problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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5. Let $\angle A, \angle B, \angle C$ be the three interior angles of $\triangle ABC$. If $\sin A=a, \cos B=b$, where $a>0, b>0$, and $a^{2}+b^{2} \leqslant 1$, then $\tan C=$ $\qquad$ | 5. $\frac{a b+\sqrt{1-a^{2}} \sqrt{1-b^{2}}}{a \sqrt{1-b^{2}}-b \sqrt{1-a^{2}}}$.
Since $\cos B=b>0$, we have
$\angle B$ is an acute angle,
$\sin B=\sqrt{1-\cos ^{2} B}=\sqrt{1-b^{2}}$.
Also, $a^{2}+b^{2} \leqslant 1$, then
$\sin A=a \leqslant \sqrt{1-b^{2}}=\sin B$.
Thus, $\sin (\pi-A) \leqslant \sin B$.
If $\angle A... | \frac{a b+\sqrt{1-a^{2}} \sqrt{1-b^{2}}}{a \sqrt{1-b^{2}}-b \sqrt{1-a^{2}}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,645 |
6. Two real numbers $a$ and $b$ are randomly chosen from $[-1,1]$. Then the probability that the equation $x^{2}+a x+b=0$ has real roots is $\qquad$ | 6. $\frac{13}{24}$.
Notice,
the equation $x^{2}+a x+b=0$ has real roots
$$
\Leftrightarrow a^{2}-4 b \geqslant 0 \Leftrightarrow b \leqslant \frac{1}{4} a^{2} \text {. }
$$
Thus, in the Cartesian coordinate system $a O b$, the set of real number pairs $(a, b)$ that satisfy the condition forms the shaded region in Fig... | \frac{13}{24} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,646 |
7. Given that $O$ is the circumcenter of $\triangle A B C$. If $A B=A C$, $\angle C A B=30^{\circ}$, and $\overrightarrow{C O}=\lambda_{1} \overrightarrow{C A}+\lambda_{2} \overrightarrow{C B}$, then $\lambda_{1} \lambda_{2}=$ $\qquad$ . | 7. $7 \sqrt{3}-12$.
Let's assume $AB=2$. Establish a Cartesian coordinate system with $A$ as the origin and the line $AB$ as the $x$-axis. Then
$$
A(0,0), B(2,0), C(\sqrt{3}, 1) \text {. }
$$
Let the circumcenter be $O(1, y)$.
From $|\overrightarrow{O A}|=|\overrightarrow{O C}|$, we get
$$
1+y^{2}=(\sqrt{3}-1)^{2}+(y... | 7 \sqrt{3}-12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,647 |
8. Let $m, n$ be integers, $53<m, n<100$, and let $x=(m+\sqrt{m}+1)^{n}, y=(m-\sqrt{m}+1)^{n}$.
If the integer part of $y$ is 2013, and the remainder when 2013 is divided by $m$ is 53, then the remainder when the integer part of $x$ is divided by $m$ is $\qquad$ | 8. $m-52$.
Let $x=(m+1+\sqrt{m})^{n}$
$$
=A+B \sqrt{m}\left(A, B \in \mathbf{N}_{+}\right) \text {. }
$$
Then $y=(m+1-\sqrt{m})^{n}=A-B \sqrt{m}$.
If $m$ is a perfect square, let $m=a^{2}$, we get
$$
\begin{array}{l}
y=(m+1-\sqrt{m})^{n}=\left(a^{2}+1-a\right)^{n} \\
=2013=3 \times 11 \times 61 .
\end{array}
$$
Thus... | m-52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,648 |
9. (16 points) Let $f(x)=x+\frac{1}{x}(x>0)$. If for any positive number $a$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in the interval $\left[1, a+\frac{2013}{a}\right]$, such that the inequality
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right)<f\left(a_{m+1}\right)
$$
holds, ... | II. 9. Let $a=\sqrt{2013}$, then there exist $m+1$ real numbers that meet the requirements in the interval $[1,2 \sqrt{2013}]$.
Notice that $[1,2 \sqrt{2013}] \subseteq\left[1, a+\frac{2013}{a}\right]$.
Therefore, we only need to consider the existence of real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in $[1,2 \sqrt{2013... | 44 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,649 |
10. (20 points) In an annual super football round-robin tournament, 2013 teams each play one match against every other team. Each match awards 3 points to the winner, 0 points to the loser, and 1 point to each team in the event of a draw. After the tournament, Jia told Yi the total points of his team, and Yi immediatel... | 10. Consider the case with $n$ teams.
Let the team where Jia is located win $x$ games, draw $y$ games, and lose $z$ games. Then the total score of the team is
$$
S=3 x+y \quad (x+y+z=n-1, x, y, z \geqslant 0).
$$
Consider the region $\Omega:\left\{\begin{array}{l}x+y \leqslant n-1, \\ x \geqslant 0, \\ y \geqslant 0 ... | 6034 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,650 |
11. (20 points) Given that $P$ is a moving point on the parabola $y=x^{2}-1$, and let point $Q(-1,0)$. If the circle with diameter $P Q$ intersects the parabola at only two points, find the range of the area of the circle.
untranslated text:
已知 $P$ 是抛物线 $y=x^{2}-1$上一动点, 设点 $Q(-1,0)$. 若以 $P Q$ 为直径的圆与抛物线只有两个公共点,求该圆面积的取... | 11. Consider the opposite.
Let the circle with diameter $P Q$ intersect the parabola at a point $R\left(t, t^{2}-1\right)$ other than points $P$ and $Q$.
If $t=1$, then $k_{Q R}=0$. Thus, $P R$ is perpendicular to the $x$-axis, which is a contradiction. Therefore, $t \neq 1$.
Let $P(x, y)$.
Since $k_{Q R}=\frac{t^{2}... | (0, \pi) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,651 |
One, (40 points) Find all positive rational triples $(x, y, z)(x \leqslant y \leqslant z)$ such that
$$
x+y+z 、 \frac{1}{x}+\frac{1}{y}+\frac{1}{z} 、 x y z
$$
are all integers. | Consider the polynomial with roots $x$, $y$, $z$:
$$
\begin{array}{l}
f(t)=(t-x)(t-y)(t-z) \\
=t^{3}-(x+y+z) t^{2}+(x y+y z+z x) t-x y z .
\end{array}
$$
Notice that, $x y+y z+z x=x y z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is an integer. Therefore, $f(t)$ is a monic polynomial with integer coefficients. Si... | (1,1,1),(1,2,2),(2,3,6),(2,4,4),(3,3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,653 |
II. (40 points) Draw two tangents $OA$ and $OB$ from point $O$ to circle $C$, with points of tangency at $A$ and $B$ respectively. Draw a line $l$ through $O$ that intersects circle $C$ at points $D$ and $E$, and intersects line $AB$ at point $F$. Prove:
$$
\frac{O F}{O D}+\frac{O F}{O E}=2 .
$$ | $$
\begin{array}{l}
\text { Given } \angle O A D=\angle O E A \Rightarrow \triangle O A D \sim \triangle O E A \\
\Rightarrow \frac{O D}{O A}=\frac{O A}{O E}=\frac{A D}{A E} . \\
\text { Then } \frac{O D}{O E}=\frac{O D}{O A} \cdot \frac{O A}{O E}=\frac{O A}{O E} \cdot \frac{O B}{O E} \\
=\frac{A D}{A E} \cdot \frac{B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,654 |
Three, (50 points) Given a positive integer $n$. Find $\sum_{k=1}^{n}\left[\frac{n}{2^{k}}-\frac{1}{2}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
---
Please note that the format and line breaks have been preserved as requested. | $$
\text { Three, let } n=2^{m} a_{m}+2^{m-1} a_{m-1}+\cdots+2^{1} a_{1}+a_{0}
$$
where, $a_{m} \neq 0$.
At this point, $2^{m} \leqslant n<2^{m+1}$, so, $\left[\log _{2} n\right]=m$.
If $k \geqslant m+2$, then
$$
\frac{n}{2^{k}}-\frac{1}{2}<\frac{2^{m+1}}{2^{m+2}}-\frac{1}{2}=0 \text {, }
$$
at this time, $\left[\fra... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,655 |
Four, (50 points) There are $n$ white points on a circle. First, one of them is colored black (called the first coloring). For any positive integer $k$, after the $k$-th coloring, the next point is colored in the opposite color to its original color by moving $k$ points counterclockwise (called the $(k+1)$-th coloring)... | Let $n$ points be numbered counterclockwise as $1,2, \cdots, n$. For a fixed $n$, let the number of the point colored at the $k$-th time be $a_{k}$ $(k=1,2, \cdots)$. The sequence $\left\{a_{k}\right\}(k=1,2, \cdots)$ is called the coloring sequence.
Assume $a_{1}=1$. Then
$$
a_{2}=3, a_{3}=6, a_{4}=10, \cdots \cdots
$... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,656 |
As shown in Figure 2, in the square $A B C D$, the diagonals $A C$ and $B D$ intersect at
point $O$, point $F$ is on side $C D$,
the extension of $A F$ intersects the
extension of $B C$ at
point $E$, the extension of $O F$
intersects $D E$ at point $M$. Find the measure of $\angle O M D$. | From the given conditions, we have
$$
\begin{array}{l}
A B=A D, \angle A D F=\angle A B E=90^{\circ}, \\
A D \parallel B E \Rightarrow \angle D A F=\angle B E A .
\end{array}
$$
Therefore, $\triangle A D F \sim \triangle E B A \Rightarrow \frac{A D}{B E}=\frac{D F}{A B}$
$$
\Rightarrow B E \cdot D F=A D \cdot A B=A D^... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,657 |
Given $a, b, c > 0$, and $a^{2} + b^{2} + c^{2} + abc = 4$. Prove:
$$
\begin{array}{l}
\sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}} + \sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}} + \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} = 1 .
\end{array}
$$ | $$
\begin{array}{l}
16=4 a^{2}+4 b^{2}+4 c^{2}+4 a b c \text {. } \\
\text { Then } \sqrt{\frac{(2-a)(2-b)}{(2+a)(2+b)}}+\sqrt{\frac{(2-b)(2-c)}{(2+b)(2+c)}}+ \\
\sqrt{\frac{(2-c)(2-a)}{(2+c)(2+a)}} \\
=\frac{\sqrt{(4-a)^{2}\left(4-b^{2}\right)}}{(2+a)(2+b)}+ \\
\frac{\sqrt{\left(4-b^{2}\right)\left(4-c^{2}\right)}}{(2... | 1 | Algebra | proof | Yes | Yes | cn_contest | false | 726,658 |
Given $m, n \in \mathbf{N}_{+}, n \geqslant 2$, and $m \leqslant n$. Prove: $\sum_{i=1}^{n}(-1)^{i-1} i^{m} \mathrm{C}_{n}^{i}=0$. | Prove by mathematical induction on $m$.
When $m=1$,
$$
\begin{array}{l}
\sum_{i=1}^{n}(-1)^{i-1} i \mathrm{C}_{n}^{i} \\
=\sum_{i=1}^{n}(-1)^{i-1} i \mathrm{C}_{n-1}^{i}+\sum_{i=1}^{n}(-1)^{i-1} i \mathrm{C}_{n-1}^{i-1} \\
=\sum_{i=1}^{n-1}(-1)^{i-1} i \mathrm{C}_{n-1}^{i}+\sum_{i=0}^{n-1}(-1)^{i}(i+1) \mathrm{C}_{n-1}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,659 |
As shown in Figure 3, in the acute triangle $\triangle ABC$, $\angle A = 60^{\circ}$, $AB > AC$, $O$ and $G$ are the circumcenter and centroid of $\triangle ABC$ respectively, $E$ is a trisection point of $BC$, and $CE = 2BE$. Prove: $\angle OGE = 60^{\circ}$. | Prove as shown in Figure 4, construct an equilateral $\triangle B C A^{\prime}$ on the same side of $\triangle A B C$, let the midpoint of $B C$ be $M$, connect $M A, M A^{\prime}, A A^{\prime}$. Then $M, O, A^{\prime}$ and $M, G, A$ are collinear respectively, and
$$
\frac{M O}{O A^{\prime}}=\frac{1}{2}=\frac{M G}{G A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,660 |
Example 1 Let the integer sequence $x_{11}, x_{21}, \cdots, x_{n 1}(n$ be an odd number greater than 2), and $x_{i 1}(i=1,2, \cdots, n)$ are not all equal. Let
$$
\begin{array}{l}
x_{i(k+1)}=\frac{1}{2}\left(x_{i k}+x_{(i+1) k}\right)(i=1,2, \cdots, n-1), \\
x_{n(k+1)}=\frac{1}{2}\left(x_{n k}+x_{1 k}\right) .
\end{arr... | Prove that placing $x_{11}, x_{21}, \cdots, x_{n 1}$ sequentially on a circle, denoted as circle $\Gamma_{1}$.
Placing the average of adjacent numbers on circle $\Gamma_{k}(k=1,2, \cdots)$ sequentially on another circle, denoted as circle $\Gamma_{k+1}$.
Then, by the problem statement, the numbers on circle $\Gamma_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,661 |
Example 2 Find all pairs of positive integers $(a, n)$ such that $\frac{(a+1)^{n}-a^{n}}{n}$ is an integer. | Solution: Clearly, $(a, 1)\left(a \in \mathbf{N}_{+}\right)$ is a class of solutions to the original problem.
Below, we prove: there are no other solutions.
Assume $(a, n)\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$ is a solution to the original problem.
By the problem statement, we have
$$
(a+1)^{n} \equiv a^{n}(... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,662 |
Example 1 Given $k \in \mathbf{N}, a, b, c \geqslant 0$, $c=\min \{a, b, c\}$.
Prove:
$$
f_{k}(a, b, c)=\sum a^{k}(a-b)(a-c) \geqslant 0 .
$$ | Prove that when $k=0$,
$$
g(t)=\sum(a-b)(a-c)
$$
is independent of $t$.
Thus, $g^{\prime}(t)=0$.
$$
\begin{array}{l}
\text { and } f(a, b, 0)=(a-b) a+b(b-a) \\
=a b+a^{2}+b^{2}-a b \geqslant 0,
\end{array}
$$
Therefore, $f(a, b, c) \geqslant 0$ holds.
Assume that when $k=m$, $f_{m}(a, b, c) \geqslant 0$, then
$$
\beg... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,663 |
Example 2 Let $a+b+c=3(a, b, c \geqslant 0)$. Prove: $a^{2} b+b^{2} c+c^{2} a \leqslant 4$.
| Prove that it is sufficient to prove when $a, b, c \geqslant 0$,
$$
f(a, b, c)=4\left(\sum a\right)^{2}-27 \sum a^{2} b \geqslant 0
$$
always holds.
$$
\begin{array}{l}
\text { Let } g(t)=f(a+t, b+t, c+t), \\
A=a+t, B=b+t, C=c+t .
\end{array}
$$
Differentiating, we get
$$
\begin{array}{l}
g^{\prime}(t)=36\left(\sum A... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,664 |
Example 3 Let $a, b, c \geqslant 0$. Prove:
$$
\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 3\left(a^{3} b+b^{3} c+c^{3} a\right) .
$$ | $$
\begin{array}{l}
f(a, b, c)=\left(\sum a^{2}\right)^{2}-3 \sum a^{3} b, \\
g(t)=f(a+t, b+t, c+t), \\
A=a+t, B=b+t, C=c+t .
\end{array}
$$
Without loss of generality, assume $c=\min \{a, b, c\}$. Then
$$
\begin{array}{l}
g^{\prime}(t)=2\left(\sum A^{2}\right)\left(\sum 2 A\right)-3 \sum\left(A^{3}+3 A^{2} B\right), ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,665 |
Example 4 Given $a, b, c \geqslant 0, a+b+c=3$. Prove:
$$
\prod\left(A^{2}-A B+B^{2}\right) \leqslant 12 \text {. }
$$ | Prove that
$$
f(A, B, C)=\frac{4}{3^{5}}\left(\sum A\right)^{6}-\Pi\left(A^{2}-A B+B^{2}\right) \text {. }
$$
When $C=0$, $f(A, B, 0) \geqslant 0$.
Differentiating, we get
$$
\begin{array}{l}
f^{\prime}(A, B, C) \\
=\frac{8}{3^{3}}\left(\sum A\right)^{3}-\sum(A+B)\left(B^{2}-B C+C^{2}\right)\left(C^{2}-A C+A^{2}\right... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,666 |
Question 1 Given that the circumcenter of $\triangle ABC$ is $O$, and $A', B', C'$ are points on sides $BC, CA, AB$ respectively, and satisfy that the circumcircles of $\triangle AB'C', \triangle BC'A', \triangle CA'B'$ all pass through point $O$. The radical axis of the circle with center $B'$ and radius $B'C$ and the... | Proof: Let the circumcircles of $\triangle A B^{\prime} C^{\prime}$, $\triangle B C^{\prime} A^{\prime}$, and $\triangle C A^{\{\prime} B^{\prime}$ be $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$, respectively. Let $D$ be the intersection point of $\odot O_{1}$ and $\odot O$ other than $A$, as shown in Figure 1.
Fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,667 |
Given $\triangle ABC$, points $D, E, F$ are on sides $BC, CA, AB$ respectively such that the circumcircles of $\triangle AEF, \triangle BFD, \triangle CDE$ (denoted as $\odot O_{1}, \odot O_{2}, \odot O_{3}$ respectively) pass through a point $J$ (an arbitrary point inside $\triangle ABC$). The circumcircle of $\triang... | Proof (1) Auxiliary lines as shown in Figure 3. Let $B M$ and $C N$ intersect at point $P$.
First, we prove: Quadrilateral $\mathrm{OO}_{2}^{\prime} \mathrm{BO}_{2}$ is a parallelogram.
In fact, since $\mathrm{OO}_{2} \perp \mathrm{DF}$, it suffices to prove $B O_{2}^{\prime} \perp \mathrm{DF}\left(\mathrm{BO}_{2}^{\pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,668 |
Question 1 Divide the length and width of a rectangle into three equal parts and five equal parts. | 【Analysis】As shown in Figure 1, first solve the case of trisection.
By folding, find the midpoints $E$ and $F$ of sides $AB$ and $AD$ of rectangle $ABCD$ respectively.
Connect $BF$ and $ED$ to intersect at point $N$. Draw creases $PQ$ and $XY$ through $N$ such that $PQ \parallel AD$ and $XY \parallel AB$.
By Menelaus'... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,669 |
Question 2 Divide the length and width of the rectangle into seven and nine equal parts, respectively. | 【Analysis】In the figure $4, E$ and $F$ are the midpoints of sides $AB$ and $AD$ respectively, points $G$ and $H$ are the quarter points of side $BC$, connecting $DE$, intersecting $FG$ and $FH$ at points $N$ and $M$ respectively.
Then $FN: FG=2: 7, FM: FH=2: 9$. | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,670 |
Question 3 Divide the length and width of the rectangle into 11 equal parts and 13 equal parts. | 【Analysis】In the figure, $E$ and $F$ are the midpoints of sides $AB$ and $AD$ respectively, and points $G$ and $H$ are the trisection points of side $BC$. Connecting $FG$ and $FH$ intersect $ED$ at points $N$ and $M$ respectively.
Then $FN: FG = 3: 11, FM: FH = 3: 13$.
【Conclusion】If the length and width of rectangle $... | FN: FG = 3: 11, \quad FM: FH = 3: 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,671 |
1. Prove: For any pair of positive integers $k, n$, there exist $k$ positive integers (allowing repetition) $m_{1}, m_{2}, \cdots, m_{k}$, such that
$$
1+\frac{2^{k}-1}{n}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{k}}\right) .
$$ | 1. Proof 1: Use mathematical induction on $k$.
When $k=1$, the conclusion is obvious.
Assume that the conclusion holds for $k=j-1$, we will prove the case for $k=j$.
(1) When $n$ is odd, i.e., there exists a positive integer $t$ such that $n=2 t-1$. Notice that,
$$
\begin{array}{l}
1+\frac{2^{j}-1}{2 t-1}=\frac{2\left... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,672 |
Example 3 In the infinite sequence $\left\{x_{n}\right\}$, the first term $x_{1}$ is a rational number greater than 1, and for any positive integer $n$, we have
$$
x_{n+1}=x_{n}+\frac{1}{\left[x_{n}\right]},
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Prove: The sequence contains an ... | Prove that obviously, $\left\{x_{n}\right\}$ is a strictly increasing sequence of rational numbers.
If the conclusion does not hold, placing $\left\{x_{n}\right\}$ on the number line, the integers $1,2, \cdots$ will divide $\left\{x_{n}\right\}$ into several segments. The segment of $\left\{x_{n}\right\}$ between inte... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,673 |
2. A set of 4027 points in the plane is called a "Colombian point set", where no three points are collinear, and 2013 points are red, 2014 points are blue. Draw a set of lines in the plane, which can divide the plane into several regions. If a set of lines for a Colombian point set satisfies the following two condition... | 2. $k=2013$.
Solution 1 First, give an example to show that $k \geqslant 2013$.
Mark 2013 red points and 2013 blue points alternately on a circle, and color another point in the plane blue. This circle is divided into 4026 arcs, each with endpoints of different colors. If the requirement of the problem is to be met, ea... | 2013 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,674 |
3. Let the excircle opposite to vertex $A$ of $\triangle A B C$ touch side $B C$ at point $A_{1}$. Similarly, define points $B_{1}$ and $C_{1}$ on sides $C A$ and $A B$ using the excircles opposite to vertices $B$ and $C$, respectively. Assume that the circumcenter of $\triangle A_{1} B_{1} C_{1}$ lies on the circumcir... | 3. Auxiliary lines as shown in Figure 1.
Let the circumcircles of $\triangle ABC$ and $\triangle A_{1}B_{1}C_{1}$ be circles $\Gamma$ and $\Gamma_{1}$, respectively. Let the midpoint of arc $\overparen{BC}$ (containing point $A$) on circle $\Gamma$ be $A_{0}$, and similarly define $B_{0}$ and $C_{0}$.
By the problem's... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,675 |
4. Let $\triangle ABC$ be an acute triangle, with orthocenter $H$. Let $W$ be a point on side $BC$, not coinciding with vertices $B$ and $C$. Let $M$ and $N$ be the feet of the altitudes from vertices $B$ and $C$, respectively. Denote the circumcircle of $\triangle BWN$ as circle $\omega_{1}$, and let $X$ be a point on... | 4. As shown in Figure 2, let $AL$ be the altitude from $A$ to side $BC$, and let $Z$ be the other intersection point of circles $\omega_{1}$ and $\omega_{2}$, different from point $W$.
Next, we prove that points $X$, $Y$, $Z$, and $H$ are collinear.
Since $\angle BNC = \angle BMC = 90^{\circ}$, points $B$, $C$, $M$, an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,676 |
5. Let $Q_{+}$ be the set of all positive rational numbers. Suppose the function $f: \mathbf{Q}_{+} \rightarrow \mathbf{R}$ satisfies the following three conditions:
(1) For all $x, y \in \mathbf{Q}_{+}$, we have
$$
f(x) f(y) \geqslant f(x y) \text {; }
$$
(2) For all $x, y \in \mathbf{Q}_{+}$, we have
$$
f(x+y) \geqsl... | 5. Substituting $x=1, y=a$ into inequality (1) yields
$$
f(1) \geqslant 1 \text {. }
$$
Starting from inequality (2), using mathematical induction on $n$, we get that for any $n \in \mathbf{Z}_{\text {, and }}^{x} \in \mathbf{Q}_{\text {+ }}$,
$$
f(n x) \geqslant n f(x) \text {. }
$$
In particular,
$$
f(n) \geqslant ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,677 |
6. Let the integer $n \geqslant 3$, and there are $n+1$ equally spaced points on the circumference of a circle. Label these points with the numbers $0,1, \cdots, n$, using each number exactly once. Consider all possible labeling methods. If one labeling method can be obtained from another by rotating the circle, then t... | \
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,678 |
2. Let $\mathbf{Z}$ and $\mathbf{Q}$ be the sets of integers and rational numbers, respectively.
(1) Can $\mathbf{Z}$ be partitioned into three non-empty subsets $A$, $B$, and $C$, such that $A+B$, $B+C$, and $C+A$ are pairwise disjoint?
(2) Can $\mathbf{Q}$ be partitioned into three non-empty subsets $A$, $B$, and $C$... | 2. (1) $\mathbf{Z}$ can be decomposed into three non-empty subsets:
$$
\begin{array}{l}
A=\{3 k \mid k \in \mathbf{Z}\}, \\
B=\{3 k+1 \mid k \in \mathbf{Z}\}, \\
C=\{3 k+2 \mid k \in \mathbf{Z}\} .
\end{array}
$$
(2) It cannot.
Assume $\mathbf{Q}$ can be decomposed into three non-empty subsets $A, B, C$ and satisfy th... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,679 |
4. Given that $f$ and $g$ are two non-zero polynomials with integer coefficients, and $\operatorname{deg} f > \operatorname{deg} g$. If for infinitely many primes $p$, the polynomial $p f + g$ has a rational root, prove: $f$ has a rational root. | 4. Since $\operatorname{deg} f > \operatorname{deg} g$, for sufficiently large $x$, we have $\left|\frac{g(x)}{f(x)}\right| < \frac{1}{2}$. Therefore, for any $p > 2R$, we have $\left|\frac{g(x)}{f(x)}\right| < \frac{1}{2}$ for all $x \in \mathbb{R} \setminus [-R, R]$. This implies that for all $x \in \mathbb{R} \setmi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,680 |
5. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all $x, y \in \mathbf{R}$, we have
$$
f(1+x y)-f(x+y)=f(x) f(y),
$$
and $f(-1) \neq 0$. | 5. The unique solution that satisfies the condition is the function
$$
f(x)=x-1(x \in \mathbf{R}) \text {. }
$$
Let $g(x)=f(x)+1$.
First, prove that for all real numbers $x$, $g(x)=x$.
The original condition is transformed into: for all $x, y \in \mathbf{R}$,
$$
\begin{array}{l}
g(1+x y)-g(x+y) \\
=(g(x)-1)(g(y)-1),
\... | f(x)=x-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,681 |
6. Let the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ satisfy that for each $n \in$ $\mathbf{Z}_{+}$, there exists a $k \in \mathbf{Z}_{+}$ such that $f^{2 k}(n)=n+k$, where $f^{m}$ is the $m$-fold composition of $f$. Let $k_{n}$ be the smallest $k$ satisfying the above condition. Prove: The sequence $k_{1... | 6. Let $S=\left\{1, f(1), f^{2}(1), \cdots\right\}$. For each positive integer $n \in S$, there exists a positive integer $k$, such that $f^{2 k}(n) = n + k \in S$. Therefore, $S$ is unbounded, and $f$ maps $S$ to $S$. Moreover, $f$ is injective on $S$. Indeed, if $f^{i}(1) = f^{j}(1) (i \neq j)$, then $f^{m}(1)$ start... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,682 |
Example 4 For a finite set $X$ of positive integers, define
$$
\sum(X)=\sum_{x \in X} \arctan \frac{1}{x} \text {. }
$$
Let $S$ be a finite set of positive integers, satisfying
$$
\sum(S)<\frac{\pi}{2} \text {. }
$$
Prove: There exists at least one finite set $T$ of positive integers such that
$$
S \subset T \text {,... | Prove that from $\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$, we know the following two conclusions.
(i) If $\tan \alpha, \tan \beta \in \mathbf{Q}$, then $\tan (\alpha+\beta) \in \mathbf{Q}$.
(ii) When $\alpha, \beta(\alpha>\beta)$ are acute angles,
$$
\begin{array}{l}
\tan (\alp... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,684 |
1. Let $a$, $b$, $c$ be given distinct real numbers. Prove: The equations
$$
\begin{array}{l}
(x-a)(x-b)=x-c, \\
(x-b)(x-c)=x-a, \\
(x-c)(x-a)=x-b
\end{array}
$$
have at least two real roots. | 1. Let's assume $a<b<c$. Then
$$
\begin{array}{l}
f_{1}(x)=(x-b)(x-c)-(x-a), \\
f_{2}(x)=(x-c)(x-a)-(x-b), \\
f_{3}(x)=(x-a)(x-b)-(x-c) .
\end{array}
$$
Notice that, their quadratic coefficients are positive, and
$$
f_{2}(c)=b-c<0, f_{1}(b)=a-b<0 \text {. }
$$
Therefore, $f_{1}=0, f_{2}=0$ both have real roots. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,685 |
2. In the acute triangle $\triangle ABC$, the circumcircle $\Gamma$ of the triangle has tangents at points $B$ and $C$ that intersect at point $P$. Points $D$ and $E$ are the projections of $P$ onto the lines $AB$ and $AC$, respectively. Prove that the orthocenter of $\triangle ADE$ is the midpoint of segment $BC$. | 2. Let $M$ be the midpoint of $BC$.
From $\triangle B P C$ being an isosceles triangle, we know $P M \perp B C$.
Since $\angle P M C=\angle P E C=90^{\circ}$, quadrilateral $M C E P$ has a circumcircle, so,
$$
\angle M E P=\angle M C P \text {. }
$$
Since $C P$ is a tangent to circle $\Gamma$,
$$
\begin{array}{l}
\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,686 |
3. Given $a_{1}, a_{2}, \cdots, a_{100}$ are 100 distinct positive integers. For any positive integer $i \in\{1,2, \cdots, 100\}, d_{i}$ represents the greatest common divisor of the 99 numbers $a_{j}(j \neq i)$, and $b_{i}=a_{i}+$ $d_{i}$. Question: How many different positive integers are there at least in $b_{1}, b_... | 3. Contains at least 99 different positive integers.
If we let $a_{100}=1, a_{i}=2 i(1 \leqslant i \leqslant 99)$; then $b_{1}=b_{100}=3$.
This indicates that there are at most 99 different positive integers in $b_{i}$.
Next, we prove: there are at least 99 different positive integers in $b_{i}$.
Without loss of gene... | 99 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,687 |
4. Given $n$ lines in the plane, where no two are parallel and no three are concurrent. Prove: There exists a non-closed, non-self-intersecting broken line $A_{0} A_{1} \cdots A_{n}$ composed of $n$ segments, such that each of the $n$ lines contains exactly one segment of the broken line. | 4. First, prove a strengthened conclusion using mathematical induction:
Given any point $A_{0}$ on an arbitrary line, as long as $A_{0}$ does not belong to any other line, there exists a broken line starting from $A_{0}$ that satisfies the conclusion.
When $n=1$, the conclusion is obvious.
Assume $n \geqslant 2$.
Let ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,688 |
$5.2 n$ real numbers are placed at $2 n$ distinct positions on a circle. It is known that the sum of these $2 n$ numbers is positive. Prove: there exists one of these positions such that: starting from this position (inclusive), the sum of the numbers at the next $n$ consecutive positions in both the clockwise and coun... | 5. Let the $2n$ numbers in clockwise order be $a_{1}, a_{2}$, $\cdots, a_{2 n}$, with their sum $S>0$.
Let $S_{i}=a_{i}+a_{i+1}+\cdots+a_{i+n-1}$, where $a_{2 n+i}=a_{i}, S_{2 n+i}=S_{i}$.
It suffices to prove that there exists $i$ such that $S_{i}$ and $S_{i+1-n}$ are both greater than 0. Since $S_{i}+S_{n+i}=S>0$, th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,689 |
6. Let $A$ be a set of ten real-coefficient quadratic polynomials. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible valu... | 6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that for any quadratic polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most two real roots. Therefore, each polynomial in $A$ appears at... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,690 |
7. For an acute triangle $\triangle ABC$, construct squares $CAKL$ and $CBMN$ outwardly on sides $CA$ and $CB$ respectively. The line $CN$ intersects segment $AK$ at point $X$, and the line $CL$ intersects segment $BM$ at point $Y$. The circumcircle of $\triangle KXN$ and the circumcircle of $\triangle LYM$ intersect a... | 7. Let $Q$ be the intersection of lines $K L$ and $M N$.
From $\angle Q L C=\angle N M Y=90^{\circ}$, we know that points $Q, L, Y, M$ are concyclic.
Similarly, points $Q, N, X, K$ are concyclic.
Let $Q, N, X, K$ and $Q, L, Y, M$ be concyclic in circles $\Gamma_{1}$ and $\Gamma_{2}$, respectively. Then $Q$ is the seco... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,691 |
8. A $2 \times 2$ square with a $1 \times 1$ corner removed is called a "corner shape". On a $55 \times 55$ grid paper, draw 400 corner shapes and 500 $1 \times 1$ unit squares, totaling 900 shapes. It is known that the interiors of these 900 shapes are pairwise disjoint, and the boundaries of all shapes lie on the ori... | 8. Assume that no two of the 900 figures have a common boundary of positive length.
On the boundary of each triangular figure, at the midpoints of the two sides of length 2, construct line segments of length $\frac{1}{2}$ perpendicular to these sides, resulting in a "large triangle" with two "antennae". Each large tri... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,692 |
1. Calculate: $\frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}=(\quad$.
(A) -1
(B) $\frac{5}{36}$
(C) $\frac{7}{12}$
(D) $\frac{49}{20}$
(E) $\frac{43}{3}$ | 1. C.
$$
\frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}=\frac{4}{3}-\frac{3}{4}=\frac{7}{12} \text{. }
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,693 |
2. Mr. Green measured his rectangular garden by walking, and found the two sides to be 15 steps and 20 steps. Mr. Green's step length is 2 feet. If he plants potatoes in the garden, expecting to produce 0.5 pounds per square foot, then Mr. Green can harvest ( ) pounds of potatoes from the garden.
(A) 600
(B) 800
(C) 10... | 2. A.
$$
(15 \times 2) \times(20 \times 2) \times 0.5=600 \text {. }
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,694 |
Example 5 Given a rational-coefficient polynomial $f$, with degree $d \geqslant 2$.
The sequence of sets $f^{0}(\mathbf{Q}), f^{1}(\mathbf{Q}), \cdots$ is defined as follows:
$$
f^{0}(\mathbf{Q})=\mathbf{Q}, f^{n+1}=f\left(f^{n}(\mathbf{Q})\right)(n \geqslant 0),
$$
where, for a given set $S$, we have
$$
f(S)=\{f(x) \... | First, we prove: there exists a positive integer $N_{0}$, such that when $q > N_{0}$ or $\left|\frac{q}{p}\right| > N_{0}\left(q \in \mathbf{N}_{+}, p \in \mathbf{Z}, (p, q) = 1\right)$, $f\left(\frac{p}{q}\right) \neq 0$, and when $\left|\frac{p}{q}\right| > N_{0}$,
$$
\left|f\left(\frac{p}{q}\right)\right| > \left|\f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,695 |
3. On a certain day in January, the highest temperature in Lincoln, Nebraska, was $16^{\circ} \mathrm{C}$ higher than the lowest temperature, and the average temperature was $3^{\circ} \mathrm{C}$. Then the temperature range in Lincoln on that day was ( ).
(A) $13^{\circ} \mathrm{C}$
(B) $8^{\circ} \mathrm{C}$
(C) $-5^... | 3. C.
Let the highest temperature of the day be $x^{\circ} \mathrm{C}$, and the lowest be $y^{\circ} \mathrm{C}$. Then
$$
\left\{\begin{array} { l }
{ x - y = 1 6 , } \\
{ \frac { x + y } { 2 } = 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=11 \\
y=-5 .
\end{array}\right.\right.
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,696 |
4. When counting from 3 to 201 in order, 53 is the 51st number, and when counting from 201 to 3 in reverse order, 53 is the ( )th number.
(A) 146
(B) 147
(C) 148
(D) 149
(E) 150 | 4. D.
Notice that, in the sequence, if an item is the $i$-th term in order, then it should be the $199+1-i$-th term in reverse order. Therefore,
$$
199+1-51=149 \text {. }
$$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,697 |
5. Let positive integers $a, b$ be less than 6. Then the minimum possible value of $2a - ab$ is ( ).
(A) -20
(B) -15
(C) -10
(D) 0
(E) 2 | 5. B.
From the problem, we know that $a=1,2,3,4,5 ; b=1,2,3,4,5$. Therefore, $2a-ab=a(2-b) \geqslant 5(2-5)=-15$. At this point, $a=b=5$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,698 |
6.33 The average age of the students is 11 years old, and the average age of the 55 parents of these students is 33 years old. Then the overall average age of all students and parents is ( ) years old.
(A) 22
(B) 23.25
(C) 24.75
(D) 26.25
(E) 28 | 6. C.
$$
\bar{x}=\frac{33 \times 11+55 \times 33}{33+55}=24.75 .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,699 |
7. A circle with a radius of 1 has six points, these six points divide the circle into six equal parts. Take three of these points as vertices to form a triangle. If the triangle is neither equilateral nor isosceles, then the area of this triangle is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{2}$
(C) 1
(D) $\sq... | 7. B.
Consider three cases:
(1) If two vertices of the triangle are adjacent, let's assume they are $A_{1}$ and $A_{2}$ (as shown in Figure 2), then the third vertex can only be $A_{4}$ or $A_{5}$. Therefore, the triangle is a right triangle.
(2) If two vertices of the triangle are separated, let's assume they are $A_... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,700 |
8. Ryan's car averages 40 miles per gallon of gasoline, Tom's car averages 10 miles per gallon, Ryan and Tom drive their respective cars the same distance. Considering both cars together, on average, they can travel ( ) miles per gallon of gasoline.
(A) 10
(B) 16
(C) 25
(D) 30
(E) 40 | 8. B.
$$
\frac{2}{\frac{1}{40}+\frac{1}{10}}=16
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,701 |
9. Three pairwise coprime positive integers, all greater than 1, have a product of 27000. Then their sum is ( ).
(A) 100
(B) 137
(C) 156
(D) 160
(E) 165 | 9. D.
Notice that, $27000=2^{3} \times 3^{3} \times 5^{3}$. Therefore, the three numbers are $2^{3} 、 3^{3} 、 5^{3}$, and their sum is 160. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,702 |
10. A basketball player has a probability of making a two-point shot and a three-point shot of $50\%$ and $40\%$, respectively. The player scored a total of 54 points, and the number of two-point shots attempted was $50\%$ more than the number of three-point shots attempted. How many three-point shots did the player at... | 10. C.
$$
\begin{array}{l}
\text { Given }\left\{\begin{array}{l}
y=x(1+50 \%), \\
2 y \cdot 50 \%+3 x \cdot 40 \%=54
\end{array}\right. \\
\Rightarrow(x, y)=(20,30) .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,703 |
11. Let $x, y \in \mathbf{R}$, satisfy the equation
$$
x^{2}+y^{2}=10 x-6 y-34 \text {. }
$$
Then $x+y=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 6
(E) 8 | 11. B.
$$
\begin{array}{l}
\text { Given } x^{2}-10 x+25+y^{2}+6 y+9=0 \\
\Rightarrow(x-5)^{2}+(y+3)^{2}=0 \\
\Rightarrow x=5, y=-3 \Rightarrow x+y=2 \text {. }
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,704 |
12. Let $S$ be the set composed of the sides and diagonals of a regular pentagon. If any two elements are chosen from it, the probability that these two elements have the same length is ( ).
(A) $\frac{2}{5}$
(B) $\frac{4}{9}$
(C) $\frac{1}{2}$
(D) $\frac{5}{9}$
(E) $\frac{4}{5}$ | 12. B.
$$
\frac{C_{5}^{2}+C_{5}^{2}}{C_{10}^{2}}=\frac{20}{45}=\frac{4}{9} .
$$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,705 |
13. Joe and Blair take turns counting from 1, each person counting one more number than the last number counted in the previous round. If Joe starts with 1, Blair then counts 1, 2, Joe then counts 1, 2, 3, and so on, then the 53rd number counted is ( ).
(A) 2
(B) 3
(C) 5
(D) 6
(E) 8 | 13. E.
Write the sequence $1,1,2,1,2,3, \cdots$ as
(1) , $(1,2),(1,2,3), \cdots$
Then the last number of the $k$-th group is $k$.
Thus, $a_{1+2+\cdots+k}=a_{\frac{k(k+1)}{2}}=k$.
Also, $\frac{9 \times 10}{2}=45<53<55=\frac{10 \times 11}{2}$, so the 53rd number is the 8th number in the 10th group, which is 8. | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,707 |
14. Define $a * b=a^{2} b-a b^{2}$. Then the set of points
$$
\{(x, y) \mid x * y=y * x\}
$$
represents $(\quad)$.
(A) a finite set of points
(B) a line
(C) two parallel lines
(D) two intersecting lines
(E) three lines | 14. E.
$$
\begin{array}{l}
\{(x, y) \mid x * y=y * x\} \\
=\left\{(x, y) \mid x^{2} y-y^{2} x=y^{2} x-x^{2} y\right\} \\
=\{(x, y) \mid x y(x-y)=0\} \\
=\{(x, y) \mid x=0 \text { or } y=0 \text { or } x=y\} .
\end{array}
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,708 |
15. A string is divided into two segments of lengths $a$ and $b$. The segments of lengths $a$ and $b$ are used to form an equilateral triangle and a regular hexagon, respectively. If their areas are equal, then $\frac{a}{b}=$ ( ).
(A) 1
(B) $\frac{\sqrt{6}}{2}$
(C) $\sqrt{3}$
(D) 2
(E) $\frac{3 \sqrt{2}}{2}$ | 15. B.
$$
\begin{array}{l}
\frac{\sqrt{3}}{4}\left(\frac{a}{3}\right)^{2}=\frac{3 \sqrt{3}}{2}\left(\frac{b}{6}\right)^{2} \\
\Rightarrow \frac{a^{2}}{9}=\frac{b^{2}}{6} \Rightarrow \frac{a}{b}=\frac{\sqrt{6}}{2} .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,709 |
16. As shown in Figure 1, in $\triangle A B C$, it is known that $A D$ and $C E$ are medians of $\triangle A B C$, and $A D$ intersects $C E$ at point $P$. If $P E=1.5, P D=2, D E=2.5$, then the area of quadrilateral $A E D C$ is ( ).
(A) 13
(B) 13.5
(C) 14
(D) 14.5
(E) 15 | 16. B.
From the problem, we know that $P$ is the centroid of $\triangle ABC$. Therefore, $AP = 2PD = 4$, and $CP = 2PE = 3$.
Also, $DE$ is the midline of $\triangle ABC$, and $DE = 2.5$, so $AC = 5$.
Notice that, $AC^2 = AP^2 + CP^2$.
Thus, $AD \perp CE$.
Therefore, $S_{\text{quadrilateral } AEDC} = \frac{1}{2} AD \cd... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,710 |
17. Alex has 75 red cards and 75 blue cards. It is known that Alex can exchange 2 red cards for 1 silver card and 1 blue card at one stall, and can exchange 3 blue cards for 1 silver card and 1 red card at another stall. If he continues to exchange according to the above methods until he can no longer exchange any card... | 17. E.
If Alex has $a$ red cards, $b$ blue cards, and $c$ silver cards, denoted as $(a, b, c)$, then
$$
\begin{array}{l}
(75,75,0) \rightarrow(1,112,37) \\
\rightarrow(38,1,74) \rightarrow(0,20,93) \\
\rightarrow(6,2,99) \rightarrow(0,5,102) \\
\rightarrow(1,2,103) .
\end{array}
$$
Therefore, Alex has 103 silver card... | 103 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,711 |
18. The characteristic of the positive integer 2013 is that the unit digit is the sum of the other digits, i.e., $2 +$ $0 + 1 = 3$. Among the numbers greater than 1000 and less than 2013, the number of numbers with the above characteristic is ( ) .
(A) 33
(B) 34
(C) 45
(D) 46
(E) 58 | 18. D.
Consider two cases.
(1) When the thousand's digit is 1, the unit's digit is $a(a=$ $1,2, \cdots, 9)$, so the ten's digit can only be $b=0,1, \cdots, a-1$, giving $a$ choices.
At this point, consider the digit in the hundred's place, which has only $a-1-b$ one choice.
By the multiplication principle and the add... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,712 |
19. Given real numbers $a, b, c$ form an arithmetic sequence, and $a \geqslant b \geqslant c \geqslant 0$. If the quadratic trinomial $a x^{2}+b x+c$ has only one real root, then the real root is ( ).
(A) $-7-4 \sqrt{3}$
(B) $-2-\sqrt{3}$
(C) 1
(D) $-2+\sqrt{3}$
(E) $-7+4 \sqrt{3}$ | 19. D.
From the problem, we know
$$
\Delta=b^{2}-4 a c=0 \text {. }
$$
Also, $a, b, c$ form an arithmetic sequence, so $b=\frac{a+c}{2}$.
Substituting the above equation into equation (1) yields
$$
\begin{array}{l}
\frac{(a+c)^{2}}{4}=4 a c \\
\Rightarrow\left(\frac{a}{c}\right)^{2}-14\left(\frac{a}{c}\right)+1=0 \\
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,713 |
20. Express the integer 2013 as
$$
2013=\frac{a_{1}!\cdot a_{2}!\cdots \cdots a_{m}!}{b_{1}!\cdot b_{2}!\cdots \cdots \cdot b_{n}!},
$$
where, $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m}, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, a_{i} 、 b_{j}$ $\in \mathbf{Z}_{+}(i=1,2, \cdots, m, j=1,2, \cd... | 20. B.
Given $2013=61 \times 11 \times 3$, and 61 is in the numerator, hence $a_{1}=61$.
Notice that, there does not exist a prime $p$ such that: $p$ is not a prime factor of 2013, $b_{1}<p<61$.
Since primes represent the numerator, not the denominator, the largest $p<61$ is 59, so 59 must be in the numerator, $b_{1... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,714 |
21. Two non-decreasing sequences of non-negative integers have different first terms. Each sequence has the following property: from the third term onwards, each term is equal to the sum of the two preceding terms, and the seventh term of both sequences is $N$. Then the smallest possible value of $N$ is ( ).
(A) 55
(B)... | 21. C.
If we denote two non-decreasing non-negative integer sequences as
$$
\begin{array}{l}
0,13,13,26,39,65,104, \cdots ; \\
8,8,16,24,40,64,104, \cdots
\end{array}
$$
In this case, the seventh term of both sequences is 104.
Next, we prove that 89 and 55 do not meet the conditions.
Let the first two terms of the se... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,715 |
22. The center of the regular octagon $A B C D E F G H$ is $J$. Each vertex and the center, these nine points, correspond to one of the numbers $1 \sim 9$, and each point corresponds to a different number. The sums of the numbers corresponding to the diagonals $A J E$, $B J F$, $C J G$, and $D J H$ are equal. How many ... | 22. C.
From the problem, let
$$
\begin{array}{l}
A+J+E=B+J+F=C+J+G=D+J+H=k \\
\Rightarrow A+B+C+D+E+F+G+H+4 J=4 k \\
\Rightarrow 1+2+\cdots+9+3 J=4 k \\
\Rightarrow 45+3 J=4 k(1 \leqslant J \leqslant 9, J \in \mathbf{Z}) \\
\Rightarrow(J, k)=(1,12),(5,15),(9,18) .
\end{array}
$$
Therefore, the number of corresponding... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,716 |
1. Given $p$ is a prime number, and satisfies $p \equiv 3(\bmod 8)$. Find all integer solutions $(x, y)$ of the equation
$$
y^{2}=x^{3}-p^{2} x
$$ | If $p \nmid x$, then $x, x-p, x+p$ are pairwise coprime, so $x, x-p, x+p$ are all perfect squares. Then
$$
2 p=(x+p)-(x-p) \equiv 0(\bmod 8),
$$
which is a contradiction.
If $p \mid x$, let $x=a p$. Then
$$
(x, x-p, x+p)=p \Rightarrow p^{2} \mid y \text {. }
$$
Let $y=p^{2} b$. Then from equation (1) $\Rightarrow p b^... | (a, b)=(0,0),(-1,0),(1,0) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,717 |
23. In $\triangle A B C$, it is known that $A B=13, B C=14$, $C A=15$, points $D, E, F$ are on sides $B C, C A, D E$ respectively, satisfying $A D \perp B C, D E \perp A C, A F \perp B F$, the length of segment $D F$ is a reduced fraction $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$. Then $m+n=(\quad)$.
(... | 23. B.
As shown in Figure 5.
Let $p$ be the semi-perimeter of $\triangle ABC$. Then
$$
p=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21.
$$
Thus, $S_{\triangle ABC}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{21 \times 8 \times 7 \times 6}=84 \\
\Rightarrow AD=\frac{2 S_{\triangle ABC}}{BC}=\frac{2 \times 84}{14}=12.
... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,718 |
24. If a positive integer $m$ has exactly four positive divisors (including 1 and itself), and the sum of these four positive divisors is $n$, then $n$ is called a "good number". In the set $\{2010, 2011, \cdots, 2019\}$, there are ( ) good numbers.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 24. A.
Let $m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{r}^{\alpha_{r}}$, where $p_{i}$ is a prime number, $\alpha_{i} \in \mathbf{Z}_{+}(i=1,2, \cdots, r), p_{1}<p_{2}<\cdots<p_{r}$.
By the divisor count theorem, we know that $m$ has
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{r}+1... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,719 |
25. Bernardo chose a three-digit positive integer $N$, and wrote its base-5 and base-6 numbers on the blackboard. Later, Leroy only saw the two numbers written by Bernardo, and mistakenly took them as decimal positive integers. He added them to get the positive integer $S$. For example, if $N=749$, the numbers Bernardo... | 25. E.
$$
\begin{array}{l}
\text { Let } N \equiv 6 c + a \left(\bmod 6^{2}\right), \\
N \equiv 5 d + b \left(\bmod 5^{2}\right),
\end{array}
$$
where, $a, c \in (0,1, \cdots, 5), b, d \in \{0,1, \cdots, 4\}$.
$$
\begin{aligned}
& \text { From } N \equiv b(\bmod 5) \\
& \Rightarrow 2 N \equiv 2 b(\bmod 10) \\
\Rightar... | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,720 |
1. Given vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}|=1$. Then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is | $$
\begin{array}{l}
\text {-1. } \frac{1}{4} \\
a \cdot b=\frac{(a+b)^{2}}{4}-\frac{(a-b)^{2}}{4} \\
=\frac{1}{4}-\frac{(a-b)^{2}}{4} \leqslant \frac{1}{4}
\end{array}
$$
When $a$ and $b$ are equal and $|a|=\frac{1}{2}$, the equality holds. | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,721 |
2. The maximum surface area of a cylinder inscribed in a sphere with radius $R$ is $\qquad$ .
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. $(1+\sqrt{5}) \pi R^{2}$.
Let the height of the cylinder $h=2 R \sin \theta$, and the base radius $r=$
$R \cos \theta$. Then the surface area is
$$
\begin{array}{l}
S=2 \pi r^{2}+2 \pi r h \\
=\pi R^{2}(1+\cos 2 \theta+2 \sin 2 \theta) \\
\leqslant(1+\sqrt{5}) \pi R^{2} .
\end{array}
$$
Equality holds if and only ... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 726,722 |
3. The set of positive integer solutions $(x, y)$ for the indeterminate equation $3 \times 2^{x}+1=y^{2}$ is $\qquad$ | 3. $\{(3,5),(4,7)\}$.
Obviously, $y$ is odd. Taking both sides of the equation modulo 8, we know $x \geqslant 3$. Transforming the original equation, we get
$$
3 \times 2^{x-2}=\frac{y-1}{2} \cdot \frac{y+1}{2} \text {. }
$$
Since $\left(\frac{y-1}{2}, \frac{y+1}{2}\right)=1$, we have $x=3$ or 4. Therefore, the origi... | (3,5),(4,7) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,723 |
4. Given $x, y, z \in \mathbf{R}_{+}$,
$$
\begin{array}{l}
S=\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}, \\
T=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1} .
\end{array}
$$
Then the minimum value of $S^{2}-T^{2}$ is | 4. 36 .
$$
\begin{array}{l}
S^{2}-T^{2}=(S+T)(S-T) \\
=(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+ \\
\quad \sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1}) \cdot \\
\quad\left(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{4}{\sqrt{y+5}+\sqrt{y+1}}+\frac{9}{\sqrt{z+10}+\sqrt{z+1}}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$
The equality hold... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,724 |
5. Given that the even function $f(x)$ satisfies for any $x \in$ $\mathbf{R}$,
$$
f(1+x)=f(3-x),
$$
and $f(x)=\left\{\begin{array}{ll}m \sqrt{1-x^{2}}, & x \in[0,1] ; \\ x-1, & x \in(1,2] \text {; }\end{array}\right.$
If the equation $3 f(x)=x$ has exactly 5 real solutions, then the range of the real number $m$ is | 5. $\left(-\sqrt{7},-\frac{\sqrt{15}}{3}\right) \cup\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right)$.
It is easy to see that the function $f(x)$ is a periodic function with a period of 4.
We discuss three cases for $m$.
(1) When $m=0$, the original equation has exactly three distinct real roots, which does not meet the req... | \left(-\sqrt{7},-\frac{\sqrt{15}}{3}\right) \cup\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,725 |
6. $P$ is a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$ in the first quadrant, $Q$ is the point symmetric to $P$ with respect to the origin $O$, $P H \perp x$-axis at point $H$, and the line $H Q$ intersects the hyperbola at point $M$ (different from $Q$). If the slope of the angle bisect... | 6. $\frac{\sqrt{6}}{2}$.
From the slope property of the hyperbola, we have
$$
k_{M P} k_{M Q}=\frac{b^{2}}{a^{2}} \text {. }
$$
From the given conditions, we get
$$
k_{M Q}=\frac{1}{2} k_{P Q}, k_{P M} k_{P Q}=1 \text {. }
$$
Thus, $\frac{b^{2}}{a^{2}}=\frac{1}{2}$, and it is easy to find that the eccentricity is $\... | \frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,726 |
7. Repeatedly tossing a fair die, each time after tossing, the number of the face up is recorded. The tossing stops when there are four different numbers recorded. Then the total number of different recording results that stop tossing exactly after six tosses is $\qquad$
Translate the above text into English, please k... | 7.9000 .
In the first five throws, exactly three different numbers appear. By the principle of inclusion-exclusion, the total number of different record results is
$$
\mathrm{C}_{6}^{3}\left(3^{5}-2^{5} \mathrm{C}_{3}^{2}+\mathrm{C}_{3}^{1}\right)=3000 .
$$
The number on the sixth throw is one of the remaining three ... | null | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,727 |
2. Let $a, b$ be positive integers, satisfying
$$
(4 a b-1) \mid\left(4 a^{2}-1\right)^{2} \text {. }
$$
Prove: $a=b$. | First prove: if $(a, b)$ satisfies equation (1), then $(b, a)$ also satisfies equation (1). Then prove: if $(a, b)(a<b)$ satisfies equation (1), then $\left(a, \frac{4 a^{3}-2 a+b}{4 a b-1}\right)$ also satisfies equation (1), and $\frac{4 a^{3}-2 a+b}{4 a b-1}$ $<a$. Finally, derive a contradiction using the method of... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,728 |
8. From $1,2, \cdots, 2013$, select $3 k$ different numbers to form $k$ triples $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2$, $\cdots, k)$. If $a_{i}+b_{i}+c_{i}(i=1,2, \cdots, k)$ these $k$ numbers are all distinct and less than 2013, then the maximum value of $k$ is | 8. 402.
On the one hand,
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \leqslant \sum_{i=1}^{k}(2013-i) \\
=2013 k-\frac{k(k+1)}{2},
\end{array}
$$
and
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \geqslant 1+2+\cdots+3 k \\
=\frac{3 k(3 k+1)}{2} .
\end{array}
$$
Thus, $2013 k-... | 402 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,729 |
9. (16 points) Given the parabola $C_{1}: y=x^{2}, M$ is a moving point in the plane. Two tangent lines are drawn from $M$ to the parabola $C_{1}$, touching it at points $A$ and $B$. If the area of $\triangle M A B$ is a constant $2 c^{3}$, find the equation of the locus of point $M$. | 9. Let $A\left(x_{1}, x_{1}^{2}\right), B\left(x_{2}, x_{2}^{2}\right)$.
Then $l_{A B}: y=\left(x_{1}+x_{2}\right) x-x_{1} x_{2}$.
The equations of the tangents $M A$ and $M B$ are
$$
\begin{array}{l}
y=2 x_{1}\left(x-x_{1}\right)+x_{1}^{2}=2 x_{1} x-x_{1}^{2}, \\
y=2 x_{2}\left(x-x_{2}\right)+x_{2}^{2}=2 x_{2} x-x_{2}... | x^{2}-y-c^{2}=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,730 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}(n \in \mathbf{N})$ satisfies: $a_{1}=1$, and for any non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Find the value of $\left[\frac{a_{2013}}{2012}\right]$ (where $[x]$ denotes ... | 10. Let $m=n$, we get $a_{0}=1$.
Let $n=0$, we get $a_{2 m}=4 a_{m}+2 m-3$.
In particular, $a_{2}=3$.
Let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right)=2 a_{m}+m \\
\Rightarrow a_{m+1}-a_{m}=a_{m}-a_{m-1}+2 .
\end{array}
$$
Thus, $a_{m+1}-a_{m}=2 m(m \geqslant 1)$,
$$
\be... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,731 |
11. (20 points) Given
$$
f(x)=\frac{1+\ln (x+1)}{x}, g(x)=\frac{k}{x+1} .
$$
Find the largest positive integer $k$, such that for any positive number $c$, there exist real numbers $a$ and $b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) .
$$ | 11. For positive integer $k$, it is clear that $g(x)=\frac{k}{x+1}$ is a decreasing function on the interval $(-1,+\infty)$.
Thus, for any positive number $c$,
$$
f(c)=g(b)>g(c) \text {. }
$$
When $x>0$, the inequality
$$
\begin{array}{l}
f(x)>g(x) \\
\Leftrightarrow k0) \text {. }
$$
Then $h^{\prime}(x)=\frac{x-1-\l... | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,732 |
一、(40 points) As shown in Figure 1, given that the circle $\odot O$ with diameter $BC$ intersects the sides $AC$ and $AB$ of $\triangle ABC$ at points $D$ and $E$ respectively, $BD$ and $CE$ intersect at point $F$, and $G$ is a point on segment $DE$. $AI \perp FG$, intersecting $FG$ at point $H$, and intersecting $OD$ ... | Connect $B I$, $E I$, and $C I$. Then,
$B$, $G$, and $I$ are collinear $\Leftrightarrow \frac{S_{\triangle B E I}}{S_{\triangle B D I}}=\frac{E G}{G D}$.
Since $O$ is the midpoint of $B C$, we have
$S_{\triangle B D I}=S_{\triangle C D I}$.
And since $B C$ is the diameter, hence
$$
\angle C E A=\angle B D A=90^{\circ}.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,733 |
II. (40 points) Given $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0$, and $\prod_{i=1}^{k} b_{i} \geqslant \prod_{i=1}^{k} a_{i}(k=1,2, \cdots, n)$.
Prove: $\sum_{i=1}^{n} b_{i} \geqslant \sum_{i=1}^{n} a_{i}$. | $$
\text { II. Let } t_{i}=\frac{b_{i}}{a_{i}}(i=1,2, \cdots, n) \text {. }
$$
From the problem, for $k=1,2, \cdots, n$, we have
$$
\begin{array}{l}
\prod_{i=1}^{k} t_{i} \geqslant 1 \Rightarrow t_{i}>0 \\
\Rightarrow \sum_{i=1}^{k} t_{i} \geqslant k \sqrt[k]{\prod_{i=1}^{k} t_{i}} \geqslant k \\
\Rightarrow \sum_{i=1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,734 |
Three, (50 points) Find the smallest positive integer $n$, such that the sum of the squares of all positive divisors of $n$ is $(n+3)^{2}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part of the problem statement, so it is provided in its original form as it is ... | Three, let $16 n+8, \\
\end{array}
$$
Contradiction.
Thus, $k \leqslant 5$.
And $k=0$ clearly does not meet the problem's conditions, so $1 \leqslant k \leqslant 5$.
Let $n=\prod_{i=1}^{i} p_{i}^{\alpha_{i}}$. Then
$$
\prod_{i=1}^{t}\left(\alpha_{i}+1\right)=k+2 \in[3,7] .
$$
Thus, $1 \leqslant t \leqslant 2$.
We dis... | 287 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,735 |
In $\triangle A B C$ with unequal sides, as shown in Fig. 1, the incircle touches the sides $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively. $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ are the midpoints of sides $B C$, $C A$, and $A B$ respectively. $D^{\prime}$, $E^{\prime}$, and $F^{\prime}$ are the... | Proof As shown in Figure 2, connect $B^{\prime} C^{\prime}$ and $A^{\prime} D^{\prime}$, intersecting at point $A^{\prime \prime}$. Similarly, define points $B^{\prime \prime}$ and $C^{\prime \prime}$. Draw lines through $D^{\prime}$ parallel to $A C$ and $A B$, intersecting $A B$ and $A C$ at points $M$ and $N$, respe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,737 |
Given $a, b, c$ are positive numbers satisfying $abc=1$. Prove:
$$
\begin{array}{l}
\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(a+1)}{1+ca+a}+ \\
\frac{(c-1)(b+1)}{1+ab+b} \geqslant 0 .
\end{array}
$$ | Proof Let
$$
a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right) \text {. }
$$
Then the original inequality
$$
\begin{array}{l}
\Leftrightarrow \sum \frac{(x-y)(z+x)}{y(x+y+z)} \geqslant 0 \\
\Leftrightarrow \sum \frac{z x+x^{2}-y z-x y}{y} \geqslant 0 \\
\Leftrightarrow \sum\left(\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,738 |
3. Let $p_{1}, p_{2}, \cdots$ be the prime numbers in increasing order, and let $x_{0}$ be a real number between 0 and 1. For a positive integer $k$, define
$$
x_{k}=\left\{\begin{array}{ll}
0, & x_{k-1}=0 ; \\
\left\{\frac{p_{k}}{x_{k-1}}\right\}, & x_{k-1} \neq 0,
\end{array}\right.
$$
where $\{x\}$ denotes the frac... | Prompt: $x_{0}$ is a rational number.
The necessity is obvious from the definition.
Next, we prove the sufficiency.
Assume $x_{0}$ is a rational number. Then the sequence $x_{0}, x_{1}, \cdots$ is a sequence of rational numbers. Suppose there is no 0 in it.
Let $x_{k-1}=\frac{m}{n}(0<m<n,(m, n)=1)$. Then
$$
x_{k}=\left... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,739 |
Let $p$ be an odd prime. Prove: the equation
$$
x^{2}+2 y^{2}=p
$$
has a solution if and only if the remainder of $p$ divided by 8 is 1 or 3. | First, if equation (1) has a solution, then $\left(\frac{-2}{p}\right)=1$, which means there exists an integer $a$ satisfying
$$
a^{2} \equiv-2(\bmod p) .
$$
Second, $\left(\frac{-2}{p}\right)=1$ if and only if $p \equiv 1,3(\bmod 8)$.
Indeed, consider $1,2, \cdots, \frac{p-1}{2}$.
Multiply each of these numbers by -2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,740 |
354 Divide the sides of the equilateral $\triangle A B C$ into four equal parts, and draw lines parallel to the other two sides through each division point. The 15 points formed by the intersections of the sides of $\triangle A B C$ and these parallel lines are called lattice points. Among these 15 lattice points, if $... | Solve for the minimum value of $n$ being 6.
Let the three equal division points from point $A$ to $B$ on side $AB$ be $L, F, W$; the three equal division points from point $B$ to $C$ on side $BC$ be $V, D, U$; and the three equal division points from point $C$ to $A$ on side $CA$ be $N, E, M$. Denote the intersection p... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,741 |
4. Prove: The sequence
$$
a_{n}=\frac{1}{4}\left[(1+\sqrt{2})^{2 n+1}+(1-\sqrt{2})^{2 n+1}+2\right](n>1)
$$
contains no perfect squares. | Prompt: Proof by contradiction.
If there exists $a_{n}=M^{2}\left(M \in \mathbf{N}_{+}\right)$, then
$$
2 M^{4}-2 M^{2}+1=\left(M^{2}\right)^{2}+\left(M^{2}-1\right)^{2}
$$
is also a perfect square.
By the properties of Pythagorean triples, this is equivalent to $x^{4}+4 y^{4}=z^{2}$ having a positive integer solution... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,742 |
Example 1 As shown in Figure 1, find the size of $\angle 1+\angle 2+\cdots+\angle 7$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Analysis】It is easy to notice that $\angle 2$, $\angle 3$, and $\angle 4$ are located in one quadrilateral, and $\angle 5$, $\angle 6$, and $\angle 7$ are located in another quadrilateral. This leads us to consider the sum of the interior angles of these two quadrilaterals. As shown in Figure 1, we add $\angle 8$ and ... | 540^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,743 |
Example 2 As shown in Figure 2, in $\triangle A B C$, it is known that $\angle A B C=50^{\circ}$. Rotate $\triangle A B C$ counterclockwise around point $B$ to $\triangle A^{\prime} B C^{\prime}$, such that $A A^{\prime} / / B C$. Find the size of $\angle C B C^{\prime}$ at this time. | 【Analysis】Obviously, $\angle C B C^{\prime}=\angle A B A^{\prime}$.
Furthermore, we find that $\triangle A B A^{\prime}$ is an isosceles triangle, thus it is easy to find $\angle A B A^{\prime}$, which can be done using $\angle 2=\angle 1=50^{\circ}$.
In fact, by rotation we know
$$
\begin{array}{l}
B A=B A^{\prime} \\... | 80^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,744 |
3. Let $A D$ be the altitude of $\triangle A B C$, and $K$ a point on $A D$. $B K$ intersects $A C$ at point $E$, and $C K$ intersects $A B$ at point $F$. Prove: $\angle F D A=\angle E D A$. | As shown in Figure 12, draw a line through point $A$ parallel to $BC$, intersecting $DE$, $DF$, $BE$, and $CF$ at points $Q$, $P$, $N$, and $M$ respectively.
Obviously, $\frac{BD}{AN}=\frac{KD}{KA}=\frac{DC}{AM}$, $\frac{AP}{BD}=\frac{AF}{FB}=\frac{AM}{BC}$,
$$
\frac{AQ}{DC}=\frac{AE}{EC}=\frac{AN}{BC} \text{. }
$$
Co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,745 |
4. Given that $D$ is any point inside $\triangle A B C$. Prove:
$$
\angle B D C>\angle B A C .
$$ | As shown in Figure 13, extend $BD$ to intersect $AC$ at point $E$. At this time, $\angle BDC$ is the exterior angle of $\triangle EDC$.
Therefore, $\angle BDC > \angle DEC$.
Similarly, $\angle DEC > \angle BAC$.
Thus, $\angle BDC > \angle BAC$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,746 |
Example 1 What is the maximum number of rational points (points with both coordinates being rational numbers) that can lie on a circle in the plane, given that the center of the circle is not a rational point. | 【Analysis】If $A, B, C$ are three rational points on a circle,
then the midpoint $D$ of $AB$ is a rational point, the slope of $AB$ is a rational number or infinite, so the equation of the perpendicular bisector of $AB$ is a linear equation with rational coefficients. Similarly, the equation of the perpendicular bisec... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,748 |
Example 2 Let $A, B, C$ be three non-collinear lattice points on a plane, and the side lengths of $\triangle ABC$ are all positive integers. Find the minimum value of $AB$ and the minimum perimeter. | 【Analysis】If $A B=1$, we might as well set $A(0,0), B(1,0)$. Then $|A C-B C|<A B=1$, which can only be $A C=B C$.
Thus, point $C$ lies on the perpendicular bisector of $A B$.
Therefore, the x-coordinate of point $C$ is $\frac{1}{2}$, meaning $C$ cannot be a lattice point.
Hence, $A B$ cannot be 1.
If $A B=2$, we might ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,749 |
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