problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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3. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{6}=26, a_{7}=2$, then the maximum value of $n S_{n}$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result direct... | 3. 338 .
$$
\begin{array}{l}
\text { Given } S_{6}=26, a_{7}=2 \\
\Rightarrow a_{1}=6, d=-\frac{2}{3} \\
\Rightarrow S_{n}=-\frac{1}{3} n^{2}+\frac{19}{3} n \\
\Rightarrow n S_{n}=-\frac{1}{3} n^{3}+\frac{19}{3} n^{2} . \\
\text { Let } f(x)=-\frac{1}{3} x^{3}+\frac{19}{3} x^{2}(x \in \mathbf{R}) .
\end{array}
$$
Then... | 338 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,704 |
Example 1 Try to find the triple $(a, b, c)$ that satisfies the following conditions:
(1) $a<b<c<100$ ($a, b, c$ are prime numbers);
(2) $a+1, b+1, c+1$ form a geometric sequence.
(4th China Southeast Mathematical Olympiad) | ```
Solution: From condition (2), we have
(a+1)(c+1)=(b+1)^2.
```
$$
\text { Let } a+1=n^{2} x, c+1=m^{2} y \text {, where } x \text { and } y \text { do not }
$$
```
contain any square factors greater than 1.
Then it must be that } x = y \text{. This is because from equation (1) we get
```
Thus, $m... | f_{1}, f_{2}, \cdots, f_{11} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,705 |
4. If the complex coefficient equation with respect to $x$
$$
(1+2 \mathrm{i}) x^{2}+m x+1-2 \mathrm{i}=0
$$
has real roots, then the minimum value of the modulus of the complex number $m$ is $\qquad$ | 4.2.
Let $\alpha$ be a real root of the original equation, $m=p+q$ i.
$$
\begin{array}{l}
\text { Then }(1+2 \mathrm{i}) \alpha^{2}+(p+q \mathrm{i}) \alpha+1-2 \mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
\alpha^{2}+p \alpha+1=0, \\
2 \alpha^{2}+q \alpha-2=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,706 |
5. A point $P$ in space is at distances of $2$ and $3$ from the vertices $A$ and $B$ of a regular tetrahedron $ABCD$. When the edge length of the tetrahedron changes, the maximum distance from point $P$ to the line $CD$ is $\qquad$ | 5. $\frac{5 \sqrt{3}}{2}$.
When the tetrahedron is determined, the trajectory of point $P$ is also determined. The plane of the trajectory is perpendicular to $AB$, and take the midpoint $E$ of $CD$. Only when point $P$ is located in the plane $ABE$ and on the opposite side of $E$ with respect to $AB$, can the distanc... | \frac{5 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,707 |
6. In a match without ties, the game ends when one person wins 2 more games than the other, and the one with more wins is the winner. It is known that in the odd-numbered games, the probability of A winning is $\frac{3}{5}$; in the even-numbered games, the probability of B winning is $\frac{3}{5}$. Then the expected nu... | 6. $\frac{25}{6}$.
Let the expected number of games played until the end of the match be $E$.
The probability that player A wins the first game and player B wins the second, or player B wins the first and player A wins the second, is
$$
\frac{3}{5} \times \frac{3}{5}+\frac{2}{5} \times \frac{2}{5}=\frac{13}{25} .
$$
... | \frac{25}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,708 |
7. Let
$$
f(x)=\ln x-\frac{1}{2} a x^{2}-2 x(a \in[-1,0)) \text {, }
$$
and $f(x)<b$ holds for all $x$ in the interval $(0,1]$. Then the range of the real number $b$ is . $\qquad$ | 7. $\left(-\frac{3}{2},+\infty\right)$.
From the problem, we know
$$
b>\max _{a \in[-1,0)}\left\{\ln x-\frac{1}{2} a x^{2}-2 x\right\} .
$$
Since the function $h(a)=\ln x-\frac{1}{2} a x^{2}-2 x$ is monotonically decreasing on the interval $[-1,0)$, we have
$$
\begin{array}{l}
h(a)_{\max }=h(-1)=\frac{1}{2} x^{2}-2 x... | \left(-\frac{3}{2},+\infty\right) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,709 |
9. (16 points) When $x \in [1,2017]$, find
$$
f(x)=\sum_{i=1}^{2017} i|x-i|
$$
the minimum value. | When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2016)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2017} i(i-x) \\
=\left(k^{2}+k-2017 \times 1009\right) x+ \\
\frac{2017 \times 2018 \times 4035}{6}-\frac{k(k+1)(2 k+1)}{3}
\end{array}
$$
is a linear function, and its minimum value is attain... | 801730806 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,710 |
II. (40 points) Given a $2016 \times 2016$ grid. Find the smallest positive integer $M$, such that it is possible to draw $M$ rectangles (with their sides on the grid lines) in the grid, and each small square in the grid has its sides included in the sides of one of the $M$ rectangles. | Second, $M=2017$.
First, we prove: All grid lines of a $2016 \times 2016$ square grid can be covered by the edges of 2017 rectangles.
In fact, all horizontal grid lines can be covered by the edges of 1008 $1 \times 2016$ rectangles and one $2016 \times 2016$ rectangle; all vertical grid lines can be covered by the edg... | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,711 |
Example 2 (Pascal's Theorem) As shown in Figure 1, let $A, B, C, D, E, F$ be six distinct points on circle $\Gamma$. Prove: the intersections of lines $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$ are collinear. | Proof: Let $a=CE$,
$$
\begin{array}{l}
b=EA, c=AC, \\
A(1,0,0), \\
\quad C(0,1,0), E(0,0,1), B\left(x_{1}: y_{1}: z_{1}\right), \\
\quad D\left(x_{2}: y_{2}: z_{2}\right), F\left(x_{3}: y_{3}: z_{3}\right) .
\end{array}
$$
Since the circle $\Gamma$ is the circumcircle of $\triangle ACE$, its equation is
$$
-a^{2} y z ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,713 |
Title: A quadruple of positive integers $(p, a, b, c)$ that satisfies the following conditions is called a "Leyden quadruple":
(i) $p$ is an odd prime;
(ii) $a, b, c$ are distinct;
(iii) $p \mid (ab + 1), p \mid (bc + 1), p \mid (ca + 1)$.
(1) Prove that for each Leyden quadruple $(p, a, b, c)$, we have $p + 2 \leq \fr... | From the form, the Leiden quadruples that satisfy the conditions have a certain rotational symmetry. To facilitate expression and discovery of its inherent laws, it is extended as follows:
Definition If the $(k+1)$-tuple of positive integers $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$ satisfies the following three p... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false | 729,714 |
Given that $a$, $b$, and $c$ are three distinct real numbers. If the quadratic equations:
$$
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
$$
each have exactly one common root with any other, find the value of $a^{2} + b^{2} + c^{2}$.
(The 4th Chen Shengshen Cup Natio... | From [1] we know
$$
x_{1}=\frac{a-b}{a-c}, x_{2}=\frac{b-c}{b-a}, x_{3}=\frac{c-a}{c-b},
$$
where $x_{1}$ is the common root of equations (1) and (3), $x_{2}$ is the common root of equations (1) and (2), and $x_{3}$ is the common root of equations (2) and (3).
Since $x_{1}$ and $x_{2}$ are the two roots of equation (1... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,715 |
Question 1 In $\triangle B C F$, $\angle B=90^{\circ}$. On the line $C F$, take point $A$ such that $F A=F B$, and $F$ is between points $A$ and $C$; take point $D$ such that $D A=D C$, and $A C$ is the angle bisector of $\angle D A B$; take point $E$ such that $E A=E D$, and $A D$ is the angle bisector of $\angle E A ... | 【Analysis】After seeing the three lines concurrent, the author thought of using the converse of the trigonometric form of Ceva's Theorem in $\triangle F M D$.
Thus, the following proof is given.
As shown in Figure 1, connect $M D$ and $C X$.
Let $\angle A B F=\angle B A F=\angle D A C$
$$
\begin{array}{l}
=\angle A C D=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,716 |
Question 2 As shown in Figure 2, let the circumcircle of $\triangle A B C$ be $\Gamma$. Construct the tangents to circle $\Gamma$ at points $B$ and $C$, and let these tangents intersect at point $D$. Construct squares $B A G H$ and $A C E F$ outwardly on the sides $A B$ and $C A$ of $\triangle A B C$, respectively. Let... | Prove as shown in Figure 2, let $A D$ intersect the circle $\Gamma$ at point $I$. Connect $B I, C I, G F$. Since $B D, C D$ are tangent to the circle $\Gamma$ at points $B, C$ respectively, quadrilateral $A B I C$ is a harmonic quadrilateral.
Therefore, $A B \cdot C I = A C \cdot B I$.
Also, $A B = A G, A C = A F$, the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,717 |
2. As shown in Figure 1, in the acute triangle $\triangle ABC$, $AB > AC$, $\odot O$ and $\odot I$ are the circumcircle and incircle of $\triangle ABC$ respectively. $\odot I$ is tangent to side $BC$ at point $D$. Line $AO$ intersects side $BC$ at point $X$. $AY$ is the altitude from $A$ to side $BC$. The tangents to $... | 2. As shown in Figure 3.
Let $a = BC, b = CA, c = AB, p = \frac{1}{2}(a + b + c)$
be the semi-perimeter of $\triangle ABC$.
Thus, $BD = p - b, CD = p - c$.
First, $A, D, L$ are collinear $\Leftrightarrow \frac{S_{\triangle ABL}}{S_{\triangle ACL}} = \frac{BD}{CD}$.
Notice that,
$\frac{BD}{CD} = \frac{p - b}{p - c} = \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,718 |
3. Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". A line segment on the side of a small rectangle, if both of its endpoints are nodes and its interior does not contain any other nodes... | 3. Construct a graph $G$ with all nodes as vertices and basic segments as edges. Let there be $N$ basic segments, which is the number of edges in graph $G$. It is easy to see that the degree of vertices in graph $G$ can only be $2$, $3$, or $4$.
Vertices with degree 2 are exactly the four vertices of rectangle $R$.
L... | N_{\max} = 6049, N_{\min} = 4122 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,719 |
2. Proposition A: In $\triangle A B C$, $A B=A C, D$ is the midpoint of side $B C$, then $\angle A B C+\angle C A D=90^{\circ}$;
Proposition B: In Rt $\triangle A B C$, $A D$ is the median on side $B C$, then $\angle A B C+\angle C A D=90^{\circ}$. These two propositions ( ).
(A) A true, B false
(B) A false, B true
(C... | 2. A.
It is easy to prove: Proposition A is true; for Proposition B, when $\angle ABC$ is a right angle, it is clearly not a true proposition. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,722 |
3. As shown in Figure 2, the inverse proportion function $y=\frac{k}{x}(x>0)$
passes through the midpoint of the diagonal $OB$ of rectangle $OABC$ with an area of 4.
$P$ is any point on the function's graph. Then the minimum value of $OP$ is ( ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2 | 3. B.
It is known that $k=x y=1$.
Let point $P(x, y)$. Then
$$
O P^{2}=x^{2}+y^{2} \geqslant 2 x y=2 \Rightarrow O P \geqslant \sqrt{2}
$$
Thus, the minimum value of $O P$ is $\sqrt{2}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,723 |
5. A triangular paper piece with a side length of 1 is placed on a regular 12-sided paper piece with a side length of 1, such that one side of the triangle coincides with one side of the 12-sided polygon, and both are flat on the table. Then the number of sides of the resulting convex polygon cannot be ( ).
(A) 11
(B) ... | 5. D.
Since the dodecagon and the triangle together have 15 sides, and the two overlapping sides are definitely not part of the new convex polygon, the new convex polygon can have at most 13 sides.
It is easy to know that each interior angle of the regular dodecagon is
$$
\frac{(12-2) \times 180^{\circ}}{12}=150^{\cir... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,724 |
1. If $\sqrt{24-t^{2}}-\sqrt{8-t^{2}}=2$, then
$$
\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=
$$
$\qquad$ | II. 1.8.
From the known equation and using the formula $a+b=\frac{a^{2}-b^{2}}{a-b}$, we get $\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=\frac{24-8}{2}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,725 |
3. A certain unit distributes a year-end bonus of 1 million yuan, with first prize at 15,000 yuan per person, second prize at 10,000 yuan per person, and third prize at 5,000 yuan per person. If the difference in the number of people between third prize and first prize is no less than 93 but less than 96, then the tota... | 3. 147.
Let the number of first prize winners be $x$, the number of second prize winners be $y$, and the number of third prize winners be $z$.
Then $1.5 x + y + 0.5 z = 100$
$$
\begin{array}{l}
\Rightarrow (x + y + z) + 0.5 x - 0.5 z = 100 \\
\Rightarrow x + y + z = 100 + 0.5(z - x). \\
\text{Given } 93 \leqslant z - ... | 147 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,726 |
(A) 19
(B) 20
(C) 21
(D) 22
Three. (10 points) Given positive integers $a, b, c, d$ satisfying $ab = cd$. Prove: $a^{2016} + b^{2016} + c^{2016} + d^{2016}$ is a composite number. | $$
\begin{array}{l}
a=x y, b=z t, c=x z, d=y t . \\
\text { Then } a^{2016}+b^{2016}+c^{2016}+d^{2016} \\
=(x y)^{2016}+(z t)^{2016}+(x z)^{2016}+(y t)^{2016} \\
=\left(x^{2016} y^{2016}+y^{2016} t^{2016}\right)+\left(x^{2016} z^{2016}+z^{2016} t^{2016}\right) \\
=y^{2016}\left(x^{2016}+t^{2016}\right)+z^{2016}\left(x^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,727 |
2. Given the curve
$$
(x-20)^{2}+(y-16)^{2}=r^{2} \text { and } y=\sin x
$$
have exactly one common point $P\left(x_{0}, y_{0}\right)$. Then
$$
\frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=
$$
$\qquad$ | 2. 20 .
From the given information, there is a common tangent line at point $P\left(x_{0}, y_{0}\right)$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sin x_{0}-16}{x_{0}-20} \cos x_{0}=-1 \\
\Rightarrow \frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=20 .
\end{array}
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,728 |
3. Given the set $M=\{(a, b) \mid a \leqslant-1, b \leqslant m\}$. If for any $(a, b) \in M$, it always holds that $a \cdot 2^{b}-b-3 a \geqslant 0$, then the maximum value of the real number $m$ is $\qquad$. | 3. 1 .
Notice that,
$$
a \cdot 2^{b}-b-3 a \geqslant 0 \Leftrightarrow\left(2^{b}-3\right) a-b \geqslant 0
$$
holds for any $a \leqslant-1$.
$$
\text { Then }\left\{\begin{array}{l}
2^{b}-3 \leqslant 0, \\
2^{b}+b \leqslant 3
\end{array} \Rightarrow b \leqslant 1\right. \text {. }
$$ | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,729 |
6. In tetrahedron $ABCD$, $\triangle ABD$ is an equilateral triangle, $\angle BCD=90^{\circ}, BC=CD=1, AC=\sqrt{3}, E$ and $F$ are the midpoints of $BD$ and $AC$ respectively. Then the cosine of the angle formed by line $AE$ and $BF$ is | 6. $\frac{\sqrt{2}}{3}$.
Given $\angle B C D=90^{\circ}, B C=C D=1$
$$
\begin{array}{l}
\Rightarrow B D=B A=A D=\sqrt{2} \\
\Rightarrow \angle A B C=\angle A D C=90^{\circ} \text {. }
\end{array}
$$
Thus, $\overrightarrow{A E}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A D}), \overrightarrow{B F}=\frac{1}{2} \o... | \frac{\sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,730 |
7. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b$ $>0)$, and $P$ is a point on the ellipse $C$, and the incenter of $\triangle F_{1} P F_{2}$ is $I$. If there exists a real number $\lambda$ such that
$$
(1+\lambda) \overrightarrow{P F_{1}}+(1-\lam... | 7. $\frac{1}{2}$.
Since the incenter of $\triangle F_{1} P F_{2}$ is $I$, we have
$$
\left|F_{1} F_{2}\right| \overrightarrow{I P}+\left|P F_{1}\right| \overrightarrow{I F_{2}}+\left|P F_{2}\right| \overrightarrow{I F_{1}}=\mathbf{0} \text {. }
$$
Combining with the given conditions, we get
$$
\begin{array}{l}
(1+\la... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,731 |
11. (20 points) Given the function
$$
f(x)=\left(1-x^{2}\right)\left(x^{2}+b x+c\right)(x \in[-1,1]) \text {. }
$$
Let the maximum value of $\mid f(x)$ | be $M(b, c)$. When $b$ and $c$ vary, find the minimum value of $M(b, c)$. | 11. Since for any $x \in[-1,1],|f(x)|$ $\leqslant M(b, c)$, thus, taking $x= \pm \lambda, 0$, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
|c| \leqslant M(b, c), \\
\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}+b \lambda+c\right)\right| \leqslant M(b, c), \\
\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}-b... | 3-2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,732 |
One, (40 points) Given integers $a>b>1$, and satisfying $(a+b)|(a b+1),(a-b)|(a b-1)$.
Prove: $a<\sqrt{3} b$. | $$
\begin{array}{l}
(a+b) \mid(b(a+b)-(a b+1)) \\
\Rightarrow(a+b) \mid\left(b^{2}-1\right), \\
(a-b) \mid(b(a-b)-(a b-1)) \\
\Rightarrow(a-b) \mid\left(b^{2}-1\right) .
\end{array}
$$
Notice that, $a>b>1$.
Then $b^{2}-1>0, a-b>0$
$$
\begin{array}{l}
\Rightarrow[a+b, a-b] \mid\left(b^{2}-1\right) \\
\Rightarrow[a+b, a... | a<\sqrt{3} b | Number Theory | proof | Yes | Yes | cn_contest | false | 729,733 |
Three. (50 points) There are 20 different colors of balls, with at least 10 balls of each color, and a total of 800 balls. Now, these balls are placed into several boxes, such that each box contains at least 10 balls of the same color. Question: Is there a way to distribute the balls into boxes so that, after giving th... | Three, generally, it can be proven by mathematical induction: For $n$ different colors of balls (a total of $40 n$ balls), if each color has at least 10 balls, and each box must contain at least 10 balls of the same color, then the balls can be distributed to $n$ students so that each student has exactly 40 balls.
When... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,734 |
Example 6 Let points $O, I_{B}, I_{C}$ be the circumcenter of acute $\triangle ABC$, the excenter of $\angle B$, and the excenter of $\angle C$, respectively. On side $AC$, take points $E, Y$ such that $\angle ABY = \angle CBY$ and $BE \perp AC$. On side $AB$, take points $F, Z$ such that $\angle ACZ = \angle BCZ$ and ... | Proof: Let $A(1,0,0), B(0,1,0), C(0,0,1)$,
$$
\begin{array}{l}
Y\left(\frac{a}{a+c}, 0, \frac{c}{a+c}\right), Z\left(\frac{a}{a+b}, \frac{b}{a+b}, 0\right) . \\
\text { Hence } \overrightarrow{Y Z}=\left(\frac{a}{a+b}-\frac{a}{a+c}, \frac{b}{a+b}, \frac{-c}{a+c}\right) \\
=\frac{1}{(a+b)(a+c)}(a c-a b, b a+b c,-c a-c b... | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 729,735 |
2. In $\triangle A B C$, $\angle A=60^{\circ}$, points $M$, $N$, and $K$ are on sides $B C$, $A C$, and $A B$ respectively, such that $B K=K M=M N=$ $N C$. If $A N=2 A K$, find the measures of $\angle B$ and $\angle C$. | 2. As shown in Figure 6, take the midpoint $P$ of $AN$. Then $AK = AP = PN$. Therefore, $\triangle APK$ is an equilateral triangle.
Hence $\angle ANK$
$$
=\frac{1}{2} \angle KPA = 30^{\circ} \text{.}
$$
Let $\angle ACB = \angle NMC = \alpha$. Then
$$
\angle ABC = \angle KMB = 120^{\circ} - \alpha \text{.}
$$
Therefor... | \angle C = 45^{\circ}, \angle B = 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,736 |
1. As shown in Figure 4, points $P, A, B$ are on the circle, and point $Q$ is inside the circle such that $\angle P A Q=90^{\circ}$, and $P Q = B Q$. Prove:
$$
\begin{array}{l}
\angle A Q B - \angle P Q A \\
=\overparen{A B}^{\circ} .
\end{array}
$$ | 1. Let $M$ be the midpoint of $PB$. Then $\angle PMQ=90^{\circ}$.
Another $\angle PAQ=90^{\circ}$, thus,
$A, P, Q, M$ are concyclic
$$
\begin{aligned}
\Rightarrow & \angle APM=\angle AQM \\
\Rightarrow & \angle AQB-\angle PQA \\
& =\angle PQM+\angle AQM-\angle PQA \\
& =2 \angle AQM=2 \angle APB=\overparen{AB}^{\circ}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,737 |
3. In $\triangle A B C$, $M$, $N$, $K$ are the midpoints of sides $B C$, $C A$, $A B$ respectively, $\Gamma_{B}$, $\Gamma_{C}$ are the semicircles with diameters $A C$, $A B$ respectively, constructed outside the triangle. $M K$ intersects the semicircle $\Gamma_{C}$ at point $X$, and $M N$ intersects the semicircle $\... | 3. As shown in Figure 10, draw $A H \perp B C$, with the foot of the perpendicular being $H$. Then $H$ is also the intersection point of the two circles.
By $K M / / A C, M N / / A B$
$$
\begin{array}{l}
\Rightarrow \angle A K X=\angle A N Y=\angle B A C \\
\Rightarrow \angle A B X=\angle A C Y=\frac{1}{2} \angle B A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,738 |
5. (1) Does there exist five circles in the plane such that each circle passes through the centers of exactly three of the other four circles?
(2) Same as the 5th problem in the basic section. | 5. (1) There do not exist five circles such that each circle passes through the centers of exactly three of the other four circles.
Assume there exist five circles satisfying the problem's conditions, then the five centers satisfy that each point is at the same distance from three of the other four points, and at a di... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,739 |
2. The system of equations
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+x-2 y-6=0, \\
2 x y+2 x+y-4=0
\end{array}\right.
$$
has the real solutions $(x, y)$ as $\qquad$ . | 2. $(2,0),(-3,-2)$.
(2) $\times \mathrm{i}+$ (1) gives
$$
\begin{array}{l}
(x+y \mathrm{i})^{2}+(1+2 \mathrm{i})(x+y \mathrm{i})-(6+4 \mathrm{i})=0 \\
\Rightarrow((x+y \mathrm{i})-2)((x+y \mathrm{i})+3+2 \mathrm{i})=0 \\
\Rightarrow x+y \mathrm{i}=2 \text { or }-3-2 \mathrm{i} .
\end{array}
$$
Since $x, y$ are real nu... | (2,0),(-3,-2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,740 |
8. As shown in Figure 1, $P$ is a point inside $\triangle ABC$, with the perimeter and area of $\triangle ABC$ being $l$ and $s$ respectively. The perpendicular segments from point $P$ to $AB$, $BC$, and $CA$ are $PD$, $PE$, and $PF$ respectively, and
$$
\frac{AB}{PD}+\frac{BC}{PE}+\frac{CA}{PF} \leqslant \frac{l^{2}}{... | 8. Inner.
From the area formula we get
$$
A B \cdot P D+B C \cdot P E+A C \cdot P F=2 s \text {. }
$$
Using the Cauchy-Schwarz inequality we get
$$
\begin{array}{l}
(A B \cdot P D+B C \cdot P E+A C \cdot P F) \cdot \\
\left(\frac{A B}{P D}+\frac{B C}{P E}+\frac{C A}{P F}\right) \\
\geqslant(A B+B C+C A)^{2}=l^{2} .
\... | P \text{ is the incenter of } \triangle ABC | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,741 |
$-、$ (40 points) As shown in Figure $2, O$ is the intersection point of the diagonals of trapezoid $A B C D$, $A B / / C D$,
$P$ is a point on the leg $A D$, and $\angle A P B=\angle C P D$.
Prove:
$$
\frac{1}{A B}+\frac{1}{C D}=\frac{1}{O P} .
$$ | Let point $B$ be reflected over $AD$ to point $F$, and connect $AF, FP$.
From $\angle CPD = \angle APB$, we know that points $C, P, F$ are collinear.
Using the Law of Sines in $\triangle PAB$ and $\triangle PCD$, we get
$$
\begin{array}{l}
\frac{PB}{\sin \angle PAB} = \frac{AB}{\sin \angle APB}, \\
\frac{PC}{\sin \angl... | \frac{1}{AB} + \frac{1}{CD} = \frac{1}{OP} | Geometry | proof | Yes | Yes | cn_contest | false | 729,742 |
Three, (50 points) Find all positive integer triples $(a, b, c)$ such that $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!-c!.
$$ | Let's assume $a \geqslant b$.
When $a=1$, $b=1$, then $c!=1 \Rightarrow c=1$.
When $a=2$, $b=1$ or 2, only when $b=1, c=1$, the original equation holds.
When $a \geqslant 3$, $a!$ is even, the original equation becomes
$$
a!=\frac{a!}{b!}+1-\frac{c!}{b!}.
$$
Since all three terms in the above equation are integers, th... | (a, b, c)=(m, 1,1),(1, m, 1)\left(m \in \mathbf{Z}_{+}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,743 |
Four. (50 points) The International Mathematical Olympiad Chief Examination Committee has $n$ countries participating, with each country being represented by a team leader and a deputy team leader. Before the meeting, the participants shake hands with each other, but the team leader does not shake hands with their own ... | The team leader of the host country asks $2 n-1$ people, and since the answers are all different, the possible set of answers is $A=\{0,1, \cdots, 2 n-2\}$. For each $k \in A$, there is a unique person $P(k)$ who has shaken hands with exactly $k$ people. Therefore, $P(0)$ is the only person who has not shaken hands wit... | m=0, n=50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,744 |
510 For a regular $n$-prism $A_{1} A_{2} \cdots A_{n}-A_{1}^{\prime} A_{2}^{\prime} \cdots A_{n}^{\prime}$, each vertex is colored with one of four colors. If adjacent vertices must have different colors, how many different coloring methods are there? | Consider unfolding a regular $n$-prism into $n$ columns, as shown in Figure 2. The point at the top of the $i$-th column is denoted as $i_{1}$, and the point at the bottom of the $i$-th column is denoted as $i_{2}$.
First, consider Figure 3. The points $A$ and $B$ in the first column can be colored in $4 \times 3 = 12... | 7^{n} - (-3)^{n+1} - 3(-1)^{n+1} + 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,745 |
Let $n$ be a positive integer, and $z_{j}(j=1,2, \cdots, n)$ be complex numbers, with
$\arg z_{j} \in[0, \pi],\left|z_{1}\right| \leqslant\left|z_{2}\right| \leqslant \cdots \leqslant\left|z_{n}\right|$.
If $\left|\prod_{j=1}^{k} z_{j}-1\right| \leqslant \frac{1}{2016}(k=1,2, \cdots, n)$,
prove: $\sum_{j=1}^{n}\left|z... | Proof: Let $z_{j}=r_{j}\left(\cos \theta_{j}+\mathrm{i} \sin \theta_{j}\right)$, where $\theta_{j} \in[0, \pi]$, and $r_{1} \leqslant r_{2} \leqslant \cdots \leqslant r_{n}$.
From the condition, we have
$$
\begin{array}{l}
\left|\left(\prod_{j=1}^{k} r_{j}\right)\left(\cos \sum_{j=1}^{k} \theta_{j}+\mathrm{i} \sin \sum... | \frac{4}{2015} | Algebra | proof | Yes | Yes | cn_contest | false | 729,746 |
Example 1 As shown in Figure $1, \triangle A B C$ is inscribed in $\odot O, \angle A < 90^{\circ}$, and the tangents $X B, Y C$ are drawn to $\odot O$ through points $B$ and $C$ respectively. Draw $O P / / A B$, intersecting $X B$ at point $P$; draw $O Q / / A C$, intersecting $Y C$ at point $Q$. Prove: the line $P Q$ ... | Proof 1 As shown in Figure 1, draw $OD \perp PQ$ at point $D$. We will prove that $OD$ is the radius of $\odot O$.
On the extension of $PB$, take point $R$ such that $BR = CQ$, and connect $OR$, $OB$, and $OC$. It is easy to see that,
$$
\begin{array}{l}
\triangle OBR \cong \triangle OCQ \\
\Rightarrow OR = OQ, \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,747 |
Prove: There exist infinitely many sets of positive integers $(a, b, c)$, such that $a$, $b$, and $c$ are pairwise coprime, and $ab+c$, $bc+a$, $ca+b$ are pairwise coprime.
(2016, China Western Mathematical Invitational) | 【Thought 1】In fact, any two consecutive positive integers are coprime. Furthermore, for three consecutive positive integers $x-1, x, x+1$, as long as $x$ is even, these three positive integers are pairwise coprime. Based on this, consider taking
$(a, b, c)=(x-1, x, x+1)$ ( $x$ is even).
At this point, the condition “$a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,750 |
2. The integer solution $x$ of the equation $(\lg x)^{\lg (\lg x)}=10000$ is $x$ $=$ . $\qquad$ | 2. $10^{100}$.
Let $a=\lg x$. Then
$$
\begin{array}{l}
a^{\operatorname{Ig} a}=10^{4} \Rightarrow(\lg a)(\lg a)=\lg 10^{4}=4 \\
\Rightarrow \lg a= \pm 2 \Rightarrow \lg (\lg x)= \pm 2 .
\end{array}
$$
If $\lg (\lg x)=2$, then
$$
\lg x=10^{2} \Rightarrow x=10^{100} \text {; }
$$
If $\lg (\lg x)=-2$, then $\lg x=10^{-... | 10^{100} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,752 |
5. If $P$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, \angle C=120^{\circ}$. Then the value of the real number $\lambda$ is $\qquad$. | 5. -1 .
Let the circumradius of $\triangle ABC$ be $R$.
From the given information, we have
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} ... | -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,753 |
9. Let $\alpha, \beta$ satisfy the equations respectively
$$
\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-4=0, \\
\beta^{3}-3 \beta^{2}+5 \beta-2=0
\end{array}
$$
then $\alpha+\beta=$ $\qquad$ | 9. 2 .
We have
$$
\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-1=0, \\
(1-\beta)^{3}+2(1-\beta)-1=0,
\end{array}
$$
which means $\alpha-1$ and $1-\beta$ are solutions to the equation $x^{3}+2 x-1=0$.
Since $x^{3}+2 x-1=0$ has only one solution, then
$$
\alpha-1=1-\beta \Rightarrow \alpha+\beta=2
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,754 |
10. Make $\frac{a b^{2}}{a+b}(a \neq b)$ a prime number, the positive integer pair $(a, b)=$ $\qquad$ | 10. $(6,2)$.
Let $d=(a, b)$, $a=d u, b=d v\left(u 、 v \in \mathbf{Z}_{+},(u, v)=1\right)$.
Let $p=\frac{a b^{2}}{a+b}$. Then $d^{2} u v^{2}=p(u+v)$.
Since $(u+v, u)=(u+v, v)=(u, v)=1$, we have $u v^{2} \mid p$.
Also, $p$ is a prime number, so $v=1$, and $u=1$ or $p$.
If $u=1$, then $d^{2}=2 p$.
Since $p$ is a prime nu... | (6,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,755 |
14. (20 points) Given sets $A$ and $B$ are both sets of positive integers, and $|A|=20,|B|=16$. Set $A$ satisfies the following condition: if $a, b, m, n \in A$, and $a+b=$ $m+n$, then $\{a, b\}=\{m, n\}$. Define
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Determine the minimum value of $|A+B|$. | 14. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{20}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{16}\right\}, \\
C_{j}=\left\{a_{i}+b_{j} \mid i=1,2, \cdots, 20\right\},
\end{array}
$$
where $j=1,2, \cdots, 16$.
Thus, $A+B=\bigcup_{j=1}^{16} C_{j}$.
We now prove: $\left|C_{m} \cap C_{n}\right| \leqslant ... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,756 |
2. Given a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with a height of 2 and a base edge length of 1, the center of the upper base $\triangle A_{1} B_{1} C_{1}$ is $P$. A plane $B C D \perp A P$ is made through the lower base edge $B C$, intersecting the edge $A A_{1}$ at point $D$. Then the area of the section... | 2. $\frac{\sqrt{13}}{8}$.
As shown in Figure 1, let the plane $A A_{1} P$ intersect the edges $B C$ and $B_{1} C_{1}$ at points $E$ and $E_{1}$, respectively.
As shown in Figure 2, in the rectangle $A E E_{1} A_{1}$, it is easy to know that $A A_{1}=2, A E=A_{1} E_{1}=\frac{\sqrt{3}}{2}, A_{1} P=\frac{\sqrt{3}}{3}$. L... | \frac{\sqrt{13}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,757 |
3. Let $n$ be a positive integer such that $\sqrt{3}$ lies between $\frac{n+3}{n}$ and $\frac{n+4}{n+1}$. Then $n=$ $\qquad$ | 3. 4 .
Notice that,
$$
\frac{n+4}{n+1}=1+\frac{3}{n+1}<1+\frac{3}{n}=\frac{n+3}{n} \text {. }
$$
Since $\sqrt{3}$ is an irrational number, then
$$
\begin{array}{l}
1+\frac{3}{n+1}<\sqrt{3}<1+\frac{3}{n} \\
\Rightarrow \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow n<\frac{3}{\sqrt{3}-1}<n+1 .
\end{array}
$$
Thus... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,758 |
4. If the sum of 12 distinct positive integers is 2016, then the maximum value of the greatest common divisor of these positive integers is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. 24 .
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d\left(\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1\right)$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then, $2016=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$$
S \... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,759 |
7. If the distance between the two directrices of an ellipse is twice the distance between the two foci, then its eccentricity $e=$ $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 7. $\frac{\sqrt{2}}{2}$.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, with foci at $(-c, 0),(c, 0)$, and the equations of the directrices be $x=-\frac{a^{2}}{c}, x=\frac{a^{2}}{c}$. The distance between the two directrices is $2 \cdot \frac{a^{2}}{c}$, and the focal distance i... | \frac{\sqrt{2}}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,760 |
Example 5 In a cyclic quadrilateral $ABCD$, $BC = CD$, and $\odot C$ is tangent to line $BD$. $I$ is the incenter of $\triangle ABD$. Prove: The line through point $I$ and parallel to $AB$ is tangent to $\odot C$.
(31st Iranian Mathematical Olympiad) | Proof As shown in Figure 6, let line $m$ be tangent to the circumcircle of quadrilateral $ABCD$ at point $D$.
Since $BC = CD$, point $C$ is the midpoint of arc $\overparen{BD}$. Denote $\angle(CD, m)$ as the angle between $CD$ and $m$ (the same notation applies below).
By the tangent-secant theorem, we have
$$
\angle(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,761 |
In the Cartesian coordinate system, the system of inequalities
$$
\left\{\begin{array}{l}
\sqrt{3} x-y \leqslant 0, \\
x-\sqrt{3} y+2 \geqslant 0 \\
y \geqslant 0
\end{array}\right.
$$
represents a planar region whose area is equal to ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\sqrt{3}$
(C) 2
(D) $2 \sqrt{3}$ | 2. B.
The system of inequalities represents the interior (including the boundary) of a triangular plane region, with the coordinates of the three vertices being $A(-2,0), O(0,0), B(1, \sqrt{3})$.
It is easy to see that, $S_{\triangle A O B}=\frac{1}{2} \times 2 \times \sqrt{3}=\sqrt{3}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,762 |
Example 6 Given that $\odot O_{1}$ and $\odot O_{2}$ intersect at two points, let one of the intersection points be $Q$. A point $P$ on $\odot O_{2}$ lies inside $\odot O_{1}$. The line $P Q$ intersects $\odot O_{1}$ at point $X$ (different from point $Q$). A tangent to $\odot O_{1}$ at point $X$ intersects $\odot O_{2... | Proof as shown in Figure 7.
From the given, $O O_{2}$ is perpendicular to $A B$ and intersects $A B$ at its midpoint $M$.
Let $O A=R, A B=2 t$, and the distance from line $l$ to $A B$ be $h$.
Then, in the right triangle $\triangle O A M$,
$$
\begin{array}{l}
R^{2}=t^{2}+(R-h)^{2} \\
\Rightarrow t^{2}+h^{2}=2 R h \\
\R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,763 |
8. Let non-negative real numbers $a, b, c$ satisfy
$$
a b + b c + c a = a + b + c > 0.
$$
Then the minimum value of $\sqrt{a b} + \sqrt{b c} + \sqrt{c a}$ is ( ).
(A) 2
(B) 3
(C) $\sqrt{3}$
(D) $2 \sqrt{2}$ | 8. A.
Assume $a \geqslant b \geqslant c$. By the AM-GM inequality, we have
$$
\begin{array}{l}
(a+b+c)(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) \\
\geqslant(a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a} \\
\geqslant 2 \sqrt{a b} \sqrt{a b}+2 \sqrt{b c} \sqrt{b c}+2 \sqrt{c a} \sqrt{c a} \\
=2(a b+b c+c a),
\end{array}
$$... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,764 |
9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15. Then $a_{2016}=$ | Ni, 9.5.
According to the problem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15 \\
\Rightarrow a_{n+3}=a_{n} .
\end{array}
$$
Then $a_{1}=a_{4}=1, a_{2}=a_{11}=9$,
$$
a_{3}=15-a_{1}-a_{2}=5 \text {. }
$$
Therefore, $a_{2016}=a_{3 \times 672}=a_{3}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,765 |
10. Let $m$ and $n$ be positive integers, and satisfy $24 m=n^{4}$. Then the minimum value of $m$ is $\qquad$ | 10. 54 .
$$
\begin{array}{l}
\text { Given } n^{4}=24 m=2^{3} \times 3 m \text {, we know that } \\
m_{\min }=2 \times 3^{3}=54 \text {. }
\end{array}
$$ | 54 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,766 |
12. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|
$$
has a minimum value of $\qquad$ | 12. 1 .
Notice,
$$
\begin{array}{l}
f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+ \\
4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right| \\
=2\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{4}{5}\right|\right)+ \\
\quad 3\left(\left|x-\frac{2}{3}\right|+\left|x-\frac{4}{5}\right|\right)+4\left|x-\fr... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,767 |
One. (20 points) Let $x, y$ be non-zero real numbers, and satisfy $\frac{x \sin \frac{\pi}{5} + y \cos \frac{\pi}{5}}{x \cos \frac{\pi}{5} - y \sin \frac{\pi}{5}} = \tan \frac{9 \pi}{20}$.
(1) Find the value of $\frac{y}{x}$;
(2) In $\triangle ABC$, if $\tan C = \frac{y}{x}$, find the maximum value of $\sin 2A + 2 \cos... | (1) Let $\frac{y}{x}=\tan \theta$. Then from the given information,
$$
\begin{array}{c}
\frac{\tan \frac{\pi}{5}+\tan \theta}{1-\tan \theta \cdot \tan \frac{\pi}{5}}=\tan \frac{9 \pi}{20} \\
\Rightarrow \tan \left(\theta+\frac{\pi}{5}\right)=\tan \frac{9 \pi}{20} \\
\Rightarrow \theta+\frac{\pi}{5}=k \pi+\frac{9 \pi}{2... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,768 |
Four. (30 points) Let the function
$$
f(x)=\ln x+a\left(\frac{1}{x}-1\right)(a \in \mathbf{R}),
$$
and the minimum value of $f(x)$ is 0.
(1) Find the value of $a$;
(2) If the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=f\left(a_{n}\right)+2\left(n \in \mathbf{Z}_{+}\right),
$$
let $S_{n}=\left[a_{1}... | (1) $f^{\prime}(x)=\frac{1}{x}-\frac{a}{x^{2}}=\frac{x-a}{x^{2}}(x>0)$.
When $a \leqslant 0$, $f^{\prime}(x)>0$, then $f(x)$ is monotonically increasing in the interval $(0,+\infty)$, with no minimum value, which does not meet the requirement.
When $a>0$,
if $0<a$, then $f^{\prime}(x)>0$.
Therefore, the function $f(x)$... | 2n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,769 |
Five. (30 points) Let “ $\sum$ ” denote the cyclic sum. Suppose $a, b, c$ are positive real numbers, and satisfy $abc=1$. For any integer $n \geqslant 2$, prove:
$$
\sum \frac{a}{\sqrt[n]{b+c}} \geqslant \frac{3}{\sqrt[n]{2}}.
$$ | Let's assume $a \leqslant b \leqslant c$. Then
$$
\begin{array}{l}
\sqrt[n]{b+c} \geqslant \sqrt[n]{c+a} \geqslant \sqrt[n]{a+b}, \\
\frac{a}{\sqrt[n]{b+c}} \leqslant \frac{b}{\sqrt[n]{c+a}} \leqslant \frac{c}{\sqrt[n]{a+b}} .
\end{array}
$$
By Chebyshev's inequality, we have
$$
\begin{array}{l}
\sum a=\sum\left(\sqrt... | \sum \frac{a}{\sqrt[n]{b+c}} \geqslant \frac{3}{\sqrt[n]{2}} | Inequalities | proof | Yes | Yes | cn_contest | false | 729,770 |
2. As shown in Figure 2, in $\triangle ABC$, $AC > AB$, and circle $\Gamma$ is the circumcircle of $\triangle ABC$. Let $X$ be a point on segment $AC$, and $Y$ be a point on circle $\Gamma$ (points $A$ and $Y$ are on opposite sides of line $BC$), such that $CX = CY = AB$. Line $XY$ intersects circle $\Gamma$ again at p... | 2. Since $C X=C Y$, then,
$$
\begin{array}{l}
\angle Y X C=\angle X Y C . \\
\text { This indicates, } \overparen{A P}+\overparen{C Y}=\overparen{P C} . \\
\text { Also, } A B=C Y, \text { then } \overparen{A B}=\overparen{C Y} \\
\Rightarrow \overparen{A P}+\overparen{C Y}=\overparen{A P}+\overparen{A B}=\overparen{P ... | PB = PC | Geometry | proof | Yes | Yes | cn_contest | false | 729,771 |
3. Let convex quadrilateral $A B C D$ have two pairs of non-parallel opposite sides. Construct a parallelogram using any pair of adjacent sides, and each parallelogram contains a new vertex different from points $A, B, C, D$. Prove: Among the four new vertices, exactly one is inside quadrilateral $A B C D$. | 3. Obviously, the necessary and sufficient condition for a line passing through point $B$ and parallel to side $A D$ to pass through quadrilateral $A B C D$ is
$$
\angle D A B + \angle A B C > 180^{\circ} \text {. }
$$
The proof below shows that there exists a parallelogram whose newly constructed two adjacent sides b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,772 |
5. Given a convex quadrilateral $A B C D$ satisfying
$$
\begin{array}{l}
\angle A D C=135^{\circ}, \\
\angle A D B-\angle A B D=2 \angle D A B=4 \angle C B D .
\end{array}
$$
If $B C=\sqrt{2} C D$, prove: $A B=B C+A D$. | 5. As shown in Figure 9.
From the given conditions, let $\angle C B D=\alpha$.
Then $\angle D A B=2 \alpha, \angle A D B-\angle A B D=4 \alpha$.
In $\triangle A B D$, we have
$$
\begin{array}{l}
\angle A D B+\angle A B D=180^{\circ}-\angle A=180^{\circ}-2 \alpha \\
\Rightarrow \angle A D B=90^{\circ}+\alpha, \angle A B... | A B=B C+A D | Geometry | proof | Yes | Yes | cn_contest | false | 729,773 |
Example 8 Given that the incircle $\Gamma$ of $\triangle ABC$ touches sides $BC$ and $CA$ at points $P$ and $Q$ respectively, and the excenter of $\angle C$ is $I_{C}$. The circumcircle of $\triangle I_{C}BP$ and the circumcircle of $\triangle I_{C}AQ$ intersect at a second point $T$. Prove that the circumcircle of $\t... | Proof As shown in Figure 9, let circle $\Gamma$ be tangent to $AB$ at point $R$.
Let $\angle BAC = \alpha, \angle ABC = \beta, \angle ACB = \gamma$.
$$
\begin{array}{l}
\angle PTI_{C} = 180^{\circ} - \angle PBI_{C} \\
\quad = 180^{\circ} - \left(90^{\circ} + \frac{\beta}{2}\right) = 90^{\circ} - \frac{\beta}{2} = \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,774 |
5. Does there exist six points $X_{1}, X_{2}, Y_{1}, Y_{2},$ $Z_{1}, Z_{2}$ in the plane, such that all $\triangle X_{i} Y_{j} Z_{k}(1 \leqslant i, j, k \leqslant 2)$ are similar to each other? | 4. Let the circumcenter of $\triangle X Q Y$ be $O$. With $P$ as the inversion center, perform an inversion transformation about a certain circle, resulting in the figure as shown in Figure 16.
Thus, point $X^{\prime}$ is the image of point $X$ under the inversion transformation, and similarly for the others.
Next, w... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,776 |
1. Let the elements of set $A$ be three real numbers, and by appropriately adjusting the order of these three numbers, they can form both a geometric sequence and an arithmetic sequence. Then the common ratio of the geometric sequence is $\square$ $\qquad$ | 1. $-\frac{1}{2}$ or -2.
From the problem, let the three numbers be $a, ra, r^{2}a (r>0, r \neq 1)$.
Since $a, b, c$ can form an arithmetic sequence in some order, we have,
$a+r^{2}a=2ra$ or $a+ra=2r^{2}a$ or $r^{2}a+ra=2a$.
Solving these, we get $r=1$ or $-\frac{1}{2}$ or -2.
Thus, the required common ratio is $-\frac... | -\frac{1}{2} \text{ or } -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,777 |
2. In the Cartesian coordinate system, $O$ is the origin, $A(1,2)$, $B(3,4)$; in the oblique projection, $A B$ becomes $A^{\prime} B^{\prime}$. Then $\overrightarrow{O A^{\prime}} \cdot \overrightarrow{O B^{\prime}}=$ $\qquad$ | $2.5+\frac{5 \sqrt{2}}{2}$.
Note that, in the oblique dimetric projection, the correspondence from the Cartesian coordinate system to the Cartesian coordinate system for coordinates $(x, y)$ is $\left(x+\frac{\sqrt{2} y}{4}, \frac{\sqrt{2} y}{4}\right)$.
Thus, the corresponding points for $(1,2)$ and $(3,4)$ are
$$
\be... | 5+\frac{5 \sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,778 |
3. Please write a continuous even function whose domain and range are both the set of real numbers $\qquad$ . | 3. $y=x \sin x$.
There are infinitely many solutions, but they generally involve trigonometric functions. For example, $y=x \sin x, y=x^{2} \sin x^{2}$. | y=x \sin x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,779 |
4. In space, four lines satisfy that the angle between any two lines is $\theta$. Then $\theta=$ $\qquad$ . | 4. $\arccos \frac{1}{3}$.
As shown in Figure 2, the four lines are translated to the same point $O$. By symmetry, we can take one point $A$, $B$, $C$, and $D$ on each of the four lines, such that
$$
\begin{array}{l}
O A=O B=O C \\
=O D=1 .
\end{array}
$$
Thus, $O$ is the center of the regular tetrahedron $A B C D$.
C... | \arccos \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,780 |
Example 1 Let $p$ be an odd prime, and $a, b$ be positive integers less than $p$. Prove: $a+b=p$ if and only if for any positive integer $n$ less than $p$, $\left[\frac{2 a n}{p}\right]+\left[\frac{2 b n}{p}\right]$ is a positive odd number. | Prove the necessity.
If $a+b=p, n$ is any positive integer less than $p$, let
$$
\left[\frac{2 a n}{p}\right]=u,\left[\frac2 b n}{p}\right]=v \text {. }
$$
Since $p$ is a prime, $\frac{2 a n}{p}$ and $\frac{2 b n}{p}$ are not integers.
Therefore, there exist $\alpha, \beta (0<\alpha, \beta<1)$ such that
$$
\frac{2 a n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,781 |
5. Let four complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy
$$
\begin{array}{l}
\left|z_{1}-z_{2}\right|=1,\left|z_{3}-z_{4}\right|=2, \\
\left|z_{1}-z_{4}\right|=3,\left|z_{2}-z_{3}\right|=4, \\
z=\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right) .
\end{array}
$$
Then the maximum value of $|z|$ is | 5.14 .
Notice,
$$
\begin{array}{l}
|z|=\left|\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right)\right| \\
=\left|z_{1} z_{2}-z_{1} z_{4}-z_{3} z_{2}+z_{3} z_{4}\right| \\
=\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)+\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
\leqslant\left|\left(z_{1}-z_{2}\ri... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,782 |
6. Given the parabola $y^{2}=2 p x(p>0), P$ is a moving point on the negative half-axis of the $x$-axis, $P A 、 P B$ are tangent lines to the parabola, and $A 、 B$ are the points of tangency. Then the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$ . | 6. $-\frac{p^{2}}{4}$.
Let the equation of the tangent line be $x=t y+m$.
Substituting into the parabola equation yields
$$
y^{2}-2 p t y-2 p m=0 \text {. }
$$
From the condition that the line is tangent to the parabola, we get
$$
\begin{array}{l}
\Delta=4 p^{2} t^{2}+8 p m=0 \\
\Rightarrow m=-\frac{p t^{2}}{2}, y=p ... | -\frac{p^{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,783 |
7. Given that when $x$ is in radians, we have
$$
(\sin x)^{\prime}=\cos x \text {. }
$$
then when $x$ is in degrees, the derivative of $\sin x$ is | 7. $\frac{\pi}{180} \cos x$.
When $x$ is in degrees,
$$
\begin{array}{l}
\left(\sin x^{\circ}\right)^{\prime}=\left(\sin \frac{\pi x}{180} \text{rad}\right)^{\prime} \\
=\cos \left(\frac{\pi x}{180} \text{rad}\right) \cdot\left(\frac{\pi x}{180} \text{rad}\right)^{\prime} \\
=\frac{\pi}{180} \cos x^{\circ} .
\end{arra... | \frac{\pi}{180} \cos x | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,784 |
8. The accuracy of HIV testing technology is quite astonishing. If a person is truly HIV positive, blood testing methods have a $99 \%$ chance of detecting it. If a person does not carry HIV, the accuracy of the test is even higher, reaching $99.99 \%$, which means there is only a $0.01 \%$ chance of a false positive. ... | 8.0.50.
Let event $A=\{$ a person carries HIV $\}$,
$B=\{$ a person is diagnosed as carrying HIV through a blood test $\}$.
Then $P(B \mid A)=99\%, P(B \mid \bar{A})=0.01\%$,
$P(A)=0.01\%$.
As shown in Figure 3, let the total area be 1, and the areas of the four regions be $a, b, c, d$.
$$
\begin{array}{r}
\text { The... | 0.50 | Other | math-word-problem | Yes | Yes | cn_contest | false | 729,785 |
9. (16 points) Given the function
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {, }
$$
$g(x)$ is a quadratic function, and it always holds that
$$
f(g(x))=g(f(x)) .
$$
Find $g(x)$. | 9. Let $g(x)=p x^{2}+q x+r(p \neq 0)$.
From $f(g(x))=g(f(x))$, we get
$$
\begin{array}{l}
a\left(p x^{2}+q x+r\right)^{2}+b\left(p x^{2}+q x+r\right)+c \\
=p\left(a x^{2}+b x+c\right)^{2}+q\left(a x^{2}+b x+c\right)+r .
\end{array}
$$
Notice that, $a p \neq 0$.
By comparing the coefficients of $x^{4}$, $x^{3}$, and $x... | g(x)=f(x) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,786 |
10. (20 points) Find the number of all positive integer solutions $(x, y, z)$ to the equation $\arctan \frac{1}{x}+\arctan \frac{1}{y}+\arctan \frac{1}{z}=\frac{\pi}{4}$. | 10. By symmetry, let $x \leqslant y \leqslant z$.
Taking the tangent of both sides of the given equation, we get
$\frac{\frac{1}{y}+\frac{1}{z}}{1-\frac{1}{y z}}=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$
$\Rightarrow \frac{y+z}{y z-1}=\frac{x-1}{x+1}=1-\frac{2}{x+1}$.
If $x \geqslant 5$, then
$1-\frac{2}{x+1} \geqslant 1-\... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,787 |
11. (20 points) For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, do there exist two fixed points $A$ and $B$ such that for any point $M$ on the ellipse, if the slopes $k_{1}$ and $k_{2}$ of $AM$ and $BM$ exist, then $k_{1} k_{2}$ is a constant? If they exist, find all such pairs of points; if not, ex... | 11. Suppose there exist fixed points $A, B$, and let
$$
\begin{array}{l}
A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M(x, y). \\
\text { Then } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \text { and } \\
k_{1} k_{2}=\frac{y-y_{1}}{x-x_{1}} \cdot \frac{y-y_{2}}{x-x_{2}} \\
=\frac{y^{2}-\left(y_{1}+y_{2}\righ... | k_{1} k_{2}=-\frac{b^{2}}{a^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,788 |
一、(40 points) As shown in Figure $1$, the circumcenter of $\triangle A B C$ is $O$, and $D, E$ are arbitrary points on $C A, A B$ respectively. $F, G, H$ are the midpoints of segments $B D, C E, D E$ respectively, and $D E$ intersects the circumcircle of $\triangle F G H$ at another point $I$. Prove that: $O I \perp D ... | As shown in Figure 4, connect $IF$, $IG$, $HF$, $HG$, and $FG$.
From the given, $DC \Perp_{2} HG$, $EB \Perp_{2} FH$.
Also, since $I$, $H$, $G$, and $F$ are concyclic,
$$
\angle IGF = \angle IHF = \angle AED.
$$
Similarly, $\angle IFG = \angle ADE$.
Thus, $\triangle IFG \backsim \triangle ADE$
$$
\Rightarrow IF : FG : ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,789 |
Sure, here is the translated text:
```
II. (40 points) Let $b, c$ be given integers, and the set
$$
M=\left\{x^{2}+b x y+c y^{2} \mid x, y \in \mathbf{Z}\right\} \text {. }
$$
Prove: For any $m, n \in M$, we have $m n \in M$.
``` | $$
\begin{array}{l}
x^{2}+b x y+c y^{2}=\left(x+\frac{b}{2} y\right)^{2}-\frac{b^{2}-4 c}{4} y^{2} \\
=\left(a x+\frac{b}{2} y+\frac{\sqrt{b^{2}-4 c}}{2} y\right)\left(a x+\frac{b}{2} y-\frac{\sqrt{b^{2}-4 c}}{2} y\right) .
\end{array}
$$
For any $m, n \in M$, let
$$
m=p^{2}+b p q+c q^{2}, n=u^{2}+b u v+c v^{2} \text ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,790 |
Three, (50 points) A magician performs a magic trick: letting an audience member write any two numbers on a blackboard. Starting from the third number, each subsequent number is the sum of the two preceding numbers. When enough numbers are written (more than 10), the audience member tells the magician the last number, ... | Three, let the number said by the audience be $x$, the magician only needs to calculate $\frac{1+\sqrt{5}}{2} x$, and the integer obtained by rounding to the nearest whole number is the next number.
In fact, by $a_{n}=a_{n-1}+a_{n-2}$, we know that $\left\{a_{n}\right\}$ is an increasing sequence of integers.
Let $a_{... | \frac{1+\sqrt{5}}{2} x | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,791 |
Four, (50 points) Using equilateral triangles and squares of equal side lengths, how many non-similar convex 11-sided polygons can be formed?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Four, countless convex 11-gons can be assembled.
According to the problem, if a convex 11-gon can be formed, then its interior angles can only be $90^{\circ} 、 120^{\circ} 、 150^{\circ}$, and the corresponding exterior angles are $90^{\circ} 、 60^{\circ} 、 30^{\circ}$. Let the number of these angles be $x 、 y 、 11-x-y$... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,793 |
Given $0<a_{i} \leqslant \frac{1}{2}(i=1,2, \cdots, n)$. Prove:
$$
\begin{array}{l}
\frac{\sum_{i=1}^{n} a_{i}^{2}}{\sum_{i=1}^{n} a_{i}}-\sqrt[n]{\prod_{i=1}^{n} a_{i}} \\
\geqslant \frac{\sum_{i=1}^{n}\left(1-a_{i}\right)^{2}}{\sum_{i=1}^{n}\left(1-a_{i}\right)}-\sqrt[n]{\prod_{i=1}^{n}\left(1-a_{i}\right)} .
\end{ar... | Prove that the original inequality is equivalent to
$$
\begin{array}{c}
\frac{n \sum_{i=1}^{n} a_{i}^{2}}{\sum_{i=1}^{n} a_{i}}-\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} a_{i}-n \sqrt[n]{\prod_{i=1}^{n} a_{i}} \\
\geqslant \frac{n \sum_{i=1}^{n}\left(1-a_{i}\right)^{2}}{\sum_{i=1}^{n}\left(1-a_{i}\right)}-\sum_{i=1}^{n}\left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,794 |
Given $CD$ is the altitude on the hypotenuse $AB$ of the right triangle $\triangle ABC$, and $\odot O_{1}$ and $\odot O_{2}$ are the incircles of $\triangle ACD$ and $\triangle BCD$ respectively. The other common external tangent of the two circles intersects $BC$ and $AC$ at points $P$ and $Q$ respectively. Prove:
$$
... | Proof: Let $PQ$ and $CD$ intersect at point $S$, and $\odot O_{1}$ is tangent to $AD$ at point $E$. Auxiliary lines are as shown in Figure 3. Let $BC = a$, $CA = b$, $AB = c$.
It is easy to know that $AD = \frac{b^2}{c}$, $CD = \frac{ab}{c}$.
From the given conditions,
$$
\begin{array}{l}
O_{1}E = DE = \frac{1}{2}(AD +... | PQ = AC + BC - AB | Geometry | proof | Yes | Yes | cn_contest | false | 729,795 |
4. As shown in Figure $12, A B>B C>A C, \triangle A B C$ has three excircles which are internally tangent to the circumcircle of $\triangle A B C$ at points $D, E, F$, respectively. Line $F D$ intersects $A C$ at point $Y$, and line $F E$ intersects $B C$ at point $X$. $O, I$ are the circumcenter and incenter of $\tria... | Let $B$ be a pseudo-incircle tangent to $AB$ and $BC$ at points $U$ and $V$, respectively. Then $UV$ passes through point $Y$. Let $YE$ intersect $\odot O$ at point $S$. Thus, $B$, $I$, and $S$ are collinear.
Therefore, $OY^2 - IY^2 = YE \cdot YS + R^2 - YE \cdot YS = R^2$.
Similarly, $OX^2 - IX^2 = R^2$.
Thus, $OX^2 -... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,796 |
Example 1 Find the solutions in the set of integers for
$$
x^{2010}-2006=4 y^{2009}+4 y^{2008}+2007 y
$$
$^{[3]}$
(2009, Macedonian Mathematical Olympiad) | 【Analysis】It seems that there is no term of $x^{2}+1$, but we find that $x^{2010}=\left(x^{1005}\right)^{2}$ is also in the form of a square. Therefore, by using factorization, we get
$$
x^{2010}+1=(y+1)\left(4 y^{2008}+2007\right) \text {. }
$$
Since $4 y^{2008}+2007$ is a number of the form $4 k+3$, it must contain ... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,797 |
Example 2 Find all prime numbers $p$ such that there exists a positive odd number $n$ and an integer-coefficient polynomial $Q(x)$, satisfying that the polynomial $1+p n^{2}+\prod_{i=1}^{2 p-2} Q\left(x^{i}\right)$ has at least one integer root. [4] (2009, Turkey National Team Selection Exam) | 【Analysis】This problem does not have the desired $x^{2}+1$ type expression at first glance, but by using the properties of integer coefficient polynomials, we have
$$
\prod_{i=1}^{2 p-2} Q\left(x^{i}\right) \equiv\left(\prod_{i=1}^{p-1} Q\left(x^{i}\right)\right)^{2}(\bmod p) .
$$
This naturally leads us to analyze th... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,798 |
Example 3 Let $n$ be a positive integer, $p$ be a prime, and
$$
p=3(\bmod 4) \text {. }
$$
Question: Can the set of numbers $\{n, n+1, \cdots, n+p-2\}$ be partitioned into two non-empty disjoint subsets such that the product of the numbers in each subset is equal? | 【Analysis】Notice that, among $p-1$ consecutive positive integers, there is at most one multiple of $p$.
We discuss in two cases.
(1) If there is exactly one multiple of $p$ among these $p-1$ numbers, then it is impossible to divide them into two sets satisfying the condition.
(2) If there is no multiple of $p$ among th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,799 |
Example 1 Find all positive integers $a, b, c$, such that for any real numbers $u, v (0 \leqslant u < v \leqslant 1)$, there exists a positive integer $n$, such that
$$
\left\{\sqrt{a n^{2} + b n + c}\right\} \in (u, v) \text{. }
$$
(2016, Peking University Mathematics Autumn Camp) | Notice,
$$
\begin{array}{l}
\sqrt{a n^{2}+b n+c}-\sqrt{a n^{2}} \\
=\frac{b n+c}{\sqrt{a n^{2}+b n+c}+\sqrt{a n^{2}}}.
\end{array}
$$
When $n \rightarrow+\infty$,
$$
\begin{array}{l}
\lim _{n \rightarrow+\infty}\left(\sqrt{a n^{2}+b n+c}-n \sqrt{a}\right) \\
=\lim _{n \rightarrow+\infty} \frac{b n+c}{\sqrt{a n^{2}+b n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,801 |
Example 2 Given $\alpha>0$. Prove: There exist infinitely many positive integers $n$, such that $\left[n^{2} \alpha\right]$ is even, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
(2015, Peking University Mathematics Summer Camp) | Proof (1) If $\alpha$ is a rational number, let's assume
$$
\alpha=\frac{p}{q}\left(p, q \in \mathbf{Z}_{+},(p, q)=1\right) \text {. }
$$
Take $n=2 k q\left(k \in \mathbf{Z}_{+}\right)$. At this point,
$$
\left[n^{2} \alpha\right]=\left[4 k^{2} q^{2} \frac{p}{q}\right]=\left[4 k^{2} p q\right]=4 k^{2} p q \text {, }
$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,802 |
Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$, the following holds:
$$
f(x+y)+f(x) f(y)=f(x y)+2 x y+1 .
$$
(46th IMO Shortlist) | Let $x=y=0$, we get $f^{2}(0)=1$.
If $f(0)=1$, let $y=0$, we get $f(x)=1$.
Substituting into equation (1) yields
$$
1+1 \times 1=1+2 x y+1 \Rightarrow x y=0 \text {. }
$$
This does not meet the problem's requirements.
Therefore, $f(0)=-1$.
Let $x=1, y=-1$, we get
$$
f(1) f(-1)=f(-1) \text {. }
$$
(1) If $f(1)=1$, let ... | f(x)=2x-1, f(x)=-x-1, f(x)=x^2-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,803 |
Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all real numbers $x, y$, we have
$$
f\left(f(y)+x^{2}+1\right)+2 x=y+f^{2}(x+1) \text {. }
$$
(2014, Turkey National Team Selection Exam) | Let $x=0$, we get
$$
f(f(y)+1)=y+f^{2}(1) \text {. }
$$
In equation (2), replace $y$ with $f(y)+1$ to get
$$
f\left(y+f^{2}(1)+1\right)=f(y)+1+f^{2}(1) \text {. }
$$
Replace $y$ with $x^{2}+y$ in the above equation to get
$$
\begin{array}{l}
f\left(x^{2}+y+f^{2}(1)+1\right)=f(y)+1+f^{2}(1) \\
\Rightarrow f\left(x^{2}... | f(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,804 |
Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$, we have
$$
f(1+x y)-f(x+y)=f(x) f(y),
$$
and $f(-1) \neq 0 .{ }^{[3]}$
(53rd IMO Shortlist) | Let $x=0$, we get
$f(1)-f(y)=f(0) f(y)$;
Let $x=1$, we get
$$
f(1+y)-f(1+y)=f(1) f(y) \text {. }
$$
Thus $f(1)=0$ (the case $f(y)=0$ is discarded). Then $f(0)=-1$.
Let $y=-1$, we get
$$
f(1-x)-f(x-1)=f(-1) f(x) \text {. }
$$
Replace $x$ with $1-x$ in equation (1), we get
$$
f(x)-f(-x)=f(-1) f(1-x) \text {. }
$$
Repl... | f(x)=x-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,805 |
Property 2 As shown in Figure 3, let the A-excircle $\odot Q$ of $\triangle ABC$ touch the sides $AB$ and $AC$ at points $E$ and $F$, respectively, and be internally tangent to the circumcircle $\odot O$ of $\triangle ABC$ at point $P$. The incircle $\odot I$ of $\triangle ABC$ touches side $BC$ at point $D$, and the o... | Proof of Property 2 As shown in Figure 3, draw the tangent of $\odot I$ through point $S$, intersecting $A B$ and $A C$ at points $J$ and $K$ respectively. It is easy to see that,
$\triangle J S I \backsim \triangle I D B \Rightarrow J S \cdot B D=S I \cdot D I$.
Similarly, $S K \cdot D C=S I \cdot D I$.
Thus, $\frac{J... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,806 |
1. If the function
$$
f(x)=3 \cos \left(\omega x+\frac{\pi}{6}\right)-\sin \left(\omega x-\frac{\pi}{3}\right)(\omega>0)
$$
has the smallest positive period of $\pi$, then the maximum value of $f(x)$ in the interval $\left[0, \frac{\pi}{2}\right]$ is . $\qquad$ | $-1.2 \sqrt{3}$.
Notice,
$$
\begin{aligned}
f(x) & =3 \cos \left(\omega x+\frac{\pi}{6}\right)-\sin \left(\omega x-\frac{\pi}{3}\right) \\
& =3 \cos \left(\omega x+\frac{\pi}{6}\right)-\sin \left(\omega x+\frac{\pi}{6}-\frac{\pi}{2}\right) \\
& =3 \cos \left(\omega x+\frac{\pi}{6}\right)+\cos \left(\omega x+\frac{\pi}{... | 2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,807 |
2. Given the sets
$$
A=\left\{x \mid x^{2}-3 x+2 \leqslant 0\right\}, B=\left\{x \left\lvert\, \frac{1}{x-3}<a\right.\right\} \text {. }
$$
If $A \subseteq B$, then the range of real number $a$ is
$\qquad$ | 2. $\left(-\frac{1}{2},+\infty\right)$.
It is easy to see that $A=\{x \mid 1 \leqslant x \leqslant 2\}$.
When $x \in A$, $\frac{1}{x-3} \in\left[-1,-\frac{1}{2}\right]$.
Therefore, $a>-\frac{1}{2}$.
Thus, the range of values for $a$ is $\left(-\frac{1}{2},+\infty\right)$. | \left(-\frac{1}{2},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,808 |
3. The number of zeros of the function $f(x)=x^{2} \ln x+x^{2}-2$ is . $\qquad$ | 3.1 .
From the condition, we have
$$
f^{\prime}(x)=2 x \ln x+x+2 x=x(2 \ln x+3) \text {. }
$$
When $0\mathrm{e}^{-\frac{3}{2}}$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is a decreasing function on the interval $\left(0, \mathrm{e}^{-\frac{3}{2}}\right)$ and an increasing function on the interval $\left(\mathrm{e}^{-\frac{3}{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,809 |
4. As shown in Figure 1, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the size of the dihedral angle $B-A_{1} C-D$ is $\qquad$ . | 4. $120^{\circ}$.
Let the edge length of the cube be 1.
As shown in Figure 4, draw $B E \perp A_{1} C$ at point $E$, and connect $D E$.
By the properties of the cube, we know that
$$
\triangle A_{1} D C \cong \triangle A_{1} B C \text {. }
$$
Then $D E \perp A_{1} C, \angle B E D$
is the plane angle of the dihedral a... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,810 |
7. Given $z \in \mathbf{C}$. If the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (where $\mathrm{i}$ is the imaginary unit) has real roots, then the minimum value of $|z|$ is $\qquad$ . | 7.1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be a real root of the original equation.
$$
\begin{array}{l}
\text { Then } x_{0}^{2}-2(a+b \mathrm{i}) x_{0}+\frac{3}{4}+\mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
x_{0}^{2}-2 a x_{0}+\frac{3}{4}=0, \\
-2 b x_{0}+1=0
\end{array}\right. \\
\Rightarrow ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,811 |
8. Distribute 16 identical books to four classes, with each class getting at least one book, and the number of books each class receives being different. The number of different distribution methods is $\qquad$ (answer with a number).
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 8. 216 .
Decomposing 16 into the sum of four distinct positive integers has 9 different ways:
$$
\begin{array}{l}
16=1+2+3+10, 16=1+2+4+9, \\
16=1+2+5+8, 16=1+2+6+7, \\
16=1+3+4+8, 16=1+3+5+7, \\
16=1+4+5+6, 16=2+3+4+7, \\
16=2+3+5+6 .
\end{array}
$$
Therefore, the number of different allocation methods that meet the... | null | Other | math-word-problem | Yes | Yes | cn_contest | false | 729,812 |
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