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Example 7 For an integer $n(n \geqslant 2)$, let real numbers $x_{1}, x_{2}$, $\cdots, x_{n}$ satisfy
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=0, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1 .
\end{array}
$$
For any set $A \subseteq\{1,2, \cdots, n\}$, define $S_{A}=\sum_{i \in A} x_{i}\left(\right.$ if $A$ is empty,... | Proof: For any positive real number $\lambda$, the number of sets $A$ satisfying $S_{A} \geqslant \lambda$ is at most $\frac{2^{n-3}}{\lambda^{2}}$, and determine all $x_{1}, x_{2}, \cdots, x_{n}$ that make the equality hold.
(2012, USA Mathematical Olympiad)
Proof Let the set $\mathscr{F}$ be the family of all subset... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,608 |
2. Given the function
$$
f(x)=\sin x+\sqrt{1+\cos ^{2} x}(x \in \mathbf{R}) \text {. }
$$
Then the range of the function $f(x)$ is $\qquad$ | 2. $[0,2]$.
From the given information, we have
$$
f(x) \leqslant \sqrt{2\left(\sin ^{2} x+1+\cos ^{2} x\right)}=2,
$$
and $|\sin x| \leqslant 1 \leqslant \sqrt{1+\cos ^{2} x}$.
Therefore, $0 \leqslant \sin x+\sqrt{1+\cos ^{2} x} \leqslant 2$.
Hence, when $\sqrt{1+\cos ^{2} x}=\sin x$, that is,
$$
x=2 k \pi+\frac{\pi... | [0,2] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,609 |
$$
\begin{array}{l}
\text { 4. If } x_{1}>x_{2}>x_{3}>x_{4}>0, \text { and the inequality } \\
\log _{\frac{x_{1}}{2}} 2014+\log _{\frac{x_{2}}{x_{3}}} 2014+\log _{x_{4} \frac{1}{4}} 2014 \\
\geqslant k \log _{\frac{x_{1}}{}} 2014
\end{array}
$$
always holds, then the maximum value of the real number $k$ is . $\qquad$ | 4.9.
From the given, we have
$$
\begin{array}{l}
\frac{\ln 2014}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln 2014}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln 2014}{\ln \frac{x_{3}}{x_{4}}} \\
\geqslant k \frac{\ln 2014}{\ln \frac{x_{1}}{x_{4}}}.
\end{array}
$$
Since $x_{1}>x_{2}>x_{3}>x_{4}>0$, we have
$$
\begin{array}{l}
\ln \frac{x... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,610 |
5. As shown in Figure 1, given the right triangle $\triangle ABC$ with the two legs $AC=2, BC=3$, and $P$ is a point on the hypotenuse $AB$. The triangle is folded along $CP$ to form a right dihedral angle $A-CP-B$. When $AB = \sqrt{7}$, the value of the dihedral angle $P-AC-B$ is . $\qquad$ | 5. $\arctan \sqrt{2}$.
As shown in Figure 2, draw $P D \perp A C$ at point $D$, draw $P E \perp C P$ intersecting $B C$ at point $E$, and connect $D E$.
Since $A-C P-B$ is a right dihedral angle, i.e., plane $A C P \perp$ plane $C P B$, therefore,
$P E \perp$ plane $A C P$.
Also, $P D \perp A C$, so by the theorem of... | \arctan \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,611 |
6. Given the sequence $\left\{a_{n}\right\}$
satisfies
$$
a_{n}=\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}(n \geqslant 1),
$$
and its first $n$ terms sum is $S_{n}$. Then $\left[S_{n}\right]=$ $\qquad$ ( $[x]$ represents the greatest integer not exceeding the real number $x$). | 6. n.
From the given, we have
$$
\begin{array}{l}
a_{n}=\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\frac{[n(n+1)]^{2}+(n+1)^{2}+n^{2}}{n^{2}(n+1)^{2}}} \\
=\frac{n(n+1)+1}{n(n+1)}=1+\frac{1}{n}-\frac{1}{n+1} .
\end{array}
$$
Then, $S_{n}=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{... | n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,612 |
8. A and B take turns rolling a die, with A starting first. The rule is: if A rolls a 1, A continues to roll; otherwise, B rolls. If B rolls a 3, B continues to roll; otherwise, A rolls. The two always follow this rule. Then the probability that the $n$-th roll is made by A is
$$
p_{n}=
$$
$\qquad$ | 8. $\frac{1}{2}-\frac{1}{3}\left(-\frac{2}{3}\right)^{n-2}$.
If Player A rolls a 1 (Player B rolls a 3), the probability is $\frac{1}{6}$; if Player A does not roll a 1 (Player B does not roll a 3), the probability is $\frac{5}{6}$.
Let the probability that Player A rolls on the $k$-th turn be $p_{k}$. Then the proba... | \frac{1}{2}-\frac{1}{3}\left(-\frac{2}{3}\right)^{n-2} | Other | math-word-problem | Yes | Yes | cn_contest | false | 728,613 |
9. Let the positive integer $n$ satisfy $31 \mid\left(5^{n}+n\right)$. Then the minimum value of $n$ is $\qquad$ . | 9. 30 .
Given $5^{3} \equiv 1(\bmod 31)$, when $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $5^{n}+n \equiv 1+n \equiv 0(\bmod 31)$, at this time, the minimum value of $n$ is $n_{\text {min }}=30$;
When $n=3 k+1\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 5+n \equiv 0(\bmod 31),
$$
at this time, the minimum val... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,614 |
10. Given
$$
S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }
$$
where, $n \in \mathbf{Z}_{+}$. Then the minimum value of $S_{n}$ is $\qquad$ | 10. 112 .
From the problem, we know
$$
\begin{array}{l}
S_{n+1}-S_{n} \\
=|n|+2|n-1|+\cdots+10|n-9|- \\
{[|n-1|+2|n-2|+\cdots+10|n-10|] } \\
=|n|+|n-1|+\cdots+|n-9|-10|n-10| .
\end{array}
$$
When $n \geqslant 10$, $S_{n+1}-S_{n}>0$, thus, $S_{n}$ is monotonically increasing; $\square$
When $n=0$, $S_{1}-S_{0}>0$;
Wh... | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,615 |
Example 8 In a certain club, any two people are either friendly or hostile. Let the club have $n$ people and $q$ friendly pairs, and suppose that among any three people, at least one pair is hostile. Prove: there is at least one member in the club whose set of enemies contains no more than $q\left(1-\frac{4 q}{n^{2}}\r... | Prove that by randomly selecting a person $v$, and using the random variable $X$ to represent the number of friendly pairs among those who are enemies with this person. Using the double counting method, we get
$$
\begin{array}{l}
E(X)=\sum_{\text {friendly pair } e} E\left(I_{e}\right) \\
=\sum_{\text {friendly pair } ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,616 |
1. If $15 \mid \overline{\text { xaxax }}$, then the sum of all five-digit numbers $\overline{x a x a x}$ that satisfy the requirement is ( ).
(A) 50505
(B) 59595
(C) 110100
(D) 220200 | - 1. D.
Notice that, $15| \overline{\text { xaxax }} \Leftrightarrow\left\{\begin{array}{l}5 \mid \overline{\text { xaxax }}, \\ 3 \mid \overline{\text { xaxax }} .\end{array}\right.$
From $51 \overline{\text { xaxax }} \Rightarrow x=5$;
From $31 \overline{x a x a x} \Rightarrow 31(x+a+x+a+x)$
$$
\Rightarrow 3|2 a \Ri... | 220200 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,617 |
2. Given $x=\frac{1-\sqrt{5}}{2}$ is a solution to the equation $x^{4}+a x-b=0$, where $a$ and $b$ are rational numbers. Then the values of $a$ and $b$ are $(\quad)$.
(A) 3,2
(B) $-3,2$
(C) $3, -2$
(D) $-3,-2$ | 2. B.
Given $x=\frac{1-\sqrt{5}}{2}$, we know $x^{2}-x-1=0$.
Then $x^{4}+a x-b=0$
$$
\begin{array}{l}
\Rightarrow(x+1)^{2}+a x-b=0 \\
\Rightarrow x^{2}+2 x+1+a x-b=0 \\
\Rightarrow(3+a) x+2-b=0 .
\end{array}
$$
Combining the fact that $a$ and $b$ are rational numbers, and $x$ is an irrational number, we have
$$
a=-3,... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,618 |
3. The number of all different positive integer solutions to the equation $\frac{x}{y}-\frac{2 y}{x}=\frac{2015}{x y}-1$ is $(\quad$.
(A) 0
(B) 1
(C) 2
(D) 4 | 3. A.
$$
\begin{array}{l}
\text { Given } \frac{x}{y}-\frac{2 y}{x}=\frac{2015}{x y}-1 \\
\Rightarrow x^{2}-2 y^{2}=2015-x y \\
\Rightarrow(x-y)(x+2 y)=2015 .
\end{array}
$$
Obviously, $x>y>0$.
Therefore, there are only four possibilities:
$$
\begin{array}{l}
\left\{\begin{array}{l}
x-y=1, \\
x+2 y=2
\end{array} 015 ;... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,619 |
4. If the equation $s x^{2}+t x+s-1=0$ with respect to $x$ has real roots for any real number $t$, then the range of the real number $s$ is ( ).
(A) $s<0$
(B) $s<0$ or $s \geqslant 1$
(C) $0 \leqslant s<1$
(D) $0<s \leqslant 1$ | 4. D.
If $s=0$, the equation becomes $t x-1=0$. When $t=0$, it clearly has no solution, which contradicts the problem statement.
Therefore, $s \neq 0$, and the equation must be a quadratic equation.
Consider the discriminant
$$
\Delta=t^{2}-4 s(s-1) \geqslant 0 \Rightarrow s(s-1) \leqslant \frac{t^{2}}{4} \text {. }
$... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,620 |
5. As shown in Figure 1, given $\angle B=30^{\circ}, \angle C=$ $45^{\circ}, \angle B D C=150^{\circ}$, and $B D=C D=5$. Then $A B=(\quad)$.
(A) 7.5
(B) $5 \sqrt{2}$
(C) $5 \sqrt{3}$
(D) $5+\sqrt{5}$ | 5. C.
Connect $A D$. It is easy to know that
$$
\angle B D C=\angle B A C+\angle B+\angle C=2 \angle B A C \text {, }
$$
and $B D=C D$.
Therefore, points $A, B, C$ lie on the circle $\odot D$ with radius $B D$.
Thus, $A B=2 B D \cos B=5 \sqrt{3}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,621 |
Example 9 In a graph $G$ with $n$ vertices, if it does not contain a complete graph $K_{p}$ of order $p$, then the number of edges in graph $G$
$$
|E| \leqslant\left(1-\frac{1}{p-1}\right) \frac{n^{2}}{2},
$$
where a complete graph $K_{p}$ of order $p$ refers to a graph with $p$ vertices and edges between every pair o... | Let the vertices of graph $G$ be $v_{1}, v_{2}, \cdots, v_{n}$, with corresponding degrees $d_{1}, d_{2}, \cdots, d_{n}$. Let $\omega(G)$ denote the number of vertices in the largest complete subgraph of $G$, and let permutation $\tau$ represent a random permutation of the $n$ vertices of $G$, of which there are clearl... | |E| \leqslant\left(1-\frac{1}{p-1}\right) \frac{n^{2}}{2} | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,622 |
1. Let $a_{1}, a_{2}, \cdots, a_{2015}$ be a sequence of numbers taking values from $-1, 0, 1$, satisfying
$$
\sum_{i=1}^{2015} a_{i}=5 \text {, and } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040,
$$
where $\sum_{i=1}^{n} a_{i}$ denotes the sum of $a_{1}, a_{2}, \cdots, a_{n}$.
Then the number of 1's in this sequenc... | $=, 1.510$.
Let the number of -1's be $x$, and the number of 0's be $y$.
Then, from $\sum_{i=1}^{2015} a_{i}=5$, we know the number of 1's is $x+5$. Combining this with
$$
\begin{array}{c}
\text { the equation } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, \text { we get } \\
\left\{\begin{array}{l}
x+(x+5)+y=2015, ... | 510 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,623 |
3. If the integer $n$
satisfies
$$
(n+1)^{2} \mid\left(n^{2015}+1\right) \text {, }
$$
then the minimum value of $n$ is $\qquad$ | 3. -2016 .
From $n^{2015}+1=(n+1) \sum_{i=0}^{2014}(-1)^{i} n^{2014-i}$, we know
$$
\begin{array}{l}
(n+1)^{2} \mid\left(n^{2015}+1\right) \\
\left.\Leftrightarrow(n+1)\right|_{i=0} ^{2014}(-1)^{i} n^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{i}(-1)^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2... | -2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,624 |
4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$
then $k=$ | 4. 4 .
Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\ri... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,625 |
One, (20 points) If $x, y \in [0,1]$, try to find the maximum value of
$$
x \sqrt{1-y} + y \sqrt{1-x}
$$ | $$
\begin{array}{l}
\text { I. Let } s=\sqrt{1-x}, t=\sqrt{1-y}, \\
z=x \sqrt{1-y}+y \sqrt{1-x} .
\end{array}
$$
Then $z=\left(1-s^{2}\right) t+\left(1-t^{2}\right) s=(s+t)(1-s t)$.
Also, $(1-s)(1-t)=1-(s+t)+s t \geqslant 0$, which means $s+t \leqslant 1+s t$,
thus $z \leqslant(1+s t)(1-s t)=1-s^{2} t^{2} \leqslant 1$... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,626 |
II. (25 points) As shown in Figure 4, in $\triangle ABC$, it is given that $\angle A=60^{\circ}$, and $P$ is a point on the line segment $BC$ (excluding the endpoints). Points $E$ and $F$ are on the rays $AB$ and $AC$, respectively, such that $BP=BE$ and $CP=CF$. The tangents to the circumcircle of $\triangle ABC$ at p... | II. As shown in Figure 5, connect $O C, O E, O F, O P, T Y$.
Since $B S, C S$ are tangents to the circumcircle of $\triangle A B C$,
$\Rightarrow B S = S C$.
Given $\angle A = 60^{\circ}$, thus $\triangle S B C$ is an equilateral triangle, i.e., $\angle B S C = 60^{\circ}$.
Since $O$ is the circumcenter of $\triangle P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,627 |
1. Given the function $y=\frac{|x-1|}{x-1}$, its graph intersects with the graph of the function $y=k x-2$ at exactly two points. Then the range of the real number $k$ is $\qquad$ | $$
-1 .(0,1) \cup(1,4)
$$
From the given information, we have
$$
y=\left\{\begin{array}{ll}
x+1, & x \in(-\infty,-1) \cup(1,+\infty) ; \\
-x-1, & x \in[-1,1) .
\end{array}\right.
$$
Combining with the graph, we know $k \in(0,1) \cup(1,4)$. | k \in (0,1) \cup (1,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,628 |
2. In the triangular prism $A B C-A_{1} B_{1} C_{1}$, the base edge lengths and the side edge lengths are all equal, $\angle B A A_{1}=\angle C A A_{1}=60^{\circ}$. Then the angle formed by the skew lines $A B_{1}$ and $B C_{1}$ is $\qquad$ | 2. $\arccos \frac{\sqrt{6}}{6}$.
In the plane $B B_{1} C_{1} C$, translate $B C_{1}$ to $D B_{1}$, then $\angle A B_{1} D$ is the angle we need to find.
Assume the base edge length and the side edge length of the triangular prism are both 1.
In $\triangle A B_{1} D$,
$$
\begin{array}{l}
A B_{1}=\sqrt{3}, B_{1} D=\sqrt... | \arccos \frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,629 |
3. Given the ellipse $x^{2}+4(y-a)^{2}=4$ and the parabola $x^{2}=2 y$ have common points. Then the range of values for $a$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $\left[-1, \frac{17}{8}\right]$.
Notice that, the parametric equation of the ellipse is
$$
\left\{\begin{array}{l}
x=2 \cos \theta, \\
y=a+\sin \theta
\end{array}(\theta \in \mathbf{R})\right. \text {. }
$$
Substituting into the parabola equation, we get
$$
\begin{array}{l}
4 \cos ^{2} \theta=2(a+\sin \theta) \\
\... | \left[-1, \frac{17}{8}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,630 |
4. As shown in Figure 1, in a regular hexagon $A B C D E F$ with side length 2, a moving circle $\odot Q$ has a radius of 1, and its center moves on line segment $C D$ (including endpoints). $P$ is a moving point on and inside $\odot Q$. Let vector $\overrightarrow{A P}=m \overrightarrow{A B}+n \overrightarrow{A F}(m, ... | 4. $[2,5]$.
From the given information, we have
$$
\begin{array}{l}
\overrightarrow{A P} \cdot \overrightarrow{A B}=4 m-2 n, \\
\overrightarrow{A P} \cdot \overrightarrow{A F}=-2 m+4 n . \\
\text { Then } m+n=\frac{1}{2}(\overrightarrow{A P} \cdot \overrightarrow{A B}+\overrightarrow{A P} \cdot \overrightarrow{A F}) \... | [2,5] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,631 |
Example 10 If a set of lines in the plane has no two lines parallel and no three lines concurrent, it is called "in general position". A set of lines in general position divides the plane into several regions, and the regions with finite area are called "bounded regions" of this set of lines. Prove that for sufficientl... | Proof: First, count the maximum number of triangles and polygons (figures with more than three sides) that can be formed by $n$ lines in general position. Since there are $\mathrm{C}_{n}^{2}$ intersection points, each intersection point can be a vertex of at most two triangles, and each triangle has three vertices, the... | \frac{2\sqrt{n}}{3} | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,632 |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0$,
$$
a_{n+1}=\frac{8}{5} a_{n}+\frac{6}{5} \sqrt{4^{n}-a_{n}^{2}}(n \geqslant 0, n \in \mathbf{N}) \text {. }
$$
Then $a_{10}=$ | 7. $\frac{24576}{25}$.
Notice, $a_{n+1}=2\left(\frac{4}{5} a_{n}+\frac{3}{5} \sqrt{4^{n}-a_{n}^{2}}\right)$.
Let $a_{n}=2^{n} \sin \alpha_{n}$.
Let $\theta$ satisfy $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $\sin \alpha_{n+1}=\sin \alpha_{n} \cdot \cos \theta+\sin \theta \cdot\left|\cos \alpha_{n}\right... | \frac{24576}{25} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,633 |
8. To color the eight vertices of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with four different colors, such that the two endpoints of the same edge have different colors, there are a total of coloring methods. | 8. 2652 .
First, color the four points $A, B, C, D$ above, with 84 coloring methods. Then consider the four points below, using the principle of inclusion-exclusion, the total number of methods is
$$
\begin{array}{l}
84\left\{84-C_{4}^{1}[3 \times(3+2 \times 2)+\right. \\
(3+2 \times 2)]+2\left(\frac{36}{84} \times 9+... | 2652 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,634 |
Example 2 As shown in Figure 2, given that the $B$-excircle $\odot O_{B}$ and the $C$-excircle $\odot O_{C}$ of $\triangle A B C$ touch the line $B C$ at points $E$ and $D$ respectively, $\odot O_{B}$ touches $A C$ at point $F$, and $\odot O_{C}$ touches $A B$ at point $G$. The line $D G$ intersects $E F$ at point $P$.... | Prove as shown in Figure 2, extend $A P$ to intersect $B C$ at point $H$. By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{A P}{P H} \cdot \frac{H D}{D B} \cdot \frac{B G}{G A}=1, \frac{A P}{P H} \cdot \frac{H E}{E C} \cdot \frac{C F}{F A}=1 . \\
\text { Combining } G B=D B=C E=C F, \text { we get } \\
\frac{A P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,636 |
Example 4 Let $p$ be a prime, $n \in Z_{+}$, and
$$
n=n_{0}+n_{1} p+\cdots+n_{t} p^{t},
$$
where, $n_{i} \in \mathbf{N}, 0 \leqslant n_{i} \leqslant p-1, i=0,1, \cdots, t$.
Let $S_{n}$ denote the set of ordered triples $(a, b, c)$ that satisfy the following conditions:
(1) $a, b, c \in \mathbf{N}$;
(2) $a+b+c=n$;
(3) ... | 【Analysis】By the lemma, imitating the proof of Corollary 1, we have
$$
\begin{array}{l}
v_{p}\left(\frac{n!}{a!\cdot b!\cdot c!}\right)=0 \\
\Leftrightarrow \tau_{p}(a)+\tau_{p}(b)+\tau_{p}(c)-\tau_{p}(n)=0
\end{array}
$$
$\Leftrightarrow a, b, c$ do not produce a carry in any digit when performing column addition in $... | \prod_{i=0}^{t} \mathrm{C}_{n_{i}+2}^{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,637 |
Example 5 Let $k, l$ be given positive integers. Prove: there are infinitely many positive integers $n>k$, such that $\mathrm{C}_{n}^{k}$ is coprime with $l$.
(2009, National High School Mathematics Joint Competition) | 【Analysis】When $l=1$, the conclusion is obviously true.
When $l>1$, take any prime factor $p$ of $l$ to study the problem.
To make $v_{p}\left(\mathrm{C}_{n}^{k}\right)=0$, by Corollary 3, it is only necessary that there is no carry in each digit when performing vertical addition of $k$ and $n-k$ in $p$-ary representa... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,638 |
Example 6 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[5]}$
$(2012$, Girls' Mathematical Olympiad) | 【Analysis】Perform prime factorization $2012=2^{2} \times 503$.
First consider 503 I $\mathrm{C}_{2012}^{k}$.
By Corollary 3, we know that when and only when $k$ and $2012-k$ do not produce a carry when added in base 503, $\left(503, \mathrm{C}_{2012}^{k}\right)=1$.
Writing 2012 in base 503 gives $2012=(40)_{503}$.
Let ... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,639 |
Example 7 For a positive integer $n$ and an integer $i(0 \leqslant i \leqslant n)$, let $\mathrm{C}_{n}^{i} \equiv c(n, i)(\bmod 2)(c(n, i) \in\{0,1\})$.
Denote $f(n, q)=\sum_{i=0}^{n} c(n, i) q^{i}$.
Suppose $m, n, q \in \mathbf{Z}_{+}$, and $q+1$ is not a power of 2. Prove: If $f(m, q) \mid f(n, q)$, then for any pos... | 【Analysis】First, try to establish the expression for $f(n, q)$. According to the problem,
$$
\mathrm{C}_{n}^{i} \equiv c(n, i) \equiv 1(\bmod 2)
$$
only then is $q^{i}$ counted by $f(n, q)$.
We only need to consider the $i$ that satisfies $c(n, i) \equiv 1(\bmod 2)$.
By Corollary 3,
$$
\mathrm{C}_{n}^{i} \equiv 1(\b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,640 |
In $\triangle A B C$, it is known that
$$
\angle B A C=40^{\circ}, \angle A B C=60^{\circ} \text {, }
$$
If $D$ and $E$ are points on sides $A C$ and $A B$ respectively, such that $\angle C B D=40^{\circ}, \angle B C E=70^{\circ}$, and $F$ is the intersection of $B D$ and $C E$, connect $A F$. Prove: $A F \perp B C .{... | Prove as shown in Figure 1, draw the diagram.
Take $B$ as the origin and $B C$ as the $x$-axis to establish a Cartesian coordinate system. Then, it is sufficient to prove that the $x$-coordinates of points $A$ and $F$ are the same.
Assume $B C=1$. Then, in this coordinate system,
$$
l_{A B}: y=x \tan 60^{\circ},
$$
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,642 |
Let $n$ be an integer greater than 1, and let positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}-x_{i+1}^{3}} \geqslant \frac{n^{3}}{n^{2}-1}\left(x_{n+1}=x_{1}\right) \text {. }
$$
(11th China Southeast Mathematical Olympiad) | Prove that
$$
\begin{array}{l}
n+1=\sum_{i=1}^{n}\left(1+x_{i}\right) \geqslant n \sqrt[n]{\prod_{i=1}^{n}\left(1+x_{i}\right)} \\
\Rightarrow \frac{1}{\sqrt[n]{\prod_{i=1}^{n}\left(1+x_{i}\right)}} \geqslant \frac{n}{n+1} \text {. } \\
\end{array}
$$
Similarly, $\frac{1}{\sqrt[n]{\prod_{i=1}^{n}\left(1-x_{i}\right)}}... | \frac{n^{3}}{n^{2}-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,643 |
Question 1 As shown in Figure 1, through a point $P$ inside $\triangle A B C$, draw $D E / / B C$, intersecting
$A B, A C$ at points $D$,
$E$; draw $F G / / A C$, intersecting
$A B, B C$ at points
$F, G$; draw $H K / / A B$, intersecting
$A C, B C$ at points
$H, K, F K, H G$ intersecting
$D E$ at points
$M, N$. Then
(1... | Prove (1) In $\triangle F B G$ and $\triangle F K G$, from $D E / / B C \Rightarrow \frac{P D}{B G}=\frac{F P}{F G}=\frac{P M}{K G}$ $\Rightarrow \frac{P D}{B K+K G}=\frac{P M}{K G}$.
It is easy to prove that quadrilateral $D B K P$ is a parallelogram.
Thus, $B K=P D$.
Substituting into equation (1) yields $K G=\frac{P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,644 |
Example 3 As shown in Figure 3, given that the excircle of $\triangle A B C$ opposite to $B$, $\odot O_{B}$, and the excircle opposite to $C$, $\odot O_{C}$, touch the line $B C$ at points $E$ and $D$ respectively, $\odot O_{B}$ touches $A C$ at point $F$, and $\odot O_{C}$ touches $A B$ at point $G$. The extensions of... | Proof As shown in Figure 3, let $\odot O_{B}$ and $\odot O_{C}$ be tangent to the extensions of $BA$ and $CA$ at points $K$ and $L$, respectively. The line $DL$ intersects $EK$ at point $M$, and the extensions of $DG$ and $EF$ intersect at point $P$.
According to Example 1 and Example 2, we know that points $M$, $A$, a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,645 |
1. As shown in Figure 1, in isosceles $\triangle ABC$, $AB=AC>BC$, $D$ is a point inside $\triangle ABC$ such that $DA=DB+DC$. The perpendicular bisector of side $AB$ intersects the external angle bisector of $\angle ADB$ at point $P$, and the perpendicular bisector of side $AC$ intersects the external angle bisector o... | 1. First, prove that points $A$, $B$, $D$, and $P$ are concyclic.
In fact, as shown in Figure 2, take the midpoint $P'$ of arc $\overparen{A D B}$. Then $P'$ lies on the perpendicular bisector of segment $A B$.
Take any point $X$ on the extension of $B D$. Since $P' A = P' B$ and points $A$, $B$, $D$, and $P'$ are co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,647 |
2. Let $X$ be a non-empty finite set, and $A_{1}, A_{2}, \cdots, A_{k}$ be $k$ subsets of $X$, satisfying
(1) $\left|A_{i}\right| \leqslant 3 \ (i=1,2, \cdots, k)$;
(2) Any element in $X$ belongs to at least four of the sets $A_{1}, A_{2}, \cdots, A_{k}$.
Prove: It is possible to select $\left[\frac{3}{7} k\right]$ se... | 2. Select as many disjoint triples as possible from $A_{1}, A_{2}, \cdots, A_{k}$, and suppose $x$ sets are selected, which we can assume to be $A_{1}, A_{2}, \cdots, A_{x}$.
For $i(x<i \leqslant k)$, let $B_{i}=A_{i} \backslash\left(\bigcup_{j=1}^{x} A_{j}\right)$. Thus, $\left|B_{i}\right| \leqslant 2$ (otherwise, i... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,648 |
4. Given an integer $n \geqslant 2$, let $x_{1}, x_{2}, \cdots, x_{n}$ be a non-decreasing sequence of positive numbers, and let $x_{1}, \frac{x_{2}}{2}, \cdots, \frac{x_{n}}{n}$ form a non-increasing sequence. Prove:
$$
\frac{A_{n}}{G_{n}} \leqslant \frac{n+1}{2 \sqrt[n]{n!}},
$$
where $A_{n}$ and $G_{n}$ denote the ... | 4. From the conditions, we know $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, and
$$
x_{1} \geqslant \frac{x_{2}}{2} \geqslant \cdots \geqslant \frac{x_{n}}{n} \Rightarrow \frac{1}{x_{1}} \leqslant \frac{2}{x_{2}} \leqslant \cdots \leqslant \frac{n}{x_{n}} \text {. }
$$
By Chebyshev's inequality, we have
$... | \frac{A_{n}}{G_{n}} \leqslant \frac{n+1}{2 \sqrt[n]{n!}} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,650 |
5. Color each edge of the complete graph $G$ of order 2015 with one of two colors: red or blue. For any two-element subset $\{u, v\}$ of the vertex set $V$ of graph $G$, define
$$
L(u, v)=\{u, v\} \cup\{w \in V \mid \text { the triangle with vertices } u, v, w \text { has exactly two red edges }\}.
$$
Prove that when ... | 5. Take any point $v \in V$.
On one hand, if there are 120 vertices $u_{i}$ $(1 \leqslant i \leqslant 120)$ in $V \backslash\{v\}$ such that $v u_{i}$ is a blue edge, then for any $i$ $(1 \leqslant i \leqslant 120)$, by definition, $u_{i} \in L\left(v, u_{i}\right)$.
On the other hand, since for any $i, j (1 \leqslan... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,651 |
6. For a positive integer $n$, define
$$
f(n)=\tau(n!)-\tau((n-1)!),
$$
where $\tau(a)$ denotes the number of positive divisors of the positive integer $a$. Prove that there are infinitely many composite numbers $n$ such that for any positive integer $m<n$, we have $f(m)<f(n)$. (Yuhong Bing) | 6. First, prove four lemmas.
Lemma 1 For integer $n>1$, we have
$$
\tau((n-1)!)(n-1)!$, then $n!$ has at least one positive divisor that is not a divisor of $(n-1)!$ (such as $n!$).
Therefore, $\tau((n-1)!)f(m)(m=1,2, \cdots, p-1) \text {. }
$$
Proof of Lemma 2 Since $\tau$ is a multiplicative function, we have
$$
\b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,652 |
2. Given the lengths of line segments $A B$ and $C D$ are $a$ and $b$ $(a, b > 0)$. If line segments $A B$ and $C D$ slide on the $x$-axis and $y$-axis respectively, and such that points $A, B, C, D$ are concyclic, then the equation of the locus of the centers of these circles is $\qquad$ . | 2. $4 x^{2}-4 y^{2}-a^{2}+b^{2}=0$.
Let the center of the required circle be $O^{\prime}(x, y)$. Then
$$
\begin{array}{l}
A\left(x-\frac{a}{2}, 0\right), B\left(x+\frac{a}{2}, 0\right), \\
C\left(0, y-\frac{b}{2}\right), D\left(0, y+\frac{b}{2}\right) .
\end{array}
$$
Notice that,
$A, B, C, D$ are concyclic
$$
\begin... | 4 x^{2}-4 y^{2}-a^{2}+b^{2}=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,653 |
3. If $x \in(-1,1)$, then
$$
f(x)=x^{2}-a x+\frac{a}{2}
$$
is always positive, the range of the real number $a$ is $\qquad$ $ـ$. | 3. $(0,2]$.
Notice that, the graph of the quadratic function $y=x^{2}-a x+\frac{a}{2}$ is a parabola, with its axis of symmetry at $x=\frac{a}{2}$.
When $-10 \Leftrightarrow 0<a<2$;
When $a \geqslant 2$, $f(x)$ is a decreasing function in the interval $(-1,1)$,
The problem statement $\Leftrightarrow f(1)=1-\frac{a}{2}... | (0,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,654 |
4. The sequence $\left\{x_{n}\right\}$ is defined as follows:
$$
x_{1}=\frac{2}{3}, x_{n+1}=\frac{x_{n}}{2(2 n+1) x_{n}+1}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then $x_{1}+x_{2}+\cdots+x_{2014}=$ | 4. $\frac{4028}{4029}$.
By definition, $\frac{1}{x_{n+1}}=2(2 n+1)+\frac{1}{x_{n}}$.
Let $y_{n}=\frac{1}{x_{n}}$.
Thus, $y_{1}=\frac{3}{2}$, and $y_{n+1}-y_{n}=2(2 n+1)$.
$$
\begin{array}{l}
\text { Hence } y_{n}=\left(y_{n}-y_{n-1}\right)+\left(y_{n-1}-y_{n-2}\right)+\cdots+ \\
\left(y_{2}-y_{1}\right)+y_{1} \\
=2[(... | \frac{4028}{4029} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,655 |
6. Let $a_{1}, a_{2}, \cdots, a_{2014}$ be a permutation of the positive integers $1,2, \cdots$, 2014. Denote
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, 2014) \text {. }
$$
Then the maximum number of odd numbers in $S_{1}, S_{2}, \cdots, S_{2014}$ is $\qquad$ | 6.1511.
If $a_{i}(2 \leqslant i \leqslant 2014)$ is odd, then $S_{i}$ and $S_{i-1}$ have different parities. From $a_{2}$ to $a_{2014}$, there are at least $1007-1=1006$ odd numbers. Therefore, $S_{1}, S_{2}, \cdots, S_{2014}$ must change parity at least 1006 times.
Thus, $S_{1}, S_{2}, \cdots, S_{2014}$ must have at... | 1511 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,656 |
8. Print 90000 five-digit numbers
$$
10000,10001, \cdots, 99999
$$
on cards, with one five-digit number on each card. Some cards (such as 19806, which reads 90861 when flipped) have numbers that can be read in two different ways, causing confusion. The number of cards that will not cause confusion is $\qquad$ cards. | 8. 88060 .
Among the ten digits $0 \sim 9$, the digits that can still represent numbers when inverted are $0, 1, 6, 8, 9$.
Since the first digit cannot be 0 and the last digit cannot be 0, the number of such five-digit numbers that can be read when inverted is $4 \times 5 \times 5 \times 5 \times 4=2000$.
Among these... | 88060 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,657 |
1. Given
$$
\begin{array}{l}
A \cup B \cup C=\{a, b, c, d, e, f\}, \\
A \cap B=\{a, b, c, d\}, c \in A \cap B \cap C .
\end{array}
$$
Then the number of sets $\{A, B, C\}$ that satisfy the conditions is (
) groups $\square$
(A) 100
(B) 140
(C) 180
(D) 200 | - 1. D.
As shown in Figure 3, the set $A \cup B \cup C$ can be divided into seven mutually disjoint regions, denoted as $I_{1}, I_{2}, \cdots, I_{7}$.
Given that $c \in I_{7}$, the elements $a, b, d$ belong to either $I_{4}$ or $I_{7}$, with a total of 8 possible combinations. The elements $e, f$ belong to one of the... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,658 |
2. Given the set
$$
S_{1}=\left\{(x, y) \| \log _{2}\left(1+x^{2}+y^{2}\right) \leqslant 1+\log _{2}(x+y)\right\}
$$
and the set
$$
S_{2}=\left\{(x, y) \log _{\frac{1}{2}}\left(2+x^{2}+y^{2}\right) \geqslant-2+\log _{\frac{1}{2}}(x+y)\right\} \text {. }
$$
Then the ratio of the area of $S_{2}$ to the area of $S_{1}$ ... | 2. C.
From the problem, we know that $S_{1}$ represents the interior of the circle (including the circumference)
$$
(x-1)^{2}+(y-1)^{2} \leqslant 1 \text {, }
$$
with an area of $\pi$;
$S_{2}$ represents the interior of the circle (including the circumference)
$$
(x-2)^{2}+(y-2)^{2} \leqslant 6 \text {, }
$$
with an... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,659 |
3. Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the angles opposite to $a, b, c$ are $\angle A, \angle B, \angle C$ respectively. Then the range of $\sin B+\cos B$ is ( ).
(A) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(B) $(1, \sqrt{2}]$
(C) $\left(1,1+\frac{\sqrt{3}}{... | 3. B.
From the known information and the cosine rule, we have
$$
\begin{array}{l}
a c=b^{2}=a^{2}+c^{2}-2 a c \cos B \\
\geqslant 2 a c-2 a c \cos B .
\end{array}
$$
Then $\cos B \geqslant \frac{1}{2}$.
Therefore, $0<\angle B \leqslant \frac{\pi}{3}$.
Hence, $\sin B+\cos B$
$$
=\sqrt{2} \sin \left(B+\frac{\pi}{4}\rig... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,660 |
6. Given that line $l$ intersects sphere $O$ at exactly one point $P$, and two half-planes $\alpha$ and $\beta$ starting from line $l$ intersect sphere $O$ to form two circular sections with radii $1$ and $\sqrt{3}$, respectively. The dihedral angle $\alpha-l-\beta$ has a plane angle of $\frac{5 \pi}{6}$. Then the radi... | 6. B.
From the problem and the symmetry of the sphere, we know that the cross-sectional diagram is as shown in Figure 4.
Let $O_{1} P$ and $O_{2} P$ be the radii of the two cross-sections, with $O_{1} P=1, O_{2} P=\sqrt{3}$, $O_{1} O=d_{1}, O_{2} O=d_{2}$, and the radius of the sphere be $R$.
Then, from the problem a... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,661 |
7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$ | $=, 7 .-1$. Therefore $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,662 |
8. The solution set of the inequality
$$
\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9
$$
is . $\qquad$ | 8. $\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{45}{8}\right)$.
Obviously, $x \geqslant-\frac{1}{2}, x \neq 0$.
The original inequality simplifies to
$$
x^{2}(8 x-45)<0 \Rightarrow x<\frac{45}{8} \text {. }
$$
Therefore, $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$. | \left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{45}{8}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,663 |
9. Two people, A and B, are playing a table tennis match, agreeing that the winner of each game gets 1 point, and the loser gets 0 points. The match stops when one person is 2 points ahead or after 6 games have been played. Suppose the probability of A winning each game is $\frac{2}{3}$, and the probability of B winnin... | 9. $\frac{266}{81}$.
According to the problem, all possible values of $\xi$ are $2, 4, 6$. If we consider every two games as a round, then the probability that the match ends at the end of this round is $\left(\frac{2}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}=\frac{5}{9}$.
If the match is to continue after this rou... | \frac{266}{81} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,664 |
10. As shown in Figure 1, in $\triangle A B C$, it is known that
$$
\begin{array}{l}
\cos C=\frac{2 \sqrt{5}}{5}, \\
\overrightarrow{A H} \cdot \overrightarrow{B C}=0, \\
\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 .
\end{array}
$$
Then the eccentricity of the hyperbola passing through poin... | 10. $2+\sqrt{5}$.
Take the midpoint $D$ of side $AB$. Then $\overrightarrow{CA}+\overrightarrow{CB}=2\overrightarrow{AD}$.
From $\overrightarrow{AB} \cdot (\overrightarrow{CA}+\overrightarrow{CB})=0$
$$
\Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AD}=0 \Rightarrow AB \perp AD.
$$
Thus, the altitude from $C$... | 2+\sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,665 |
11. Let $A$ be a set composed of any 100 distinct positive integers, and let
$$
B=\left\{\left.\frac{a}{b} \right\rvert\, a 、 b \in A \text { and } a \neq b\right\},
$$
$f(A)$ denotes the number of elements in set $B$. Then the sum of the maximum and minimum values of $f(A)$ is $\qquad$ . | 11. 10098.
From the problem, when the elements in set $B$ are pairwise coprime, $f(A)$ reaches its maximum value $\mathrm{A}_{100}^{2}=9900$; when the elements in set $B$ form a geometric sequence with a common ratio not equal to 1, $f(A)$ reaches its minimum value of $99 \times 2=198$.
Therefore, the sum of the maxim... | 10098 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,666 |
13. (20 points) The domain of the function $f(x)$ is the set of real numbers $\mathbf{R}$. It is known that when $x>0$, $f(x)>0$, and for any $m, n \in \mathbf{R}$, we have
$$
f(m+n)=f(m)+f(n).
$$
(1) Discuss the odd/even nature and monotonicity of the function $f(x)$;
(2) Let the sets be
$$
\begin{array}{l}
A=\left\{(... | (1) From the given, let $m=n=0$, we get
$$
f(0)=0 \text {; }
$$
Then let $m=x, n=-x$, we get
$$
f(-x)=-f(x) .
$$
Therefore, $f(x)$ is an odd function on $\mathbf{R}$.
Let $x_{1}, x_{2} \in \mathbf{R}$, and $x_{2}>x_{1}$. Then $x_{2}-x_{1}>0$. Since $f(x)$ is an odd function and from the given, we have
$$
\begin{array... | \left[-\frac{13}{6},-\frac{\sqrt{15}}{3}\right] \cup\left[\frac{\sqrt{15}}{3}, 2\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,667 |
2. The function $f(x)=x+\frac{1}{(x+1)^{3}}+1(x>0)$. Then, the value of $x$ when the function reaches its minimum is
保留了源文本的换行和格式。 | 2. $\sqrt[4]{3}-1$.
Notice that,
$$
\begin{array}{l}
f(x)=x+\frac{1}{(x+1)^{3}}+1 \\
=\frac{1}{3}(x+1)+\frac{1}{3}(x+1)+\frac{1}{3}(x+1)+\frac{1}{(x+1)^{3}} \\
\geqslant 4 \sqrt[4]{\left(\frac{1}{3}\right)^{3}} .
\end{array}
$$
Equality holds if and only if $\frac{1}{3}(x+1)=\frac{1}{(x+1)^{3}}$, i.e., $x=$ $\sqrt[4]... | \sqrt[4]{3}-1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,668 |
2. As shown in Figure 8, given that the excircle of $\triangle A B C$ opposite to $B$ is $\odot O_{B}$ and the excircle opposite to $C$ is $\odot O_{C}$, which are tangent to line $B C$ at points $E$ and $D$ respectively. The incircle of $\triangle A B C$ is $\odot I$, which is tangent to $A C$ and $A B$ at points $G$ ... | ```
By Menelaus' theorem, we have
\[
\begin{array}{l}
\frac{A P}{P T} \cdot \frac{T D}{D B} \cdot \frac{B F}{F A}=1, \frac{A P}{P T} \cdot \frac{T E}{E C} \cdot \frac{C G}{A G}=1 \\
\Rightarrow \frac{E H}{D H}=\frac{D T}{T E}=\frac{C G}{B F}=\frac{A O_{c}}{A O_{B}} \\
\Rightarrow O_{c} D \parallel A H \parallel O_{B} E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,670 |
II. (16 points) Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms as $S_{n}, a_{1} \neq 0, 2 S_{n+1}-3 S_{n}=2 a_{1}\left(n \in \mathbf{Z}_{+}\right)$.
(1) Prove: The sequence $\left\{a_{n}\right\}$ is a geometric sequence;
(2) If $a_{1}$ and $a_{n}(p \geqslant 3)$ are both positive integers, ... | (1) From the problem, we get $2 a_{2}-3 a_{1}=0$.
Also, $a_{1} \neq 0$, so $\frac{a_{2}}{a_{1}}=\frac{3}{2}$.
From $2 S_{n+1}-3 S_{n}=2 a_{1}$,
$2 S_{n+2}-3 S_{n+1}=2 a_{1}$,
we get $2 a_{n+2}-3 a_{n+1}=0\left(n \in \mathbf{Z}_{+}\right)$.
Since $a_{1} \neq 0$, we know $a_{n+1} \neq 0$.
Thus, $\frac{a_{n+2}}{a_{n+1}}=\... | a_{n}=2^{p-1}\left(\frac{3}{2}\right)^{n-1} | Algebra | proof | Yes | Yes | cn_contest | false | 728,671 |
一、(40 points) As shown in Figure 1, given that $\odot O_{1}$ is externally tangent to $\odot O$ at point $P$, $\odot O_{2}$ is internally tangent to $\odot O$ at point $Q$, line segment $O_{1} O_{2}$ intersects $P Q$ at point $G$, and a tangent line $G N$ is drawn from $G$ to $\odot O_{1}$, with the point of tangency b... | As shown in Figure 2, draw the tangent line $G M$ of $\odot O_{2}$ through point $G$, with the point of tangency being $M$. Connect $O_{1} O$, $O Q$, $O_{1} N$, and $O_{2} M$. Then point $P$ lies on $O_{1} O$.
Draw $O_{2} H \parallel P Q$ through point $O_{2}$, intersecting $O O_{1}$ at point $H$. Let the radii of $\od... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,672 |
1. Let $f(x)=\mathrm{e}^{2 x}-1, g(x)=\ln (x+1)$. Then the solution set of the inequality $f(g(x))-g(f(x)) \leqslant 1$ is | $$
-1 .(-1,1] \text {. }
$$
Notice,
$$
f(g(x))=x^{2}+2 x, g(f(x))=2 x \text {. }
$$
Therefore, $f(g(x))-g(f(x))=x^{2}$.
The domain is defined as $(-1,+\infty)$, thus, the solution set of the inequality is $(-1,1]$. | (-1,1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,673 |
4. Bivariate function
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{\cos 4 x+7}+\sqrt{\cos 4 y+7}+ \\
\quad \sqrt{\cos 4 x+\cos 4 y-8 \sin ^{2} x \cdot \sin ^{2} y+6}
\end{array}
$$
The maximum value of the function is | $4.6 \sqrt{2}$.
Let $\cos ^{2} x=a, \cos ^{2} y=b$.
Then $0 \leqslant a, b \leqslant 1$,
$$
\begin{aligned}
f= & 2 \sqrt{2}\left(\sqrt{a^{2}-a+1}+\sqrt{b^{2}-b+1}+\right. \\
& \left.\sqrt{a^{2}-a b+b^{2}}\right) .
\end{aligned}
$$
From $0 \leqslant a \leqslant 1 \Rightarrow a^{2} \leqslant a \Rightarrow \sqrt{a^{2}-a+... | 6 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,675 |
5. Label the 2014 small squares of a $1 \times 2014$ grid from left to right as $1,2, \cdots, 2014$. Now, use three colors $g, r, y$ to color each small square, such that even-numbered squares can be colored with any of $g, r, y$, while odd-numbered squares can only be colored with $g, y$, and adjacent squares must be ... | 5. $S_{2014}=4 \times 3^{1006}$.
Let $g_{n} 、 r_{n} 、 y_{n}$ represent the number of ways to color the $n$-th cell in a $1 \times n$ grid with colors $g 、 r 、 y$, respectively.
Then $\left\{\begin{array}{l}g_{1}=g_{2}=1, \\ r_{1}=0, r_{2}=2, \\ y_{1}=y_{2}=1,\end{array}\right.$
and $\square$
$\left\{\begin{array}{l}g... | 4 \times 3^{1006} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,676 |
6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4028} x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4026}-a_{4007}-a_{4028}
$$
is $\qquad$ | 6.2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$.
Then $x^{2}+x+2=1$.
Therefore, $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{0}+a_{1} \omega... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,677 |
8. Let
$$
\begin{array}{l}
A=\{1,2, \cdots, 2014\}, \\
B_{i}=\left\{x_{i}, y_{i}\right\}(i=1,2, \cdots, t)
\end{array}
$$
be $t$ pairwise disjoint binary subsets of $A$, and satisfy the conditions
$$
\begin{array}{l}
x_{i}+y_{i} \leqslant 2014(i=1,2, \cdots, t), \\
x_{i}+y_{i} \neq x_{j}+y_{j}(1 \leqslant i<j \leqslan... | 8. 805 .
On the one hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\left\{x_{i}, y_{i}\right\} \cap\left\{x_{j}, y_{j}\right\}=\varnothing \text {. }
$$
Thus, $x_{1}, x_{2}, \cdots, x_{t}, y_{1}, y_{2}, \cdots, y_{t}$ are $2 t$ distinct integers. Therefore,
$$
\sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \geqslant... | 805 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,678 |
Example 1 Let $k$ be a given positive integer, $r=k+\frac{1}{2}$.
$$
\begin{array}{l}
\text { Let } f^{(1)}(x)=f(x)=x\lceil x\rceil, \\
f^{(l)}(x)=f\left(f^{(l-1)}(x)\right), x \in \mathbf{R}_{+},
\end{array}
$$
where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
Prove: There exists ... | 【Analysis】First, try applying rule $f$ to $r$ a few times:
$$
f(r)=(k+1)\left(k+\frac{1}{2}\right)=k^{2}+k+\frac{k+1}{2} \text {. }
$$
Obviously, if $v_{2}(k)=0$, then $f(r) \in \mathbf{Z}$.
If $k \equiv 0(\bmod 2)$, continue applying $f$ to get
$$
\begin{array}{l}
f^{(2)}(r)=\left(k^{2}+k+\frac{k}{2}+\frac{1}{2}\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,679 |
One. (40 points) In $\triangle ABC$, $AD$, $BE$, and $CF$ are the altitudes to sides $BC$, $CA$, and $AB$ respectively. A circle $\Gamma$ with diameter $AD$ intersects $AC$ and $AB$ at points $M$ and $N$ respectively. Tangents to circle $\Gamma$ at points $M$ and $N$ intersect at point $P$. $O$ is the circumcenter of $... | Let the intersection of $A Q$ and $E F$ be $H$, and $P N, P M$ intersect $B C$ at points $T, S$ respectively. Connect $D E, D F, D H$.
Notice,
$$
\begin{array}{l}
\angle D E F=\angle B E F+\angle B E D \\
=180^{\circ}-2 \angle A B C=\angle B T N=\angle P T S .
\end{array}
$$
Similarly, $\angle D F E=\angle P S T$.
The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,680 |
$$
\begin{array}{l}
\text { Three. (50 points) Let } \\
D=\left\{(i, j) \in \mathbf{Z}^{2} \mid 1 \leqslant i, j \leqslant n\right\} .
\end{array}
$$
Let $[x]$ denote the greatest integer not exceeding the real number $x$.
Prove: there exists $S \subset D$, satisfying
$$
|S| \geqslant\left[\frac{3}{5} n(n+1)\right],
$... | Three, take $S=\{(i, j) \mid k \leqslant i+j \leqslant 2 k-1\}$, where $k \in \mathbf{Z}_{+}, \frac{n}{2}2 k-1$.
Thus, $\left(x_{1}+x_{2}, y_{1}+y_{2}\right) \notin S$.
Now we estimate $|S|$.
Since $i+j \leqslant 2 k-1$, we have
$$
|S| \leqslant \sum_{i=0}^{n} \min \{n, 2 k-1-i\} \leqslant \sum_{i=0}^{n} \min \{n, 2 k-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,681 |
For prime $p$, define the sets
$$
\begin{array}{l}
S_{1}(p)=\left\{(a, b, c) \in \mathbf{Z}^{3} \mid a^{2} b^{2}+b^{2} c^{2}+\right. \\
\left.c^{2} a^{2}+1 \equiv 0(\bmod p)\right\}
\end{array}
$$
and $S_{2}(p)=\left\{(a, b, c) \in \mathbf{Z}^{3} \mid a^{2} b^{2} c^{2}\left(a^{2} b^{2} c^{2}+\right.\right.$
$$
\left.... | All prime numbers $p$ that satisfy the conditions are $2, 3, 5, 13, 17$.
(1) First, verify that when $p=2, 3, 5, 13, 17$, the conditions are met.
(i) When $p=2$, for any $(a, b, c) \in S_{1}(p)$, $a, b, c$ are all odd or two are odd and one is even. In this case,
$(a, b, c) \in S_{2}(p)$.
Thus, $S_{1}(p) \subseteq S_{2... | 2, 3, 5, 13, 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,682 |
Example 2 Proof: For any integer $n \geqslant 4$, there exists an $n$-degree polynomial
$$
f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0},
$$
with the following properties:
(1) $a_{0}, a_{1}, \cdots, a_{n-1}$ are all positive integers;
(2) For any positive integer $m$, and any $k(k \geqslant 2)$ distinct positive int... | 【Analysis】Starting from the consideration of satisfying condition (2), to meet this inequality, many variables are involved, a natural idea is to consider from the perspective of number theory. Starting from the simple, if there is a prime $p$, such that
$$
f(x) \equiv t(\bmod p)(x, t \in \mathbf{N}),
$$
then the righ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,683 |
Theorem If the three points $X, Y, Z$ on the sides $BC, AC, AB$ (or their extensions) of $\triangle ABC$ are collinear, then
$$
\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB} = 1 .{ }^{[4]}
$$
Proof Draw $AD \parallel BC$ through point $A$, intersecting the line $XYZ$ at point $D$.
By $\frac{CY}{YA} = \frac{XC}... | Proof: Let the extension of $X Y$ intersect $A B$ at point $Z^{\prime}$. By Menelaus' theorem, we have
$$
\frac{B X}{X C} \cdot \frac{C Y}{Y A} \cdot \frac{A Z^{\prime}}{Z^{\prime} B}=1 .
$$
Combining this with $\frac{B X}{X C} \cdot \frac{C Y}{Y A} \cdot \frac{A Z}{Z B}=1$, we know $\frac{A Z^{\prime}}{Z^{\prime} B}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,684 |
6. Given a positive integer $n$. Find the largest real number $\mu$ such that: for any set $C$ of $4 n$ points in the "open" unit square $U$, there exists an "open" rectangle $V$ in $U$ satisfying the following properties:
(1) The sides of the open rectangle $V$ are parallel to the sides of $U$;
(2) The open rectangle ... | 6. $\mu_{\max }=\frac{1}{2 n+2}$.
Without loss of generality, let $U$ be the set of points in the Cartesian coordinate system $\left\{\begin{array}{l}01$, otherwise swap the horizontal and vertical coordinates.
First, we provide a lemma.
Lemma If segment $A_{0} A_{m}$ has $m(m \geqslant 2)$ points inside, denoted as $... | \frac{1}{2 n+2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,686 |
2. A line $l$ is drawn through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersecting the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly three lines $l$, then $\lambda=$ $\qquad$. | 2.4.
Since there are an odd number of lines that satisfy the condition, by symmetry, the line perpendicular to the $x$-axis satisfies the condition, at this time,
$$
x=\sqrt{3}, y= \pm 2, \lambda=|A B|=4 \text {. }
$$ | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,687 |
4. Given the set $A=\{1,2,3\}, f$ and $g$ are functions from set $A$ to $A$. Then the number of function pairs $(f, g)$ whose image sets intersect at the empty set is $\qquad$ . | 4. 42 .
When the image set of function $f$ is 1 element, if the image set of function $f$ is $\{1\}$, at this time the image set of function $g$ is a subset of $\{2,3\}$, there are $2^{3}=8$ kinds, so there are $3 \times 8=24$ pairs of functions $(f, g)$ that meet the requirements.
When the image set of function $f$ ... | 42 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,688 |
5. Given $f(x)=\left(x^{2}+3 x+2\right)^{\cos \pi x}$. Then the sum of all $n$ that satisfy the equation
$$
\left|\sum_{k=1}^{n} \log _{10} f(k)\right|=1
$$
is | 5. 21 .
It is known that for integer $x$, we have
$$
f(x)=[(x+1)(x+2)]^{(-1)^{x}} \text {. }
$$
Thus, when $n$ is odd,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2-\log _{10}(n+2) \text {; }
$$
When $n$ is even,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2+\log _{10}(n+2) \text {. }
$$
Therefore, the $n$ that... | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,689 |
7. Given that two people, A and B, are playing a game, the probability of A winning is $\frac{2}{3}$, and the probability of B winning is $\frac{1}{3}$. If one of them wins two more games than the other, the game ends. Then, the expected number of games that need to be played is $\qquad$ | 7. $\frac{18}{5}$.
Let the required mathematical expectation be $\mathrm{E} \xi$. Note that, the probability of the game ending in two rounds is equal to
$$
\left(\frac{2}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}=\frac{5}{9} \text {. }
$$
If the game does not end in two rounds, then exactly one round must have been... | \frac{18}{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,690 |
10. (20 points) Given real numbers $a_{0}, a_{1}, \cdots, a_{2015}$, $b_{0}, b_{1}, \cdots, b_{2011}$ satisfy
$$
\begin{array}{l}
a_{n}=\frac{1}{65} \sqrt{2 n+2}+a_{n-1}, \\
b_{n}=\frac{1}{1009} \sqrt{2 n+2}-b_{n-1},
\end{array}
$$
where $n=1,2, \cdots, 2015$.
If $a_{0}=b_{2015}$, and $b_{0}=a_{2015}$, find the value ... | 10. Notice that, for any $k=1,2 \cdots, 2015$, we have
$$
\begin{array}{l}
a_{k}-a_{k-1}=\frac{1}{65} \sqrt{2 k+2}, \\
b_{k}+b_{k-1}=\frac{1}{1009} \sqrt{2 k+2} .
\end{array}
$$
Multiplying the two equations, we get
$$
\begin{array}{l}
a_{k} b_{k}-a_{k-1} b_{k-1}+a_{k} b_{k-1}-a_{k-1} b_{k} \\
=\frac{2 k+2}{65 \times ... | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,691 |
一、(40 points) As shown in Figure 1, the incircle of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. $AD$ intersects $BE$ at point $P$. Let the reflections of point $P$ over lines $EF$, $FD$, and $DE$ be $X$, $Y$, and $Z$ respectively. Prove that lines $AX$, $BY$, and $CZ$ ar... | Let the intersection of line $A X$ and $E F$ be $M$, the intersection of line $B Y$ and $F D$ be $N$, the intersection of line $C Z$ and $D E$ be $L$, the intersection of line $P A$ and $E F$ be $R$, the intersection of line $P B$ and $F D$ be $S$, and the intersection of line $P C$ and $D E$ be $T$.
Since the pencil ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,692 |
```
In an acute triangle $\triangle A B C$, prove:
\[
\begin{array}{l}
(\sin A+\sin B+\sin C)\left(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}\right) \\
\leqslant \pi\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right) .
\end{array}
\]
``` | Let's assume $A \geqslant B \geqslant C$.
From $A+B+C=\pi$, we know that equation (1) is equivalent to
$$
\begin{array}{l}
\left(\sqrt{\frac{A}{B}}-\sqrt{\frac{B}{A}}\right)^{2}+\left(\sqrt{\frac{B}{C}}-\sqrt{\frac{C}{B}}\right)^{2}+\left(\sqrt{\frac{C}{A}}-\sqrt{\frac{A}{C}}\right)^{2} \\
\geqslant\left(\sqrt{\frac{\s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,693 |
Given the equation $x^{2}-x-1=0$ with roots $x_{1}$ and $x_{2}$, let $a_{n}=\frac{x_{1}^{n}-x_{2}^{n}}{x_{1}-x_{2}}\left(n \in \mathbf{Z}_{+}\right)$. Prove:
(1) For any positive integer $n$, we have
$$
a_{n+2}=a_{n+1}+a_{n} \text {; }
$$
(2) For any positive integer $m(m \geqslant 2)$, there exists
a positive integer... | (1) Note that for the roots $x_{i}$ $(i=1,2)$ of the known equation, we have $x_{i}^{2}=x_{i}+1$. Therefore,
$$
\begin{array}{l}
a_{n+1}+a_{n}=\frac{x_{1}^{n+1}-x_{2}^{n+1}}{x_{1}-x_{2}}+\frac{x_{1}^{n}-x_{2}^{n}}{x_{1}-x_{2}} \\
=\frac{x_{1}^{n}\left(x_{1}+1\right)-x_{2}^{n}\left(x_{2}+1\right)}{x_{1}-x_{2}} \\
=\frac... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,694 |
431 There is a card game called "Twelve Months" that is only played during the Spring Festival. The rules are as follows:
First, take a brand new deck of cards, remove the big joker, the small joker, and the 4 Ks, leaving 48 cards, which are then shuffled.
Second, lay out the shuffled cards face down in 12 columns, w... | Solve the generalization, construct an $m \times n$ number table, where $a_{ij}$ represents the number in the $i$-th row and $j$-th column. Extract some or all of these numbers according to a predetermined rule to form a sequence $\left\{b_{n}\right\}$. The rule is as follows:
(i) First, extract $a_{11}$, and let $b_{1... | \frac{1}{12} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,695 |
Example 7 Find all positive integers $n$ such that there exist two complete residue systems modulo $n$, $a_{i}$ and $b_{i} (1 \leqslant i \leqslant n)$, for which $a_{i} b_{i} (1 \leqslant i \leqslant n)$ is also a complete residue system modulo $n$. ${ }^{[2]}$ | 【Analysis and Solution】When $n=1,2$, let $a_{i}=b_{i}=i$ $(1 \leqslant i \leqslant n)$, so $n=1,2$ satisfy the condition.
The following proves that $n \geqslant 3$ does not satisfy the condition.
(1) An odd prime $p$ does not satisfy the condition.
In fact, suppose $a_{i} 、 b_{i} 、 a_{i} b_{i}(1 \leqslant i \leqslant p... | n=1,2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,696 |
1. Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $1,2, \cdots, n$. It is known that $0, a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{n}$ leave distinct remainders when divided by $n+1$. Find all possible values of the positive integer $n$.
(45th Canadian Mathematical Olympiad) | Prompt: In $0, a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{n}$, any two numbers are not congruent modulo $n+1$, especially,
$$
a_{1}+a_{2}+\cdots+a_{n}=\frac{n(n+1)}{2} \neq 0(\bmod n+1) \text {. }
$$
Therefore, $n$ is odd.
For $n=2 k-1$, construct an example,
$$
\begin{array}{l}
a_{2 t-1}=2 t-1(1 \leqslant t \l... | n \text{ is odd} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,697 |
2. Let $P_{1}, P_{2}, \cdots, P_{2 n}$ be a permutation of the vertices of a regular $2 n$-gon. Prove: In every closed broken line $P_{1} P_{2}, P_{2} P_{3}, \cdots, P_{2 n-1} P_{2 n}, P_{2 n} P_{1}$, there is at least one pair of parallel segments.
(2011, Croatian National Mathematical Competition) | Hint: Label the vertices of a regular $2 n$-gon in sequence as 1, $2, \cdots, 2 n$, and let $a_{k}$ represent the number corresponding to vertex $P_{k}$. Then
$P_{i} P_{i+1} / / P_{j} P_{j+1}$
$\Leftrightarrow a_{i}+a_{i+1} \equiv a_{j}+a_{j+1}(\bmod 2 n)$.
Agree that $P_{2 n+1}=P_{1}$.
If $P_{i} P_{i+1}(1 \leqslant i ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,698 |
6. In an All-Star basketball game, 27 players participate, each wearing a jersey with their favorite number, which is a non-negative integer. After the game, they line up in a 3-row, 9-column formation for fans to take photos. An eccentric fan only takes photos where the players in the frame form a rectangle of $a$ row... | Prompt: $s_{\text {min }}=2$.
Assume at most one player is photographed.
By symmetry, without loss of generality, assume no players in the first row are photographed.
For $i=1,2, \cdots, 9$, let the players in the $i$-th column of the 1st, 2nd, and 3rd rows have jersey numbers $a_{i}, b_{i}, c_{i}$, respectively, and ... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,699 |
Example 4 If
$$
\cos ^{5} \theta-\sin ^{5} \theta<7\left(\sin ^{3} \theta-\cos ^{3} \theta\right)(\theta \in[0,2 \pi)),
$$
then the range of values for $\theta$ is $\qquad$ [3]
(2011, National High School Mathematics Joint Competition) | 【Analysis】The inequality has different degrees on both sides, unlike Example 3 where the function to be constructed can be clearly seen. However, a simple transformation of the inequality will reveal the solution.
The transformation is as follows:
$$
\begin{array}{l}
\cos ^{5} \theta-\sin ^{5} \theta \cos ^{3} \theta+\... | \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,700 |
Example 6 Let $f(x)$ be a function defined on $\mathbf{R}$. If $f(0)=2008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+2)-f(x) \leqslant 3 \times 2^{x}, \\
f(x+6)-f(x) \geqslant 63 \times 2^{x},
\end{array}
$$
then $f(2008)=$ $\qquad$
$(2008$, National High School Mathematics League Competiti... | 【Analysis】The problem provides two inequalities about $f(x)$. The key to solving the problem lies in appropriately transforming these two inequalities. Since it is a value-finding problem, we should strive to convert the inequality relations into equality relations.
Transform the two inequalities as follows:
$$
\begin{... | 2^{2008} + 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,701 |
Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \). ${ }^{[4]}$
(2011, Hebei Province High School Mathematics Competition) | 【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.
First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,702 |
Example 8 Let real numbers $a, b, c$ satisfy $a+b+c=1$, $abc>0$. Prove:
$$
ab+bc+ca<\frac{\sqrt{abc}}{2}+\frac{1}{4}.
$$
(2014, National High School Mathematics Joint Competition) | 【Analysis】There are many ways to prove this problem. We might as well try constructing a function.
Notice that, $a b c>0$.
By the Pigeonhole Principle, we know that among $a, b, c$, there exist two numbers with the same sign, let's assume they are $b, c$. Hence, $b c>0$, and consequently, $a>0$.
Let $t=\sqrt{b c}>0$.
B... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,703 |
Let $A, B, D, E, F, C$ be six points on a circle in that order, satisfying $AB = AC$. Line $AD$ intersects $BE$ at point $P$, line $AF$ intersects $CE$ at point $R$, line $BF$ intersects $CD$ at point $Q$, line $AD$ intersects $BF$ at point $S$, and line $AF$ intersects $CD$ at point $T$. Point $K$ lies on segment $ST$... | Prove as shown in Figure 1, construct $TH \parallel KQ$, intersecting $BF$ at point $H$, and connect $RH$, $CF$, and $FD$.
By Pascal's theorem, points $P$, $Q$, and $R$ are collinear.
By $AB = AC \Rightarrow \angle AFB = \angle ADC$
$\Rightarrow S$, $D$, $F$, and $T$ are concyclic
$\Rightarrow \angle TSF = \angle CDF =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,704 |
Example 3 As shown in Figure $2, D$ is a point inside $\triangle A B C, B D$ intersects $A C$ and $C D$ intersects $A B$ at points $E, F$ respectively, and $A F=B F$ $=C D, C E=D E$. Find the degree measure of $\angle B F C$. ${ }^{[6]}$ (2003, Japan Mathematical Olympiad) | As shown in Figure 2, let $G$ be a point on segment $FC$ such that $FG = AF$, and connect $AG$.
Since $CE = DE$, we have
$\angle ECD = \angle EDC = \angle FDB$.
Also, $FG = AF = CD$, so
$$
GC = CD + DG = FG + GD = DF \text{. }
$$
Applying Menelaus' Theorem to $\triangle ABE$ and the transversal $FDC$, we get
$$
\frac{... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,705 |
Question: Let $n>2$ be an integer, and $P_{1}, P_{2}, \cdots, P_{n}$ be any $n$ distinct points on a plane. $S$ represents the set of all points on the segments $P_{1} P_{2}$, $P_{2} P_{3}$, $\cdots$, $P_{n-1} P_{n}$. Ask: For (1) $k=\frac{5}{2}$; (2) $k=3$, can we always find two points $A$ and $B$ in the set $S$ such... | (1) Let $P_{1} P_{2}, P_{2} P_{3}, \cdots, P_{n-1} P_{n}$ be either perpendicular or parallel, so that the range of $A B$ can be as small as possible.
When $n=10$, set
$\left|P_{1} P_{10}\right|=1,\left|P_{2} P_{3}\right|=\left|P_{8} P_{9}\right|=\frac{1}{2.6}$,
$\left|P_{4} P_{5}\right|=\left|P_{6} P_{7}\right|$.
As s... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,706 |
For $n \in \mathbf{Z}_{+}$, prove: any $k$-order linear recurrence sequence satisfying the linear recurrence relation
$$
a_{n+k}=c_{1} a_{n+k-1}+c_{2} a_{n+k-2}+\cdots+c_{k} a_{n}
$$
when each term is divided by any positive integer $m$, the remainders form a periodic sequence. | Proof Consider the $k$-tuple ordered array
$$
\left(a_{n}^{\prime}, a_{n+1}^{\prime}, \cdots, a_{n+k-1}^{\prime}\right)(n=1,2, \cdots),
$$
where, $a_{i}^{\prime} \equiv a_{i}(\bmod m)$, and
$$
a_{i}^{\prime} \in\{0,1, \cdots, m-1\}(i=1,2, \cdots) .
$$
Since there are at most $m^{k}$ such $k$-tuple arrays, there must ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,707 |
Example 3 Let the sequence of real numbers $\left\{a_{n}\right\}$ satisfy
$$
a_{n+2}=\left|a_{n+1}\right|-a_{n}(n=1,2, \cdots) \text {. }
$$
Prove: For each sufficiently large $m \in \mathbf{Z}_{+}$, we have
$$
a_{m+9}=a_{m} .
$$ | Obviously, $\left\{a_{n}\right\}$ cannot be all positive, nor can it be all negative.
Take $a_{n} \geqslant 0 \geqslant a_{n-1}(n \geqslant 3)$.
If $a_{n-2} \leqslant 0$, then
$$
\begin{array}{l}
a_{n+1}=a_{n}-a_{n-1} \geqslant 0, \\
a_{n+2}=-a_{n-1} \geqslant 0, \\
a_{n+3}=-a_{n} \geqslant 0, \\
a_{n+4}=a_{n}+a_{n-1}=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,708 |
Example 4 Given the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$
Determine the periodicity of the sequence $\left\{a_{n}\right\}$. | Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$
which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,709 |
Example 5 Given that $\left\{x_{n}\right\}$ is a bounded integer sequence, satisfying the recurrence relation
$$
x_{n+5}=\frac{5 x_{n+4}^{3}+x_{n+3}-3 x_{n+2}+x_{n}}{2 x_{n+2}+x_{n+1}^{2}+x_{n+1} x_{n}} .
$$
Prove: $\left\{x_{n}\right\}$ is eventually periodic.
Proof Since $\left\{x_{n}\right\}$ is bounded, there exis... | At most there are $a^{5}$.
Thus, in $a^{5}+1=k$ quintuples
\[
\begin{array}{l}
\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right),\left(x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right), \cdots, \\
\left(x_{k}, x_{k+1}, \cdots, x_{k+4}\right)
\end{array}
\]
there must be two that are the same. Let them be
\[
\left(x_{i}, x_{i+1}, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,710 |
Example 6 (1) If the $k$-order linear recursive sequence $\left\{a_{n}\right\}$ with constant coefficients has $k$ distinct characteristic roots $x_{1}, x_{2}, \cdots, x_{k}$, and there exist $T_{j} \in \mathbf{Z}_{+}(j=1,2, \cdots, k)$, such that $x_{j}^{T_{j}}=1$. Prove: $\left\{a_{n}\right\}$ is a purely periodic se... | (1) Let the general term formula of $\left\{a_{n}\right\}$ be
$$
a_{n}=\sum_{i=1}^{k} c_{i} x_{i}^{n} \text {. }
$$
Take $T=\left[T_{1}, T_{2}, \cdots, T_{k}\right]$.
It is easy to see that for all $n \in \mathbf{Z}_{+}$, we have $a_{n+T}=a_{n}$.
(2) The characteristic equation of $\left\{y_{n}\right\}$ is
$$
\begin{a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,711 |
2. As shown in Figure 1, in square $A B C D$, it is known that $E$ and $F$ are points on the extensions of $A B$ and $B C$ respectively, and $A E=E F+F C$. Then $\angle E D F=(\quad)$.
(A) $75^{\circ}$
(B) $60^{\circ}$
(C) $45^{\circ}$
(D) $30^{\circ}$ | 2. C.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,712 |
Example 4 A line intersects a triangle's three sides or their extensions at three points. Construct the symmetric points of these three points with respect to the midpoints of their respective sides. Prove: These three symmetric points lie on a straight line. ${ }^{[4]}$ | Proof As shown in Figure 3, let the line $D E F$ intersect the extensions of the sides $B C$, $A B$, and $C A$ of $\triangle A B C$ at points $D$, $E$, and $F$ respectively. By Menelaus' theorem, we have
\[
\frac{C D}{D B} \cdot \frac{B E}{E A} \cdot \frac{A F}{F C}=1.
\]
Let $D^{\prime}$, $E^{\prime}$, and $F^{\prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,713 |
6. In an opaque bag, there are several red, yellow, and blue glass balls that are identical except for their color. Among them, there are 6 red glass balls and 9 yellow glass balls. It is known that the probability of randomly drawing 1 blue glass ball from the bag is $\frac{2}{5}$. Then, the probability of randomly dr... | $$
\text { II. 6. } \frac{6}{25} \text {. }
$$
Let the number of blue glass balls in the bag be $x$.
$$
\text { Then } \frac{x}{15+x}=\frac{2}{5} \Rightarrow x=10 \text {. }
$$
Therefore, the required probability is $\frac{6}{6+9+10}=\frac{6}{25}$. | \frac{6}{25} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,715 |
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