^{4} \theta+4=0$ has its focus on the line $x-y=3$, then the value of $\theta$ in the interval $\left[\frac{3 \pi}{2}, 2 \pi\right]$ is $\qquad$ $\bullet$ | 12. $\frac{11}{6} \pi$. | \frac{11}{6} \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,214 |
14. Calculate: $\operatorname{arcctg} \frac{1}{\sqrt{2}}$ $+\frac{1}{2} \arcsin \frac{2 \sqrt{2}}{3}=$ $\qquad$ | 14. $\frac{\pi}{2}$. | \frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,216 |
15. If each asymptote of the hyperbola \((x+m)^{2}-\frac{(y+2 m)^{2}}{9}=1\) intersects the parabola \(y=x^{2}+1\) at two distinct points, then the range of the real number \(m\) is \(\qquad\). | 15. $-\frac{5}{4}<m<\frac{1}{4}$ | -\frac{5}{4}<m<\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,217 |
16. Calculate: $\frac{C_{11}^{0}}{1}+\frac{C_{11}^{1}}{2}+\frac{C_{11}^{2}}{3}+\cdots$ $+\frac{C_{11}^{k}}{k+1}+\cdots+\frac{C_{11}^{11}}{12}=$ $\qquad$ | 16. $\frac{1365}{4}$ | \frac{1365}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,218 |
17. Let $A=\left\{(x, y) \mid 2-x^{2}-y^{2}-\right.$ $\left.\sqrt{\left(1-x^{2}\right)^{2}+\left(1-y^{2}\right)^{2}} \geqslant 0\right\}$ represent a set of points on the Cartesian plane, then the area of $A$ is | 17. 4.
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 | 704,219 |
19. Given the quadratic equation $z^{2}-2(4 \operatorname{tg} \theta+3) z$ $+25 \sec ^{2} \theta=0$ with roots corresponding to points $F_{1}, F_{2}$ on the complex plane, then when $\theta$ takes all real values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the ordinary equation of the trajectory of the endpoints o... | 19. $\frac{y^{2}}{25}-\frac{(x-3)^{2}}{16}=1$ | \frac{y^{2}}{25}-\frac{(x-3)^{2}}{16}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,222 |
There is no other measuring tool, but it was found that by laying this carpet flat in each of the two store rooms and securing each corner to the room's different measurement points - if the dimensions of the two rooms are 38 feet $\times 55$ feet and 50 feet $\times 55$ feet, find the area of the carpet. | 2. Let the side lengths of the carpet be $x, y$.
It is easy to see that
$\triangle A E H \cong \triangle C G F \backsim \triangle B F E \cong \triangle D H G$.
Let $\frac{y}{x}=k$, which is the similarity ratio of the two sets of triangles. Let $A E=a, A H=b$. By similarity, $B E=k b$, $D H=k a$. Therefore,
$$
\begin{a... | 25 \times 50 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,232 |
．The system of equations
$$
\left\{\begin{array}{l}
\operatorname{tg} x+\frac{\pi}{\operatorname{tg} x}=2 \sin \left(y+\frac{\pi}{4}\right), \\
\operatorname{tg} y+\frac{1}{\operatorname{tg} y}=2 \sin \left(x-\frac{\pi}{4}\right)
\end{array}\right.
$$
has the solution $\qquad$ $\bullet$ | 2. The solution to the system of equations
is $\left\{\begin{array}{l}x=2 k \pi+\frac{3 \pi}{4}, \\ y=2 k \pi-\frac{3 \pi}{4},\end{array}(k \in Z)\right.$, | \left\{\begin{array}{l}x=2 k \pi+\frac{3 \pi}{4}, \\ y=2 k \pi-\frac{3 \pi}{4},\end{array}(k \in Z)\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,234 |
3. Arrange the positive rational numbers in the following sequence,
$$
\begin{array}{l}
\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,
\end{array}
$$
Then the position number of the number $\frac{1989}{1949}$ is $\qquad$ | 3. The number's position in the original
sequence is $(1+2+\cdots+3937)+$ $1949=7753902$, | 7753902 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,235 |
4. Let the set $A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}x=\sec \theta, \\ y=\tan \theta,\end{array}\right.\right.\right.$
$$
\begin{array}{l}
0 \leqslant \theta < \pi\right.\right\}, B= \\
\left\{(x, y) \mid y^{2} \leqslant 3(x+3)\right\}, D=A \cap B \cap C .
\end{array}
$$
When $(x, y) \in D$, then the ma... | 4. $y-2x$ attains its maximum value $4+\sqrt{\overline{3}}$ at point $M(-2, \sqrt{\overline{3}})$, | 4+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,236 |
5. Through the side $B C$ of the base of a regular triangular prism $A B C-$ $A_{1} B_{1} C_{1}$ with edge length 1, a section is made forming a dihedral angle $\theta$ with the base. Find the functional relationship of the section area $S$ with $\theta$. $S(\theta)=$ $\qquad$ | $\begin{array}{l}\text { 5. } S(\theta) \\ =\left\{\begin{array}{l}\frac{\sqrt{3}}{4} \sec \left(0 \leqslant \theta \leqslant \operatorname{arctg} \frac{2 \sqrt{3}}{3}\right), \\ \frac{2 \sqrt{3}}{3} \sin \left(\theta-\frac{\pi}{6}\right) \csc ^{2} \theta\left(\operatorname{arctg} \frac{2 \sqrt{3}}{3}\right. \\ \left.<... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,237 |
7, A moving point starts from the origin and first moves along the $x$-axis. After completing each unit length, it turns left by an angle $\theta$. Find the distance of this moving point from the origin after it has traveled $n$ unit lengths, which equals
| $\begin{array}{l}7 . \overline{O P_{n} \mid} \\ =\left|\frac{\sin \frac{n}{2} \theta}{\sin \frac{1}{2} \theta}\right|\end{array}$ | \left|\frac{\sin \frac{n}{2} \theta}{\sin \frac{1}{2} \theta}\right| | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,239 |
8. A hotel has 90 vacant rooms, each with a unique key. 100 guests arrive, and keys need to be distributed so that any 90 of them can stay in the 90 rooms, with each person getting one room (assuming there is no limit to the number of keys that can be issued for each room or the number of keys each person can receive).... | 8. At least 990 keys should be prepared. | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,240 |
Let $A=\left\{x \mid x=3^{\mathrm{n}} x_{\mathrm{a}}+3^{\mathrm{n}-1} x_{\mathrm{n}-1}+\cdots\right.$ $+3 x_{1}+x_{0}, x_{1}=-1,0$ or $1, i=0$, $1,2, \cdots, n\}, H=\frac{3^{n+1}-1}{3-1}$, prove: it is possible to use $n+1$ specially designed weights to measure any weight from 1 to $H$ on a balance scale. | First, use $H$ to form the set $E$. $B=\{-H, \cdots, -1, 0, 1, \cdots, H\}$. It can be proven that each element in $B$ can be uniquely represented in the form: $3^{\mathrm{n}} x_{\mathrm{n}} + 3^{n-1} x_{1} + \cdots + 3 x_{1} + x_{0}$, where $x_{1} = -1, 0, 1$. This proves that $A = B$. Now, take specially designed wei... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,242 |
3. The carpet owner has a rectangular carpet, the size of which is unknown. Unfortunately, his measuring tape is broken, and he has no other measuring tools. However, he discovers that if he lays the carpet flat in either of his two store rooms, each corner of the carpet exactly meets a different wall. He knows that th... | 3. Let the unknown side of the room be \( q \) chi. From the previous problem, we have
\[
\begin{array}{l}
(k q-50)^{2}+(50 k-q)^{2} \\
=(k q-38)^{2}+(38 k-q)^{2}
\end{array}
\]
Thus,
\[
\begin{aligned}
& k^{2} q^{2}-100 k q+2500+2500 k^{2}-100 k q+q^{2} \\
= & k^{2} q^{2}-76 k q+1444+1444 k^{2}-76 k q+q^{2} .
\end{al... | 25 \text{ chi}, 50 \text{ chi} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,243 |
Five, divide a circle into $n(n \geqslant 2)$ sectors, sequentially denoted as $S_{1}, S_{2}, \cdots, S_{n}$. Each sector can be painted with any of the three different colors: red, white, and blue, with the requirement that adjacent sectors must have different colors. How many ways are there to color the sectors? | Let the total number of coloring methods be $a_{n}(n \geqslant 2)$.
When $n=2$, first color $S_{1}$, there are three ways to color it. After coloring $S_{1}$, continue to color $S_{2}$, there are only 2 ways, thus $a_{2}=2 \times 3=6$.
Now, let's determine the recursive relationship.
If we first color $S_{1}$, there ar... | 2 \left[2^{n-1} - (-1)^{n-1}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,244 |
Six, there are 6000 points inside a circle, and no three points are collinear:
(1) Can this circle be divided into 2000 parts, each containing exactly three points? How to divide it?
(2) If the three points in each part satisfy: the distance between any two points is an integer and does not exceed 9, then using the thr... | (1) 6000 points inside a circle can determine $C_{6000}^{2}$ lines. Since $C_{6000}^{2}$ is a finite number, there must exist a tangent line to the circle that is not parallel to any of the $C_{5000}^{2}$ lines, denoted as $l$. Moving $l$ parallel within the circle, it is clear that the 6000 points will be crossed (if ... | 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,245 |
3. Let $x, y, z$ be three distinct natural numbers, and the product of any two of them is divisible by the third. Prove: the equation $x-y+z=1$ has infinitely many solutions. | Prove that the solution can be expressed in the form $x=m n, y=n k, z=m k$, where $m, n, k$ are natural numbers, which satisfies the required divisibility conditions. Substituting into the equation, we get $m n - n k + m k = 1$, i.e., $n(k - m) = m k - 1$. We only need to consider the solutions when $k - m = 1$. Thus, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,248 |
1. Find the integer solution to the equation
$$
\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1088}
$$ | Answer $m=-1989$. | -1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,250 |
3. A school needs to organize duty shifts as follows: one student from Class 1, Grade 9 and one student from Class 2, Grade 9 will be on duty together each day, and each day exactly one pair of students will be on duty. Students from each class will take turns according to the list in the class notebook, and after the ... | Assume that the required duty schedule can be formulated. According to this schedule, let Class 9(1) rotate $a$ times, and Class 9(2) rotate $b$ times. Then, from the given conditions, we get the equation $a+b=32 \cdot 29$. At the same time, according to the conditions, the first pair of students is on duty again at th... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,252 |
4. On the side $BC$ of the right-angled triangle $ABC$ with $\angle C$ as the right angle, take a point $D$ between points $B$ and $C$. On the segment $BC$, there is another point $M$ different from point $D$. Draw a line $AM$ through $M$, intersecting the circumcircle $S$ of triangle $ABC$ at point $N$. Draw a circle ... | We denote the smallest angle required to rotate line $l$ counterclockwise to be parallel to line $n$ as $\angle(l, n)$.
Lemma: Four non-collinear points $P, Q, R, S$ are concyclic if and only if $\angle(Q P, Q R)=\angle(S P, S R)$ (Figure 1). This can be proven by the properties of angles subtended by the same arc in ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,253 |
3. All vertices of a broken line lie on the faces of a cube with an edge length of 2, and each segment of the broken line is 3 units long. This broken line connects two farthest vertices of the cube. How many segments does such a broken line have at least?
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:... | Solve: Consider the cube circumscribing a sphere with center $A$ and radius 3, intersecting the cube's faces at three arcs: $K \hat{L}, \widehat{L N}, \widehat{N K}$ (Figure 3). The points $K, L, N$ on the edges bisect these three edges. In fact, $A D_{1}=\sqrt{8}$, so $L D_{1}=\sqrt{A} L^{\overline{2}-\overline{D_{1}^... | 6 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,256 |
6. On the blackboard, there are numbers 1 and 2. It is stipulated that new numbers can be written according to the following method: If there are numbers $a$ and $b$ on the blackboard, then the number $a b + a + b$ can be written. Using this method, can the following numbers be obtained:
(a) Number 131213
(b) Number 12... | Let the new number $ab + a + b$ be $c$. This means $c + 1 = ab + a + b + 1 = (a + 1)(b + 1)$. This implies that if the number written on the blackboard is replaced by a number that is 1 greater, then each new number will be the product of two existing numbers. Starting with the numbers 2 and 3, after several multiplica... | 13121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,258 |
7. Let rational numbers $x, y$ satisfy the equation $x^{5}+y^{5}$ $=2 x^{2} y^{2}$. Prove: $1-xy$ is the square of a rational number. | Prove that if $xy=0$, then $1-xy=1^2$; if $xy \neq 0$, then square both sides of the given equation and subtract $4x^5y^5$, to get the equation $\left(x^5-y^5\right)=4x^4y^4-4x^5y^5$. From this, we obtain: $1-xy=\left(\frac{x^5-y^5}{2x^2y^2}\right)^2$ | 1-xy=\left(\frac{x^5-y^5}{2x^2y^2}\right)^2 | Algebra | proof | Yes | Yes | cn_contest | false | 704,259 |
8. A country has 21 cities, and some airlines can implement air transportation between these cities. Each airline connects pairs of cities with non-stop flights (and several airlines can operate flights between the same two cities at the same time). Every two cities are connected by at least one non-stop flight. How ma... | To make this country form an aviation network that meets the conditions required by the problem, there must be at least 21 airlines, because the total number of non-stop routes is no less than $20+19+\cdots+3+2+1=210$, and each airline provides $4+3+2+1=10$ non-stop routes. Figure 5 is an example of a service route map... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,260 |
5. Prove that in the sequence with the general term $a_{n}=1+2^{2}+3^{3}+\cdots+n^{n}$, there are infinitely many odd composite numbers.
Hint: It is sufficient to prove that there are infinitely many terms in the sequence $\left\{a_{n}\right\}$ for $n=4 m+1$ that are divisible by 3. | Let $p$ be a natural number, the terms $\left\{a_{n}\right\}$ of the sequence at positions $36 p + 17, 36 p + 25, 36 p + 33$ are odd composite numbers. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,261 |
6. For the acute $\triangle A B C$, construct the circumcircle. The tangents to the circle at points $A$ and $C$ intersect the line through point $B$ at points $M$ and $N$ respectively: In $\triangle A B C$, draw $B P$ (where point $P$ is on side $A C$). Try to prove: Line $B P$ is the angle bisector of $\angle M P N$. | From the similarity of right triangles $\triangle A M M_{1}$ and right $\triangle C N N_{1}$ (Figure 6), we get
$$
\frac{A M_{1}}{C N_{1}}=\frac{A M}{C N}=\frac{a}{b} .
$$
By the intercept theorem for parallel lines:
$$
\frac{M_{1} P}{P N_{1}}=\frac{M B}{B N}=\frac{a}{b} \text {. }
$$
From (1) and (2), we obtain:
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,262 |
8. In an $n \times n$ square grid, real numbers are recorded, and the sum of numbers in any row and any column is zero. The following operation can be performed on the grid: add all elements of any row to another column, and subtract another column (the $i$-th element of the row is added to (subtracted from) the $i$-th... | To prove: If we number the rows of the table from top to bottom and the columns from left to right, the operation of adding the $i$-th column and subtracting the $k$-th column is denoted as $\mathrm{O},{ }^{1}, \mathrm{k}$.
For the entire table, the operations should be performed in the following order: $\mathrm{O}_{\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,264 |
8. Prove: For any tetrahedron, the inequality holds
$$
r < \frac{ab}{2(a+b)}.
$$
where \(a, b\) are the lengths of two opposite edges, and \(r\) is the radius of the exsphere. | Given $A D=a, B C=1$, prove that in the plane containing $A D$ and $B C$, the intersection forms a parallelogram $K L M N$. Let $m=K L, n=L M$. Since $K L \| A D$, $L M \| B C$, we have,
$$
\frac{m}{a}=\frac{B L}{B D}, \quad \frac{n}{b}=\frac{D L}{B D},
$$
which implies $-\frac{m}{a}+\frac{n}{b}=1$. From this, we can ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,267 |
3. Given the decimal 0.123456789101112131415 $161718192021 \cdots$, the 1988th digit after the decimal point is ( ) .
(A) 0 .
(B) 3 .
(C) 4 .
( D) 8 .
(E) 9 . | 3. Noting that there are 9 one-digit numbers (1 to 9), 90 two-digit numbers (10 to 99), and 900 three-digit numbers (100 to 999), and that $1988=9+90 \times 2+599 \times 3+2$;
we have
$$
\begin{array}{l}
\underbrace{0.1 \cdots 9}_{9 \uparrow} \underbrace{10 \cdots 99}_{90 \uparrow} 1 \underbrace{100 \cdots 698}_{599 \u... | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 704,270 |
4. $\frac{1}{2 \sqrt{ } 1}+1 \sqrt{2}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}$ $+\cdots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}}$ The value is ( ).
(A) $\frac{3}{4}$.
(B) $\frac{9}{10}$.
(C) 1 .
(D) $\sqrt{2}$.
(E) None of the above. | 4. $\sum_{n=1}^{n} \frac{1}{(n+1)} \sqrt{n}+n \sqrt{n+1}$
$$
\begin{array}{l}
=\sum_{n=1}^{\infty} \frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1) n} \\
=\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) \\
=1-\frac{1}{\sqrt{100}}=\frac{9}{10} .
\end{array}
$$
Therefore, the correct choice is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 704,271 |
5. The maximum value of the function $y=\sqrt{1}+\sin x+\sqrt{1-\sin x}$ is ().
(A) 2.
(B) $\sqrt{5}$.
(C) $\sqrt{6}$.
(D) $\sqrt{2}$.
(E) None of the above. | $$
\text { 5. Since } \begin{aligned}
y^{2} & =\left(\sqrt{1+\sin x}-\sqrt{1-\sin x)^{2}}\right. \\
& =2-2 \sqrt{(1+\sin x)(1-\sin x)} \\
& =2-2|\cos x|,
\end{aligned}
$$
if $|\cos x| \leqslant 1$,
therefore, when $x=k \pi$ ( $k$ is an integer), $y$ reaches its maximum value of 2. Hence, the correct choice is (A).
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 704,272 |
7. Below are four conditional operations:
(1) $1=2 \rightarrow 10^{2}=100$,
(2) $1=2 \rightarrow 10^{\prime} \neq 100$,
(3) $1 \neq 2 \rightarrow 10^{2}=100$,
(4) $1 \neq 2 \rightarrow 10^{2} \neq 100$.
The number of correct ones is ( ).
(A) None.
(B) 1.
(C) 2.
(D) 3. (E) 4. | False, therefore among these four proposition operations, only (2) is not true, so the answer should be (D).
Translate the text into English, please keep the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 704,274 |
Example 1. (Qicheng City Junior High School Competition Question) Prove that the discriminant $\Delta \neq 1986$ for the quadratic equation $a x^{2}+b x+c=0$ with integer coefficients. | Proof Assume $\Delta=1986$, since 1986 can be expressed in the form $4k + 2$, where $k$ is an integer (here $k=496$), let $\Delta=b^{2} - 4ac = 4k + 2$, then $b^{2}$ is even, hence $b$ is even. Let $b=2t$, where $t$ is an integer, substituting and simplifying we get
$$
2\left(t^{2} - ac\right) = 2k + 1 \text{.}
$$
The... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,276 |
Example 2 - (1961 Moscow Mathematical Olympiad problem) Prove: there do not exist integers $a, b, c, d$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
a b c d-a=1961, \\
a b c d-b=961, \\
a b c d-c=61, \\
a b c d-d=1
\end{array}\right.
$$ | Proof: Assuming such integers exist, then from 1961 being an odd number and $1961=a(b c d-1)$, we can deduce that $a$ is odd, and $b c d-1$ is a composite number.
Similarly, it can be proven that $b, c, d$ are all odd. Therefore, $b c d$ is odd, making $b c d-1$ even, which is a contradiction.
Therefore, there do not... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,277 |
Example 11. (Adapted from a Soviet competition problem) A rectangular prism is formed using identical cubes, and the three faces of the rectangular prism that meet at a common vertex are painted. Can it be ensured that at least one face of each of the 1988 cubes is painted? Prove your conclusion. | Let the edge length of the cube be 1, and the dimensions of the rectangular prism be $m n k(m \geqslant n \geqslant k)$. At this time, the total number of cubes is $m n k$, and the number of uncolored cubes is $(m-1)(n-1)(k-1)$. According to the problem,
$$
(m-1)(n-1)(k-1)=\left(1-\frac{1921}{1988}\right) m n k,
$$
wh... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,278 |
Example 1. Prove that in tetrahedron $ABCD$, there must be a vertex from which the three edges emanating can form a triangle. (68th IMO Problem 4) | Proof one Consider the sum of the lengths of the three edges emanating from each vertex. Without loss of generality, assume that the sum of the edge lengths from vertex $A$ is the largest, then $A B, A C, A D$ can form a triangle.
Otherwise, there must be a sum of two edge lengths not greater than the third edge lengt... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,283 |
Example 2. At a party, $n(\geqslant 2)$ pairs of young men and women dance together. Suppose no man has danced with all the women, and each woman has danced with at least one man. Prove that there must be two men $b_{1}, b_{2}$ and two women $g_{1}, g_{2}$, such that $b_{1}$ has danced with $g_{1}$, $b_{2}$ has danced ... | Proof Let one of the male youths who danced with the most female youths be $b_{1}$. Since $b_{1}$ has not danced with all the female youths, there exists a female youth $g_{2}$ who has not danced with $b_{1}$. Since $g_{2}$ has danced with at least one male youth, there exists $b_{2}$ who has danced with $g_{2}$. If ev... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,284 |
In a tournament, each match must determine a winner. The excellent player is determined through matches. Player $A$ is designated as an excellent player under the condition that for any other player $B$, either $A$ beats $B$, or $A$ indirectly beats $B$, i.e., there exists a player $C$, such that $\boldsymbol{A}$ beats... | Prove that there must exist an excellent player. Since the number of participants is finite, there must be a player with the most wins. Let $A$ be one of the players with the most wins. If $A$ wins all other players, then $A$ is certainly an excellent player. Otherwise, let $A$ win $B_{1}, \cdots, B_{k}$ and lose to $B... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 704,285 |
Example 4. Given a sequence of real numbers $\left\{a_{k}\right\}_{k=1}^{\infty}$ with the following properties: there exists a natural number $n$, such that
$$
a_{1}+a_{2}+\cdots+a_{a}=0
$$
and $a_{a+k}=a_{k}, k=1,2, \cdots$.
Prove that there exists a natural number $N$, such that for $k=0,1,2, \cdots$, we always hav... | Let
$$
S_{\mathrm{j}}=a_{1}+a_{2}+\cdots+a_{\mathrm{j}}, \quad i=1,2, \cdots .
$$
According to the given information, we have
$$
\begin{array}{l}
S_{\mathrm{PA}}=0, \quad p=1,2, \cdots, \\
S_{\mathrm{O}+\mathrm{i}}=S_{\mathrm{j}}, \quad j=1,2, \cdots .
\end{array}
$$
This indicates that the sequence $\left\{S_{i}\rig... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,286 |
Example 3 - (Beijing Initial Number Competition Question) A cinema has a total of 985 seats, with performances in the morning and afternoon. Schools A and B each have 1985 students watching the movie (either the morning or the afternoon session). Prove: There must be such a seat in the cinema where a student from a dif... | Proof: Assuming no seat is occupied by students from different schools in the morning and afternoon, let's say students from School A sit in $n$ seats in the morning, then students from School B sit in $1985-n$ seats. The remaining $n$ students from School B must sit in the $1985-n$ seats occupied by School B in the mo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,288 |
Example 6. Given $a_{1}=1, a_{2}=2$,
$$
a_{n+2}=\left\{\begin{array}{l}
5 a_{n+1}-3 a_{n}, \text { when } a_{n} \cdot a_{n+1} \text { is a composite number, } \\
a_{n+1}-a_{n}, \text { when } a_{n} \cdot a_{n+1} \text { is an odd number. }
\end{array}\right.
$$
Prove that for all natural numbers $n, a_{n} \neq 0$. | Proof one: Given $a_{1}=1, a_{2}=2$ and the recursive formula, we know that the parity of $a_{0}, a_{\mathrm{a}+1}, a_{\mathrm{a}+2}$ can only be one of the following three:
Odd, Even, Odd, Even, Odd, Odd; Odd, Odd, Even. Notice that $a_{1}=1, a_{2}=2, a_{3}=7, a_{4}=29, a_{5}=22$ are not multiples of 4, and we will p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,289 |
Example 8. Prove that the equation $x^{3}-2 y^{3}-4 z^{3}=0$ has no positive integer solution set. | Proof: If the equation has a set of positive integer solutions, let $\left(x_{0}, y_{0}, z_{0}\right)$ be the set with the smallest $x$ value among all positive integer solutions. From the equation, we know that $x_{0}$ is even, and we set $x_{0}=2 x_{1}$. Thus, the equation becomes
i.e., $\square$
$$
\begin{array}{l}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,290 |
Example 9. Let $f(n)$ be a strictly increasing function defined on the set of natural numbers and taking natural number values, with $f(2)=2$, and when $m$, $n$ are coprime, $f(m n)=f(m) f(n)$. Prove that for all natural numbers $n$, $f(n)=n$.
| Proof Given that
$$
\begin{aligned}
f(3) f(7) & =f(21)f(2)=2, \text{ hence } f(3)=3.
\end{aligned}
$$
If the proposition is not true, let the smallest positive integer for which $f(n) \neq n$ be $n_{0} \geqslant 4$. Since $f\left(n_{0}\right)>f\left(n_{0}-1\right)=n_{0}-1$, it follows that $f\left(n_{0}\right)>n_{0}$. ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,291 |
Example 10. Positive integers $a$ and $b$ make $a b+1$ divide $a^{2} + b^{2}$. Prove that $\frac{a^{2}+b^{2}}{a b+1}$ is the square of some positive integer. | Prove that if the positive integer
$$
k=\frac{a^{2}+b^{2}}{a b+1}
$$
is not a perfect square, consider the indeterminate equation
$$
a^{2}+b^{2}-k a b=k, \text{ where } k \text{ is a constant. }
$$
Obviously, the solution $(a, b)$ of this indeterminate equation will not make $a b$ $0, b>0$ and. The solution that make... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,292 |
Example 11. On an $n \times n$ square chessboard, placing pieces follows the following conditions: if a certain small square is empty, then the total number of pieces placed on the horizontal line and vertical line passing through this square is no less than $n$. Prove that the total number of pieces placed on the boar... | Proof: Suppose in $n$ rows and $n$ columns, the row with the fewest chess pieces has $k$ pieces. For each of these $k$ pieces, the number of chess pieces in the column where they are located is $\geqslant k$. On the other hand, this row has $n-k$ empty cells, and the number of chess pieces in the column corresponding t... | \frac{n^{2}}{2} | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,293 |
Example 12. On the plane, there is an infinite set of rectangles, each with vertices at coordinates $(0,0), (0, m)$, $(n, m), (n, 0)$, where $n$ and $m$ are positive integers (different rectangles correspond to different $m$, $n$ values). Prove that from these rectangles, two can be selected such that one rectangle is ... | Prove that obviously, the horizontal side length of the rectangle with vertices $(0,0), (0, m), (n, m), (n, 0)$ is $n$ and the vertical side length is $m$. From the known rectangles, select the rectangle $R_{1}$ with the smallest horizontal side length, and denote its vertical side length as $m_{1}$. If there exists a ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,294 |
Example 1. (Wenzhou Junior High School Math Competition Question, 1987) Let the natural number $n$ have the following property: from 1, 2, ..., $n$, any 51 different numbers chosen will definitely have two numbers whose sum is 101. The largest such $n$ is $\qquad$ | Consider $\{1,2, \cdots, n\}$ as the vertex set. When the sum of two numbers is 101, connect the corresponding two vertices to form a graph $G$. Clearly, the original problem is equivalent to finding the largest $n$ such that any selection of 51 vertices in $G$ must include two adjacent vertices.
i) When $51 \leqslant ... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,297 |
Example 2. (Sichuan Province Junior High School Mathematics Joint Competition Question, 1987) Five teams are participating in a round-robin tournament. The rules state that the winner of each match gets 2 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. It is known that the total poin... | Let the 5 teams be denoted as $v_{1}, v_{2}$, $\cdots, v_{5}$. Consider the teams as vertices, and connect edges according to the following rules:
If team A beats team B, connect a directed edge from the former to the latter; if team A draws with team B, connect an edge between the two vertices.
$\rightarrow$ By (2), ... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,298 |
Example 4. (1978 Shanghai Mathematics Competition Question) Prove: There do not exist two irreducible fractions whose sum and product are both integers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove that if there exist two irreducible fractions $\frac{m_{1}}{n_{1}}, \frac{m_{2}}{n_{2}}$,
such that $\left\{\begin{array}{l}\frac{m_{1}}{n_{1}}+\frac{m_{2}}{n_{2}}=p, \\ \frac{m_{1}}{n_{1}} \cdot \frac{m_{2}}{n_{2}}=q,\end{array}\right.$ where $p, q$ are integers, then $\frac{m_{1}}{n_{1}}$ and $\frac{m_{2}}{n_{2... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,299 |
Example 3. (1956 Beijing Mathematical Competition Question) In space, it is impossible for polyhedra to exist that have an odd number of faces, each of which also has an odd number of edges. | Prove that for a polyhedron, if the faces are considered as vertices, and an edge is drawn between two vertices when the corresponding two faces share an edge, the resulting graph $G$ satisfies:
i) $|V|$ is odd (because the number of faces is odd)
ii) $d\left(v_{1}\right)$ is odd (because each face has an odd number of... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,300 |
Example 4. (85 Provincial Six Autonomous Regions High School Mathematics Joint Competition Question) A football invitational tournament involves sixteen cities, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and teams from the same city do not play a... | Prove that if two teams have played against each other, then an edge is connected between the corresponding two vertices to form graph $G$. Let the team from City A be $v^{*}$, then the original problem is equivalent to: find the degree of the vertex corresponding to the other team from City A.
Obviously, $\max d\left(... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,301 |
Example 5. (Hefei Mathematical Competition Question in 1983) A new station is opened, and several bus routes are planned to serve the community. Their wishes are: (1) to open as many routes as possible; (2) each route must have at least one bus stop; (3) ensure that each bus stop is served by at least two different rou... | Let $S$ be the number of lines that can be opened, and consider the lines as vertices to form a graph $K_{\mathrm{s}}$. Label the 1983 stations as $A_{1}, A_{2}, \cdots, A_{19}$. If two lines have a common station, color the edge between the corresponding two vertices with color $C$.
From (2), every edge of $K_{\mathrm... | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,302 |
Example 1. As shown in Figure 4. Let the edge length of the cube be $2, M, N$ be the centers of faces $A_{1} C_{1}, C_{1} B$, respectively. Find the distance between $A N$ and $D M$, the angle they form, and the position of their common perpendicular. | It is known that $|A N|=|D M|=\sqrt{6}$, hence
$$
\left(\begin{array}{l}
x^{2}+y^{2}+d^{2}-2 x y \cos \theta=|M N|^{2}=2, \\
(\sqrt{6}-x)^{2}+(\sqrt{6}-y)^{2}+d^{2}-2 \\
\cdot(\sqrt{6}-x)(\sqrt{6}-y) \cos \theta=|A D|^{2} \\
=4, \\
x^{2}+(\sqrt{6}-y)^{2}+d^{2}+2 x(\sqrt{6}-y) \\
\cdot \cos \theta=|A M|^{2}=6, \\
(\sqrt... | \left\{\begin{array}{l}
x=\frac{\sqrt{6}}{3} \\
y=\frac{\sqrt{6}}{3} \\
\cos \theta=\frac{1}{2} \\
d=\frac{2 \sqrt{3}}{3}
\end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,306 |
Example 3. As shown in Figure 6. Let the right prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ have a height of 1, and the base be an isosceles trapezoid, with the longer base $|A B|=7$, the leg length being 2, and the base angle being $60^{\circ}$. Let $E$ be on $A B$ such that $|E B|=1$. Find the distance between $A_{1} D_{1... | It is evident that $\left|B_{1} E\right|=\sqrt{2}$, hence
$$
\left(\begin{array}{l}
x^{2}+y^{2}+d^{2}-2 x y \cos \theta=\left|A_{1} E\right|^{2}=37, \\
(\sqrt{2}-x)^{2}+(2-y)^{2}+d^{2}-2(\sqrt{2} \\
-x)(2-y) \cos \theta=\left|B_{1} D_{1}\right|^{2}=39, \\
(\sqrt{2}-x)^{2}+y^{2}+d^{2}+2(\sqrt{2}-x) y \\
\cdot \cos \thet... | d=\sqrt{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,308 |
For example, placing $n$ identical balls into $m$ different boxes arbitrarily, with no limit on the number of balls in each box, how many different ways are there to do this?
Putting the above text into English while preserving the original text's line breaks and format, the translation result is as follows: | If we arrange $m$ boxes in a row and place $n$ balls into these $m$ boxes, also arranged in a row, the problem can be transformed into an occupancy problem. That is, treating $n$ balls and $m-1$ box dividers as elements, one way to arrange $n$ balls (or one way to arrange $m-1$ dividers) is one way to place $n$ balls i... | C_{\mathrm{n}+\mathrm{m}-1}^{\mathrm{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,309 |
2 2. Let integers $a, b$ satisfy $2 a^{2}+a=3 b^{2}+b$, try to prove that $a-b, 2 a+2 b+1$ are both perfect squares.
The text has been translated while preserving the original line breaks and format. | To clarify the relationship between $a-b$, $2a+2b+1$, and
$$
2a + a = 3b^2 + b,
$$
we perform the following factorization:
$$
\begin{array}{l}
2\left(a^2 - b^2\right) + a - b = b^2, \\
(a - b)(2a + 2b + 1) = b^2.
\end{array}
$$
Since we cannot guarantee that $a - b$ and $2a + 2b + 1$ are coprime, we cannot directly c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,312 |
Example 3. Prove that the product of 5 consecutive integers is not a perfect square. | Proof: Let $n \geqslant 3$, and
$$(n-2)(n-1) n(n+1)(n+2)=a^{2}$$
where $a \in \mathbb{Z}$. Note that
$$
\begin{array}{l}
(n, n \pm 1)=(n, \pm 1)=1, \\
(n, n \pm 2)=(n, \pm 2)=1 \text { or } 2 .
\end{array}
$$
Let $b=(n-2)(n-1)(n+1)(n+2)$, then $(n, b)=1, 2$ or 4.
Thus, from (6) it follows that $n$ can only be $m^{2}$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,313 |
Example 5. Prove that for any integer $a$, the number $1990 a^{12} + 6$ is not a perfect square.
保留源文本的换行和格式，直接输出翻译结果。 | Prove that if $(a, 13) \div 1$, then $13 \mid a$, hence $1990 a^{12}+6 \equiv 6 \pmod{13}$. $a^{12} \equiv 1 \pmod{13}$.
$$
1990 a^{2}+6 \leq 1+6 \equiv 7 \pmod{13} .
$$
[if $x \equiv 0, \pm 1, \pm 2, \cdots, \pm 6 \pmod{13} \cdots$, only $x^{2} \equiv 0,1,4,9,3,12,10 \pmod{13}$. Therefore, from (11) and (12),
$$
190 a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,315 |
Example 6. Find all positive integers $m, n$ such that $2^{m}+3^{n}$ is a perfect square. | Proof: Let $2^{m}+3^{v}=x^{2}$, then it is easy to prove $2+x, 3+x$. At this point, $\quad x^{2}=(3 a \pm 1)^{2} \equiv 1(\bmod 3)$.
Therefore, it must be that $2^{m}=1(\bmod 3)$, hence $2 \mid m$. Let $m=2 s$, then from $2^{2 s}=x^{2}-3^{\mathrm{n}}$ we know
$$
x^{2}-3^{\mathrm{n}} \equiv 0(\bmod 4),
$$
Thus, $\quad ... | 2^{4}+3^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,316 |
Find the range of the function $y=\frac{1}{(x-1)(2 x-1)}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
The translation is as follows:
Find the range of the function $y=\frac{1}{(x-1)(2 x-1)}$. | "Solve for $x$ first, from the original equation we get
$$
\begin{array}{l}
21 x_{2}^{2}-3 y x+y-1=0, \\
\therefore \quad x=3 y \pm \sqrt{y(y+8)} .
\end{array}
$$
Since $x$ is a real number, we have
$$
\left\{\begin{array}{l}
y \neq 0, \\
y(y+8) \geqslant 0 .
\end{array}\right.
$$
Solving this, we get $y \leqslant-8$... | y \leqslant-8 \text{ or } y>0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,317 |
Find the range of the function $y=x+\sqrt{1-2 x}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solving for $x$, we get
$$
x=y-1 \pm \sqrt{2-2 y},
$$
from which we deduce $y \leqslant 1$.
Therefore, the range of the function is $(-\infty, 1]$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,318 |
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