problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. If $a, b, c$ are real numbers, then for any real number $x$, the necessary and sufficient condition for the inequality $a \sin x + b \cos x + c > 0$ to always hold is
(A) $a, b$ are both 0, and $c > 0$
(B) $\sqrt{a^{2}+b^{2}}=c$
(C) $\sqrt{a^{2}+b^{2}}<c$ | $a \sin x+b \cos x+c$
$$
=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)+c.
$$
Here, $\operatorname{tg} \varphi=\frac{b}{a}$.
However, when making the equivalent transformation
$$
\begin{array}{l}
a \sin x+b \cos x+c>0 \\
\Leftrightarrow \sin (x+\varphi)>-\frac{c}{\sqrt{a^{2}+b^{2}}}
\end{array}
$$
some students feel confused a... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 708,035 |
1. Given that the origin is inside the circle $k^{2} x^{2}+y^{2}-4 k x+2 k y+k^{2}$ $-1=0$. Then, the range of values for the number $k$ is ( ).
(A) $|k|>1$
(B) $|k| \neq 1$
(C) $-1<k<1$
(D) $0<|k|<1$ | To find the "criterion", we can use the "specialization" method: Given that the origin $(0,0)$ is inside the ellipse $f(x, y)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1=0$, we have $f(0,0)=-1 < 0$ (and vice versa); thus, we know that the origin being inside the ellipse $f(x, y)=0$ means $f(0,0)<0$. Let $f_{k}(x, y)$ be ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,036 |
1. Let the arithmetic sequence $\left\{a_{n}\right\}$ satisfy $3 a_{8}=5 a_{13}$ and $a_{1}>0, S_{n}$ be the sum of its first $n$ terms. Then the largest $S_{n}(n \in N)$ is ( ).
(A) $S_{10}$
(B) $S_{11}$
(C) $S_{20}$
(D) $S_{21}$ | $-1 .(C)$
Let the common difference of the arithmetic sequence be $d$. Then, from $3 a_{8}=5 a_{13}$, we get $3\left(a_{1}+7 d\right)=5\left(a_{1}+12 d\right)$, which simplifies to $a_{1}=-19.5d$. Therefore, $a_{20}$ $=a_{1}+19 d>0$, and $a_{21}=a_{1}+20 d<0$. Hence, the sum of the first 20 terms of this arithmetic seq... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,038 |
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of different points corresponding to the complex numbers $z_{1}^{1995}$, $z_{2}^{1905}, \cdots, z_{20}^{1195}$ is ( ).
(A) 4
(B) 5
(C) ... | 2. (A).
Let $z_{1}=\cos \theta+i \sin \theta$, then
$$
\begin{aligned}
z_{t}= & (\cos \theta+i \sin \theta)\left(\cos \frac{2(k-1) \pi}{20}\right. \\
& \left.+i \sin \frac{2(k-1) \pi}{20}\right), 1 \leqslant k \leqslant 20 .
\end{aligned}
$$
From $1995=20 \times 99+15$, we get
$$
\begin{aligned}
z k^{995}= & (\cos 19... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,039 |
3. If at least one of A's height or weight is greater than B's, then A is said to be no less than B. Among 100 young men, if someone is no less than the other 99, he is called a great young man. Then, the maximum number of great young men among 100 young men could be ( ).
(A) 1
(B) 2
(C) 50
(D) $100 \uparrow$ | 3. (D).
Take 100 young men as a special case, where their heights and weights are all different, and the tallest is also the lightest, the second tallest is also the second lightest, $\cdots$, the $k$-th tallest is also the $k$-th lightest $(k=1,2, \cdots, 100)$. Clearly, these 100 young men are all excellent young me... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 708,040 |
4. Given the equation $|x-2 n|=k \sqrt{x}(n \in N)$ has two distinct real roots in the interval $(2 n-1,2 n+1)$. Then the range of $k$ is ( ).
(A) $k>0$
(B) $0<k \leqslant \frac{1}{\sqrt{2 n+1}}$
(C) $\frac{1}{2 n+1}<k \leqslant \frac{1}{\sqrt{2 n+1}}$
(D) None of the above | 4. (B).
Obviously, $k \geqslant 0$, and $k=0$ leads to $x=2 n$, so the original equation has only one root, hence $k>0$. Also, from $(x-2 n)^{2}=k^{2} x$, we know that the parabola $y=(x-2 n)^{2}$ intersects the line $y=k^{2} x$ at two different points in the interval $(2 n-1,2 n+1)$. Therefore, when $x=2 n-1$, we hav... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,041 |
6. Let $O$ be the center of the base $\triangle A B C$ of the regular tetrahedron $P-A B C$. A moving plane through $O$ intersects the edges or their extensions of $P-A B C$ at points $Q, R, S$. The expression $\frac{1}{P Q}+\frac{1}{P R}+\frac{1}{P S}$
(A) has a maximum value but no minimum value
(B) has a minimum val... | 6. (D).
Tetrahedron $P Q R S$ can be divided into three pyramids with a common vertex $O$ and bases $\triangle P Q R, \triangle P R S, \triangle P Q S$.
Given that $\angle Q P R=\angle R P S=\angle P Q S=\alpha$.
Also, $O$ is the center of the base $\triangle A B C$ of the pyramid $P-A B C$, and the distance from $O$ ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,043 |
Example 4. In the acute triangle $\triangle A B C$, the lengths of the three sides $a, b, c$ are all integers, and $a>b>c, a+b+c=20$. Find $\angle A + \angle C$. | First, determine the range of the maximum side length $a$.
From $b+c>a>b>c$, we get $20=a+b+c>2a$, i.e., $a<\frac{20}{2}=10$ and $a>\frac{20}{3}$.
$\therefore \frac{20}{3}<a<10$.
(1) When $a=7$, $b+c=13$, then only $b=8, c=5$.
At this time, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{2}>0$, so $\triangle ABC$ is a... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,044 |
3. Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 3 .
From $[\lg x] \leqslant \lg x$, we get $\lg ^{2} x-\lg x-2 \leqslant 0$, which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$. Substituting into the original equation, we get $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqs... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,047 |
4. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y \leqslant 3 x, \\
y \geqslant \frac{1}{3} x, \\
x+y \leqslant 100
\end{array}\right.
$$
is | $$
1+2+3+\cdots+101=\frac{102 \times 101}{2}=5151 \uparrow .
$$
The region enclosed by $y=\frac{1}{3} x, x+y=100$, and the $x$-axis (excluding the boundary $y=\frac{1}{3} x$) contains
$$
\begin{aligned}
& (1+1+1+1)+(2+2+2+2)+\cdots \\
& +(25+25+25+25) \\
= & 4 \times(1+2+\cdots+25)=1300 \text { points. }
\end{aligned}... | 2551 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,048 |
5. Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, the total number of different coloring methods is $\qquad$ | $5,4,20$.
Suppose, for the pyramid $S-ABCD$, the vertices $S, A, B$ are colored with different colors, and they have a total of $5 \times 4 \times 3=60$ coloring methods.
When $S, A, B$ are already colored, assume their colors are $1, 2, 3$ respectively; if $C$ is colored with color 2, then $D$ can be colored with one... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,049 |
6. Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A$, then $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$ . | 6. 1870.
Let $n(A)$ denote the number of elements in the set $A$.
By the problem's setup, at least one of the numbers $k$ and $15k$ (where $k=9,10, \cdots, 133$) does not belong to $A$, so at least $125(=133-9+1)$ numbers do not belong to $A$, i.e., $n(A) \leqslant 1995-125=1870$. On the other hand, we can take $A=\{1... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,050 |
$$
2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0,
$$
where $\theta$ is a parameter.
Find the maximum value of the chord length of this family of curves on the line $y=2 x$. | Obviously, the family of curves always passes through the origin, and the line $y=2 x$ also passes through the origin, so the length of the chord intercepted by the family of curves on $y=2 x$ depends only on the coordinates of the other intersection point of the family of curves with $y=2 x$.
Substituting $y=2 x$ into... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,051 |
II. (This question is worth 25 points) Find all real numbers $p$ such that the cubic equation $5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three roots that are all natural numbers. | From observation, it is easy to see that $x=1$ is a root of the original cubic equation. By synthetic division, the original cubic equation can be reduced to a quadratic equation:
$$
5 x^{2}-5 p x+66 p-1=0 \text {. }
$$
The three roots of the original cubic equation are natural numbers, and the two roots of the quadra... | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,052 |
Three. (This question is worth 35 points) As shown in the figure, the incircle $O$ of rhombus $ABCD$ touches the sides at points $E, F, G, H$. Tangents to $\odot O$ are drawn on $\overparen{EF}$ and $\overparen{GH}$, intersecting $AB$ at $M$, $BC$ at $N$, $CD$ at $P$, and $DA$ at $Q$. Prove that $MQ \parallel NP$.
---... | Three, connect $A C, B D$, their intersection is the incenter $O$. Let $M N$ be tangent to $\odot O$ at $L$, and connect $O E, O M, O L, O N, O F$, and set $\angle M O L=\alpha$, $\angle L O N=\beta, \angle A B O=\varphi$.
Then it is easy to know
$$
\begin{array}{l}
\angle E O M=\alpha, \\
\angle F O N=\beta, \\
\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,053 |
Example 5. Prove: There does not exist an isosceles triangle with integer sides such that its perimeter equals 1995 and its area is an integer.
Prove: There does not exist an isosceles triangle with integer sides such that its perimeter equals 1995 and its area is an integer. | Assume such a triangle exists, and the base and the side lengths are $a, b(a, b$ are integers $)$, then the height on the base $L=\sqrt{b^{2}-\left(\frac{a}{2}\right)^{2}}=\frac{1}{2} \sqrt{4 b^{2}-a^{2}}$.
$$
\therefore S_{\Delta}=\frac{1}{2} a h=\frac{1}{4} a \sqrt{4 b^{2}-a^{2}} \text {. }
$$
From $a+2 b=1995$ we k... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,055 |
1. Find the smallest prime $p$ that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers. | It has been verified that $2,3,5,7,11,13,17,19,23,29,31$, and 37 can all be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers:
$$
\begin{array}{l}
2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, \\
7=2^{3}-3^{0}, 11=3^{3}-2^{4}, 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, 19=3^{3}-2^{3},... | 41 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,056 |
2. Given an acute angle $\theta$ and two internally tangent circles. Draw a fixed line $l$ (not passing through the centers) through the common tangent point $A$, intersecting the outer circle at another point $B$. Let point $M$ move on the major arc of the outer circle, $N$ be the other intersection of $M A$ with the ... | When point $M$ is at different positions on the major arc of the larger circle, the corresponding figures are slightly different (refer to the text for a unified explanation).
Let the other intersection point of line $l$ with the inner circle be $C$, and connect $N C$. Draw the tangents to the outer circle from points... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,057 |
3. 21 people participate in an exam, with the test paper containing 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most questions correctly? Please explain your reasoning. | For the $i$-th problem, $a_{i}$ people answered correctly, thus there are exactly
$$
b_{i}=C_{s_{i}}^{z}
$$
pairs of people who answered the $i$-th problem correctly $(i=1,2, \cdots, 15)$.
Below, we focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we have
$$
15 C... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,058 |
4. Let $S=\left\{A=\left(a_{1}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $1, i=1, \cdots$, 8\}. For two elements $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}\right.$, $\cdots, b_{8}$ ) in $S$, denote
$$
d(A, B)=\sum_{i=1}^{\delta}\left|a_{i}-b_{i}\right|,
$$
and call it the distance between $A$ and $B$... | Solution One
(I) First, we point out that the sum of the weights of any two codewords in $\mathscr{D}$ does not exceed 11; otherwise, if
$$
\omega(X)+\omega(Y) \geqslant 12,
$$
then because $12-8=4$, these two codewords must share at least four positions with 1s, and the distance between them
$$
a^{\prime}\left(X, y^{... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,059 |
5. Two players, A and B, play a game of filling in the coefficients of a polynomial of at least 4th degree:
$$
x^{2 n}+\square x^{2 n-1}+\square x^{2 n-2}+\cdots+\square x+1
$$
They take turns to choose one of the empty slots and fill in a real number as the coefficient of that term, until all slots are filled. If the... | Yi has a winning strategy. The specific approach is as follows.
There are $2n-1$ coefficients to be filled, including $n$ coefficients of odd terms and $n-1$ coefficients of even terms. It is agreed that each round of filling by Jia and Yi is called a "round." In each round, as long as possible, Yi tries to fill the ev... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,060 |
6. Prove that the interval $[0,1]$ can be divided into several black and white subintervals such that for any quadratic polynomial $P(x)$, the sum of the increments on all black subintervals equals the sum of the increments on all white subintervals. (The increment of $P(x)$ on the interval $[a, b]$ is defined as $P(b)... | First, prove a more general conclusion.
Theorem: Let $l$ be a positive real number, and $k$ be a positive integer. The interval $\left[0,2^{k} l\right]$ can be divided into several subintervals, which are then alternately colored as black subintervals and white subintervals. After doing this, the sum of the increments ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,061 |
2. Given $a=\frac{-1}{2+\frac{1}{3}}$. Then, $\frac{a}{a+1}-$ $\frac{\sqrt{a^{2}-2 a+1}}{a^{2}-a}$ is ( ).
(A) $-(1+2 \sqrt{3})$
(B) -1
(C) $2-\sqrt{3}$
(D) 3 | 2. D (Hint: $a<1$, so $\sqrt{a^{2}-2 a+1}=1$ $-a$.
| D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,063 |
3. Given three propositions:
(i) Rt $\triangle A B C, \angle A C B=90^{\circ}, C D$ is the altitude. Then $C D^{2}$ $=A D \cdot D B_{1}$
(ii) Rt $\triangle A B C, \angle A C B=90^{\circ}, D$ is a point on $A B$, $C D^{2}=A D \cdot D B$. Then $C D$ is the altitude of $\triangle A B C$;
(iii) $\triangle A B C, C D$ is th... | 3. $\mathrm{B}$ (Hint: For proposition (ii), consider the case where $D$ is the midpoint; for proposition (iii), consider the case where $\angle A B C>90^{\circ}$.) | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,064 |
4. Given that $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ are both equilateral triangles, and the bases of the two triangles $B C=B^{\prime} C^{\prime}$, and $A B>A^{\prime} B^{\prime}$. The orthocenters of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ are denoted as $H$ and ... | 4. $\mathrm{B}$ (Hint ${ }_{2} M$
$$
\left.=N=\frac{1}{16} \cdot B C^{2} .\right)
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,065 |
Example 6. Given a right triangle $\triangle A B C$ with the lengths of the two legs $a, b$ being integers, and $a$ being a prime number. If the length of the hypotenuse $c$ is also an integer, prove that $2(a+b+1)$ is a perfect square.
---
The translation maintains the original text's line breaks and format. | Prove $\because a^{2}=c^{2}-b^{2}=(i+b)(c-b)$, and $a$ is an odd number, and $c+b>c-b$.
$$
\therefore\left\{\begin{array}{l}
c+b=a^{2}, \\
c-b=1 .
\end{array}\right.
$$
Eliminating $c$, we get $2 b=a^{2}-1$.
$$
\begin{array}{l}
\therefore 2(a+b+1)=2 a+\left(a^{2}-1\right)+2 \\
=(a+1)^{2}
\end{array}
$$ | (a+1)^2 | Number Theory | proof | Yes | Yes | cn_contest | false | 708,066 |
5. A, B, C, and D four people put up the same amount of money, to partner in buying ducks. A, B, and C each got 3, 7, and 14 more items of goods than D. At the final settlement, B paid
(A) $2 x ;$
(b) 56
(C) 70
(D) 12 | 5. C (Hint: Since $(3+7+14) \div 4=$ 6 (items), that is, Yi only has 1 more item of goods than the average, which means each item costs 14 yuan, and Bing has $14-6=8$ (items). But Jia is short of $6-3=3$ (items), and Yi has already paid 1 item's worth of money to Ding, so, Bing should pay Jia 3 items' worth of money, a... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,067 |
2. Given a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n>4)$ where all interior angles are integer multiples of $15^{\circ}$, and $\angle A_{1}+\angle A_{2}+\angle A_{3}=$ $285^{\circ}$. Then, $n=$ | 2. 10 (Hint: The sum of the $n-3$ interior angles other than the male one is $(n-2) \cdot$ $180^{\circ}-285^{\circ}$, it can be divided by $n-3$, and its quotient should also be an integer multiple of $15^{\circ}$.） | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,070 |
3. As shown in the right figure, given $\frac{A D}{D B}=\frac{5}{2}, \frac{A C}{C E}=\frac{4}{3}$. Then $\frac{B F}{F C}=$ | 3. $\frac{14}{15}$ (Hint, use Mei's Theorem.) | \frac{14}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,071 |
4. A football made of leather in black and white colors is shown as follows. It is known that this football has 12 black leather pieces. Therefore, this football has white leather pieces.
保留源文本的换行和格式，直接输出翻译结果。 | 4. 20 (Hint: Each black piece is a pentagon, and the total number of edges for 12 pieces is: $5 \times 12=60$ (edges). Each white piece is a hexagon, and three of its edges are adjacent to the edges of black pieces. That is, three edges of black pieces determine one white piece. Therefore, the number of white pieces is... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,072 |
7. For real numbers $x, y$, define a new operation: $x * y=a x+b y+$ $c$, where $a, b, c$ are constants, and the right side of the equation is the usual addition and multiplication operations. It is known that $3 * 5=15,4 * 7=28$. Then, 1*1= $\qquad$ | 7. -11
Brief solution: According to the definition, we know that $3 * 5=3a+5b+c=15, 4 * 7$ $=4a+7b+c=28$, and $1 * 1=a+b+c=3(3a+5b+$c) $-2(4a+7b+c)=3 \times 15-2 \times 28=-11$. | -11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,075 |
Given a convex $(4n+2)$-sided polygon $A_{1} A_{2} \cdots A_{4n+2}$ (where $n$ is a positive integer) with each interior angle being an integer multiple of $30^{\circ}$. Consider the equations in $x$:
$$
\begin{array}{l}
x^{2}+2 x \sin A_{1}+\sin A_{2}=0, \\
x^{2}+2 x \sin A_{2}+\sin A_{3}=0, \\
x^{2}+2 x \sin A_{3}+\s... | Analyzing according to the given conditions, the degree measures of the internal angles can only be $30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}$, or $150^{\circ}$, and their sine values can only be $\frac{1}{2}, \frac{\sqrt{3}}{2}$, or 1.
$$
\begin{array}{l}
\Delta=4 \sin ^{2} A_{1}-4 \sin A_{2} \\
\leqslant 4 \si... | \angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}, \angle A_{4}=\angle A_{5}=\angle A_{6}=150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,077 |
Three, (12 points) Given points $P_{1}\left(x_{1}, 1994\right), P_{2}\left(x_{2}, 1994\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+7(a \neq 0)$. Try to find the value of the quadratic function $y=x_{1}+x_{2}$. | Three, since $P_{1}, P_{2}$ are two points on the graph of a quadratic function, we have
$$
\begin{array}{l}
a x_{1}^{2}+b x_{1}+7=1994, \\
a x_{2}^{2}+b x_{2}+7=1994 .
\end{array}
$$
Subtracting the two equations and simplifying, we get
$$
\left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0 .
$$
Since... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,078 |
Four, (14 points) As shown in the figure, the angle bisector $B E$ of $\triangle A B C$ and the median $A D$ of side $B C$ are perpendicular and equal. Given $B E = A D = 4$. Find the three sides of $\triangle A B C$. | Since $B E$ is the bisector of $\angle B$, and $A D \perp B E$, it is known that points $A$ and $D$ are symmetric about line $B E$.
Taking $B E$ as the axis of symmetry, construct the symmetric point $F$ of point $C$ about $B E$. Clearly, line segment $D F$ is symmetric to $A C$, and their intersection point is $\pm$.... | AB = \sqrt{13}, BC = 2\sqrt{13}, AC = 3\sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,079 |
Six. (10 points) A mall distributes 9999 shopping coupons to customers. Each coupon has a four-digit number, ranging from 0001 to 9999. If the sum of the first two digits equals the sum of the last two digits, the coupon is called a "lucky coupon." For example: the number 0734, since $0+7=3+4$, this number is a lucky c... | Six, obviously, the ticket with number 9999 is a lucky ticket. Apart from this lucky ticket, if a number $n$ is a lucky ticket, then the ticket with number $m = 9999 - n$ is also a lucky ticket. Since 9 is an odd number, $m \neq n$.
Since $m + n = 9999$, there is no carry when adding, which means that, apart from the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,081 |
2. (Kalei) Let $\triangle ABC$ have a circumradius $R=1$, an inradius $r$, and the inradius of its pedal triangle $A' B' C'$ be $p$. Prove that $p \leqslant 1-\frac{1}{3}(1+r)^{2}$. | Let the orthocenter of $\triangle A R C$ be $H$, and the incenter of its pedal triangle $A^{\prime} B^{\prime} \mathbf{C}^{\prime}$. Draw $H D \perp A^{\prime} C^{\prime}$ at $D$, then $H D=p$. Since $C^{\prime}$, $B, A^{\prime}, H$ are concyclic, we have $\angle H A^{\prime} D=\angle H B C^{\prime}$ $=90^{\circ}-\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,082 |
3. (West Dent) Given that the circumcircle $k$ of $\triangle ABC$ has center $O$ and radius $R$, and the incircle has center $I$ and radius $r$. Another circle $k_{0}$ is tangent to sides $CA$, $CB$ at points $D$, $E$ respectively, and is internally tangent to circle $k$. Prove that the incenter $I$ is the midpoint of ... | Let the center of circle $k_{0}$ be $O_{1}$, and its radius be $\rho$. Then,
$$
C I=\frac{r}{\sin \frac{C}{2}}, C O_{1}=\frac{\rho}{\sin \frac{C}{2}}, I O_{1}=\frac{\rho-r}{\sin \frac{C}{2}} .
$$
Therefore, we have
$$
\frac{I O_{1}}{C O_{1}}=\frac{\rho-r}{\rho}=1-\frac{r}{\rho} .
$$
For $\triangle C O O_{1}$ and the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,083 |
4. (Spain) Let points $D$ and $E$ be on the side $BC$ of $\triangle ABC$ such that $\angle BAD = \angle CAE$. Let $M$ and $N$ be the points where the incircles of $\triangle ABD$ and $\triangle ACE$ touch $BC$, respectively. Prove that:
$$
\frac{1}{MB} + \frac{1}{MD} = \frac{1}{NC} + \frac{1}{NE}.
$$ | To prove that
$$
\begin{array}{l}
\frac{1}{M B}+\frac{1}{M D}=\frac{B D}{M B \cdot M D}, \\
\frac{1}{N C}+\frac{1}{N E}=\frac{C E}{N C \cdot N E},
\end{array}
$$
it suffices to prove that
$$
B D \cdot N E \cdot N C=C E \cdot M B \cdot M D .
$$
Since \( M \) and \( N \) are the points of tangency of the incircles of \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,084 |
7. (GEO) Let $a, b, c$ be integers, and $a>0, a c-b^{2}=p$ $=p_{1} \cdots p_{n}$, where $p_{1}, \cdots, p_{n}$ are distinct primes. Let $M(n)$ denote the number of integer solutions $(x, y)$ to the equation
$$
a x^{2}+2 b x y+c y^{2}=n
$$
Prove that $M(n)$ is finite and for each non-negative integer $k$ we have $M\lef... | Let the integer pair $(x, y)$ satisfy the equation
$$
a x^{2}+2 b x y+c y^{2}=p^{k} n \text {. }
$$
Multiplying (1) by $a$ and noting that $a c-b^{2}=p$, we get
$$
(a x+b y)^{2}+p y^{2}=a p^{k} n .
$$
Similarly, multiplying (1) by $c$ yields
$$
(b x+c y)^{2}+p x^{2}=c p^{k} n \text {. }
$$
From (2) and (3), we know ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,085 |
8. (India) The sequence of positive integers $\left\{f_{n}\right\}_{m=1}^{m}$ is defined as follows:
$f(1)=1$, and for $n \geqslant 2$,
$$
f(n)=\left\{\begin{array}{l}
f(n-1)-n, \text { when } f(n-1)>n ; \\
f(n-1)+n, \text { when } f(n-1) \leqslant n .
\end{array}\right.
$$
Let $S=\{n \in \mathbb{N} \mid f(n)=1993\}$.... | (i) We point out that if $f(n)=1$, then $f(3n+3)=1$. From $f(n)=1$, we can sequentially obtain according to the definition:
$$
\begin{array}{l}
f(n+1)=n+2, f(n+2)=2n+4, \\
f(n+3)=n+1, f(n+4)=2n+5, \\
f(n+5)=n, \quad f(n+6)=2n+6 \\
f(n+7)=n-1, \cdots
\end{array}
$$
It is evident that the sequence $f(n+1), f(n+3), f(n+5... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,086 |
9. (India) (a) Prove that the set of all positive rational numbers $Q^{+}$ can be divided into 3 mutually disjoint subsets $A, B, C$, such that $B A=B, B^{2}=C, B C=A$, where for any two subsets $H$ and $K$ of $Q^{+}$, $H K=\{h k \mid h \in H, k \in K\}$, and $H^{2}$ denotes $H H_{1}$.
(b) Prove that all positive ratio... | (a) First, let $1 \in A$ and arbitrarily distribute the prime numbers into $A, B, C$. Then, for any rational number $x$, write it in its prime factorization form:
$$
x=p_{1}^{\beta} p_{2}^{\dot{j}} \cdots p_{n}^{\beta} q_{1}^{\beta} 1 q_{2}^{\beta} \cdots q_{1}^{\beta} s_{1}^{\gamma} 1 s_{2}^{\gamma} \cdots s_{m}^{\gam... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,087 |
Nephew 8. Find the number of lattice points inside and on the boundary of the triangle formed by the line $y=\frac{2}{3} x-\frac{1}{2}, x=10$ and the $x$-axis. | Given the function $y=\frac{2}{3} x-\frac{1}{2}$, when $x$ takes the values $1,2, \cdots$, 10 (note: when $x=\frac{3}{4}$, $y=0$), the corresponding $y$ values are $\frac{1}{6}, \frac{5}{6}, \frac{3}{2}, \frac{13}{6}, \frac{17}{6}, \frac{7}{2}, \frac{25}{6}, \frac{29}{6}, \frac{11}{2}, \frac{37}{6}$. The number of inte... | 37 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,088 |
10. (India) For a natural number $n$, if for any integer $a$, whenever $n \mid a^{*}-1$, it follows that $n^{2} \mid a^{*}-1$, then $n$ is said to have property $P$.
(a) Prove that every prime number $n$ has property $P$;
(b) Prove that there are infinitely many composite numbers that also have property $P$. | (a) Let $n=p$ be a prime and $p \mid\left(a^{p}-1\right)$, then
$$
\begin{aligned}
(a ; p)=1 & \text { write } \\
a^{p}-1 & =a\left(a^{p-1}-1\right)+(a-1),
\end{aligned}
$$
By Fermat's Little Theorem, we know $p \mid\left(a^{p-1}-1\right)$. Therefore, $p \mid(a-1)$,
which means $a \equiv 1(\bmod p)$. Hence,
$$
a^{i} ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,089 |
12. (Ireland) Let $n, k \in \mathbb{N}$ and $k \leqslant n$ and let $S$ be a set containing $n$ distinct real numbers. Let $T$ be the set of all real numbers of the form $x_{1}+x_{2}+\cdots+x_{k}$, where $x_{1}, x_{2}, \cdots, x_{k}$ are $k$ distinct elements of $S$. Prove that $T$ has at least $k(n-k)+1$ distinct elem... | Let $s_{1}<s_{2}<\cdots<s_{n}$ be $n$ elements of $S$ and prove by mathematical induction. First, when $k=1$ and $k=n$, the conclusion is trivial. Suppose $k \leqslant n-1$, and the conclusion holds for $S_{0}=\left\{s_{1}, s_{2}, \cdots, s_{n-1}\right\}$, and let $T_{0}$ be the corresponding set when $S$ is replaced b... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,090 |
$-\sqrt{3-2 \sqrt{2}}$, then $N=(\quad)$.
(A) 1
(B) $2 \sqrt{2}-1$
(C) $\frac{\sqrt{5}}{2}$
(D) $\sqrt{\frac{5}{2}}$ | \begin{array}{l}-1 . \ A. \\ \because\left(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\right)^{2} \\ =\frac{2 \sqrt{5}+2}{\sqrt{5}+1}=2 . \\ \therefore N=\sqrt{2}-\sqrt{3-2 \sqrt{2}} \\ =\sqrt{2}-(\sqrt{2}-1)=1 . \\\end{array} | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,091 |
2. If $a$ and $b$ are positive numbers, and the equations $x^{2}+a x+2 b=0$ and $x^{2}+2 b x+a=0$ both have real roots, then the minimum value of $a+b$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 2. C.
According to the problem, we have
$$
\left\{\begin{array}{l}
\Delta_{1}=a^{2}-8 b \geqslant 0, \\
\Delta_{2}=(2 b)^{2}-4 a \geqslant 0 .
\end{array}\right.
$$
From (1), $a^{2} \geqslant 8 b$.
From (2), $b^{2} \geqslant a$.
Since $a, b$ are positive numbers,
$$
\therefore b^{4} \geqslant a^{2} \text {. }
$$
Fro... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,092 |
3. Given that $x, y$ are distinct natural numbers, and $x^{2}+19 y$ $=y^{3}+19 x$. Then the integer part of $\sqrt{x^{2}+y^{2}}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3.C.
Assume $x>y$.
$\because x^{3}-y^{3}=19(x-y)$, and $x, y$ are natural numbers, $\therefore x^{2}+x<x^{2}+xy+y^{2}=19<3x^{2}$. From this, we get $x=3$, and then $y=m$. Therefore, $\sqrt{x^{2}+y^{2}}=\sqrt{13}$, and its integer part is 3. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,093 |
4. Given that $A D, B E, C F$ are the altitudes of $\triangle A B C$ (with $D, E, F$ being the feet of the altitudes), $\angle B=45^{\circ}, \angle C=60^{\circ}$. Then the value of $\frac{D F}{D F}$ is
(A) $\frac{2}{\sqrt{3}}$
(B) $\frac{1}{2}$
(C) $\frac{1}{\sqrt{2}}$
(D) $\frac{\sqrt{3}}{2}$ | 4. D.
$\because A, B, D, E$ are concyclic,
$$
\therefore \angle C E D=\angle C B A \text {. }
$$
Therefore, $\triangle C E D \sim \triangle C B A$.
Thus, $\frac{D E}{A B}=\frac{D C}{A C}=$ $\cos 60^{\circ}=\frac{1}{2}, D E=\frac{1}{2} A B$.
Similarly, $D F=\frac{\sqrt{2}}{2} A C$.
Therefore, $\frac{D E}{D F}=\frac{1}... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,094 |
5. As shown in the figure, $M$ is the midpoint of arc $C A B$, and $M P$ is perpendicular to chord $A B$ at $P$. If the length of chord $A C$ is $x$, and the length of segment $A P$ is ($x$ +1), then the length of segment $P B$ is ( ).
(A) $3 x+1$
(B) $2 x+3$
(C) $2 x+2$
(D) $2 x+1$ | 5. D.
Take a point $D$ on $AB$ such that $PD=PA$, and draw $MA$, $MD$. Then $\triangle AMD$ is an isosceles triangle, $MA=MD$. Extend $CA$ to $E$, and draw $MC$, $MB$, $BC$. Since $M$ is the midpoint of $\overparen{CAB}$, we have $MC=MB$.
Thus, $\angle EAM=\angle CBM=\angle BCM=\angle MAB$ $=\angle MDA$.
Therefore, $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,095 |
6. Given that the base $BC$ and the height $AD$ of isosceles $\triangle ABC$ are both integers. Then, $\sin A$ and $\cos A$ ( ).
(A) one is a rational number, the other is an irrational number
(C) both are rational numbers
(D) both are irrational numbers, which depends on the values of $BC$ and $AD$ to determine | 6. B.
Min, $\because S_{\triangle A B C}$
$$
\begin{array}{l}
=\frac{-1}{2} A B \cdot A C \sin A \\
=\frac{1}{2} \cdot B C \cdot A D, \\
\therefore \sin A=\frac{B C \cdot A D}{A B \cdot A C}=\frac{B C \cdot A D}{A B^{2}}=\frac{B C \cdot A D}{B D^{2}+A D^{2}} \\
\quad=\frac{B C \cdot A D}{\frac{1}{4} B C^{2}+A D^{2}}=\... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,096 |
9. In the triangle $\triangle ABC$ with vertices $A(6,5), B(2,1), C(6,1)$ (including the interior and boundary), take any 11 integer points. Prove that there must exist 3 integer points, whose 6 coordinates are all different. | Prove that there are 15 integer points on $\triangle A B C$:
$$
\begin{array}{l}
(2,1),(3,1),(4,1),(5,1),(6,1),(3,2), \\
(4,2),(5,2),(6,2),(4,3),(5,3),(6,3), \\
(5,4),(6,4),(6,5) .
\end{array}
$$
According to the requirement that no two points have the same coordinates, they can be divided into 5 groups: $\square$
Fir... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,099 |
3. Let $x$ be a real number. Then the maximum value of the function $y=\sqrt{8 x-x^{2}}-$ $\sqrt{14 x-x^{2}-48}$ is | 3. $2 \sqrt{3}$.
$\because y=\sqrt{(8-x) x}-\sqrt{(8-x)(x-6)}$ is a real number if and only if $6 \leqslant x \leqslant 8$, and
$$
\begin{aligned}
y= & \sqrt{8-x}(\sqrt{x}-\sqrt{x-6})=\sqrt{8-x} \\
& \cdot \frac{(\sqrt{x}-\sqrt{x-6})(\sqrt{x}+\sqrt{x-6})}{\sqrt{x}+\sqrt{x-6}} \\
= & \frac{6 \sqrt{8-x}}{\sqrt{x}+\sqrt{x... | 2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,100 |
4. In $\triangle A B C$, $\angle C=3 \angle A, a=27, c=48$. Then $b$ $=$ ـ $\qquad$ | 4. 35 .
As shown. Draw the trisectors of $\angle C$ as $C D$ and $C E$.
$$
\begin{array}{l}
\text { Let } C D=1, A N=x, \\
A E=E C=y .
\end{array}
$$
From $\triangle C D B \sim \triangle A C B$, we have
$$
\begin{array}{l}
\frac{a}{c}=\frac{t}{b}=\frac{c-x}{a}, \\
t=\frac{a b}{c}, x=\frac{c^{2}-a^{2}}{c} .
\end{array... | 35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,101 |
Given $\odot O$ is the circumcircle of $\triangle A B C$, $P B$ is the tangent line through $B$ intersecting the extension of $A C$ at $P$, $P D \perp A B$ at $D$, $C$ is the midpoint of $\widehat{A B}$, and $E$ is the midpoint of $A B$.
Prove that $P B=2 D E$. | Given that $C$ is the midpoint of $\overparen{A B}$, we have $\angle 1=\angle A$.
Also, $P B$ is the tangent line passing through point $B$,
$$
\therefore \angle 2=\angle A \text{. }
$$
Thus, $\angle 1=\angle 2$.
Taking the midpoint $F$ of $A P$, and connecting $E F$ and $D F$. Then $E F$ is the midline of $\triangle ... | P B=2 D E | Geometry | proof | Yes | Yes | cn_contest | false | 708,102 |
How many real numbers $x$ make $y=\frac{x^{2}-2 x+4}{x^{2}-3 x+3}$ an integer, and find the values of $x$ at this time. | Let $\frac{x+1}{x^{2}-3 x-3}=k$. ( $k$ is an integer)
There is $k x^{2}-(3 x+1) x+3 k-1=0$.
(1) When $k=0$, $-x-1=0 . x=-1$, which meets the requirement.
$(3 k+1)^{2}-4 k(3 k-1) \geqslant 0$.
Thus $3 k^{2}-10 k-1 \leqslant 0$.
We have $\frac{5-\sqrt{28}}{3} \leqslant k \leqslant \frac{5+\sqrt{28}}{3}$.
Since $k$ is an ... | x_{1}=-1, x_{2}=2+\sqrt{2}, x_{3}=2-\sqrt{2}, x_{4}=1, x_{5}=\frac{5}{2}, x_{6}=2, x_{7}=\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,103 |
Three, from the 1995 natural numbers $1,2,3, \cdots$, 1995, remove some numbers so that in the remaining numbers, no number is equal to the product of any two other numbers. How many numbers must be removed at a minimum to achieve this? | (1) First, prove: When any 42 numbers are removed, there will be three numbers among the remaining numbers, such that the product of two of them equals the third.
We consider the array $(2,87,2 \times 87),(3,86,3 \times 86)$, $\cdots,(44,5,44 \times 45)$,
i.e., $(x, 89-x, x(89-x))(2 \leqslant x \leqslant 44)$.
$$
\beg... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,104 |
1. The curve corresponding to the equation $x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}= \mathrm{~ m ~ is ~}$ ( ).
The translation maintains the original text's line breaks and format. | -,1.L.
From the given, we have
$\left(x-\sqrt{1-y^{2}}\right)^{2}+\left(y+\sqrt{1-x^{2}}\right)^{2}$
$=2 \cdots 2\left(, \sqrt{1 \cdots y^{2}} \cdots \sqrt{1-x^{2}}\right)=2-2=0$.
Thus, $x \sqrt{1-y^{2}}=0, y+\sqrt{1-x^{2}}=0$,
which implies $x^{2}+y^{4}=1(x \geqslant 0, y \leqslant 0)$. | x^{2}+y^{4}=1(x \geqslant 0, y \leqslant 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,105 |
3. Given that the four sides $a, b, c, d$ of quadrilateral $A B C D$ satisfy
$$
\left\{\begin{array}{l}
a-3 b+2 c=0, \\
b-3 c+2 d=0, \\
c-3 d+2 a=0, \\
d-3 a+2 b=0 .
\end{array}\right.
$$
then the quadrilateral $A B C D$ is ().
(A) rhombus
(B) cyclic quadrilateral
(C) rectangle
(D) trapezoid | 3. D.
In the equation, we know
$$
\begin{array}{l}
a-b=2(b-c), \\
b-c=2(c \cdots d), \\
c-a=2(d-a) . \\
d-a=2(a-b) .
\end{array}
$$
From (1) and (3), substituting (2) into (1) yields
$$
\begin{aligned}
a \cdots b & =2(b-c)=2^{2}(c-d)=2^{3}(d-a) \\
& =2^{4}(a-b) .
\end{aligned}
$$
Thus, $a-b=0$, which means $a=b$. Co... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,107 |
5. Given $[x]$ denotes the greatest integer not exceeding $x$, $a, b, c \in$ $R^{+}, a+b+c=1$, let $M=\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1}$. Then the value of $[M]$ is $($. | 5. B.
Given $a \in(0,1), b \in(0,1), c \in(0,1) \Rightarrow a^{2} & \sqrt{a^{2}+2 a+1}+\sqrt{b^{2}+2 b+1} \\
& +\sqrt{c^{2}+2 c+1} \\
= & (a+b+c)+3=4 .
\end{aligned}
$$
And $M=\sqrt{(3 a+1) \cdot 1}+\sqrt{(3 b+1) \cdot 1}$
$$
\begin{aligned}
& +\sqrt{(3 c+1) \cdot 1} \\
& \frac{(3 a+1)+1}{2}+\frac{(3 b+1)+1}{2}+\frac... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,109 |
10. Given that there are exactly 600 triangles with integer sides, all of different lengths, and the longest side is exactly $n$. Find $n$.
The longest side of the triangle is $n$, and the other two sides are $a$ and $b$ with $a < b < n$. The triangle inequality theorem states that:
1. $a + b > n$
2. $a + n > b$
3. $b... | Analysis: Since one side of the triangle is fixed at length $n$, and the three side lengths $x, y, n$ are all integers, the number of such triangular shapes corresponds one-to-one with the lattice points $(x, y)$ in the coordinate plane. Therefore, the lattice point method can be used to solve this problem.
Let the le... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,110 |
‥ For the equation greater than $x$, $2 x^{2}+3 a x+a^{2}-a=0$ to have at least one root with a modulus equal to 1, the value of the real number $a$ ().
(A) does not exist
(B) has one
(C) has two
(D) has four | 6. C.
$$
2 x^{2}+3 a x+a^{2}-a=0 .
$$
(1) When the roots are real numbers
$$
\Delta=9 a^{2}-8\left(a^{2}-a\right)=a(a+8) \geqslant 0,
$$
we have $a \geqslant 0$ or $a \leqslant \cdots 8$. Substituting $x=1$ into (1) yields $a^{2}+2 a+2=0$.
Since $a \in \mathbb{R}$, this equation has no solution. Substituting $x=-1$ in... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,111 |
1. Find the value of $\operatorname{ctg} 10^{\circ}-4 \cos 10^{\circ}=$ | \[
\begin{array}{l}
=\frac{\sin 80^{\circ}-2 \sin 20^{\circ}}{\sin 10^{\circ}} \\
=\frac{\sin 80^{\circ}-3 \sin \left(80^{\circ}-60^{\circ}\right)}{\sin 10^{\circ}} \\
=\frac{\sin 80^{\circ}-2\left(\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}\right)}{\sin 10^{\circ}} \\
=\frac{2 \cos 80^{\circ}}{\sin... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,112 |
2. Given the functions $y=2 \cos \pi x(0 \leqslant x \leqslant 2)$ and $y=2(x \in$ $R$ ) whose graphs enclose a closed plane figure. Then the area of this figure is $\qquad$ . | 2. As shown in the figure, by symmetry, the area of $CDE$ = the area of $AOD$ + the area of $BCF$, which means the shaded area is equal to the area of square $OABF$, so the answer is 4. | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 708,113 |
4. Given that the area $S$ and the internal angle $A$ of $\triangle A B C$ are constants. Then the range of values for the length $a$ of side $B C$ is $\qquad$ . | 4. $\because S=\frac{1}{2} b c \sin A$,
$$
\begin{array}{l}
\therefore \quad b c=\frac{2 S}{\sin A \text {. }} \\
|B C|^{2}=a^{2}=b^{2}+c^{2}-2 b c \cos A \geqslant 2 b c-2 b c \cos A \\
=4 S \cdot \frac{1-\cos A}{\sin A}=4 S \text { tan } \frac{A}{2} \text {, } \\
\therefore a \geqslant 2 \sqrt{S \text { tan } \frac{A... | a \in\left(2 \sqrt{S \text { tan } \frac{A}{2}},+\infty\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,115 |
5. Let the set of $n(2<n<6)$ complex numbers satisfy the following 2 conditions to form a set $S$.
(1) $1 \in S$;
(2) If $x_{1} \in S, x_{2} \in S$, then $x_{1}-2 x_{2} \cos \theta \in S$, where $\theta$ $=\arg \frac{z_{1}}{x_{2}}$.
Then the set $S=$ | 5. When $x_{1}=z_{2}=z$, from (2) if $z \in S$, then
$$
\begin{array}{l}
z_{1}-2 z_{2} \cos \theta=-z \in S . \\
\because|z|=1 .
\end{array}
$$
$\therefore z \neq-z$, which means $S$ contains an even number of elements.
$$
\because 2<n<6, \therefore n=4 \text {. }
$$
(i) From (1) $1 \in S$. Thus $-1 \in S$;
(ii) Let $z... | S=\{1,-1, i,-i\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,116 |
6. Now we have 1 one-tenth yuan note, 1 one-tenth yuan note, 1 six-tenth yuan note, 5 one yuan notes, and 2 two yuan notes. The number of different amounts (excluding the case of not paying) that can be made is $\qquad$ | 6. Using: Jiao Shi Ji to match 0 Jiao, 1 Jiao, 2 Zhong different denominations. That is, $(1+i)$ kinds.
Using 5 Jiao can match 0 Jiao, 5 Jiao, which is $(1+1)$ kinds; Yuan. That is, $(5+1)$ kinds;
Using 5 Yuan can match 0 Yuan, 5 Yuan, 10 Yuan, which is $(2+1)$ kinds.
Considering the amounts to be paid are 5.5, 1.5, 2... | 127 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,117 |
One, (Total 25 points) Given $a, b, c \in R^{+}$
Prove: $\frac{a}{1+a+a b}+\frac{b}{1+b+b c}+\frac{2}{1+c+c a} \leqslant 1$. | $$
\begin{array}{l}
\frac{s}{1+b+b c} \frac{c}{1+c+c a} \frac{1+a b}{1+a+a b} \\
\Leftrightarrow \frac{b(1+c+c a)+c(1+b+b c)}{(1+b+b c)(1+c+c a)}=\frac{1+a b}{1+a+a b} \\
\Leftrightarrow \frac{b+c+2 b c+a b c+b c^{2}}{1+b+c+2 b c+a b c+b c^{2}+c a+a b c^{2}} \\
\leqslant \frac{1+a b}{1+a+a b} \\
\Leftrightarrow\left(b+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,118 |
Three, (Full marks 35 points) On the side $AB$ of $\triangle ABC$, take any point $P$. Draw a line through $P$ parallel to $AC$ intersecting $BC$ at $Q$. Draw a line through $P$ parallel to $BC$ intersecting $AC$ at $R$. Does there exist a fixed point $M$ other than $C$ such that $C, Q, R, M$ are concyclic? Prove your ... | Three, as shown in the figure. Draw a tangent $BC^{\circ} C$ through point $A$, and then draw a tangent $AC$ through point $R$ at $C$, these two tangents indicate that points $C, Q, R, M$ are concyclic.
$$
\begin{array}{l}
\because \angle M C B=\angle M A C, \\
\angle M C A=\angle M B C, \\
\therefore \triangle M B C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,120 |
Example 2. Non-zero real numbers $a, b, c$ satisfy the equation
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {. }
$$
Prove: Among $a, b, c$, there must be two numbers that are opposites of each other. | Given the equation, we have
$$
\frac{b c+c a+a b}{a b c}=\frac{1}{a+b+c} \text {. }
$$
Clearing the denominators, we get
$$
(a+b+c)(b c+c a+a b)=a b c .
$$
Expanding and rearranging terms, we obtain
$$
2 a b c+a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}=0 .
$$
Factoring the left side, we get
$$
(a+b)(b+c)(c+a)=0... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,122 |
Column 4. In the figure, $\odot O$ is the incircle of rhombus $ABCD$. $MN, PQ$ are tangent to $\odot O$, with $M, N, P, Q$ on $AB, BC, CD, DA$ respectively. Prove: $MQ \parallel NP$. | For convenience, let's denote the equal angles derived from $BA = BC$ and the tangent length theorem as $\alpha, \beta, \theta$ respectively. Observing quadrilateral $AMNC$, we easily obtain
$$
2 \alpha + 2 \beta + 2 \theta = 360^{\circ} \Rightarrow \alpha + \beta + \theta = 180^{\circ}.
$$
At this point, $\angle CON ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,123 |
13. (Shou Erland) Let $m, n \in \mathbb{N}$, $m$ and $n$ be coprime, $n$ be even, and $m < n$. Let $S$ denote the set of all such pairs $(m, n)$. For $(m, n) = s \in S$, write $n = 2^k n_0$, where $k, n_0 \in \mathbb{N}$ and $n_0$ is odd, and define $f(s) = (n_0, m + n - n_0)$. Prove that $f$ is a function from $S$ to ... | For an odd $M$, let
$$
S_{M}=\{(m, n) \in S \mid m+n=M\}.
$$
Note that when $f(m, n)=\left(m_{1}, n_{1}\right)$, we have $m_{1}+n_{1}=m+n$, so $f: S_{M} \rightarrow S_{M}$. It is easy to see that $\left|S_{M}\right| \leqslant \frac{M+1}{4}$, with equality holding if and only if $M=3 \pmod{4}$. Thus, by the pigeonhole ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,124 |
14. (Israel) On the three sides $BC, CA, AB$ of $\triangle ABC$, take points $D, E, F$ respectively, such that $\triangle DEF$ is an equilateral triangle. Let $a, b, c$ represent the lengths of the three sides of $\triangle ABC$ and $S$ represent its area. Prove that:
$$
DE \geqslant 2 \sqrt{2} S \cdot \left\{a^{2}+b^{... | Let $\angle B A C, \angle C B A, \angle A C B$ be denoted as $\alpha$, $\beta, \gamma$ respectively, and let the circle $D E F$ intersect sides $B C, A C$ at points $H, G$ respectively. Clearly, $\angle F G A=\angle F D E=60^{\circ}, \angle F H B=\angle F E D=60^{\circ}$. Therefore, we have
$$
\begin{array}{l}
\angle G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,125 |
16. (MAK) Let $n \in \mathbb{N}, n \geqslant 2$ and let $A_{0}=\left(a_{01}, a_{07}, \cdots, a_{0 m}\right)$ be an $n$-tuple of non-negative integers such that $0 \leqslant a_{0i} \leqslant i-1$, $i=1,2, \cdots, n$. The $n$-tuples $A_{1}=\left(a_{11}, a_{12}, \cdots, a_{1 n}\right), A_{2}=\left(a_{21}, a_{22}, \cdots, ... | We prove by mathematical induction on $n$. When $n=2$, $A_{0}=(0,0)$ or $(0,1)$. If $A_{0}=(0,0)$, then $A_{1}=(0, 1)$, $A_{2}=(0,0)$; if $A_{0}=(0,1)$, then $A_{1}=(0,0)$, $A_{2}=(0,1)$. It is clear that in either case, we always have $A_{2}=A_{0}$.
Assume for $m \in \mathbb{N}, m \geqslant 2$ and any $m$-tuple $A_{0... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,126 |
18. (Poland) Let $S_{n}$ be the number of sequences $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ where $a_{i} \in\{0,1\}$ and no 6 consecutive terms are all the same. Prove that as $n \rightarrow \infty$, $S_{n} \rightarrow \infty$. | Proof Let $B_{n}$ be the set of all sequences satisfying the requirements of the problem, then $S_{n}=\left|B_{n}\right|$. We will prove a strengthened proposition:
For all $n \in N$, we have $S_{n} \geqslant\left(\frac{3}{2}\right)^{n}$.
Since $S_{1}=2>\frac{3}{2}$, it only remains to prove
$$
S_{n+1} \geqslant \frac{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,127 |
20. (Romania) Let $c_{1}, \cdots, c_{n} \in \mathbb{R} (n \geqslant 2)$, such that
$$
0 \leqslant \sum_{i=1}^{n} c_{i} \leqslant n \text {. }
$$
Prove that there exist integers $k_{1}, \cdots, k_{n}$, such that $\sum_{i=1}^{n} k_{i}=0$ and for each $i \in\{1,2, \cdots, n\}$, we have
$$
1-n \leqslant c_{i}+n k_{i} \leq... | For each real number $x$, we use $[x]$ to denote the greatest integer not greater than $x$, and $(x)$ to denote the smallest integer not less than $x$. The condition $c+n k \in [1-n, n]$ implies $k \in i_{n}(c)=\left[\frac{1-c}{n}-1, 1-\frac{c}{n}\right]$. For every $c \in \mathbb{R}$ and $n \geqslant \bar{y}$, the len... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,129 |
1. Convex quadrilateral $A E C D$ has sides that are all prime numbers, and $A B /$ $C D, A B+B C=A L^{\prime}+C D=20, A B=B C$. Then $B C+$ $A D=$ ( ).
(A) 6 to 14
(B) 6
(C) 14
(D) 10 | -1. (A).
Assume $A B C D$ is not a parallelogram, and $C D B C$ as well as $A B, B C$ are all prime numbers, then we can only have $B C=3$ or 7. Therefore, $A D=3$ or 7, which means $A D+B C=6$ or 14. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,130 |
2 If $\alpha, \beta$ are two distinct real roots of the equation $x^{2}+p x+8=0$, and $|\alpha|>|\beta|$, then among the following four conclusions, the one that does not necessarily hold is ( ).
(A) $|\alpha|>2$ and $|\beta|>2$
(B) $|\alpha|+|\beta|>4 \sqrt{2}$
(C) $|\alpha|>\frac{5}{2}$ or $|\beta|>\frac{5}{2}$
(D) $... | 2. (A).
From the given $\Delta=p^{2}-32>0$, i.e., $p>4 \sqrt{2}$ or $p < -4 \sqrt{2}$. Noting that $|\alpha|+|\beta| \geqslant|\alpha+\beta|$, then $|\alpha|+|\beta|>4 \sqrt{2}$. Therefore, (B) holds, and from (B) it is easy to see that (C) holds. This is because at least one of $|\alpha|$ and $|\beta|$ is greater tha... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,131 |
3. On the line of the hypotenuse $AB$ of the isosceles right $\triangle ABC$, there is a point $P$ satisfying $S=AP^2+BP^2$, then ( ).
(A) There are infinitely many positions of $P$ such that $S<2CP^2$
(B) There are finitely many positions of $P$ such that $S<2CP^2$
(C) If and only if $P$ is the midpoint of $AB$, or $P... | 3. (D)
If $P$ coincides with $A$ or $B$, by the Pythagorean theorem, we get $A P^{2} + B P^{2} = 2 C P^{2}$.
If $P$ is on $A B$,
draw $P D \perp A C$ at $D$,
$P E \perp B C$ at $E$.
Then, by $A P^{2} = 2 D P^{2}, B P^{2} = 2 P E^{2} = 2 O D^{2}$, we have $A P^{2} + B P^{2} = 2 D P^{2} + 2 C D^{2} = \angle C P^{P}$.
I... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,132 |
4. A merchant bought $n$ (where $n$ is a prime number) televisions for $m$ yuan (where $m$ is a natural number). Two of these televisions were sold to a charitable organization at half the cost price, and the rest were sold in the store, each with a profit of 500 yuan. As a result, the merchant made a profit of 5500 yu... | 4. (C).
From the problem, $\frac{m}{n}+(n \cdots-2)\left(\frac{m}{n}+506\right) \cdots 5500+m$.
Solving, we get $m=500 n(n-13)$.
Since $m>0$, then $n>13$, and the smallest prime number greater than 13 is 17. Therefore, the answer is (C). | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,133 |
5. If the graphs of the equations $3 x+b y+c=0$ and $c x-2 y+12=0$ coincide, let $n$ be the number of pairs $(b, c)$ that satisfy the above conditions, then $n$ equals ( ).
(A) $0 \quad$ (B) 1
(C) 2
(D) finite but more than 2 | 5. (C).
Given the equations can be transformed into $y=\frac{3}{b} x-\frac{c}{b}, y=\frac{c}{2} x+6$. If the two lines coincide, then we should have $\left\{\begin{array}{c}\frac{3}{b}=\frac{c}{2}, \\ \frac{c}{b}=6 .\end{array}\right.$ Solving this, we get $(b, c)$ as $(-1,6),(1,-6)$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,135 |
6. In an acute triangle $ABC$, the altitudes from $AC$ and $AB$ are $BE$ and $CF$ respectively, with $E$ and $F$ being the feet of the perpendiculars, and $BE$ and $CF$ intersect at $H$. If $\angle BAC=60^{\circ}, BC=2$, then the length of $AK$ is ( ).
(A) $\frac{3}{2}$.
(B) $\frac{\pi}{3}-\sqrt{3}$
(C) $\frac{\sqrt{2}... | 6. (B).
In the right-angled $\triangle A B E$, $A E$ $=A B \cdot \cos \angle B A C$. Similarly, $A F=A C \cdot \cos \angle B A C$.
In $\triangle A E F$,
$$
\begin{array}{l}
E F^{2}= A E^{2}+A F^{2} \\
-2 A E \cdot A F \cdot \cos \angle B A C \\
= \cos ^{2} \angle B A C\left(A B^{2}+A C^{2}\right. \\
-2 A B \cdot A C ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,136 |
1. If the natural numbers $a, x, y$ satisfy $\sqrt{a-2 \sqrt{6}}=\sqrt{x} -\sqrt{y}$, then the maximum value of $a$ is $\qquad$. | $$
=, 1, a_{\max }=7 \text {. }
$$
Squaring both sides of the known equation yields $a-2 \sqrt{6}=x+y-$ number, it must be that $x+y=\alpha$ and $x y=6$. From the known equation, we also know that $x>y$, so it can only be $x=6, y=1$ or $x=3, y=2$. Therefore, $a=7$ or 2, the maximum value is 7. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,137 |
2. For the quadratic function $y=x^{2}-4 a x+5 a^{2}$ $-3 a$ in terms of the independent variable $x$, the minimum value $m$ is a function of $a$, and $a$ satisfies the inequality $0 \leqslant$ $a^{2}-4 a-2 \leqslant 10$. Then the maximum value of $m$ is $\qquad$ $ـ$ | 2. 18 .
Solving $0 \leqslant a^{2}-4 a-2 \leqslant 10$ yields $-2 \leqslant a \leqslant 2-\sqrt{6}$ or $2+\sqrt{6} \leqslant a \leqslant 6$. Also, $m=a^{2}-3 a$, considering the quadratic function, $m=a^{2}-3 a$ when $-2 \leqslant a \leqslant 2-\sqrt{6}$ and $2+\sqrt{6} \leqslant a \leqslant 6$, we can see that the ma... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,138 |
3. Person A and Person B start from the same point $A$ on a circular track at the same time and run in opposite directions. Person A's speed is $5 \mathrm{~m}$ per second, and Person B's speed is $7 \mathrm{~m}$ per second. They stop running when they meet again at point $A$ for the first time. During this period, they... | 3. 12 .
Two $\lambda \mu A$ start from $A$ and meet once (not at point $A$). At the moment of their first meeting, the two have run a certain distance. From the ratio of their speeds, it can be deduced that person A has run $\frac{5}{12}$ of a lap, and person B has run $\frac{7}{12}$ of a lap. Therefore, after the $n$... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 708,139 |
4. If the inequality $|x-a|+|x|<2$ has no real solution, then the range of real number $a$ is $\qquad$ . | 4. $a \leqslant-2$ or $a \geqslant 2$.
Transform the original inequality into $|x-a|0) \\ 2+x,(x<0)\end{array}\right.$ Draw the graphs of $y_{1}$ and $y_{2}$.
If the inequality has no solution, then it should be $y_{1} \geqslant y_{2}$, i.e., the graph of $y_{1}$ coincides with or is above the graph of $y_{2}$. Clearl... | a \leqslant-2 \text{ or } a \geqslant 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,140 |
In an acute triangle $ABC$, the angle bisectors of $\angle BAC$, $\angle ABC$, and $\angle ACB$ intersect the circumcircle of $\triangle ABC$ at $D$, $E$, and $F$ respectively. Connecting $EF$, $FD$, and $DE$ intersect $AD$, $BE$, and $CF$ at $A_1$, $B_1$, and $C_1$ respectively. Prove that the incenter $I$ of $\triang... | Let $I$ be the incenter of $\triangle A B C$, and connect $A F$, then
$$
\begin{array}{l}
\angle 1=\angle 2 \\
=\frac{1}{2} \angle A B C, \\
\angle 3=\angle 4=\frac{1}{2}
\end{array}
$$
$\angle A C B$.
$$
\text { Also, } \angle 5=-\frac{1}{2} \angle B A C \text {, }
$$
Thus, $\angle 1+\angle 3+\angle 5$
$$
=\frac{1}{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,141 |
ニ、(Full score 25 points) Write 1995 equations of the form $A_{i} x^{2}+B_{i} x+C_{i}=0(i=1,2, \cdots, 1995)$ on the blackboard. Two people, A and B, take turns to play a game. Each time, only one non-zero real number is allowed to replace one of the $A_{i}, B$ or $C$ in the $n$ equations $(i=1,2, \cdots, 1995)$. Once a... | Second, the answer is that Jia can get at most 998 rootless equations.
If Jia first fills in the coefficient of the linear term $B_{1}$ with 1, and tries to fill in the linear term coefficient of each equation with 1 as much as possible, at this time, Jia has at least $\frac{1995+1}{2}=998$ chances.
If Yi fills in the... | 998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,142 |
Three. (Full marks 25 points) If the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}-x-2>0, \\
2 x^{2}+(5+2 k) x+5 k<0
\end{array}\right.
$$
has only the integer solution $x=-2$, find the range of the real number $k$. | Three, because $x=-2$ is a solution to the system of inequalities, substituting $x=-2$ into the second inequality yields
$$
(2 x+5)(x+k)=[2 \cdot(-2)+5](-2+k)2$. Therefore, $-k>-2>-\frac{5}{2}$,
which means the solution to the second inequality is $-\frac{5}{2}<x<-k$.
These two inequalities have only the integer soluti... | -3 \leqslant k < 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,143 |
1. Given the set $P=\left\{a_{1}, a_{2}, \cdots, a_{n}, \cdots\right\}$, and it satisfies $a_{1}=$ $1, a_{n}=a_{n-1}+4(n-1)(n \in N, n \neq 1)$. Also, the set $Q=$ $\left\{n \mid(1+i)^{2 n}=2^{n} i, n \in N\right\}$. Then the relationship between $P, Q$ is ( ).
(A) $P \subset Q$
(B) $P=Q$
(C) $P \supset Q$
(D) $P \cap ... | $-1 .(\mathrm{A})$.
$$
\begin{aligned}
a_{n} & =a_{n-1}+4(n-1)=a_{n-2}+4(n-2)+4(n-1) \\
& =\cdots=a_{1}+4[(n-1)+(n-2)+\cdots+1] \\
& =1+2 n(n-1) .
\end{aligned}
$$
Since $n$ and $(n-1)$ are consecutive integers, $n(n-1)$ is an even number. Therefore, $a_{n}$ is a number of the form $4k+1$.
From $(1+i)^{2 n}=2^{*} i$ ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,144 |
Example 6. In $\square A B C D$ there is a point $P$
satisfying $\angle P B A = \angle P D A$. Take the circumcenters of $\triangle P A B$, $\triangle P B C$, $\triangle P C D$, $\triangle P D A$ as $O_{1}, O_{2}, O_{3}, O_{4}$.
Prove: $S_{\text {quadrilateral } O_{1} O_{2} O_{3} O_{4}}=\frac{1}{2} \cdot S_{\square... | Analyzing the translation of $\triangle P D A$ to the position of $\triangle Q C B$, and connecting $P Q$. It is easy to see that $P A B Q$ and $P Q C D$ are both parallelograms. From $\angle Q P B = \angle P B A = \angle P D A = \angle Q C B \Rightarrow P, B, Q, C$ are concyclic. In other words, the circumcircles of $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,145 |
4. The number of positive integer solutions to the equation $7^{x}-3 \cdot 2^{y}=1$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 4. (C).
From the given equation, we have $\frac{7^{x}-1}{7-1}=2^{y-1}$,
which simplifies to $\underbrace{7^{x-1}+7^{x-2}+\cdots+1}_{x}=2^{y-1}$.
If $y=1$, then $x=1$.
If $y \neq 1$, then $x$ is even. Factoring the left side of (1) yields
$$
\begin{array}{l}
(7+1)\left(7^{x-2}+7^{x-4}+\cdots+1\right)=2^{y-1}, \\
7^{x-2... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,148 |
6. Given that $x, y$ are natural numbers, and $x^{2}+y^{2}+6$ is a multiple of $x y$. Then $\frac{x^{2}+y^{2}+6}{x y}$ is a ( ).
(A) Perfect square
(B) Perfect cube
(C) Perfect fourth power
(D) Prime number | 6. (B).
Given $x y \mid x^{2}+y^{2}+6$, we can set $k=\frac{x^{2}+y^{2}+6}{x y}$. Then we have $x^{2}-k y \cdot x+y^{2}+6=0$.
This is a quadratic equation in $x$, with its two roots denoted as $x_{1}$ and $x_{2}$. Thus, $\left\{\begin{array}{l}x_{1}+x_{2}=k y, \\ x_{1} \cdot x_{2}=y^{2}+6 .\end{array}\right.$
Therefo... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,150 |
1. The Chinese Football League A has 12 football clubs participating, with a home and away double round-robin system, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point each, and a loss earns 0 points. At the end of the league, teams are ranked according ... | $=1.46$ points.
Suppose the score difference between the $k$-th and the $(k+1)$-th is the largest. Since the total score of each match between any two teams is at most 3 points and at least 2 points, the maximum total score of the first $k$ teams playing against each other is $3 \cdot 2 C_{k}^{2}=6 C_{k}^{2}$. Addition... | 46 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,151 |
3. In the plane, a parabola divides the plane into two parts, two parabolas can divide the plane into at most seven parts. Then 10 parabolas can divide the plane into $\qquad$ parts. | 3. 191 .
Let $n$ parabolas divide the plane into $a_{n}$ regions. When adding another parabola to maximize the number of regions, this new parabola intersects each of the $n$ given parabolas at 4 points, thus the $n+1$ parabolas are divided into $4n+1$ segments, each of which splits an original region into two. This l... | 191 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,153 |
4. Consider the hyperbola \((x-2)^{2}-\frac{y^{2}}{2}=1\). A line \(l\) is drawn through the right focus of the hyperbola, intersecting the hyperbola at points \(A\) and \(B\). If \(|A B|=4\), then the number of such lines is \(\qquad\). | 4. 3 lines.
The coordinates of the right focus are $(2+\sqrt{3}, 0)$. The length of the chord perpendicular to the real axis through the right focus is exactly 4. The length of the chord $d$ formed by passing through the right focus and intersecting both branches of the hyperbola has the range $d>2$, so there are two ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,154 |
Example 7. As shown in the figure, on the sides $BC$, $CA$ of the equilateral $\triangle ABC$, there are points $E$, $F$ respectively, with $BE = CF = a$, $EC = FA = b$.
When $BF$ bisects $AE$, try to compare the sizes of $\frac{2a+b}{b}$, $\frac{a+b}{a-b}$, $\frac{(a+b)^2}{ab}$, $\frac{a^3}{b^3}$. | To compare the sizes of the above four expressions, it is necessary to determine what kind of relationship exists between $a$ and $b$. Draw $E K!$ perpendicular to $C A$ at $K$, easily obtaining $E K = A F = b$. From $\triangle B E K \sim \triangle B C F \Rightarrow \frac{a}{a+b} = \frac{b}{a} \Rightarrow a^2 = b(a+b)$... | \frac{2a + b}{b} = \frac{a + b}{a - b} = \frac{(a + b)^2}{ab} = \frac{a^3}{b^3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,156 |
6. A square piece of paper is divided into two parts by a straight line, one of which is then divided into two parts by another straight line, and then one of the three pieces is divided into two parts by another straight line, and so on. In the end, 19 96-sided polygons and some other polygons are obtained. Then the m... | \%. 1766 . After each division, the sum of the angles of the resulting polygons increases by $2 \pi$. Thus, after $k$ divisions, $k+1$ polygons are obtained, the sum of whose interior angles is $2 \pi(k+1)$. Since among these $k+1$ polygons, there are 19 ninety-six-sided polygons, the sum of their interior angles is $1... | 1766 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,157 |
One, (25 points) In the tetrahedron $A-BCD$,
\[
\begin{array}{l}
\angle DAB + \angle BAC + \angle DAC = 90^{\circ}, \\
\angle ADB = \angle BDC = \angle ADC = 90^{\circ}.
\end{array}
\]
Prove: The dihedral angle $A-BC-D$ is greater than $70^{\circ}$. | Given $D B=a, D C=b, D A=x$.
Since $\angle D A B+\angle B A C+\angle D A C=90^{\circ}, \angle A D B=90^{\circ}, \angle A D C=90^{\circ}$, unfolding the tetrahedron as shown, we have $D_{1} A=D_{2} A=x$.
Extending $D_{1} B$ and $D_{2} C$ to intersect at $K$, the quadrilateral $A D_{1} K D_{2}$ is a square. Thus, $B K=x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,158 |
II. (25 points) Let $a_{i}>0, i=1,2, \cdots, n, n \geqslant 2$. Prove:
$$
\begin{array}{l}
\frac{3}{a_{1}}+\frac{5}{a_{1}+a_{2}}+\frac{7}{a_{1}+a_{2}+a_{3}}+\cdots \\
+\frac{2 n+1}{a_{1}+a_{2}+\cdots+a_{n}}<4 \sum_{i=1}^{n} \frac{1}{a_{i}} .
\end{array}
$$ | Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, n)$. Then
$$
\begin{aligned}
\text { Left }= & \frac{2^{2}-1^{2}}{S_{1}}+\frac{3^{2}-2^{2}}{S_{2}}+\cdots+\frac{(n+1)^{2}-n^{2}}{S_{n}} \\
= & \frac{2^{2}}{S_{1}}+\left(\frac{3^{2}}{S_{2}}-\frac{1^{2}}{S_{1}}\right)+\left(\frac{4^{2}}{S_{3}}-\frac{2^{2}}{S_{2}}\right)+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,159 |
Three. (35 points) Given $n$ ($n \geqslant 5$) distinct real numbers $a_{1} < a_{2} < \cdots < a_{n}$, the necessary and sufficient condition for all $C_{n}^{2}$ sums $a_{i} + a_{j} (1 \leqslant i, j \leqslant n, i \neq j)$ to have exactly $2n-3$ distinct numbers is that $a_{1}, a_{2}, \cdots, a_{n}$ form an arithmetic... | Three, first note that when $a_{1}n$, $a_{i}+a_{j}=a_{i+j-n}+a_{n}$.
Thus, all $a_{i}+a_{j}$ are equal to some number in the inequality chain, meaning $a_{i}+a_{j}$ has exactly $2 n-3$ distinct numbers.
(2) Next, prove that if $a_{i}+a_{j}$ has $2 n-3$ distinct numbers, then $a_{1}<a_{2}<\cdots<a_{n}$ forms an arithmet... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,160 |
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