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1. $F$ is a point on the axis of symmetry of a non-degenerate quadratic curve $G$, $H$ is the intersection of the polar line $l$ of $F$ with respect to curve $G$ and the axis of symmetry (i.e., $F$ and $H$ are a pair of conjugate poles on the axis of symmetry of curve $G$), $AB$ is the chord of curve $G$ intercepted by... | If $A B$ intersects with $l$, let the intersection point be $K$.
By the definition of the polar line, we have $\frac{A K}{K B}=\frac{A F}{F B}$.
Also, $\angle F G K=90^{\circ}$, and by the Apollonius circle theorem, we get
$$
\frac{A H}{H B}=\frac{A K}{K B}=\frac{A F}{F B} \text {. }
$$
Thus, $H F$ and $H K$ bisect $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,384 |
3. $P$ is a fixed point inside the non-degenerate conic section $G$ and on the axis of symmetry of $G$. Line $l$ is the polar of $P$ with respect to the curve $G$. Through $P$, draw chords $B F$ and $A D$ of the curve $G$, and let $A F$ intersect $l$ at point $E$, and $A B$ intersect $l$ at point $C$. Prove:
(1) $C, D,... | (1) As shown in Figure 8, let $F D$ intersect $A B$ at point $C^{\prime}$, $B D$ intersect $A F$ at point $E^{\prime}$, and $A P$ intersect $C^{\prime} E^{\prime}$ at point $P^{\prime}$.
By Proposition 2, $C^{\prime} E^{\prime}$ is the polar line of point $P$ with respect to curve $G$. Therefore, $C^{\prime}$ is the i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,385 |
4. In the Cartesian coordinate system $x O y$, draw any line through point $C(0, c)$ on the positive direction of the $y$-axis, intersecting the parabola $y=x^{2}$ at points $A$ and $B$. A line perpendicular to the $x$-axis intersects the line segment $A B$ and the line $l: y=-c$ at points $P$ and $Q$, respectively.
(1... | $l$ is the polar line of $P$. It is easy to prove: the intersection point $Q_{1}$ of the tangents at $A$ and $B$ lies on $l$, and $Q_{1} P \perp x$-axis, so $Q_{1}$ is $Q$. Therefore, (2) and (3) are solved simultaneously. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,386 |
5. Given the parabola $x^{2}=4 y$ with focus $F$, and $A, B$ are two moving points on the parabola, and $\overrightarrow{A F}=\lambda \overrightarrow{F B}(\lambda>0)$. Tangents to the parabola are drawn through points $A$ and $B$, and their intersection point is $M$. Prove: $\overrightarrow{F M} \cdot \overrightarrow{A... | The moving chord $A B$ is the chord of contact of $M$, and $A B$ always passes through the focus $F$, so point $M$ always lies on the polar line of the focus $F$, which is the directrix $y=-1$.
Similarly, from Corollary 1, we can get $M F \perp A B$, i.e., $\overrightarrow{F M} \cdot \overrightarrow{A B}=0$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,387 |
Example 1 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$. Find the extremum of $S=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$. | By symmetry, we can assume
$$
a \leqslant b \leqslant c\left(0 \leqslant a \leqslant \frac{1}{3} \leqslant c \leqslant 1\right) \text {. }
$$
First, arbitrarily set \( a = a_{0} \). Then \( c + b = 1 - a_{0} \).
Let \( x = c - b \geqslant 0 \). Then
$$
c = \frac{1 - a_{0} + x}{2}, \quad b = \frac{1 - a_{0} - x}{2} .
$... | \frac{9}{4} \leqslant S \leqslant \frac{5}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,388 |
Example 2 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$. Find the
$$
S=\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}
$$
extreme values. | Let $a \leqslant b \leqslant c$, for any given $a_{0}\left(0 \leqslant a_{0} \leqslant \frac{1}{3}\right)$.
Then $b+c=1-a_{0}$.
Let $x=c-b$. Then
$$
c=\frac{1-a_{0}+x}{2}, b=\frac{1-a_{0}-x}{2} .
$$
Substitute into $S$ to get
$$
S=\sqrt{4 a_{0}+1}+\sqrt{3-2 a_{0}-2 x}+\sqrt{3-2 a_{0}+2 x} \text {. }
$$
Differentiate ... | 2+\sqrt{5} \leqslant S \leqslant \sqrt{21} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,389 |
Example 3 Let $a, b, c$ be positive real numbers, and $a+b+c=3$. Prove:
$$
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant a^{2}+b^{2}+c^{2} \text {. }
$$
(2006, Romanian National Training Team Problem) | Let $S=\frac{1}{a^{2}}-a^{2}+\frac{1}{b^{2}}-b^{2}+\frac{1}{c^{2}}-c^{2}$. If we can find that the minimum value of $S$ is 0, then the conclusion is proven.
Let $a \leqslant b \leqslant c$, and for any given $c=c_{0}\left(1 \leqslant c_{0} \leqslant 3\right)$. Then $b+a=3-c_{0}$.
Let $x=b-a \geqslant 0$. Then
$b=\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,390 |
Example 1 Given that $a, b, c$ are three real numbers, satisfying $a+b+c>0, ab+bc+ca>0, abc>0$. Prove: $a>0, b>0, c>0$.
| 【Analysis】Obviously, $a$, $b$, and $c$ are the three roots of the equation
$$
x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c=0
$$
Since the constant term $-a b c \neq 0$, $x=0$ is not a root of the equation.
Assume the equation has a negative real root, i.e., $x<0$, while $a>0, b>0, c>0$.
$$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,391 |
Example 2 Given that the volume of a rectangular prism is 1, the sum of its length, width, and height is $k$, and its surface area is $2k$. Find the range of the real number $k$.
| 【Analysis】Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$ respectively. Then
$$
a b c=1, a+b+c=k, a b+b c+c a=k .
$$
According to Vieta's formulas, $a$, $b$, and $c$ are the three roots of the equation
$$
x^{3}-k x^{2}+k x-1=0
$$
We have
$$
\begin{array}{l}
\text { Also, } x^{3}-k x^{2... | [3,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,392 |
Example 4 Given that the three sides of $\triangle A B C$ are $a$, $b$, and $c$, and they satisfy
$$
a b c=2(a-1)(b-1)(c-1) .
$$
Does there exist a $\triangle A B C$ with all sides being integers? If it exists, find the lengths of the three sides; if not, explain the reason. | Let's assume the integers $a \geqslant b \geqslant c$. Clearly, $c \geqslant 2$.
If $c \geqslant 5$, then, $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5}$.
From $a b c=2(a-1)(b-1)(c-1)$, we get
$$
\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) ... | 4,5,6 \text{ or } 3,7,8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,393 |
Example 3 Given that $a, b, c$ are three distinct real numbers. Try to solve the system of equations about $x, y, z$
$$
\left\{\begin{array}{l}
\frac{x}{a^{3}}-\frac{y}{a^{2}}+\frac{z}{a}=1, \\
\frac{x}{b^{3}}-\frac{y}{b^{2}}+\frac{z}{b}=1, \\
\frac{x}{c^{3}}-\frac{y}{c^{2}}+\frac{z}{c}=1 .
\end{array}\right.
$$ | 【Analysis】This is a classic problem involving the application of Vieta's formulas to a univariate cubic equation. Solving the system of equations directly would involve a large amount of computation. Observing the characteristics of the three equations, we know that $a$, $b$, and $c$ are the roots of the equation $\fra... | x=abc, y=ab+bc+ca, z=a+b+c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,394 |
Example 4 Evaluate:
(1) $\cos 40^{\circ}+\cos 80^{\circ}+\cos 160^{\circ}$;
(2) $\cos 40^{\circ} \cdot \cos 80^{\circ}+\cos 80^{\circ} \cdot \cos 160^{\circ}+$ $\cos 160^{\circ} \cdot \cos 40^{\circ}$;
(3) $\cos 40^{\circ} \cdot \cos 80^{\circ} \cdot \cos 160^{\circ}$. | 【Analysis】If we look at a single equation, it is quite difficult to find the value, and it requires a high level of proficiency in the application of trigonometric formulas. However, when considering the three equations together, their structure perfectly meets the requirements of Vieta's formulas, allowing us to const... | 0, -\frac{3}{4}, -\frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,395 |
1. The median and mode of the data set $4,5,6,7,7,8$ are ( ).
(A) 7,7
(B) $7,6.5$
(C) $5.5,7$
(D) $6.5,7$ | 1. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Other | MCQ | Yes | Yes | cn_contest | false | 723,396 |
2. If the polynomial $P=a^{2}+4 a+2014$, then the minimum value of $P$ is ( ).
(A) 2010
(B) 2011
(C) 2012
(D) 2013 | 2. A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,397 |
3. If $100 x^{2}-k x y+49 y^{2}$ is a perfect square, then the value of $k$ is ().
(A) $\pm 4900$
(B) $\pm 9800$
(C) $\pm 140$
(D) $\pm 70$ | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,398 |
4. Place a three-digit number $m$ in front of a two-digit number $n$ to form a five-digit number. It can be represented as ( ).
(A) $m n$
(B) $m+n$
(C) $10 m+n$
(D) $100 m+n$ | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,399 |
5. If $-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}$, then
$$
\sqrt{4 x^{2}+4 x+1}+\sqrt{4 x^{2}-4 x+1}=(\quad) \text {. }
$$
(A) $4 x$
(B) 2
(C) -2
(D) $2-4 x$ | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,400 |
6. Among the following real numbers, the largest one is ( ).
(A) $5 \times \sqrt{0.039}$
(B) $\frac{3.141}{\pi}$
(C) $\frac{7}{\sqrt{14}+\sqrt{7}}$
(D) $\sqrt{0.3}+\sqrt{0.2}$ | 6. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,401 |
7. Define the operation symbol “ * " to mean $a * b=\frac{a+b}{a b}$ (where $a$ and $b$ are not 0).
Given the following two conclusions:
(1) The operation “ * " satisfies the commutative law;
(2) The operation “ * " satisfies the associative law.
Which of the following is correct? ( ).
(A)(1)
(B)(2)
(C)(1)(2)
(D) None... | 7. $\mathrm{A}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,402 |
Example 5 Let the three sides of a right-angled triangle be $a$, $b$, and $c$, all positive integers, and the hypotenuse $c$ satisfies $87 \leqslant c \leqslant 91$. Find the lengths of the three sides of such a right-angled triangle. | 【Analysis】According to the representation of Pythagorean triples
$$
\left(m^{2}-n^{2}, 2 m n, m^{2}+n^{2}\right) \text {, }
$$
we can determine the values of $m$ and $n$.
Solution Since Pythagorean triples have the form
$$
\left(m^{2}-n^{2}, 2 m n, m^{2}+n^{2}\right) \text {, }
$$
we set the hypotenuse length as
$$
k... | (63,60,87),(39,80,89),(54,72,90),(35,84,91) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,404 |
18. As shown in Figure 8, in the right triangle $\triangle ABC$, $\angle ABC = 90^{\circ}$, $AB = 8$, $BC = 6$. Circles are drawn with $A$ and $C$ as centers and $\frac{AC}{2}$ as the radius, cutting out two sectors from the right triangle $\triangle ABC$. The area of the remaining (shaded) part is ( ).
(A) $24 - \frac... | 18. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,414 |
1. In $\triangle A B C$, it is known that $A B=A C, \angle C$'s bisector $C D$ intersects $A B$ at point $D, B D, B C, C D$ are three consecutive integers. Find the perimeter of $\triangle A B C$. | Let $A B=A C=b, B C=a$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
B D=\frac{a b}{a+b}, \\
B C-B D=a-\frac{a b}{a+b}=\frac{a^{2}}{a+b} \\
=k \in\{1,2\} .
\end{array}
$$
Following Example 1, we get
$$
C D^{2}=B D(B D+B C)=\frac{a^{2} b(a+2 b)}{(a+b)^{2}} \text {. }
$$
If $k=1$, then $b=a^{2}-a$.
Thus, ... | 45 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,415 |
19. Arrange the numbers $1,2,3,4,5$ in a row such that the last number is odd, and the sum of any three consecutive numbers is divisible by the first of these three numbers. How many arrangements satisfy these conditions?
(A) 2
(B) 3
(C.) 4
(D) 5 | 19. D | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,416 |
20. If the lengths of two sides of $\triangle A B C$ are $a$ and $b$, then the area of $\triangle A B C$ cannot be equal to ( ).
(A) $\frac{1}{4}\left(a^{2}+b^{2}\right)$
(B) $\frac{1}{2}\left(a^{2}+b^{2}\right)$
(C) $\frac{1}{8}(a+b)^{2}$
(D) $\frac{1}{4} a b$ | 20. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,417 |
27. Let positive real numbers $x, y$ satisfy $x y=1$. Then the minimum value of $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{5}{8}$
(C) 1
(D) $\sqrt{2}$ | 27. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The note above is a clarification and should not be included in the final translation. Here is the direct translation:
27. C | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,424 |
28. Let $a, b$ be real numbers, and $\frac{1}{1+a}-\frac{1}{1+b}=\frac{1}{b-a}$. Then $\frac{1+b}{1+a}=(\quad$.
(A) $\frac{1 \pm \sqrt{5}}{2}$
(B) $\pm \frac{1+\sqrt{5}}{2}$
(C) $\pm \frac{3-\sqrt{5}}{2}$
(D) $\frac{3 \pm \sqrt{5}}{2}$ | 28. D | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,425 |
3. Given a right-angled triangle with the lengths of the two legs being $l$ and $m$, and the hypotenuse being $n$, where $l$, $m$, and $n$ are positive integers, and $l$ is a prime number. Prove that $2(l+m+1)$ is a perfect square. | By the Pythagorean theorem, we have
$$
l^{2}=n^{2}-m^{2}=(n-m)(n+m) .
$$
Since $l$ is a prime number, we have
$$
\begin{array}{l}
(n+m, n-m)=\left(l^{2}, 1\right) \\
\Rightarrow(n, m)=\left(\frac{l^{2}+1}{2}, \frac{l^{2}-1}{2}\right) .
\end{array}
$$
Therefore, $2(l+m+1)=(l+1)^{2}$. | (l+1)^2 | Number Theory | proof | Yes | Yes | cn_contest | false | 723,426 |
1. Assuming the Earth rotates once around the axis connecting the North Pole and the South Pole in 23 hours 56 minutes 4 seconds, and the Earth's equatorial radius is $6378.1 \mathrm{~km}$. Then, when you stand on the equator and rotate with the Earth, the linear velocity around the axis is $\qquad$ meters/second (roun... | 1. 465 | 465 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,429 |
3. In astronomy, "parsec" is commonly used as a unit of distance. If in a right triangle $\triangle ABC$,
$$
\angle ACB=90^{\circ}, CB=1.496 \times 10^{8} \text{ km},
$$
i.e., the length of side $CB$ is equal to the average distance between the Moon $(C)$ and the Earth $(B)$, then, when the size of $\angle BAC$ is 1 s... | $\begin{array}{l}\text { 3. } 3.086 \times \\ 10^{13}\end{array}$ | 3.086 \times 10^{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,431 |
7. Given that the line $y=x$ intersects the cosine curve $y=\cos x$ at point $A$. Then the distance from the origin $O$ to point $A$, $|O A|=$ (accurate to $10^{-4}$ ). | 7. 1.0452 | 1.0452 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,435 |
4. Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
(2010, National Junior High School Mathematics League, Tianjin Preliminary ... | By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Substituting into the given equation and simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\Rightarrow(a-6)(b-6)=18 .
\end{array}
$$
Solving for $(a, b, c)$, we get
$$
=(7,24,25),(8,15,17),(9,12,15) .
$$ | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,437 |
II. (20 points) As shown in Figure 1, it is known that $M$ is a moving point on the line where side $DC$ of the square $ABCD$ lies. Find the maximum value of $\frac{MA}{MB}$.
保留源文本的换行和格式，翻译结果如下：
```
II. (20 points) As shown in Figure 1, it is known that $M$ is a moving point on the line where side $DC$ of the square ... | Let's assume the side length of the square is 2, and establish a Cartesian coordinate system with $DC$ as the $x$-axis and the midpoint of $DC$ as the origin $O$.
Then $A(-1,2)$ and $B(1,2)$.
Let $M(x, 0)$.
When $M A$ reaches its maximum value, it is clear that $x>0$.
$$
\begin{array}{l}
\text { Also, }\left(\frac{M A}... | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,438 |
Three. (20 points) Given the parabola $C: y=\frac{1}{2} x^{2}$, $A_{1}\left(x_{1}, 0\right)$ and $A_{2}\left(x_{2}, 0\right)$ are two points on the $x$-axis $\left(x_{1}+x_{2} \neq 0, x_{1} x_{2} \neq 0\right)$. Perpendicular lines to the $x$-axis are drawn through points $A_{1}$ and $A_{2}$, intersecting the parabola ... | Three, from the given conditions, we have
$$
A_{1}^{\prime}\left(x_{1}, \frac{1}{2} x_{1}^{2}\right), A_{2}^{\prime}\left(x_{2}, \frac{1}{2} x_{2}^{2}\right) \text {. }
$$
Then \( k_{x / \dot{k} / 2}=\frac{\frac{1}{2}\left(x_{2}^{2}-x_{1}^{2}\right)}{x_{2}-x_{1}}=\frac{1}{2}\left(x_{2}+x_{1}\right) \). Therefore, \( l... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,439 |
Four. (20 points) Let
$a_{1}=1, a_{n+1}=\sqrt{a_{n}+n^{2}} \quad(n=1,2, \cdots)$.
(1) Prove: $\left[a_{n}\right]=n-1 \quad(n=2,3, \cdots)$;
(2) Find the sum: $\left[a_{1}^{2}\right]+\left[a_{2}^{2}\right]+\cdots+\left[a_{n}^{2}\right]$.
where, $[x]$ denotes the greatest integer not exceeding the real number $x$. | (1) For integer $n(n \geqslant 2)$, we have
$$
a_{n}=\sqrt{a_{n-1}+(n-1)^{2}}>n-1 \text {. }
$$
We will prove: $a_{n}<n(n=2,3, \cdots)$.
When $n=2$, $a_{2}=\sqrt{a_{1}+1}=\sqrt{2}<2$.
When $n \geqslant 3$, if $a_{n-1}<n-1$, then
$$
\begin{aligned}
a_{n} & =\sqrt{a_{n-1}+(n-1)^{2}}<\sqrt{(n-1)+(n-1)^{2}} \\
& =\sqrt{n(... | \frac{1}{3}\left(n^{3}-4 n+9\right) | Algebra | proof | Yes | Yes | cn_contest | false | 723,440 |
One, (20 points) On the Cartesian plane, it is known that the line $y=x+a(-1<a<1)$ intersects the parabola $y=1-x^{2}$ at points $A$ and $B$, and the coordinates of point $C$ are $(1,0)$. Question: For what value of $a$ is the area of $\triangle A B C$ maximized? Find the maximum area of $\triangle A B C$. | Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=1-x^{2}, \\ y=x+a\end{array}\right.$ eliminating $y$ gives
$$
x^{2}+x+a-1=0 \text {. }
$$
Thus, $x_{1}+x_{2}=-1, x_{1} x_{2}=a-1$.
Then $|A B|^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$
$$
\begin{array}{l}
... | \frac{3 \sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,441 |
II. (20 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n}=\frac{a_{n-1}^{2}}{a_{n-2}}(n=3,4, \cdots) \text {. }
$$
Suppose $a_{2} 、 a_{5}$ are positive integers, and $a_{5} \leqslant 2010$. Find all possible values of $a_{5}$. | From the given, $\frac{a_{n}}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}}$. Therefore, the sequence $\left\{a_{n}\right\}$ is a geometric sequence.
Let $a_{2}=x$. Then the common ratio of the sequence $\left\{a_{n}\right\}$ is $\frac{x}{2}$.
Thus, $a_{5}=2\left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{8}$.
Since $a_{5}$ is a positive ... | 2, 32, 162, 512, 1250 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,442 |
Three, (20 points) The numbers $1,2, \cdots, 666$ are written on a blackboard. In the first step, the first eight numbers: $1,2, \cdots, 8$, are erased, and the sum of these numbers, 36, is written after 666; in the second step, the next eight numbers: $9,10, \cdots, 16$, are erased, and the sum of these numbers, 100, ... | (1) Since each step reduces the numbers by seven, after $\frac{666-1}{7}=95$ steps, only one number remains.
(2) From $666-512=154$, it follows that after $\frac{154}{7}=22$ steps, there are 512 numbers left.
In 22 steps, a total of $22 \times 8=176$ numbers are crossed out, and their sum is
$1+2+\cdots+176=88 \times ... | 904020 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,443 |
1. Given the inequality about $x$
$$
\sqrt{x}+\sqrt{2-x} \geqslant k
$$
has real solutions. Then the range of the real number $k$ is ( ).
(A) $(0,2]$
(B) $(-\infty, 0]$
(C) $(-\infty, 0)$
(D) $(-\infty, 2]$ | -.1. D.
Let $y=\sqrt{x}+\sqrt{2-x}(0 \leqslant x \leqslant 2)$. Then $y^{2}=x+(2-x)+2 \sqrt{x(2-x)} \leqslant 4$. Therefore, $0<y \leqslant 2$, and when $x=1$, the equality holds. Hence, the range of real number $k$ is $(-\infty, 2]$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 723,444 |
2. The set of positive even numbers $\{2,4, \cdots\}$ is divided into groups in ascending order, with the $n$-th group containing $2 n-1$ numbers:
$$
\{2\},\{4,6,8\},\{10,12,14,16,18\}, \cdots \cdots .
$$
Question: In which group is 2010 located?
(A) 30
(B) 31
(C) 32
(D) 33 | 2. C.
Obviously, 2010 is the 1005th term of the sequence $a_{n}=2 n$. Suppose 2010 is in the $n$-th group. Then
$$
\begin{array}{l}
\sum_{i=1}^{n-1}(2 i-1)<1005 \leqslant \sum_{i=1}^{n}(2 i-1) \\
\Rightarrow(n-1)^{2}<1005 \leqslant n^{2} \\
\Rightarrow n=32 .
\end{array}
$$
Therefore, 2010 is in the 32nd group. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,445 |
3. In the tetrahedron $S-ABC$, the three pairs of opposite edges are equal, and are $5$, $4$, and $x$ respectively. Then the range of values for $x$ is ( ).
(A) $(2, \sqrt{41})$
(B) $(3,9)$
(C) $(3, \sqrt{41})$
(D) $(2,9)$ | 3. C.
A tetrahedron can be embedded in a rectangular prism, in which case the three pairs of opposite edges of the tetrahedron are the face diagonals of the rectangular prism. Therefore, 5, 4, and $x$ are the side lengths of an acute triangle.
Thus, $x \in\left(\sqrt{5^{2}-4^{2}}, \sqrt{5^{2}+4^{2}}\right)$, which mea... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,446 |
4. For any integer $n(n \geqslant 2)$, satisfying
$$
a^{n}=a+1, b^{2 n}=b+3 a
$$
the size relationship of positive numbers $a$ and $b$ is ( ).
(A) $a>b>1$
(B) $b>a>1$
(C) $a>1,01$ | 4. A.
From the problem, we know that $a>1, b>1$.
Also, $a^{2 n}-a=a^{2}+a+1>3 a=b^{2 n}-b$, so $a>b>1$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,447 |
Example 1 Consider a complete graph with $n$ vertices. The vertices and edges of this complete graph are colored according to the following rules:
(1) Two edges emanating from the same vertex have different colors;
(2) The color of a vertex is different from the colors of the edges emanating from it.
For each fixed $n... | 【Analysis】Since the same vertex can lead to $n-1$ edges of different colors, and the color of this vertex is different from the colors of these $n-1$ edges, thus, at least $n$ colors are needed.
Solution The minimum value is $n$.
Let the $n$ vertices be $v_{0}, v_{2}, \cdots, v_{n-1}$, and the $n$ colors be $C_{0}, C_{... | n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,448 |
5. The graph of the function $f(x)=x^{3}-3 x^{2}+3 x+1$ has a center of symmetry at ( ).
(A) $(-1,2)$
(B) $(1,2)$
(C) $(-1,-2)$
(D) $(1,-2)$ | 5. B.
Since $f(x)=(x-1)^{3}+2$, the center of symmetry of the function's graph is $(1,2)$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,449 |
6. From the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+a_{n}(n \geqslant 1)
$$
the terms that are divisible by 3 are extracted to form the sequence $\left\{b_{n}\right\}$. Then $b_{100}=(\quad)$.
(A) $a_{100}$
(B) $a_{200}$
(C) $a_{300}$
(D) $a_{400}$ | 6. D.
It is easy to know that $a_{4 k}(k \geqslant 1)$ is divisible by 3, hence the answer is D. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,450 |
7. Given the inverse function of $y=f(x+1)$ is
$$
\begin{array}{c}
y=f^{-1}(x+1) \text {, and } f(1)=4007 \text {. Then } \\
f(1998)=
\end{array}
$$ | II, 7.2010.
From $y=f^{-1}(x+1)$, we get $x+1=f(y)$, which means $x=f(y)-1$.
Thus, the inverse function of $y=f^{-1}(x+1)$ is $y=f(x)-1$. Therefore, $f(x+1)-f(x)=-1$.
Let $x=1,2, \cdots, 1997$, add up all the equations and simplify to get $f(1998)-f(1)=-1997$.
Hence, $f(1998)=2010$. | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,451 |
8. For a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with all edges of length 3, a line segment $M N$ of length 2 has one endpoint $M$ moving on $A A_{1}$ and the other endpoint $N$ moving on the base $A B C$. Then, the trajectory (surface) of the midpoint $P$ of $M N$ and the three faces of the regular triangul... | 8. $\frac{\pi}{9}$.
From the problem, we know that the distance from the midpoint $P$ of $M N$ to point $A$ is always 1. Therefore, the locus of point $P$ is the part of the sphere with $A$ as the center and 1 as the radius that lies within the triangular prism.
Thus, the volume of the enclosed geometric body is $\fra... | \frac{\pi}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,452 |
9. Given the figure
$$
\begin{array}{l}
C_{1}:(x+3)^{2}+y^{2}=4, \\
C_{2}: x^{2}+(y-5)^{2}=4,
\end{array}
$$
There are infinitely many pairs of perpendicular lines $l_{1} 、 l_{2}$ passing through a point $P$ in the plane, which intersect circles $C_{1}$ and $C_{2}$ respectively, and the lengths of the chords intercept... | 9. $P(1,1)$ and $P(-4,4)$.
Let $P(a, b)$, the equations of lines $l_{1}$ and $l_{2}$ are respectively
$$
y-b=k(x-a), y-b=-\frac{1}{k}(x-a),
$$
i.e., $k x-y-k a+b=0, x+k y-a-b k=0$.
It is easy to see that the distance from the center $C_{1}(-3,0)$ to $l_{1}$ is equal to the distance from the center $C_{2}(0,5)$ to $l_... | P(1,1) \text{ and } P(-4,4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,453 |
10. The sequence $\left\{a_{n}\right\}$ of $n$ terms, formed by the permutation of $1,2, \cdots, n$, satisfies: each term is greater than all the terms before it or less than all the terms before it. Then the number of sequences $\left\{a_{n}\right\}$ that satisfy this condition is $\qquad$. | 10. $2^{n-1}$.
Let the number we are looking for be $A_{n}$. Then $A_{1}=1$.
For $n>1$, if $n$ is in the $i$-th position, then the $n-i$ positions after it are completely determined, and can only be $n-i, n-i-1$, $\cdots, 2,1$. The $i-1$ positions before it, $n-i+1, n-i+2$, $\cdots, n-1$ have $A_{i-1}$ ways of arrange... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,454 |
11. Given the quadratic function
$$
y=a x^{2}+b x+c \geqslant 0(a<b) \text {. }
$$
Then the minimum value of $M=\frac{a+2 b+4 c}{b-a}$ is $\qquad$ | 11. 8.
From the conditions, it is easy to see that $a>0, b^{2}-4 a c \leqslant 0$.
Notice that
$$
\begin{array}{l}
M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\
\geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} .
\end{array}
$$
Let $t=\frac{b}{a}$. Then $t>1$. Thus,
$$
\begin{array}{l}
M \geqslant \frac{a^{2}+2... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,455 |
12. A movie theater sells tickets at 5 yuan each. There are 10 people, 5 of whom have 5-yuan bills, and the other 5 have 10-yuan bills. Assuming the ticket booth has no money at the start, these 10 people queue up to buy tickets in a random order. The probability that the ticket booth will not encounter a situation whe... | 12. $\frac{1}{6}$.
Consider the positions of 5 people holding 5-yuan bills in the queue, there are $\mathrm{C}_{10}^{\mathrm{s}}=252$ equally probable ways to queue up.
Let $p(m, n)$ denote the number of valid queue arrangements for $m$ people holding 5-yuan bills and $n$ people holding 10-yuan bills.
Then $p(m, 0)=1... | \frac{1}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,456 |
13. (10 points) Given $a, b \in [1,3], a+b=4$.
Prove:
$$
\sqrt{10} \leqslant \sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}<\frac{4 \sqrt{6}}{3} .
$$ | Three, 13. Given $a, b \in [1,3], a+b=4$, we have
$$
a b=a(4-a)=-(a-2)^{2}+4 \in [3,4] \text{. }
$$
Let $u=\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}$. Then
$$
\begin{aligned}
u^{2}=a & +\frac{1}{a}+b+\frac{1}{b}+2 \sqrt{\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)} \\
& =4+\frac{4}{a b}+2 \sqrt{a b+\frac{1}{a ... | \sqrt{10} \leqslant u < \frac{4 \sqrt{6}}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 723,457 |
14. (10 points) As shown in Figure 1, the quadrilateral pyramid $P-ABCD$, $PA \perp$ plane $ABCD$, and $PA=4$. The base $ABCD$ is a right trapezoid,
$$
\begin{array}{l}
\angle CDA = \angle BAD \\
= 90^{\circ}, AB = 2, CD \\
= 1, AD = \sqrt{2}, M, N
\end{array}
$$
are the midpoints of $PD$ and $PB$ respectively, and th... | 14. (1) Take the midpoint $E$ of $AP$, and connect $ED$. Then $ED \parallel CN$.
Take the midpoint of $EP$ as point $Q$, since $MQ \parallel ED$, hence $MQ \parallel CN$.
Therefore, points $M, N, C, Q$ are coplanar, and the intersection point $Q$ of plane $MCN$ with $AP$ is the quarter point of $AP$.
Thus, $PQ=1$.
(2)... | 1, \frac{\pi}{3}, \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,458 |
Example 2 An office is equipped with 2004 telephones, any two of which are connected by a wire of one of four colors, and it is known that all four colors are used. Is it necessarily possible to find some telephones such that the wires connecting them among themselves are of exactly three different colors? ?2]
(2004, R... | Consider a complete graph where the vertices represent telephones and the edges represent wires.
Consider the smallest set of vertices $N$ in the graph: the edges connecting them contain all four colors. If any vertex (denoted as $A$) is removed from $N$, the edges connecting the remaining vertices will not contain al... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,459 |
15. (10 points) Let $a_{1}=3, a_{n+1}=a_{n}^{2}+a_{n}-1$ $\left(n \in \mathbf{N}_{+}\right)$. Prove:
(1) For all $n, a_{n}=3(\bmod 4)$;
(2) When $m \neq n$, $\left(a_{m}, a_{n}\right)=1$ (i.e., $a_{m} 、 a_{n}$ are coprime). | 15. (1) When $n=1$, $a_{1} \equiv 3(\bmod 4)$.
Assume $a_{n} \equiv 3(\bmod 4)$. Then
$$
a_{n+1}=a^{2}+a-1 \equiv 3^{2}+3-1 \equiv 3(\bmod 4) \text {. }
$$
Therefore, for all $n$, $a_{n} \equiv 3(\bmod 4)$.
(2) From the recurrence relation, it is easy to get
$$
a_{n+1}+1=4 a_{n} a_{n-1} \cdots a_{1} \text {. }
$$
As... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,460 |
16. (15 points) As shown in Figure 2, it is known that the ellipse $C$ passes through the point $M(2,1)$, with the two foci at $(-\sqrt{6}, 0)$ and $(\sqrt{6}, 0)$. $O$ is the origin, and a line $l$ parallel to $OM$ intersects the ellipse $C$ at two different points $A$ and $B$.
(1) Find the maximum value of the area o... | 16. (1) Let the equation of the ellipse $C$ be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
From the problem, we have
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - b ^ { 2 } = 6 , } \\
{ \frac { 4 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,461 |
17. (15 points) Given the function
$$
f(x)=\frac{1}{2} m x^{2}-2 x+1+\ln (x+1)(m \geqslant 1) \text {. }
$$
(1) If the curve $C: y=f(x)$ has a tangent line $l$ at point $P(0,1)$ that intersects $C$ at only one point, find the value of $m$;
(2) Prove that the function $f(x)$ has a decreasing interval $[a, b]$, and find ... | 17. (1) Note that the domain of the function $f(x)$ is $(-1,+\infty)$,
$$
f^{\prime}(x)=m x-2+\frac{1}{x+1}, f^{\prime}(0)=-1 .
$$
Therefore, the slope of the tangent line $l$ at the point of tangency $P(0,1)$ is -1.
Thus, the equation of the tangent line is $y=-x+1$.
Since the tangent line $l$ intersects the curve $C... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,462 |
1. If $f(x)$ is an odd function with a period of 5 on $\mathbf{R}$, and satisfies $f(1)=8$, then
$$
f(2010)-f(2009)=(\quad) .
$$
(A) 6
(B) 7
(C) 8
(D) 9 | - 1. C.
Given that $f(x)$ is an odd function with a period of 5 on $\mathbf{R}$, then
$$
\begin{array}{l}
f(2010)-f(2009)=f(0)-f(-1) \\
=f(0)+f(1)=8 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,463 |
2. For non-zero vectors $a$ and $b$, there are two propositions.
Proposition A: $a \perp b$;
Proposition B: The function $f(x)=(x a+b) \cdot(x b-a)$ is a linear function.
Then A is B's ( ) condition.
(A) sufficient but not necessary
(B) necessary but not sufficient
(C) sufficient and necessary
(D) neither sufficient n... | 2. B.
Notice
$$
\begin{array}{l}
f(x)=a \cdot b x^{2}+\left(b^{2}-a^{2}\right) x-a \cdot b, \\
a \perp b \Leftrightarrow a \cdot b=0 .
\end{array}
$$
Thus, $f(x)$ is a linear function $\Rightarrow a \cdot b=0$.
However, when $\boldsymbol{a} \cdot \boldsymbol{b}=0$, $f(x)$ could be a constant function, not necessarily... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,464 |
3. As shown in Figure $1, \Omega$ is the geometric body obtained after the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ is cut by the plane $E F G H$, removing the geometric body $E F G H B_{1} C_{1}$. Here, $E$ and $F$ are points on the line segments $A_{1} B_{1}$ and $B B_{1}$, respectively, different from $B_... | 3. D.
Since $E H / / A_{1} D_{1}$, therefore,
$E H / / B C, E H / /$ plane $B C C_{1} B_{1}$,
$F G=$ plane $B C C_{1} B_{1} \cap$ plane $E F G H$.
Thus, $E H / / F G$.
It is also easy to see that quadrilateral $E F G H$ is a parallelogram, and $A_{1} D_{1} \perp E F$, so $E H \perp E F$.
Clearly, $\Omega$ is a prism. ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,465 |
4. As shown in Figure 2, construct a regular hexagon inside a circle with radius $r=1$, then construct the incircle of the hexagon, and again construct a regular hexagon inside this incircle, and so on, continuing infinitely. Let $S_{n}$ be the sum of the areas of the first $n$ circles. Given the positive number $\xi=3... | 4. A.
Let the radius of the $n$-th circle be $r_{n}$.
It is easy to see that $r_{n}=\frac{\sqrt{3}}{2} r_{n-1}$, the area of the circle $a_{n}=\frac{3}{4} a_{n-1}$, $a_{1}=\pi r_{1}^{2}=\pi$.
Then $S_{n}=\frac{1-\left(\frac{3}{4}\right)^{n}}{1-\frac{3}{4}} \cdot \pi r_{1}^{2}=4 \pi\left[1-\left(\frac{3}{4}\right)^{n}\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,466 |
5. Let the line $x=2$ intersect the asymptotes of the hyperbola $\Gamma: \frac{x^{2}}{4}-y^{2}=1$ at points $E_{1}$ and $E_{2}$, and let $\overrightarrow{O E_{1}}=e_{1}, \overrightarrow{O E_{2}}=e_{2}$. For any point $P$ on the hyperbola $\Gamma$, if $\overrightarrow{O P}=a e_{1}+b e_{2}$ $(a, b \in \mathbf{R})$, then ... | 5. D.
It is easy to find $E_{1}(2,1), E_{2}(2,-1)$. Then $\overrightarrow{O P}=a e_{1}+b e_{2}=(2 a+2 b, a-b)$. Since point $P$ is on the hyperbola, we have
$$
\frac{(2 a+2 b)^{2}}{4}-(a-b)^{2}=1 \text {. }
$$
Simplifying, we get $4 a b=1$.
Therefore, $a^{2}+b^{2} \geqslant 2 a b=\frac{1}{2}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,467 |
7. Given the function $y=6 \cos x$ defined on the interval $\left(0, \frac{\pi}{2}\right)$, its graph intersects with the graph of $y=5 \tan x$ at point $P$. A perpendicular line $P P_{1}$ is drawn from $P$ to the x-axis at point $P_{1}$. The line $P P_{1}$ intersects the graph of $y=\sin x$ at point $P_{2}$. Then the ... | $$
=7 . \frac{2}{3} \text {. }
$$
Let the x-coordinate of point $P$ be $x$. Then $6 \cos x=5 \tan x$.
Solving this, we get $\sin x=\frac{2}{3}$.
Given the condition, the length of $P_{1} P_{2}$ is $\frac{2}{3}$. | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,469 |
Example 1 Given real numbers $a, b (a \neq b)$, and they satisfy
$$
\begin{array}{l}
(a+1)^{2}=3-3(a+1), \\
3(b+1)=3-(b+1)^{2} .
\end{array}
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13 | 【Analysis】Transform the known two equations into
$$
\begin{array}{l}
(a+1)^{2}+3(a+1)-3=0, \\
(b+1)^{2}+3(b+1)-3=0 .
\end{array}
$$
It can be seen that $a$ and $b$ are the two roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$
Solution: From equation (1) in the analysis, simplifying and rearranging ... | -23 | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,470 |
Example 2 Let real numbers $s, t$ satisfy
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, \\
t^{2}+99 t+19=0(s t \neq 1) . \\
\text { Find the value of } \frac{s t+4 s+1}{t} \text { . }
\end{array}
$$
(1999, National Junior High School Mathematics Competition) | 【Analysis】Transform the first equation of the known equations, and combine it with the second equation to find that $\frac{1}{s} 、 t(s t \neq 1)$ are the two roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Since $s \neq 0$, the first equation can be transformed into
$$
\left(\frac{1}{s}\right)^{2}+99\left(\frac{... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,471 |
3. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$
Then the range of values for $t$ is $\qquad$ | Hint: The answer is $-3 \leqslant t \leqslant-\frac{1}{3}$. | -3 \leqslant t \leqslant -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,472 |
7. Arrange all numbers whose prime factors are only $2, 3, 5, 7$ into a sequence $n_{1}, n_{2}, \cdots$. Then $\sum_{i=1}^{\infty} \frac{1}{n_{i}}=$ $\qquad$ . | $\begin{array}{l}\text { 7. } \frac{27}{8} \\ \sum_{i=1}^{\infty} \frac{1}{n_{i}}=\left(\prod_{k=0}^{\infty} \frac{1}{2^{k}}\right)\left(\prod_{k=0}^{\infty} \frac{1}{3^{k}}\right)\left(\prod_{k=0}^{\infty} \frac{1}{5^{k}}\right)\left(\prod_{k=0}^{\infty} \frac{1}{7^{k}}\right)-1 \\ =\frac{1}{1-\frac{1}{2}} \cdot \frac... | \frac{27}{8} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,473 |
8. The maximum volume of a cone inscribed in a sphere with radius 1 is . $\qquad$ | 8. $\frac{32 \pi}{81}$.
Let the radius of the base of the inscribed cone be $r$, and the height be $h$.
To find the maximum volume, by symmetry, it is clear that we should take $h \geqslant 1$, then $h=1+\sqrt{1-r^{2}}$.
Thus, $V=\frac{\pi}{3} r^{2} h=\frac{\pi}{3} r^{2}\left(1+\sqrt{1-r^{2}}\right)$.
Let $d=\sqrt{1-r... | \frac{32 \pi}{81} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,474 |
9. (16 points) Let $a, b, c$ be positive real numbers. Prove:
$$
\sum \sqrt{\frac{a b}{(b+c)(c+a)}} \leqslant \frac{3}{2} .
$$
Here, “ $\sum$ ” denotes the cyclic sum. | $$
\begin{array}{l}
\sum \sqrt{\frac{a b}{(b+c)(c+a)}}=\sum \sqrt{\frac{b}{b+c} \cdot \frac{a}{c+a}} \\
\leqslant \frac{1}{2}\left(\frac{b}{b+c}+\frac{a}{c+a}+\frac{c}{c+a}+\frac{b}{a+b}+\frac{a}{a+b}+\frac{c}{b+c}\right) \\
=\frac{3}{2} .
\end{array}
$$
The equality holds if and only if $a=b=c$. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,475 |
10. (20 points) Given
$$
\lim _{x \rightarrow 0} f(x)=f(0)=1, f(2 x)-f(x)=x^{2}
$$
for any real number $x$. Find the analytical expression of $f(x)$. | 10. When $x \neq 0$,
$$
f\left(\frac{x}{2^{k-1}}\right)-f\left(\frac{x}{2^{k}}\right)=\frac{x^{2}}{2^{2 k}}(k=1,2, \cdots, n) \text {. }
$$
Adding these $n$ equations, we get
$$
f(x)-f\left(\frac{x}{2^{n}}\right)=x^{2} \cdot \frac{\frac{1}{4}\left[1-\left(\frac{1}{4}\right)^{n}\right]}{1-\frac{1}{4}} \text {. }
$$
Le... | f(x)=1+\frac{x^{2}}{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,476 |
11. (20 points) How many inscribed isosceles right triangles $\triangle ABC$ with the right angle vertex $A(0,1)$ are there on the ellipse $\frac{x^{2}}{a^{2}}+y^{2}=1$? | 11. Let $l_{A B}: y=k x+1(k>0)$.
Substituting into the ellipse equation, we get
$$
\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0 \text {. }
$$
Thus, $x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$,
$$
|A B|=\sqrt{1+k^{2}}\left|x_{A}-x_{B}\right|
$$
$$
=\frac{2 a^{2} k \sqrt{1+k^{2}}}{1+a^{2} k^{2}} \text {. }
$$
Similarly... | 3 \text{ triangles when } a^{2}>3, 1 \text{ triangle when } a^{2} \leqslant 3 \text{ and } a^{2} \neq 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,477 |
One, (40 points) Given that $I$ is the incenter of $\triangle ABC$, and $P$ is a point inside the triangle satisfying
$$
\angle P B A + \angle P C A = \angle P B C + \angle P C B.
$$
Prove that: $A P \geqslant A I$, and the equality holds if and only if point $P$ coincides with $I$. | As shown in Figure 1, it is easy to see that
$$
\begin{array}{l}
\angle P B C + \angle P C B \\
= \frac{1}{2} (\angle A B C + \angle A C B) \\
= \angle I B C + \angle I C B.
\end{array}
$$
Therefore, $\angle B P C$
$$
= \angle B I C.
$$
Thus, points $P, B, C, I$ are concyclic.
By the property that the internal and ex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,478 |
II. (40 points) Let $n \in \mathbf{N}$, and suppose $n$ positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy that for any $i, j \in\{1,2, \cdots, n\}$, we have $x_{i} x_{j} \leqslant t^{|i-j|}(t \in(0,1))$. Prove:
$$
\sum_{i=1}^{n} x_{i}<\frac{1}{1-\sqrt{t}} .
$$ | ```
Prove by mathematical induction: $x_{1}, x_{2}, \cdots, x_{n}$ are each no greater than $1, \sqrt{t}, \cdots,(\sqrt{t})^{n-1}$.
When $n=1$, let $i=j=1$, we know $x_{1}^{2} \leqslant 1$.
Thus, $x_{1} \leqslant 1=(\sqrt{t})^{1-1}$, the conclusion holds.
Assume the conclusion holds when $n=k$, then when $n=k+1$, $x_{1... | \frac{1}{1-\sqrt{t}} | Inequalities | proof | Yes | Yes | cn_contest | false | 723,479 |
Three. (50 points) Determine all positive integers $n$ $(n \geqslant 2)$, such that $\mathrm{C}_{n}^{1}, \mathrm{C}_{n}^{2}, \cdots, \mathrm{C}_{n}^{n-1}$ contain a prime number.
Determine all positive integers $n$ $(n \geqslant 2)$, such that $\mathrm{C}_{n}^{1}, \mathrm{C}_{n}^{2}, \cdots, \mathrm{C}_{n}^{n-1}$ cont... | Three, first, from $\mathrm{C}_{p}^{\mathrm{I}}=p$, we know that all prime numbers $p$ satisfy the condition.
Secondly, we prove: all composite numbers do not satisfy the condition, that is, for composite number $n, \mathrm{C}_{n}^{k}(1 \leqslant k \leqslant n-1)$ is a composite number.
(1) If $(n, k)=1$, by $\mathrm{... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,480 |
Four, for a $7 \times 7$ grid of small squares, how many of the small squares can be shaded such that no two shaded squares are adjacent? Adjacent means they share a side.
How many of the small squares can be shaded? | Four, color at most 26 small squares.
The coloring in Figure 2 satisfies the conditions.
The following proves: At most 26 small squares can be colored.
First, according to the problem, for any $2 \times 2$ square grid, at most two of the small squares can be colored; for a $3 \times 3$ square grid, at most 5 of the s... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,481 |
Let $\triangle ABC$ be an acute-angled triangle with altitudes $AD$, $BE$, and $CF$. The angle bisectors of $\angle A$, $\angle B$, and $\angle C$ intersect $EF$, $FD$, and $DE$ at points $A'$, $B'$, and $C'$, respectively. Prove:
$$
S_{\triangle H' B' C'} \leqslant \frac{1}{4} S_{\triangle U E F} .
$$ | Prove the following algebraic inequality first.
Let $x, y, z$ all be positive numbers. Then we have
$$
\begin{array}{l}
\frac{x y}{(x+z)(y+z)}+\frac{y z}{(y+x)(z+x)}+\frac{z x}{(z+y)(x+y)} \\
\geqslant \frac{3}{4},
\end{array}
$$
with equality holding if and only if $x=y=z$.
In fact,
$$
\begin{array}{l}
\text { Equati... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,482 |
4. Given real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value? | Prompt: Example 6. From the problem, we know that $x$ and $y$ are the two real roots of the equation
$$
t^{2}-(z-1) t+z^{2}-7 z+14=0
$$
By the discriminant $\Delta \geqslant 0$, we get
$$
3 z^{2}-26 z+55 \leqslant 0 \text {. }
$$
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\text {... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,483 |
the integer part.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Let
then $2<\sqrt[3]{23}=a_{1}<3, a_{n+1}=\sqrt[3]{23+a_{n}}$.
By mathematical induction, it is easy to prove
$$
2<a_{n}<3\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
In fact,
the integer part is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,484 |
Given 291 As shown in Figure $2, \angle X O Y<90^{\circ}, \triangle A B C$ has its vertex $A$ on $O Y$ and side $B C$ on $O X$, and $A B = A C, B E, C F$ are two altitudes of $\triangle A B C$, $\odot M$ is the incircle of $\triangle A O B$ with radius $m$, $\odot N$ is the excircle of $\triangle A O C$ with radius $n$... | Prove that obviously, $O$, $M$, and $N$ are collinear.
As shown in Figure 3, draw the altitude $AD$.
Then $BD = DC = \frac{1}{2} BC$.
Let $\angle BAD = \angle CAD = \alpha$.
Since $\angle BEC = 90^\circ = \angle BFC$, points $B$, $C$, $E$, and $F$ are concyclic. Thus, $\angle AEF = \angle ABC$.
Therefore, $\triangle AE... | R + r = m + n | Geometry | proof | Yes | Yes | cn_contest | false | 723,485 |
292 Given $P$ is a point inside $\triangle ABC$, points $D$, $E$, $F$ are on sides $BC$, $CA$, $AB$ respectively, and $PD \parallel AB$, $PE \parallel BC$, $PF \parallel CA$. Let the areas of $\triangle AEF$, $\triangle BFD$, $\triangle CDE$, and $\triangle DEF$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{0}$ respectively. P... | Prove as shown in Figure 4, extend $A P$, $B P$, and $C P$ to intersect the opposite sides at points $M$, $N$, and $G$ respectively.
Without loss of generality, let the area of $\triangle A B C$ be 1, and set
\[
\begin{array}{l}
\frac{B D}{B C}=x, \\
\frac{C E}{C A}=y, \frac{A F}{A B}=z .
\end{array}
\]
From $P D \par... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,486 |
5. Given real numbers $x, y$ satisfy
$$
x^{2}+3 y^{2}-12 y+12=0 \text {. }
$$
then the value of $y^{x}$ is $\qquad$ | Hint: Treat the known equation as a quadratic equation in $y$ (the main variable)
$$
3 y^{2}-12 y+\left(12+x^{2}\right)=0 \text {. }
$$
From $\Delta=-12 x^{2} \geqslant 0$, and since $x^{2} \geqslant 0$, then $x=0$.
Thus, $y=2$. Therefore, $y^{x}=2^{0}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,487 |
6. Given real numbers $a, b, c$ satisfy
$$
a=2 b+\sqrt{2}, a b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 \text {. }
$$
Then $\frac{b c}{a}=$ $\qquad$ . | Hint: Eliminate $a$ from the two known equations, then treat $c$ as a constant, to obtain a quadratic equation in $b$
$$
2 b^{2}+\sqrt{2} b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 .
$$
From the discriminant $\Delta \geqslant 0$, we get $c=0$.
Therefore, $\frac{b c}{a}=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,488 |
Example 1 As shown in Figure 5, in the complete quadrilateral $A B C D E F$, $G J \perp E F$ at point J. Then
$$
\begin{array}{l}
\angle B J A \\
=\angle D J C .
\end{array}
$$ | $$
\begin{array}{l}
\angle B J G=\angle D J G, \angle A J G=\angle C J G . \\
\text { Then } \angle B J A=\angle D J C .
\end{array}
$$ | \angle B J A=\angle D J C | Geometry | proof | Yes | Yes | cn_contest | false | 723,489 |
Example 2 As shown in Figure $6, \triangle A B C$ has internal angle bisectors $B E$ and $C F$ intersecting at point $I, I Q \perp E F$ intersects $B C$ at point $P$, and $I P=2 I Q$. Prove: $\angle B A C=60^{\circ}$. | Prove as shown in Figure 6, construct $AX \perp EF$ intersecting $BC$ at point $Y$.
By Property 4, $A, D', I, D$ form a harmonic range.
Thus, $\frac{IQ}{AX}=\frac{D'I}{D'A}=\frac{DI}{DA}=\frac{PI}{YA}$.
Since $IP = 2IQ$, then $AX = XY$, meaning $EF$ is the median of $AY$.
By the Law of Sines,
$$
\begin{array}{l}
\frac{... | 60^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 723,490 |
Example 4 As shown in Figure $8, \triangle A B C$ has an incircle that touches side $B C$ at point $D, A D$ intersects the circle at point $E$, and $C F = C D, C F$ intersects $B E$ at point $G$. Prove: $G F = F C .{ }^{[6]}$ | Proof As shown in Figure 8, let the other two points of tangency be $H$ and $I$, and let $HI$ intersect $BD$ at point $J$. Connect $JE$.
By Property 6, $A, E, K, D$ form a harmonic range. By Theorem 3, the pole of $AD$ lies on $HI$.
Since the pole of $AD$ also lies on $BD$, $J$ is the pole of $AD$. Therefore, $JE$ is ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,492 |
Example 6 Let $D$ be a point on side $BC$ of $\triangle ABC$, satisfying $\angle CAD = \angle CBA$. $\odot O$ passes through points $B, D$, and intersects line segments $AB, AD$ at points $E, F$ respectively. $BF$ intersects $DE$ at point $G$, and $M$ is the midpoint of $AG$. Prove: $CM \perp AO.^{[8]}$ Proof As shown ... | From property 3, $A, K, G, L$ form a harmonic range.
From property 2(4), we have
$L K \cdot G M = L G \cdot K A$.
Also, $\angle C A D = \angle A B D = \angle J F D$, so $E J \parallel C A$.
Thus, $\frac{L J}{J C} = \frac{L K}{K \Lambda} = \frac{L G}{G M}$, which means $J G \parallel C M$.
And from Example 5, we have $J... | null | Geometry | proof | Yes | Yes | cn_contest | false | 723,494 |
Example 3 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \\
\frac{x}{5^{3}+4^{3}}+\frac{y}{5^{3}+6^{3}}=1,
\end{array}
$$
then $x+y=$ $\qquad$
(2005, National Junior High School Mathematics Competition) | 【Analysis】It is easy to notice that the denominators of the two given equations contain $3^{3}$ and $5^{3}$ respectively. Therefore, we can consider $3^{3}$ and $5^{3}$ as the roots of a certain equation.
Solution From the given conditions, it is easy to see that $3^{3}$ and $5^{3}$ are the roots of the equation in ter... | 432 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,495 |
Example 7 As shown in Figure 12, let the circumscribed quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ of $\odot O$ have its opposite sides intersecting at points $E^{\prime} 、 F^{\prime}, A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ intersecting at point $G^{\prime}$. Then $O G^{\prime} \perp E^{\pri... | Proof As shown in Figure 12, let $\odot O$ be the incircle of the circumscribed quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, with the points of tangency on the sides being $A, B, C, D$ respectively. Let $AC$ and $BD$ intersect at point $G$, $AB$ and $CD$ intersect at point $E$, and $AD$ and $BC$ interse... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,496 |
Example 8 As shown in Figure 13, quadrilateral $A B C D$ is a circumscribed quadrilateral of $\odot O$, and $O E \perp A C$ at point $E$. Then
$$
\angle B E C=\angle D E C .
$$ | Prove as shown in Figure 13, draw auxiliary lines.
From Example 7, we know that $FI$, $GH$, and $BD$ intersect at point $M$, and $M$ is the pole of $AC$.
Thus, $OE$ also passes through point $M$, and $B$, $L$, $D$, and $M$ form a harmonic range.
By Property 5, we get $\angle BEC = \angle DEC$.
Finally, let's look at a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,497 |
For example, in $\triangle ABC$, the incircle $\odot I$ touches $BC$ at point $D$. $AD$ intersects $\odot I$ at point $K$, and $BK$, $CK$ intersect $\odot I$ at points $E$ and $F$ respectively. Prove that $BF$, $AD$, and $CE$ are concurrent. | Proof As shown in Figure 14, let the other two points of tangency be $M$ and $N$, and let $MN$ intersect $BC$ at point $J$. From Example 4, we know that $B, D, C, J$ form a harmonic range. Therefore, for point $K$ on $AD$, by Property 1, $EF$ must pass through point $J$; by Property 4 for the complete quadrilateral $BE... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,498 |
2 Prove the inequality $f(a, b, c) \geqslant \frac{7}{8}$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Suppose $a, b, c$ with $c$ being the largest. Let $r=\frac{a+b}{2}$. Then $c \geqslant r$.
The proof is divided into three steps.
First, we prove:
$$
\begin{array}{l}
f(a, b, c) \\
\geqslant \frac{2 r}{1+c+r}+\frac{c}{1+2 r}+(1-c)(1-r)^{2} \text {. } \\
\text { Equation (7) } \Leftrightarrow\left(\frac{a}{1+b+c}-\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,499 |
Example 1 Let $a, b, c > 0, a + b + c = abc$. Prove:
$$
\sum \frac{1}{\sqrt{1+a^{2}}} \leqslant \frac{3}{2} \text {. }
$$
where, “$\sum$” denotes the cyclic sum. | Given the problem, we can set
$$
a=\tan A, b=\tan B, c=\tan C,
$$
where $\angle A, \angle B, \angle C$ are the interior angles of the acute triangle $\triangle ABC$. Then,
$$
\sum \frac{1}{\sqrt{1+a^{2}}}=\sum \cos A.
$$
Since the function $y=\cos x$ is concave on $\left(0, \frac{\pi}{2}\right)$, by Jensen's inequali... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,500 |
Example 3 Let $n \in \mathbf{N}_{+}, x_{0}=0, x_{i}>0(i=1,2$, $\cdots, n)$, and $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$
\begin{aligned}
1 & \leqslant \sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}}} \\
& <\frac{\pi}{2}
\end{aligned}
$$ | Prove that if $\sin \theta_{i}=x_{0}+x_{1}+\cdots+x_{i}$, then
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}}} \\
=\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+\sin \theta_{i-1}} \cdot \sqrt{1-\sin \theta_{i-1}}} \\
=\sum_{i=1}^{n} \frac{x_{i}}{\left|\cos \the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,502 |
Example 4 Prove the inequality:
$$
-1<\sum_{k=1}^{n} \frac{k}{k^{2}+1}-\ln n \leqslant \frac{1}{2}(n=1,2, \cdots) .
$$ | Proof First, prove:
$$
\sum_{k=1}^{n-1} \frac{1}{k+1}1) \text {. }
$$
Since the function $y=\frac{1}{x}$ is concave on $\mathbf{R}_{+}$, the area $S$ enclosed by the graph of $y=\frac{1}{x}$, $y=0$, $x=1$, and $x=n$ satisfies:
(1) It is less than the sum of the areas of rectangles with heights $1, \frac{1}{2}, \cdots,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,503 |
1. Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ be real numbers. Prove:
$$
\sum_{i=1}^{n} a_{i}^{2}-\sum_{i=1}^{n} a_{i} a_{i+1} \leqslant\left[\frac{n}{2}\right](M-m)^{2} \text {, }
$$
where, $a_{n+1}=a_{1}, M=\max _{1 \leqslant i \leqslant n} a_{i}, m=\min _{1 \leqslant i \leqslant n} a_{i}$, $[x]$ denotes the g... | 1. If $n=2 k\left(k \in \mathbf{N}_{+}\right)$, then
$$
\begin{array}{l}
2\left(\sum_{i=1}^{n} a_{i}^{2}-\sum_{i=1}^{n} a_{i} a_{i+1}\right)=\sum_{i=1}^{n}\left(a_{i}-a_{i+1}\right)^{2} \\
\leqslant n(M-m)^{2} . \\
\text { Hence } \sum_{i=1}^{n} a_{i}^{2}-\sum_{i=1}^{n} a_{i} a_{i+1} \leqslant \frac{n}{2}(M-m)^{2} \\
=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,504 |
2. As shown in Figure 1, let $D$ be the midpoint of arc $\overparen{B C}$ on the circumcircle $\Gamma$ of acute $\triangle A B C$, and let point $X$ be on arc $\overparen{B D}$. Let $E$ be the midpoint of arc $\overparen{A B X}$, and $S$ be a point on arc $\overparen{A C}$. The line $S D$ intersects $B C$ at point $R$,... | 2. As shown in Figure 2, connect $A D$ and $R T$ intersecting at point $I$.
Since $D$ is the midpoint of arc $\overparen{B C}$, $A I$ is the angle bisector of $\angle B A C$.
Connect $A S$ and $S I$. Since $R T \parallel D E$, we have $\angle S T I = \angle S E D = \angle S A I$.
Therefore, points $A$, $T$, $I$, and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,505 |
3. Let $A$ be a finite set of real numbers, and $A_{1}, A_{2}, \cdots, A_{n}$ be non-empty subsets of $A$, satisfying:
(1) The sum of all elements in $A$ is 0;
(2) For any $x_{i} \in A_{i} (i=1,2, \cdots, n)$, we have $x_{1}+x_{2}+\cdots+x_{n}>0$.
Prove: There exist $1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant n$, s... | 3. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, a_{1}>a_{2}>\cdots>a_{m}$. Then, by condition (1), we have
$$
a_{1}+a_{2}+\cdots+a_{m}=0 .
$$
Consider the smallest number in each $A_{i}$, and let $A_{1}, A_{2}$, $\cdots, A_{n}$ contain exactly $k_{i}$ sets whose smallest number is $a_{i}(i=1$, $2, \cdots, m$ ).
... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,506 |
4. Let $n$ be a given positive integer, and let the set $S=\{1,2$, $\cdots, n\}$. For non-empty finite sets of real numbers $A$ and $B$, find the minimum value of $|A \otimes S|+|B \otimes S|+|C \otimes S|$, where $C=A+B=\{a+b \mid a \in A, b \in B\}$, $X \otimes Y=\{x \mid x$ belongs to exactly one of $X$ and $Y\}$, a... | 4. The sought minimum value is $n+1$.
First, take $A=B=S$, then we know
$$
|A \otimes S|+|B \otimes S|+|C \otimes S|=n+1 \text {. }
$$
Next, we prove:
$$
l=|A \otimes S|+|B \otimes S|+|C \otimes S| \geqslant n+1 .
$$
Let $X \backslash Y=\{x \mid x \in X, x \notin Y\}$. Clearly,
$$
\begin{aligned}
l= & |A \backslash ... | n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,507 |
5. Given an integer $n(n \geqslant 4)$, for any non-zero real numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}>0
$$
find the maximum value of
$$
\frac{\sum_{i=1}^{n} a_{i}\left(a_{i}+b_{i}\right)}{\sum_{i=1}^{n} b_{i}\left(a_{i}+b_{i}\ri... | 5. The maximum value sought is $n-1$.
By homogeneity, we may assume
$$
\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1 \text {. }
$$
First, consider
$$
\begin{array}{l}
a_{1}=1, a_{2}=a_{3}=\cdots=a_{n}=0, \\
b_{1}=0, b_{2}=b_{3}=\cdots=b_{n}=\frac{1}{n-1}
\end{array}
$$
In this case, we have
$$
\begin{array}{l}
\sum_{i... | n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,508 |
6. Prove: For any given positive integers $m, n$, there exist infinitely many pairs of coprime positive integers $a, b$ such that
$$
(a+b) \mid (a m^{a} + b n^{b}).
$$ | 6. If $m n=1$, then the conclusion holds.
Suppose $m n \geqslant 2$. Since
$$
\begin{array}{l}
n^{a}\left(a m^{a}+b n^{b}\right) \\
=(a+b) n^{a+b}+a\left[(m n)^{a}-n^{a+b}\right],
\end{array}
$$
it suffices to prove that there exist infinitely many pairs of coprime positive integers $a, b$ such that
$(a+b) \mid\left[... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,509 |
1. Let $x$ and $y$ be real numbers, satisfying
$$
x+y=1, x^{4}+y^{4}=\frac{7}{2} \text {. }
$$
Then the value of $x^{2}+y^{2}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | -1 . A.
Let $x^{2}+y^{2}=a$. Then
$$
\begin{array}{l}
x y=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2}=\frac{1-a}{2} . \\
\text { Also } x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2} \\
=a^{2}-2\left(\frac{1-a}{2}\right)^{2}=\frac{1}{2} a^{2}+a-\frac{1}{2}=\frac{7}{2} \\
\Rightarrow a^{2}+2 a-1=7 \\
\Rightarro... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,510 |
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