problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Question 2: Prove that if $2n-1$ is a prime number, then for any $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, there exist $i, j \in \{1,2, \cdots, n\}$, such that
$$
\frac{a_{i}+a_{j}}{(a_{i}, a_{j})} \geqslant 2n-1,
$$
where $(x, y)$ denotes the greatest common divisor of the positive integers $x$ and... | Let $2 n-1=p$, without loss of generality, assume
$\left(a_{1}, a_{2}, \cdots, a_{n}\right)=1$.
If there exists $i(1 \leqslant i \leqslant n)$, such that $p \mid a_{i}$, then there must exist $j \neq i$, such that $p \nmid a_{j}$.
Since $p \nmid\left(a_{i}, a_{j}\right)$, we have
$$
\frac{a_{i}+a_{j}}{\left(a_{i}, a_{j... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,785 |
Given a positive integer $n$. Prove: for any positive integers $a, b, c$ not exceeding $3 n^{2}+4 n$, there exist integers $x, y, z$ with absolute values not exceeding $2 n$ and not all zero, such that $a x+b y+c z=0$.
---
The translation maintains the original text's formatting and line breaks. | Prove that if $a=b$, let $z=0, x=1, y=-1$, the problem holds.
Assume $a, b, c$ are not equal, without loss of generality, let $a2 n$.
From equation (1), without loss of generality, let $z>2 n$, otherwise, replace $x, y, z$ with $-x, -y, -z$.
Since positive integers $a, b$ are both less than $c$, we have,
$$
x=u_{1}-u_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,786 |
Question 1 Given an acute triangle $\triangle ABC$ with its circumcircle $\odot O$, the feet of the altitudes from $B C$, $C A$, and $A B$ are $D$, $E$, and $F$ respectively. The line $E F$ intersects the arcs $\overparen{A B}$ and $\overparen{A C}$ of $\odot O$ at points $G$ and $H$ respectively. The line $D F$ inters... | Proof As shown in Figure 1, let the orthocenter of $\triangle ABC$ be $P$.
From $CF \perp AB, BE \perp AC, AD \perp BC$
$$
\Rightarrow \angle BFD = \angle ACB = \angle AHB
$$
$\Rightarrow A, F, L, H$ are concyclic
$$
\Rightarrow \angle LAH = \angle LFH = 2(90^\circ - \angle ACB)
$$
$\Rightarrow \triangle ALH$ is an iso... | \sqrt{2(b^2 + c^2 - a^2)} | Geometry | proof | Yes | Yes | cn_contest | false | 730,787 |
Question 5 Given an acute triangle $\triangle ABC$ satisfying $AB > AC$, the incenter is $I$, the excenter opposite to $\angle A$ is $I_{1}$, and the midpoint of $BC$ is $L$. The line $LI$ intersects side $AC$ at point $M$, and the line $I_{1}L$ intersects side $AB$ at point $N$. Let $BC = a$, $CA = b$, and $AB = c$. P... | Proof as shown in Figure 2.
It is well-known that points $A, I, I_{1}$ are collinear. Let $A I$ intersect $B C$ at point $K$.
For $\triangle A K C$ and the secant $L I M$, by Menelaus' theorem, we have
$$
\frac{A M}{M C} \cdot \frac{C L}{L K} \cdot \frac{K I}{I A}=1 \quad \text{(since $I C$ bisects $\angle K C A$)}.
$... | \frac{a(c-b)}{(a+c-b)^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 730,789 |
2. Given that $n$ is a positive integer, define a chameleon as any sequence of $3 n$ letters, and it has exactly $n$ $a$s, $n$ $b$s, and $n$ $c$s. Swapping the positions of two adjacent letters in a chameleon is called an operation. Prove: For any chameleon $X$, there exists another chameleon $Y$ such that $X$ cannot b... | 2. Since the chameleon does not change when the same two letters are operated on once, we can assume that there is no such operation.
For any two chameleons $X, Y$, define the "distance" $d(X, Y)$ as the minimum number of operations required to transform $X$ into $Y$ (or $Y$ into $X$). Then, for any three chameleons $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,790 |
3. Mr. Alex plays the following game: At the beginning, there are nine empty boxes arranged in a row. Each time, Alex chooses one of the following two situations:
(1) Choose any number of the form $2^{j}(j \in \mathbf{N})$ and place this number in an empty box;
(2) Choose two boxes that contain the same number (they do... | $$
\text { 3.2 } 2 \sum_{j=0}^{8} \mathrm{C}_{n}^{j}-1 .
$$
Consider a more general problem: replace 9 with a positive integer $k$. Let $m(n, k)$ be the maximum number of operations Alex can perform such that starting with $k$ empty boxes, one box ends up with the number $2^n$ and the other $k-1$ boxes are empty. Oper... | 2 \sum_{j=0}^{k-1} \mathrm{C}_{n}^{j}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,791 |
7. For any finite sets $X, Y$ composed of positive integers, define $f_{X}(k)$ as the $k$-th smallest positive integer not in $X$,
$$
X * Y=X \cup\left\{f_{X}(y) \mid y \in Y\right\} .
$$
Let the set of positive integers $A$ contain $a (a>0)$ positive integers, and the set of positive integers $B$ contain $b (b>0)$ po... | 7. For any function $g: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ and any subset $X \subset \mathbf{Z}_{+}$, define
$$
g(X)=\{g(x) \mid x \in X\} .
$$
Then the image of $f_{X}$ is $f_{X}\left(\mathbf{Z}_{+}\right)=\mathbf{Z}_{+} \backslash X$.
First, prove two lemmas.
Lemma 1 Let $X, Y$ be finite sets of positive int... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,792 |
2. Prove: The equation $x^{3}+3 x y^{2}=\underset{2018 \text { ones }}{11 \cdots 1}$ has no integer solutions $(x, y)$.
(Goodlin provided the problem) | 2. The original equation can be transformed into
$$
2 x^{3}+6 x y^{2}=\underset{2018 \uparrow}{22 \cdots 2} \text {, }
$$
which is equivalent to $(x+y)^{3}+(x-y)^{3}=\underset{2018 \text { 个 }}{22 \cdots 2}$.
Notice that, $\underset{2018 \uparrow}{22 \cdots 2} \equiv 4(\bmod 9)$,
$$
n^{3} \equiv 0,1,-1(\bmod 9) \text ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,794 |
3. Given positive integers $m, n$. Each cell of an $m \times n$ grid is colored either black or white. A "L-shape" is defined as the remaining part of a $2 \times 2$ grid after removing the top-right cell (rotations and reflections are not allowed). It is known that the number of L-shapes with all three cells white is ... | 3. Proof by contradiction. Assume that there does not exist a $2 \times 2$ square with all four cells being white.
For convenience, we call an L-shaped figure with three white cells a "white L-shape." If a cell is at the bottom-left corner of a white L-shape, we call it an A-class cell; if a cell is at the top-left or... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,795 |
6. As shown in Figure 2, hexagon $A B C D E F$ is inscribed in $\odot O$, the tangents to $\odot O$ at points $A$ and $D$ intersect at point $P$, line $B C$ intersects $P D$ at point $M$, line $E F$ intersects $P A$ at point $N$, $\angle A P F = \angle D P C$, and $A B \parallel D E$. Prove: $A M \parallel D N$. | 6. Let the intersection of $P C$ and $\odot O$ be $Q$.
By symmetry, we know $P Q = P F$.
By the power of a point theorem, we have
$$
\begin{array}{l}
P F \cdot P C = P Q \cdot P C = P A^{2} \\
= P D^{2} = P A \cdot P D .
\end{array}
$$
Since $\angle A P F = \angle D P C$, we have
$$
\begin{array}{l}
\triangle A P F ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,797 |
7. For a positive integer $n \geqslant 4$, if a convex $n$-gon $A_{1} A_{2} \cdots A_{n}$ satisfies that for any $2 \leqslant k \leqslant n-2$,
$$
A_{1} A_{k+1}=A_{2} A_{k+2}=\cdots=A_{n-1} A_{k-1}=A_{n} A_{k} \text {, }
$$
then it is called a "regular $n$-gon". For example, rectangles and isosceles trapezoids are bot... | 7. (1) When $n=6$, the following construction can be made:
Take an equilateral triangle and rotate it around its center by any angle that is not a multiple of $30^{\circ}$. Connect the six vertices of the two triangles in sequence, and the resulting hexagon is a regular hexagon that meets the conditions.
(2) When $n \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,798 |
8. (1) Prove: There exists a one-to-one mapping $f: \mathbf{Z}_{+} \times \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ that satisfies the following conditions:
(i) $f(1,1)=1$;
(ii) For any integer $i>1, f(i, j)(j=1,2, \cdots)$ has a common divisor greater than 1;
(iii) For any integer $j>1, f(i, j)(i=1,2, \cdots)$ has a ... | 8. (1) Take $f(i, j)=2^{i-1}(2 j-1)$ to satisfy the requirement.
(2) Such a one-to-one mapping exists.
Let $S_{n}=\{1,2, \cdots, n\}$.
Construct $\left.f\right|_{S_{n} \times S_{n}}$ inductively, thus, when $n \rightarrow+\infty$, we get $f$.
Let $f(1,1)=2$.
If $\left.f\right|_{S_{n-1} \times S_{n-1}}$ has been constru... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,799 |
Example 5 defines an "operation": cutting a rope of positive integer length into three ropes of positive integer length, with the longest rope being exactly one. Given a rope of length $a^{b}$ (where $a$ and $b$ are integers greater than 1), two players, A and B, take turns, with A starting first by performing one oper... | 【Analysis】Consider the general case. Let the initial length of the rope be $n\left(n \in \mathbf{Z}_{+}\right)$.
Obviously, if $1 \leqslant n \leqslant 3$, A must fail;
if $4 \leqslant n \leqslant 7=3+2+2$, then A has a winning strategy;
if $8 \leqslant n \leqslant 9$, the length of the longest rope after A's operation... | a=2, b=3 \text{ or } a \text{ is a power of 3, B has a winning strategy; otherwise, A has a winning strategy.} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,800 |
5. As shown in Figure 2, the orthocenter of acute triangle $\triangle ABC$ is $H$, and points $D$ on side $AB$ and $E$ on side $AC$ satisfy $DE \parallel CH$, and the circumcircle of $\triangle BDH$ passes through the midpoint of segment $DE$. Prove: $\angle ABM = \angle ACM$. | 5. Connect $D H$.
Since $D E / / C H$ and $C H \perp A B$, therefore, $D E \perp A B$.
From $B 、 D 、 M 、 H$ being concyclic $\Rightarrow M H \perp B H$. Also, $B H \perp A C$, then $M H / / A C$. Hence, quadrilateral $C H M E$ is a parallelogram $\Rightarrow C H / / E M$, and $C H=E M$.
Notice that, $M$ is the midpoi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,801 |
8. Prove: There exist infinitely many positive integers $n$, such that 2018 divides $\left(1+2^{n}+3^{n}+4^{n}\right)$. (Zhang Lei) | 8. Let $n=9 \times 7 \times 4 k$. Then $2 ! \left(1+2^{n}+3^{n}+4^{n}\right)$.
To prove: $1009 \mid \left(1+2^{n}+3^{n}+4^{n}\right)$. Since $2^{9} \equiv 3^{9}(\bmod 1009)$, it suffices to prove: $1009 \mid \left(1+2 \times\left(2^{9}\right)^{4 \times 7 \times k}+\left(\left(2^{9}\right)^{4 \times 7 \times k}\right)^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,802 |
1. Let $a_{1}, a_{2}, \cdots$ be an increasing infinite sequence of positive integers, and $p_{1}, p_{2}, \cdots$ be a sequence of prime numbers such that for any $n$, $p_{n} \mid a_{n}$. Also, for any positive integers $n, k$, $a_{n}-a_{k}=p_{n}-p_{k}$. Prove: Each term in the sequence $a_{1}, a_{2}, \cdots$ is a prim... | 1. Let $a_{1}-p_{1}=c$.
According to the problem, for any $n$, we have
$$
\begin{array}{l}
a_{n}-a_{1}=p_{n}-p_{1} \\
\Rightarrow a_{n}-p_{n}=a_{1}-p_{1}=c .
\end{array}
$$
Assume $c>0$.
Since the sequence $\left\{p_{n}\right\}$ is increasing, there exists $n$ such that
$$
a_{n}=p_{n}+c>2 c \text {. }
$$
Thus, $a_{n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,803 |
Example 6 A grade has $3 N$ students. In a math test, the possible scores are $60, 61, \cdots, 100$. These scores each appear at least twice, and the average score is 82.4. Prove: the students of this grade can be divided into three classes, each with $N$ students, and the average score of each class is 82.4.
(Ninth Ch... | 【Analysis】The original solution is a bit complicated. In fact, the adjustment method can be used.
Arrange $3N$ students in a line, with the scores from the end of the line to the front being $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{3 N}$. Give a red flower to each of the $N$ students at the end of the line... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,804 |
3. Let $a_{1}, a_{2}, \cdots, a_{25}$ be non-negative integers, and $k$ be the smallest one among them. Prove:
$$
\sum_{i=1}^{25}\left[\sqrt{a_{i}}\right] \geqslant\left[\sqrt{\sum_{i=1}^{25} a_{i}+200 k}\right] \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. | 3. Let $n_{i}=\left[\sqrt{a_{i}}\right]$. Then $a_{i}<\left(n_{i}+1\right)^{2}$.
Since $a_{i}$ is an integer, we have
$$
a_{i} \leqslant n_{i}^{2}+2 n_{i} \text {. }
$$
If we can prove that
$$
\sqrt{\sum_{i=1}^{25} a_{i}+200 k}<\sum_{i=1}^{25} n_{i}+1,
$$
then equation (1) holds.
For definiteness, let $k=a_{1}$.
Thu... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,805 |
5. 99 points are marked at equal intervals on the circumference. Betya and Vasya take turns coloring these points, with Betya starting first. Betya can arbitrarily color one point red or blue, and subsequently, each person on their turn colors an uncolored point adjacent to an already colored point red or blue. If, aft... | 5. No.
First, provide a strategy to ensure Vasya's victory.
Before the number of colored points reaches 33, Vasya can act arbitrarily.
Assume that 33 points have already been colored. Let point $A$ be colored and at the edge of the already colored points, and point $B$ be uncolored and adjacent to the point at the ot... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,806 |
6. Given positive integers $a, b$. Prove: There exist infinitely many positive integers $n$, such that $\left(n^{b}+1\right) \nmid\left(a^{n}+1\right)$. | 6. If $\left(n^{b}+1\right) \nmid\left(a^{n}+1\right)$, then the positive integer $n$ is called "bad".
It is sufficient to prove: For any even $n$, at least one of $n$ and $n^{3}$ is bad, which would then establish the conclusion.
Assume otherwise. Then for any even $n$, we have
$$
\begin{array}{l}
\left(n^{b}+1\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,807 |
2. Given positive integers $x_{1}, x_{2}, \cdots, x_{10}$ are all distinct, satisfying $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2} \leqslant 2019$. Then
$$
x_{8}-\left(x_{5}+x_{4}+x_{3}+x_{2}+x_{1}\right)
$$
The maximum value is ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | 2. C.
From the conditions, we know $x_{i} \geqslant i(i=1,2, \cdots, 10)$, and $x_{9} \geqslant x_{8}+1, x_{10} \geqslant x_{8}+2$.
Then $1^{2}+2^{2}+\cdots+7^{2}+x_{8}^{2}+\left(x_{8}+1\right)^{2}+\left(x_{8}+2\right)^{2}$ $\leqslant 2019$
$$
\Rightarrow x_{8}^{2}+2 x_{8}-624 \frac{2}{3} \leqslant 0 \Rightarrow x_{8}... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,808 |
6. As shown in Figure 1, in $\triangle ABC$, $AD$, $BE$, and $CF$ intersect at point $P$, $\frac{AF}{BF}=\frac{AE}{CE}$, and $AD \perp BC$. A line parallel to $AB$ is drawn through point $C$, intersecting the extension of $BE$ at point $G$. If $PE=1$, $EG=2$, then the value of $BP$ is ( ).
(A) 3
(B) 2
(C) $\sqrt{3}$
(D... | 6. C.
By Ceva's Theorem,
$$
\frac{B D}{D C}=\frac{E A}{C E} \cdot \frac{F B}{A F}=1
$$
$\Rightarrow D$ is the midpoint of side $B C$
$$
\Rightarrow A B=A C \text {. }
$$
It is easy to see that, $\angle A B P=\angle A C P, B P=C P$.
By $\angle G=\angle A B P=\angle A C P$
$\Rightarrow \triangle P C E \sim \triangle P ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,809 |
1. The system of equations in $x, y, z$
$$
\left\{\begin{array}{l}
x y+y z+z x=1, \\
5 x+8 y+9 z=12
\end{array}\right.
$$
all real solutions $(x, y, z)$ are $\qquad$ | $$
\text { II. 1. }\left(1, \frac{1}{2}, \frac{1}{3}\right) \text {. }
$$
Substituting equation (2) into equation (1) yields
$$
\begin{array}{l}
5 x^{2}+8 y^{2}+4 x y-12 x-12 y+9=0 \\
\Rightarrow(4 y+x-3)^{2}+9(x-1)^{2}=0 \\
\Rightarrow x-1=0,4 y+x-3=0 \\
\Rightarrow(x, y, z)=\left(1, \frac{1}{2}, \frac{1}{3}\right) .... | \left(1, \frac{1}{2}, \frac{1}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,810 |
2. As shown in Figure 2, person A at point $A$ on the shore notices person B in distress at point $B$ in the water. Point $B$ is 30 meters away from the shore at $C$, and $\angle BAC=15^{\circ}$. Person A runs on the shore at a speed that is $\sqrt{2}$ times their swimming speed in the water. Given that A's swimming sp... | 2. $15 \sqrt{2}+5 \sqrt{6}$
As shown in Figure 3, construct
$\angle C A D=45^{\circ}$, and let point $P$ be where person 甲 enters the water from the shore. Draw $P H \perp A D$ at point $H$, at this time,
$$
A P=\sqrt{2} P H .
$$
According to the problem, the time it takes for 甲 to travel from point $A$ to $B$ is the... | 15 \sqrt{2}+5 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,811 |
1. Let the set $P_{n}=\{1,2, \cdots, n\}\left(n \in \mathbf{Z}_{+}\right)$. Denote $f(n)$ as the number of sets $A$ that simultaneously satisfy the following conditions:
(1) $A \subseteq P_{n}, \bar{A}=P_{n} \backslash A$;
(2) If $x \in A$, then $2 x \notin A$;
(3) If $x \in \bar{A}$, then $2 x \notin \bar{A}$.
Then $f... | $-1.2^{1009}$
For a geometric sequence with a common ratio of 2, $a, 2a, 2^{2}a, \cdots, 2^{k-1}a\left(a\right.$ is an odd number, $k$ such that $\left.2^{k-1}a \leqslant 2018 < 2^{k}a\right)$, the terms have only two cases:
If $a \in A$, then $2a \in \bar{A}, 2^{2}a \in A, \cdots$ and so on; if $a \in \bar{A}$, then ... | 2^{1009} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,812 |
2. Let the side length of the regular octagon $A_{1} A_{2} \cdots A_{8}$ be 1. For any two points $A_{i} 、 A_{j}$, the maximum value of $\overrightarrow{A_{i} A_{j}} \cdot \overrightarrow{A_{1} A_{2}}$ is $\qquad$ . | 2. $\sqrt{2}+1$.
Notice that, the four sides $A_{1} A_{2}$, $A_{3} A_{4}$, $A_{5} A_{6}$, and $A_{7} A_{8}$ of the regular octagon $A_{1} A_{2} \cdots A_{8}$ form a square with a side length of $\sqrt{2}+1$.
Also, $\overrightarrow{A_{i} A_{j}} \cdot \overrightarrow{A_{1} A_{2}}$ is the projection of $\overrightarrow{... | \sqrt{2}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,813 |
3. Given $0<\alpha \leqslant \beta \leqslant \gamma, \alpha+\beta+\gamma=\pi$. Then the range of $\min \left\{\frac{\sin \beta}{\sin \alpha}, \frac{\sin \gamma}{\sin \beta}\right\}$ is . $\qquad$ | 3. $\left[1, \frac{\sqrt{5}+1}{2}\right)$.
Let $q=\min \left\{\frac{\sin \beta}{\sin \alpha}, \frac{\sin \gamma}{\sin \beta}\right\}$.
Then $q \leqslant \frac{\sin \beta}{\sin \alpha}, q \leqslant \frac{\sin \gamma}{\sin \beta}$
$$
\begin{aligned}
\Rightarrow & \frac{1}{q} \geqslant \frac{\sin \alpha}{\sin \beta}, \fr... | \left[1, \frac{\sqrt{5}+1}{2}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,814 |
5. Given that the length of the major axis of an ellipse is 4, the left vertex is on the parabola $y^{2}=x-1$, and the left directrix is the $y$-axis. Then the maximum value of the eccentricity of such an ellipse is $\qquad$
Translate the above text into English, please retain the original text's line breaks and forma... | 5. $\frac{2}{3}$.
Since the x-coordinate of the point on the parabola is not less than 1, that is, the distance to the y-axis is not less than 1, thus, the distance from the left vertex of the ellipse to the left directrix is not less than 1.
$$
\begin{array}{l}
\text { Then } \frac{a^{2}}{c}-a \geqslant 1 \Rightarrow... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,815 |
7. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the minimum value of the function $y=\frac{1}{\sin ^{2} x}+\frac{12 \sqrt{3}}{\cos x}$ is $\qquad$ . | 7. 28 .
Notice that,
$$
\begin{array}{l}
y=16\left(\frac{1}{16 \sin ^{2} x}+\sin ^{2} x\right)+16\left(\frac{3 \sqrt{3}}{4 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16\left(\frac{3 \sqrt{3}}{8 \cos x}+\frac{3 \sqrt{3}}{8 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16 \times ... | 28 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,816 |
8. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=$ $A D=5, A A_{1}=1, M$ is a point on edge $A B$ and satisfies $3 \overrightarrow{A M}+2 \overrightarrow{B M}=\mathbf{0}$. A point $P$ is randomly selected within the square base $A B C D$, and the distance from point $P$ to line $A_{1} D_{1}$ is $d_{1... | 8. $\frac{32}{75}$.
As shown in Figure 2, let point $P$ have projections $R$ and $Q$ on $AD$ and $A_1D_1$, respectively.
Then $\triangle PQR$ is a right
triangle,
$$
\begin{array}{l}
\angle PRQ=90^{\circ}, \\
RQ=1 .
\end{array}
$$
From $1 \leqslant d_1^2 - d_2^2$
$$
\begin{array}{l}
= PQ^2 - PM^2 = (PR^2 + RQ^2) - P... | \frac{32}{75} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,817 |
10. (20 points) The sequence $\left\{a_{n}\right\}$ has the sum of the first $n$ terms $S_{n}$ satisfying $a_{n}=\frac{1}{2}+\frac{1}{2} \sqrt{1+8 S_{n-1}}(n \geqslant 2)$.
Also, $a_{1}=1$, find the general term formula of the sequence. | 10. From $S_{n}-S_{n-1}=\frac{1}{2}+\frac{1}{2} \sqrt{1+8 S_{n-1}}$
$$
\begin{aligned}
\Rightarrow & 2 S_{n}=2 S_{n-1}+1+\sqrt{1+8 S_{n-1}} \\
\Rightarrow & 8 S_{n}+1=\left(8 S_{n-1}+1\right)+4 \sqrt{1+8 S_{n-1}}+4 \\
& =\left(\sqrt{8 S_{n-1}+1}+2\right)^{2} \\
\Rightarrow & \sqrt{8 S_{n}+1}=\sqrt{8 S_{n-1}+1}+2 .
\end... | a_{n}=n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,818 |
Example 1 As shown in Figure 1, let the circumcircle of $\triangle ABC$ be $\Gamma$. Construct the tangents to circle $\Gamma$ at points $B$ and $C$, and let these tangents intersect at point $D$. Construct squares $BAGH$ and $ACEF$ outwardly on sides $AB$ and $CA$ of $\triangle ABC$, respectively. Let $EF$ and $HG$ in... | 【Analysis】Let $\Gamma$ be the unit circle, with the center $O$ of circle $\Gamma$ as the origin, and the $y$-axis // $BC$ with point $C$ in the first quadrant and point $B$ in the fourth quadrant, establishing a Cartesian coordinate system, as shown in Figure 2.
Let $C(\cos \alpha, \sin \alpha), B(\cos \alpha, -\sin \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,819 |
Three. (50 points) As shown in Figure 1, points $A$ and $B$ are on line $l$, and the midpoint of segment $AB$ is $M$. Points $P$ and $Q$ are taken on opposite sides of line $l$. Point $C$ on ray $PA$ satisfies $PA \cdot PC = PM^2$, and point $D$ on ray $QB$ satisfies $QB \cdot QD = QM^2$. If $\angle PCQ = 90^{\circ}$, ... | Three, as shown in Figure 5, take point $S$ on ray $P Q$ such that
$$
P M^{2}=P S \cdot P Q \text {; }
$$
Take point $T$ on ray $Q P$ such that
$$
\begin{array}{l}
Q M^{2}=Q T \cdot Q P . \\
\text { Hence } \triangle P S M \backsim \triangle P M Q \backsim \triangle M T Q \\
\Rightarrow \angle P S M=\angle P M Q=\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,820 |
Given $x \geqslant y \geqslant z>0, z>x y$, and $x+y+z=1$.
Prove: $x y z+2 \sqrt{x y z}>x y+y z+z x$.
---
The problem statement and the proof requirement have been translated into English while maintaining the original format and line breaks. | Prove that since $x+y+z=1$, it is sufficient to prove
$$
\begin{array}{l}
x y z+2 \sqrt{x y z}>(x y+y z+z x)(x+y+z) \\
\Leftarrow 2 \sqrt{x y z}>(x+y)(y+z)(z+x) \\
\Leftarrow 4 x y z>(x+y)^{2}(y+z)^{2}(z+x)^{2} \\
=((x+y)(y+z))((y+z)(z+x)) . \\
((z+x)(x+y)) \\
=(y+z x)(z+x y)(x+y z) .
\end{array}
$$
Since $z>x y$, th... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,821 |
Given a convex quadrilateral $ABCD$ inscribed in $\odot O$, $AB$ intersects $DC$ at point $E$, $BC$ intersects $AD$ at point $F$, and $OH \perp EF$ at point $H$. Prove: $AC$, $BD$, and $OH$ are concurrent. | Prove that in Figure 2, draw the tangents to $\odot O$ through points $E$ and $F$, with the points of tangency being $X$ and $Y$.
First, prove a lemma.
Lemma: $E F^{2}=E X^{2}+F Y^{2}$.
Proof: Find a point $K$ on $E F$ such that points $E, B, C, K$ are concyclic.
Since $\angle C K F=\angle E B C=\angle A D C$,
$\Righta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,822 |
Hunter 591 placed $n(n \geqslant 2)$ nets at different positions on a number line, the initial positions of the nets are denoted as $a_{1}, a_{2}$, $\cdots, a_{n}$. There is a bug between each pair of adjacent nets, and each net moves in the positive or negative direction at speeds $v_{1}, v_{2}, \cdots, v_{n}$. If two... | Let the maximum number of bugs caught be denoted as $f(n)$.
It is easy to see that $f(2)=2$.
It is evident that any two nets can meet at most once. Therefore, adding the $(n+1)$-th net can capture at most $n$ new bugs, meaning that during the movement, at most $n$ new bugs can be added, and one more bug can be added in... | \frac{n(n+1)}{2}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,823 |
Given a prime number $p$ satisfying $p^{2} \mid\left(2^{p-1}-1\right)$, the set $S=\{0,1,1+2, \cdots, 1+2+\cdots+(p-2)\}$. Prove: There exist two elements in the set $S$ that are congruent modulo $p-1$. | Proof: Let $p-1=2^{k}$, i.e., $p=2^{k}+1$.
If $s \geqslant 3$ is an odd divisor of $k$, then
$$
p=2^{s t}+1=\left(2^{t}+1\right) A \text {, }
$$
which contradicts the fact that $p$ is a prime.
Thus, $k=2^{t}$.
Therefore, $2^{p-1}-1=2^{2^{k}}-1$
$$
=\left(2^{2^{t}}-1\right) \prod_{u=t}^{k-1}\left(2^{2^{u}}+1\right) \te... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,824 |
Example 2 As shown in Figure 3, given $\triangle A_{1} A_{2} A_{3}$ and a point $P$ inside it, let the circumcenters of $\triangle A_{1} A_{2} A_{3}$, $\triangle P A_{2} A_{3}$, $\triangle P A_{3} A_{1}$, and $\triangle P A_{1} A_{2}$ be $O$, $O_{1}$, $O_{2}$, and $O_{3}$, respectively. The line $O_{1} O$ intersects $O... | 【Analysis】If the positions of points $A_{2}$ and $A_{3}$ are swapped, then the two fractions in the problem will both take their reciprocals. To compare the sizes of the two fractions, their values must be equal. Therefore, it is sufficient to prove that the values of these two fractions are equal.
First, observe $\fra... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,825 |
Example 2 Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a, b, c$ in $S$ such that $a+b, b+c, c+a$ are all in the subset $A$.
(2012, China Mathematical Olympiad) | 【Analysis】The condition of the problem is equivalent to being able to find three numbers $x, y, z$ in the subset $A$ such that
$$
\frac{-x+y+z}{2}, \frac{x-y+z}{2}, \frac{x+y-z}{2}
$$
are three distinct positive integers. For this, $x, y, z$ must be distinct, satisfy the triangle inequality, and their sum must be even... | 1008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,827 |
Example 1 Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{0}=b_{0}=1, \\
a_{n+1}=\alpha a_{n}+\beta b_{n}, b_{n+1}=\beta a_{n}+\gamma b_{n},
\end{array}
$$
where $\alpha, \beta, \gamma$ are positive integers, $\alpha<\gamma$, and $\alpha \gamma=\beta^{2}+1$.
Prove:... | Example 1 Proof First prove two lemmas.
Lemma 1 If positive integers $m, n, k$ satisfy $mn=k^2+1$,
then there exist $x_1, x_2, y_1, y_2 \in \mathbf{N}$, such that
$$
m=x_1^2+y_1^2, n=x_2^2+y_2^2, k=x_1 x_2+y_1 y_2
$$
all hold simultaneously.
Proof of Lemma 1 Without loss of generality, assume $m \leqslant n$, and use ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,828 |
Question 3 Prove the identity:
$$
(x+n)^{n}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k}(x+k)(x+n)^{n-k-1} k!.
$$ | Prove that in Theorem 3 (2) of [4], taking \( f(x) = x^n \), we get
\[
\sum_{k=0}^{n}(-1)^{n-k} \mathrm{C}_{n}^{k}(x+k)^{n}=n!.
\]
Substituting \( -x \) for \( x \) gives
\[
\sum_{k=0}^{n} \mathrm{C}_{n}^{k}(x-k)^{n-k}(-x+k)^{k}=n!.
\]
Then the original identity is equivalent to
\[
\begin{aligned}
& (-x+n)^{n} \\
= &... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,829 |
Question 1 In $\triangle ABC$, $AB < AC$. The angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ABC$ at a second point $D$. The perpendicular bisector of side $AC$ intersects the external angle bisector of $\angle BAC$ at point $Z$. Prove that the midpoint $M$ of segment $AB$ lies on the circumci... | Proof 1: Take the midpoint $N$ of side $AC$. Connect the auxiliary lines as shown in Figure 1.
Since $MN$ is the midline of $\triangle ABC$, we have
$$
\begin{array}{l}
\angle MNZ = \angle ANZ + \angle ANM \\
= 90^{\circ} + \angle ACB.
\end{array}
$$
It is also easy to see that
$$
\begin{array}{l}
\angle DCZ = \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,830 |
Question 2 As shown in Figure 3, in the acute triangle $\triangle ABC$, $AB \neq AC$, $K$ is the midpoint of the median $AD$, $DE \perp AB$ at point $E$, $DF \perp AC$ at point $F$, the lines $KE$ and $KF$ intersect $BC$ at points $M$ and $N$ respectively, and the circumcenters of $\triangle DEM$ and $\triangle DFN$ ar... | Given that
$E K=F K=A K=D K=\frac{1}{2} A D$
$\Rightarrow A, E, D, F$ lie on the circle $\odot K$ with radius $A K$.
As shown in Figure 3, extend $A B$ and $A F$ to intersect the circumcircle $\odot O_{1}$ of $\triangle D E M$ and the circumcircle $\odot O_{2}$ of $\triangle D F N$ at points $P$ and $Q$, respectively. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,831 |
Question 3 Let $\triangle ABC$ be an acute-angled triangle with orthocenter $H$, and let $W$ be a point on side $BC$, not coinciding with vertices $B$ and $C$. $CN \perp AB$ at point $N$, and $BM \perp AC$ at point $M$. $WX$ is the diameter of the circumcircle of $\triangle BWN$, and $WY$ is the diameter of the circumc... | Proof: From the given, we know that points $A, N, H, M$ are concyclic.
Let the second intersection point of the circumcircles of $\triangle B W N$ and $\triangle C W N$ be $Z$, and connect the auxiliary lines as shown in Figure 4.
Then $\angle B N Z = \angle Z W C = \angle A M H$, and since $A, N, H, M$ are concyclic,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,832 |
Question 4 Let $\Gamma$ be the circumcircle of acute-angled $\triangle ABC$, and let points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, such that $AD = AE$. The perpendicular bisectors of segments $BD$ and $CE$ intersect the minor arcs $\overparen{AB}$ and $\overparen{AC}$ of circle $\Gamma$ at points $F$ ... | Proof: Let the perpendicular bisectors of $BD$ and $CE$ intersect $BD$ and $CE$ at points $M$ and $N$, respectively, and intersect the circle $\Gamma$ at the second points $J$ and $I$. Extend $JD$ and $IE$ to intersect the circle at points $X$ and $Y$. Draw the auxiliary lines as shown in Figure 5.
$$
\begin{array}{l}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,833 |
Example 1 Let convex pentagon $A B C D E$ satisfy:
$$
\begin{array}{l}
\angle B A C=\angle C A D=\angle D A E, \\
\angle A B C=\angle A C D=\angle A D E,
\end{array}
$$
Diagonals $B D$ and $C E$ intersect at point $P$. Prove: $A P$ bisects segment $C D$.
(47th IMO Shortlist) | Proof: From the problem, we know that $\triangle A B C$ and $\triangle A D E$ are rotationally similar figures centered at $A$. Therefore, $B C$ can be transformed into $D E$ through rotational similarity.
From Lemmas 1 and 2, we know that points $P, A, B, C$ are concyclic on circle $\Gamma_{1}$, and points $P, A, E, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,834 |
Example 2 As shown in Figure 2, $E$ and $F$ are points on the sides $AD$ and $BC$ of quadrilateral $ABCD$, respectively, satisfying $\frac{AE}{ED}=\frac{BF}{FC}$. Ray $FE$ intersects $BA$ and $CD$ at points $S$ and $T$, respectively. Prove that the circumcircles of $\triangle SAE$, $\triangle SBF$, $\triangle TCF$, and... | Proof: Let $X$ be the second intersection point of the circumcircles of $\triangle SAE$ and $\triangle SBF$.
By Lemma 1, we know that $\triangle XBF$ is rotationally similar to $\triangle XAE$.
Then, by $\frac{AE}{ED}=\frac{BF}{FC}$, we know that $\triangle XFC \backsim \triangle XED$.
By Lemmas 1 and 2, we know that $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,835 |
Example 3 As shown in Figure 3, $P$, $Q$, $R$ are points on the sides $BC$, $CA$, $AB$ of $\triangle ABC$, respectively. Denote the circumcircles of $\triangle AQR$, $\triangle BRP$, $\triangle CPQ$ as $\Gamma_{A}$, $\Gamma_{B}$, $\Gamma_{C}$, respectively. It is known that segment $AP$ intersects circles $\Gamma_{A}$,... | Proof: Let the second intersection point of circles $\Gamma_{B}$ and $\Gamma_{C}$ be $I$. Then $\angle A R I = \angle I P B = \angle I Q C$.
Therefore, point $I$ lies on circle $\Gamma_{A}$ and
$$
\angle I X Z = \angle I Q A = \angle I P C.
$$
For circles $\Gamma_{B}$ and $\Gamma_{C}$, by Lemma 1, we know that $\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,836 |
Example 4 Given that the incircle of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively, and the circumcircle of $\triangle ABC$, $\odot O$, intersects the circumcircles of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$, denoted as $\Gamma_{A}$, $\Gamma_{B}$, and $\Gamma_... | For circle $\Gamma_{A}$ and $\odot O$, by Lemma 1, we know that $\triangle P F E$ is rotationally similar to $\triangle P B C$.
Combining the tangent length theorem, we have
$$
\frac{P B}{P C}=\frac{B F}{C E}=\frac{B D}{C D} \text {. }
$$
Therefore, $P D$ is the angle bisector of $\angle B P C$.
Thus, $P D$ passes thr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,837 |
Example 5 Given a convex quadrilateral $ABCD$, $BC = AD$, and $BC$ is not parallel to $AD$. Let points $E$ and $F$ be on the sides $BC$ and $AD$ respectively, such that $BE = DF$. The line $AC$ intersects $BD$ at point $P$, the line $BD$ intersects $EF$ at point $Q$, and the line $EF$ intersects $AC$ at point $R$. Prov... | Proof As shown in Figure 6.
Here, $A D = B C, D F = B E$. We can handle it similarly to Example 2, by finding the rotation center $O$ that transforms $A D$ into $C B$.
By Lemma 1, we know that $O$ is the second intersection point of the circumcircles of $\triangle P A D$ and $\triangle P B C$, which is a fixed point. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,838 |
3. Given the non-isosceles acute triangle $\triangle ABC$ with circumcenter $O$, the line $OA$ intersects the altitudes from $B$ and $C$ at points $P$ and $Q$ respectively, and these altitudes intersect at point $H$. Prove: The circumcenter of $\triangle PQH$ lies on a median of $\triangle ABC$.
---
The translation i... | 3. Suppose $A B<A C$, as shown in Figure 2.
Then $\angle P Q H=90^{\circ}-\angle Q A B$
$=90^{\circ}-\angle O A B=\angle A C B$.
Similarly, $\angle Q P H=\angle A B C$.
Thus, $\triangle A B C \backsim \triangle H P Q$.
Let the circumcircle of $\triangle H P Q$ be $\odot T$.
Since $\angle A H P=90^{\circ}-\angle H A C$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,839 |
5. Given a convex hexagon $A B C C_{1} B_{1} A_{1}$ satisfying $A B = B C$, and $A A_{1}, B B_{1}, C C_{1}$ have a common perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ intersect at point $D$, and the circumcircle $\Gamma$ of $\triangle A B C$ intersects the circumcircle of $\triangle A_{1} B C_{1}$ ... | 5. If $A A_{1}=C C_{1}$, then the convex hexagon $A B C C_{1} B_{1} A_{1}$ is symmetric about the line $B B_{1}$. Thus, the circumcircle $\Gamma$ of $\triangle A B C$ and the circumcircle of $\triangle A_{1} B C_{1}$ are tangent at point $B$, which is a contradiction. Therefore, $A A_{1} \neq C C_{1}$. Without loss of ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,840 |
$$
\begin{array}{l}
A=\{1,2, \cdots, 99\}, \\
B=\{2 x \mid x \in A\}, \\
C=\{x \mid 2 x \in A\} .
\end{array}
$$
Then the number of elements in $B \cap C$ is $\qquad$ | 1. 24 .
From the conditions, we have
$$
\begin{array}{l}
B \cap C \\
=\{2,4, \cdots, 198\} \cap\left\{\frac{1}{2}, 1, \frac{3}{2}, 2, \cdots, \frac{99}{2}\right\} \\
=\{2,4, \cdots, 48\} .
\end{array}
$$
Therefore, the number of elements in $B \cap C$ is 24 . | 24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,843 |
Example 4 Given that $S$ is a subset of the set $\{1,2, \cdots, n\}$, $S$ contains no number that divides another, and no two numbers are coprime. Find the maximum value of $S$. | 【Analysis】First, guess the maximum value $k$ by construction. To construct a set without divisibility relations, we can choose $\{m, m+1, \cdots, n\}\left(m>\frac{n}{2}\right)$. To construct a set of numbers that are pairwise not coprime, we can choose $\left\{2,4, \cdots, 2\left[\frac{n}{2}\right]\right\}([x]$ represe... | \left[\frac{n+2}{4}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,844 |
2. Let the distance from point $P$ to plane $\alpha$ be $\sqrt{3}$, and point $Q$ lies on plane $\alpha$ such that the angle between line $P Q$ and plane $\alpha$ is no less than $30^{\circ}$ and no more than $60^{\circ}$. Then the area of the region formed by such points $Q$ is $\qquad$. | 2. $8 \pi$.
Let point $P$ have its projection on plane $\alpha$ as $O$. From the given condition, we have
$$
\frac{O P}{O Q}=\tan \angle O Q P \in\left[\frac{\sqrt{3}}{3}, \sqrt{3}\right] \text {, }
$$
which means $O Q \in[1,3]$.
Therefore, the area of the required region is $\pi \cdot 3^{2}-\pi \cdot 1^{2}=8 \pi$. | 8 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,845 |
3. Arrange $1, 2, 3, 4, 5, 6$ randomly in a row, denoted as $a, b, c, d, e, f$. Then the probability that $a b c + d e f$ is an even number is $\qquad$ | 3. $\frac{9}{10}$.
Consider the case where $a b c+\operatorname{def}$ is odd. In this case, $a b c$ and $d e f$ are one odd and one even. If $a b c$ is odd, then $a, b, c$ are permutations of 1, 3, 5, and thus, $d, e, f$ are permutations of 2, 4, 6. This results in $3! \times 3! = 36$ cases.
By symmetry, the number o... | \frac{9}{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,846 |
4. In the Cartesian coordinate system $x O y$, the ellipse $C$: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Chords $S T$ and $U V$ are parallel to the $x$-axis and $y$-axis, respectively, and intersect at point $P$. It is known that the lengths of segmen... | 4. $\sqrt{15}$.
By symmetry, without loss of generality, let $P\left(x_{P}, y_{P}\right)$ be in the first quadrant. Then, by the given conditions,
$$
\begin{array}{l}
x_{P}=\frac{1}{2}(|P T|-|P S|)=2, \\
y_{P}=\frac{1}{2}(|P V|-|P U|)=1,
\end{array}
$$
which means point $P(2,1)$.
Also, from $x_{P}=|P U|=1,|P S|=2$, w... | \sqrt{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,847 |
6. Let the complex number $z$ satisfy $|z|=1$, such that the equation $z x^{2}+2 z x+2=0$ has real roots. Then the sum of such complex numbers $z$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6. $-\frac{3}{2}$.
Let $z=a+b \mathrm{i}\left(a, b \in \mathbf{R}, a^{2}+b^{2}=1\right)$.
Change the original equation to
$$
\left(a+b \text { i) } x^{2}+2(a-b \text { i) } x+2=0 \text {. }\right.
$$
Separating the real and imaginary parts, it is equivalent to
$$
\left\{\begin{array}{l}
a x^{2}+2 a x+2=0, \\
b x^{2}-2... | -\frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,848 |
8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy:
$$
a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5} \text {, }
$$
and $a_{i+1} \in\left\{1+a_{i}, 2+a_{i}\right\}(i=1,2, \cdots, 9)$. Then the number of such sequences is $\qquad$ | 8. 80 .
$$
\begin{array}{l}
\text { Let } b_{i}=a_{i+1}-a_{i} \in\{1,2\}(i=1,2, \cdots, 9) \text {. } \\
\text { Then } 2 a_{1}=a_{10}-a_{1}=b_{1}+b_{2}+\cdots+b_{9}, \\
b_{2}+b_{3}+b_{4}=a_{5}-a_{2}=a_{8}-a_{5} \\
=b_{5}+b_{6}+b_{7} .
\end{array}
$$
Let $t$ represent the number of terms with value 2 among $b_{2}, b_{... | 80 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,849 |
Example 5 Color some cells black in a $2013 \times 2013$ grid such that any $19 \times 19$ subgrid contains at least 21 black cells. Find the minimum number of black cells. | 【Analysis】Notice that, $2013=19 \times 105+18$.
If we color the cells black based on a certain condition of coordinates modulo 19, then the number of black cells in each $19 \times 19$ square is the same. By classifying the cells in the $2013 \times 2013$ grid according to the coordinates modulo 19, the category contai... | 105 \times 106 \times 21 - 105 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,850 |
One, (40 points) Let $n$ be a positive integer, $a_{1}, a_{2}, \cdots, a_{n}$, $b_{1}, b_{2}, \cdots, b_{n}$, and $A, B$ be positive real numbers, satisfying:
$$
\begin{array}{c}
a_{i} \leqslant b_{i}, a_{i} \leqslant A \quad (i=1,2, \cdots, n), \\
\text { and } \frac{b_{1} b_{2} \cdots b_{n}}{a_{1} a_{2} \cdots a_{n}}... | Given the conditions $k_{i}=\frac{b_{i}}{a_{i}} \geqslant 1(i=1,2, \cdots, n)$.
Let $\frac{B}{A}=K$.
Then $\frac{b_{1} b_{2} \cdots b_{n}}{a_{1} a_{2} \cdots a_{n}} \leqslant \frac{B}{A}$ can be rewritten as $\prod_{i=1}^{n} k_{i} \leqslant K$.
We need to prove: $\prod_{i=1}^{n} \frac{k_{i} a_{i}+1}{a_{i}+1} \leqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,851 |
II. (40 points) As shown in Figure 1, $\triangle ABC$ is an acute triangle with $AB < AC$, and $M$ is the midpoint of side $BC$. Points $D$ and $E$ are the midpoints of the arcs $\overparen{BAC}$ and $\overparen{BC}$ of the circumcircle of $\triangle ABC$, respectively. Point $F$ is the point where the incircle of $\tr... | Given the conditions, we know that $DE$ is the diameter of the circumcircle of $\triangle ABC$, $DE \perp BC$ at point $M$, and $AE \perp AD$.
As shown in Figure 2, let $I$ be the incenter of $\triangle ABC$.
Then point $I$ lies on $AE$, and $IF \perp AB$.
Since $NB \perp AB$
$$
\begin{array}{c}
\Rightarrow \angle NBE... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,852 |
2. As shown in Figure 1, points $D$ and $E$ are on the sides $AB$ and $AC$ of $\triangle ABC$, respectively, and $DE \parallel BC$. $O_{1}$ and $O_{2}$ are the circumcenters of $\triangle ABE$ and $\triangle ACD$, respectively. The line $O_{1}O_{2}$ intersects $AB$ and $AC$ at points $P$ and $Q$, respectively. $O$ is t... | 2. Let the second intersection point of $\odot O_{1}$ and $\odot O_{2}$ be $T$, and connect $A T, B T, C T, D T, E T$.
Since $A, C, T, D$ are concyclic,
$\Rightarrow \angle B D T = \angle E C T$.
Similarly, $\angle C E T = \angle D B T$.
Thus, $\triangle B D T \backsim \triangle E C T \Rightarrow \frac{B T}{E T} = \fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,854 |
3. Let the sequence of real numbers $\left\{x_{n}\right\}(n \geqslant 1)$ satisfy $x_{1}=1$. Prove: For any integer $n \geqslant 2$, we have
$$
\sum_{i \mid n} \sum_{j \mid n} \frac{x_{i} x_{j}}{[i, j]} \geqslant \prod_{\text {prime } p \mid n}\left(1-\frac{1}{p}\right),
$$
where $\sum_{i \mid n}$ denotes the sum over... | $$
\begin{array}{l}
\text { Note that, for any positive integer } n, \text { we have } \\
n=\sum_{d \mid n} \varphi(d) \text {, } \\
\text { where } \varphi(d) \text { denotes the Euler's totient function of } d. \\
\text { Hence, } \sum_{i \backslash n} \sum_{j \backslash n} \frac{x_{i} x_{j}}{[i, j]}=\sum_{d \backsla... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,855 |
4. $n$ students with distinct names exchange greeting cards. Each person prepares $n-1$ envelopes, writing the names and addresses of the other $n-1$ students, and prepares at least one card signed with their own name. Each day, one student will send a card according to the following requirements: first, they choose a ... | 4. Let these $n$ students be denoted as $P_{1}, P_{2}, \cdots, P_{n}$, with their initial number of greeting cards being $p_{1}, p_{2}, \cdots, p_{n}$, and their final number of greeting cards being $q_{1}, q_{2}, \cdots, q_{n}$.
First, we prove: If at the final moment $P_{i}$ has never sent a greeting card to $P_{j}$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,856 |
6. Given a positive integer $k$. If a sequence of subsets of the set of integers $\mathbf{Z}$, $I_{1} \supseteq I_{2} \supseteq \cdots \supseteq I_{k}$, satisfies that for $i=1,2, \cdots, k, 168$ $\in I_{i}$, and for any $x, y \in I_{i}$, we have $x-y \in I_{i}$ $(x, y$ can be equal $)$, then $\left(I_{1}, I_{2}, \cdot... | 6. Let $I$ be a non-empty subset of $\mathbf{Z}$ and for any $x, y \in I$, we have $x - y \in I$. Define $I$ as a "good set".
If $I = \{0\}$ or $I = \{a \mid a \in \mathbf{Z}$ and $b \mid a\}$, where $b$ is a positive integer, then $I$ is a good set.
Furthermore, if $I \neq \{0\}$ is a good set, let $b$ be the smalle... | \frac{(k+1)^{3}(k+2)(k+3)}{6} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,857 |
Example 6 Let $n, k, m$ be positive integers, satisfying $k \geqslant 2$, and $n \leqslant m<\frac{2 k-1}{k} n$. Let $A$ be an $n$-element subset of $\{1,2, \cdots, m\}$. Prove: each integer in the interval $\left(0, \frac{n}{k-1}\right)$ can be expressed as $a-a^{\prime}\left(a, a^{\prime} \in A\right)$.
$(2018$, Nati... | 【Analysis】Let $t \in\left(0, \frac{n}{k-1}\right)$,
$$
m=2 t a+b(0 \leqslant b(k-1) t \geqslant a t+\min \{b, t\} \text {. }
$$
If $a \geqslant k-1$, then
$$
\begin{array}{l}
(2 k-1) n>k m=2 k a t+k b \\
=(2 k-1) a t+a t+k b \\
\geqslant(2 k-1) a t+(k-1) t+k b \\
\geqslant(2 k-1)(a t+\min \{b, t\}) .
\end{array}
$$
T... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,858 |
2. In an acute triangle $\triangle ABC$, $AB < AC$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and $D$ be the foot of the perpendicular from vertex $A$. Take a point $K$ on segment $MN$ such that $BK = CK$, and let ray $KD$ intersect the circumcircle $\Gamma$ of $\triangle ABC$ at point $Q$.... | 2. Let $l$ be the perpendicular bisector of segment $B C$.
As shown in Figure 1, note that
$l$ passes through point $K$. Let $A^{\prime}$ be the reflection of point $A$ over $l$.
Clearly, point $A^{\prime}$ lies on circle $\Gamma$, and by symmetry,
$A A^{\prime} / / B C / / M N$.
Since point $D$ is the reflection of
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,859 |
8. The game board is divided into two parts. In each part, there are some squares, and some segments connect the squares, with each segment connecting two squares from different parts. From any square, it is possible to reach any other square by following the segments. At the start, a purple checkers piece is placed in... | 8. Does not exist.
Construct a strategy that allows Liao Sha to win.
Let $L$ and $P$ denote the initial positions of the purple piece $(l)$ and the cyan piece $(p)$, respectively.
According to the problem, a path can be traced from $L$ to $P$. Suppose Liao Sha chooses such a path:
$$
L=L_{1}, P_{1}, L_{2}, P_{2}, \c... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,860 |
2. Given positive numbers $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$. Prove:
$$
\sum_{i=1}^{n} \frac{1+x_{i}^{2}}{1+x_{i} x_{i+1}} \geqslant n\left(x_{n+i}=x_{i}\right) .
$$ | 2. First, prove that for all $i=1,2, \cdots, n$, we have
$$
\left(1+x_{i}^{2}\right)\left(1+x_{i+1}^{2}\right) \geqslant\left(1+x_{i} x_{i+1}\right)^{2} \text {. }
$$
In fact,
$$
\begin{array}{l}
\left(1+x_{i}^{2}\right)\left(1+x_{i+1}^{2}\right)-\left(1+x_{i} x_{i+1}\right)^{2} \\
=\left(x_{i}-x_{i+1}\right)^{2} \geq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,861 |
6. In an $n$-prism, a line segment connecting two vertices not lying on the same face is called a "diagonal" of the prism. If three diagonals of a regular $n$-prism intersect at an internal point $O$, prove: $O$ is the center of the prism. | 6. As shown in Figure 5, let the diagonals $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect at point $O$, with points $A$, $B$, and $C$ located on one base of the prism, and points $A_{1}$, $B_{1}$, and $C_{1}$ located on the opposite base.
Thus, points $A$, $A_{1}$, $B$, and $B_{1}$ lie in the same plane $\alpha$, which... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,862 |
8. At the beginning, there is a knight of each color, red and blue, on the bottom-left and bottom-right squares of a $2018 \times 2018$ chessboard, respectively. Kolya and Sasha take turns moving their pieces, with Kolya controlling the red knight and Sasha controlling the blue knight, and Kolya starting first. Each mo... | 8. With the correct strategy, Sasha has a winning strategy. The solution to this problem is similar to the 8th problem of the 10th grade.
Below is the construction of Sasha's winning strategy.
Color all the squares alternately black and white like a chessboard. Let $K$ and $S$ represent the squares where the red knight... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,863 |
2. Let the complex number $z=5 \cos \theta+2 \mathrm{i} \sin \theta$. If the function
$$
f(\theta)=\theta-\arg z\left(0<\theta \leqslant \frac{\pi}{4}\right)
$$
has a maximum value of $\gamma$, then $\tan \gamma=$ | 2. $\frac{3}{7}$.
Let $t=\tan \theta(t \in(0,1])$. Then $\tan (\theta-\arg z)=\frac{t-\frac{2}{5} t}{1+t \cdot \frac{2}{5} t}=\frac{3}{\frac{5}{t}+2 t} \leqslant \frac{3}{7}$.
When $t=1$, the equality holds. | \frac{3}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,864 |
3. Let non-zero real numbers $a, b$ satisfy $a^{2}+b^{2}=25$. If the function $y=\frac{a x+b}{x^{2}+1}$ has a maximum value $y_{1}$ and a minimum value $y_{2}$, then $y_{1}-y_{2}=$ $\qquad$. | 3.5.
From
$$
\begin{aligned}
y & =\frac{a x+b}{x^{2}+1} \Rightarrow y x^{2}-a x+y-b=0 \\
& \Rightarrow \Delta=a^{2}-4 y(y-b) \geqslant 0 .
\end{aligned}
$$
Thus, $y_{2}$ and $y_{1}$ are the two roots of $a^{2}-4 y(y-b)=0$, at this point,
$$
\Delta_{1}=16 b^{2}+16 a^{2}=400>0 \text {. }
$$
Therefore, $y_{2}$ and $y_{... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,865 |
4. Given that one focus of the ellipse $C$ is $F_{1}(1,1)$, the length of the major axis is 2, and the line $l: y=x+2$ is tangent to the ellipse. Then the equation of the trajectory of the other focus $F_{2}$ of the ellipse $C$ is $\qquad$ . | 4. $(x+1)^{2}+(y-3)^{2}=4(-1<x<1$, $1<y<3)$.
Let the point of tangency between the ellipse $C$ and the line $l$ be $M$, and the symmetric point of $F_{1}$ with respect to the line $l$ be $F_{1}^{\prime}$.
Then $F_{1}^{\prime}, F_{2}, M$ are collinear, and
$$
\begin{array}{l}
\left|F_{1}^{\prime} F_{2}\right|=\left|F_... | (x+1)^{2}+(y-3)^{2}=4(-1<x<1,1<y<3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,866 |
5. If a hollow sphere with radius $R$ contains two solid spheres with radius $r$ and one solid sphere with radius $2r$, then the maximum value of $r$ is $\qquad$ . | 5. $\frac{4 \sqrt{2}-1}{4} R$.
Consider the centers of these four spheres lying on the same plane.
Let the centers of the spheres with radii $R$, $2r$, and two radii $r$ be $O$, $O_{1}$, $O_{2}$, and $O_{3}$, respectively, where
$$
\begin{array}{l}
\mathrm{OO}_{1}=R-2 r, \mathrm{O}_{1} \mathrm{O}_{2}=\mathrm{O}_{1} \m... | \frac{4 \sqrt{2}-1}{4} R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,867 |
2. For an integer $n(n \geqslant 4)$, find the smallest integer $f(n)$ such that for any positive integer $m$, in any $f(n)$-element subset of the set $\{m, m+1, \cdots, m+n-1\}$, there are at least three pairwise coprime elements.
$(2004$, National High School Mathematics Joint Competition) | Prompt: Prove: Among any five numbers chosen from six consecutive numbers, there must be three numbers that are pairwise coprime.
Answer: $n=6k+m(0 \leqslant m<6)$,
$$
f(n)=4k+m+1-\left[\frac{m}{4}\right] .
$$ | f(n)=4k+m+1-\left[\frac{m}{4}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,868 |
6. Let the side length of rhombus $A_{1} A_{2} A_{3} A_{4}$ be $1, \angle A_{1} A_{2} A_{3}=$ $\frac{\pi}{6}, P$ be a point in the plane of rhombus $A_{1} A_{2} A_{3} A_{4}$. Then the minimum value of $\sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}}$ is $\qquad$ | 6. -1 .
Let the center of the rhombus be $O$. Then
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}} \\
= & \mathrm{C}_{4}^{2}|\overrightarrow{P Q}|^{2}+\overrightarrow{P O} \cdot 3 \sum_{1 \leqslant i \leqslant 4} \overrightarrow{O A_{i}}+\sum_{1 \leqslant... | -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,869 |
7. Let $P(x)=x^{5}-x^{2}+1$ have five roots $r_{1}$, $r_{2}, \cdots, r_{5}$, and $Q(x)=x^{2}+1$. Then
$$
\begin{array}{l}
Q\left(r_{1}\right) Q\left(r_{2}\right) Q\left(r_{3}\right) Q\left(r_{4}\right) Q\left(r_{5}\right) \\
=
\end{array}
$$ | 7.5.
Given that $P(x)=\prod_{j=1}^{5}\left(x-r_{j}\right)$.
$$
\begin{array}{l}
\text { Then } \prod_{j=1}^{5} Q\left(r_{j}\right)=\left(\prod_{j=1}^{5}\left(r_{j}+\mathrm{i}\right)\right)\left(\prod_{j=1}^{5}\left(r_{j}-\mathrm{i}\right)\right) \\
=P(\mathrm{i}) P(-\mathrm{i}) \\
=\left(\mathrm{i}^{5}-\mathrm{i}^{2}+... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,870 |
Given $k(k \geqslant 2)$ distinct square-free positive integers $a_{1}, a_{2}, \cdots, a_{k}$, and satisfying
$$
\min \left\{\sqrt{a_{1}}, \sqrt{a_{2}}, \cdots, \sqrt{a_{k}}\right\}>k .
$$
Prove: There exists a positive constant $c$, such that for any positive integer $n$, we have
$$
\left\{n \sqrt{a_{1}}\right\}+\lef... | Let \( t_{n}=\sum_{i=1}^{k}\left[n \sqrt{a_{i}}\right] \). Then,
where, \( A_{k}=\prod_{\substack{\left(\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{k}\right) \\ s_{i}= \pm 1(1 \leqslant i \leqslant k)}}\left|t_{n}+\sum_{i=1}^{k} \varepsilon_{i} n \sqrt{a_{i}}\right| \).
$$
\text { Let } t_{n}=x, \sqrt{a_{1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,871 |
4. Let set $S \subset\{1,2, \cdots, 200\}, S$ such that the difference between any two elements is not 4, 5, or 9. Find the maximum value of $|S|$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note to the translator and should not be included in the translated text. Here is ... | Hint Consider the maximum number of numbers from $S$ that can be contained in any sequence of 13 consecutive numbers. Answer: 64. | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,872 |
Example 2 Calculate
$$
\left(1+\cos \frac{\pi}{7}\right)\left(1+\cos \frac{3 \pi}{7}\right)\left(1+\cos \frac{5 \pi}{7}\right)
$$
the value.
$(2017$, Peking University Boya Talent Program Exam) | 【Analysis】Using the double angle formula, the original expression can be transformed into $8 \sin \frac{\pi}{7} \cdot \sin \frac{2 \pi}{7} \cdot \sin \frac{3 \pi}{7} \cdot \sin \frac{4 \pi}{7} \cdot \sin \frac{5 \pi}{7} \cdot \sin \frac{6 \pi}{7}$.
Let the $n$ complex roots of $x^{n}=1$ be $1, \omega, \omega^{2}, \cdot... | \frac{7}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,873 |
3. Given an integer $n \geqslant 4$. Prove: there does not exist a regular $n$-gon $A_{1} A_{2} \cdots A_{n}$, such that for any $1 \leqslant i<j \leqslant n, A_{i} A_{j}$ has an integer length.
(2014, Tsinghua University Mathematics Autumn Camp Test) | Hint: The problem of cosine law and roots of unity is transformed into a problem of trigonometric functions being irrational.
Proof by contradiction.
Assume $\cos \frac{k \pi}{n} \in \mathbf{Q}(1 \leqslant k \leqslant n-2)$.
Using $\cos n \theta+\mathrm{i} \sin n \theta=(\cos \theta+\mathrm{i} \sin \theta)^{n}$, combin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,874 |
5. Let four points $Z_{1}, Z_{2}, Z_{3}, Z_{4}$ on the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$, which satisfy $z_{1}+z_{2}+z_{3}+z_{4}=0$. Prove: $Z_{1}, Z_{2}, Z_{3}, Z_{4}$ form a rectangle.
(2000, Fudan University Independent Admission Examination) | Prompt: It is only necessary to prove:
$$
Z_{1} Z_{2} // Z_{3} Z_{4}, Z_{1} Z_{4} // Z_{3} Z_{2} \text{. }
$$
Let $Z_{i}\left(a_{i}, b_{i}\right)(i=1,2,3,4)$. Then
$$
\begin{array}{l}
\sum_{i=1}^{4} a_{i}=\sum_{i=1}^{4} b_{i}=0, \\
a_{i}^{2}+b_{i}^{2}=1(i=1,2,3,4) .
\end{array}
$$
From this, we can get $\frac{a_{2}-a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,876 |
Question 1 Define the sequence $\left\{a_{n}\right\}$:
(1) $a_{1}=p, a_{2}=q(p, q$ are primes, and $p \neq q)$;
(2) For any $n \in \mathbf{Z}_{+}$, if there exists $m \in \mathbf{Z}_{+}$ such that $a_{n}+a_{n+1}=2^{m}$, then $a_{n+2}=2$, otherwise, $a_{n+2}$ is the smallest odd prime factor of $a_{n}+a_{n+1}$.
Prove: T... | Prove: Assume there does not exist \( k \in \mathbf{Z}_{+} \) such that \( a_{k}=2 \).
Then, by definition, for any \( n \in \mathbf{Z}_{+} \), \( a_{n} \) is an odd prime, and \( a_{n+2} \) is the smallest odd prime factor of \( a_{n+1} + a_{n} \).
First, prove: \( a_{n} \neq a_{n+1} \left( n \in \mathbf{Z}_{+} \righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,877 |
Question 2 Define the sequence $\left\{a_{n}\right\}$ :
(1) $a_{1}=p, a_{2}=q(p, q$ are prime numbers, and $p<q)$ ;
(2) For any $n \in \mathbf{Z}_{+}$, if there exists $m \in$ $\mathbf{Z}_{+}$, such that $a_{n}+a_{n+1}=2^{m}$, then $a_{n+2}=2$, otherwise, $a_{n+2}$ is the smallest odd prime factor of $a_{n}+a_{n+1}$. | Proof: There exists a positive integer $M$, such that when $n>M$, $\{a_{n}\}$ is a periodic sequence, and the cycle is $2,7,3,5$.
Proof First, we provide two lemmas.
Lemma 1 \quad $a_{n} \neq a_{n+1} (n \in \mathbf{Z}_{+})$.
Lemma 2 If $\{a_{n}\}$ contains only odd primes, then the sequence $\{\max \{a_{n}, a_{n+1}\}\}... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,878 |
Question 3 Define the sequence $\left\{a_{n}\right\}$:
(1) $a_{1}=p, a_{2}=q(p, q$ are prime numbers, and $p<q)$;
(2) For any $n \in \mathbf{Z}_{+}$, if there exists $m \in$ $\mathbf{Z}_{+}$, such that $a_{n}+a_{n+1}=2^{m}$, then $a_{n+2}=1$, otherwise, $a_{n+2}$ is the smallest odd prime factor of $a_{n}+a_{n+1}$. | Prove: There exists a positive integer $M$, such that when $n>M$, $\left\{a_{n}\right\}$ is a periodic sequence.
After such modification, the terms of the sequence from the third term onward must be odd.
Thus, when $n \geqslant 3$, $a_{n}+a_{n+1}$ is even,
$$
a_{n+2} \leqslant \frac{a_{n}+a_{n+1}}{2} \leqslant \max \l... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,879 |
"Finding all positive integer solutions to the equation $x^{y}=y^{x}+1$ is a problem I studied over twenty years ago. At that time, I provided a solution, but there was actually an issue with one step in the process. It was pointed out by readers after I included this problem in the 2015 Special Issue of *High School M... | ```
Solution 1 (1) If }x\text{ is even; }y\text{ is odd}
Let }x=2m(m\in\mathbb{Z}^+)\text{. Then
}ym=(2m)^y-
# y^2m-1=(2m)^y-2
# (ym+1)(y^m-1)=(2m)^y-
and }ym+1, y^m-1\text{ are two consecutive even numbers, so
4|((2m)^y -2).
Therefore, }y=1, \text{ substituting into the original equatio... | (2,1),(3,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,880 |
Second question As shown in Figure $1, \triangle A B C$ is an acute triangle, $A B$ $<A C, M$ is the midpoint of side $B C$, $D$ and $E$ are the midpoints of the arcs $\overparen{B A C}$ and $\overparen{B C}$ of the circumcircle of $\triangle A B C$, $F$ is the point where the incircle of $\triangle A B C$ touches side... | Proof 1 Let $I$ be the incenter of $\triangle ABC$, and the circle passing through points $A$, $G$, $M$, and $D$ intersects side $AC$ at a second point $S$. Connect auxiliary lines as shown in Figure 2.
By the properties of the incenter of a triangle, we have
$\angle BIE = \angle ABI + \angle BAI$
$= \angle CBI + \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,882 |
Example 3 Calculate $\tan \frac{\pi}{7} \cdot \tan \frac{2 \pi}{7} \cdot \tan \frac{3 \pi}{7}$. (2016, Harvard-MIT Mathematics Tournament) | 【Analysis】Consider the polynomial $P(z)=z^{7}-1$.
Let $z=\mathrm{e}^{\mathrm{i} x}=\cos x+\mathrm{i} \sin x$.
Then $z^{7}-1$
$$
\begin{aligned}
= & \left(\cos ^{7} x-\mathrm{C}_{7}^{2} \cos ^{5} x \cdot \sin ^{2} x+\right. \\
& \left.\mathrm{C}_{7}^{4} \cos ^{3} x \cdot \sin ^{4} x-\mathrm{C}_{7}^{6} \cos x \cdot \sin ... | \sqrt{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,883 |
For $n \geqslant 2, \lambda \in(-1,0) \cup(0,1)$, define
$$
\begin{array}{l}
a_{n, k}=\mathrm{C}_{n}^{k} \lambda^{k(n-k)}, \\
P_{n}(z)=a_{n, 0} z^{n}+\cdots+a_{n, n-1} z+a_{n, n} .
\end{array}
$$
Prove: The zeros of $P_{n}(z)$ are all on the unit circle. | Proof of Proposition 1. Mathematical Induction.
When $n=2$,
$$
a_{2,0}=a_{2,2}=1, a_{2,1}=2 \lambda \text {, }
$$
the discriminant of $P_{2}(z)$ is $\Delta=(2 \lambda)^{2}-4<0$.
Thus, $P_{2}(z)$ has a pair of conjugate complex roots, whose product is 1.
Therefore, both roots lie on the unit circle.
Assume the propositi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,884 |
For positive integer $n$ and real number $\lambda \in[0,1)$, the polynomial
$$
f_{n, \lambda}(x)=\left\{\begin{array}{ll}
\sum_{k=0}^{n} \mathrm{C}_{n}^{k} \lambda^{k(n-k)} x^{k}, & \lambda>0 ; \\
1+x^{n}, & \lambda=0
\end{array}\right.
$$
each root has a modulus of 1. | The proof of Proposition 2 is by induction on $n$.
When $n=1$, $f_{n, \lambda}(x)=1+x$, and the conclusion holds.
Assume the conclusion holds for $n-1$. Next, consider the case for $n$.
Proof by contradiction.
Let $S \subseteq [0,1)$ be the set of all $\lambda \in [0,1)$ for which $f_{n, \lambda}(x)$ has a root with mo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,885 |
2. Given that $p$ is a prime number. Two people, A and B, play a game, alternately performing the following operations: each person selects an index $i$ from the set $\{0,1, \cdots, p-1\}$ (where $i$ has not been chosen by either of them before), and then selects an element $a_{i}$ from the set $\{0,1, \cdots, 9\}$. A ... | 2. If a person chooses the index $i$, and then selects $a_{i}$ from $\{0,1, \cdots, 9\}$, this operation is denoted as $\left(i, a_{i}\right)$.
If $p=2$ or $p=5$, then player A chooses $i=0$ and $a_{0}=0$ in the first operation. Since no matter how the subsequent selections are made, $M$ will always be divisible by 10... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,886 |
8. Given that $p$ is an odd prime, the function $f: \mathbf{Z}_{+} \times \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ satisfies the following conditions:
(i) $f(1,1)=0$;
(ii) For any pair of positive integers $(a, b)$, both not equal to 1 and coprime,
$$
f(a, b)+f(b, a)=1 \text {; }
$$
(iii) For any pair of coprime pos... | 8. Let the set of all pairs of coprime positive integers be denoted as $A$. For any $(a, b) \in A$, there exists $(u, v) \in \mathbf{Z}^{2}$ such that
$$
u a + v b = 1.
$$
If $\left(u_{0}, v_{0}\right)$ is an integer pair satisfying equation (1), then all integer pairs satisfying equation (1) are of the form
$$
(u, v)... | \sqrt{2p} - 2 | Number Theory | proof | Yes | Yes | cn_contest | false | 730,887 |
2. Given a large regular tetrahedron with an edge length of 6, a smaller regular tetrahedron is placed inside it. If the smaller tetrahedron can rotate freely within the larger one, the maximum edge length of the smaller tetrahedron is . $\qquad$ | 2. 2 .
Given that a smaller regular tetrahedron can rotate freely inside a larger regular tetrahedron, the maximum edge length of the smaller tetrahedron occurs when it is inscribed in the insphere of the larger tetrahedron.
Let the circumradius of the larger tetrahedron be $R$, and the circumradius of the smaller te... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,888 |
Example 4 Let $n \in N$. Prove: The equation
$$
z^{n+1}-z^{n}-1=0
$$
has a complex root of modulus 1 if and only if $6 \mid (n+2)$. (2nd China Mathematical Olympiad) | 【Analysis】Suppose the equation (1) has a root $z$, and $|z|=1$.
Then $z^{n}(z-1)=1$
$$
\begin{array}{l}
\Rightarrow\left|z^{n}(z-1)\right|=1 \\
\quad \Rightarrow|z-1|=1 .
\end{array}
$$
Therefore, $z$ is the intersection point of the circle $|z|=1$ and the circle $|z-1|=1$ (or, $z, 0, 1$ form an equilateral triangle).... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,889 |
3. In the expansion of $(\sqrt{3}+i)^{10}$, the sum of all odd terms is $\qquad$ . | 3. 512 .
It is known that in the expansion of $(\sqrt{3}+\mathrm{i})^{10}$, the sum of all odd terms is the real part of the complex number.
$$
\begin{array}{l}
\text { Therefore, }(\sqrt{3}+i)^{10}=\left((-2 i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^{10} \\
=(-2 i)^{10}\left(-\frac{1}{2}+\frac{\sqrt{3}}... | 512 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,890 |
4. Given point $P$ inside $\triangle A B C$, and satisfies
$$
\overrightarrow{A P}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} \text {, }
$$
and let the areas of $\triangle P B C$, $\triangle P C A$, and $\triangle P A B$ be $S_{1}$, $S_{2}$, and $S_{3}$, respectively. Then $S_{1}: S_{2}: S_{3}=$... | 4. $5: 4: 3$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{A P}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} \\
=\frac{1}{3}(\overrightarrow{P B}-\overrightarrow{P A})+\frac{1}{4}(\overrightarrow{P C}-\overrightarrow{P A})
\end{array}
$$
Then $5 \overrightarrow{P A}+4 \overrightarrow{P B}+3 \... | 5: 4: 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,891 |
6. If $(2 \dot{x}+4)^{2 n}=\sum_{i=0}^{2 n} a_{i} x^{i}\left(n \in \mathbf{Z}_{+}\right)$, then the remainder of $\sum_{i=1}^{n} a_{2 i}$ when divided by 3 is $\qquad$ | 6.1.
Let $x=0$, we get $a_{0}=4^{2 n}$.
By setting $x=1$ and $x=-1$ respectively, and adding the two resulting equations, we get
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right) . \\
\text { Therefore, } \sum_{i=1}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n} \\
\equiv(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,892 |
8. Divide a circle into a group of $n$ equal parts and color each point either red or blue. Starting from any point, record the colors of $k(k \leqslant n)$ consecutive points in a counterclockwise direction, which is called a “$k$-order color sequence” of the circle. Two $k$-order color sequences are considered differ... | 8. 8 .
In a 3rd-order color sequence, since each point has two color choices, there are $2 \times 2 \times 2=8$ kinds of 3rd-order color sequences.
Given that $n$ points can form $n$ 3rd-order color sequences, we know $n \leqslant 8$.
Thus, $n=8$ can be achieved.
For example, determining the colors of eight points in ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,893 |
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