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1. There are five cities in a line connected by semi-circular roads, as shown in Figure 1. Each segment of the journey is from one city to another along a semi-circle. If the journey can be repeated, the total number of possible ways to start from city 5 and return to city 5 after four segments is $\qquad$.
保留源文本的换行和格... | ,- 1.80 .
After four segments, there are five possible ways to start from city 5 and return to city 5:
$$
\begin{array}{l}
5 \rightarrow 1 \rightarrow 5 \rightarrow 1 \rightarrow 5, \\
5 \rightarrow 1 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarr... | 80 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,106 |
2. Let positive integers $m, n$ satisfy
$$
m(n-m)=-11 n+8 \text {. }
$$
Then the sum of all possible values of $m-n$ is $\qquad$ | 2. 18 .
From the problem, we have
$$
n=\frac{m^{2}+8}{m+11}=m-11+\frac{129}{m+11} \in \mathbf{Z}_{+} \text {. }
$$
Then $(m+11) \mid 129$
$$
\Rightarrow m+11=1,3,43,129 \text {. }
$$
Also, $m \in \mathbf{Z}_{+}$, checking we find that when $m=32,118$, the corresponding $n$ is a positive integer.
$$
\text { Hence }(m... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,107 |
3. Person A tosses a fair coin twice, and Person B tosses the same coin three times. If the probability that they end up with the same number of heads is written as a simplified fraction, the sum of the numerator and the denominator is $\qquad$ . (Romania) | 3.21.
Let the outcomes of a coin landing heads up and tails up be denoted as $\mathrm{H}$ and $\mathrm{T}$, respectively.
Jia has four equally probable outcomes: HH, HT, TH, TT;
Yi has eight equally probable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
The outcomes that match $\mathrm{HH}$ are $\mathrm{HHT}$, $... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,108 |
6. The product $1!\cdot 2!\cdot 3!\cdot \cdots \cdot 99!\cdot 100!$ ends with $\qquad$ consecutive 0s.
(Hong Kong, China, provided) | 6. 1124.
Since in the product $1!\cdot 2!\cdot 3!\cdots \cdots 99!\cdot 100!$, there are a large number of factor 2s, the number of consecutive 0s at the end is determined by the number of factor 5s. The number of factor 5s in each factorial $x$! is shown in Table 1.
Thus, the total number of factor 5s is
$$
\begin{ar... | 1124 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,109 |
Example 3 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[3]}$
(2012, Girls' Mathematical Olympiad) | Notice that, $2012=4 \times 503$, where $p=503$ is a prime number, and
$$
\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1} .
$$
If $p \nmid k$, then $p \mid \mathrm{C}_{2012}^{k}$;
If $p \mid k$, then $k \in\{0, p, 2 p, 3 p, 4 p\}$.
Notice that, $\mathrm{... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,110 |
7. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are real coefficients. Assume that
$$
P(1)=7, P(2)=52, P(3)=97 \text {, }
$$
then $\frac{P(9)+P(-5)}{4}=$ $\qquad$ . (Vietnam) | 7. 1202.
Notice that, $52-7=97-52=45$,
$$
\begin{array}{l}
7=45 \times 1-38,52=45 \times 2-38, \\
97=45 \times 3-38 .
\end{array}
$$
Let $Q(x)=P(x)-45 x+38$. Then $Q(x)$ is a fourth-degree polynomial with a leading coefficient of 1, and
$$
Q(1)=Q(2)=Q(3)=0 \text {. }
$$
Thus, for some $r$,
$$
\begin{array}{l}
Q(x)=(... | 1202 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,111 |
11. Given that $a$ and $b$ are real numbers, satisfying:
$$
\sqrt[3]{a}-\sqrt[3]{b}=12, \quad a b=\left(\frac{a+b+8}{6}\right)^{3} \text {. }
$$
Then $a-b=$ $\qquad$ (Proposed by Thailand) | 11. 468 .
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}$. Then $x-y=12, 6xy=x^{3}+y^{3}+8$.
Thus, $x^{3}+y^{3}+2^{3}-3 \times 2xy=0$
$$
\Rightarrow(x+y+2)\left(x^{2}+y^{2}+2^{2}-xy-2x-2y\right)=0 \text {. }
$$
Therefore, $x+y+2=0$, or
$$
x^{2}-xy+y^{2}-2x-2y+4=0 \text {. }
$$
Equation (1) can be rewritten as
$$
\frac{1}{2}(x-y)... | 468 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,112 |
1. Given the equation $\left|x^{2}-2 a x+b\right|=8$ has exactly three real roots, and they are the side lengths of a right triangle. Find the value of $a+b$.
(Bulgaria) | II. 1. Note that, for the equation
$$
x^{2}-2 a x+b-8=0
$$
the discriminant $\Delta_{1}=4\left(a^{2}-b+8\right)$; for the equation
$$
x^{2}-2 a x+b+8=0
$$
the discriminant $\Delta_{2}=4\left(a^{2}-b-8\right)$.
According to the problem, the original equation has exactly three real roots, so one of the discriminants is... | 264 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,113 |
2. Color the six vertices of a regular pentagonal pyramid using at most six different colors, such that the two endpoints of any edge are colored differently. If one coloring method can be obtained from another by rotation, they are considered the same method. How many different coloring methods are there?
(Romania) | 2. Since the problem does not specify the number of colors to choose from, we will calculate the coloring methods based on the actual number of colors used.
If fewer than four colors are used, the bottom face can have at most two colors. Since 5 is an odd number, the colors of the bottom face vertices cannot be altern... | 288 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,114 |
Example 4 It is known that country $K$ consists of 1024 cities (denoted by $0,1, \cdots, 1023$), and cities $m$ and $n$ are connected by exactly one road if and only if the binary representations of $m$ and $n$ differ in exactly one digit. Now, eight roads are closed. Prove: There exists a tour route such that each cit... | Proposition: Let there be $2^{n}$ vertices, each represented by an $n$-bit binary number $(\underbrace{00 \cdots 0}_{n \uparrow})_{2}, \cdots,(\underbrace{11 \cdots 1}_{n \uparrow})_{2}$. An edge $(u, v)$ is drawn between vertices $u$ and $v$ if and only if $u$ and $v$ differ in exactly one bit. Denote this graph as $G... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 731,115 |
5. For what positive integer $k$ does $\frac{20^{k}+18^{k}}{k!}$ achieve its maximum value?
(Hong Kong, China, Contributed) | 5.19.
Let $A_{k}=\frac{20^{k}+18^{k}}{k!}$. Then
$A_{k+1}=\frac{20^{k+1}+18^{k+1}}{(k+1)!}$,
$\frac{A_{k+1}}{A_{k}}=\frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}$.
Notice that,
$$
\begin{array}{l}
\frac{A_{k+1}}{A_{k}}>1 \Leftrightarrow \frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}>1 \\
\Leftri... | 19 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,116 |
1. The imaginary part of the complex number $z=(1+\mathrm{i})^{2}(2+\mathrm{i})$ is ( ).
(A) -2 i
(B) -2
(C) 4 i
(D) 4 | \begin{array}{l}\text {-.1. D. } \\ z=(1+\mathrm{i})^{2}(2+\mathrm{i})=2 \mathrm{i}(2+\mathrm{i})=4 \mathrm{i}-2 \text {. }\end{array} | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,118 |
Example $\mathbf{5}$ Given $n\left(n \in \mathbf{Z}_{+}, n \geqslant 2\right)$ lamps, each lamp has only two states: on and off. For each non-empty subset $A$ of these lamps, there is a button that can operate the lamps in set $A$. When this button is operated, all lamps in set $A$ will change their states (on becomes ... | 【Analysis】The original solution uses a set method, which is somewhat cumbersome. It is better to use binary digits $0, 1$ to represent the lights being off and on, respectively.
Proof: Number the $n$ lights as $1, 2, \cdots, n$. Each button corresponds to an $n$-digit binary string $\left(a_{1} a_{2} \cdots a_{n}\righ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 731,119 |
2. Given sets:
$$
\begin{array}{l}
A=\left\{(x, y) \mid x+a^{2} y+6=0\right\}, \\
B=\{(x, y) \mid(a-2) x+3 a y+2 a=0\} .
\end{array}
$$
If $A \cap B=\varnothing$, then the value of $a$ is ( ).
(A) 3 or -1
(B) 0
(C) -1
(D) 0 or -1 | 2. D.
From the problem, we know
$$
x+a^{2} y+6=0 \text { and }(a-2) x+3 a y+2 a=0
$$
are both equations of lines.
If $A \cap B=\varnothing$, then these two lines have the same slope and do not coincide.
From the same slope, we get $a^{2}=\frac{3 a}{a-2}$.
Thus, $a^{3}-2 a^{2}-3 a=0$.
(1) When $a=0$, it is verified to... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,120 |
3. The coefficient of $a b c^{2}$ in the expansion of $(a+2 b-3 c)^{4}$ is ( ).
(A) 208
(B) 216
(C) 217
(D) 218 | 3. B.
According to the problem, the coefficient of $a b c^{2}$ is $C_{4}^{2}(-3)^{2} C_{2}^{1} \times 2 \times 1=216$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,121 |
4. A company's income and expenditure from 2013 to 2017 are shown in Table 1.
According to the data in the table, the regression line equation is
$$
\hat{y}=0.8 x+\hat{a} \text {, }
$$
Based on this estimate, the expenditure of the company in 2018 when the income is 7 billion yuan is $(\quad)$ billion yuan.
(A) 4.5
(B... | 4. B.
$$
\begin{array}{l}
\text { Given } \frac{0.2+1.5+2.0+2.5+3.8}{5} \\
=0.8 \times \frac{2.2+2.6+4.0+5.3+5.9}{5}+\hat{a} \\
\Rightarrow \hat{a}=-1.2 .
\end{array}
$$
When $x=7$, $y=0.8 \times 7-1.2=4.4$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,122 |
5. In the square in Figure 1, 10000 points are randomly thrown. Then the estimated number of points that fall into the shaded area (the equation of curve $C$ is $x^{2}-y=0$) is ( ).
(A) 5000
(B) 6667
(C) 7500
(D) 7854 | 5. B.
According to the problem, the area of the blank region is
$$
\int_{0}^{1} x^{2} \mathrm{~d} x=\left.\frac{1}{3} x^{3}\right|_{0} ^{1}=\frac{1}{3} \text {. }
$$
Therefore, the area of the shaded part is $\frac{2}{3}$, and the ratio of the area of the shaded part to the area of the blank part is $2: 1$.
Thus, the... | 6667 | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,123 |
9. The mathematician Qin Jiushao (approximately 1202-1261 AD) of the Southern Song Dynasty in China proposed Qin Jiushao's algorithm for evaluating polynomials in his work "Mathematical Treatise in Nine Sections." The flowchart in Figure 3 provides an example of using Qin Jiushao's algorithm to evaluate a polynomial. I... | 9. A.
Substituting $n=5$ gives $i=4 \geqslant 0$, at this point,
$$
v=2 \times 1+1=2+1 \text {. }
$$
When $i=3 \geqslant 0$,
$$
v=2(2+1)+1=2^{2}+2+1 \text {; }
$$
When $i=2 \geqslant 0$,
$$
v=2\left(2^{2}+2+1\right)+1=2^{3}+2^{2}+2+1 ;
$$
When $i=1 \geqslant 0$,
$$
\begin{array}{c}
v=2\left(2^{3}+2^{2}+2+1\right)+1... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,124 |
10. Let point $O$ be inside $\triangle A B C$, satisfying $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0}$.
Then the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$ is $(\quad)$.
(A) 2
(B) 3
(C) $\frac{3}{2}$
(D) $\frac{5}{3}$ | 10. B.
Take the midpoints $D, E$ of $AC, BC$ respectively.
$$
\begin{array}{l}
\text { By } \overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0} \\
\Rightarrow \overrightarrow{O A}+\overrightarrow{O C}=-2(\overrightarrow{O B}+\overrightarrow{O C}) \\
\Rightarrow 2 \overrightarrow{O D}=-4 \over... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,125 |
11. Given the parabola $C: y^{2}=2 p x(p>0)$ intersects with the moving line $l: y=k x+b(k, b$ are parameters, and $k \neq 0$, $b \neq 0)$ at points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, with the origin of the Cartesian coordinate system being $O$, and the slopes of lines $O A$ and $O B$ being ... | 11. D.
By combining the equation of the line with the equation of the parabola, and eliminating $y$, we get
$$
\begin{array}{l}
k^{2} x^{2}+(2 k b-2 p) x+b^{2}=0 . \\
\text { Hence } x_{1}+x_{2}=\frac{-2 k b+2 p}{k^{2}}, x_{1} x_{2}=\frac{b^{2}}{k^{2}} .
\end{array}
$$
Since $k_{O A} k_{O B}=\sqrt{3}$, we have
$$
y_{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,126 |
1. The sequence $\left\{x_{n}\right\}$ is defined as follows: $x_{1}=1, x_{2}=a$, where $a$ is a given real number greater than 1; if for some integer $n \geqslant 1$, the first $2^{n}$ terms of the sequence have been defined, then for $2^{n}+1 \leqslant k \leqslant 2^{n+1}$, define $x_{k}=a x_{k-2^{n}}$. Thus, the ter... | Let $x_{k}=X(k), S_{k}=S(k)$, and the number of 1s in the binary representation of $n$ be $E(n)$. Then we have the following conclusions:
(1) $X(n)=a^{E(n-1)}$;
(2) $X(2 n+1)=X(n+1), X(2 n)=a X(n)$;
(3) $S(2 n)=(1+a) S(n)$,
$S(2 n+1)=a S(n)+S(n+1)$.
Let $n_{t}=\sum_{s=t}^{r} 2^{k_{s}}(t=1,2, \cdots, r)$.
Then $S(n)=\su... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 731,127 |
12. Given the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{1}{2} x+\frac{3}{2}, & x \leqslant 1 ; \\
\ln x, & x>1
\end{array}\right.
$$
$(\ln x$ is the natural logarithm with base $\mathrm{e}$, $\mathrm{e}=2.71828 \cdots)$. If there exist real numbers $m, n (m<n)$, such that $f(m)=f(n)$, then the range of $n-m$ is (... | 12. C.
It is known that when the slopes are equal, $n-m$ reaches its minimum value, i.e., $(\ln x)' = \frac{1}{2} \Rightarrow x = 2$.
At this point, $f(x) = \ln x$ passes through the point $(2, \ln 2)$.
If $f(x) = \ln 2 = \frac{1}{2} x + \frac{3}{2}$, then $x = 2 \ln 2 - 3$.
At this point, $n-m = 5 - 2 \ln 2$.
Also, w... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,128 |
13. Let real numbers $x y$ satisfy the constraint conditions $\left\{\begin{array}{l}x+2 y \geqslant 2, \\ 2 x+y \leqslant 4, \\ 4 x-y \geqslant 1 .\end{array}\right.$ Then the range of the objective function $z=3 x+y$ is $\qquad$ | 二、13. $\left[\frac{19}{9}, 6\right]$.
According to the problem, the constraints form a triangle with three vertices at $(2,0)$, $\left(\frac{4}{9}, \frac{7}{9}\right)$, and $\left(\frac{5}{6}, \frac{7}{3}\right)$. Calculating, we get $z \in\left[\frac{19}{9}, 6\right]$. | \left[\frac{19}{9}, 6\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,129 |
14. As shown in Figure 4, in rectangle $A B C D$, $A B = 2 A D$, and $E$ is the midpoint of side $A B$. Fold $\triangle A D E$ along $D E$ to form $\triangle A_{1} D E$. If $M$ is the midpoint of line segment $A_{1} C$, then during the folding process of $\triangle A D E$, which of the following statements is correct? ... | 14. (1)(3).
Take the midpoint of $CD$ as $F$. Then
$M F / / D A, B F / / D E$.
Thus, plane $M B F / /$ plane $A_{1} D E$
$\Rightarrow M B / /$ plane $A_{1} D E \Rightarrow$ statement (4) is incorrect.
Since $\angle M F B = \angle A_{1} D E, M F = \frac{1}{2} A_{1} D, F B = D E$ are all constants, it follows that $M B$... | (1)(3) | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,130 |
15. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}$ and $F_{2}$, respectively, a line passing through focus $F_{2}$ intersects the right branch of the hyperbola at points $P$ and $Q$, and $P Q \perp P F_{1}$. If $|P Q|=\frac{5}{12}\left|P F_{1}\right|$, then th... | 15. $\frac{\sqrt{37}}{5}$.
From the problem, we know in the right triangle $\triangle P F_{1} Q$,
$$
\left|Q F_{1}\right|=\sqrt{\left|P F_{1}\right|^{2}+|P Q|^{2}}=\frac{13}{12}\left|P F_{1}\right| \text {. }
$$
By the definition of a hyperbola,
$$
\begin{array}{l}
2 a=\left|P F_{1}\right|-\left|P F_{2}\right|=\left|... | \frac{\sqrt{37}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,131 |
16. Nine consecutive positive integers are arranged in an increasing sequence $a_{1}, a_{2}, \cdots, a_{9}$. If $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ is a perfect square, and $a_{2}+a_{4}+a_{6}+a_{8}$ is a perfect cube, then the minimum value of $a_{1}+a_{2}+\cdots+a_{9}$ is $\qquad$. | 16. 18000 .
Let these nine numbers be $a-4, a-3, \cdots, a+4$. Then $5 a=m^{2}, 4 a=n^{3}, S=9 a$. Thus, $a=\frac{m^{2}}{5}=\frac{n^{3}}{4} \Rightarrow 4 m^{2}=5 n^{3}$.
Clearly, $m$ and $n$ are both multiples of 10.
Let $m=10 m_{1}, n=10 n_{1}$, then $4 m_{1}^{2}=50 n_{1}^{3}$.
Take $m_{1}=10, n_{1}=2$. At this point... | 18000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,132 |
17. (10 points) In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively,
$$
\tan C=\frac{\sin A+\sin B}{\cos A+\cos B}, \sin (B-A)=\cos C \text {. }
$$
(1) Find $\angle A, \angle C$;
(2) If $S_{\triangle A B C}=3+\sqrt{3}$, find $a, c$. | Three, 17. (1) Note that,
$$
\begin{array}{l}
\tan C=\frac{\sin A+\sin B}{\cos A+\cos B} \\
\Rightarrow \frac{\sin C}{\cos C}=\frac{\sin A+\sin B}{\cos A+\cos B} \\
\Rightarrow \sin C \cdot \cos A-\cos C \cdot \sin A \\
\quad=\cos C \cdot \sin B-\sin C \cdot \cos B \\
\Rightarrow \sin (C-A)=\sin (B-C) \\
\Rightarrow \a... | a=2\sqrt{2}, c=2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,133 |
18. (12 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=2 a_{n}+1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n}}{a_{n+1}}<\frac{n}{2}$. | 18. (1) From $a_{n+1}=2 a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$, we know
$$
a_{n+1}+1=2\left(a_{n}+1\right)=2^{n-1}\left(a_{1}+1\right)=2^{n} \text {. }
$$
Therefore, $a_{n}=2^{n}-1$.
(2) Notice that,
$$
\begin{array}{l}
\frac{a_{k}}{a_{k+1}}=\frac{2^{k}-1}{2^{k+1}-1}<\frac{2^{k}-1}{2 \times 2^{k}-1-1} \\
=\frac{2^{... | \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n}}{a_{n+1}}<\frac{n}{2} | Algebra | proof | Yes | Yes | cn_contest | false | 731,134 |
19. (12 points) In response to the national strategy of "targeted poverty alleviation, industrial poverty alleviation," Harbin City is recruiting volunteer advocates for the "Poverty Alleviation Policy" from the public. From 500 volunteers aged between 20 and 45, 100 are randomly selected, and their age frequency distr... | 19. (1) According to the frequency distribution histogram, we have
$$
(0.01+0.02+0.04+x+0.07) \times 5=1 \text {. }
$$
Solving for $x$ gives $x=0.06$.
(2) Using stratified sampling, 10 volunteers are selected from 100, among which, 6 are under 35 years old, and 4 are 35 years old or older. Therefore, the possible valu... | 1.8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,135 |
20. (12 points) As shown in Figure 6, in $\triangle A B C$, $\angle C = 90^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $E$, and a line perpendicular to $B E$ through point $E$ intersects $A B$ at point $F$. $\odot O$ is the circumcircle of $\triangle B E F$, intersecting $B C$ at point $D$.... | 20. (1) Connect $O E$.
Since $B E$ bisects $\angle A B C$
$$
\Rightarrow \angle C B E=\angle O B E \text {. }
$$
Also, $O B=O E$, thus,
$$
\begin{array}{l}
\angle O B E=\angle O E B=\angle C B E \\
\Rightarrow O E / / B C \\
\Rightarrow \angle A E O=\angle C=90^{\circ}
\end{array}
$$
$\Rightarrow A C$ is a tangent to... | B F=10, A F=\frac{5}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 731,136 |
2. Laura has 2010 lamps and 2010 switches in front of her, with different switches controlling different lamps. She wants to find the correspondence between the switches and the lamps. For this, Charlie operates the switches. Each time Charlie presses some switches, and the number of lamps that light up is the same as ... | (1) Charlie selects a pair of switches $(A, B)$, each operation involves pressing them simultaneously or leaving them untouched, while other switches can be chosen arbitrarily, resulting in $2^{2009}$ different operations, but Laura cannot determine which lights are controlled by switches $A$ and $B$.
If Charlie perfo... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,137 |
22. (12 points) Given the function
$$
f(x)=\frac{m x-n}{x}-\ln x \quad (m, n \in \mathbf{R}) \text{.}
$$
(1) If the tangent line to the function $f(x)$ at the point $(2, f(2))$ is parallel to the line $x-y=0$, find the value of the real number $n$;
(2) Discuss the maximum value of the function $f(x)$ on the interval $[... | 22. (1) Notice,
$$
f^{\prime}(x)=\frac{n-x}{x^{2}}, f^{\prime}(2)=\frac{n-2}{4} \text {. }
$$
Since the tangent line of the function $f(x)$ at the point $(2, f(2))$ is parallel to the line $x-y=0$, we have,
$$
\frac{n-2}{4}=1 \Rightarrow n=6 \text {. }
$$
(2) It is easy to see that, $f^{\prime}(x)=\frac{n-x}{x^{2}}(x>... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 731,138 |
1. Given sets:
$$
\begin{array}{l}
A=\left\{x \mid x=2 n-1, n \in \mathbf{Z}_{+}\right\}, \\
B=\{y \mid y=3 x-1, x \in A\} .
\end{array}
$$
Then $A \cap B=$ | $-1 . \varnothing$.
Since the elements in set $A$ are positive odd numbers, and the elements in set $B$ are even numbers, therefore, $A \cap B=\varnothing$. | \varnothing | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,139 |
2. In the Cartesian coordinate system $x O y$, if the asymptotes of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ do not intersect the circle $x^{2}+y^{2}-$ $6 y+5=0$, then the range of the eccentricity of the hyperbola is $\qquad$. | 2. $\left(1, \frac{3}{2}\right)$.
From the problem, we know that the distance from the center of the circle $(0,3)$ to the asymptote is greater than 2, i.e., $\frac{3 a}{\sqrt{a^{2}+b^{2}}}>2$.
Therefore, the eccentricity of the hyperbola $e \in\left(1, \frac{3}{2}\right)$. | \left(1, \frac{3}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,140 |
4. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
b_{n}=\left\{\begin{array}{ll}
a_{\frac{n+1}{2}}, & n \text { is odd; } \\
\sqrt{a_{n+1}}, & n \text { is even. }
\end{array}\right.
$$
If $\left\{b_{n}\right\}$ is a geometric sequence, and $a_{2}+b_{2}=108$, then the general term fo... | 4. $a_{n}=9^{n}$.
Since $\left\{b_{n}\right\}$ is a geometric sequence, its subsequence $\left\{b_{2 n-1}\right\}$ is also a geometric sequence.
And $b_{2 n-1}=a_{n}$, so $\left\{a_{n}\right\}$ is also a geometric sequence, with the common ratio $q$.
$$
\begin{array}{l}
\text { By } b_{1}=a_{1}, b_{2}=\sqrt{a_{3}}, b... | a_{n}=9^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,141 |
5. Given that $a$ and $b$ are positive real numbers, and
$$
2 a+2 b \leqslant 15, \frac{4}{a}+\frac{3}{b} \leqslant 2 \text {. }
$$
Then the range of $3 a+4 b$ is $\qquad$ | 5. $[24,27]$.
From $\frac{4}{a}+\frac{3}{b} \leqslant 2 \Rightarrow b \geqslant \frac{3 a}{2 a-4}$.
Thus, the point $(a, b)$ is in the first quadrant and lies within the region bounded by the line $2 a+2 b=15$ and the hyperbola $b=\frac{3 a}{2 a-4}$.
Therefore, when $a=4, b=3$, $3 a+4 b$ reaches its minimum value of ... | [24,27] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,142 |
6. As shown in Figure 1, in the octahedron $P A B C D Q$ inscribed in sphere $O$, vertices $P$ and $Q$ are on opposite sides of plane $A B C D$, and the pyramids $P-A B C D$ and $Q-A B C D$ are both regular square pyramids. Let the size of the dihedral angle $P-A B-Q$ be $\theta$. Then the range of $\tan \theta$ is | 6. $(-\infty,-2 \sqrt{2}]$.
Let the dihedral angle $P-A B-C$ be $\alpha$, and the dihedral angle $Q-A B-C$ be $\beta$. The distance from the center of the sphere to the plane $A B C D$ is $d$, and the radius of the sphere is 1. Then,
$$
\tan \alpha=\frac{\sqrt{2}(1-d)}{\sqrt{1-d^{2}}}, \tan \beta=\frac{\sqrt{2}(1+d)}{... | -2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,143 |
7. Given real numbers $a>b>0$. Then the maximum value of the function
$$
f(x)=\frac{x}{\sqrt{a-x^{2}}-\sqrt{b-x^{2}}}
$$
is | 7. $\frac{\sqrt{a b}}{a-b}$.
Obviously, when $x=0$, $f(x)>0$.
Therefore, we only need to consider the case when $x>0$.
Notice that,
$$
\begin{array}{l}
f(x)=\frac{x}{\sqrt{a-x^{2}}-\sqrt{b-x^{2}}} \\
=\frac{x\left(\sqrt{a-x^{2}}+\sqrt{b-x^{2}}\right)}{a-b} \\
=\frac{1}{a-b} \sqrt{x^{2}(a+b)-2 x^{4}+2 x^{2} \sqrt{\left... | \frac{\sqrt{a b}}{a-b} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 731,144 |
II. (40 points) As shown in Figure 2, the large circle and the small circle are concentric circles with the center $O$. A tangent line $AC$ is drawn from a point $A$ on the large circle to the small circle, with the point of tangency at $B$. Point $C$ is on the large circle, and $D$ is the midpoint of $AB$. The vertex ... | II. As shown in Figure 4, connect $O B, P E, P F$.
Since $A C$ is a tangent to the smaller circle, we have $O B \perp A C$.
Also, since $A C$ is a chord of the larger circle, then $A B = B C$.
By the secant-tangent theorem, we get
$$
A B^{2} = A F \cdot A E \text{. }
$$
Since $D$ is the midpoint of $A B$, we know $A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,145 |
Three. (50 points) Let positive integers
$$
1<a_{1}<a_{2}<\cdots<a_{1009}<2018 \text {, }
$$
where, for any $i \neq j$, the least common multiple of $a_{i}$ and $a_{j}$ is greater than 2018. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{1009}}<\frac{7}{6} .
$$ | Three, let $[x]$ denote the greatest integer not exceeding the real number $x$.
First, prove: $a_{1} \geqslant\left[\frac{2018}{3}\right]+1=673$.
If not, then
$$
\begin{array}{l}
a_{1} \leqslant\left[\frac{2018}{3}\right]2018$ leads to a contradiction.
Therefore, $a_{1} \geqslant 673$.
For each $a_{i}$, construct the s... | \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{1009}}<\frac{7}{6} | Number Theory | proof | Yes | Yes | cn_contest | false | 731,146 |
Four, (50 points) Mr. Wanda often forgets the numbers he should remember, such as his friends' phone numbers, the password of the safe, etc. For this reason, the manufacturer specially designed a password lock for his office safe, with keys labeled $0 \sim 9$. It is known that the password of the safe is a three-digit ... | Four, 50 times.
First, introduce several notations. Let
$E=\{1,3,5,7,9\}, F=\{0,2,4,6,8\}$
$\Phi=E \cup F=\{0,1, \cdots, 9\}$,
$\Omega=\{x y z \mid x, y, z \in \Phi\}$ (the set of all three-digit codes),
$\Omega_{11}=\{x a b \in \Omega \mid x \in \Phi, a, b \in E\}$,
$\Omega_{12}=\{a y b \in \Omega \mid y \in \Phi, a, ... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,147 |
1. Given the function
$$
f(x)=\left|8 x^{3}-12 x-a\right|+a
$$
has a maximum value of 0 on the interval $[0,1]$. Then the maximum value of the real number $a$ is $\qquad$ | $-1 .-2 \sqrt{2}$.
Let $g(x)=8 x^{3}-12 x$. Then $g^{\prime}(x)=24 x^{2}-12$.
Thus, $g(x)_{\min }=g\left(\frac{\sqrt{2}}{2}\right)=-4 \sqrt{2}$, $g(x)_{\max }=\max \{g(0), g(1)\}=0$.
Therefore, $\max _{1 \leqslant x \leqslant 2}\{f(x)\}$
$$
\begin{array}{l}
=\max \{|-4 \sqrt{2}-a|+a,|0-a|+a\}=0 \\
\Rightarrow\left\{\be... | -2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,148 |
3. Given the complex number $z$ satisfies $(1-z)^{7}=-z^{7}$. Then $z=$ $\qquad$ | 3. $-\frac{1}{2}+\frac{1}{2} \mathrm{i} \cot \frac{k \pi}{7}(k=0,1, \cdots, 6)$.
Since $\left(1-\frac{1}{z}\right)^{7}=1$, we have,
$$
\begin{array}{l}
\frac{1}{z_{k}}=1-\cos \frac{2 k \pi}{7}-\mathrm{i} \sin \frac{2 k \pi}{7}(k=0,1, \cdots, 6) \\
\Rightarrow z_{k}=\frac{1}{1-\cos \frac{2 k \pi}{7}-\mathrm{i} \sin \fr... | -\frac{1}{2}+\frac{1}{2} \mathrm{i} \cot \frac{k \pi}{7}(k=0,1, \cdots, 6) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,149 |
4. Given $\sin ^{10} x+\cos ^{10} x=\frac{11}{36}$. Then $\sin ^{14} x+\cos ^{14} x=$ $\qquad$ . | 4. $\frac{41}{216}$.
Let $\sin ^{n} x+\cos ^{n} x=a_{n}, \sin ^{2} x \cdot \cos ^{2} x=y \leqslant \frac{1}{4}$.
Then $a_{n+2}=a_{n}-y a_{n-2}(n \geqslant 2)$, where $a_{0}=2, a_{2}=1, a_{4}=1-2 y, a_{6}=1-3 y$, $a_{10}=5 y^{2}-5 y+1=\frac{11}{36}$.
Solving for $y$ gives $y=\frac{1}{6}$.
Therefore, $\sin ^{14} x+\cos ... | \frac{41}{216} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,150 |
5. Given planar vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfy:
$$
\begin{array}{l}
|\boldsymbol{a}|=|\boldsymbol{b}|=|\boldsymbol{c}|=2, \boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}=\mathbf{0} . \\
\text { If } 0 \leqslant x \leqslant \frac{1}{2} \leqslant y \leqslant 1 \text {, then } \\
|x(\boldsym... | 5. $\frac{1}{2}$.
Since $|a|=|b|=|c|, a+b+c=0$, therefore, the angles between $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are all $120^{\circ}$.
Thus, $\boldsymbol{a} \cdot \boldsymbol{b}=\boldsymbol{b} \cdot \boldsymbol{c}=\boldsymbol{a} \cdot \boldsymbol{c}=-2$.
Let $z=1-x-y$. Then
$$
\begin{array}{l}
|x(\bolds... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,151 |
6. As shown in Figure 1, the cube $A B C D-E F G H$ has an edge length of 1, and point $M$ is on the surface $E F G H$ of the cube. Define a path as a route where every point lies on the surface of the cube. It is known that the shortest path length from point $M$ to $A$, $l(M, A)$, is equal to the shortest path length... | 6. $\frac{5 \sqrt{2}}{6}$.
By symmetry, without loss of generality, assume point $M$ is inside $\triangle E F G$ (including the boundary).
Unfold the square $A B F E$ and the square $E F G H$ into a plane, as shown in Figure 3. Then point $M$ lies on the perpendicular bisector of $A G$.
When $M$ is the intersection ... | \frac{5 \sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,152 |
7. Given a regular 2019-gon, then, the maximum number of diagonals such that any two of them are either perpendicular or do not intersect except at endpoints. | 7.2016 .
On one hand, no two diagonals of the regular 2019-gon $\Gamma$ are perpendicular.
Assume there exist two diagonals of the regular 2019-gon $\Gamma$ that are perpendicular, say $AB \perp CD$, and $A, B, C, D$ are four vertices of $\Gamma$. Let $E$ be a point on the perpendicular bisector of $AB$, and let $E'$... | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,153 |
8. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ with the general terms $a_{n}=2^{n}, b_{n}=5 n-2$. Then the sum of all elements in the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{2019}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{2019}\right\}
$$
is $\qquad$ | 8.2184 .
Let the elements of $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be arranged in ascending order to form the sequence $\left\{c_{n}\right\}$.
Then the problem is reduced to finding the number of solutions to the equation
$$
2^{n}=5 m-2\left(n, m \in \mathbf{Z}_{+}... | 2184 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,154 |
One, (40 points) Let real numbers $a_{1}<a_{2}<\cdots<a_{n}$ satisfy $\sum_{1 \leqslant i, j \leqslant n}\left|a_{i}-a_{j}\right|=1$. Prove:
$$
\sum_{1 \leqslant i, j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \geqslant \frac{3}{2\left(n^{2}-1\right)} .
$$ | By the Cauchy inequality, we have
$$
\begin{array}{l}
\left(\sum_{1 \leqslant i, j \leqslant n}|i-j|\left|a_{i}-a_{j}\right|\right)^{2} \\
\leqslant\left(\sum_{1 \leqslant i, j \leqslant n}(i-j)^{2}\right)\left(\sum_{1 \leqslant i, j \leqslant n}\left(a_{i}-a_{j}\right)^{2}\right),
\end{array}
$$
Equality holds if and... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 731,155 |
Question 1 As shown in Figure 1, the two pairs of opposite sides of the convex quadrilateral $ABCD$ intersect at points $E$ and $F$, and the diagonals intersect at point $G$. Through a point $P$ inside the quadrilateral $ABCD$, perpendiculars are drawn to each side, with the feet of the perpendiculars being $Q, R, S, T... | 【Question $\mathbf{1}$ Analysis】As shown in Figure 4.
Obviously, $A, Q, P, T$, $B, Q, P, R$, $C, R, P, S$, $D$, $S, P, T$ are respectively concyclic. Therefore,
$$
\begin{array}{l}
\angle R Q T=\angle R Q P+\angle P Q T \\
=\angle R B P+\angle P A T=\angle A P B-\angle A F B, \text { (2) } \\
\angle Q R S=\angle Q R P+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,157 |
For a positive odd number $n$, let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ non-negative real numbers, and define
$$
a_{i}=x_{i}^{2}+x_{i+1}^{2}, b_{i}=2 x_{i} x_{i+1},
$$
where $i=1,2, \cdots, n$, and define $x_{n+i}=x_{i}$.
Define $A=\min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$,
$$
B=\max \left\{b_{1}, b_{2}, \cdots... | Proof by contradiction.
If $A>B$, that is, for any $i, j \in [1, n]$, we have $a_{i} > b_{j}$.
If there exists an $i$ such that $x_{i} = x_{i+1}$, then
$$
x_{i}^{2} + x_{i+1}^{2} = 2 x_{i} x_{i+1} \Rightarrow a_{i} = b_{i},
$$
which is a contradiction.
Next, we prove that there exists an $i$ such that
$$
x_{i} x_{i+1} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 731,158 |
615 As shown in Figure 2, let $O T$ bisect $\angle X O Y$ and $\angle R O S$. Point $P$ is on $O T$. A line $P M$ is drawn through point $P$, intersecting $O X$, $O Y$, and $O S$ at points $M$, $A$, and $B$ respectively; a line $P N$ is drawn through point $P$, intersecting $O Y$, $O X$, and $O R$ at points $N$, $C$, a... | Prove (1) Connect $B C$, intersecting $O T$ at point $E$.
Let $\angle X O R=\alpha, \angle R O T=\beta$. Then
$$
\angle Y O S=\alpha, \angle S O T=\beta.
$$
Using the ratio of areas, we get
$$
\frac{C D}{D P}=\frac{S_{\triangle O C D}}{S_{\triangle O D P}}=\frac{O C \sin \alpha}{O P \sin \beta}.
$$
Similarly,
$$
\beg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,159 |
Example 1 If $a, b \in \mathbf{R}^{+}$, prove: $\frac{a^{2}}{b}+\frac{b^{2}}{a} \geqslant a+b$. | Proof: Using the basic inequality, we can obtain $\frac{a^{2}}{b}+b \geqslant 2 a$, $\frac{b^{2}}{a}+a \geqslant 2 b$. Adding the two inequalities can prove that
$$\frac{a^{2}}{b}+\frac{b^{2}}{a} \geqslant a+b$$
Reflection and conjecture: (1) The condition for equality is $a=$ $b$, is this helpful for the proof? (2) I... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,160 |
Example 2 Given positive numbers $a, b$ satisfying $a+b=1$, prove:
$$(a+2)^{2}+(b+2)^{2} \geqslant \frac{25}{2}$$ | The proof methods for this problem are numerous, but the author only uses the idea elaborated above: from $(a+2)^{2}=(b+2)^{2}=\frac{25}{4}$, we conjecture that the condition for equality is $a=b=\frac{1}{2}$. Verifying this conjecture: using the inequality $(a+2)^{2}+\frac{25}{4} \geqslant 5(a+2)$, similarly $(b+2)^{2... | (a+2)^{2}+(b+2)^{2} \geqslant \frac{25}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,161 |
Example 3 (1990, 1st "Hope Cup"): Given $x, y, z > 0$, and $\frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}=2$, prove that: $\frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}} \leqslant \sqrt{2}$. | Proof: The condition for the equality to hold is $x=y=z=\sqrt{2}$
$$\begin{array}{l}
\because \frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}=2 \\
\Longleftrightarrow \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}=1 \\
\because \frac{x}{1+x^{2}}=\frac{\sqrt{2}}{2} \frac{\sqrt{2} x}{1+x^{2}} \\
... | \sqrt{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,162 |
Example 4 (36th IMO Problem) Let $a, b, c > 0$, and $abc = 1$, prove that: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}$ $\geqslant \frac{3}{2}$ | Proof: The condition for the equality to hold in the conjecture is $a=b=c=$
1.
$$\begin{array}{l}
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \Longleftrightarrow \\
\frac{b c}{a^{2}(b+c)}+\frac{c a}{b^{2}(c+a)}+\frac{a b}{c^{2}(a+b)} \geqslant \frac{3}{2}
\end{array}$$
Accordin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,163 |
Example 5 (2005 China National Training Team Test Question) Let $a, b, c > 0, ab + bc + ca = \frac{1}{3}$, prove: $\frac{1}{a^{2} - bc + 1}$ $+\frac{1}{b^{2} - ca + 1}+\frac{1}{c^{2} - ab + 1} \leqslant 3$. | $$\begin{array}{l}
\text { Prove: The condition for equality in the conjecture is } \\
a=b=c=\frac{1}{3} \text {. } \\
\because a^{2}-b c+1=a^{2}-b c+3(a b+b c+c a)= \\
a(a+b+c)+\frac{2}{3} \text {, let } q=a+b+c \text {, } \\
\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-c a+1}+\frac{1}{c^{2}-a b+1} \\
=\frac{1}{a q+\frac{2}{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,164 |
Prove: $\left(1-\frac{1}{a_{1}}\right)\left(1-\frac{1}{a_{2}}\right) \cdots \cdots\left(1-\frac{1}{a_{n}}\right) \geqslant\left[\frac{1}{2}\left(\frac{a_{1}}{a_{1}-1}+\frac{a_{n}}{a_{n}-1}\right)\right]^{-n}$. | Proof: Let $\{a_n\}$ have a common difference of $d$, $i \in \mathbf{N}$, and $1 < i < n$, then
$$
a_i + a_{n+1-i} = a_1 + (i-1)d + a_1 + (n-i)d = 2a_1 + (n-1)d = a_1 + a_n.
$$
$$
a_1 a_{n+1-i} = [a_1 + (i-1)d][a_1 + (n-i)d] = a_1^2 + (n-1)a_1d + (i-1)(n-i)d^2 \geq a_1^2 + (n-1)a_1d = a_1 a_n.
$$
$$
(a_1 - 1)(a_{n+1-i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,165 |
1. (Albania) Let $a, b, c$ be the lengths of the sides of a triangle, and consider the inequality
$$a^{3}+b^{3}+c^{3}<k(a+b+c)(a b+b c+c a) \text {. }$$
(a) Prove that the inequality holds when $k=1$.
(b) Find the smallest constant $k$ such that the inequality always holds. | Given that $a, b, c$ are the lengths of the sides of a triangle, we have
$$a+b>c, \quad b+c>a, \quad c+a>b$$
Thus,
$$\begin{aligned}
& (a+b+c)(a b+b c+c a) \\
= & a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+3 a b c \\
> & a^{2} \cdot a+b^{2} \cdot b+c^{2} \cdot c=a^{3}+b^{3}+c^{3}
\end{aligned}$$
Therefore, the inequality holds... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 731,166 |
2. (Greek) Given $x, y>0$, and $x+y=2 a$, prove:
$$x^{3} y^{3}\left(x^{2}+y^{2}\right)^{2} \leqslant 4 a^{10}$$ | Proof. By the AM-GM inequality, we have
$$4 a^{2}=(x+y)^{2}=\left(x^{2}+y^{2}\right)+2 x y \geqslant 2 \sqrt{2 x y\left(x^{2}+y^{2}\right)},$$
Therefore,
$$x y\left(x^{2}+y^{2}\right) \leqslant 2 a^{4} .$$
Also, since $4 a^{2}=(x+y)^{2} \geqslant 4 x y$, we have $x y \leqslant a^{2}$. Thus,
$$x^{3} y^{3}\left(x^{2}+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,167 |
11. (Indonesia) Given $a, b, c \geqslant 0$ and $x, y, z > 0$ and $a+b+c=x+y+z$, prove: $\frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \geqslant a+b+c$. | Proof. By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{a^{3}}{x^{2}}+x+x \geqslant 3 \sqrt[3]{\frac{a^{3}}{x^{2}} \cdot x \cdot x}=3 a \\
\frac{b^{3}}{y^{2}}+y+y \geqslant 3 \sqrt[3]{\frac{b^{3}}{y^{2}} \cdot y \cdot y}=3 b \\
\frac{c^{3}}{z^{2}}+z+z \geqslant 3 \sqrt[3]{\frac{c^{3}}{z^{2}} \cdot z \cdot z}=3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,168 |
12. (Indonesia) The real-coefficient polynomial $f(x)=a x^{3}+b x^{2}+c x+d$ has 3 positive real roots, and $f(0)<0$. Prove that: $2 b^{3}+9 a^{2} d-7 a b c \leqslant 0$. | Proof. Let the three positive real roots of $f(x)$ be $p, q, r$, then $f(x)=a(x-p)(x-q)(x-r)$. By $f(0)=-a p q r$. By Vieta's theorem, $p+q+r=-\frac{b}{a}, p q+q r+r p=\frac{c}{a}$, $p q r=-\frac{d}{a}$. Since $a>0$, the original inequality is equivalent to
$$2\left(\frac{b}{a}\right)^{3}+9 \frac{d}{a}-7 \frac{b}{a} \c... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,169 |
13. (Kazakhstan) Given a real number $A$, find the largest real number $M$ such that
$$\frac{1}{x}+\frac{1}{y}+\frac{A}{x+y} \geqslant \frac{M}{\sqrt{x y}}$$
holds for all $x, y > 0$. | Solution. Since both sides of the inequality are homogeneous, we may assume $xy=1, x+y=t$, then by the AM-GM inequality we have $t=x+y \geqslant 2 \sqrt{xy}=2$. The problem is transformed into, for a given real number $A$, finding the largest $M$ such that for any $t \geqslant 2$ we have $t+\frac{A}{t} \geqslant M$. Le... | 2+\frac{A}{2} \text{ if } A \leqslant 4, \text{ and } 2\sqrt{A} \text{ if } A > 4 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,170 |
14. (Kazakhstan) Given $x, y \geqslant 0$, prove:
$$\sqrt{x^{2}-x+1} \sqrt{y^{2}-y+1}+\sqrt{x^{2}+x+1} \sqrt{y^{2}+y+1} \geqslant 2(x+y) .$$ | Proof. As shown in the figure, let $A O=O C=1, O D=x, O B=y, \angle A O D=120^{\circ}$, then by the cosine rule
$$\begin{array}{ll}
A D=\sqrt{x^{2}+x+1}, & C D=\sqrt{x^{2}-x+1} \\
A B=\sqrt{y^{2}-y+1}, & C B=\sqrt{y^{2}+y+1}
\end{array}$$
By Ptolemy's inequality, we have
$$A D \times B C+C D \times B A \geqslant A C \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,171 |
15. (Switzerland) Given $x, y, z > 0$, and $xyz = 1$, prove:
$$\frac{(x+y-1)^{2}}{z}+\frac{(y+z-1)^{2}}{x}+\frac{(z+x-1)^{2}}{y} \geqslant x+y+z$$ | Proof. First, by the AM-GM inequality, we have
$$x+y+z \geqslant 3 \sqrt[3]{x y z}=3 .$$
Second, by the AM-GM inequality, we have
$$\begin{array}{l}
\frac{(x+y-1)^{2}}{z}+z \geqslant 2(x+y-1) \\
\frac{(y+z-1)^{2}}{x}+x \geqslant 2(y+z-1) \\
\frac{(z+x-1)^{2}}{y}+y \geqslant 2(z+x-1)
\end{array}$$
Adding the above thr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,172 |
16. (Turkey) Given $a, b, c > 0$, prove:
$$\sum_{c y c} \sqrt[4]{\frac{\left(a^{2}+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{2}} \leqslant \frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)$$ | Proof. First, prove
$$\sqrt[4]{\frac{\left(a^{2}+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{2}} \leqslant \frac{a^{2}+b^{2}}{a+b}$$
In fact, since
$$\begin{aligned}
& \left(a^{2}-a b+b^{2}\right) \cdot\left(\frac{a+b}{2}\right)^{2} \cdot\left(\frac{a+b}{2}\right)^{2} \\
\leqslant & \left(\frac{\left(a^{2}-a b+b^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,173 |
17. (Turkey) Positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1} a_{2} \cdots a_{n}=1$, prove that
$$\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{i}^{4}+3}} \leq \sum_{i=1}^{n} \frac{1}{2 a_{i}} .$$ | Proof. Notice that $x^{4}+3 \geqslant 2 x^{2}+2=(1+1)\left(1+x^{2}\right) \geqslant(1+x)^{2}$, and
$$\sum_{i=1}^{n} \frac{1}{a_{i}} \geqslant n \sqrt[n]{\prod_{i=1}^{n} \frac{1}{a_{i}}}=n$$
Therefore, it is sufficient to prove the stronger inequality
$$\sum_{i=1}^{n} \frac{1}{a_{i}}+n \geqslant 4 \sum_{i=1}^{n} \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,174 |
18. (USA) Positive real numbers $a, b, c$ satisfy $a b c=1$. Prove:
$$\frac{1}{a^{5}(b+2 c)^{2}}+\frac{1}{b^{5}(c+2 a)^{2}}+\frac{1}{c^{5}(a+2 b)^{2}} \geqslant \frac{1}{3} .$$ | Proof. Since $a b c=1$, we have
$$\begin{array}{l}
\frac{1}{a^{5}(b+2 c)^{2}}=\frac{b^{3} c^{3}}{(a b+2 a c)^{2}} \\
\frac{1}{b^{5}(c+2 a)^{2}}=\frac{c^{3} a^{3}}{(b c+2 b a)^{2}} \\
\frac{1}{c^{5}(a+2 b)^{2}}=\frac{a^{3} b^{3}}{(c a+2 c b)^{2}}
\end{array}$$
Therefore, we only need to prove
$$\frac{b^{3} c^{3}}{(a b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,175 |
19. (USA) Given that $P$ is a point inside $\triangle ABC$, prove that:
$$\frac{AP}{h_{b}+h_{c}}+\frac{BP}{h_{c}+h_{a}}+\frac{CP}{h_{a}+h_{b}} \geqslant 1$$ | Proof. Lemma: If $P$ is a point inside $\triangle ABC$, and the distances from $P$ to sides $AB$ and $AC$ are $x$ and $y$ respectively, then we have $AP \cdot a \geqslant cx + by$, and $AP \cdot a \geqslant bx + cy$.
Proof of the lemma: As shown in the figure, since
$$AP \cdot a \geqslant 2 S_{\text{quadrilateral } AB... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,176 |
20. (Balkan) Given $a, b, c > 0$, prove:
$$\frac{a^{2} b(b-c)}{a+b}+\frac{b^{2} c(c-a)}{b+c}+\frac{c^{2} a(a-b)}{c+a} \geqslant 0 .$$ | Proof. By transforming the inequality, we can see that
$$\begin{aligned}
& \frac{a^{2} b(b-c)}{a+b}+\frac{b^{2} c(c-a)}{b+c}+\frac{c^{2} a(a-b)}{c+a} \geqslant 0 \\
\Longleftrightarrow & a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \geqslant a^{3} b^{2} c+a b^{3} c^{2}+a^{2} b c^{3}
\end{aligned}$$
Notice that, by the AM-GM in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,177 |
3. (India) Given $a, b, c > 0$, and $a b + b c + c a \leqslant 3 a b c$. Prove:
$$\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+\sqrt{\frac{a^{2}+b^{2}}{a+b}}+3 \leqslant \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}) .$$ | Proof. First, prove that $\sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{2 a b}{a+b}} \leqslant \sqrt{2(a+b)}$. In fact, this inequality is equivalent to $2 \sqrt{2 a b\left(a^{2}+b^{2}\right)} \leqslant(a+b)^{2}$, and the last inequality has already been proven in Problem 2.
Similarly, $\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,178 |
21. (China-Europe) For any given integer $n \geqslant 2$, find the largest $C_{n}$ such that for any positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$\frac{a_{1}^{2}+\cdots+a_{n}^{2}}{n} \geqslant\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)^{2}+C_{n}\left(a_{1}-a_{n}\right)^{2}$$ | Proof. Taking $a_{1}=3, a_{2}=\cdots=a_{n-1}=2, a_{n}=1$, it is easy to see that $C_{n} \leqslant \frac{1}{2 n}$. If we can prove that for any positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$\frac{a_{1}^{2}+\cdots+a_{n}^{2}}{n} \geqslant\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)^{2}+\frac{1}{2 n}\left(a_{... | \frac{1}{2 n} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,179 |
22. (Mediterranean) Given $n>2$, positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that:
$$\frac{a_{2} a_{3} \cdots a_{n}}{a_{1}+n-2}+\frac{a_{1} a_{3} \cdots a_{n}}{a_{2}+n-2}+\cdots+\frac{a_{1} a_{2} \cdots a_{n-1}}{a_{n}+n-2} \leqslant \frac{1}{(n-1)^{2}}$$
and find the... | Sure, here is the translation:
```
Solution. We first prove that the inequality holds for $n=3$, i.e.,
$$\frac{a_{2} a_{3}}{a_{1}+1}+\frac{a_{3} a_{1}}{a_{2}+1}+\frac{a_{1} a_{2}}{a_{3}+1} \leqslant \frac{1}{4}$$
Given the condition $a_{1}+a_{2}+a_{3}=1$, the above inequality is equivalent to
$$\frac{a_{2} a_{3}}{2 a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,180 |
23. (Singapore) Let $a_{1} \geqslant 1$, and for any $k=1,2, \cdots, n$, $a_{k+1} \geqslant a_{k}+1$. Prove:
$$a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} \geqslant\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}$$ | Proof. We use mathematical induction to prove this problem.
When $n=1$, by $a_{1} \geqslant 1$, we know that the inequality holds at this time.
Assume that when $n=k$, the inequality also holds, that is,
$$a_{1}^{3}+a_{2}^{3}+\cdots+a_{k}^{3} \geqslant\left(a_{1}+a_{2}+\cdots+a_{k}\right)^{2}$$
Then, when $n=k+1$, by ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,181 |
24. (Northern) Let positive real numbers $a, b, c$ satisfy $(a+2 b)(b+2 c)=9$. Prove:
$$\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant 3$$ | Proof. By the Power Mean Inequality, we have
$$\sqrt{\frac{a^{2}+b^{2}}{2}} \geqslant \frac{a+b}{2}, \quad \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant \frac{b+c}{2}$$
Therefore,
$$\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant \frac{a+b}{2}+2 \frac{b+c}{2}$$
Since $\frac{a+b}{2}+2 \frac{b+c}{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,182 |
25. (Northern) Let $x, y, z \in [0,1], |x-y| \leqslant \frac{1}{2}, |z-y| \leqslant \frac{1}{2}, |x-z| \leqslant \frac{1}{2}$, find
$$W(x, y, z)=x+y+z-xy-yz-zx$$
the maximum and minimum values. | Given that $x, y, z \in [0,1]$, we have $W = x(1-y) + y(1-z) + z(1-x) \geqslant 0$. By setting $x = y = z = 0$, we get $W = 0$, hence $W_{\min} = 0$.
Next, we find $W_{\max}$. Notice that $W(x, y, z) = W(1-x, 1-y, 1-z)$, so we can assume that at least two of $x, y, z$ are not less than $\frac{1}{2}$. Therefore, withou... | \frac{5}{6} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,183 |
26. (Southeast) Let $n$ be a positive integer, and real numbers $a_{1}, a_{2}, \cdots, a_{n}, r_{1}, r_{2}, \cdots, r_{n}$ satisfy:
$$a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \quad \text { and } \quad 0 \leqslant r_{1} \leqslant r_{2} \leqslant \cdots \leqslant r_{n} \text {, }$$
Prove:
$$\sum_{i=1}^{n} ... | Notice that when $0 \leqslant r_{1} \leqslant r_{2} \leqslant \cdots \leqslant r_{n}$, the identity holds:
$$\begin{aligned}
& \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j} \min \left(r_{i}, r_{j}\right) \\
= & \left(r_{n}-r_{n-1}\right) a_{n}^{2}+\left(r_{n-1}-r_{n-2}\right)\left(a_{n}+a_{n-1}\right)^{2}+\cdots \\
& +\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,184 |
27. (Female) Let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{i=1}^{n} x_{i}^{2}=1, n \geqslant 2$, and let $S=\sum_{i=1}^{n} i x_{i}^{2}$. Prove:
$$\sum_{k=1}^{n}\left(1-\frac{k}{S}\right)^{2} \frac{x_{k}^{2}}{k} \leqslant\left(\frac{n-1}{n+1}\right)^{2} \sum_{k=1}^{n} \frac{x_{k}^{2}}{k}$$
and find the ... | Proof. First, perform an identity transformation on the left side of the inequality:
$$\begin{aligned}
\sum_{k=1}^{n}\left(1-\frac{k}{S}\right)^{2} \frac{x_{k}^{2}}{k} & =\sum_{k=1}^{n} \frac{x_{k}^{2}}{k}-2 \frac{1}{S} \sum_{k=1}^{n} x_{k}^{2}+\frac{1}{S^{2}} \sum_{k=1}^{n} k x_{k}^{2} \\
& =\sum_{k=1}^{n} \frac{x_{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,185 |
28. (Western Mathematical Olympiad) Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ satisfy the following conditions:
$$\sum_{i=1}^{n}\left(a_{i}+b_{i}\right)=1 ; \quad \sum_{i=1}^{n} i\left(a_{i}-b_{i}\right)=0 ; \quad \sum_{i=1}^{n} i^{2}\left(a_{i}+b_{i}\right)=10$$
Prove th... | Proof. By utilizing the known conditions, we have
$$\left(k a_{k}\right)^{2}=\left(\sum_{i \neq k} i\left(b_{i}-a_{i}\right)+k b_{k}\right)^{2}=\left(\sum_{i \neq k} i b_{i}-\sum_{i \neq k} i a_{i}+k b_{k}\right)^{2}$$
By the Cauchy inequality, we get
$$\left(\sum_{i \neq k} i b_{i}-\sum_{i \neq k} i a_{i}+k b_{k}\rig... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,186 |
29. (League) Given a positive integer $n>2$, let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{k} \leqslant 1, k=1,2, \cdots, n$. Define
$$A_{k}=\frac{a_{1}+\cdots+a_{k}}{k}, \quad k=1,2, \cdots, n$$
Prove:
$$\left|\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} A_{i}\right|<\frac{n-1}{2}$$ | Proof. When $00$, $|x-y|<\max \{x, y\}$, thus for any $1 \leqslant k \leqslant n-1$, we have
$$\begin{aligned}
\left|A_{n}-A_{k}\right| & =\left|\left(\frac{1}{n}-\frac{1}{k}\right) \sum_{i=1}^{k} a_{i}+\frac{1}{n} \sum_{i=k+1}^{n} a_{i}\right| \\
& =\left|\frac{1}{n} \sum_{i=k+1}^{n} a_{i}-\left(\frac{1}{k}-\frac{1}{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,187 |
30. (Training Team) Given a positive integer $n \geqslant 2$ and a positive real number $a$, positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\prod_{i=1}^{n} x_{i}=1$. Find the smallest real number $M=M(n, a)$, such that
$$\sum_{i=1}^{n} \frac{1}{a+S-x_{i}} \leqslant M$$
always holds, where $S=x_{1}+x_{2}+... | Let $x_{i}=y_{i}^{n}, y_{i}>0$, then $y_{1} y_{2} \cdots y_{n}=1, i=1,2, \cdots, n$. By the power mean and the arithmetic mean inequality, we have
$$S-x_{i}=\sum_{j \neq i} y_{j}^{n} \geqslant(n-1)\left(\frac{\sum_{j \neq i} y_{j}}{n-1}\right)^{n} \geqslant(n-1)\left(\frac{\sum_{j \neq i} y_{j}}{n-1}\right) \prod_{j \n... | \left\{\begin{array}{ll}
\frac{n}{a-1+n} & \text { if } a \geqslant 1 , \\
\frac{1}{a} & \text { if } 0<a<1
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,188 |
4. (India) Let $P$ be the Brocard point of $\triangle A B C$, prove that:
$$\left(\frac{A P}{B C}\right)^{2}+\left(\frac{B P}{C A}\right)^{2}+\left(\frac{C P}{A B}\right)^{2} \geqslant 1$$ | Proof. We prove a stronger proposition, for any point $P$ in the plane of $\triangle ABC$, we have
$$\frac{A P}{B C} \cdot \frac{B P}{C A}+\frac{B P}{C A} \cdot \frac{C P}{A B}+\frac{C P}{A B} \cdot \frac{A P}{B C} \geqslant 1$$
To this end, we introduce the complex plane, and let $A, B, C, P$ correspond to points $z_... | proof | Geometry | proof | Yes | Yes | inequalities | false | 731,189 |
5. (Iran) Given $a, b, c > 0$, prove:
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}} \geqslant \frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^{2}$$ | Proof. First, by the Cauchy inequality, we have
$$\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}}\right)(9+9+9+1) \geqslant\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}+\frac{1}{a+b+c}\right)^{2}$$
Therefore, it suffices to prove
$$\frac{1}{28}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,190 |
6. (Iran) Given $x, y, z > 0$, and $x y + y z + z x = 1$. Prove:
$$3 - \sqrt{3} + \frac{x^{2}}{y} + \frac{y^{2}}{z} + \frac{z^{2}}{x} \geqslant (x + y + z)^{2}$$ | Proof. By the mean inequality $(x+y+z)^{2} \geqslant 3(x y+y z+z x)=3$, so $x+y+z \geqslant \sqrt{3}$. Therefore, we only need to prove
$$\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}-x-y-z \geqslant(x+y+z)^{2}-3(x y+y z+z x)$$
Rearranging and completing the square, we get
$$\frac{(x-y)^{2}}{y}+\frac{(y-z)^{2}}{z}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,191 |
7. (Bulgaria) Let positive real numbers $a, b, c$ satisfy $a+b+c=3$, prove that:
$$a b c\left(a^{2}+b^{2}+c^{2}\right) \leqslant 3 .$$ | Proof. By the AM-GM inequality, we have
$$\begin{aligned}
(a+b+c)^{2} & =\left(a^{2}+b^{2}+c^{2}\right)+2(a b+b c+c a) \\
& \geqslant 3 \sqrt[3]{\left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a)^{2}} \\
& \geqslant 3 \sqrt[3]{\left(a^{2}+b^{2}+c^{2}\right) \cdot 3 a b c(a+b+c)}
\end{aligned}$$
Using the condition $a+b+c=3$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,192 |
8. (Korea) Given any triangle $A B C$, let $P, Q, R$ be the points where the incircle of triangle $A B C$ touches the sides $B C, C A, A B$ respectively. Let $T$ be the area of triangle $A B C$ and $L$ be its perimeter. Prove that:
$$\left(\frac{A B}{P Q}\right)^{3}+\left(\frac{B C}{Q R}\right)^{3}+\left(\frac{C A}{R P... | Proof. The original inequality is transformed into
$$\sum\left(\sin ^{3} \frac{A}{2} \sin ^{3} \frac{B}{2}\right)^{-1} \geqslant \frac{32}{\sqrt{3}} \frac{(\sin A+\sin B+\sin C)^{2}}{\sin A \sin B \sin C}$$
By the AM-GM inequality, we have
$$\sum\left(\sin ^{3} \frac{A}{2} \sin ^{3} \frac{B}{2}\right)^{-1} \geqslant 3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,193 |
9. (Portugal) Prove: For any triangle, there exist two sides of lengths $a, b$ such that
$$\frac{\sqrt{5}-1}{2}<\frac{a}{b}<\frac{\sqrt{5}+1}{2}$$ | Proof. Without loss of generality, let the three sides be $x, y, z$ satisfying $x \leqslant y \leqslant z$, and set $k=\min \left(\frac{y}{x}, \frac{z}{y}\right)$. Clearly, $k \geqslant 1$, so it suffices to prove that $k<\frac{\sqrt{5}+1}{2}$. By assumption, we have $y \geqslant k x, z \geqslant k y$; and in the trian... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,194 |
10. (Vietnam) Given $a, b, c > 0, 16(a + b + c) \geqslant \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. Prove:
$$\sum_{c y c}\left(\frac{1}{a + b + \sqrt{2a + 2c}}\right)^{3} \leqslant \frac{8}{9} .$$ | Proof. By the AM-GM inequality,
$$a+b+\sqrt{2 a+2 c}=(a+b)+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}} \geqslant 3 \sqrt[3]{\frac{(a+b)(a+c)}{2}}$$
Thus,
$$\left(\frac{1}{a+b+\sqrt{2 a+2 c}}\right)^{3} \leqslant \frac{2}{27(a+b)(a+c)}$$
Similarly, we have
$$\left(\frac{1}{b+c+\sqrt{2 b+2 a}}\right)^{3} \leqslant \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,195 |
Theorem 1 Let $a, b, c$ and $r$ be the lengths of the sides and the inradius of a triangle, respectively. Prove that
$$a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geqslant 8 r^{2}(a-b)^{2} .$$ | Prove that for $a=y+z, b=z+x, c=x+y, x, y, z \in \mathbf{R}_{+}$, we have $r^{2}=\frac{x y z}{x+y+z}$. Thus, inequality (2) is equivalent to
$$x y^{3}+y z^{3}+z x^{3}-x y z(x+y+z) \geqslant \frac{4 x y z(x-y)^{2}}{x+y+z}$$
That is, for $x, y, z \in \mathbf{R}_{+}$, we have
$$\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,196 |
Theorem 2 Let $a, b, c$ and $p$ be the lengths of the three sides and the semi-perimeter of a triangle, respectively. Then the following identity holds:
$$\begin{array}{c}
a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \\
=2(p-c)(p-a)(a-b)^{2}+2(p-a)(p-b) \cdot \\
(b-c)^{2}+2(p-b)(p-c)(c-a)^{2} .
\end{array}$$ | $$\begin{array}{l}
\text { Prove that, note the algebraic identity (4) } \\
\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}=x+y+z+\frac{(x-y)^{2}}{y}+\frac{(y-z)^{2}}{z} \\
+\frac{(z-x)^{2}}{x}
\end{array}$$
can be transformed into
$$\begin{array}{l}
x y^{3}+y z^{3}+z x^{3}=x y z(x+y+z)+z x(x- \\
y)^{2}+x y(y-z)^{2}+y... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,197 |
Theorem 2 Let $A_{1}\left(x_{1}, y_{1}\right), A_{2}\left(x_{2}, y_{2}\right), \cdots$, $A^{n}\left(x^{n}, y_{n}\right)$ be $n(n \geqslant 3)$ points in a Cartesian coordinate system, and $A_{1} A_{2} \ldots A_{n}$ form a convex $n$-sided polygon with $A_{1}, A_{2}, \ldots, A_{n}$ arranged in a counterclockwise directi... | Prove for $n$ using mathematical induction.
By Theorem 1, the proposition holds when $n=3$.
Assume $n>3$ and the proposition holds for $n-1$, then $S_{A_{1} \cdot A_{n}}=$ $S_{A_{1} A_{2} A_{3}}+S_{A_{1} A_{3} \cdots A_{n}}$.
By Theorem 1
$$S_{A_{1} A_{2} A_{3}}=\frac{1}{2}\left|\begin{array}{llll}
x_{1} & x_{2} & x_{... | proof | Geometry | proof | Yes | Yes | inequalities | false | 731,198 |
Example 1 Given three points $A(2,1), B(3,2), C(1,4)$, find
$$S_{\triangle A B C}$$ | $\begin{array}{l} \text { Solution: By Theorem } 1 \\ S_{\triangle A B C}=\frac{1}{2}\left|\begin{array}{llll}2 & 3 & 1 & 2 \\ 1 & 2 & 4 & 1\end{array}\right|=\frac{1}{2}(2 \times 2+3 \times \\ 4+1 \times 1-3 \times 1-1 \times 2-2 \times 4)=2 .\end{array}$ | 2 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 731,199 |
Example 2 Given points $A_{1}(-1,0), A_{2}(1, -1), A_{3}(2,0), A_{4}(3,3), A_{5}(0,5)$ on a plane, find the area of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$. | Solve: Since points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are arranged counterclockwise, by Theorem 2, the area of pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ is
$$\begin{array}{l}
S=\frac{1}{2}\left|\begin{array}{rrrrrr}
1 & 1 & 2 & 3 & 0 & -1 \\
0 & -1 & 0 & 3 & 5 & 0
\end{array}\right| \\
\quad=\frac{1}{2}[(-1) \times(-1... | 14.5 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 731,200 |
Example 1 Let $n$ be a positive integer greater than 1, and $a_{1}, a_{2}$, $\cdots, a_{n}$ be $n$ distinct integers. Prove that the polynomial
$$f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1$$
cannot be divisible by any positive degree and less than $n$ monic integer coefficient polynomia... | Prove: It is only necessary to prove that the polynomial $f(x)$ cannot be decomposed into the product of two polynomials with integer coefficients and degrees greater than zero.
Assume $f(x)$ has a proper factor $g(x) \in \mathbf{Z}[x]$ (the set of all integer-coefficient polynomials with the leading coefficient 1), t... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,201 |
Theorem 1 Polynomial
$$\begin{aligned}
F(x)= & x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots \cdots \\
& \left(x-a_{n-1}\right)+1
\end{aligned}$$
(where $0<a_{1}<a_{2}<\cdots<a_{n-1}$ are integers), is reducible only in the following two cases:
$$\begin{array}{l}
n=4, a_{1}=1, a_{2}=2, a_{3}=3 ; \\
n=2, a_{1}=2
\end... | Proof: Suppose $\varphi(x)$ and $\psi(x)$ are integer-coefficient polynomials with degrees no greater than $n-1$, and
$$\begin{array}{l}
F(x)=x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n-1}\right)+1 \\
=\varphi(x) \psi(x) .
\end{array}$$
Using the proof method from Example 1, we have
$$\varphi(x)=\psi(... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,202 |
Theorem 2 Polynomial
$$\left(x-a_{1}\right)^{2}\left(x-a_{2}\right)^{2} \cdots\left(x-a_{n}\right)^{2}+1$$
is irreducible. | Prove: If $\varphi(x)$ and $\psi(x)$ are integer-coefficient polynomials of degrees $k$ and $l$ respectively, and the leading coefficient of $\varphi(x)$ is positive. Also,
$$\begin{array}{l}
\left(x-a_{1}\right)^{2}\left(x-a_{2}\right)^{2} \cdots\left(x-a_{n}\right)^{2}+1 \\
=\varphi(x) \psi(x)
\end{array}$$
Then, fo... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,203 |
J.Wolstenholme-embedded inequality: For $\triangle A B C$ and any real numbers $x, y, z$, we have $x^{2}+y^{2}+z^{2} \geq 2 y z \cos A+2 z x \cos B+2 x y \cos C$, (1) where equality holds if and only if $x: y: z=\sin A: \sin B: \sin C$. | Proof: In fact, since in $\triangle ABC$, $\cos A = -\cos (B+C)$, we have
$$\begin{array}{l}
x^{2}+y^{2}+z^{2}-(2 y z \cos A + 2 z x \cos B + 2 x y \cos C) \\
=(x - y \cos C - z \cos B)^{2} + (y \sin C)^{2} + (z \sin B)^{2} - 2 y z \cos B \cos C + 2 y z \cos A \\
=(x - y \cos C - z \cos B)^{2} + (y \sin C)^{2} + (z \si... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,206 |
Example 1 (MOSP2000) Let $\triangle A B C$ be an acute triangle.
Prove:
$$\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2}+8 \cos A \cos B \cos C \geq 4$$ | Prove that the original problem is equivalent to
$$\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2} \geq 4-8 \cos A \cos B \cos C$$
Because
$$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cdot \cos B \cdot \cos C=1$$
So the original problem is equiva... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,207 |
Example $2(2007$ National Team Training Test Question) Let positive numbers $u, v, w$ satisfy $u+v+w+\sqrt{uvw}=4$.
Prove: $\sqrt{\frac{vw}{u}}+\sqrt{\frac{uw}{v}}+\sqrt{\frac{uv}{w}} \geq u+v+w$.
Lemma: The equation $x^{2}+y^{2}+z^{2}+xyz=4$ has positive real solutions $x, y, z$ if and only if there exists an acute $\... | Prove: Since
$$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cdot \cos B \cdot \cos C=1$$
Therefore,
$$(2 \cos A)^{2}+(2 \cos B)^{2}+(2 \cos C)^{2}+(2 \cos A) \cdot(2 \cos B) \cdot(2 \cos C)=4$$
Thus, the triplet $(2 \cos A, 2 \cos B, 2 \cos C)$ is a solution to the equation $x^{2}+y^{2}+z^{2}+x y z=4$.
Conversely, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,208 |
Example 3 (1998 Indian Mathematics Competition) Given that $x, y, z$ are positive real numbers, satisfying $x y + y z + z x + x y z = 4$. | To prove: $x+y+z \geq x y+y z+z x$. (See [7])
Analysis: Observing the condition $x y+y z+z x+x y z=4$, it naturally leads to the lemma in Example 2. Therefore, we can transform the condition to match the form of the lemma, laying the groundwork for applying J.Wolstenholme's embedding inequality. Observing the inequalit... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,209 |
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