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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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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2. $\frac{[\sqrt{2013}]+[\sqrt{2014}]+[\sqrt{2015}]+[\sqrt{2016}]}{[\sqrt{2014}] \times[\sqrt{2015}]}$
$=$ ( $[x]$ represents the greatest integer not exceeding the real number $x$). | 2. $\frac{1}{11}$.
Notice that, $44^{2}=1936<2013$,
$$
\begin{array}{l}
2016<2025=45^{2} . \\
\text { Then }[\sqrt{2013}]=[\sqrt{2014}]=[\sqrt{2015}] \\
=[\sqrt{2016}]=44 . \\
\text { Therefore, } \frac{[\sqrt{2013}]+[\sqrt{2014}]+[\sqrt{2015}]+[\sqrt{2016}]}{[\sqrt{2014}] \times[\sqrt{2015}]} \\
=\frac{44+44+44+44}{4... | \frac{1}{11} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,061 |
3. In quadrilateral $A B C D$, it is known that $B C=8, C D=12$, $A D=10, \angle A=\angle B=60^{\circ}$. Then $A B=$ | $$
3.9+\sqrt{141} \text {. }
$$
Extend $A D$ and $B C$ to intersect at point $P$.
Let $A B=x$. Then
$$
\begin{array}{l}
D P=x-10, \\
P C=x-8 .
\end{array}
$$
Draw $D H \perp$
$P C$ at point $H$.
In Rt $\triangle P H D$ and Rt $\triangle C H D$, we have respectively
$$
\begin{array}{l}
D H^{2}=D P^{2}-P H^{2}, D H^{2}... | 9+\sqrt{141} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,062 |
4. Given that $M$ is the least common multiple of 15 consecutive natural numbers $1,2, \cdots, 15$. If a divisor of $M$ is divisible by exactly 14 of these 15 natural numbers, it is called a "good number" of $M$. Then the number of good numbers of $M$ is $\qquad$.
| 4.4.
It is known that $M=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$.
Since $2 \times 11, 2 \times 13$ are both greater than 15, therefore,
$$
\begin{array}{l}
\frac{M}{11}=2^{3} \times 3^{2} \times 5 \times 7 \times 13, \\
\frac{M}{13}=2^{3} \times 3^{2} \times 5 \times 7 \times 11,
\end{array}
$$
each... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,063 |
5. Let the sequence of natural numbers from $1 \sim 8$ be $a_{1}, a_{2}$, $\cdots, a_{8}$. Then
$$
\begin{array}{l}
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+\left|a_{4}-a_{5}\right|^{\prime}+ \\
\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+\left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1... | 5. 32 .
From the problem, we have
$$
\begin{aligned}
S= & \left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+ \\
& \left|a_{4}-a_{5}\right|+\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+ \\
& \left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1}\right| .
\end{aligned}
$$
Removing the absolute value s... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,064 |
Three, (10 points) Given
$$
a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=0 \text {. }
$$
Prove: Among $a$, $b$, and $c$, at least two of them are equal. | $$
\begin{array}{l}
a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b) \\
=(b-c) a^{2}+b^{2} c-b^{2} a+c^{2} a-c^{2} b \\
=(b-c) a^{2}-\left(b^{2}-c^{2}\right) a+b c(b-c) \\
=(b-c)\left[a^{2}-(b+c) a+b c\right] \\
=(b-c)(a-b)(a-c)=0 .
\end{array}
$$
Therefore, at least one of $b-c$, $a-b$, and $a-c$ is equal to 0, which means that at l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,065 |
Example 7 Find all positive integers $n$, such that there exists an integer-coefficient polynomial $P(x)$, satisfying $P(d)=\left(\frac{n}{d}\right)^{2}$, where, $d$ is each divisor of $n$. [3] | For a polynomial $P(x)$ with integer coefficients, and positive integers $a, b (a \neq b)$, we always have
$$
(a-b) \mid (P(a)-P(b)).
$$
This is an invariant.
If $n=1$, then $P(1)=1$. Hence, we can take the polynomial $P(x)=x$.
If $n$ is a prime number, then $n$ has only two factors, 1 and $n$. The polynomial $P(x)$ ... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,066 |
Four. (15 points) In the convex quadrilateral $ABCD$, it is known that $\angle BAC=30^{\circ}, \angle ADC=150^{\circ}$, and $AB=DB$. Prove: $AC$ bisects $\angle BCD$.
保留源文本的换行和格式，直接输出翻译结果如下：
Four. (15 points) In the convex quadrilateral $ABCD$, it is known that $\angle BAC=30^{\circ}, \angle ADC=150^{\circ}$, and $AB... | As shown in Figure 4, construct the symmetric point $E$ of point $B$ with respect to $AC$, and connect $AE$, $BE$, and $DE$. Then $\triangle ABE$ is an equilateral triangle.
Let $\angle DBE = \theta$. Then $\angle ABD = 60^{\circ} + \theta$.
Since $AB = DB$, we have
$$
\angle ADB = \frac{180^{\circ} - (60^{\circ} + \th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,067 |
Five, (15 points) A school assigns numbers to the contestants participating in a math competition, with the smallest number being 0001 and the largest number being 2014. No matter which contestant steps forward to calculate the average of the numbers of all other contestants in the school, the average is always an inte... | Let the school have a total of $n$ participants, whose admission numbers are
$$
1=x_{1}<x_{2}<\cdots<x_{n-1}<x_{n}=2014 .
$$
According to the problem, we have
$$
S_{k}=\frac{x_{1}+x_{2}+\cdots+x_{n}-x_{k}}{n-1}(k=1,2, \cdots, n) \in \mathbf{Z}_{+} .
$$
For any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
S_{i}-S_{... | 34 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,068 |
In the Cartesian coordinate system, the equation
$$
x^{2}+2 x \sin x y+1=0
$$
represents the figure ( ).
(A) a straight line
(B) a parabola
(C) a point
(D) none of the above | -,1. D.
It is known that, $x=-1, \sin x y=1$, or $x=1, \sin x y=-1$.
Therefore, the graph represented by this equation consists of the point sequences $\left(-1,2 k \pi-\frac{\pi}{2}\right)$ $(k \in \mathbf{Z})$ and the point sequences $\left(1,2 k \pi-\frac{\pi}{2}\right)(k \in \mathbf{Z})$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,069 |
2. Given the radius of the base of a cylinder is $r$, the height is $h$, the volume is 2, and the surface area is 24. Then $\frac{1}{r}+\frac{1}{h}=(\quad)$.
(A) 6
(B) 8
(C) 12
(D) 24 | 2. A. From the conditions, we know $\pi r^{2} h=2, 2 \pi r^{2}+2 \pi r h=24$. Dividing the two equations gives $\frac{1}{r}+\frac{1}{h}=6$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,070 |
3. Given a geometric sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and for any positive integer $n$ we have $S_{n+2}=4 S_{n}+3$. Then $a_{2}=$ ( ).
(A) 2
(B) 6
(C) 2 or 6
(D) 2 or -6 | 3. C.
Let the common ratio be $q$. Since
$$
\begin{array}{l}
q S_{n}=q\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
=a_{2}+a_{3}+\cdots+a_{n+1},
\end{array}
$$
we have, $S_{n+1}=q S_{n}+a_{1}$.
Thus, $S_{n+2}=q\left(q S_{n}+a_{1}\right)+a_{1}$
$$
=q^{2} S_{n}+a_{1}(q+1) \text {. }
$$
Comparing with the given condition, w... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,071 |
4. If the inequality $\frac{4 x}{a}+\frac{1}{x} \geqslant 4$ holds for all $x$ in the interval $[1,2]$, then the range of the real number $a$ is ( ).
(A) $\left(0, \frac{4}{3}\right]$
(B) $\left(1, \frac{4}{3}\right)$
(C) $\left[1, \frac{4}{3}\right]$
(D) $\left[\frac{16}{7}, \frac{4}{3}\right]$ | 4. A.
Let $x=1$, we get
$$
\frac{4}{a} \geqslant 3 \Rightarrow 0<a \leqslant \frac{4}{3}\left(a \in \mathbf{R}_{+}\right) \text {. }
$$
Thus, the original inequality becomes that for any $x \in[1,2]$, we have
$$
a \leqslant \frac{4 x^{2}}{4 x-1} \text {. }
$$
Therefore, we only need to find the minimum value of the ... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 728,072 |
5. Line $l$ lies on plane $\alpha$, line $m$ is parallel to plane $\alpha$, and is skew to line $l$. A moving point $P$ is on plane $\alpha$, and is equidistant from lines $l$ and $m$. Then the locus of point $P$ is ( ).
(A) Line
(B) Ellipse
(C) Parabola
(D) Hyperbola | 5. D.
Let the projection of $m$ on plane $\alpha$ be $m^{\prime}$, and $m^{\prime}$ intersects line $l$ at point $O$.
In plane $\alpha$, establish a Cartesian coordinate system with $O$ as the origin and line $l$ as the $y$-axis. Then, the equation of $m^{\prime}$ is $y=k x$.
Let point $P(x, y)$.
The distance from po... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,073 |
1. If positive real numbers $a, b$ satisfy
$$
\log _{8} a+\log _{4} b^{2}=5, \log _{8} b+\log _{4} a^{2}=7 \text {, }
$$
then $\log _{4} a+\log _{8} b=$ $\qquad$ | $=1.4$.
Let $a=2^{x}, b=2^{y}$. Then $\frac{x}{3}+y=5, \frac{y}{3}+x=7$.
Thus, $x=6, y=3$.
Therefore, $\log _{4} a+\log _{8} b=\frac{x}{2}+\frac{y}{3}=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,075 |
2. Let $x=\sin ^{2} \alpha+\sin \left(\alpha+\frac{2 \pi}{3}\right) \cdot \sin \left(\alpha+\frac{\pi}{3}\right)$. When $\alpha=\frac{67 \pi}{2014}$, the first digit after the decimal point of $x$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the transla... | 2.7.
Notice that,
$$
\begin{array}{l}
\sin \left(\alpha+\frac{2 \pi}{3}\right)=-\frac{1}{2} \sin \alpha+\frac{\sqrt{3}}{2} \cos \alpha, \\
\sin \left(\alpha+\frac{\pi}{3}\right)=\frac{1}{2} \sin \alpha+\frac{\sqrt{3}}{2} \cos \alpha .
\end{array}
$$
Multiplying the two equations gives
$$
\sin \left(\alpha+\frac{2 \pi... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 728,076 |
Example 8 Let $p \geqslant 3$ be a prime number, and $r_{i}$ be the remainder when the integer $\frac{i^{p-1}-1}{p}$ is divided by $p$, where $i=1,2, \cdots, p-1$. Prove:
$$
r_{1}+2 r_{2}+\cdots+(p-1) r_{p-1} \equiv \frac{p+1}{2}(\bmod p) .
$$ | Since $(i, p)=1$, we have $\frac{i^{p-1}-1}{p}$ is a positive integer.
Let $\frac{i^{p-1}-1}{p}=a_{i} p+r_{i}\left(a_{i}\right.$ be an integer, $\left.i=1,2, \cdots, p-1\right)$.
Then $\frac{i^{p}-i}{p}=i p a_{i}+i r_{i}$.
Using its dual form
$$
\frac{(p-i)^{p}-(p-i)}{p}=(p-i) p a_{i}+(p-i) r_{i},
$$
Adding the two eq... | r_{1}+2 r_{2}+\cdots+(p-1) r_{p-1} \equiv \frac{p+1}{2}(\bmod p) | Number Theory | proof | Yes | Yes | cn_contest | false | 728,077 |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}+a_{n-1}(n \geqslant 2) \text {. }
$$
If $a_{7}=8$, then $a_{1}+a_{2}+\cdots+a_{10}=$ $\qquad$ | 3.88.
From the problem, we know that $a_{7}=8 a_{2}+5 a_{1}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{10}$
$$
=88 a_{2}+55 a_{1}=11 a_{7}=88 .
$$ | 88 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,078 |
$$
\begin{array}{l}
\text { 4. If } a=1+\mathrm{i}, b=2+\mathrm{i}, c=3+\mathrm{i}, \\
x=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i},
\end{array}
$$
then $\left|a+b x+c x^{2}\right|=$ | 4. $\sqrt{3}$.
Notice that $x$ satisfies $x^{2}+x+1=0$.
Thus, $x^{3}=1,|x|=1, x \bar{x}=1$.
Also, the imaginary parts of $a, b, c$ are equal, and combining this with $x^{2}+x+1=0$, we only need to compute for $a=1, b=2, c=3$.
$$
\begin{array}{l}
\text { Hence }\left|a+b x+c x^{2}\right|^{2}=\left(a+b x+c x^{2}\right)... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,079 |
5. Arrange the numbers in the set $\left\{2^{x}+2^{y}+2^{z} \mid x 、 y 、 z \in \mathbf{N}, x<y<z\right\}$ in ascending order. The 100th number is $\qquad$ (answer with a number).
| 5.577.
Notice that the number of combinations $(x, y, z)$ such that $0 \leqslant x<y<z \leqslant n$ is $\mathrm{C}_{n+1}^{3}$.
Since $\mathrm{C}_{9}^{3}=84<100<120=\mathrm{C}_{10}^{3}$, the 100th number must satisfy $z=9$.
Also notice that the number of combinations $(x, y)$ such that $0 \leqslant x<y \leqslant m$ i... | 577 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,080 |
6. Given the function $f(x)$ satisfies
$$
f(x)=\left\{\begin{array}{ll}
x-3, & x \geqslant 1000 ; \\
f(f(x+5)), & x<1000 .
\end{array}\right.
$$
Then $f(84)=$ . $\qquad$ | 6.997.
Let $f^{(n)}(x)=\underbrace{f(f(\cdots f(x)))}_{n \uparrow}$. Then
$$
\begin{aligned}
& f(84)=f(f(89))=\cdots=f^{(184)}(999) \\
= & f^{(185)}(1004)=f^{(184)}(1001)=f^{(183)}(998) \\
= & f^{(184)}(1003)=f^{(183)}(1000)=f^{(182)}(997) \\
= & f^{(183)}(1002)=f^{(182)}(999)=f^{(183)}(1004) \\
= & f^{(182)}(1001)=f^... | 997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,081 |
1. Let $A$ and $B$ be two moving points on the ellipse $\frac{x^{2}}{2}+y^{2}=1$, and $O$ be the origin. Also, $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$. Let point $P$ be on $AB$, and $O P \perp A B$. Find the value of $|O P|$. | Three, 1. Let $A(a \cos \alpha, a \sin \alpha), B(-b \sin \alpha, b \cos \alpha)$.
Substituting into the ellipse equation, we get
$$
\begin{array}{l}
a^{2}\left(\frac{1}{2} \cos ^{2} \alpha+\sin ^{2} \alpha\right)=1, b^{2}\left(\frac{1}{2} \sin ^{2} \alpha+\cos ^{2} \alpha\right)=1 \\
\Rightarrow \frac{1}{a^{2}}+\frac{... | \frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,082 |
2. Inside the tetrahedron $ABCD$ there is a point $O$, satisfying $OA=OB=OC=4, OD=1$.
Find the maximum volume of the tetrahedron $ABCD$. | 2. First, fix points $A$, $B$, $C$, and $O$. To maximize the volume of tetrahedron $ABCD$, the distance from point $D$ to plane $ABC$ should be maximized.
Since point $D$ moves on a sphere with center $O$ and radius 4, the volume is maximized when $OD \perp$ plane $ABC$.
Let the projection of $O$ on plane $ABC$ be po... | 9 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,083 |
3. Let the function $f(x)=1-\mathrm{e}^{-x}$. Prove:
(1) When $x>0$, $f(x)>\frac{x}{x+1}$;
(2) If the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n} \mathrm{e}^{-a_{n+1}}=f\left(a_{n}\right)$, then the sequence $\left\{a_{n}\right\}$ is decreasing, and $a_{n}<\frac{1}{2^{n}}$. | $$
\begin{array}{l}
f(x)>\frac{x}{x+1} \\
\Leftrightarrow \mathrm{e}^{-x} < \frac{1}{x+1} \\
\Leftrightarrow (x+1)\mathrm{e}^{-x} < 1 \\
\Leftrightarrow \mathrm{e}^{-x} < \frac{1}{x+1} \\
\Leftrightarrow \mathrm{e}^{-x} < \frac{1}{x+1} \\
\text{Let } h(x) = (x+1)\mathrm{e}^{-x}. \text{ Then } h'(x) = -x\mathrm{e}^{-x} ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,084 |
1. (50 points) As shown in Figure 1, given an acute triangle $\triangle ABC$ satisfying $AB > AC$, $O$ and $H$ are the circumcenter and orthocenter of $\triangle ABC$, respectively. Line $BH$ intersects $AC$ at point $B_1$, and line $CH$ intersects $AB$ at point $C_1$. If $OH \parallel B_1C_1$, prove: $\cos 2B + \cos 2... | 1. Connect $A O$ and $A H$.
Since $A, B_{1}, H, C_{1}$ are concyclic, and $O H \parallel B_{1} C_{1}$, we have,
$$
\begin{array}{l}
\angle O H C_{1}=\angle H C_{1} B_{1}=\angle H A B_{1} \\
=90^{\circ}-\angle A C B .
\end{array}
$$
Also, since $B, C, B_{1}, C_{1}$ are concyclic, we have,
$$
\begin{array}{l}
\angle A ... | \cos 2B + \cos 2C + 1 = 0 | Geometry | proof | Yes | Yes | cn_contest | false | 728,085 |
2. (50 points) Let $0 < x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} (n \geqslant 3)$. Prove:
$$
\begin{array}{l}
\frac{x_{1} x_{2}}{x_{3}}+\frac{x_{2} x_{3}}{x_{4}}+\cdots+\frac{x_{n-2} x_{n-1}}{x_{n}}+\frac{x_{n-1} x_{n}}{x_{1}}+\frac{x_{n} x_{1}}{x_{2}} \\
\geqslant x_{1}+x_{2}+\cdots+x_{n},
\end{array}
$$... | 2. Let $f_{n}=\frac{x_{1} x_{2}}{x_{3}}+\frac{x_{2} x_{3}}{x_{4}}+\cdots+\frac{x_{n-2} x_{n-1}}{x_{n}}+$
$$
\frac{x_{n-1} x_{n}}{x_{1}}+\frac{x_{n} x_{1}}{x_{2}}-x_{1}-x_{2}-\cdots-x_{n} .
$$
We will prove $f_{n} \geqslant 0$ using mathematical induction.
When $n=3$,
$$
\begin{aligned}
f_{3}= & \frac{x_{1} x_{2}}{x_{3... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,086 |
3. (50 points) Prove: In the first octant of the Cartesian coordinate system $(x, y, z$ coordinates are all greater than 0$)$, there exists a cube with a side length of 2014, which contains exactly 2014 prime points $(x, y, z$ coordinates on all three axes are prime numbers).
| 3. First, prove a lemma.
Lemma Let $f(n)$ be the number of primes in the closed interval $[2, n+1]$. Then for any $0 \leqslant k \leqslant f(n)$, there exist $n$ consecutive positive integers, among which exactly $k$ are primes.
Proof Let $F(n, l)$ be the number of primes in the closed interval $[l, n+l-1]$.
When $l$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,087 |
1. Find the unit digit of $(2+\sqrt{3})^{2013}$. | Notice,
$$
\begin{array}{l}
{\left[(2+\sqrt{3})^{2013}\right]=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}-1 .} \\
\text { Let } a_{n}=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}=4 a_{n-1}-a_{n-2}, \\
a_{0}=2, a_{1}=4 .
\end{array}
$$
Then $2 \mid a_{n}$,
$$
\begin{array}{l}
a_{n} \equiv 2,4,4,2,4,4,2, \cdots(\bmod 5) \\
\Righ... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,088 |
4. (50 points) To save the planet Cybertron, the Autobots established a Space Bridge on Earth consisting of $n(n \geqslant 3)$ energy pillars, which are located at $n$ points on a plane, with no three points being collinear. The activation method is as follows: arbitrarily select an energy pillar, and emit a laser from... | 4. Let the points be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Suppose the path of the laser is $a_{1}, a_{2}, \cdots$, where $a_{i} \in \{A_{1}, A_{2}, \cdots, A_{n}\}$.
Consider the set of pairs of adjacent points
$$
\{(a_{i}, a_{i+1}) \mid i=1,2, \cdots\},
$$
which is an infinite set, but its values can be at most ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,089 |
5. (50 points) As shown in Figure 2, given an acute triangle $\triangle ABC$ with its circumcircle $\odot O$, a tangent line $l$ is drawn through point $A$ to $\odot O$, and $l$ intersects line $BC$ at point $D$. Point $E$ is on the extension of $DA$, and point $F$ is on the minor arc $\overparen{BC}$. Line $EF$ inters... | 5. Let $E F$ intersect $B C$ at point $H$.
Applying Menelaus' theorem to line $G C Q$ and $\triangle H E D$, we get $\frac{H G}{G E} \cdot \frac{E Q}{Q D} \cdot \frac{D C}{C H}=1$.
Applying Menelaus' theorem to line $P B F$ and $\triangle H E D$, we get
$$
\frac{H F}{F E} \cdot \frac{E P}{P D} \cdot \frac{D B}{B H}=1 ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,090 |
6. (50 points) Given that for any $x, y, z \geqslant 0$ there is $x^{3}+y^{3}+z^{3}-3 x y z \geqslant c|(x-y)(y-z)(z-x)|$. Find the maximum value of $c$.
保留源文本的换行和格式，直接输出翻译结果如下：
6. (50 points) Given that for any $x, y, z \geqslant 0$ there is $x^{3}+y^{3}+z^{3}-3 x y z \geqslant c|(x-y)(y-z)(z-x)|$. Find the maximum ... | 6. The left side of the inequality
$$
\begin{array}{l}
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) \\
=\frac{(x+y+z)\left[(x-z)^{2}+(y-z)^{2}+(z-x)^{2}\right]}{2} .
\end{array}
$$
Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$.
If $x=y$ or $y=z$, the inequality holds for any real... | \left(\frac{\sqrt{6}+3 \sqrt{2}}{2}\right) \sqrt[4]{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,091 |
7. (50 points) Let $r(n)$ denote the sum of the remainders when $n$ is divided by $1, 2, \cdots, n$. Find all positive integers $m (1 < m \leqslant 2014)$ such that
$$
r(m) = r(m-1).
$$ | 7. Let the remainder of $n$ divided by $k(1 \leqslant k \leqslant n)$ be $r_{k}(n)$.
Thus, $r_{k}(n)=n-k\left[\frac{n}{k}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
$$
\begin{array}{l}
\text { Then } r(m)=\sum_{k=1}^{m} r_{k}(m) \\
=\sum_{k=1}^{m}\left(m-k\left[\frac{m}{k}\rig... | m=2^{s}(s \in \mathbf{Z}, s \leqslant 10) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,092 |
8. (50 points) $n(n \geqslant 3)$ boxes are arranged in a circle, and 1 ball, 2 balls, $\cdots, n$ balls are placed in them in a counterclockwise direction, respectively. Then the following operation is performed: choose three adjacent boxes, and either place 1 ball in each or, if all three boxes are non-empty, remove ... | 8. Let the box number with exactly $i(i=1,2, \cdots, n)$ balls in the initial state be $i$, and the number of balls in this box after a certain operation be $x_{i}$.
When $n=3 l$, divide the $n$ boxes into three groups according to their numbers:
$\{1,4, \cdots, 3 l-2\},\{2,5, \cdots, 3 l-1\},\{3,6, \cdots, 3 l\}$.
Obv... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,093 |
1. Place 99 positive integers on a circle. It is known that any two adjacent numbers differ by 1 or 2 or one is twice the other. Prove: Among these 99 numbers, there is a multiple of 3. | 1. Suppose there are no multiples of 3 among the 99 numbers. Then, by the condition, it is easy to see that any two adjacent numbers are not congruent modulo 3.
Therefore, they would alternately be congruent to $1$ and $2$ modulo 3, which contradicts the fact that there are 99 numbers (an odd number). | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,094 |
2. Given that $a$ and $b$ are two different positive integers. Ask:
$$
\begin{array}{l}
a(a+2), a b, a(b+2), (a+2) b, \\
(a+2)(b+2), b(b+2)
\end{array}
$$
Among these six numbers, what is the maximum number of perfect squares? | 2. At most two numbers are perfect squares (such as $a=2, b=16$). Notice that,
$$
\begin{array}{l}
a(a+2)=(a+1)^{2}-1, \\
b(b+2)=(b+1)^{2}-1
\end{array}
$$
cannot be perfect squares;
$$
a b \cdot a(b+2)=a^{2}\left(b^{2}+2 b\right)
$$
is not a perfect square, so at most one of $a b$ and $a(b+2)$ is a perfect square.
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,095 |
3. Let $A$ be a set composed of several diagonals of a convex $n$-gon. If a diagonal in set $A$ intersects with exactly one other diagonal inside the convex $n$-gon, then it is called a "good" diagonal. Find the maximum possible number of good diagonals. | 3. The maximum possible number of good diagonals is $2\left[\frac{n}{2}\right]-2$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
Using induction, it is easy to prove a lemma.
Lemma: In a convex $n$-gon, at most $n-3$ diagonals can exist such that no two intersect internally.
The proof is ... | 2\left[\frac{n}{2}\right]-2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,096 |
4. In the acute triangle $\triangle ABC$, it is known that $AB > BC$, $M$ is the midpoint of side $AC$, circle $\Gamma$ is the circumcircle of $\triangle ABC$, the tangents to circle $\Gamma$ at points $A$ and $C$ intersect at point $P$, line segment $BP$ intersects $AC$ at point $S$, $AD$ is the altitude of $\triangle... | 4. Since $\angle A M P=\angle A D P=90^{\circ}$, points $M$ and $D$ lie on the circle $\Gamma_{1}$ with diameter $A P$.
Also, since $P A$ is a tangent to circle $\Gamma$, we have
$\angle K A P=\angle A C K$.
Since $C$, $K$, $D$, and $S$ all lie on the circumcircle $\Gamma_{2}$ of $\triangle C S D$, we have $\angle A C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,097 |
5. Let $N>1$ be a positive integer, and $m$ denote the largest divisor of $N$ that is less than $N$. If $N+m$ is a power of 10, find $N$.
| 5. $N=75$.
Let $N=m p$. Then $p$ is the smallest prime factor of $N$.
By the problem, we know $m(p+1)=10^{k}$.
Since $10^{k}$ is not a multiple of 3, therefore, $p>2$.
Hence, $N$ and $m$ are both odd.
Thus, $m=5^{*}$.
If $s=0, N=p=10^{k}-1$ is a multiple of 9, which is a contradiction.
Then $s \geqslant 1,5 \mid N$.
T... | 75 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,098 |
2. Find all integer-coefficient polynomials $f(x)$ such that for all positive integers $n$, we have $f(n) \mid (2^n - 1)$.
untranslated text:
2. 求所有的整系数多项式 $f(x)$, 使得对所有的正整数 $n$, 均有 $f(n) \mid\left(2^{n}-1\right)$.
translated text:
2. Find all integer-coefficient polynomials $f(x)$ such that for all positive intege... | Assume $f(x)$ is not a constant, and without loss of generality, let the leading coefficient of $f(x)$ be positive.
Then there exists an integer $N$, such that when $x \geqslant N$, $f(x) \geqslant 2$.
Take any positive integer $n, n \geqslant N$, and take a prime factor $p$ of $f(n)$.
Since $f(n) \mid\left(2^{n}-1\rig... | f(x) = \pm 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,099 |
6. Given a trapezoid $A B C D$ inscribed in a circle $\Gamma$ with bases $A B$ and $C D$, a circle $\Gamma_{1}$ passing through points $C$ and $D$ intersects line segments $C A$ and $C B$ at points $A_{1}$ (distinct from point $C$) and $B_{1}$ (distinct from point $D$), respectively. If $A_{2}$ and $B_{2}$ are the poin... | 6. The conclusion is equivalent to $C A_{2} \cdot C A=C B_{2} \cdot C B$. Since $A A_{1}=C A_{2}, B B_{1}=C B_{2}$, it suffices to prove $A A_{1} \cdot A C=B B_{1} \cdot B C$.
Let $D_{1}$ be the second intersection point of circle $\Gamma_{1}$ and $A D$. By symmetry, we know $A D=B C, A D_{1}=B B_{1}$. Therefore, $A A_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,100 |
7. The central bank of Mathland decides to issue coins with denominations of $\boldsymbol{\alpha}^{k}$ $(k=0,1, \cdots)$. The bank governor hopes to find a positive real number $\alpha$, such that for any $k \geqslant 1, \alpha^{k}$ is an irrational number greater than 2, and for any positive integer $n$, theoretically... | 7. Can be achieved.
Let $\alpha=\frac{\sqrt{29}-1}{2}$. Then $\alpha^{2}+\alpha=7$.
Obviously, $\alpha^{k}>2(k=1,2, \cdots)$ is an irrational number.
We will prove: Any positive integer $n$ can be expressed as a polynomial in $\alpha$ with coefficients being non-negative integers not greater than 6.
In fact, let $n=\... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,101 |
8. A country has $n$ cities, and there are bidirectional direct flights between any two cities. It is known that for any two cities, the ticket prices in both directions are the same, and the flight ticket prices between different city pairs are all different. Prove: there exists a sequence of $n-1$ consecutive flights... | 8. Let $A B$ denote the flight route from city $A$ to city $B$. For any city $A$, construct a route consisting of flight routes starting from city $A$ as follows.
Let $A_{0}=A, A_{0} A_{1}$ be the flight with the highest ticket price $a_{1}$ among all flights departing from city $A_{0}$, $A_{1} A_{2}$ be the flight wi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,102 |
2. If $\sqrt{\cos \theta}-\sqrt{\sin \theta}<5\left(\sin ^{3} \theta-\cos ^{3} \theta\right)$, $\theta \in[0,2 \pi)$, then the range of values for $\theta$ is . $\qquad$ | 2. $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right]$.
To make the original inequality meaningful, we need $\cos \theta$ and $\sin \theta$ to be greater than or equal to 0, thus, $\theta \in\left[0, \frac{\pi}{2}\right]$.
$$
\text { Let } f(x)=5 x^{3}+\sqrt{x}(x \geqslant 0) \text {. }
$$
Then the original inequal... | \theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,104 |
3. Given $\log _{10} \sin x+\log _{10} \cos x=-1$. Then $2 \log _{10}(\sin x+\cos x)=$ $\qquad$ | 3. $\lg 6-\lg 5$.
Notice that, $\sin x, \cos x$ are both positive, and
$$
\begin{array}{l}
\sin x \cdot \cos x=\frac{1}{10} . \\
\text { Then } 2 \log _{10}(\sin x+\cos x) \\
=\log _{10}(\sin x+\cos x)^{2} \\
=\log _{10}(1+2 \sin x \cdot \cos x) \\
=\log _{10} \frac{6}{5}=\lg 6-\lg 5 .
\end{array}
$$ | \lg 6-\lg 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,105 |
4. In tetrahedron $ABCD$, it is known that
$$
\angle ADB = \angle BDC = \angle CDA = \frac{\pi}{3} \text{, }
$$
the areas of $\triangle ADB$, $\triangle BDC$, and $\triangle CDA$ are $\frac{\sqrt{3}}{2}$, $2$, and $1$ respectively. Then the volume of this tetrahedron is $\qquad$ | 4. $\frac{2 \sqrt{6}}{9}$.
Let $D A, D B, D C$ be $x, y, z$ respectively. Then
$$
\frac{x y \sin \frac{\pi}{3}}{2}=\frac{\sqrt{3}}{2}, \frac{y z \sin \frac{\pi}{3}}{2}=2, \frac{x z \sin \frac{\pi}{3}}{2}=1 \text {. }
$$
Multiplying the three equations gives $x y z=\frac{8}{\sqrt{3}}$.
Let the angle between $D C$ and ... | \frac{2 \sqrt{6}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,106 |
5. Xiao Ming and Xiao Hong independently and repeatedly roll a fair die until the first 6 appears. The probability that the number of rolls by Xiao Ming and Xiao Hong differs by no more than 1 is $\qquad$ | 5. $\frac{8}{33}$.
Let the number of throws by Xiao Ming and Xiao Hong be $\xi$ and $\eta$, respectively. Then the required probability is
$$
\begin{aligned}
\sum_{i=1}^{+\infty}[P(\xi=\eta=i)+P(\xi=i, \eta=i+1)+ \\
P(\xi=i+1, \eta=i)] .
\end{aligned}
$$
By independence, the required probability is
$$
\begin{array}{l... | \frac{8}{33} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,107 |
6. Given $x^{2}+y^{2}+z^{2}=3$. Then the minimum value of $x y+y z+z x$ is $\qquad$ . | 6. $-\frac{3}{2}$.
$$
\begin{array}{l}
\text { From }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \geqslant 0,
\end{array}
$$
we know that the minimum value is $-\frac{3}{2}$, which is achieved when $x+y+z=0$ and $x^{2}+y^{2}+z^{2}=3$. For example, $(x, y, z)=\left(\sqrt{\frac{3}{2}},-\sqrt{\frac{3}{2}}, 0\right)$... | -\frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,108 |
7. In the Cartesian coordinate system, it is known that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P(3,2)$ and $Q$, and the product of the radii of the two circles is $\frac{13}{2}$. If both circles are tangent to the line $l: y=k x$ and the $x$-axis, then the equation of line $l$ is | 7. $y=2 \sqrt{2} x$.
By the problem, let the equations of the two circles be
$$
\left(x-a_{i}\right)^{2}+\left(y-r_{i}\right)^{2}=r_{i}^{2}(i=1,2) \text {. }
$$
Then $r_{1} r_{2}=\frac{13}{2}, k>0$.
Since both circles pass through the point $P(3,2)$, we have
$$
\begin{array}{l}
\left(3-a_{i}\right)^{2}+\left(2-r_{i}\... | y=2 \sqrt{2} x | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,109 |
Example 1 Let $a, b, c > 0$, and $abc + a + c = b$. Find the maximum value of
$$
p=\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}+\frac{3}{c^{2}+1}
$$
(1999, Vietnam Mathematical Olympiad) | From the given conditions, we have
$$
a+c=(1-a c) b \text{. }
$$
Obviously, $1-a c \neq 0$. Therefore, $b=\frac{a+c}{1-a c}$.
Let $\alpha=\arctan a, \beta=\arctan b, \gamma=\arctan c$, where $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$.
Then $\tan \beta=\frac{\tan \alpha+\tan \gamma}{1-\tan \alpha \cdot \t... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,110 |
Example 2 Let $a, b, c$ be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{8 a^{2}+2 a b}{(b+\sqrt{6 a c}+3 c)^{2}}+\frac{2 b^{2}+3 b c}{(3 c+\sqrt{2 a b}+2 a)^{2}}+ \\
\frac{18 c^{2}+6 a c}{(2 a+\sqrt{3 b c}+b)^{2}} \geqslant 1 .
\end{array}
$$
(2013, Taiwan Mathematical Olympiad Training Camp) | 【Analysis and Proof】The coefficients in equation (1) are quite scattered, so we can first attempt a substitution. Let $x=2a, y=b, z=3c$, transforming equation (1) into a cyclic symmetric inequality:
$$
\sum \frac{2 x^{2}+x y}{(y+z+\sqrt{z x})^{2}} \geqslant 1,
$$
where “$\sum$” denotes the cyclic sum.
The denominator ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,111 |
Example 1 Given integers $n \geqslant k \geqslant 0$, define the number $c(n, k)$:
$$
\left\{\begin{array}{ll}
c(n, 0)=c(0,0)=1, & n \geqslant 0 ; \\
c(n+1, k)=2^{k} c(n, k)+c(n, k-1), & n \geqslant k \geqslant 1 .
\end{array}\right.
$$
Prove: For all integers $n, k$ satisfying $n \geqslant k \geqslant 0$, we have $c(... | 【Analysis】Starting from the definition to seek proof
$$
c(n, k)=c(n, n-k)
$$
is a massive "project". At this moment, finding a "substitute" for $c(n, k)$ might be the only way out.
Proof For $m \geqslant 1$, let
$$
\begin{array}{l}
f(m)=\prod_{k=1}^{m}\left(2^{k}-1\right), f(0)=1 . \\
\text { Let } a(n, k)=\frac{f(n)}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,112 |
Example 2 There are three piles of stones. Each time, A moves one stone from one pile to another, and A can receive a reward from B for each move, which is equal to the difference between the number of stones in the pile to which A moves the stone and the number of stones in the pile from which A moves the stone. If th... | 【Analysis】Due to the uncertainty of A's operations, it is necessary to start from the whole and establish a "substitute" for the increase or decrease of A's reward each time, which must be simple to calculate.
Solution A's reward is 0.
In fact, the three piles of stones can be imagined as three complete graphs (each st... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,113 |
1. Given that $n$ is a positive integer. Find the smallest positive integer $k$ such that for any real numbers $a_{1}, a_{2}, \cdots, a_{d}$, and
$$
a_{1}+a_{2}+\cdots+a_{d}=n\left(0 \leqslant a_{i} \leqslant 1, i=1,2, \cdots, d\right) \text {, }
$$
it is always possible to partition these numbers into $k$ groups (some... | 1. $k=2 n-1$.
If $d=2 n-1$, and
$$
a_{1}=a_{2}=\cdots=a_{2 n-1}=\frac{n}{2 n-1} \text {. }
$$
Since $\frac{2 n}{2 n-1}>1$, for any partition satisfying the condition, there is at most one number in each group.
Therefore, $k \geqslant 2 n-1$.
Next, we prove that it is possible to partition into $2 n-1$ groups satisfyi... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,116 |
3. A fanatical physicist discovered a particle in his laboratory that he called "imon". After the appearance of this mysterious particle, the physicist found that a pair of imons in the laboratory could become entangled, and each imon could participate in multiple entanglement relationships. The physicist performed the... | 3. Consider the imons as points in a graph $G$. If two imons are entangled, then a line segment is drawn between the corresponding points. Coloring the points in graph $G$ such that adjacent points have different colors is called a "proper coloring".
First, prove a lemma.
Lemma Assume that graph $G$ has a proper colori... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,117 |
6. In a certain country, some pairs of cities (i.e., two cities in the country) are connected by bidirectional flight routes, and it is possible to travel from any city to any other city through a series of flights. The distance between two cities is defined as the minimum number of flights required to travel from one ... | 6. Let $d(a, b)$ denote the distance between cities $a$ and $b$,
$$
S_{i}(a)=\{c \mid d(a, c)=i\},
$$
i.e., the set of cities whose distance from city $a$ is exactly $i$.
Assume there exists a city $x$ such that the number of elements in the set $D=S_{4}(x)$ is at least 2551.
Let $A=S_{1}(x)$.
If every city in set $D$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,120 |
8. Two players $A$ and $B$ are playing a coloring game on a real line. Player $A$ has a bucket containing 4 units of black ink. An amount $p$ of black ink is sufficient to color a closed interval of length $p$ on the real line. In each round, player $A$ chooses a positive integer $m$, and extracts an amount of $\frac{1... | 8. Player $A$ does not have a winning strategy.
Below is player $B$'s strategy to ensure that the interval $[0,1]$ is completely black once the black ink in the bucket is used up.
Before the $r$-th round, let the maximum value of $x$ in the already blackened interval $[0, x]$ be $x_{r}$. For completeness, define $x_{... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,121 |
Example 3 Let $f(n)$ be a function from $\mathbf{Z}_{+}$ to $\mathbf{Z}_{+}$, $f(1)=$ 1, and for any $n \in \mathbf{Z}_{+}, \varepsilon \in\{0,1\}$, we have
$$
f(2 n+\varepsilon)=3 f(n)+\varepsilon \text {. }
$$
Find the range of the function $f(n)$. | Solve: First, calculate some specific values of $f(n)$, see Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ \\
\hline$f(n)$ & 1 & 3 & 4 & 9 & 10 & 12 & $\cdots$ \\
\hline
\end{tabular}
If $n$ and $f(n)$ are represented in binary and ternary respectively, the table is re... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,122 |
1. Let $p$ be an odd prime, and $a, b, c, d$ be positive integers less than $p$, such that $a^{2}+b^{2}$ and $c^{2}+d^{2}$ are both multiples of $p$. Prove: $ac + bd$ and $ad + bc$ are exactly one of them a multiple of $p$.
（Ben Shenghong supplied the problem） | 1. From $a^{2}+b^{2} 、 c^{2}+d^{2}$ both being multiples of $p$, we know
$$
(a c+b d)(a d+b c)=\left(a^{2}+b^{2}\right) c d+\left(c^{2}+d^{2}\right) a b
$$
is also a multiple of $p$.
Since $p$ is a prime, at least one of $a c+b d 、 a d+b c$ must be a multiple of $p$.
On the other hand, assume $a c+b d 、 a d+b c$ are ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,123 |
2. Let $n(n \geqslant 4)$ be a positive integer. $n$ players each play a table tennis match against every other player (each match has a winner and a loser). Find the minimum value of $n$ such that after all the matches, there always exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, satisfying that w... | 2. First, prove: when $n=8$, there always exists an ordered quartet that satisfies the problem's conditions.
Since 8 players have played a total of $\mathrm{C}_{8}^{2}=28$ matches, there must be at least one player who has won at least $\left\lceil\frac{28}{8}\right\rceil=4$ matches (where $\left\lceil x \right\rceil$... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,124 |
3. As shown in Figure 1, in obtuse triangle $\triangle ABC$, $AB > AC$, point $O$ is its circumcenter, the midpoints of sides $BC$, $CA$, and $AB$ are $D$, $E$, and $F$ respectively, median $AD$ intersects lines $OF$ and $OE$ at points $M$ and $N$ respectively, and lines $BM$ and $CN$ intersect at point $P$. Prove: $OP... | 3. From the given information, we have
$$
\begin{array}{l}
B M=A M, C N=A N, \\
\angle A M P=2 \angle B A M, \angle P N D=2 \angle C A M .
\end{array}
$$
Connecting $O B$ and $O C$.
Since $O$ is the circumcenter of $\triangle A B C$, then
$$
\begin{array}{l}
\angle B O C=2 \angle B A C=2 \angle B A M+2 \angle C A M \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,125 |
4. Let $n$ be a positive integer, and non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{i} x_{j} \leqslant 4^{-|i-j|}(1 \leqslant i, j \leqslant n)$.
Prove: $x_{1}+x_{2}+\cdots+x_{n}<\frac{5}{3}$.
(Jin Mengwei) | 4. Let $\max _{1 \leq i \in n} x_{i}=M$.
First, consider the case $0 \leqslant M \leqslant \frac{2}{3}$.
We need to prove: For any $s, k (1 \leqslant s \leqslant s+k \leqslant n)$, we have
$$
\sum_{i=s}^{s+k} x_{i} \leqslant \frac{2}{3}+\sum_{i=1}^{k} \frac{1}{2^{i}} \text {, }
$$
with the convention $\sum_{i=1}^{0} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,126 |
5. Let $\triangle A B C$ and $\triangle X Y Z$ both be acute triangles. Prove:
$$
\begin{array}{l}
\cot A \cdot(\cot Y+\cot Z) 、 \cot B \cdot(\cot Z+\cot X) \text { 、 } \\
\cot C \cdot(\cot X+\cot Y)
\end{array}
$$
The maximum of these three numbers is not less than $\frac{2}{3}$. (Zhang Si Hui) | 5. Let $\cot A, \cot B, \cot C, \cot X, \cot Y, \cot Z$ be denoted as $a, b, c, x, y, z$ respectively. Then
$$
a b + b c + c a = x y + y z + z x = 1,
$$
where $a, b, c, x, y, z > 0$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(a+b+c)^{2}(x+y+z)^{2} \\
=\left(a^{2}+b^{2}+c^{2}+2\right)\left(x^{2}+y^{... | \frac{2}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,127 |
6. Let integers $a, b, c$ and real number $r$ satisfy
$$
a r^{2}+b r+c=0, a c \neq 0 \text {. }
$$
Prove: $\sqrt{r^{2}+c^{2}}$ is an irrational number.
(Supplied by He Yijie) | 6. From the condition, we have $b^{2}-4 a c \geqslant 0$.
Let $r=\frac{-b+m}{2 a}\left(m^{2}=b^{2}-4 a c\right)$.
Given $a c \neq 0$, we know
$m \neq \pm b$.
We use proof by contradiction.
Assume $\sqrt{r^{2}+c^{2}}$ equals some rational number $q$, denoted as
$s=2 a q \in \mathbf{Q}$.
Then $s^{2}=4 a^{2} q^{2}=4 a^{2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,128 |
8. In the figure shown in Figure 3, on both sides of square $P$, there are $a$ and $b$ squares to the left and right, and $c$ and $d$ squares above and below, where $a$, $b$, $c$, and $d$ are positive integers, satisfying
$$
(a-b)(c-d)=0 \text {. }
$$
The shape formed by these squares is called a "cross star".
There i... | 8. For a cross, the cell $P$ referred to in the problem is called the "center block" of the cross.
When $a=b$, the cross is called "standing"; when $c=d$, it is called "lying" (some crosses are both standing and lying).
If the union of a row and a column of a rectangle $R$ is exactly a cross $S$, then $R$ is called t... | 13483236 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,130 |
1. As shown in Figure 4, in the acute triangle $\triangle ABC$, $AB > AC$, $M$ is the midpoint of side $BC$, $I$ is the incenter, $MI$ intersects side $AC$ at point $D$, and $BI$ intersects the circumcircle of $\triangle ABC$ at another point $E$. Prove: $\frac{ED}{EI} = \frac{IC}{IB}$. | 1. Let $B E$ intersect $A C$ at point $F$. Connect $A I, C E$. Applying Menelaus' theorem to $\triangle B C F$ and the transversal $M I D$ yields $\frac{B M}{M C} \cdot \frac{C D}{D F} \cdot \frac{F I}{I B}=1$.
Since $B M=M C$, we have $\frac{C D}{D F}=\frac{I B}{F I}$.
Given that $A I$ bisects $\angle B A C$, we know ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,131 |
3. Let $p$ be a prime, and positive integers $x, y, z$ satisfy
$$
x<y<z<p \text {, and }\left\{\frac{x^{3}}{p}\right\}=\left\{\frac{y^{3}}{p}\right\}=\left\{\frac{z^{3}}{p}\right\} \text {, }
$$
where $\{a\}$ denotes the fractional part of the real number $a$. Prove:
$(x+y+z) \mid\left(x^{5}+y^{5}+z^{5}\right)$. (Yang... | 3. Clearly, prime $p>3$, and by the condition, $x-y$, $y-z$, and $z-x$ are not multiples of $p$.
$$
\begin{array}{l}
\text { By }\left\{\frac{x^{3}}{p}\right\}=\left\{\frac{y^{3}}{p}\right\}=\left\{\frac{z^{3}}{p}\right\} \\
\Rightarrow p\left|\left(x^{3}-y^{3}\right) \Rightarrow p\right|(x-y)\left(x^{2}+x y+y^{2}\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,132 |
Example 4 Let $f(x)$ be a function defined on the set of non-negative real numbers, $f(0)=0$, and for any $x \geqslant y \geqslant 0$, we have
$$
|f(x)-f(y)| \leqslant(x-y) f(x) \text {. }
$$
Find the expression for $f(x)$. | Prove by mathematical induction: For any positive integer $n$, when $\frac{n-1}{2} \leqslant x<\frac{n}{2}$, we have $f(x)=0$.
When $n=1$, take $0 \leqslant x<\frac{1}{2}, y=0$ in the condition, we get $|f(x)| \leqslant x f(x) \leqslant x|f(x)| \leqslant \frac{1}{2}|f(x)|$.
Thus, $|f(x)|=0$, i.e., $f(x)=0$ holds on $\l... | f(x) \equiv 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,133 |
5. Let $n$ be an integer greater than 1, and let positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}-x_{i+1}^{3}} \geqslant \frac{n^{3}}{n^{2}-1}\left(x_{n+1}=x_{1}\right) \text {. }
$$
(Li Shenghong) | 5. Clearly, $0<x_{i}<1(i=1,2, \cdots, n)$. By the Cauchy-Schwarz inequality and the AM-GM inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}-x_{i+1}^{3}}\right)\left[\sum_{i=1}^{n}\left(1-x_{i+1}^{2}\right)\right] \\
\geqslant\left(\sum_{i=1}^{n} \sqrt{\frac{x_{i}}{x_{i+1}-x_{i+1}^{3}}} \... | \sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}-x_{i+1}^{3}} \geqslant \frac{n^{3}}{n^{2}-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,134 |
7. Prove: The equation $a^{2}+b^{3}=c^{4}$ has infinitely many sets of positive integer solutions $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2, \cdots)$, such that for each positive integer $n$, $c_{n}$ and $c_{n+1}$ are coprime.
(Tao Pingsheng) | 7. Transform the original equation into $b^{3}=\left(c^{2}-a\right)\left(c^{2}+a\right)$.
Consider the solutions $(a, b, c)$ that satisfy $c^{2}-a=b, c^{2}+a=b^{2}$. In this case, $2 c^{2}=b(b+1), 2 a=b(b-1)$.
Let $b$ be an odd number. Then $c^{2}=b \cdot \frac{b+1}{2}\left(b, \frac{b+1}{2} \in \mathbf{Z}_{+}\right)$.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,135 |
1. If a positive integer has exactly two prime factors among its positive divisors, then it is called a "good number". Question: Does there exist a sequence of 18 consecutive positive integers that are all good numbers? | 1. Does not exist.
If there exist 18 consecutive good numbers, then among them there are three multiples of 6, i.e., $6n, 6(n+1), 6(n+2)$, where $n, n+1, n+2$ can only have prime factors of 2 or 3.
Since two of these three numbers are not multiples of 3, they must be powers of 2. However, two powers of 2 that differ ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,136 |
3. In the bank's safe, there are $n$ drawers numbered $1,2, \cdots, n$, each containing a file numbered $1,2, \cdots, n$. Each drawer holds one file, and the $i(i=1,2, \cdots, n)$-th drawer contains the file numbered $a_{i}$. Bethya performs the following operations on these files: each time he can choose any two files... | 3. First, prove a lemma.
Lemma Let $b_{1}, b_{2}, \cdots, b_{n}$ be a permutation of $1,2, \cdots, n$, and satisfy
$$
\left|b_{1}-1\right|+\left|b_{2}-2\right|+\cdots+\left|b_{n}-n\right|>0 .
$$
Then there exists $1 \leqslant i0$, we have
$$
\left|c_{1}-1\right|+\left|c_{2}-2\right|+\cdots+\left|c_{n}-n\right|>0 \tex... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,138 |
4. Given $\triangle A B C(A B>B C)$ with the circumcircle $\Gamma$, $M$ and $N$ are points on sides $A B$ and $B C$ respectively, such that $A M=C N$. Line $M N$ intersects $A C$ at point $K$, $P$ is the incenter of $\triangle A M K$, and $Q$ is the excenter of $\triangle C N K$ opposite to side $C N$. Prove: The midpo... | 4. Let $S$ be the midpoint of arc $\overparen{A B C}$, $R$ and $T$ be the midpoints of $A C$ and $M N$ respectively. Then
$$
\begin{array}{l}
\triangle A M S \cong \triangle C N S \Rightarrow S M=S N \\
\Rightarrow \angle S R K=\angle S T K=90^{\circ},
\end{array}
$$
i.e., $R$ and $T$ both lie on the circle $\Gamma_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,139 |
6. Given that $M$ is the midpoint of side $A C$ of $\triangle A B C$, points $P$ and $Q$ lie on segments $A M$ and $C M$ respectively, and satisfy $P Q=\frac{A C}{2}$. The circumcircle of $\triangle A B Q$ intersects side $B C$ at point $X$ (different from point $B$), and the circumcircle of $\triangle B C P$ intersect... | 6. Take point $Z$ on segment $P Q$ such that $C Q=Q Z$.
From $A C=2 P Q \Rightarrow A P+Q C=A C-P Q=P Q$
$\Rightarrow P Z=P Q-Q Z=P Q-Q C=A P$.
Since $B, Y, P, C$ are concyclic, then
$$
A B \cdot A Y=A P \cdot A C=2 A P \cdot \frac{A C}{2}=A Z \cdot A M \text {. }
$$
Similarly, $C X \cdot C B=C Z \cdot C M$.
If point... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,140 |
8. Let the set composed of a polygon and its interior points be called a closed polygon. On the plane, $n$ closed convex $k$-gons satisfy: the intersection of any two closed polygons is non-empty; any two polygons are similar, and the similarity coefficient is positive. Prove: there exists a point on the plane that bel... | 8. First, we prove a lemma.
Lemma: Let polygons \( P \) and \( P' \) be two closed convex polygons in the plane with a non-empty intersection. If polygons \( P \) and \( P' \) are similar with a positive similarity ratio, then a vertex of polygon \( P \) or a vertex of polygon \( P' \) lies within the other closed pol... | 1 + \frac{n-1}{2k} | Geometry | proof | Yes | Yes | cn_contest | false | 728,141 |
1. Find all real constants $t$ such that if $a, b, c$ are the side lengths of some triangle, then $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$ are also the side lengths of some triangle. | 1. $t \in\left[\frac{2}{3}, 2\right]$.
If $t > 2$, consider a triangle with side lengths $b=c=1, a=\varepsilon$ for some positive $\varepsilon$ (such as $0 < \varepsilon < 2$). In this case, $a < b+c$.
Combining the inequality $(b+c)^{2} \geqslant 4 b c$, we get
$$
\begin{array}{l}
b^{2}+c a t+c^{2}+a b t-a^{2}-b c t... | t \in\left[\frac{2}{3}, 2\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,142 |
2. In $\triangle A B C$, points $D$ and $E$ are on sides $A B$ and $A C$ respectively, and satisfy $D B=B C=C E$. Let the line $C D$ intersect $B E$ at point $F$. Prove: the incenter $I$ of $\triangle A B C$, the orthocenter $H$ of $\triangle D E F$, and the midpoint $M$ of the arc $\overparen{B A C}$ of the circumcirc... | 2. As shown in Figure 1.
From $D B=B C=C E$, we know $B I \perp C D, C I \perp B E$.
Therefore, $I$ is the orthocenter of $\triangle B F C$.
Let the intersection of line $B I$ and $C D$ be point $K$, and the intersection of line $C I$ and $B E$ be point $L$. Then, by the power of a point theorem, we have
$$
I B \cdot I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,143 |
Example 5 If real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{1} x_{5}=-1, \\
x_{2} x_{1}+x_{2} x_{3}+x_{2} x_{4}+x_{2} x_{5}=-1, \\
x_{3} x_{1}+x_{3} x_{2}+x_{3} x_{4}+x_{3} x_{5}=-1, \\
x_{4} x_{1}+x_{4} x_{2}+x_{4} x_{3... | Let $S=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$.
From the given, $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all satisfy the equation $x^{2}-S x-1=0$.
Thus, $x_{i} \in\left\{\frac{S+\sqrt{S^{2}+4}}{2}, \frac{S-\sqrt{S^{2}+4}}{2}\right\}(i=1,2, \cdots, 5)$.
Therefore, $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$
$$
=\frac{5 S+\varepsilon \sqrt{S^{2}+... | \pm \sqrt{2}, \pm \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,144 |
3. Let the number of positive divisors of a positive integer $m$ be denoted by $d(m)$, and the number of distinct prime factors by $\omega(m)$. Let $k$ be a positive integer. Prove: there exist infinitely many positive integers $n$, such that $\omega(n)=k$, and for any positive integers $a, b$ satisfying $a+b=n$,
$$
d(... | 3. Positive integers of the form $n=2^{p-1} \mathrm{~m}$ meet the requirements, where $m$ has exactly $k-1$ prime factors, and each prime factor is greater than $3$, and $p$ is a prime number satisfying $\left(\frac{5}{4}\right)^{\frac{2-1}{2}}>m$.
Assume positive integers $a, b$ satisfy
$$
a+b=n, d(n) \mid d\left(a^{2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,145 |
4. Find all positive integers $n(n \geqslant 2)$, such that there exist integers $x_{1}, x_{2}, \cdots, x_{n-1}$, satisfying if $0<i 、 j<n(i \neq j)$, and $n \mid(2 i+j)$, then $x_{i}<x_{j}$. | 4. $n=2^{k}(k \geqslant 1)$ or $3 \times 2^{k}(k \geqslant 0)$.
Assume $n$ has the aforementioned form. For a positive integer $i$, let $x_{i}$ be the largest positive integer such that $2^{x_{i}} \mid i$.
Assuming for $0n \text {, }
$$
is also impossible.
Now assume $n$ does not have the aforementioned form, and the... | n=2^{k}(k \geqslant 1) \text{ or } 3 \times 2^{k}(k \geqslant 0) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,146 |
5. Let $n$ be a positive integer. There are $n$ boxes, each containing some crystal balls (empty boxes are allowed). In each operation, you are allowed to select a box, take out two crystal balls from it, discard one, and place the other in another selected box. For a certain initial distribution of crystal balls, if a... | 5. The desired distribution is: there are a total of $2 n-2$ crystal balls in all the boxes, and the number of crystal balls in each box is even.
Number the boxes from $1 \sim n$, and use the array
$$
x=\left(x_{1}, x_{2}, \cdots, x_{n}\right)
$$
to represent a distribution state, where $x_{i}(i=1,2, \cdots, n)$ repre... | N(x)=2 n-2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,147 |
6. Find all functions \( f: \mathbf{R} \rightarrow \mathbf{R} \), such that for any real numbers \( x, y \) we have
\[
f\left(y^{2}+2 x f(y)+f^{2}(x)\right)=(y+f(x))(x+f(y)).
\] | 6. The functions that are easy to verify as meeting the conditions are
$$
f(x)= \pm x \text { or } \frac{1}{2}-x \text {. }
$$
Next, we prove that only the above functions meet the conditions.
Let $y=-f(x)$, for any $x \in \mathbf{R}$, substituting into the original equation, we get
$$
f\left(2 f^{2}(x)+2 x f(-f(x))\r... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,148 |
1. Given distinct complex numbers $a$ and $b$ satisfying $a b \neq 0$, the set $\{a, b\}=\left\{a^{2}, b^{2}\right\}$. Then $a+b=$ $\qquad$ . | $-1 .-1$.
If $a=a^{2}, b=b^{2}$, by $a b \neq 0$, we get $a=b=1$, which is a contradiction. If $a=b^{2}, b=a^{2}$, by $a b \neq 0$, we get $a^{3}=1$.
Clearly, $a \neq 1$.
Thus, $a^{2}+a+1=0 \Rightarrow a=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
Similarly, $b=-\frac{1}{2} \mp \frac{\sqrt{3}}{2} \mathrm{i}$.
Ther... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,149 |
2. Given a positive integer $a$ such that the function
$$
f(x)=x+\sqrt{13-2 a x}
$$
has a maximum value that is also a positive integer. Then the maximum value of the function is $\qquad$ . | 2.7.
Let $t=\sqrt{13-2 a x} \geqslant 0$. Then
$$
\begin{aligned}
y= & f(x)=\frac{13-t^{2}}{2 a}+t \\
& =-\frac{1}{2 a}(t-a)^{2}+\frac{1}{2}\left(a+\frac{13}{a}\right) .
\end{aligned}
$$
Since $a$ is a positive integer, $y_{\max }=\frac{1}{2}\left(a+\frac{13}{a}\right)$ is also a positive integer, so, $y_{\max }=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,150 |
3. In the positive geometric sequence $\left\{a_{n}\right\}$,
$$
a_{5}=\frac{1}{2}, a_{6}+a_{7}=3 \text {. }
$$
Then the maximum positive integer $n$ that satisfies $a_{1}+a_{2}+\cdots+a_{n}>a_{1} a_{2} \cdots a_{n}$ is $\qquad$ | 3. 12 .
According to the problem, $\frac{a_{6}+a_{7}}{a_{5}}=q+q^{2}=6$.
Since $a_{n}>0$, we have $q=2, a_{n}=2^{n-6}$.
Thus, $2^{-5}\left(2^{n}-1\right)>2^{\frac{n(n-11)}{2}} \Rightarrow 2^{n}-1>2^{\frac{n(n-11)}{2}+5}$.
Estimating $n>\frac{n(n-11)}{2}+5$, we get $n_{\max }=12$.
Upon verification, $n=12$ meets the re... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,151 |
4. Given the function $f(x)=\sin \frac{\pi}{2} x$, for any $t \in \mathbf{R}$, denote the maximum value of the function $f(x)$ on the interval $[t, t+1]$ as $M_{t}$, the minimum value as $m_{t}$, and $h(t)=M_{t}-m_{t}$. Then the range of the function $h(t)$ is $\qquad$ | 4. $\left[1-\frac{\sqrt{2}}{2}, \sqrt{2}\right]$.
Notice that, for $k \in \mathbf{Z}$, $h(t)$ is given by
$$
=\left\{\begin{array}{ll}
\sin \frac{\pi}{2}(t+1)-\sin \frac{\pi}{2} t, & t \in[2 k-1,2 k] ; \\
1-\sin \frac{\pi}{2} t, & t \in\left[2 k, 2 k+\frac{1}{2}\right] ; \\
1-\sin \frac{\pi}{2}(t+1), & t \in\left[2 k+... | \left[1-\frac{\sqrt{2}}{2}, \sqrt{2}\right] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,152 |
5. In the rectangular coordinate plane, given points $A(0,2)$, $B(0,1)$, $D(t, 0)(t>0)$, and $M$ is a moving point on line segment $AD$. If $|AM| \leqslant 2|BM|$ always holds, then the minimum value of the positive real number $t$ is $\qquad$ | 5. $\frac{2 \sqrt{3}}{3}$.
Let $M(x, y)$.
From $|A M| \leqslant 2|B M|$, we get $x^{2}+\left(y-\frac{2}{3}\right)^{2} \geqslant \frac{4}{9}$.
Therefore, the line segment $A D$ is always outside the Apollonius circle $x^{2}+\left(y-\frac{2}{3}\right)^{2}=\frac{4}{9}$.
When $t$ is at its minimum, the line segment $A D$ ... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,153 |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set
Set
$$
\begin{array}{l}
A=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 1\right\}, \\
B=\left\{(x, y) \mid[x]^{2}+[y]^{2} \leqslant 1\right\} .
\end{array}
$$
Then the area of the plane region represented by $A \cup B$ is
$\qquad$ | $6.5+\frac{\pi}{4}$.
When $x \in[-1,0)$, $[x]=-1$. Thus,
$[y]=0, y \in[0,1)$;
When $x \in[0,1)$, $[x]=0$.
Thus, $[y]=-1$ or 0 or $1, y \in[-1,2)$;
When $x \in[1,2)$, $[x]=1$.
Thus, $[y]=0, y \in[0,1)$.
Therefore, the plane region represented by $A \cup B$ consists of five unit squares and one quarter of a unit circle i... | 5+\frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,154 |
Example 6 Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, b_{1}=2, \\
a_{n+1}=\frac{1+a_{n}+a_{n} b_{n}}{b_{n}}, \\
b_{n+1}=\frac{1+b_{n}+a_{n} b_{n}}{a_{n}} .
\end{array}
$$
Prove: $a_{2008}<5$. ${ }^{|3|}$
(2008, Russian Mathematical Olympiad) | Proof Given
$$
\begin{array}{l}
1+a_{n+1}=\frac{\left(1+a_{n}\right)\left(1+b_{n}\right)}{b_{n}}, \\
1+b_{n+1}=\frac{\left(1+a_{n}\right)\left(1+b_{n}\right)}{a_{n}} .
\end{array}
$$
According to the recursive relations, it is clear that $a_{n}>0, b_{n}>0$.
Thus, $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}=\frac{b_{n}-a_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,155 |
7. As shown in Figure 1, given a regular tetrahedron $P-A B C$ with all edge lengths equal to 4, points $D, E, F$ are on edges $P A, P B, P C$ respectively. Then the number of $\triangle D E F$ that satisfy $D E = E F = 3, D F = 2$ is $\qquad$. | 7.3.
Let $P D=x, P E=y, P F=z$. Then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-x y=9, \\
y^{2}+z^{2}-y z=9, \\
z^{2}+x^{2}-z x=4 .
\end{array}\right.
$$
(1) - (2) gives $x=z$ or $x+z=y$.
When $x=z$, we get $x=z=2, y=1+\sqrt{6}$;
When $x+z=y$, $x z=\frac{5}{2}, x^{2}+z^{2}=\frac{13}{2}$, there are two sets of solutions. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,156 |
8. Given $1 \leqslant x, y, z \leqslant 6$.
The number of cases where the product of the positive integers $x, y, z$ is divisible by 10 is
$\qquad$ kinds. | 8. 72 .
(1) The number of ways to choose $x, y, z$ is $6^{3}$;
(2) The number of ways to choose $x, y, z$ without taking $2, 4, 6$ is $3^{3}$; (3) The number of ways to choose $x, y, z$ without taking 5 is $5^{3}$;
(4) The number of ways to choose $x, y, z$ without taking $2, 4, 5, 6$ is $2^{3}$. Therefore, the number ... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,157 |
9. (16 points) Given the function $f(x)=\left|x^{2}-a\right|$, where $a$ is a positive constant. If there are exactly two pairs of solutions $(m, n)$ such that the range of $f(x)$ on the domain $[m, n]$ is also $[m, n]$, find the range of values for $a$.
---
The translation maintains the original text's formatting an... | II. 9. Since $f(x)=\left|x^{2}-a\right| \geqslant 0$, therefore, $n>m \geqslant 0$.
(1) It is easy to find that the x-coordinate of the intersection point of $y=x$ and $y=x^{2}-a(x \geqslant \sqrt{a})$ is $x=\frac{1+\sqrt{1+4 a}}{2}$.
Obviously, the interval
$[m, n]=\left[0, \frac{1+\sqrt{1+4 a}}{2}\right]$ satisfies t... | \frac{3}{4}<a<2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,158 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$ satisfies $a_{0}=0$, $a_{1}=1$, and for all positive integers $n$,
$$
a_{n+1}=2 a_{n}+2013 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2014 \mid a_{n}$. | 10. Below are $a_{n}$ modulo 2014.
Then $a_{n+1} \equiv 2 a_{n}-a_{n-1} \Rightarrow a_{n+1}-a_{n} \equiv a_{n}-a_{n-1}$.
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2014.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, the smallest positive integer ... | 2014 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,159 |
11. (20 points) Given that the three vertices of $\triangle A B C$ lie on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the origin $O$ is the centroid of $\triangle A B C$. Prove: The area of $\triangle A B C$ is a constant. | 11. Let $A(a \cos \alpha, b \sin \alpha), B(a \cos \beta, b \sin \beta)$.
Then $C(-a(\cos \alpha+\cos \beta),-b(\sin \alpha+\sin \beta))$.
Since point $C$ is on the ellipse, substituting it in we get
$$
\begin{array}{l}
(\cos \alpha+\cos \beta)^{2}+(\sin \alpha+\sin \beta)^{2}=1 \\
\Rightarrow \cos (\alpha-\beta)=-\fr... | \frac{3 \sqrt{3}}{4} a b | Geometry | proof | Yes | Yes | cn_contest | false | 728,160 |
For a composite number $n$, let $f(n)$ denote the sum of its smallest three positive divisors, and $g(n)$ denote the sum of its largest two positive divisors. Find all composite numbers $n$ such that $g(n)$ is a positive integer power of $f(n)$. | II. Notice that when $n$ is odd, all divisors of $n$ are also odd.
Given that $f(n)$ is odd and $g(n)$ is even, it follows that $g(n)$ cannot be a positive integer power of $f(n)$. Therefore, we only need to consider the case where $n$ is even. In this case, 1 and 2 are the smallest two positive divisors of $n$, and $... | n=4 \times 6^{l}\left(l \in \mathbf{Z}_{+}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,162 |
Three. (50 points) Let $n \geqslant 2$ be a positive integer. Find the maximum value of $f(n)$ such that for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in(0,1)(i=1,2, \cdots, n)$, and
$$
\left(1-x_{i}\right)\left(1-x_{j}\right) \geqslant \frac{1}{4}(1 \leqslant i<j \leqslant n)
$$
we have
$$
\su... | When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{2}$,
$$
\begin{array}{l}
\frac{n}{2} \geqslant f(n) \sum_{1 \leqslant i<j \leqslant n} 1 \Rightarrow \frac{n}{2} \geqslant f(n) \mathrm{C}_{n}^{2} \\
\Rightarrow f(n) \leqslant \frac{1}{n-1} .
\end{array}
$$
Next, we prove: $\sum_{i=1}^{n} x_{i} \geqslant \frac{1}{n-1} \sum_{1 \... | \frac{1}{n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,163 |
Four, (50 points) In a round-robin tournament with $2 n+1$ teams, each team plays exactly one match against every other team, and there are no ties. If three teams $A, B, C$ satisfy: $A$ beats $B, B$ beats $C, C$ beats $A$, then they form a “cyclic triplet”. Find:
(1) the minimum possible number of cyclic triplets;
(2)... | (1) The minimum value is 0.
For two participating teams $T_{i}$ and $T_{j}$, if and only if $i>j$, $T_{i}$ defeats $T_{j}$, at which point the number of cyclic triples is minimized.
(2) Any three participating teams either form a cyclic triple or a "dominant" triple (i.e., one team defeats the other two). Let the forme... | \frac{1}{6} n(n+1)(2 n+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,164 |
1. Given the function defined on the set of complex numbers $f(z)=(4+\mathrm{i}) z^{2}+p z+q(p, q$ are complex numbers $)$. If $f(1)$ and $f(\mathrm{i})$ are both real numbers, then the minimum value of $|p|+|q|$ is $\qquad$ . | $-1 . \sqrt{2}$.
$$
\begin{array}{l}
\text { Let } p=a+b i, q=c+d i(a, b, c, d \in \mathbf{R}) . \\
\text { From } f(1)=(4+a+c)+(1+b+d) i, \\
f(i)=(-4-b+c)+(-1+a+d) i
\end{array}
$$
being real, we know $a=1-d, b=-1-d$.
$$
\begin{array}{l}
\text { Then }|p|+|q|=\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \\
=\sqrt{2+2 d^{2}}... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,165 |
Example 7 Proof: For any $n \in \mathbf{Z}_{+}$, we have
$$
\frac{1}{2} \times \frac{3}{4} \times \frac{7}{8} \times \cdots \times \frac{2^{n}-1}{2^{n}}>\frac{1}{4} .
$$ | Prove that for $\frac{2^{k}-1}{2^{k}}(k=2,3, \cdots, n)$, the following inequality holds after appropriate bounding:
$$
\begin{array}{l}
\frac{2^{k}-1}{2^{k}}=1-\frac{1}{2^{k}} \geqslant 1-\frac{1}{2^{k-1}+2}=\frac{2^{k-1}+1}{2\left(2^{k-2}+1\right)} \\
\Rightarrow \frac{1}{2} \times \frac{3}{4} \times \frac{7}{8} \tim... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,166 |
2. Given the function
$$
f(x)=a \sin x+b \cos x \quad(a, b \in \mathbf{Z}),
$$
and it satisfies
$$
\{x \mid f(x)=0\}=\{x \mid f(f(x))=0\} .
$$
Then the maximum value of $a$ is . $\qquad$ | 2.3.
Let $A=\{x \mid f(x)=0\}, B=\{x \mid f(f(x))=0\}$.
Obviously, set $A$ is non-empty.
Take $x_{0} \in A$, i.e., $x_{0} \in B$, hence
$$
b=f(0)=f\left(f\left(x_{0}\right)\right)=0 \text {. }
$$
Thus, $f(x)=a \sin x(a \in \mathbf{Z})$.
When $a=0$, obviously, $A=B$.
Now assume $a \neq 0$, in this case,
$$
\begin{array... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,167 |
3. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, and $SA=SB=SC=AB=2$. Find the surface area of the circumscribed sphere of the tetrahedron $S-ABC$ is $\qquad$ . | 3. $\frac{16 \pi}{3}$.
As shown in Figure 3, let the circumcenter of the tetrahedron $S-A B C$ be $O$, and the radius be $R$.
Given $S A=S B=S C$ $=A B=2$, we know that the projection of $S$ on the plane $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $H$ of $A B$.
Since $O A=O B=O C$, we kno... | \frac{16 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,168 |
4: Let $F(x, y)=(x-y)^{2}+\left(\frac{x}{3}+\frac{3}{y}\right)(y \neq 0)$. Then the minimum value of $F(x, y)$ is $\qquad$ | 4. $\frac{18}{5}$.
Let $P\left(x, \frac{x}{3}\right), Q\left(y,-\frac{3}{y}\right)$.
Then $F(x, y)=|P Q|^{2}$, where $P$ is a moving point on the line $l_{1}: y=\frac{1}{3} x$, and $Q$ is a moving point on the curve $y=-\frac{3}{x}$.
Suppose the line $l_{2}: y=\frac{1}{3} x+m$ is tangent to the curve $y=-\frac{3}{x}$.... | \frac{18}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,169 |
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