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
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class | __index_level_0__ int64 0 742k |
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2. In isosceles trapezoid $ABCD$, $BC \parallel AD$, and $AB$ is not parallel to $CD$. The circle $\Gamma$ passing through points $B$ and $C$ intersects line segments $AB$ and $BD$ at points $X$ and $Y$, respectively. The tangent to circle $\Gamma$ at point $C$ intersects ray $AD$ at point $Z$. Prove that points $X$, $... | 2. From $B C / / A D, Z C$ being the tangent of circle $\Gamma$, we know $\angle A D B=\angle Y B C=\angle Y C Z$.
Therefore, $\angle Y D Z+\angle Y C Z=180^{\circ}$.
Thus, points $C, Y, D, Z$ are concyclic. See Figure 1.
Hence, $\angle C Y Z=\angle C D Z$.
Also, from the problem statement,
$$
\angle X B C=\angle B C D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,133 |
5. There are $n(n>3)$ distinct positive integers on the blackboard, all of which are less than $(n-1)!$. For any pair of numbers $a, b(a>b)$, Peter writes $\left[\frac{a}{b}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) in his notebook. Prove: There must be two equal numbers among... | 5. Proof by Contradiction.
Assume that all numbers written in the notebook are distinct.
Let the $n$ numbers on the blackboard be $a_{1}<a_{2}<\cdots<a_{n}$.
Let $q_{i}=\left[\frac{a_{i+1}}{a_{i}}\right](i=1,2, \cdots, n-1)$.
Then, $a_{i+1} \geqslant q_{i} a_{i}$, and all $q_{i}$ are distinct.
Thus, $q_{1} q_{2} \cdot... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,134 |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 1511
(B) 2015
(C) 22017
(D) 2018 | $\begin{array}{l}\text { I. 1.C. } \\ \text { Given } S=a+b+c=1014+b+497 \text {, and } a>b \\ \Rightarrow 1014=a+b \geqslant b+1+b \\ \Rightarrow b \leqslant 506.6 \Rightarrow b_{\max }=506 \\ \Rightarrow S_{\text {max }}=1014+506+497=2017 .\end{array}$ | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,135 |
2. As shown in Figure 1, in $\triangle A B C$, a semicircle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$, respectively. If $D E=E C=4, B C-B D$ $=\frac{16}{5}$, then the value of $\sqrt{\frac{B D-A D}{B C}}$ is ( ).
(A) $\frac{3}{5}$
(B) $\frac{2}{3}$
(C) $\frac{4}{5}$
(D) $\frac{5}{6}$ | 2. A.
Since $B C$ is the diameter of the semicircle, therefore,
$$
\begin{array}{l}
\angle B E C=90^{\circ} . \\
\text { By } D E=E C \Rightarrow \angle A B E=\angle C B E \\
\Rightarrow A B=B C, A E=E C=D E \\
\Rightarrow A D=A B-B D=B C-B D=\frac{16}{5} .
\end{array}
$$
$$
\begin{array}{l}
\text { By } A E=D E=E C=4... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,136 |
5. Figure 2 shows the inside of a sewage purification tower. Sewage enters from the top and flows over the surface of purification materials shaped like isosceles right triangles, as indicated by the arrows in the figure. Each time the water flows, it has an equal chance of flowing along either leg of the triangle. Aft... | 5. C.
According to the differences between the outlets shown in the diagram, we know (1) is incorrect; by symmetry, we know (2) is correct;
Based on the first outlet having a flow rate of $\frac{1}{16}$, the second outlet having a flow rate of $\frac{1}{4}$, and the third outlet having a flow rate of $\frac{3}{8}$, th... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,137 |
1. Let $a, b$ be positive integers, and $a b \mid\left(a^{2}+b^{2}\right)$. Prove: $a=b$. $\quad$. | Notice that,
$$
\begin{array}{l}
a b \mid\left(a^{2}+b^{2}\right) \\
\Leftrightarrow a^{2}+b^{2}=k a b\left(k \in \mathbf{Z}_{+}, k \geqslant 2\right),
\end{array}
$$
i.e., $a^{2}-(k b) a+b^{2}=0$.
If $a \neq b$, then $k>2$.
By symmetry, assume $a>b$, and $a+b$ is minimal. Then $x_{1}=a$ is a positive integer root of ... | a=b | Number Theory | proof | Yes | Yes | cn_contest | false | 730,138 |
4. The height of a ball thrown vertically upwards from the ground is a quadratic function of its motion time. Xiao Hong throws two balls vertically upwards, 1 second apart. Assuming the two balls are thrown from the same height above the ground, and they reach the same maximum height above the ground 1.1 seconds after ... | 4.1.6.
As shown in Figure 7, $AB=1$.
Let $C(1.1, h)$.
Then $D(2.1, h)$.
By symmetry, we know that the x-coordinate of point
$P$ is
$$
\begin{array}{l}
1.1+\frac{2.1-1.1}{2} \\
=1.6 .
\end{array}
$$ | 1.6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,140 |
One, (20 points) Let the quadratic function be $y=a x^{2}+b x+c$. The graph is above the line $y=-2 x$ if and only if $1<x<3$.
(1) If the equation $a x^{2}+b x+6 a+c=0$ has two equal real roots, find the expression of the quadratic function;
(2) If the maximum value of the quadratic function is positive, find the range... | (1) From the problem, we know that $x=1$ and 3 are the roots of the equation $a x^{2} +b x+c=-2 x$, and $a0$.
Also $a0$.
Then $(-4 a-2)^{2}-4 a(3 a)>0$
$$
\Rightarrow a<-2-\sqrt{3} \text { or }-2+\sqrt{3}<a<0 \text {. }
$$ | a<-2-\sqrt{3} \text { or }-2+\sqrt{3}<a<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,141 |
II. (25 points) Several boxes are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each box does not exceed 1 ton. To ensure that these boxes can be transported away in one go, the question is: what is the minimum number of trucks with a carrying capacity of 3 tons needed? | II. First, notice that the weight of each box does not exceed 1 ton. Therefore, the weight of boxes that each vehicle can transport at once will not be less than 2 tons. Otherwise, another box can be added.
Let $n$ be the number of vehicles needed, and the weights of the boxes transported by each vehicle be $a_{1}, a_... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 730,142 |
2. Let the complex number $z \neq 1, z^{11}=1$. Then $z+z^{3}+z^{4}+z^{5}+z^{9}=$ $\qquad$ | 2. $\frac{-1 \pm \sqrt{11} \mathrm{i}}{2}$.
Given $z \neq 1, z^{11}=1$
$$
\begin{array}{l}
\Rightarrow 0=\frac{z^{11}-1}{z-1}=1+z+z^{2}+\cdots+z^{10} \\
\Rightarrow z+z^{2}+\cdots+z^{10}=-1 .
\end{array}
$$
Let $x=z+z^{3}+z^{4}+z^{5}+z^{9}$.
$$
\begin{array}{c}
\text { Then } x^{2}=z^{2}+z^{6}+z^{8}+z^{10}+z^{18}+ \\... | \frac{-1 \pm \sqrt{11} \mathrm{i}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,143 |
6. Let $x, y \in \mathbf{R}_{+}$. Then the function
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{x^{2}-x y+y^{2}}+\sqrt{x^{2}-9 x+27}+\sqrt{y^{2}-15 y+75}
\end{array}
$$
has the minimum value of | 6. $7 \sqrt{3}$.
As shown in Figure 3.
Let $A B=3 \sqrt{3}, A C=5 \sqrt{3}, A M=x, A N=y$,
$$
\angle B A M=30^{\circ}, \angle M A N=60^{\circ}, \angle C A N=30^{\circ} \text {. }
$$
By the Law of Cosines, we get
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{x^{2}-x y+y^{2}}+\sqrt{x^{2}-9 x+27}+\sqrt{y^{2}-15 y+75} \\
=B M+M ... | 7 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,144 |
7. Let $D_{1}, D_{2}, \cdots, D_{2 n}$ be $2 n(n \geqslant 1)$ points on the hypotenuse $B C$ of the right triangle $\triangle A B C$. Denote
$$
\angle D_{i-1} A D_{i}=\alpha_{i}(i=1,2, \cdots, 2 n+1) \text {, }
$$
satisfying $D_{i-1} D_{i}=D_{i} D_{i+1}\left(i=1,2, \cdots, 2 n, D_{0}=B\right.$, $D_{2 n+1}=C$ ). Then
... | 7. $\frac{1}{2 n+1}$.
Since $B D_{1}=D_{1} D_{2}=\cdots=D_{2 n} C$, therefore, the areas of $\triangle B A D_{1}, \triangle D_{1} A D_{2}, \cdots, \triangle D_{2 n} A C$ are equal, each being $\frac{1}{2 n+1}$ of the area of $\triangle A B C$.
$$
\begin{array}{l}
\text { Hence } \prod_{i=0}^{n} S_{\triangle D_{2 i} A ... | \frac{1}{2 n+1} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,145 |
$\begin{array}{c}\text { 10. (20 points) Let } \\ f(x)=x^{5}-3 x^{3}+2 x^{2}+3 x+6 \text {, } \\ \text { and let } A_{n}=\prod_{k=1}^{n} \frac{(4 k-3) f(4 k-3)}{(4 k-1) f(4 k-1)} \text {. Find the value of } A_{25} \text { . }\end{array}$ | 10. Notice,
$$
\begin{array}{l}
x f(x)=x\left(x^{5}-3 x^{3}+2 x^{2}+3 x+6\right) \\
=x(x+2)\left(x^{2}+x+1\right)\left(x^{2}-3 x+3\right) \\
=\left(x^{3}+3 x^{2}+3 x+2\right)\left(x^{3}-3 x^{2}+3 x\right) \\
=\left((x+1)^{3}+1\right)\left((x-1)^{3}+1\right) \\
\Rightarrow A_{n}=\prod_{k=1}^{n} \frac{\left((4 k-2)^{3}+1... | \frac{1}{1000001} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,146 |
11. (20 points) Let the sequence $\left\{A_{n}\right\}$ satisfy
$$
A_{1}=1, A_{2}=3, A_{n}=4 A_{n-1}-A_{n-2} \text {. }
$$
Find the value of $\sum_{n=1}^{+\infty} \operatorname{arccot} 2 A_{n}^{2}$. | 11. From the given conditions, we have
$$
\begin{array}{l}
A_{3}=4 A_{2}-A_{1}=4 \times 3-1=11, \\
\frac{A_{0}}{A_{1}}=\frac{4 A_{1}-A_{2}}{A_{1}}=1 .
\end{array}
$$
Solving the characteristic equation \(x^{2}-4 x+1=0\), we get
$$
x_{1}=2+\sqrt{3}, x_{2}=2-\sqrt{3} \text {. }
$$
Thus, the general term formula is
$$
\... | \frac{\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,147 |
One. (40 points) Let $H$ be the orthocenter of acute $\triangle ABC$, and $E$, $F$ be the midpoints of sides $AB$, $AC$ respectively. $EP \perp HE$ intersects $AC$ at point $P$, $FQ \perp HF$ intersects $AB$ at point $Q$, and $Y$ is the midpoint of $PQ$. Prove: $2 \angle BAC + \angle EYF = 180^{\circ}$. | As shown in Figure 5, let $O$ be the circumcenter of $\triangle ABC$, and let $M$, $N$, $Z$, and $D$ be the midpoints of $BH$, $CH$, $OH$, and $BC$ respectively. Then $Z$ is the nine-point circle center of $\triangle ABC$. The auxiliary lines are as shown in the figure.
Then $E M \parallel A H \parallel F N, E M = F N,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,148 |
II. (40 points) Let $n \geqslant 3$ be a positive integer, and $\alpha_{i} \in \left(0, \arccos \frac{\sqrt{6}}{3}\right)$. If $\sum_{i=1}^{n} \cos 2 \alpha_{i}=n-2$, prove: $\sum_{i=1}^{n} \cot \alpha_{i} \geqslant(n-1) \sum_{i=1}^{n} \tan \alpha_{i}$. | ```
Because $\alpha_{i} \in\left(0, \arccos \frac{\sqrt{6}}{3}\right)$, so, $\tan \alpha_{i} \in\left(0, \frac{1}{\sqrt{2}}\right), \cot \alpha_{i} \in(\sqrt{2},+\infty)$. From $n-2=\sum_{i=1}^{n} \cos 2 \alpha_{i}=\sum_{i=1}^{n}\left(1-2 \sin ^{2} \alpha_{i}\right)$ $=n-2 \sum_{i=1}^{n} \frac{1}{\cot ^{2} \alpha_{i}+1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,149 |
Four. (50 points) Let $A=\{0,1, \cdots, 2016\}$. If a surjective function $f: \mathbf{N} \rightarrow A$ satisfies: for any $i \in \mathbf{N}$,
$$
f(i+2017)=f(i),
$$
then $f$ is called a "harmonious function".
$$
\begin{array}{l}
\text { Let } f^{(1)}(x)=f(x), \\
f^{(k+1)}(x)=f\left(f^{(k)}(x)\right)\left(k \in \mathbf... | On the one hand, note that 2017 is a prime number.
Let $g$ be a primitive root modulo 2017, then the half-order of $g$ modulo 2017 is 1008.
$$
\text{Let } f(i) \equiv g(i-1)+1(\bmod 2017) \text{.}
$$
Since $(g, 2017)=1$, $g(i-1)+1$ runs through a complete residue system modulo 2017.
Thus, the mapping $f: \mathbf{N} \r... | 1008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,150 |
Given positive integers $k, n, k \leqslant n-1, n \neq 2 k$. Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $\{1,2, \cdots, n\}$. Define
$$
S_{i}=a_{i}+a_{i+1}+\cdots+a_{i+k-1},
$$
where $1 \leqslant i \leqslant n$, and the indices are considered modulo $n$.
Let $S=\min \left\{S_{1}, S_{2}, \cdots, S_{n}\right\... | Proof: Assume $S \geqslant\left[\frac{k(n+1)}{2}\right]$.
By $k \leqslant n-1 \Rightarrow a_{1} \neq a_{k+1} \Rightarrow S_{1} \neq S_{2}$
$\Rightarrow$ There must be $S_{1}>S$ or $S_{2}>S$
$$
\Rightarrow n S<\sum_{i=1}^{n} S_{i}=k \sum_{i=1}^{n} a_{i}=\frac{k n(n+1)}{2} \text {. }
$$
When $k$ is even or $n$ is odd,
$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,151 |
Given 544 As shown in Figure $2, \odot O_{1}$ and $\odot O_{2}$ are externally tangent, $P$ is a point on $\odot O_{1}$, $P A$ and $P B$ are tangent to $\odot O_{2}$ at points $A$ and $B$ respectively, $M$ is the midpoint of $A B$, $O_{1} C \perp P A$, intersecting $\odot O_{1}$ at point $C$, and $P B$ intersects $\odo... | Proof: Let $\odot O_{1}$ and $\odot O_{2}$ be externally tangent at point $S$. Then $O_{1}, S, O_{2}, P, M, O_{2}$ are collinear.
From $O_{1} C \perp A P, O_{2} A \perp A P$, we get $A, S, C$ are collinear.
To prove $C, D, M$ are collinear, it suffices to prove:
$$
\begin{array}{l}
\angle S D M=\frac{1}{2} \angle C O_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,152 |
$CD$ intersects at point $P, \triangle PBC$'s circumcircle and $\triangle PAD$'s circumcircle intersect at points $P, Q$. Prove: $OQ \perp PQ$. | Proof: Let $K$ and $L$ be the midpoints of $AB$ and $CD$ respectively, and let $AD$ and $BC$ intersect at point $E$.
By property 2, we know that $Q$ is the Miquel point of the complete quadrilateral $ABCDEP$, and $\triangle ABQ \sim \triangle CDQ$, with $K$ and $L$ being corresponding points.
Thus, $\angle QKB = \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,153 |
Example 2 (Steiner's Theorem) Given two circles intersecting at points $P$ and $Q$, $AC$ and $BD$ are two chords passing through $P$, and $H_{1}$ and $H_{2}$ are the orthocenters of $\triangle PCD$ and $\triangle PAB$, respectively. Prove: The point symmetric to $Q$ with respect to $AC$ lies on $H_{1}H_{2}$. ${ }^{[2]}... | Proof As shown in Figure 4, let $D H_{1}$ intersect the larger circle at point $E$, and $Q E$ intersect the smaller circle at point $F$.
Then $\angle D E Q=\angle D P Q=\angle B A Q=\angle B F E$ $\Rightarrow B F / / D E \Rightarrow B F \perp A C$
$\Rightarrow$ line $B F$ passes through point $H_{2}$. By the perpendicu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,154 |
5. In quadrilateral $ABCD$, $AB=AC$, the circumcircle $\odot O_{1}$ of $\triangle ABD$ intersects $AC$ at point $F$, and the circumcircle $\odot O_{2}$ of $\triangle ACD$ intersects $AB$ at point $E$, $BF$ intersects $CE$ at point $G$. Prove: $\frac{BG}{CG}=\frac{BD}{CD}$. | By property 2, we know that the four circles intersect at point $D$.
Combining $A B=A C$, we have
$$
\begin{array}{l}
\frac{B G}{B D}=\frac{\sin \angle B E G}{\sin \angle B E D}=\frac{\sin \angle A E C}{\sin \angle A C D}=\frac{C A}{A D}=\frac{B A}{A D} \\
=\frac{\sin \angle A F B}{\sin \angle A B D}=\frac{\sin \angle ... | \frac{BG}{CG}=\frac{BD}{CD} | Geometry | proof | Yes | Yes | cn_contest | false | 730,155 |
Example 2 Given that $X$ is an $n$-element set, $\Phi$ is a family of 3-element subsets of the $n$-element set $X$ with the following property: any two subsets have at most one common element. Prove: There exists a subset $M$ of the $n$-element set $X$ that satisfies:
(1) $M$ does not contain any set from the family $\... | First, a subset of $X$ that does not contain any set from the family $\Phi$ exists, for example, each one-element, two-element subset of $X$, etc.
Let $M$ be the largest subset of $X$ (the subset with the most elements) that does not contain any subset from the family $\Phi$. Then $M \subset X$, and for any $x \in X \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,156 |
Example 4 In a sports competition, there are $2 n(n \geqslant 2)$ events, and each participant registers for exactly two of them, while any two participants have at most one event in common. Assume that for each $k \in\{1,2, \cdots, n-1\}$, the number of events with no more than $k$ registrants is less than $k$. Prove:... | Prove: Represent the $2n$ items with $2n$ points. If two items are selected by the same participant, connect the corresponding two points (i.e., an edge represents a participant). Thus, we obtain a simple graph $G$ of order $2n$, and graph $G$ satisfies property $P$: For any $k \in \{1, 2, \cdots, n-1\}$, the number of... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,157 |
Question 3 Given $H$ is the orthocenter of acute $\triangle ABC$, point $G$ satisfies that quadrilateral $ABGH$ is a parallelogram, $I$ is a point on line $GH$ such that $AC$ bisects segment $HI$. If line $AC$ intersects the circumcircle of $\triangle GCI$ at points $C, J$, prove: $IJ = AM$. ${ }^{[1]}$
(56th IMO Short... | Proof As shown in Figure 5.
From $I, J, G, C$ being concyclic
$\Rightarrow \angle J I G=\angle J C G, \angle J=\angle C G I$
$\Rightarrow \triangle I M J \backsim \triangle C M G$
$\Rightarrow I J=\frac{I M \cdot G C}{C M}$.
From $\square A H G B$ and orthocenter $H$
$$
\begin{array}{l}
\Rightarrow H G / / A B, C H \pe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,158 |
Given an integer $N \geqslant 2 . N(N+1)$ football players, all of different heights, stand in a row. The team coach wants to remove $N(N-1)$ players so that the remaining $2 N$ players satisfy the following $N$ conditions:
(1) The two tallest players among them have no other players between them;
(2) The third and fou... | This is a very flexible and interesting problem, with a flavor of Russian contest questions, requiring a thorough understanding of the conditions before proceeding with the thinking. Let's first analyze the conditions of this problem.
First, there are a total of $N(N+1)$ players, all with different heights, standing i... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,159 |
3. Given two points $B(-1,0)$ and $C(1,0)$ on a Cartesian plane. A non-empty bounded subset $S$ of this plane is called "good" if:
(1) There exists a point $T$ in subset $S$ such that for every point $Q$ in $S$, the line segment $TQ$ is entirely within $S$;
(2) For any triangle $\triangle P_{1} P_{2} P_{3}$, there exis... | 3. If for $\triangle A B C \backsim \triangle P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}$, $B C$ corresponds to the longest side of $\triangle P_{1} P_{2} P_{3}$, then
$$
B C \geqslant A B \geqslant A C \text {. }
$$
Since for every point $A$ in the first quadrant, $A B \geqslant A C$, thus, $B C \geqslant A B$, i.e.,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,160 |
8. Given that $A_{1} 、 B_{1} 、 C_{1}$ are points on the sides $B C 、 C A 、 A B$ of the acute triangle $\triangle A B C$, and $A A_{1} 、 B B_{1} 、 C C_{1}$ are the angle bisectors of $\angle B A C 、 \angle C B A 、 \angle A C B$ respectively, $I$ is the incenter of $\triangle A B C$, and $H$ is the orthocenter of $\trian... | 8. Let's assume
$$
\alpha=\angle B A C \leqslant \beta=\angle C B A \leqslant \gamma=\angle A C B \text {. }
$$
Let the lengths of sides $BC$, $CA$, and $AB$ of $\triangle ABC$ be $a$, $b$, and $c$ respectively. Let $D$ and $E$ be points on side $BC$ such that $B_{1} D \parallel AB$ and $B_{1} E$ bisects $\angle B B_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,161 |
2. Given real numbers $x, y$ satisfy $x^{2}+2 \cos y=1$. Then the range of $x-\cos y$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. $[-1, \sqrt{3}+1]$.
Since $x^{2}=1-2 \cos y \in[-1,3]$, we have $x \in[-\sqrt{3}, \sqrt{3}]$.
From $\cos y=\frac{1-x^{2}}{2}$, we know
$$
x-\cos y=\frac{1}{2}(x+1)^{2}-1 \text {. }
$$
Therefore, when $x=-1$, $x-\cos y$ has a minimum value of -1, at which point, $y$ can be $\frac{\pi}{2}$;
When $x=\sqrt{3}$, $x-\c... | [-1, \sqrt{3}+1] | Number Theory | proof | Yes | Yes | cn_contest | false | 730,162 |
Example 4 As shown in Figure $6, \triangle A B C$ contains point $A$ opposite to the excircle $\odot O$ which is tangent to $A B, B C, C A$ at points $D, E,$ and $F$ respectively. $E Z$ is the diameter of $\odot O$. Perpendiculars from points $B$ and $C$ to $B C$ intersect $D F$ at points $B_{1}$ and $C_{1}$ respective... | Proof: Let $XH$ and $YH$ intersect the opposite sides at points $P$ and $Q$, and $DF$ intersects $BC$ at point $N$.
For the line $NDF$ and $\triangle ABC$, applying Menelaus' theorem, we get
\[ 1 = \frac{AD}{DB} \cdot \frac{BN}{NC} \cdot \frac{CF}{FA} = \frac{BN}{EB} \cdot \frac{CE}{NC} \]
\[ \Rightarrow N, B, E, C \t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,163 |
5. In a regular tetrahedron $P-ABC$, $AB=1, AP=2$, a plane $\alpha$ passing through $AB$ bisects its volume. Then the cosine of the angle formed by edge $PC$ and plane $\alpha$ is $\qquad$ | 5. $\frac{3 \sqrt{5}}{10}$.
Let the midpoints of $AB$ and $PC$ be $K$ and $M$, respectively. Then it is easy to prove that the plane $ABM$ is the plane $\alpha$.
By the median length formula, we have
$$
\begin{array}{l}
A M^{2}=\frac{1}{2}\left(A P^{2}+A C^{2}\right)-\frac{1}{4} P C^{2} \\
=\frac{1}{2}\left(2^{2}+1^{2... | \frac{3 \sqrt{5}}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,164 |
7. In $\triangle A B C$, $M$ is the midpoint of side $B C$, and $N$ is the midpoint of segment $B M$. If $\angle A=\frac{\pi}{3}, S_{\triangle A B C}=\sqrt{3}$, then the minimum value of $\overrightarrow{A M} \cdot \overrightarrow{A N}$ is $\qquad$ | 7. $\sqrt{3}+1$
From the conditions, we have
$$
\begin{array}{l}
\overrightarrow{A M}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{A N}=\frac{3}{4} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} . \\
\text { Therefore, } \overrightarrow{A M} \cdot \overrightarrow{A N}=\frac{1}{8}\lef... | \sqrt{3}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,165 |
8. Let two strictly increasing sequences of positive integers $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{10}=b_{10}<2017$, for any positive integer $n$, there is $a_{n+2}=a_{n+1}+a_{n}, b_{n+1}=2 b_{n}$.
Then all possible values of $a_{1}+b_{1}$ are $\qquad$ | 8. 13, 20.
From the conditions, we know that \(a_{1}, a_{2}, b_{1}\) are all positive integers, and \(a_{1}b_{10} = 2^9 \times b_{1} = 512 b_{1}\), so \(b_{1} \in \{1, 2, 3\}\).
By repeatedly applying the recurrence relation of the sequence \(\{a_{n}\}\), we have
\[
\begin{array}{l}
a_{10} = a_{9} + a_{8} = 2 a_{8} + ... | 13, 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,166 |
Sure, here is the translated text:
```
II. (40 points) Let the sequence $\left\{a_{n}\right\}$ be defined as $a_{1}=1$,
$$
a_{n+1}=\left\{\begin{array}{ll}
a_{n}+n, & a_{n} \leqslant n ; \\
a_{n}-n, & a_{n}>n
\end{array}(n=1,2, \cdots) .\right.
$$
Find the number of positive integers $r$ that satisfy $a_{r}<r \leqsla... | From the definition of the sequence, we know $a_{1}=1, a_{2}=2$.
Assume for some integer $r \geqslant 2$, we have $a_{r}=r$, it is sufficient to prove that for $t=1,2, \cdots, r-1$, we have
$$
\left\{\begin{array}{l}
a_{r+2 t-1}=2 r+t-1>r+2 t-1, \\
a_{r+2 t}=r-tr+1, \\
a_{r+2}=a_{r+1}-(r+1)=2 r-(r+1) \\
=r-1r+2 t+1, \\... | \frac{3^{2017}-2019}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,167 |
2. As shown in Figure 1, in the convex quadrilateral $ABCD$, $\angle BAD + 2 \angle BCD = 180^{\circ}$, the angle bisector of $\angle BAD$ intersects the line segment $BD$ at point $E$, and the perpendicular bisector of line segment $AE$ intersects the lines $CB$ and $CD$ at points $X$ and $Y$ respectively. Prove that ... | 2. Let $L$, $M$, $N$ be the feet of the perpendiculars from point $A$ to lines $XY$, $BC$, and $CD$, respectively, and $U$, $V$ be the reflections of point $A$ across lines $BC$ and $CD$, respectively.
To prove that points $A$, $X$, $C$, and $Y$ are concyclic, by the converse of the Simson line theorem, it suffices to... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,168 |
3. Let $a_{i} \geqslant 0, x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$. Prove:
$$
\begin{array}{l}
\left(\left(1-\sum_{i=1}^{n} a_{i} \cos x_{i}\right)^{2}+\left(1-\sum_{i=1}^{n} a_{i} \sin x_{i}\right)^{2}\right)^{2} \\
\geqslant 4\left(1-\sum_{i=1}^{n} a_{i}\right)^{3} .
\end{array}
$$ | 3. If $1-\sum_{i=1}^{n} a_{i} \leqslant 0$, then the proposition holds.
If $1-\sum_{i=1}^{n} a_{i} \geqslant 0$, then
$$
\begin{array}{l}
\left(1-\sum_{i=1}^{n} a_{i} \sin x_{i}\right)^{2} \\
\geqslant\left(1-\sum_{i=1}^{n} a_{i}\right)\left(1-\sum_{i=1}^{n} a_{i} \sin ^{2} x_{i}\right), \\
\left(1-\sum_{i=1}^{n} a_{i... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,169 |
8. Given a positive integer $n \geqslant 2$, let the square grid
$A=\left(\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right)$,
$B=\left(\begin{array}{cccc}b_{11} & b_{12} & \cdots & b_... | 8. For any $x, y$, without loss of generality, let
$b_{ij} = (i-1)n + j$.
First, we prove two lemmas.
Lemma 1: Any permutation $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of $(1, 2, \cdots, n)$ can be transformed into $(1, 2, \cdots, n)$ with at most $n-1$ swaps.
Proof of Lemma 1: Move $i = 1, 2, \cdots, n-1$ to positi... | 2n(n-1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,170 |
1. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {.... | 1. (1) The total number of all possible values of $x_{1}, x_{2}, \cdots, x_{n}$ is $2^{n}$, and each corresponding function value can be either 0 or 1. Therefore, the number of all different $n$-ary Boolean functions is $2^{2^{n}}$.
(2) Let $\mid D_{10}(g)$ | denote the number of elements in the set $D_{10}(g)$. Below,... | 1817 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,171 |
2. As shown in Figure 1, in the acute triangle $\triangle ABC$, $AB \neq AC$, $K$ is the midpoint of the median $AD$, $DE \perp AB$ at point $E$, $DF \perp AC$ at point $F$, the lines $KE$ and $KF$ intersect $BC$ at points $M$ and $N$ respectively, the circumcenters of $\triangle DEM$ and $\triangle DFN$ are $O_{1}$ an... | 2. As shown in Figure 4, with $A D$ as the diameter, construct $\odot K$. Draw $A G \parallel B C$, intersecting $\odot K$ at point $G$, and connect $G E$ and $G F$.
Then $\angle G A E = 180^{\circ} - \angle A B C$,
$\angle G A F = \angle A C B$.
Thus, $D E \cdot G F = D B \sin \angle A B C \cdot A D \sin \angle G A F$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,172 |
4. Let real numbers $a_{1}, a_{2}, \cdots, a_{2017}$ satisfy
$$
\begin{array}{l}
a_{1}=a_{2017}, \\
\left|a_{i}+a_{i+2}-2 a_{i+1}\right| \leqslant 1(i=1,2, \cdots, 2015) .
\end{array}
$$
Let $M=\max _{1 \leqslant i<j \leqslant 2017}\left|a_{i}-a_{j}\right|$. Find the maximum value of $M$. | 4. Let $\left|a_{i_{0}}-a_{j_{0}}\right|=\max _{1 \leqslant i1008$ when, then
$$
\begin{array}{l}
\left|a_{j_{0}}-a_{i_{0}}\right|=\left|a_{i_{0}}-a_{1}\right|+\left|a_{2017}-a_{j_{0}}\right| \\
\leqslant \frac{\left(i_{0}-1\right)^{2}}{2}+\frac{\left(2017-j_{0}\right)^{2}}{2} \\
\leqslant \frac{\left(2016-j_{0}+i_{0}\... | \frac{1008^2}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,173 |
5. As shown in Figure 2, in the cyclic quadrilateral $ABCD$ inscribed in $\odot O$, the diagonals $AC$ and $BD$ are perpendicular to each other. The midpoints of arcs $\overparen{ADC}$ and $\overparen{ABC}$ are $M$ and $N$, respectively. The diameter through point $D$ intersects the chord $AN$ at point $G$. $K$ is a po... | 5. From $G K / / N C$ and the fact that $A, N, C, D$ are concyclic, we have
$$
\begin{array}{l}
\angle A G K=\angle A N C=180^{\circ}-\angle A D C \\
\Rightarrow A, G, K, D \text{ are concyclic } \\
\Rightarrow \angle A K G=\angle A D G .
\end{array}
$$
Also, since $D B \perp A C$, then
$$
\angle B D C=90^{\circ}-\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,174 |
6. Let the sequence of real numbers $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=\frac{1}{2}, a_{2}=\frac{3}{8},
$$
and $a_{n+1}^{2}+3 a_{n} a_{n+2}=2 a_{n+1}\left(a_{n}+a_{n+2}\right)(n=1,2, \cdots)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove that for any positive integer $n$, we... | 6. (1) From
$$
\begin{array}{l}
a_{n+1}^{2}+3 a_{n} a_{n+2}=2 a_{n+1}\left(a_{n}+a_{n+2}\right) \\
\Rightarrow a_{n} a_{n+2}-a_{n+1}^{2} \\
\quad=2 a_{n+1}\left(a_{n}+a_{n+2}\right)-2 a_{n} a_{n+2}-2 a_{n+1}^{2} \\
\Rightarrow a_{n+1}\left(a_{n}-a_{n+1}\right)-a_{n}\left(a_{n+1}-a_{n+2}\right) \\
\quad=2\left(a_{n}-a_{... | 0<a_{n}<\frac{1}{\sqrt{2 n+1}} | Algebra | proof | Yes | Yes | cn_contest | false | 730,175 |
2. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {.... | 2. (1) Same as Question 1 (1) of Grade 1.
(2) Let $|D_{n}(g)|$ denote the number of elements in the set $D_{n}(g)$.
Obviously, $|D_{1}(g)|=1, |D_{2}(g)|=1$.
$$
\begin{array}{l}
\text { Also, } g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j} \\
=\left(1+x_{1}\left(1+\sum_{i=2}^{n}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,176 |
4. For any positive integer $n$, let $D_{n}$ be the set of all positive divisors of $n$, and $f_{i}(n)(i=0,1,2,3)$ be the number of elements in the set
$$
F_{i}(n)=\left\{a \in D_{n} \mid a \equiv i(\bmod 4)\right\}
$$
Find the smallest positive integer $m$ such that
$$
f_{0}(m)+f_{1}(m)-f_{2}(m)-f_{3}(m)=2017
$$ | 4. Represent the positive integer $m$ in its standard form:
$$
m=2^{\alpha}\left(\prod_{i=1}^{k} p_{i}^{\beta_{i}}\right)\left(\prod_{j=1}^{l} q_{j}^{\gamma_{j}}\right),
$$
where $\alpha$ can be $0, p_{1}, p_{2}, \cdots, p_{k}$ are all the prime factors of $m$ that are congruent to 1 modulo 4, and $q_{1}, q_{2}, \cdot... | 2^{34} \times 3^{6} \times 7^{2} \times 11^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,178 |
3. Arrange $1,2, \cdots, k$ in a row so that each number is strictly greater than all the numbers preceding it, or strictly less than all the numbers preceding it. Let the number of different arrangements be $a_{k}(k=1,2, \cdots)$. Then $a_{n}=$ $\qquad$ . | 3. $2^{n-1}$.
For the $n$-th position being 1 or $n$, regardless of which it is, the remaining $n-1$ positions have $a_{n-1}$ arrangements, thus
$$
\begin{array}{l}
a_{n}=2 a_{n-1} \cdot \\
\text { and } a_{1}=1 \text {, therefore, } a_{n}=2^{n-1} .
\end{array}
$$ | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,180 |
6. Let $a, b, c$ be non-negative real numbers. Then
$$
\begin{array}{l}
S= \sqrt{\frac{a b}{(b+c)(c+a)}}+ \\
\sqrt{\frac{b c}{(a+c)(b+a)}}+\sqrt{\frac{c a}{(b+c)(b+a)}}
\end{array}
$$
The minimum value of $S$ is $\qquad$, and the maximum value of $S$ is $\qquad$ | 6. $1, \frac{3}{2}$.
$$
\text { Let } \frac{a}{b+c}=u^{2}, \frac{b}{c+a}=v^{2}, \frac{c}{a+b}=w^{2} \text {. }
$$
Assume $a+b+c=1$.
$$
\begin{array}{l}
\text { Then } a=\frac{u^{2}}{1+u^{2}}, b=\frac{v^{2}}{1+v^{2}}, c=\frac{w^{2}}{1+w^{2}} \\
\Rightarrow \frac{u^{2}}{1+u^{2}}+\frac{v^{2}}{1+v^{2}}+\frac{w^{2}}{1+w^{2... | 1, \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,182 |
8. For a finite set
$$
A=\left\{a_{i} \mid 1 \leqslant i \leqslant n, i \in \mathbf{Z}_{+}\right\}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
let $S=\sum_{i=1}^{n} a_{i}$, then $S$ is called the "sum" of set $A$, denoted as
$|A|$. Given the set $P=\{2 n-1 \mid n=1,2, \cdots, 10\}$,
all the subsets of $P$ containin... | 8. 3600.
Since $1+3+\cdots+19=100$, and each element in $1,3, \cdots, 19$ appears in the three-element subsets of set $P$ a number of times equal to $\mathrm{C}_{9}^{2}=36$, therefore, $\sum_{i=1}^{k}\left|P_{i}\right|=3600$. | 3600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,183 |
10. (20 points) The rules of a card game are as follows: Arrange nine cards labeled $1,2, \cdots, 9$ randomly in a row. If the number on the first card (from the left) is $k$, then reverse the order of the first $k$ cards, which is considered one operation. The game stops when no operation can be performed (i.e., the n... | 10. Obviously, for a secondary terminating permutation, the first card must be 1. Let the number marked on the card at the $i$-th position be $a_{i}$, then $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$ is a permutation of $1,2, \cdots, 9$, and $a_{1}=1$.
Since there is a unique permutation $\left(a_{1}, a_{2}, \cdots, a_... | \frac{103}{2520} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,184 |
11. (20 points) Take a point $K$ on the major axis of an ellipse (where $K$ is not the center), and draw two chords $A C$ and $B D$ through $K$. Extend the opposite sides $A B$ and $D C$ of quadrilateral $A B C D$ to intersect at point $M$, and extend $A D$ and $B C$ to intersect at point $N$. When the chords $A C$ and... | 11. (1) Let $A\left(a \cos \theta_{1}, b \sin \theta_{1}\right)$, $B\left(a \cos \theta_{2}, b \sin \theta_{2}\right)$, $C\left(a \cos \theta_{3}, b \sin \theta_{3}\right)$, $D\left(a \cos \theta_{4}, b \sin \theta_{4}\right)$, $K(u, 0)$.
Since $A$, $K$, and $C$ are collinear,
$$
\begin{array}{l}
\Rightarrow b \sin \th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,185 |
Three, (50 points) Prove: Given any positive integer $k$, there are infinitely many prime numbers $p$, for each $p$, there exists a positive integer $n$, such that $p \mid \left(2017^{n}+k\right)$. | Let's assume 2017, otherwise, let $2017^{t} \| k$. Then, $p=2017$ or $p \nmid\left(2017^{n-t}+k\right)$, which does not affect the infinitude of $p$.
Use proof by contradiction.
Assume there are only finitely many such $p$ that satisfy the original proposition. Let these primes be $p_{1}, p_{2}, \cdots, p_{s}$.
Clearly... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,186 |
3. Let a line $l$ intersect the sides $BC$, $CA$, and $AB$ of $\triangle ABC$ at points $D$, $E$, and $F$, respectively. Let $G$, $H$, and $I$ be the circumcenters of $\triangle AEF$, $\triangle BDF$, and $\triangle CDE$, respectively. Prove that the circumcenter of $\triangle GHI$ lies on the line $l$.
(2008, China Na... | Hint: This problem is essentially Steiner's Theorem. By Property 2, we know that $\odot G$, $\odot H$, and $\odot I$ intersect at the Miquel point $M$ and that $G$, $H$, $I$, and $M$ are concyclic. Clearly, the reflections of point $M$ over the sides of $\triangle GHI$ are $D$, $E$, and $F$. By Steiner's Theorem, we kn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,187 |
Prove: The irreducible proper fraction $\frac{2}{m}\left(l \in \mathbf{Z}_{+}\right)$ can be expressed as the sum of several different unit fractions, and the denominators of these unit fractions are even.
Translate the above text into English, please retain the original text's line breaks and format, and output the t... | By Euler's theorem, there exists a positive integer $t$ such that $m \mid (2^{l+t}-1)$.
Let $s=\frac{2^{l+t}-1}{m}$.
Then $\frac{2^{l}}{m}=\frac{2^{l+t}}{2^{t} m}=\frac{1}{2^{t} m}+\frac{2^{l+t}-1}{2^{t} m}=\frac{1}{2^{t} m}+\frac{s}{2^{t}}$.
Write $s$ in binary. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,188 |
4. In $\triangle A B C$, $D$ and $E$ are points on sides $A B$ and $A C$ respectively, and $D E / / B C, B E$ intersects $C D$ at point $F$, the circumcircle of $\triangle B D F$ $\odot O$ intersects the circumcircle of $\triangle C E F$ $\odot P$ at point $G$. Prove: $\angle B A F=\angle C A G$.
(2009, Balkan Mathemat... | ```
Let $\angle F A B=\alpha, \angle F A G=\beta$, $\angle G A C=\gamma$.
By parallelism and Ceva's theorem, it is easy to get $D I=I E$.
\[
\begin{array}{l}
\text { Hence } \frac{\sin \alpha}{\sin (\beta+\gamma)}=\frac{A E}{A D}=\frac{C E}{B D} \\
=\frac{G E}{G B}=\frac{\sin \gamma}{\sin (\alpha+\beta)} \\
\Rightarrow... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,189 |
Example 1 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$ (2016, National High School Mathematic... | Let the graph that satisfies the conditions be $G(V, E)$.
First, we prove a lemma.
Lemma In any $n(n \leqslant 5)$-order subgraph $G^{\prime}$ of graph $G(V, E)$, there can be at most five edges.
Proof It suffices to prove the case when $n=5$.
If there exists a vertex $A$ in $G^{\prime}$ with a degree of 4, then no edg... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,190 |
3. Let $S$ be a 68-element subset of the set $\{1,2, \cdots, 2015\}$. Prove that there exist disjoint non-empty subsets $A, B, C$ of $S$ such that
$$
|A|=|B|=|C|, \sum_{a \in A} a=\sum_{b \in B} b=\sum_{c \in C} c .
$$
(2015, China National Training Team Test) | Consider all three-element subsets of $S$. $S$ has exactly $\mathrm{C}_{68}^{3}=50116$ three-element subsets, and the sum of elements in each three-element subset does not exceed
$$
2013+2014+2015=6042 .
$$
Let $[x]$ denote the greatest integer not exceeding the real number $x$.
By the pigeonhole principle, there must... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,191 |
Example 1 Given 2015 circles of radius 1 in the plane. Prove: Among these 2015 circles, there exists a subset $S$ of 27 circles such that any two circles in $S$ either both have a common point or both do not have a common point. ${ }^{[1]}$
(The 28th Korean Mathematical Olympiad) | Proof that there do not exist 27 circles such that any two circles have a common point.
Select a line $l$ such that $l$ is neither parallel to the line connecting the centers of any two of the 2015 circles nor perpendicular to these lines.
Let this line be the $x$-axis, and the 2015 circles be denoted as $C_{1}, C_{2... | 27 | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,193 |
Example 2 Sets $A_{1}, A_{2}, \cdots, A_{35}$ satisfy $\left|A_{i}\right|=27$ $(1 \leqslant i \leqslant 35)$, and the intersection of any three of these sets contains exactly one element. Prove: The sets $A_{1}, A_{2}, \cdots, A_{35}$ have a common element.
(2006, Iran National Team Selection Exam (Round 2)) | To prove that for set $A_{1}$, $A_{2}, A_{3}, \cdots, A_{35}$ can form $\mathrm{C}_{34}^{2}$ pairs of sets.
Since the intersection of any three sets contains exactly one element, each set pair corresponds to a unique element in $A_{1}$.
By the pigeonhole principle, we know that at least $\left[\frac{\mathrm{C}_{34}^{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,194 |
Example 3 Given that $X$ is a finite set and $|X|>10$. $A_{1}, A_{2}, \cdots, A_{2016}$ are subsets of set $X$, and $2\left|X_{i}\right|>|X|$ $(i=1,2, \cdots, 2016)$. Prove: there exists a ten-element subset $B$ of $X$, such that
$$
A_{i} \cap B \neq \varnothing(i=1,2, \cdots, 2016) .
$$ | Let $X=\left\{x_{1}, x_{2}, \cdots, x_{m}\right\}(m>10)$.
If $x_{i} \in A_{j}$, then write 1 in the cell where the column of $x_{i}$ intersects with the row of $A_{j}$, and write 0 in all other cells.
Thus, the sums of the rows are $\left|A_{1}\right|,\left|A_{2}\right|, \cdots, \left|A_{2016}\right|$, and the sums of... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,195 |
Example 4 Let $S$ be the set of $m$ pairs of positive integers $(a, b)$ $(1 \leqslant a<b \leqslant n)$. Prove: there are at least $\frac{4 m^{2}-m n^{2}}{3 n}$ triples $(a, b, c)$ such that
$$
(a, b) 、(b, c) 、(a, c) \in S \text {. }
$$
(1989, Asia Pacific Mathematical Olympiad) | Consider the positive integers $1,2, \cdots, n$ as $n$ points $A_{1}, A_{2}, \cdots, A_{n}$, and assume that no three points are collinear. If $(a, b) \in S(1 \leqslant a<b \leqslant n)$, then draw an edge between the points corresponding to $a$ and $b$.
Let the degree of $A_{i}(1 \leqslant i \leqslant n)$ be $d\left(A... | \frac{4 m^{2}-m n^{2}}{3 n} | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,196 |
Example 5 Let $A$ denote the set of all sequences (of arbitrary finite or infinite length) $\left\{a_{1}, a_{2}, \cdots\right\}$ formed by the elements of $\{1,2, \cdots, 2017\}$. If the first several consecutive terms of sequence $M$ are the terms of sequence $T$, then we say that sequence $M$ "starts with" sequence $... | 【Analysis】First, give an example where (2) does not hold.
Take 2017 sequences each with only one term: $1,2, \cdots$, 2017. Then each sequence in set $A$ must start with one of these 2017 sequences, so (2) does not hold.
First, in set $S$, there cannot exist two different sequences $T_{1}$ and $T_{2}$ such that $T_{2}... | 1 | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,197 |
1. Given that 31 students participated in an exam, the exam consists of ten questions, and each student solved at least six questions. Prove: There exist two students who have solved at least five of the same questions.
$(2015$, National High School Mathematics League Anhui Province Preliminary Contest) | Prompt: Form a ten-element universal set with ten questions. Assume that each student answers exactly six questions correctly, so the questions answered correctly by each student form a six-element subset, totaling 31 six-element subsets. Next, prove: Among these 31 six-element subsets, there exist two subsets whose co... | null | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,198 |
2. Given $X_{1}, X_{2}, \cdots, X_{100}$ as a sequence of non-empty subsets of a set $S$, and all are distinct. For any $i \in \{1,2, \cdots, 99\}$, we have $X_{i} \cap X_{i+1}=\varnothing, X_{i} \cup X_{i+1} \neq S$.
Find the minimum number of elements in the set $S$.
(45th United States of America Mathematical Olympi... | First use mathematical induction to prove: when $n \geqslant 4$, a subset sequence of $2^{n-1}+1$ subsets that meets the requirements can be constructed for $S=\{1,2, \cdots, n\}$; then prove that when $|S|=7$, the number of subsets in a subset sequence that meets the requirements does not exceed 100.
The minimum numbe... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,199 |
Example 3 Let $S=\{1,2, \cdots, 100\}$. Find the largest integer $k$ such that the set $S$ has $k$ distinct non-empty subsets with the property: for any two different subsets among these $k$ subsets, if their intersection is non-empty, then the smallest element in their intersection is different from the largest elemen... | First, we prove a lemma.
Lemma If the $t_{n}$ distinct subsets of the set $T_{n}=\{1,2, \cdots, n\}$ are such that the intersection of any two subsets is non-empty, then the maximum value of $t_{n}$ is $2^{n-1}$.
Proof On one hand, the $2^{n}$ subsets of the set $T_{n}$ can be paired into $2^{n-1}$ pairs, each pair be... | 2^{99}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,201 |
4. Let the set of positive integers be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{1000}\right\},
$$
where, $a_{1}<a_{2}<\cdots<a_{1000} \leqslant 2014$, and
$$
B=\left\{a_{i}+a_{j} \mid 1 \leqslant i, j \leqslant 1000, i+j \in A\right\}
$$
is a subset of $A$. Find the number of sets $A$ that satisfy the condition.
(54th Du... | Hint: First prove that the set $A$ can be represented as $B \cup C$, where $B \subseteq \{2001,2002, \cdots, 2014\}, C \subseteq \{1, 2, \cdots, 1000\}$. Then prove that when the set $B$ is determined, the set $C$ is unique. Therefore, the number of sets $A$ that satisfy the condition is equal to the number of subsets ... | 2^{14} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,202 |
5. Find the minimum number of sets that can simultaneously satisfy the following three conditions:
(1) Each set contains four elements;
(2) Any two sets have exactly two common elements;
(3) The common elements of all sets do not exceed one.
(2014, China Hong Kong Team Selection Exam) | First, choose two sets from $\{a, b, c, d\}$ and $\{a, b, e, f\}$. It is easy to prove that the number of sets containing both elements $a$ and $b$ does not exceed three. We consider two cases: when there are two sets containing both elements $a$ and $b$, the maximum number of sets satisfying the condition is seven; wh... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,203 |
The third question: In a $33 \times 33$ grid, each cell is colored with one of three colors, such that the number of cells of each color is equal. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Translate the above text into Eng... | Assume the number of separating edges is no more than 55, and denote the three colors as $A$, $B$, and $C$.
If a separating edge corresponds to two cells of colors $A$ and $B$ or $A$ and $C$, then it is called an $A$-colored separating edge. Similarly, define $B$-colored and $C$-colored separating edges.
Since $55 < ... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,205 |
Example 4 Let $S=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}(n \geqslant 2)$, where $A_{1}, A_{2}, \cdots, A_{n}$ are $n$ distinct finite sets, satisfying that for any $A_{i}, A_{j} \in S$, there is $A_{i} \cup A_{j} \in S$. If $k=\min _{1 \leqslant i \leqslant n}\left|A_{i}\right| \geqslant 2$ (where $|X|$ denotes the ... | Prove a stronger conclusion: For any set $A_{i}(1 \leqslant i \leqslant n)$, there exists $x \in A_{i}$, such that $x$ belongs to at least $\frac{n}{\left|A_{i}\right|}$ of the sets $A_{1}$, $A_{2}, \cdots, A_{n}$.
If $A_{1}, A_{2}, \cdots, A_{n}$ all have non-empty intersections with $A_{i}$, then by the principle of... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,207 |
8. Find all integer-coefficient polynomials $P(x)$ of odd degree $d$ that satisfy the following property: for each positive integer $n$, there exist $n$ distinct positive integers $x_{1}, x_{2}, \cdots, x_{n}$, such that for every pair of indices $i, j (1 \leqslant i, j \leqslant n)$, we have $\frac{1}{2}<\frac{P\left(... | ```
8. $P(x)=a(r x+s)^{d}$, where $a, r, s$ are integers, and $a \neq 0, r \geqslant 1, (r, s)=1$.
Let $P(x)=\sum_{i=0}^{d} a_{i} x^{i}, y=d a_{d} x+a_{d-1}$.
Define $Q(y)=P(x)$, then $Q$ is a polynomial with rational coefficients, and it does not contain the term $y^{d-1}$.
Let $Q(y)=b_{d} y^{d}+\sum_{i=0}... | P(x)=a(r x+s)^{d} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,208 |
2. Given that $n$ is a positive integer, such that there exist positive integers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1} x_{2} \cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)=100 n .
$$
Find the maximum possible value of $n$.
(Lin Jin, problem contributor) | 2. The maximum possible value of $n$ is 9702.
Obviously, from the given equation, we have $\sum_{i=1}^{n} x_{i} \geqslant n$. Therefore, $\prod_{i=1}^{n} x_{i} \leqslant 100$.
Since equality cannot hold, then $\prod_{i=1}^{n} x_{i} \leqslant 99$.
$$
\begin{array}{l}
\text { and } \prod_{i=1}^{n} x_{i}=\prod_{i=1}^{n}\... | 9702 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,209 |
3. As shown in Figure 1, in $\triangle A B C$, $D$ is a point on side $B C$. Let the incenter of $\triangle A B D$ and $\triangle A C D$ be $I_{1}$ and $I_{2}$, respectively. Let the circumcenters of $\triangle A I_{1} D$ and $\triangle A I_{2} D$ be $O_{1}$ and $O_{2}$, respectively. The line $\mathrm{I}_{1} \mathrm{O... | 3. From $O_{1} A=O_{1} I_{1}=O_{1} D$ and the properties of the incenter, we know that $O_{1}$ is the midpoint of the arc $\overparen{A D}$ of the circumcircle of $\triangle A B D$.
As shown in Figure 2, extend $B I_{1}$ and $D I_{2}$ to intersect at point $J_{1}$, then $J_{1}$ is the excenter of $\triangle A B D$ wit... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,210 |
4. Given integers $n, k (n \geqslant k \geqslant 2)$. Two players, A and B, play a game on an $n \times n$ grid paper where each small square is initially white: they take turns to choose a white square and color it black, with A starting first. If after a player's move, every $k \times k$ square contains at least one ... | 4. Label the rows of the grid paper from top to bottom, and the columns from left to right.
If \( n \leqslant 2 k-1 \), then player A can color the cell at the \( k \)-th row and \( k \)-th column black, ensuring that each \( k \times k \) square contains at least one black cell. Therefore, player A wins.
If \( n \geq... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,211 |
8. Given an integer $n \geqslant 2$. Prove: For any positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\sum_{i=1}^{n}\left(\max _{1 \leqslant j \leqslant i} a_{j}\right)\left(\min _{i \leqslant j \leqslant n} a_{j}\right) \leqslant \frac{n}{2 \sqrt{n-1}} \sum_{i=1}^{n} a_{i}^{2} .
$$
(Zhang Duanyang) | 8. Use the second mathematical induction on $n$.
When $n=2$,
the left side of (1) $=a_{1} \cdot \min \left\{a_{1}, a_{2}\right\}+a_{2} \cdot \max \left\{a_{1}, a_{2}\right\}$.
If $a_{1} \geqslant a_{2}$, then
(1) $\Leftrightarrow 2 a_{1} a_{2} \leqslant a_{1}^{2}+a_{2}^{2}$,
the proposition holds;
If $a_{1} \leqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,212 |
Example 5 Let $S$ be a set of $n$ points in the plane, no four of which are collinear, and let $\left\{d_{1}, d_{2}, \cdots, d_{k}\right\}$ be the set of all distinct distances between points in $S$. Denote by $m_{i}$ the multiplicity of $d_{i}$ $(i=1,2, \cdots, k)$, i.e., the number of unordered pairs $\{P, Q\} \subse... | Notice that, $\sum_{i=1}^{k} m_{i}=\mathrm{C}_{n}^{2}$.
Let $\Delta(S)$ denote the number of isosceles triangles (including degenerate cases of two points and their midpoint) formed by triples of points in the set $S$, where each equilateral triangle is counted three times. For $D \in S$, let $m_{i}(D)$ denote the numb... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,213 |
3. Given an integer $n \geqslant 2$. Find the smallest positive real number $c$, such that for any complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have
$$
\left|\sum_{i=1}^{n} z_{i}\right|+c \sum_{1 \leqslant i<j \leqslant n}\left|z_{i}-z_{j}\right| \geqslant \sum_{i=1}^{n}\left|z_{i}\right| .
$$
(Supplied by Zhang D... | 3. When $n=2 m$, take
$$
\begin{array}{l}
z_{1}=z_{2}=\cdots=z_{m}=1, \\
z_{m+1}=z_{m+2}=\cdots=z_{2 m}=-1 ;
\end{array}
$$
When $n=2 m+1$, take
$$
\begin{array}{l}
z_{1}=z_{2}=\cdots=z_{m}=1, \\
z_{m+1}=z_{m+2}=\cdots=z_{2 m+1}=-\frac{m}{m+1} .
\end{array}
$$
In both cases, we have $c \geqslant \frac{2}{n}$.
We now ... | \frac{2}{n} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,214 |
4. Let $S$ be the set of positive real numbers, satisfying:
(1) $1 \in S$, and for any $x, y \in S$, $x+y, xy \in S$.
(2) There exists a subset $P$ of $S$ such that any number in $S \backslash\{1\}$ can be uniquely represented as the product of several numbers (repetition allowed) in $P$ (two representations are consid... | 4. Not necessarily.
Let $S=\{f(\pi) \mid f$ is a polynomial with integer coefficients, and for any $x>0$ we have $f(x)>0\}$.
It is easy to see that the set $S$ satisfies condition (1).
Next, we prove that the set $S$ satisfies condition (2).
Let $P=\{f(\pi) \in S \mid f$ is irreducible over $\mathbf{Z}[x]\}$.
For any... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,215 |
5. Let $A$ be the set of all integer sequences. Find all $f: A \rightarrow \mathbf{Z}$ such that for any $x, y \in A$, we have $f(x+y)=f(x)+f(y)$.
(Zheng Ji) | 5. $f\left(x_{1}, x_{2}, \cdots, x_{n}, \cdots\right)=\sum_{i=1}^{n} a_{i} x_{i}$.
It is easy to verify that such an $f$ satisfies the conditions.
Next, we prove that all functions $f$ that satisfy the conditions must have the given form.
Let $e_{n}$ be the sequence where the $n$-th term is 1 and all other terms are 0... | f\left(x_{1}, x_{2}, \cdots, x_{n}, \cdots\right)=\sum_{i=1}^{n} a_{i} x_{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,216 |
2. In the acute $\triangle A B C$, $A B=A C, O$ is the circumcenter of $\triangle A B C$, the rays $B O$ and $C O$ intersect the sides $A C$ and $A B$ at points $B^{\prime}$ and $C^{\prime}$ respectively, and the line $l$ passes through point $C^{\prime}$ and is parallel to $A C$. Prove: the line $l$ is tangent to the ... | 2. As shown in Figure 1, connect $A O$ and extend it to intersect line $l$ at point $T$, then connect $B^{\prime} T$.
It is easy to see that $A O$ is the axis of symmetry of $\triangle A B C$, and points $B^{\prime}$ and $C^{\prime}$ are symmetric with respect to $A O$.
Thus, $\angle B^{\prime} T O = \angle C^{\prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,217 |
3. There are three piles of stones on the table, with numbers $100$, $101$, and $102$ respectively. Two people, A and B, take turns to perform the following operation: each person can take stones from any pile in the first step, starting with A, and in each step, one of them takes one stone from a pile, but cannot take... | 3. Player A has a winning strategy.
Let the piles that initially contain $100$, $101$, and $102$ stones be denoted as $A$, $B$, and $C$ respectively.
Player A first takes one stone from pile $B$, making the number of stones in the three piles $100$, $100$, and $102$, all of which are even.
(1) If Player B does not ta... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,218 |
4. On the blackboard, there are $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$. For each $i=1,2, \cdots, n$, Vissaya wants to write a number $b_{i}\left(b_{i} \geqslant a_{i}\right)$, such that for any $i, j \in\{1,2, \cdots, n\}$, $\frac{b_{i}}{b_{j}}$ or $\frac{b_{j}}{b_{i}}$ is an integer. Prove: Vissaya can wri... | For all $k \in\{1,2, \cdots, n\}$, write a set of $b_{k 1}, b_{k 2}, \cdots, b_{k n}, b_{k i}=2^{x_{k i}} a_{k}$, where $x_{k i} \in \mathbf{Z}$, and satisfy $a_{i} \leqslant b_{k i}=2^{x_{k i}} a_{k}<2 a_{i}$.
Then for any $i, j \in\{1,2, \cdots, n\}$,
$$
\frac{b_{k i}}{b_{k j}}=2^{x_{k}-x_{k i k}}, \frac{b_{k j}}{b_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,219 |
Example 6 If the family of sets $\mathscr{F}$ satisfies: for any three sets $X_{1}, X_{2}, X_{3} \in \mathscr{F}$,
$$
\left(X_{1} \backslash X_{2}\right) \cap X_{3} \text { and }\left(X_{2} \backslash X_{1}\right) \cap X_{3}
$$
at least one of them is an empty set, then the family of sets $\mathscr{F}$ is called "perf... | Prove for $|U|$ using mathematical induction.
If $|U|=0$, i.e., $U=\varnothing$, at this point, $|\mathscr{F}| \leqslant 1$.
Assume the conclusion holds for $|U|=k-1(k \geqslant 1)$.
Now consider the case when $|U|=k$.
Take the element $A$ with the most elements from the set family $\mathscr{F}$, then $|A| \geqslant 1$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,220 |
1. Let the real number $x$ be such that
$$
\begin{aligned}
s & =\sin 64 x+\sin 65 x \\
\text { and } t & =\cos 64 x+\cos 65 x
\end{aligned}
$$
are both rational numbers. Prove: one of these sums has both of its terms as rational numbers. | 1. Note that,
$$
\begin{array}{l}
s^{2}+t^{2} \\
=\left(\sin ^{2} 64 x+\cos ^{2} 64 x\right)+\left(\sin ^{2} 65 x+\cos ^{2} 65 x\right)+ \\
2(\sin 64 x \cdot \sin 65 x+\cos 64 x \cdot \cos 65 x) \\
= 2+2 \cos (65 x-64 x)=2+2 \cos x
\end{array}
$$
is a rational number.
Thus, $\cos x$ is a rational number.
By the formu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,221 |
4. A magician and his assistant have a deck of cards, all of which have the same back, and the front is one of 2017 colors (each color has 1000000 cards). The magic trick is: the magician first leaves the room, the audience arranges $n$ face-up cards in a row on the table, the magician's assistant then flips $n-1$ of t... | 4. When $n=2018$, the magician and the assistant can agree that for $i=1,2, \cdots, 2017$, if the assistant retains the $i$-th card face up, the magician will guess that the 2018-th card is of the $i$-th color. This way, the magic trick can be completed.
Assume for some positive integer $n \leqslant 2017$, the magic t... | 2018 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 730,222 |
6. On a $200 \times 200$ chessboard, some cells contain a red or blue piece, while others are empty. If two pieces are in the same row or column, we say one piece can "see" the other. Assume each piece can see exactly five pieces of the opposite color (it may also see some pieces of the same color). Find the maximum nu... | 6. First, give an example with 3800 pieces.
The intersections of rows 1 to 5 and columns 11 to 200, as well as the intersections of columns 1 to 5 and rows 11 to 200, are all placed with red pieces; the intersections of rows 6 to 10 and columns 11 to 200, as well as the intersections of columns 6 to 10 and rows 11 to ... | 3800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,223 |
7. At the starting moment, a positive integer $N$ is written on the blackboard. In each step, Misha can choose a positive integer $a>1$ that is already written on the blackboard, erase it, and write down all its positive divisors except itself. It is known that after several steps, there are exactly $N^{2}$ numbers on ... | 7. It suffices to prove: For any positive integer $N$, the numbers written on the blackboard are always no more than $N^{2}$, and when $N \geqslant 2$, the numbers written on the blackboard are fewer than $N^{2}$.
Obviously, the conclusion holds for $N=1$.
For $N \geqslant 2$, assume the conclusion holds for all positi... | N=1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,224 |
1. Let $\{x\}$ denote the fractional part of the real number $x$. Given $a=(5 \sqrt{2}+7)^{2017}$. Then $a\{a\}=$ $\qquad$ . | ,- 1.1 .
Let $b=(5 \sqrt{2}-7)^{2017}$.
Then $0<b<1$, and $a b=1$.
Notice that,
$a-b=\sum_{k=0}^{1008} 2 \mathrm{C}_{2017}^{2 k+1}(5 \sqrt{2})^{2016-2 k} \times 7^{2 k+1} \in \mathbf{Z}$.
Since $a=(a-b)+b(a-b \in \mathbf{Z}, 0<b<1)$, it follows that $b=\{a\} \Rightarrow a\{a\}=a b=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,225 |
2. A box contains 12 good items and 3 defective items. Each time an item is drawn without replacement, the number of defective items $\xi$ drawn before a good item is obtained. The mathematical expectation $\mathrm{E} \xi=$ $\qquad$ . | 2. $\frac{24}{91}$.
$\xi$ takes values $0, 1, 2, 3$, and
$$
\begin{array}{l}
P(\xi=0)=\frac{\mathrm{C}_{12}^{1}}{\mathrm{C}_{15}^{1}}=\frac{4}{5}, \\
P(\xi=1)=\frac{\mathrm{C}_{3}^{1} \mathrm{C}_{12}^{1}}{2 \mathrm{C}_{15}^{2}}=\frac{6}{35}, \\
P(\xi=2)=\frac{\mathrm{C}_{3}^{2} \mathrm{C}_{12}^{1}}{2 \mathrm{C}_{15}^{3... | \frac{24}{91} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,226 |
3. The maximum value of the function $y=\sin 2x-2(\sin x+\cos x)$ is $\qquad$ . | 3. $1+2 \sqrt{2}$.
From the condition, we know that the period of the function is $2 \pi$.
Also, $y^{\prime}=2 \cos 2 x-2 \cos x+2 \sin x$, so the function is monotonically increasing in the interval $\left[0, \frac{\pi}{4}\right]$, monotonically decreasing in the interval $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$, ... | 1+2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,227 |
4. If the digits $a_{i}(i=1,2, \cdots, 9)$ satisfy
$$
a_{9}a_{4}>\cdots>a_{1} \text {, }
$$
then the nine-digit positive integer $\overline{a_{9} a_{8} \cdots a_{1}}$ is called a "nine-digit peak number", for example 134698752. Then, the number of all nine-digit peak numbers is . $\qquad$ | 4. 11875 .
From the conditions, we know that the middle number of the nine-digit mountain number can only be 9, 8, 7, 6, 5.
When the middle number is 9, there are $\mathrm{C}_{8}^{4} \mathrm{C}_{9}^{4}$ nine-digit mountain numbers; when the middle number is 8, there are $\mathrm{C}_{7}^{4} \mathrm{C}_{8}^{4}$ nine-dig... | 11875 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,228 |
5. Given that the 2017 roots of the equation $x^{2017}=1$ are 1, $x_{1}, x_{2}, \cdots, x_{2016}$. Then $\sum_{k=1}^{2016} \frac{1}{1+x_{k}}=$ $\qquad$ . | 5.1008 .
Given $x_{k}=\mathrm{e}^{\frac{2 \pi m}{2017} \mathrm{i}}(k=1,2, \cdots, 2016)$, we know
$$
\begin{array}{l}
\overline{x_{k}}=\mathrm{e}^{\frac{-2 k \pi}{2017} \mathrm{i}}=\mathrm{e}^{\frac{2(2017-k) \pi \mathrm{i}}{2017} \mathrm{i}}=x_{2017-k} . \\
\text { Then } \frac{1}{1+x_{k}}+\frac{1}{1+x_{2017-k}} \\
=... | 1008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,229 |
7. Given the parabola $y^{2}=4 x$, with its focus at $F$, a line passing through the focus $F$ and with an inclination angle of $\theta\left(0<\theta<\frac{\pi}{2}\right)$ intersects the parabola at points $A$ and $B$, $A O$ (where $O$ is the origin) intersects the directrix at point $B^{\prime}$, and $B O$ intersects ... | 7. $\frac{8}{\sin ^{3} \theta}$.
From the given conditions, we know that the slope of line $AB$ is $k=\tan \theta$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(y_{2}<0<y_{1}\right)$.
From $\left\{\begin{array}{l}y=k(x-1) \\ y^{2}=4 x\end{array}\right.$, we get
$y^{2}-\frac{4}{k} y-4=0$
$\Rightarro... | \frac{8}{\sin ^{3} \theta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,230 |
8. For a convex 2017-gon with unequal side lengths, each side can be colored one of four colors: red, yellow, blue, or purple, but no two adjacent sides can be the same color. Then there are a total of different coloring methods.
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 repetition of the instr... | 8. $3^{2017}-3$.
Consider the case of a convex $n$-sided polygon, let the number of different coloring methods be $p_{n}$.
It is easy to know that $p_{3}=24$.
When $n \geqslant 4$, first, for side $a_{1}$, there are four different coloring methods. Since the color of side $a_{2}$ is different from that of side $a_{1}$... | 3^{2017}-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,231 |
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)\left(n \in \mathbf{Z}_{+}\right)$. Let $S_{n}=\mathrm{C}_{n}^{1} a_{1}+\mathrm{C}_{n}^{2} a_{2}+\cdots+\mathrm{C}_{n}^{... | Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$.
Then $S_{n}=\frac{1}{\sqrt{5}} \sum_{i=1}^{n} \mathrm{C}_{n}^{i}\left(\boldsymbol{\alpha}^{i}-\beta^{i}\right)$
$$
=\frac{1}{\sqrt{5}}\left((1+\alpha)^{n}-(1+\beta)^{n}\right) \text {. }
$$
Therefore, $S_{n+2}$
$$
\begin{aligned}
= & \frac{1}{\sqrt{5}}\lef... | n=4k(k=1,2,\cdots) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,232 |
Example 8 Given a graph $G$ with $e$ edges and $n$ vertices with degrees $d_{1}, d_{2}, \cdots, d_{n}$, and an integer $k$ less than
$$
\min \left\{d_{1}, d_{2}, \cdots, d_{n}\right\} .
$$
Prove: Graph $G$ contains an induced subgraph $H$ (i.e., if two vertices in $H$ are connected by an edge in graph $G$, then they a... | Let $X_{i}$ denote the random variable indicating whether the $i$-th vertex is selected in a random permutation. If selected, it takes the value 1; otherwise, it takes the value 0.
Notice that, in the permutation of $v_{i}$ and its adjacent vertices, $v_{i}$ will be selected if it is among the first $k$ positions.
$$
... | \frac{k n^{2}}{2 e+n} | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,233 |
Three. (50 points) As shown in Figure 2, the incircle $\odot I$ of $\triangle ABC$ touches sides $AB$ and $BC$ at points $D$ and $E$, respectively. $DI$ intersects $\odot I$ again at point $F$, and $CF$ intersects $AB$ at point $G$. Point $H$ lies on segment $CG$ such that $HG = CF$. Prove that if points $A$, $H$, and ... | Three, as shown in Figure 3, let the tangent line through point $F$ on $\odot I$ intersect $BC$ and $AC$ at points $M$ and $N$ respectively.
Since $MN$ and $AB$ are both perpendicular to the diameter $DF$, we know that $MN \parallel AB$.
Furthermore, since $F$ is the point of tangency of the excircle of $\triangle CMN$... | AB = AC | Geometry | proof | Yes | Yes | cn_contest | false | 730,234 |
Four, (50 points) Given that set $X$ is a finite set of points in plane $\alpha$, $T$ is an equilateral triangle in plane $\alpha$, and set $S \subseteq X$, with $|S| \leqslant 9$. If for any set $S$ that satisfies the condition, it can be covered by two translated figures of the equilateral triangle $T$, prove: set $X... | Four, first prove two lemmas.
Lemma 1 If two triangles $T_{1}$ and $T_{2}$ are positively homothetic, and the lines containing the sides of triangle $T_{2}$ intersect with the lines containing the sides of triangle $T_{1}$, then triangle $T_{1}$ is located inside triangle $T_{2}$.
This proposition is obviously true.
Le... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,235 |
Given positive numbers $a, b, c$ satisfying $a^{2}+b^{2}+c^{2}=3$. Prove:
$$
\sqrt{\frac{a}{4-a^{2}}}+\sqrt{\frac{b}{4-b^{2}}}+\sqrt{\frac{c}{4-c^{2}}} \leqslant 2 .
$$ | Let “$\sum$” denote the cyclic sum.
First, prove a lemma.
Lemma For positive numbers $a, b, c$ satisfying $a^{2}+b^{2}+c^{2}=3$,
then $(a+b+c)^{2}+9 \sum \frac{1}{4-a^{2}} \leqslant 18$.
Proof Note that,
$$
\begin{array}{l}
9-(a+b+c)^{2}=\sum(a-b)^{2}, \\
9 \sum \frac{1}{4-a^{2}}-9 \\
=\left(\sum\left(4-a^{2}\right)\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,236 |
On a plane with a side length of 550, there is an infinite grid composed of equilateral triangles with a side length of 1. There are some small tiles of the same size and shape, each covering two adjacent grids. Place these tiles into the grid without overlapping, with the tile edges coinciding with the grid. These sma... | Prove that since each exterior angle of the convex polygon formed is an integer multiple of $60^{\circ}$, the convex polygon must be a hexagon with each interior angle equal to $120^{\circ}$ or its degenerate form (some sides being 0$). The three directions of the grid are called $\alpha \backslash \beta, \gamma$ class... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,237 |
Find all integers $k$ such that there exist 2017 integers $a_{1}, a_{2}, \cdots, a_{2017}$, satisfying that for any positive integer $n$ not divisible by the prime 2017, we have
$$
\frac{n+a_{1}^{n}+a_{2}^{n}+\cdots+a_{2017}^{n}}{n+k} \in \mathbf{Z} .
$$ | First, we prove a lemma.
Lemma There exist infinitely many primes of the form $2017k + 1$.
Proof Assume there are only finitely many primes of the form $2017k + 1$, denoted as $p_{1}, p_{2}, \cdots, p_{n}$.
Let $m = 2017 p_{1} p_{2} \cdots p_{n}$,
$$
t = \frac{m^{2017} - 1}{m - 1} = m^{2016} + m^{2015} + \cdots + m + 1... | k = 0, 1, \cdots, 2017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,238 |
As shown in Figure 2, for the acute triangle $\triangle ABC$ with incenter $I$, the extensions of the three angle bisectors intersect the circumcircle at points $A_{1}, B_{1}, C_{1}$, and the incenter of $\triangle A_{1} B_{1} C_{1}$ is $I_{1}$. Prove that the line $I_{1} I \parallel AC$ if and only if the lines $AB, B... | For the cyclic hexagon $A B C A_{1} B_{1} C_{1}$, by Pascal's theorem, we know that the intersection point $X$ of $A B$ and $B_{1} C_{1}$, the intersection point $Y$ of $A_{1} B_{1}$ and $B C$, and point $I$ are collinear.
Notice that,
$I, X, Y$ being collinear $\Leftrightarrow I_{1}, X, Y$ being collinear.
Thus, the p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,239 |
1. An exam consists of $m$ questions, with $n$ students participating, where $m, n \geqslant 2$ are given integers. The scoring rule for each question is: if exactly $x$ students fail to answer the question correctly, then each student who answers the question correctly gets $x$ points, and those who fail to answer cor... | Hint: First consider the extreme case: one student answers all questions correctly, and the other $n-1$ students answer all questions incorrectly.
Then $p_{1}+p_{n}=p_{1}=\sum_{k=1}^{m}(n-1)=m(n-1)$.
Next, prove: $p_{1}+p_{n} \leqslant m(n-1)$. | m(n-1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,240 |
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