problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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II. (16 points) Given
$$
\cos (\alpha+\beta)=\cos \alpha+\cos \beta \text {. }
$$
Find the maximum value of $\cos \alpha$. | From the problem, we have $\cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta = \cos \alpha + \cos \beta$.
Then, $(\cos \alpha - 1) \cos \beta - \sin \alpha \cdot \sin \beta - \cos \alpha = 0$.
Let point $P(\cos \beta, \sin \beta)$,
line $l: (\cos \alpha - 1) x - \sin \alpha y - \cos \alpha = 0$.
Then the tra... | \sqrt{3} - 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,894 |
Example 5 Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ modulo 3 is divisible by $x^{2}+1$.
$(2015$, Harvard-MIT Mathematics Tournament) | 【Analysis】More explicitly, the polynomial $(x+1)^{n}-1$ needs to satisfy the equivalent condition:
There exist integer-coefficient polynomials $P$ and $Q$ such that
$$
(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x) \text {. }
$$
Assume without loss of generality that
$$
R(x)=(x+1)^{n}-1-P(x)\left(x^{2}+1\right) \text {.... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,895 |
2. It is known that A's soda water capacity is $50\%$ more than B's, and C's soda water capacity is $25\%$ more than B's. Then the relationship between A's and C's soda water capacities is that A's is ( ) more than C's.
(A) $20\%$
(B) $25\%$
(C) $45\%$
(D) $75\%$
(E) $100\%$ | 2. A.
Let the capacities of soda water A, B, and C be $x$, $y$, and $z$ respectively. According to the problem,
$$
\begin{array}{l}
x=(1+50 \%) y, z=(1+25 \%) y \\
\Rightarrow \frac{x}{z}=\frac{6}{5} .
\end{array}
$$
Therefore, A is $20 \%$ more than C. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,896 |
4. There are three mathematics courses: Algebra, Geometry, and Number Theory. If no two of these courses can be scheduled consecutively, then a student has ( ) different ways to arrange the 6 periods in one day.
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24 | 4. E.
If no two courses can be scheduled consecutively, the three math classes can be arranged in periods $1, 3, 5$, periods $1, 3, 6$, periods $1, 4, 6$, and periods $2, 4, 6$, for a total of four scenarios. Each scenario has
$$
3!=3 \times 2 \times 1=6 \text{ (ways). }
$$
Therefore, there are a total of $4 \times 3... | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,897 |
6. A person uploads a video to a website, and viewers can rate it on the website: those who like it add 1 point; those who dislike it deduct 1 point. Each video starts with 0 points. After some time, the person sees that the video he uploaded has scored 90 points, and 65% of the voters like the video. Therefore, ( ) pe... | 6. B.
Assuming a total of $x$ people voted on the video. Then $65 \% x$ voted.
Therefore, $65 \% x - (1 - 65 \%) x = 90 \Rightarrow x = 300$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,898 |
7. There are ( ) integers $n$ such that $400\left(\frac{2}{5}\right)^{n}$ is an integer.
(A) 3
(B) 4
(C) 6
(D) 8
(E) 9 | 7. E.
Notice,
$$
400\left(\frac{2}{5}\right)^{n}=2^{5} \times 5^{3}\left(\frac{2}{5}\right)^{n}=2^{n+5} \times 5^{3-n} \text {, }
$$
where, $n+5$ and $3-n$ are both non-negative integers.
Therefore, $n=-5,-4, \cdots, 3$, a total of 9. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,899 |
9. As shown in Figure 1, in isosceles $\triangle ABC$, $AB=AC$, all triangles in the figure are similar, the area of the smallest triangle is $1$, and the area of $\triangle ABC$ is 40. Then the area of trapezoid $DBCE$ is $(\quad)$.
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24 | 9. E.
Notice that the ratio of the area of the smallest triangle to $\triangle A B C$ is $\frac{1}{40}$, and all triangles are similar. Thus, the similarity ratio is $\frac{1}{\sqrt{40}}$.
From the diagram, $\frac{D E}{B C}=\frac{4}{\sqrt{40}} \Rightarrow \frac{S_{\triangle A D E}}{S_{\triangle A B C}}=\frac{16}{40}$.... | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,900 |
Example 6. Let $a=\sqrt{17}, b=\mathrm{i} \sqrt{19}$. Find the maximum value of $\frac{|a-z|}{|b-z|}$, where $z$ is any complex number with modulus 1.
(2015, Harvard-MIT Mathematics Tournament) | 【Analysis】Using the geometric meaning of complex numbers, we know that $z$ is a point on the unit circle centered at the origin in the complex plane, $A(\sqrt{17}, 0), B(0, \sqrt{19})$.
As shown in Figure 1.
Thus, $\frac{|a-z|}{|b-z|}=\frac{|A Z|}{|B Z|}$.
Assume point $Z(\cos \theta, \sin \theta)$. Then
$$
\begin{arr... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,901 |
12. The system of equations $\left\{\begin{array}{l}x+3 y=3, \\ || x|-| y||=1\end{array}\right.$ has $(\quad)$ solutions.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 8 | 12. C.
Graph the functions $x+3 y=3,|| x|-| y||=1$ (as shown in Figure 3), the two functions have three intersection points: $(-3,2)$, $(0,1),\left(\frac{3}{2}, \frac{1}{2}\right)$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,902 |
15. As shown in Figure 2, two equal circles with a radius of 5 are externally tangent to each other, and both are internally tangent to a larger circle with a radius of 13, with the points of tangency being $A$ and $B$. Let $AB = \frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)\right. = 1)$. Then the value of $m+n$ is (... | 15. D.
As shown in Figure 5, let the centers of the three circles be $X$, $Y$, and $Z$.
Then points $A$ and $B$ are on the extensions of $XY$ and $XZ$, respectively, satisfying
$$
\begin{array}{l}
XY = XZ \\
= 13 - 5 = 8, \\
YZ = 5 + 5 = 10, \\
XA = XB = 13.
\end{array}
$$
Since $YZ \parallel AB \Rightarrow \triangl... | 69 | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,903 |
16. In Rt $\triangle A B C$, $\angle B=90^{\circ}, A B=20$, $B C=21$. Then the number of line segments with integer lengths connecting vertex $B$ to points on the hypotenuse (including sides $A B$ and $B C$) is ( ) .
(A) 5
(B) 8
(C) 12
(D) 13
(E) 15 | 16. D.
From the problem, we know that the distance from point $B$ to $AC$ is
$$
d=\frac{AB \cdot BC}{AC}=\frac{20 \times 21}{\sqrt{20^{2}+21^{2}}}=\frac{20 \times 21}{29}=14 \frac{14}{29} \text {. }
$$
By the property that the perpendicular segment from a point to a line segment is the shortest, we know that among th... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,904 |
17. Given the set $S \subseteq\{1,2, \cdots, 12\},|S|=6$ ( $|S|$ represents the number of elements in the set $S$). If $a, b \in$ $S, a<b$, then $b$ is not a multiple of $a$. Then, the smallest element in the set $S$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
(E) 7. | 17. C.
Obviously, $1 \notin S$, and among $\{2,4,8\}$, $\{3,6,12\}$, and $\{5,10\}$, at most one from each set can be in $S$.
Thus, $\{7,9,11\} \subset S \Rightarrow 3 \notin S$
$\Rightarrow 6$ or $12$ is in $S$ $\Rightarrow 2 \notin S$.
Upon verification, $\{4,5,6,7,9,11\}$ satisfies the requirements. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,905 |
18. There are ( ) non-negative integers that can be represented as
$$
\sum_{i=0}^{7} 3^{i} a_{i}\left(a_{i} \in\{-1,0,1\}, 0 \leqslant i \leqslant 7\right) \text {. }
$$
(A) 512
(B) 729
(C) 1094
(D) 3381
(E) 59048 | 18. D.
Since $\sum_{i=0}^{7} 3^{i} a_{i}$ is a non-negative integer, the highest bit cannot be negative.
We discuss this by cases.
$a_{7}=1$, there are $3^{7}$ cases;
$a_{7}=0, a_{6}=1$, there are $3^{6}$ cases;
$a_{7}=0, a_{6}=0, a_{5}=1$, there are $3^{5}$ cases;
$\qquad$
$a_{7}=a_{6}=\cdots=a_{1}=0, a_{0}=1$, there... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,906 |
19. Given $m \in\{11,13,15,17,19\}$, $n \in\{1999,2000, \cdots, 2018\}$.
Then the probability that the unit digit of $m^{n}$ is 1 is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{4}$
(C) $\frac{3}{10}$
(D) $\frac{7}{20}$
(E) $\frac{2}{5}$ | 19. E.
Consider different cases.
When $m=11$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 20 ways to choose;
When $m=13$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose;
When $m=15$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit o... | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,907 |
25. Let $n \in \mathbf{Z}_{+}$,
$$
\begin{array}{l}
a, b, c \in\{x \mid x \in \underline{Z} \text { and } x \in[1,9]\}, \\
A_{n}=\underbrace{a a \cdots a}_{n \uparrow}, B_{n}=\underbrace{b b \cdots b}_{n \uparrow}, C_{n}=\underbrace{c c \cdots c}_{2 n \uparrow} .
\end{array}
$$
Then the maximum value of $a+b+c$ is ( )... | 25. D.
Notice,
$$
\begin{array}{l}
A_{n}=\underbrace{\overline{a a \cdots a}}_{n \uparrow}=a \cdot \frac{10^{n}-1}{9}, \\
B_{n}=\underbrace{\overline{b b \cdots b}}_{n \uparrow}=b \cdot \frac{10^{n}-1}{9}, \\
C_{n}=\underbrace{\overline{c c \cdots c}}_{2 n \uparrow}=c \cdot \frac{10^{2 n}-1}{9} .
\end{array}
$$
From ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,908 |
1. A container has 100 small balls, of which $36 \%$ are red balls, and the rest are blue balls. Then, after removing (）blue balls, the red balls account for $72 \%$ of the remaining balls.
(A) 28
(B) 32
(C) 36
(D) 50
(E) 64 | 1. D.
Let $x$ be the number of blue balls taken away. Then
$$
\frac{100 \times 36 \%}{100-x} \times 100 \% = 72 \% \Rightarrow x=50 \text {. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,909 |
2. Karl found some valuable stones in a cave: each 5-pound stone is worth 14 yuan, each 4-pound stone is worth 11 yuan, and each 1-pound stone is worth 2 yuan. It is known that there are at least 20 of each type of stone, and Karl can take out a total of 18 pounds of stones from the cave. Then the maximum value of the ... | 2. C.
$$
14 \times 2+11 \times 2=50 \text{. }
$$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,910 |
9. If the largest subset of the set $\{y \mid 0 \leqslant y \leqslant \pi\}$ satisfies $\sin (x+y) \leqslant \sin x+\sin y(x \in[0, \pi])$, then this largest subset is ( ).
(A) $y=0$
(B) $0 \leqslant y \leqslant \frac{\pi}{4}$
(C) $0 \leqslant y \leqslant \frac{\pi}{2}$
(D) $0 \leqslant y \leqslant \frac{3 \pi}{4}$
(E)... | 9. E.
Since $x \in [0, \pi], \cos x \in [0, 1]$, therefore,
$$
\begin{array}{l}
\sin (x+y) = \sin x \cdot \cos y + \cos x \cdot \sin y \\
\leqslant \sin x + \sin y.
\end{array}
$$
Thus, $0 \leqslant y \leqslant \pi$. | E | Inequalities | MCQ | Yes | Yes | cn_contest | false | 730,911 |
Example 8 Let $D$ be a point inside an acute-angled $\triangle ABC$,
$$
\begin{array}{l}
\angle A D B=\angle A C B+90^{\circ}, \text { and } A C \cdot B D=A D \cdot B C . \\
\text { Find the value of } \frac{A B \cdot C D}{A C \cdot B D} \text { . }
\end{array}
$$
(34th IMO) | 【Analysis】With $C$ as the origin and $C A$ as the real axis, establish a complex plane, and let $|\overrightarrow{C A}|=r$.
According to the geometric meaning of complex number multiplication and division, the vector $\overrightarrow{D B}$ can be regarded as $\overrightarrow{D A}$ after rotation and scaling. Therefore... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,912 |
23. In $\triangle P A T$, $\angle P=36^{\circ}, \angle A=56^{\circ}$, $P A=10$, points $U$ and $G$ are on sides $T P$ and $T A$ respectively, with $P U=$ $A G=1$. If $M$ and $N$ are the midpoints of $P A$ and $U G$ respectively, then the acute angle formed by $M N$ and $P A$ is ( ).
(A) $76^{\circ}$
(B) $77^{\circ}$
(C... | 23. E.
As shown in Figure 10, extend $P N$ to point $Q$ such that $P N = N Q$, and connect $A Q$ and $G Q$.
Figure 10
Given $P N = N Q, U N = N G$
$\Rightarrow \triangle P N U \cong \triangle Q N G$
$\Rightarrow \angle U P N = \angle G Q N, Q G = P U = A G$
$\Rightarrow P U \parallel G Q$
$\Rightarrow \angle T = \angl... | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,913 |
24. Three people, A, B, and C, play a game: each person chooses a real number from the interval $[0,1]$, and the one who selects a number between the numbers chosen by the other two wins. A randomly selects a number from the interval $[0,1]$, B randomly selects a number from the interval $\left[\frac{1}{2}, \frac{2}{3}... | 24. B.
Let the number chosen by C be $x$.
According to the problem, the probability that the number chosen by C is greater than the number chosen by A and less than the number chosen by B is $x\left(\frac{2}{3}-x\right)$.
Similarly, the probability that the number chosen by C is greater than the number chosen by B an... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,914 |
2. If $x, y \geqslant 0$, and $x+y=1$, then the range of $\sqrt[3]{7 x+1}+$ $\sqrt[3]{7 y+1}$ is $\qquad$ . | 2. $[3, \sqrt[3]{36}]$.
Let $f(x)=\sqrt[3]{7 x+1}(x \geqslant 0)$. Then
$$
\begin{array}{l}
f^{\prime}(x)=\frac{7}{3}(7 x+1)^{-\frac{2}{3}}, \\
f^{\prime \prime}(x)=-\frac{98}{9}(7 x+1)^{-\frac{5}{3}} \leqslant 0 .
\end{array}
$$
By Jensen's inequality, we have
$$
\begin{array}{l}
\frac{f(x)+f(y)}{2} \leqslant f\left... | [3, \sqrt[3]{36}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,915 |
3. In $\triangle A B C$, the lengths of the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$, respectively. Point $G$ satisfies
$$
\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=0, \overrightarrow{G A} \cdot \overrightarrow{G B}=0 \text {. }
$$
If $(\tan A+\tan B) \tan C=m... | 3. $\frac{1}{2}$.
It is known that $G$ is the centroid of $\triangle ABC$.
Since $G A \perp G B$, we have $a^{2}+b^{2}=5 c^{2}$.
Thus, $m=\frac{\sin ^{2} C}{\sin A \cdot \sin B \cdot \sin C}$
$$
=\frac{2 c^{2}}{a^{2}+b^{2}-c^{2}}=\frac{1}{2} \text {. }
$$ | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,916 |
4. Given a sequence of positive terms $\left\{a_{n}\right\}$ with the sum of the first $n$ terms being $S_{n}$. If $\left\{a_{n}\right\}$ and $\left\{\sqrt{S_{n}}\right\}$ are both arithmetic sequences with the same common difference, then $S_{n}=$ $\qquad$. | 4. $\frac{n^{2}}{4}$.
Let $a_{n}=a_{1}+(n-1) d\left(a_{1} 、 d>0\right.$ be constants $)$.
From the given information, we have
$$
\begin{array}{l}
\sqrt{S_{n}}=\sqrt{a_{1}}+(n-1) d \\
\Rightarrow a_{n}=S_{n}-S_{n-1} \\
\quad=\left(\sqrt{a_{1}}+(n-1) d\right)^{2}-\left(\sqrt{a_{1}}+(n-2) d\right)^{2} \\
\quad=d\left(2 \... | \frac{n^{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,917 |
1. Find the largest real number $\theta(\theta<\pi)$ such that
$$
\prod_{k=0}^{10} \cos 2^{k} \theta \neq 0 \text {, and } \prod_{k=0}^{10}\left(1+\frac{1}{\cos 2^{k} \theta}\right)=1
$$
(2015, Harvard-MIT Mathematics Tournament) | Let $z=\mathrm{e}^{\frac{10}{2}} \neq \pm 1$. Then
$$
\begin{array}{l}
\prod_{k=0}^{10}\left(1+\frac{1}{\cos 2^{k} \theta}\right) \\
=\frac{z+z^{-1}}{z^{2^{11}}+z^{-2^{11}}} \cdot \frac{z^{2^{11}}-z^{-2^{11}}}{z-z^{-1}} \\
=\frac{\tan 2^{10} \theta}{\tan \frac{\theta}{2}}=1
\end{array}
$$
Given that the period of $\ta... | \frac{2046 \pi}{2047} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,918 |
6. Use red and blue to color 10 squares arranged in a row, with each square being colored only one color. If no two adjacent squares can both be red, then the total number of coloring methods is $\qquad$.
Translate the above text into English, please keep the original text's line breaks and format, and output the tran... | 6. 144 .
Let $n$ squares have $a_{n}$ different coloring methods. Thus, $a_{1}=2, a_{2}=3$.
If the first square is blue, then there are $a_{n-1}$ coloring methods; if the first square is red, then there are $a_{n-2}$ coloring methods. Therefore, $a_{n}=a_{n-1}+a_{n-2}$.
From this, we derive
$$
a_{3}=5, a_{4}=8, a_{5}=... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,919 |
7. Let $x, y, z$ be complex numbers, and
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x y+y z+z x, \\
|x+y+z|=21,|x-y|=2 \sqrt{3},|x|=3 \sqrt{3} .
\end{array}
$$
Then $|y|^{2}+|z|^{2}=$ . $\qquad$ | 7. 132.
It is easy to see that the figures corresponding to $x$, $y$, and $z$ on the complex plane form an equilateral triangle, and note that
\[
\begin{array}{l}
|x-y|^{2}+|y-z|^{2}+|z-x|^{2}+|x+y+z|^{2} \\
=3\left(|x|^{2}+|y|^{2}+|z|^{2}\right),
\end{array}
\]
Combining the conditions and $|x-y|=|y-z|=|z-x|$, we ca... | 132 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,920 |
One, (40 points) Given a positive integer $n$, there are $3n$ numbers satisfying:
$$
0 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{3 n},
$$
and $\left(\sum_{i=1}^{3 n} a_{i}\right)^{3} \geqslant k_{n}\left(\sum_{i=1}^{n} a_{i} a_{n+i} a_{2 n+i}\right)$ always holds.
Find the best possible value of $k... | Lemma 1 Let $0 \leqslant x \leqslant K \leqslant y$. Then $x y \leqslant K(x+y-K)$.
Proof of Lemma 1 Clearly,
$$
(K-x)(y-K) \geqslant 0 \text {. }
$$
Lemma 2 Let $0 \leqslant x \leqslant K \leqslant y \leqslant L \leqslant z$. Then $x y z \leqslant K L(x+y+z-K-L)$.
Proof of Lemma 2 In fact,
$$
\begin{array}{l}
x y z \... | 27 n^{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,921 |
"II. (40 points) As shown in Figure 1, in $\triangle ABC$, $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$ respectively, and $P$ is a point on the circumcircle $\odot O$ of $\triangle ABC$. The circumcircle of $\triangle ODP$ intersects $AP$ at a second point $X$. Prove: $\angle XEO = \angle XFO$." | As shown in Figure 2, let $D E$ intersect $A P$ at point $M$, and take point $N$ on $A P$ such that $A E = N E$.
By $\angle O A E = \angle O D E$,
$$
\angle O D X = \angle O P X = \angle O A X,
$$
we get
$$
\begin{array}{l}
\angle X D E = \angle X A E \\
\Rightarrow \angle X M E = \angle F A X = \angle X D F . \\
\tex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,922 |
Three. (50 points) Let $p$ be an odd prime, $S$ be the set of values of $n \in \{0,1, \cdots, p-1\}$, and $p \nmid \sum_{k=0}^{p-1} k!n^{k}$. Denote $|S|$ as the number of elements in the set $S$. Prove:
$$
|S| \geqslant \frac{p+1}{2} \text {. }
$$ | $$
\begin{array}{l}
\sum_{k=0}^{p-1} k!n^{k}=\sum_{l=0}^{p-1}(p-1-l)!n^{p-1-l} \\
\equiv-\sum_{l=0}^{p-1} \frac{1}{(p-l)(p-l+1) \cdots(p-1)} n^{-l} \\
\equiv-\sum_{l=0}^{p-1}(-1)^{l} \frac{n^{-l}}{l!}(\bmod p) \\
\Rightarrow-\sum_{l=0}^{p-1} \frac{x^{l}}{l!} \equiv 0(\bmod p) .
\end{array}
$$
Let $E(z)=\sum_{l=0}^{p-1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,923 |
Four, (50 points) Let $p$ be an odd prime, and $n$ be a positive integer. On a circle with diameter $p^n$ in the coordinate plane, there are eight distinct integer points. Prove: Among these eight points, there exist three points such that the triangle formed by these three points has side lengths whose squares are int... | Four, first prove a lemma.
Lemma: Let $p$ be a prime, $n$ a positive integer. Then among any four integer points on a circle with diameter $p^n$, there exist two integer points $A, B$ such that
$$
v_{p}(A B) \geqslant n+1,
$$
where $v_{p}(A B)$ denotes the highest power of the prime $p$ that divides $A B^2$.
Proof: I... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,924 |
Given $P$ is a point inside an acute $\triangle ABC$, the symmetric point of $P$ with respect to the midpoint $M_{a}$ of side $BC$ is $P_{a}$, the symmetric point of $P_{a}$ with respect to side $BC$ is $Q_{a}$, and the midpoint of $A Q_{a}$ is denoted as $R_{a}$. Similarly, define points $R_{b}$ and $R_{c}$. The proje... | Proof As shown in Figure 2, let the midpoints of $H_{a} R_{a}$ and $H_{b} R_{b}$ be $M$ and $N$ respectively. Let the distance from point $K$ to the line containing side $BC$ be $y_{K}$.
According to the problem, $P Q_{a} \parallel BC$.
Thus, $y_{Q_{a}} = y_{P}$.
Then, $y_{M} = \frac{1}{2} y_{R_{a}} + \frac{1}{2} y_{H_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,925 |
In $1,2, \cdots, 100$ these 100 positive integers, remove 50 so that in the remaining positive integers, any two different $a, b$ have $a \nmid b$. Find the maximum possible value of the sum of all removed positive integers. | Let the remaining numbers be $a_{i}=2^{3} t_{i}$, where $i \in \{1,2, \cdots, 50\}$, $s_{i}$ is a natural number, and $t_{i}$ is an odd number.
Since for any $1 \leqslant i \neq j \leqslant 50$, we have $a_{i} \nmid a_{j}$, it follows that $t_{i} \neq t_{j}$, meaning $t_{1}, t_{2}, \cdots, t_{50}$ are 50 distinct odd ... | 2165 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,926 |
Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets of positive integers, where each set $A_{i}(1 \leqslant i \leqslant n)$ can be arranged into an increasing arithmetic sequence with a common difference of $d_{i}$, and $\bigcup_{i=1}^{n} A_{i}=\mathbf{Z}_{+}$. Define $f(m)$ as the number of distinct prime factors of $m$ $\l... | Prove that the assumption that the conclusion does not hold, i.e., for any $1 \leqslant k \leqslant n$ and $1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant n$, we have
$$
f\left(d_{i_{1}} d_{i_{2}} \cdots d_{i_{k}}\right) \geqslant k .
$$
Let the distinct prime factors of $\prod_{i=1}^{n} d_{i}$ be $p_{1}, p_{2}, \cdot... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,927 |
Example 1 Let $m, n$ be positive integers. Prove:
$$
n!(m!)^{n} \mid (m n)! \text{. }
$$ | 【Analysis】The total number of permutations of $m n$ distinct elements is $(m n)!$.
It can be divided into three steps.
(1) Divide the $m n$ elements into $n$ groups, with $m$ elements in each group, and there are $k$ ways to do this;
(2) Treat one group as one element for permutation, there are $n!$ ways to arrange the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,928 |
Example 2 Let $p$ be a prime, and $a_{1}, a_{2}, \cdots, a_{p}$ be integers.
Prove: There exists an integer $k$, such that $a_{1}+k, a_{2}+2 k, \cdots, a_{p}+p k$
have at least $\frac{p}{2}$ different remainders when divided by $p$. | 【Analysis】From the perspective of combinations (graph theory).
For $k=0,1, \cdots, p-1$, define the graph $G_{k}$: its vertex set is $\{1,2, \cdots, p\}$, and vertices $i, j$ are connected by an edge if and only if $a_{i}+i k \equiv a_{j}+j k(\bmod p)$.
Notice that, for a given $i \neq j$, we have
$(i-j, p)=1$.
Thus, t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,929 |
11. (20 points) Given
function
$$
f(x)=x^{2}+|x+a-1|+(a+1)^{2}
$$
the minimum value is greater than 5. Find the range of $a$. | 11. From the problem, we know that $f(x)$ is a piecewise quadratic function, i.e.,
$$
f(x)=\left\{\begin{array}{ll}
\left(x-\frac{1}{2}\right)^{2}+a^{2}+a+\frac{7}{4}, & x \leqslant 1-a ; \\
\left(x+\frac{1}{2}\right)^{2}+a^{2}+3 a-\frac{1}{4}, & x \geqslant 1-a .
\end{array}\right.
$$
When $1-a>\frac{1}{2}$, i.e., $a... | a \in\left(-\infty, \frac{-1-\sqrt{14}}{2}\right) \cup\left(\frac{\sqrt{6}}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,930 |
Four. (50 points) Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Color each cell of a $5 \times 5$ grid wi... | As shown in Figure 4, a $5 \times 5$ grid is divided into five parts, each colored with one of 5 colors, and at this point, there are 16 dividing edges.
Below is the proof that the number of dividing edges is at least 16.
First, let the 5 colors be denoted as $1, 2, \cdots, 5$, and let the number of rows and columns oc... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,931 |
$\begin{array}{l}\text { Given } a_{1}, a_{2}, \cdots, a_{n}>0(n \geqslant 2, \\ \left.n \in \mathbf{Z}_{+}\right) \text {, and satisfying } \sum_{i=1}^{n} a_{i}=n . \text { Prove: } \\ \sum_{i=1}^{n} \sqrt[n+1]{a_{i}} \geqslant \frac{2 n}{n+1}\left(\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}-\frac{n^{2}-2 n-1}{2}\r... | Proof Note that,
Equation (1)
$$
\begin{array}{l}
\Leftrightarrow \sum_{i=1}^{n} \sqrt[n+1]{a_{i}} \geqslant \frac{2 n}{n+1}\left(\frac{n^{2}-\sum_{i=1}^{n} a_{i}^{2}}{2}-\frac{n^{2}-2 n-1}{2}\right) \\
\Leftrightarrow(n+1) \sum_{i=1}^{n} \sqrt[n+1]{a_{i}}+n \sum_{i=1}^{n} a_{i}^{2} \geqslant 2 n^{2}+n .
\end{array}
$$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,932 |
For all positive integers $n(n \geqslant 3)$, try to find those for which, after removing any one square from an $n \times n$ grid paper, the remaining grid paper can definitely be entirely cut into L-shaped pieces. Here, an L-shaped piece refers to a $2 \times 2$ grid paper piece with one small square removed, leaving... | When $3 \mid n$, by $3 \nmid\left(n^{2}-1\right)$, it is known that it does not meet the requirements.
When $3 \nmid n$, we discuss in cases.
(1) When $n=4$, we might as well remove a small square from the upper left $2 \times 2$ grid, and cut out the L-shape in Figure 1, reducing it to a $2 \times 2$ grid with one squ... | n \neq 5 \text{ and } 3 \nmid n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,933 |
For any positive integer $k$, let
$$
a=k-1, b=(k+1)\left(k^{2}-3\right) \text {. }
$$
Prove: When $k>10^{20}$, $a x^{2}-b y^{2}= \pm 1$ has no
positive integer solutions. | Proof by case analysis.
(1) If $a x^{2}-b y^{2}=1$ has the smallest solution $\left(x_{0}, y_{0}\right)$, then
$$
\begin{aligned}
& \left(\sqrt{a} x_{0}+\sqrt{b} y_{0}\right)\left(\sqrt{a} x_{0}-\sqrt{b} y_{0}\right)=1 \\
\Rightarrow & \left(\sqrt{a} x_{0}+\sqrt{b} y_{0}\right)^{2}\left(\sqrt{a} x_{0}-\sqrt{b} y_{0}\ri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,934 |
Theorem 2 Let $a, b, c$ be integers, $a>0, a c-b^{2}=1$. Then the equation $a x^{2}+2 b x y+c y^{2}=1$ has integer solutions. | Prove that in a Cartesian coordinate system, the dot product of $\vec{A}=\left(x_{1}, y_{1}\right)$ and $\vec{B}=\left(x_{2}, y_{2}\right)$ is defined as
$$
\begin{array}{l}
\left(x_{1}, y_{1}\right)\left(x_{2}, y_{2}\right) \\
=a x_{1} x_{2}+b x_{1} y_{2}+b x_{2} y_{1}+c y_{1} y_{2} .
\end{array}
$$
Under this defini... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,935 |
Lemma 1 Let $p$ be an odd prime, $(a, p)=1, f(x)$ be a polynomial with integer coefficients. Then when $x$ runs through a complete system of residues modulo $p$, we have
(1) $\sum_{x(\bmod p)}\left(\frac{f(a x+b)}{p}\right)=\sum_{x(\bmod p)}\left(\frac{f(x)}{p}\right)$;
(2) $\sum_{x(\bmod p)}\left(\frac{a x+b}{p}\right... | Lemma 1 Proof
(1) Since $(a, p)=1$, when $x$ runs through a complete system modulo $p$, $a x+b$ also runs through a complete system modulo $p$.
Since when $a x+b \equiv x'(\bmod p)$, we have
$$
\left(\frac{f(a x+b)}{p}\right)=\left(\frac{f\left(x'\right)}{p}\right),
$$
thus equation (6) holds.
$$
\begin{array}{l}
\text... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,936 |
Lemma 2 Let $p$ be a prime of the form $4 m+1$, $(k, p)=1$, and let $S(k)=\sum_{x=0}^{p-1}\left(\frac{x\left(x^{2}+k\right)}{p}\right)$. Then
(1) $S(k)$ is even;
(2) $S\left(k t^{2}\right)=\left(\frac{t}{p}\right) S(k)$. | Lemma 2 Proof (1) Let $f(x)=x\left(x^{2}+k\right)$. Then $f(x)$ is an odd function,
$$
\begin{array}{l}
S(k)=\sum_{x=1}^{p-1}\left(\frac{x\left(x^{2}+k\right)}{p}\right) \\
=2 \sum_{x=1}^{\frac{p-1}{2}}\left(\frac{x\left(x^{2}+k\right)}{p}\right) \text { (even). }
\end{array}
$$
(2) If $t \equiv 0(\bmod p)$, then
$$
\b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,937 |
Lemma 3 If $\left(\frac{r}{p}\right)=1,\left(\frac{n}{p}\right)=-1$, then $r \cdot 1^{2}, r \cdot 2^{2}, \cdots, r \cdot\left(\frac{p-1}{2}\right)^{2}, n \cdot 1^{2}, n \cdot 2^{2}, \cdots, n \cdot\left(\frac{p-1}{2}\right)^{2}$ form a reduced residue system modulo $p$. | Lemma 3 Proof: For $t \in \left\{1,2, \cdots, \frac{p-1}{2}\right\}$, it is clear that $\left(r t^{2}, p\right)=1,\left(n t^{2}, p\right)=1$.
When $t_{1}, t_{2} \in \left\{1,2, \cdots, \frac{p-1}{2}\right\}$ and $t_{1} \neq t_{2}$, $r t_{1}^{2} \not \equiv r t_{2}^{2}(\bmod p), n t_{1}^{2} \not \equiv n t_{2}^{2}(\bmod... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,938 |
Theorem 3 If $p$ is a prime of the form $4 m+1$, then
$$
\begin{array}{c}
p=\left(\frac{1}{2} S(r)\right)^{2}+\left(\frac{1}{2} S(n)\right)^{2}, \\
\text { where } S(k)=\sum_{x=0}^{p-1}\left(\frac{x\left(x^{2}+k\right)}{p}\right)(k \in\{r, n\}), \\
\left(\frac{r}{p}\right)=1,\left(\frac{n}{p}\right)=-1 .
\end{array}
$$ | Proof Let $p=4 m+1$.
According to Lemma 3, we know that
$$
r \cdot 1^{2}, r \cdot 2^{2}, \cdots, r \cdot\left(\frac{p-1}{2}\right)^{2}, n \cdot 1^{2}, n \cdot 2^{2}, \cdots, n \cdot\left(\frac{p-1}{2}\right)^{2}
$$
form a reduced residue system modulo $p$. Then
$$
\begin{array}{l}
\sum_{k=1}^{p-1}(S(k))^{2} \\
= \sum... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,939 |
Question 2 As shown in Figure 2, in $\triangle B C F$, $\angle B$ is a right angle. On the line $C F$, take point $A$ such that $F A=F B$, and $F$ is between points $A$ and $C$; take point $D$ such that $D A=D C$, and $A C$ is the angle bisector of $\angle D A B$; take point $E$ such that $E A=E D$, and $A D$ is the an... | Proof As shown in Figure 2, from the given conditions we have
$$
\begin{array}{l}
\angle A B F=\angle B A F=\angle B A C=\angle A C D \\
=\angle D A C=\angle D A E=\angle A D E(=\alpha) .
\end{array}
$$
Thus, $C D / / A B, E D / / A M$.
Also, $E X / / A M$, so $E, D, X$ are collinear,
$$
\angle C D X=\angle A C D=\alp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,940 |
Given the integer sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{0}=1, a_{n}=\sum_{k=0}^{n-1} \mathrm{C}_{n}^{k} a_{k}(n \geqslant 1) \text {. }
$$
Let $m$ be a positive integer, $p$ be a prime, and $q, r$ be non-negative integers. Prove: the difference of two terms $a_{p^{m} q+r}-a_{p^{m-1} q+r}$ is divisible by $p... | Notice that, when $n \geqslant 1$,
$$
\begin{array}{l}
a_{n}=\sum_{k=0}^{n-1} \mathrm{C}_{n}^{k} a_{k}=\sum_{k=0}^{n-1}\left(\mathrm{C}_{n-1}^{k}+\mathrm{C}_{n-1}^{k-1}\right) a_{k} \\
=\sum_{k=0}^{n-1} \mathrm{C}_{n-1}^{k} a_{k}+\sum_{k=0}^{n-2} \mathrm{C}_{n-1}^{k} a_{k}+a_{n-1}-a_{n-1} \\
=2 \sum_{k=0}^{n-1} \mathrm... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,941 |
Example 1 As shown in Figure $1, H$ is the orthocenter of $\triangle A B C$, and $D, E$ are points on sides $A B, A C$ respectively, such that points $D, H, E$ are collinear, and $A D=A E$. Extend $D E$ to intersect the circumcircle $\odot O$ of $\triangle A B C$ at points $F, G$. Draw perpendiculars from $D, E$ to $A ... | 【Analysis】First, the complexity to be handled is $K$.
According to the order of the point generation, from $K D \perp A B, K E \perp A C$, we think of the Simson theorem, then draw a line segment $U V \perp H L$ through point $H$, intersecting $A B$ and $A C$ at points $U$ and $V$ respectively.
By the converse of the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,942 |
Example 5 As shown in Figure 5, in acute $\triangle A B C$, $O$ is the circumcenter, $M$ is the midpoint of $B C$, point $P$ lies on the arc $\overparen{A O}$ of the circumcircle determined by points $A, O, M$, points $E, F$ lie on $A C, A B$ respectively, and satisfy
$$
\angle A P B=\angle B M E, \angle A P C=\angle C... | 【Analysis】In this problem, points $E$ and $F$ are complex.
As shown in Figure 5, extend $AP$ to intersect $\odot O$ at point $Q$.
Then $\triangle QCP \sim \triangle BFM$, $\triangle QBP \sim \triangle CEM$.
Extend $FM$ and $EM$ to points $F'$ and $E'$ respectively, such that points $B, F, C, F'$ and $C, E, B, E'$ are c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,943 |
Example 6 In $\triangle A B C$, points $P_{1}$ and $P_{2}$ are on side $A B$ such that point $P_{2}$ lies on segment $B P_{1}$, and $A P_{1}=B P_{2}$. Points $Q_{1}$ and $Q_{2}$ are on side $B C$ such that point $Q_{2}$ lies on segment $B Q_{1}$, and $B Q_{1}=C Q_{2}$. Let the intersection of segments $P_{1} Q_{2}$ and... | 【Analysis】As shown in Figure 6.
In this problem, $R$ is a special point.
Directly proving “$\angle P_{1} R S = \angle Q_{1} R M$” seems infeasible, so we might as well use an indirect method to transform the conditions.
The original conclusion is equivalent to proving:
$$
\frac{\sin \angle S R Q_{1}}{\sin \angle S R P_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,944 |
Theorem As shown in Figure 1, let $O, I_{1}$ be the circumcenter and the excenter opposite to vertex $A$ of $\triangle ABC$, respectively. $OI_{1}$ intersects $BC$ at point $A_{1}$. Similarly, define points $B_{1}, C_{1}$. Then the lines $AA_{1}$, $BB_{1}$, $CC_{1}$ concur at the Schiffler point $S$ of $\triangle ABC. ... | Proof As shown in Figure 2, let $D$ be the midpoint of the arc $\overparen{BC}$ of the circumcircle $\odot O$ of $\triangle ABC$ that does not contain point $A$. Let $M$ and $N$ be the midpoints of $BC$ and $AI_1$, respectively. Let $K$ be the point symmetric to $I_1$ with respect to $M$. Let $DK$ intersect $ON$ at poi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,945 |
2. Let the function $f(x)$ be $\mathbf{R} \rightarrow \mathbf{R}$, satisfying for all $x \in \mathbf{R}$,
$$
f(x)+x f(2-x)=2 \text{. }
$$
Then the analytical expression of $f(x)$ is | 2. $f(x)=\left\{\begin{array}{ll}-\frac{2}{x-1}, & x \neq 1 ; \\ 1, & x=1 .\end{array}\right.$
Substitute $2-x$ for $x$ to get
$$
f(2-x)+(2-x) f(x)=2 \text {, }
$$
Combining this with the given equation yields the expression for $f(x)$. | f(x)=\left\{\begin{array}{ll}-\frac{2}{x-1}, & x \neq 1 ; \\ 1, & x=1 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,947 |
3. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0), F$ is the right focus of the ellipse, and $AB$ is a chord passing through the center $O$. Then the maximum value of the area of $\triangle ABF$ is $\qquad$. | 3. $b \sqrt{a^{2}-b^{2}}$.
Make the transformation $(x, y) \rightarrow\left(x, \frac{a y}{b}\right)$, we know that $S_{\triangle A B F}$ reaches its maximum value if and only if $A B \perp O F$. | b \sqrt{a^{2}-b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,948 |
7. In the tetrahedron $P-ABC$,
$$
AB=\sqrt{3}, AC=1, PB=PC=\sqrt{2} \text{.}
$$
Then the range of $S_{\triangle ABC}^{2}+S_{\triangle PBC}^{2}$ is | 7. $\left(\frac{1}{4}, \frac{7}{4}\right)$.
Reflect $\triangle P B C$ to be coplanar with $\triangle A B C$. Consider quadrilateral PBAC.
Obviously, when $B C=2$, $S_{\triangle A B C}^{2}+S_{\triangle P B C}^{2}$ reaches the maximum value $\frac{7}{4}$; when $A$, $B$, and $C$ are collinear, $S_{\triangle A B C}^{2}+S... | \left(\frac{1}{4}, \frac{7}{4}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,949 |
12. Let $n$ be a given positive integer. Consider the set of points in the Cartesian coordinate system $x O y$
$$
T=\{(x, y) \mid | x|=| y | \leqslant n, x, y \in \mathbf{Z}\} .
$$
For two points $P, Q$ in $T$, $P, Q$ are called "adjacent" if and only if $|P Q|=\sqrt{2}$ or $P Q$ is parallel to one of the coordinate a... | $$
\begin{array}{c}
\text { 12. Let } A_{k}(-k, k), B_{k}(k, k), C_{k}(k,-k), \\
D_{k}(-k,-k), \text { where } k=1,2, \cdots, n, \text { and let } \\
S_{k}=\left\{A_{k}, B_{k}, C_{k}, D_{k}\right\}, \\
T_{k}=\{O\} \cup S_{1} \cup S_{2} \cup \cdots \cup S_{k},
\end{array}
$$
In particular, $T_{n}=T$.
Considering the nu... | t_{n}=\frac{2 \sqrt{33}}{11}\left(\left(\frac{7+\sqrt{33}}{2}\right)^{n}-\left(\frac{7-\sqrt{33}}{2}\right)^{n}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,951 |
1. Let three complex numbers $1$, $\mathrm{i}$, and $z$ correspond to three collinear points in the complex plane, and $|z|=5$. Then $z=$ $\qquad$ . | 1.4-3 i or $-3+4$ i.
Let $z=x+y$ i.
From the problem, we get $x+y=1$, and $x^{2}+y^{2}=25$.
Therefore, $(x, y)=(4,-3)$ or $(-3,4)$. | 1.4-3 i \text{ or } -3+4 i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,952 |
Example 5 Let $a$ be a given positive integer. Prove: there exist infinitely many primes, such that for each of these primes, there exists at least one $n \in \mathbf{N}$, satisfying $2^{2 n}+a$ is divisible by this prime. | 【Analysis】Assume the conclusion does not hold.
Let $p_{1}, p_{2}, \cdots, p_{k}$ be all the primes that can divide at least one number of the form $2^{2 n}+a$.
Consider $k+1$ numbers $a^{2}+a(1 \leqslant i \leqslant k+1, i \in \mathbf{Z})$.
Since these numbers are finite, there exists a natural number $r$ such that non... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,954 |
6. Inside a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, there is a small ball that is tangent to the diagonal segment $A C_{1}$. Then the maximum radius of the ball is $\qquad$ | 6. $\frac{4-\sqrt{6}}{5}$.
When the radius is maximized, the small sphere is tangent to three faces of the cube. Without loss of generality, assume the small sphere is tangent to the three faces passing through point $D_{1}$. Establish a spatial rectangular coordinate system with $D_{1}$ as the origin, $\overrightarro... | \frac{4-\sqrt{6}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,955 |
7. Let $H$ be the orthocenter of $\triangle A B C$, and
$$
3 \overrightarrow{H A}+4 \overrightarrow{H B}+5 \overrightarrow{H C}=\mathbf{0} \text {. }
$$
Then $\cos \angle A H B=$ $\qquad$ | 7. $-\frac{\sqrt{6}}{6}$.
From the given, we have $\frac{\tan A}{3}=\frac{\tan B}{4}=\frac{\tan C}{5}=\lambda$.
Also, from $\tan A+\tan B+\tan C$
$$
=\tan A \cdot \tan B \cdot \tan C \text {, }
$$
we get $\lambda=\frac{1}{\sqrt{5}}, \tan C=\sqrt{5}$.
Therefore, $\cos \angle A H B=-\cos C=-\frac{\sqrt{6}}{6}$. | -\frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,956 |
Example 6 If two positive integers $a$ and $b$ satisfy $a=p b$ or $b=p a$ for some prime $p$, then $a$ and $b$ are called "prime-related". Find all positive integers $n$ that satisfy the following property: $n$ has at least three positive divisors, and all divisors of $n$ can be arranged in a circle without repetition ... | 【Analysis】Call $n$ that meets the condition a "good number".
If $n$ is a good number, then all the divisors of $n$ can be arranged on a circle without repetition, such that any two adjacent divisors are prime-related.
Let $d_{1}, d_{2}, \cdots, d_{k}$ be all the numbers arranged clockwise on the circle. Then for any $... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,957 |
3. Let $f(x)$ be a monotonic function defined on the interval $(0,+\infty)$. If for any $x \in(0,+\infty)$, we have
$$
f\left(f(x)-2 \log _{2} x\right)=4 \text {, }
$$
then the solution set of the inequality $f(x)<6$ is $\qquad$ | 3. $\{x \mid 0<x<4\}$.
From the problem, we know there exists a positive constant $c$, such that $f(c)=4$, and for any $x \in(0,+\infty)$, we have
$$
f(x)-2 \log _{2} x=c \text {. }
$$
When $x=c$, we have
$$
f(c)=2 \log _{2} c+c=4,
$$
By monotonicity, we know this equation has a unique solution $c=2$.
Thus, $f(x)=2 ... | \{x \mid 0<x<4\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,959 |
4. Given point $P$ on the hyperbola with eccentricity $\sqrt{2}$
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
$F_{1} 、 F_{2}$ are the two foci of the hyperbola, and $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ $=0$. Then the ratio of the inradius $r$ to the circumradius $R$ of $\triangle P F_... | 4. $\frac{\sqrt{6}}{2}-1$.
From $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=0 \Rightarrow \angle F_{1} P F_{2}=90^{\circ}$.
Let $\left|P F_{1}\right|=m,\left|P F_{2}\right|=n$.
Also, $\left|F_{1} F_{2}\right|=2 c$, then
$$
\begin{array}{l}
R=c, r=\frac{1}{2}(m+n-2 c), \\
m^{2}+n^{2}=4 c^{2},|m-n|=2 a .
\e... | \frac{\sqrt{6}}{2}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,960 |
6. A die is rolled four times consecutively, and from the second roll onwards, the number of points that appear each time is not less than the number of points that appeared in the previous roll. The probability of this happening is $\qquad$ . | 6. $\frac{7}{72}$.
Let $a_{1}, a_{2}, a_{3}, a_{4}$ be the numbers obtained from four dice rolls, respectively. Then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ has $6^{4}$ different possibilities.
If from the second roll onwards, each number is not less than the previous one, then $a_{1} \leqslant a_{2} \leqslant a_{3... | \frac{7}{72} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,961 |
7. Let positive real numbers $x, y$ satisfy
$$
x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4} \text {. }
$$
Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is | 7.6.
By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x} \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$
Adding th... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,962 |
8. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{3}-n$ $\left(n \in \mathbf{Z}_{+}\right)$, and the terms in this sequence whose unit digit is 0, arranged in ascending order, form the sequence $\left\{b_{n}\right\}$. Then the remainder when $b_{2} 018$ is divided by 7 is $\qquad$ . | 8. 4 .
Since $a_{n}=n^{3}-n=n(n-1)(n+1)$, therefore, $a_{n}$ has a units digit of 0 if and only if the units digit of $n$ is $1, 4, 5, 6, 9, 0$. Hence, in any consecutive 10 terms of the sequence $\left\{a_{n}\right\}$, there are 6 terms whose units digit is 0.
Since $2018=336 \times 6+2,336 \times 10=3360$, the rema... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,963 |
9. (20 points) Given $O$ as the origin, $N(1,0)$, and $M$ as a moving point on the line $x=-1$, the angle bisector of $\angle M O N$ intersects the line $M N$ at point $P$.
(1) Find the equation of the trajectory curve $E$ of point $P$;
(2) Draw a line $l$ with slope $k$ through the point $Q\left(-\frac{1}{2},-\frac{1}... | (1) Let $P(x, y), M(-1, t)$.
It is easy to know, $0 \leqslant x<1$.
Since $O P$ bisects $\angle M O N$, we have
$\overrightarrow{M P}=\frac{|O M|}{|O N|} \overrightarrow{P N}=\sqrt{1+t^{2}} \overrightarrow{P N}$.
Then $x+1=\sqrt{1+t^{2}}(1-x)$,
$y-t=\sqrt{1+t^{2}}(0-y)$.
From equations (1) and (2), we get $t=\frac{2 y}... | k \in\left(-\frac{1}{3}, 1\right] \cup\left\{\frac{1+\sqrt{3}}{2}\right\} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,964 |
2. If the function
$$
f(x)=x^{2}-2 a x-2 a|x-a|+1
$$
has exactly three zeros, then the value of the real number $a$ is $\qquad$. | 2. 1.
Let $t=|x-a|(t \geqslant 0)$.
Then the original problem is equivalent to the equation $t^{2}-2 a t+1-a^{2}=0$ having two roots $t_{1}=0, t_{2}>0$.
Upon verification, $a=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,965 |
4. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and
$$
a_{1}=3, S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right) \text {. }
$$
If $\left\{a_{n}\right\}$ contains three terms $a_{1} 、 a_{p} 、 a_{q}(p 、 q \in$ $\left.\mathbf{Z}_{+}, 1<p<q\right)$ that form an arithmetic seque... | 4. 1 .
Given $S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right)$
$$
\Rightarrow S_{n-1}=2 a_{n-1}+\frac{3}{2}\left((-1)^{n-1}-1\right)(n \geqslant 2) \text {. }
$$
Subtracting the two equations yields $a_{n}=2 a_{n-1}-3(-1)^{n}(n \geqslant 2)$.
Let $b_{n}=\frac{a_{n}}{(-1)^{n}}$. Then
$$
\begin{array}{l}
b_{n}=-2 b_{n... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,966 |
5. If the complex number $z$ satisfies
$$
\left|\frac{z^{2}+1}{z+\mathrm{i}}\right|+\left|\frac{z^{2}+4 \mathrm{i}-3}{z-\mathrm{i}+2}\right|=4 \text {, }
$$
then the minimum value of $|z-1|$ is | 5. $\sqrt{2}$.
Notice that,
$$
\begin{array}{l}
\left|\frac{z^{2}+1}{z+\mathrm{i}}\right|+\left|\frac{z^{2}+4 \mathrm{i}-3}{z-\mathrm{i}+2}\right| \\
=|z-\mathrm{i}|+|z-(2-\mathrm{i})|=4>2 \sqrt{2} .
\end{array}
$$
Then the locus of $z$ is an ellipse with foci at $(0,1)$ and $(2,-1)$, where the center of the ellipse i... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,967 |
8. A small bug crawls on the surface of a regular octahedron, moving from one vertex to one of the four adjacent vertices with equal probability each second. What is the probability that the bug returns to its initial position at the 8th second? $\qquad$ . | 8. $\frac{43}{256}$.
As shown in Figure 3, after crawling $n$ steps, let the probability of the insect being at point $A$ be $p_{n}$, the probability of being at one of $B, C, D, E$ be $q_{n}$, and the probability of being at point $F$ be $r_{n}$.
$$
\begin{array}{l}
\text { Then } p_{n+1}=\frac{1}{4} q_{n}, \\
q_{n+1... | \frac{43}{256} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,968 |
$$
\text { II. (40 points) Given } n \in \mathbf{N}, n \geqslant 2,0<\theta<\pi \text {. }
$$
Prove:
$$
\sin \frac{\theta}{2} \cdot \sum_{k=1}^{n} \frac{\sin k \theta}{k}<1 .
$$ | $\begin{array}{l}\text { II. sin } \frac{\theta}{2} \cdot \sum_{k=1}^{n} \frac{\sin k \theta}{k} \\ =\frac{1}{2} \sum_{k=1}^{n} \frac{2 \sin \frac{\theta}{2} \cdot \sin k \theta}{k} \\ =\frac{1}{2} \sum_{k=1}^{n} \frac{\cos \frac{2 k-1}{2} \theta-\cos \frac{2 k+1}{2} \theta}{k} \\ =\frac{1}{2}\left(\cos \frac{\theta}{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,969 |
5. As shown in Figure 2, in the acute triangle $\triangle ABC$, $AB < AC$, $O$ is its circumcenter, and $M$ is the midpoint of side $BC$. The circle passing through points $A$, $O$, and $M$ intersects the extension of $AB$ at point $D$, and intersects the segment $AC$ at point $E$. Prove: $DM = EC$. | 5. From the given, $B M=M C$,
$$
\begin{array}{l}
\angle B D M=\angle A D M=\angle M E C . \\
\text { Note that }, O M \perp B C . \\
\text { Then } \angle D M B=\angle D M O-\angle B M O \\
=\left(180^{\circ}-\angle O A D\right)-90^{\circ} \\
=90^{\circ}-\angle O A B=\frac{1}{2} \angle A O B \\
=\angle A C B=\angle E ... | DM = EC | Geometry | proof | Yes | Yes | cn_contest | false | 730,970 |
7. Let positive integers $p, c$ be a prime and a composite number, respectively. Prove: there exist positive integers $m, n$, satisfying
$$
0<m-n<\frac{[n+1, n+2, \cdots, m]}{[n, n+1, \cdots, m-1]}=p^{c},
$$
where, $\left[x_{1}, x_{2}, \cdots, x_{k}\right]$ denotes the least common multiple of the positive integers $x... | 7. Let $m=p^{2 c-1}, n=p^{2 c-1}-p^{c-1}$. The following proves that $m, n$ satisfy the conditions.
Obviously, $0<m-n=p^{c-1}<p^{c}$.
It is only necessary to prove:
$$
\frac{[L, m]}{[n, L]}=p^{c}(L=[n+1, n+2, \cdots, m-1]) \text {. }
$$
Let $v_{r}(l)$ denote the exponent of the prime factor $r$ in the positive integer... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,971 |
8. Given positive integers $n, k$, where $n$ is even, $k \geqslant 2$, and $n > 4k$. There are $n$ points on a circle. If $\frac{n}{2}$ non-intersecting chords inside the circle have endpoints exactly at these $n$ points, then these chords are called a "matching". Find the largest integer $m$ such that for any matching... | 8. $\left\lceil\frac{k}{3}\right\rceil$ ( $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$). Let $m=\left\lceil\frac{k}{3}\right\rceil$.
It is easy to see that $k \geqslant 2 m$, and $3 m-2 \leqslant k \leqslant 3 m$.
First, we provide an example of a matching where any $k$ consecutive po... | \left\lceil\frac{k}{3}\right\rceil | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,972 |
3. As shown in Figure 4,
$$
\triangle X Z Y \backsim \triangle D B A \backsim \triangle E A C \backsim \triangle F E D .
$$
Let $M$ be the midpoint of $B C$. Prove:
$$
\begin{array}{l}
\angle F A M=|\angle X Y Z-\angle X Z Y|, \\
\frac{A F}{A M}=\frac{2 X Y \cdot X Z}{Y Z^{2}} .
\end{array}
$$ | $$
\begin{array}{l}
\text { Then } \frac{D-B}{A-B}=\frac{E-A}{C-A}=\frac{F-E}{D-E}=z=\frac{X-Y}{Z-Y} \\
\Rightarrow D=z A+(1-z) B, E=z C+(1-z) A \\
\Rightarrow F=z D+(1-z) E \\
=\left(2 z^{2}-2 z+1\right) A+\left(z-z^{2}\right) B+\left(z-z^{2}\right) C \\
\Rightarrow F-A=2\left(z-z^{2}\right)\left(\frac{B+C}{2}-A\right... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,973 |
5. Prove: There exist infinitely many pairs of positive integers $a, b$, such that $a>b, \mathrm{C}_{a b}^{a} \equiv \mathrm{C}_{a b}^{b} \neq 0(\bmod a+b)$. | 5. For any odd prime $p$, take $a=p+1, b=p-1$. Below we verify that it meets the requirements.
Since $a+b=2 p$, it is only necessary to prove:
$$
\mathrm{C}_{a b}^{a} \equiv \mathrm{C}_{a b}^{b} \not \equiv 0(\bmod p) \text { and } \mathrm{C}_{a b}^{a} \equiv \mathrm{C}_{a b}^{b}(\bmod 2) \text {. }
$$
By Lucas's Theo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,974 |
6. Let $n$ be a positive integer, and $p_{1}<p_{2}<\cdots<p_{k}<p_{k+1}$ be consecutive primes such that $p_{k} \leqslant n < p_{k+1}$. Define
$$
P(n)=\prod_{i=1}^{k} p_{i}, G(n)=\max _{2 \leqslant i \leqslant k+1}\left(p_{i}-p_{i-1}\right) \text {. }
$$
Define $m$ as the largest integer such that for each $p_{i}(1 \l... | 6. Note that,
$$
\left\{a_{p_{1}}\left(\bmod p_{1}\right), a_{p_{2}}\left(\bmod p_{2}\right), \cdots, a_{p_{k}}\left(\bmod p_{k}\right)\right\}
$$
can cover $\{1,2, \cdots, m\}$.
By the Chinese Remainder Theorem, there exists $x$ such that
$$
x \equiv -a_{p_{i}}\left(\bmod p_{i}\right)(1 \leqslant i \leqslant k)
$$
a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,975 |
1. Let $g(x)=\ln \left(2^{x}+1\right)$. Then
$$
\begin{array}{l}
g(-4)-g(-3)+g(-2)-g(-1)+ \\
g(1)-g(2)+g(3)-g(4) \\
=\text {. }
\end{array}
$$ | $$
-1 .-2 \ln 2 \text {. }
$$
From the problem, we know that
$$
g(x)-g(-x)=\ln \frac{1+2^{x}}{1+2^{-x}}=x \ln 2 .
$$
Therefore, the required value is $-2 \ln 2$. | -2 \ln 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,976 |
2. $\prod_{i=1}^{2018} \sin \frac{i \pi}{2019}=$ $\qquad$ . | 2. $\frac{2019}{2^{2018}}$.
Since $z_{1}=\mathrm{e}^{\frac{2 \pi i}{n}}$ is a root of $z^{n}=1$, we have
$$
\begin{array}{l}
z^{n}-1=(z-1)\left(z-z_{1}\right)\left(z-z_{1}^{2}\right) \cdots\left(z-z_{1}^{n-1}\right) \\
\Rightarrow z^{n-1}+z^{n-2}+\cdots+z+1 \\
\quad=\left(z-z_{1}\right)\left(z-z_{1}^{2}\right) \cdots\... | \frac{2019}{2^{2018}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,977 |
5. Given the sides of $\triangle A B C$ are $a, b, c$ and satisfy: $a+b \leqslant 2 c, b+c \leqslant 3 a$.
Then the range of $\frac{c}{a}$ is $\qquad$ | 5. $\left(\frac{2}{3}, 2\right)$.
From the problem, we know that $a+b+c \leqslant 3 c, a+b+c \leqslant 4 a$.
Let $s=a+b+c=1$. Then
$c \geqslant \frac{1}{3}, a \geqslant \frac{1}{4}, a+c>\frac{1}{2}$.
If $c>a \Rightarrow \max \left\{\frac{1}{3}, a\right\} \leqslant c \leqslant \frac{1}{2}$,
If $c<a \Rightarrow \frac{1}... | \left(\frac{2}{3}, 2\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,979 |
6. Given $x, y, z \in \mathbf{R}_{+}$, and $\sqrt{x^{2}+y^{2}}+z=1$. Then the maximum value of $x y+2 x z$ is $\qquad$ | 6. $\frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Given } \sqrt{x^{2}+y^{2}}+z=1 \Rightarrow x^{2}+y^{2}=(1-z)^{2} \\
\Rightarrow x^{2}=(1-z-y)(1-z+y) .
\end{array}
$$
Then $(x y+2 x z)^{2}=x^{2}(y+2 z)^{2}$
$$
\begin{array}{l}
=\frac{1}{3}(3-3 z-3 y)(1-z+y)(y+2 z)(y+2 z) \\
\leqslant \frac{1}{3}\left(\frac{(3-3 z-... | \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,980 |
8. In the Cartesian coordinate system, color the set of points
$$
\left\{(m, n) \mid m, n \in \mathbf{Z}_{+}, 1 \leqslant m, n \leqslant 6\right\}
$$
red or blue. Then the number of different coloring schemes where each unit square has exactly two red vertices is $\qquad$ kinds. | 8. 126 .
Dye the first row (points with a y-coordinate of 6), there are $2^{6}$ ways to do this, which can be divided into two cases.
(1) No two same-colored points are adjacent (i.e., red and blue alternate), there are 2 ways, and the second row can only be dyed in 2 ways, each row has only 2 ways, totaling $2^{6}$ w... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,981 |
Example 1 Given that $n$ is a positive integer and $x_{0}$ is an integer. Then for each prime factor $p$ of $\Phi_{n}\left(x_{0}\right)$, either $n \mid(p-1)$ or $p \mid n$.
| Prove a lemma first.
Lemma Let $p$ be a prime, and assume that $x^{n}-1$ has a repeated root modulo $p$, i.e., there exists an integer $a$ and a polynomial $f(x) \in \mathbf{Z}[x]$, such that
$$
x^{n}-1=(x-a)^{2} f(x)(\bmod p) .
$$
Then $p \nmid n$.
Proof Clearly, $p \nmid a$.
Make the substitution $y=x-a$, to get
$$
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,982 |
Example 2 Let $p$ be a prime. Prove: there exists a prime $q$, such that for any integer $n, n^{p}-p$ is not a multiple of $q$.
(44th IMO) | Let $q$ be a prime factor of $\Phi_{p}(p)$. And $\Phi_{p}(p)=\frac{p^{p}-1}{p-1}=\sum_{j=0}^{p-1} p^{j} \not \equiv 1\left(\bmod p^{2}\right)$, $\Phi_{p}(p) \equiv 1(\bmod p)$, so $p^{2} \nmid(q-1)$, and $(p, q)=1$.
By the conclusion of Example 1, we know that only $p \mid(q-1)$.
If there exists $n^{p} \equiv p(\bmod q... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,983 |
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Let $\{a_{n}\}$ be a sequence of non-negative real numbers. Define:
$$
X_{k}=\sum_{i=1}^{2^{k}} a_{i}, \quad Y_{k}=\sum_{i=1}^{2^{k}}\left[\frac{2^{k}}{i}\right] a_{i},
$$
where $k=0,1, \cdots$. Prove that for any positive integer $n$, we have... | 5. For any positive integer $k$, we have
$$
\begin{array}{l}
Y_{k}-2 Y_{k-1} \\
=\left(\sum_{i=1}^{2^{k-1}}\left[\frac{2^{k}}{i}\right] a_{i}+\sum_{i=2^{k-1}+1}^{2^{k}}\left[\frac{2^{k}}{i}\right] a_{i}\right)- \\
\sum_{i=1}^{2^{k-1}} 2\left[\frac{2^{k-1}}{i}\right] a_{i} \\
= \sum_{i=1}^{2^{k-1}}\left(\left[2 \frac{2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,984 |
7. A meeting was attended by 24 representatives, and between any two representatives, they either shook hands once or did not shake hands at all. After the meeting, it was found that there were a total of 216 handshakes, and for any two representatives $P$ and $Q$ who shook hands, among the remaining 22 representatives... | 7. Let the 24 representatives be $v_{1}, v_{2}, \cdots, v_{24}$, and for $i=1,2, \cdots, 24$, let $d_{i}$ denote the number of people who have shaken hands with $v_{i}$.
Define the set $E=\left\{\left\{v_{i}, v_{j}\right\} \mid v_{i}\right.$ has shaken hands with $v_{j} \}$.
For any $e=\left\{v_{i}, v_{j}\right\} \in E... | 864 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,985 |
8. Let $m$ be a given positive integer. For a positive integer $l$, denote
$$
A_{l}=(4 l+1)(4 l+2) \cdots\left(4\left(5^{m}+1\right) l\right) \text {. }
$$
Prove: There exist infinitely many positive integers $l$ such that $5^{5 m l} \mid A_{l}$, and $5^{5 m l+1} \nmid A_{l}$, and find the smallest value of $l$ that s... | 8. For an integer $p \geqslant 2$, let the sum of the digits of a positive integer $n$ in base $p$ be denoted by $S_{p}(n)$. For a prime $p$, let $v_{p}(m)$ denote the exponent of the prime factor $p$ in the standard factorization of a positive integer $m$.
First, we prove a lemma.
Lemma Let $n$ be a positive integer, ... | l=\frac{5^{m+1}-1}{4} | Number Theory | proof | Yes | Yes | cn_contest | false | 730,986 |
3. As shown in Figure $3, \triangle A B C$ is inscribed in $\odot O, \angle A B C > 90^{\circ}, M$ is the midpoint of side $B C$. Point $P$ is inside $\triangle A B C$, satisfying $P B \perp P C$. A line perpendicular to $A P$ is drawn through point $P$, and $D, E$ are two points on this line different from $P$, satisf... | 3. Let $A O$ and $D E$ intersect at a common midpoint $N$. From the 3rd problem of Grade 10, we know that points $P$ and $N$ are both on the circle $\odot M$ with $B C$ as its diameter, as shown in Figure 6.
Clearly, point $P$ is also on the circle $\Gamma$ with $A N$ as its diameter, and circle $\Gamma$ is tangent to... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,987 |
4. Does there exist a set $A \subseteq \mathbf{Z}_{+}$, such that for each positive integer $n, A \cap\{n, 2 n, \cdots, 15 n\}$ contains exactly one element, and there are infinitely many positive integers $m$, such that $\{m, m+2018\} \subseteq A$? Prove your conclusion.
（He Yijie, problem contributor） | 4. As in the 4th problem of the first year of high school, define $f(n)$ and set $A$, knowing that for each positive integer $n, A \cap\{n, 2 n, \cdots, 15 n\}$ contains exactly one element.
By the Chinese Remainder Theorem, there exist infinitely many positive integers $m$ that satisfy the system of congruences
$$
\l... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,988 |
6. Given an integer $m \geqslant 2$. A meeting has a total of $3 m$ representatives attending, and any two representatives either shake hands once or do not shake hands. For a positive integer $n (n \leqslant 3 m-1)$, if there exist $n$ representatives among them whose number of handshakes are $1, 2, \cdots, n$, then t... | 6. Let the $3m$ representatives be denoted as $A_{1}, A_{2}, \cdots, A_{3 m}$.
When $1 \leqslant n \leqslant 2 m$, construct an $n$-interesting meeting as shown in Figure 7: for $i=1,2, \cdots, m$, let $A_{3 m+1-i}$ shake hands with $A_{i}, A_{i+1}, \cdots, A_{2 m}$.
At this point, the number of handshakes for represe... | 2m+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,989 |
7. For positive integers $m, n$, let $f(m, n)$ denote the number of ordered integer triples $(x, y, z)$ satisfying
$$
\left\{\begin{array}{l}
x y z = x + y + z + m, \\
\max \{|x|,|y|,|z|\} \leqslant n
\end{array}\right.
$$
Does there exist positive integers $m, n$ such that $f(m, n) = 2018$? Prove your conclusion.
（He... | 7. Suppose positive integers $m, n$ satisfy
$$
f(m, n)=2018 \text {. }
$$
For the integer solutions $(x, y, z)$ of the system of equations (1):
If $x=y=z$, it is called the 1st type solution;
If exactly two of $x, y, z$ are equal, it is called the 2nd type solution;
If $x, y, z$ are all different, it is called the 3rd... | m=2, n=168 | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,990 |
2. Let the line $y=a$ intersect the curve $y=\sin x(0 \leqslant x \leqslant \pi)$ at points $A$ and $B$. If $|A B|=\frac{\pi}{5}$, then $a=$ $\qquad$ ( to 0.0001 ). | 2.0. 9511 .
Let $A(p, \sin p), B(q, \sin q)$, and
$$
0<p<\frac{\pi}{2}<q<\pi \text {. }
$$
Since $\sin p=\sin q=a$, we have $q=\pi-p$.
Given that $|A B|=\frac{\pi}{5}$, we have
$$
q-p=\pi-2 p=\frac{\pi}{5} \text {. }
$$
Therefore, $p=\frac{2 \pi}{5}$.
Using a TI calculator, we find $\sin \frac{2 \pi}{5} \approx 0.95... | 0.9511 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 730,992 |
3. Let integers $x, y$ satisfy $x^{2}+y^{2}4$. Then the maximum value of $x^{2}-2 x y-3 y$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 3 .
Using a TI calculator, we can obtain the region satisfying
$$
\left\{\begin{array}{l}
x^{2}+y^{2}4
\end{array},\right.
$$
and since we need to find the maximum value of $x^{2}-2 x y-3 y$, the integer point is in the third quadrant.
Substituting $(-2,-3)$ and $(-3,-2)$ into $x^{2}-2 x y-3 y$ and comparing the ... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,993 |
5. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}, C A=$ $3, C B=4$, a moving point $P$ is inside $\triangle A B C$, and the sum of its distances to the three sides is $\frac{13}{5}$. Then the length of the trajectory of point $P$ is | 5. $\frac{\sqrt{5}}{2}$.
As shown in Figure 4, place the right triangle $\triangle ABC$ in the Cartesian coordinate system, where $A(3,0), B(0,4)$.
Figure 4
Let the point $P(x, y)$ satisfy the condition. Then,
$$
x+y+\frac{|4 x+3 y-12|}{5}=\frac{13}{5} \text{. }
$$
Since point $P$ is inside $\triangle ABC$, we have,
... | \frac{\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,994 |
11. Arrange 10 flowers in a row using red, yellow, and blue flowers (assuming there are plenty of each color), and yellow flowers cannot be adjacent. How many different arrangements are there (the 10 flowers can be of one color or two colors)? | 11. Let $x_{n}$ be the number of different arrangements of $n$ flowers that meet the requirements. Then
$$
x_{1}=3, x_{2}=3^{2}-1=8 \text {. }
$$
When $n \geqslant 3$, let the arrangement of $n$ flowers be $a_{1}, a_{2}, \cdots, a_{n}$. If $a_{1}$ is a red or blue flower, then $a_{2}, a_{3}, \cdots, a_{n}$ is an arran... | 24960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,996 |
1. If the set
$$
A=\{1,2, \cdots, 10\}, B=\{1,2,3,4\},
$$
$C$ is a subset of $A$, and $C \cap B \neq \varnothing$, then the number of such subsets $C$ is $(\quad)$.
(A) 256
(B) 959
(C) 960
(D) 961 | - 1. C.
From the fact that there are $2^{6}$ subsets $C$ satisfying $C \cap B=\varnothing$, we know that there are $2^{10}-2^{6}=960$ subsets $C$ satisfying $C \cap B \neq \varnothing$. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,997 |
2. In the arithmetic sequence $\left\{a_{n}\right\}$,
$$
a_{2}+a_{11}+a_{14}=-6 \text {. }
$$
Then the sum of the first 17 terms $\sum_{i=1}^{17} a_{i}$ is ( ).
(A) 0
(B) -34
(C) 17
(D) 34 | 2. B.
Let the common difference be $d$. Then
$$
a_{2}+a_{11}+a_{14}=3 a_{1}+24 d=3 a_{9} \text {. }
$$
Combining with the given condition, we get $a_{9}=-2$.
Thus, the sum of the first 17 terms is $17 a_{9}=-34$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,998 |
4. Among the following four numbers, the largest is $(\quad)$.
(A) $\ln \sqrt{2}$
(B) $\frac{1}{\mathrm{e}}$
(C) $\frac{\ln \pi}{\pi}$
(D) $\frac{\sqrt{10} \ln 10}{20}$ | 4. B.
Consider the function $f(x)=\frac{\ln x}{x}$.
Then the four options are $f(2)$, $f(\mathrm{e})$, $f(\pi)$, and $f(\sqrt{10})$.
Also, $f^{\prime}(x)=\frac{1-\ln x}{x^{2}}$, so $f(x)$ is monotonically increasing on the interval $(0, \mathrm{e})$ and monotonically decreasing on the interval $(\mathrm{e},+\infty)$.
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,999 |
5. In Figure 1, the square $ABCD$ is composed of six equilateral triangles. When it is folded along the dotted lines, it forms a hexahedron shaped like a zongzi (sticky rice dumpling) as shown in Figure 2. In this hexahedron, the cosine of the angle between $AB$ and $CD$ is $(\quad)$.
(A) 0
(B) 1
(C) $\frac{1}{2}$
(D) ... | 5. C.
Take the middle dashed line $E F$, dividing $\square A B C D$ into two parts, each consisting of three small equilateral triangles. The three small equilateral triangles on the left fold up (point $B$ coincides with point $F$) to form a bottomless regular triangular pyramid, with $A B$ and $E F$ as its two base ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,000 |
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