problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
8.1. The side lengths of a rectangle are integers. It is known that it can be divided into a series of L-shaped figures (i.e., the shape formed by removing any one unit square from a $2 \times 2$ square). Prove: The rectangle can be divided into a series of $1 \times 3$ rectangles. | 8.1. The area of each L-shaped figure is 3, so the area of the rectangle is a multiple of 3. Since the sides of the rectangle are integers, and its area equals the product of its length and width, and 3 is a prime number, at least one side of the rectangle must be a multiple of 3. Therefore, the rectangle can be divide... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,369 |
8.3. As shown in Figure 1, let quadrilateral $ABCD$ be a rhombus, with points $E$ and $F$ located on sides $AB$ and $BC$, respectively, and $AE = 5BE$, $BF = 5CF$. If $\triangle DEF$ is an equilateral triangle, find the measure of $\angle BAD$. | 8.3. As shown in Figure 6, from the problem, we know that $B E = C F$. Take a point $K$ on side $A B$ such that $A K = B E$. It is easy to see that $\triangle A K D \cong \triangle C F D$. Therefore, $D K = D F = D E$. Hence, $\triangle D K E$ is an isosceles triangle, so $\angle D K E = \angle D E K$.
Also, $A K = B E... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,371 |
8.4. There are two types of five-digit numbers as follows:
(1) The sum of the digits is 36, and it is an even number;
(2) The sum of the digits is 38, and it is an odd number.
Try to determine: which type of number is more? Explain your reasoning. | 8.4. It is easy to see that the digits of a five-digit number whose sum of digits equals 38 cannot include 0, otherwise its sum of digits would not exceed \(4 \times 9 = 36\). If we subtract 1 from each of the last two digits of each odd number whose sum of digits equals 38, we can obtain an even number whose sum of di... | There\ are\ more\ even\ numbers\ whose\ sum\ of\ digits\ equals\ 36. | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,372 |
6.1.25 students stand in a row. It is known that the student on the far left is taller than the student on the far right. Prove: there exists a student whose left neighbor is taller than his right neighbor. | 6.1. From left to right, they are student numbers. If, except for the students at the edges, each student's right neighbor is not shorter than their left neighbor, then student 3 is not shorter than student 1, student 5 is not shorter than student 3, and so on. Eventually, student 25 is not shorter than student 1, whic... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,373 |
6.2. The number 12 is written on the blackboard. Every minute, the number on the blackboard can be multiplied by 2 or 3, or divided by 2 or 3, and the result replaces the original number. Prove: After 1 hour, the number on the blackboard cannot be 54. | 6.2. Each operation changes the parity of the total number of factors 2 and 3 in the number. Since at the beginning $12=2 \times 2 \times 3$, the total number is odd (two 2s and one 3), so after 1 hour, the number remains odd. However, in 54 there is one 2 and three 3s, making the total number even, therefore, the numb... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,374 |
6.4. Find all prime number solutions $(x, y, z)$ for the equation $19 x - y z = 1995$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.4. From $y z=19 x-1995=19(x-105)$, and both $y$ and $z$ are prime numbers, so one of them must be 19, and the other is $x-105$. Let $y=19, z=x-105$. Thus, either $x$ or $z$ is even. Since they are both prime numbers, we have $x=107, z=2$. For the case where $z=19, y=x-105$, a similar discussion can be made.
In summa... | (107,19,2),(107,2,19) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,376 |
6.5. In a $9 \times 9$ grid, 19 cells are colored red. Prove: either two red cells can be found that share a common edge; or an uncolored cell can be found that shares a common edge with at least two red cells. | 6.5. Divide the original grid into 9 $3 \times 3$ grids. In one of these $3 \times 3$ grids, at least 3 cells are colored red. If a row (or column) in this $3 \times 3$ grid contains two red cells, then the required cell naturally exists. Otherwise, any two red cells are neither in the same row nor in the same column. ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,377 |
6.6. A rectangular chocolate bar is divided into a grid of $17 \times 17$ squares by grooves. Two players, A and B, play a game according to the following rules: each player, on their turn, divides one rectangular piece of chocolate into two smaller rectangles (only along the grooves). After B's turn, B immediately eat... | 6.6. Under the correct strategy, B will win.
In order to win, B should ensure that after each of their moves, all pieces of chocolate are rectangles with all sides of odd length. It should be noted: at the beginning, the $17 \times 17$ chocolate block meets this requirement. Afterwards, after each of A's moves, there ... | B | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,378 |
Example 6 As shown in Figure 9, $P$ is a point inside $\triangle A B C$ such that $\angle P A B=$ $10^{\circ}, \angle P B A=20^{\circ}$, $\angle P C A=30^{\circ}, \angle P A C$ $=40^{\circ}$. Prove: $\triangle A B C$ is an isosceles triangle. | Proof: Let $\angle A C B=x$, then $\angle B C P=x-30^{\circ}$. Applying the trigonometric form of Ceva's theorem to $\triangle A P C$ and point $B$, we have
$$
\begin{array}{l}
1=\frac{\sin \angle A P B}{\sin \angle B P C} \cdot \frac{\sin \angle P C B}{\sin \angle B C A} \cdot \frac{\sin \angle C A B}{\sin \angle B A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,379 |
7.3. In the guard of a grand duke, there are 1000 warriors. Any two warriors are either friends, enemies, or do not know each other. The warriors are all honest and only speak to their friends. However, the current situation makes every warrior unhappy, because for each warrior, any two of his friends are enemies, and ... | 7.3. We point out that any samurai has no more than 2 friends. In fact, if a certain samurai has at least 3 friends $A, B, C$, then according to the problem, $A, B, C$ must be enemies with each other. This leads to a contradiction: since $B$ and $C$ are both enemies of $A$, they should be friends, not enemies. Therefor... | 200 | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,380 |
7.4. A rectangular grid is divided into a series of $1 \times 2$ rectangles (dominoes). It is known that the number of dominoes each grid line crosses is a multiple of 4. Prove: The length of one side of the grid is a multiple of 4. | 7.4. First, write down the number of dominos each grid line crosses, then add up all the numbers written. Clearly, this sum is a multiple of 4. Since each domino is crossed by exactly one grid line, the sum is the total number of dominos. Therefore,
it is known that the area of the grid is a multiple of 8, hence, one ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,381 |
7.5. The computer's screen displays the number 1. Every second, the computer performs the following operation: if the number on the screen is divisible by \(2^k\), it adds any positive integer from 1 to \(k+1\). Prove that any power of 2 will eventually appear on the screen. | 7.5. Since $1=2^{0}$, the first operation is to add 1, thus, $2=2^{1}$ immediately appears on the screen. Assume that $2^{n}(n>1)$ never appears on the screen. Since the numbers on the screen are monotonically increasing, there must be a moment when the number on the screen is $m2^{n}$. Since $l2^{n}-m \geqslant 2^{l}>... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,382 |
7.6. A rectangular chocolate bar is divided by grooves into $1995 \times 1995$ squares. Two players, A and B, play a game according to the following rules: Each player, on their turn, divides one rectangular piece of chocolate into two smaller rectangles (only along the grooves) and can immediately eat one of the resul... | 7.6. Under the correct strategy, B will win.
If A does not eat the chocolate after the first move, or even if A does eat it but leaves 1 piece of chocolate with one side of even length, then B will turn them into 2 identical pieces of chocolate. Then, B will consistently adopt a symmetric strategy: whatever A does to ... | B | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,383 |
7.7. A multi-volume set of "The Complete Encyclopedia of Dogs" is haphazardly placed on two shelves of a bookcase, with the "German Shepherd Volume" at the far left of the top shelf. Every morning, the librarian swaps the positions of two consecutive volumes that are on different shelves. One day, it is suddenly notice... | 7.7. In fact, all the books have returned to their original positions. If the volume number of the book at position $i$ on a certain level is less than the volume number of the book at position $j$ on the same level, then after the operation, the volume number of the book at position $i$ is still less than the volume n... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 716,384 |
8.1. Connect the center of an equilateral triangle with its three vertices. Write a positive integer on each of the three connecting lines and the three sides. For any three lines that form a triangle, the numbers written on them can be increased by 1 simultaneously. Prove: Through such operations, it is possible to ma... | 8.1. Let the 4 triangles be denoted as $\triangle_{1}, \triangle_{2}, \triangle_{3}, \triangle$, and the sums of the numbers on their three sides be denoted as $S_{1}, S_{2}, S_{3}, S_{4}$, respectively. Each operation increases a certain $S_{i}$ by 3 (without affecting the remainder when divided by 3), while the other... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,385 |
8.3. Let $p(n, k)$ denote the number of divisors of the positive integer $n$ that are not less than $k$. Try to find
$$
\begin{array}{l}
p(1001,1)+p(1002,2)+\cdots+ \\
p(2000,1000) .
\end{array}
$$ | 8.3. If we write down all the divisors of each positive integer $1000+k$ that are not less than $k$ $(k=1,2, \cdots, 1000)$, then the sum we are looking for is the sum of the number of all these divisors.
Below we prove that each positive integer $n$ from 1 to 2000 is written exactly once. If $n>1000$, then it is only... | 2000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,386 |
8.4. As shown in Figure 2, in
$\triangle ABC$, $BD$ is the angle bisector of $\angle ABC$.
Take a point $E$ outside $\triangle ABC$ such that $\angle EAB = \angle ACB$, $AE = DC$, and line segment $ED$ intersects line segment $AB$, with the intersection point denoted as $K$. Prove: $KE = KD$. | 8.4. As shown in Figure 8, draw $EX \perp AB$, $DY \perp AB$, and then draw $DZ \perp BC$. Since $BD$ is the bisector of $\angle ABC$, we have $DY=DZ$. Also, given that $\angle EAB = \angle ACB$ and $AE = CD$, it follows that $EX = DZ$. Therefore, $EX = DY$. This implies that $\triangle KEX \cong \triangle KDY$.
Thus, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,387 |
8.5. An inheritance consists of several diamonds, valued at $1,000,000. It is known that it can be divided into 5 equal parts, or into 8 equal parts. Find the maximum possible value of the smallest diamond. | 8.5. First, prove that the value of the smallest diamond cannot exceed $50,000. According to the problem, this inheritance can be divided into 8 equal parts, each worth $125,000. It is easy to see that at least one of these parts consists of no fewer than 2 diamonds. Therefore, if every diamond worth $125,000 is replac... | 50,000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,388 |
8.6. Write the positive integers from 1 to 100 in any order at the vertices of a regular 100-gon. It is allowed to swap the positions of any two numbers that differ by 1. After several such operations, each number has moved to the next vertex in the clockwise direction. The endpoints of the diameter of the circumscribe... | 8.6. Since each operation only swaps the positions of two numbers with a difference of 1, if two numbers on vertices $X$ and $Y$ have never swapped positions, then the number that was larger at the beginning on one vertex will still be larger on that vertex in the end; conversely, the number that is larger on a vertex ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,389 |
Example 7 As shown in Figure 10, in $\triangle A B C$, $A B=A C, \angle A=20^{\circ}$, points $D$ and $E$ are taken on sides $A B$ and $A C$ respectively, such that $\angle E B C=60^{\circ}, \angle D C B=50^{\circ}$. Find the measure of $\angle B E D$. | Solution 1: Let $\angle B E D=x$, then $\angle C E D=40^{\circ}+x$.
Applying the trigonometric form of Ceva's theorem to $\triangle B C E$ and point $D$, we have
$$
\begin{array}{l}
1=\frac{\sin \angle B C D}{\sin \angle D C E} \cdot \frac{\sin \angle C E D}{\sin \angle D E B} \cdot \frac{\sin \angle E B D}{\sin \angl... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,390 |
8.7. A rectangular grid is divided into a series of $1 \times 2$ rectangles (dominoes). It is known that the number of dominoes crossed by each line parallel to the grid lines but not coinciding with them is even. Prove: the length of one side of the grid is a multiple of 4. | 8.7. Since a rectangular grid can be divided into a series of $1 \times 2$ rectangles, the area of the rectangle is even.
(Continued from page 35)
Thus, it must have at least one side of even length. Without loss of generality, assume its vertical side length is even. If this side length is not a multiple of 4, then it... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,391 |
1. Let $a_{n}$ denote the last two digits of $7^{n}$, then the last three digits of $a_{1}+a_{2}+\cdots+a_{2005}$ are ( ).
(A) 109
(B) 107
(C) 103
(D) 101 | -、1.B.
Since $n$ is a positive integer, the last two digits of $7^{n}$ cycle through $07, 49, 43, 01$. Also, $2005=4 \times 501+1$, so
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2005} \\
=501 \times(7+49+43+1)+7=50107 .
\end{array}
$$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,392 |
2. A mall sells three models of electric vehicles: A, B, and C. It is known that the sales revenue of model A in the first quarter accounts for $56\%$ of the total sales revenue of these three models. In the second quarter, the sales revenue of models B and C decreased by $a\%$ compared to the first quarter, but the to... | 2.D.
From the problem, we have
$$
(1-56 \%) (1-a \%) + 56 \% \times (1+23 \%) = 1+12 \% \text{.}
$$
Solving for \( a \) gives \( a=2 \). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,393 |
3. A two-digit number $\overline{x y}$ added to the two-digit number $\overline{y x}$ after swapping positions results in a perfect square. Then the number of such two-digit numbers is ( ) .
(A) 8
(B) 4
(C) 2
(D) 1 | 3. A.
Given $1 \leqslant x \leqslant 9,1 \leqslant y \leqslant 9$, and
$$
\overline{x y}+\overline{y x}=10 x+y+10 y+x=11(x+y)
$$
is a perfect square, we know that $x+y=11$.
Therefore, the two-digit numbers sought are $29,38,47,56,65,74,83,92$, a total of 8. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,394 |
4. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$. If $\angle B=2 \angle A$, then which of the following relationships is true? ( ).
(A) $\frac{a}{b}>\frac{a+b}{a+b+c}$
(B) $\frac{a}{b}<\frac{a+b}{a+b+c}$
(C) $\frac{a}{b}=\frac{a+b}{a+b+c}$
(D) Cannot be determined | 4.C.
As shown in Figure 5, extend $C B$ to $D$, such that $A B = B D$. Then
$$
\begin{array}{l}
\angle 1 = \angle 2. \\
Also, \angle C B A = 2 \angle C A B, \\
\angle C B A = \angle 1 + \angle 2 \\
= 2 \angle 1,
\end{array}
$$
Therefore, $\angle C A B = \angle 1 = \angle 2$.
Thus, $\triangle A C B \sim \triangle D C ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,395 |
5. Several tourists are taking a boat ride, requiring the same number of people on each boat. If each boat carries 12 people, there is 1 person left unable to board; if one boat leaves empty, then all tourists can be evenly distributed among the remaining boats. It is known that each boat can hold a maximum of 15 peopl... | 5.C.
Let there be $x$ boats initially, and after one empty boat leaves, the average number of tourists per boat is $y$. Thus, $12 x+1=(x-1) y$. Then $y=\frac{12 x+1}{x-1}=12+\frac{13}{x-1}$.
Since $y$ is a positive integer, $\frac{13}{x-1}$ must also be an integer.
Therefore, $x-1=1$ or $x-1=13$, which means $x=2$ or ... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,396 |
6. As shown in Figure 1, with the side $AB$ of the acute triangle $\triangle ABC$ as the diameter, draw $\odot O$, intersecting $AC$ and $BC$ at points $E$ and $D$ respectively, and $DF \perp AB$. If
$$
\begin{array}{l}
AC=14, CD=4,7 \sin C \\
=3 \tan B, \text { then } DB= \\
(\quad) .
\end{array}
$$
(A) 3
(B) 4
(C) 5
... | 6.D.
As shown in Figure 1, connect $A D$. Since $A B$ is the diameter of $\odot O$, we have $\angle A D C=\angle A D B=90^{\circ}$.
In the right triangle $\triangle A C D$, by the Pythagorean theorem, we get
$$
A D=\sqrt{A C^{2}-C D^{2}}=\sqrt{180}=6 \sqrt{5} \text {. }
$$
Then $\sin C=\frac{A D}{A C}=\frac{3 \sqrt{5... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,397 |
1. Given $x=\frac{3-\sqrt{5}}{2}$. Then $x^{3}-3 x^{2}+3 x+$ $\frac{6}{x^{2}+1}=$ . $\qquad$ | $$
=, 1.6 \text {. }
$$
Given $x=\frac{3-\sqrt{5}}{2}$, we know that
$$
\begin{array}{l}
x^{2}-3 x+1=0, x+\frac{1}{x}=3 . \\
\text { Therefore, } x^{3}-3 x^{2}+3 x+\frac{6}{x^{2}+1} \\
=x\left(x^{2}-3 x\right)+3 x+\frac{6}{x^{2}+1}=2 x+\frac{6}{3 x} \\
=2 x+\frac{2}{x}=2\left(x+\frac{1}{x}\right)=6 .
\end{array}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,398 |
3. Given that $a$ and $b$ are real numbers, and $a \geqslant 1$. If the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ has real solutions, and satisfies $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, then $a^{2}+b^{2}=$ | 3.6 .
From the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ having real solutions, we get $\Delta=4 b^{2}+4\left(a-2 b^{2}\right) \geqslant 0$, which means $a \geqslant b^{2}$.
From $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, we can derive $2 a^{2}-5 a+4=a b^{2}-b^{2}=b^{2}(a-1) \leqslant a(a-1)$. Therefore, $2 a^{2}-5 a+4 \l... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,400 |
Example 1 A mall installs an escalator from the first floor to the second floor, which moves upward at a uniform speed. A boy and a girl start walking up the escalator at the same time (the escalator is also moving). If both of their movements are considered uniform, and the boy walks twice as many steps per minute as ... | Explanation: (1) Let the girl's speed be $x$ steps/min, the escalator's speed be $y$ steps/min, the boy's speed be $2x$ steps/min, and the stairs have $s$ steps. Then
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} .
\end{array}\right.
$$
(1)
$$
\frac{3}{4}=\frac{s-27}{s-18} \te... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,401 |
Example 2 A pedestrian and a cyclist are traveling south simultaneously on a road parallel to a railway. The pedestrian's speed is $3.6 \mathrm{~km} / \mathrm{h}$, and the cyclist's speed is $10.8 \mathrm{~km} / \mathrm{h}$. If a train comes from behind and takes $22 \mathrm{~s}$ to pass the pedestrian and $26 \mathrm{... | Explanation: First, convert the two speed units to: pedestrian $1 \mathrm{~m} / \mathrm{s}$, cyclist $3 \mathrm{~m} / \mathrm{s}$. Let the length of the train be $l \mathrm{~m}$, and the speed of the train be $v \mathrm{~m} / \mathrm{s}$. According to the problem, we have
$$
\left\{\begin{array}{l}
l=(v-1) \times 22, \... | 286 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,402 |
3. A mall has an escalator moving uniformly from the bottom to the top. Both person A and person B are in a hurry to go upstairs, so while riding the escalator, they both climb the stairs at a uniform speed. Person A reaches the top after climbing 55 steps, and person B's climbing speed is twice that of person A (the n... | (Prompt: Imitate example 1. Answer: 66.)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,403 |
3. Given the function $f(n)=\frac{20^{n}+3^{n}}{n!}, n \in \mathbf{N}$. Then, the value of $n$ that maximizes $f(n)$ is, $n=$ | 3.19.
$$
\begin{array}{l}
G(n)=f(n+1)-f(n) \\
=\frac{20^{n+1}+3^{n+1}}{(n+1)!}-\frac{20^{n}+3^{n}}{n!} \\
=\frac{(19-n) \times 20^{n}-(n-2) \times 3^{n}}{(n+1)!} \\
=\frac{3^{n}}{(n+1)!}\left[(19-n) \times\left(\frac{20}{3}\right)^{n}-(n-2)\right] .
\end{array}
$$
It is easy to see that when $n \geqslant 19$, $G(n) < ... | 19 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,405 |
4. Write an arithmetic sequence in the following manner:
1st row: 2
2nd row: 5, 8
3rd row: $11, 14, 17$
4th row: $20, 23, 26, 29$
$\qquad$
$m$-th row: $a_{m 1}, a_{m 2}, a_{m 3}, \cdots, a_{m n}$ $a_{m i}$ represents the $i$-th number in the $m$-th row, $i=1,2,3, \cdots, m$.
Then, the sum of the $m$ numbers in the $m$... | 4. $\frac{1}{2} m\left(3 m^{2}+1\right)$.
From the induction, we know that $a_{m 1}-a_{(m-1) 1}=3 \times(m-1)$. Therefore,
$$
a_{m 1}=\sum_{i=1}^{m-1} 3 i+2=\frac{3 m(m-1)}{2}+2 \text {. }
$$
Also, $a_{m m}=a_{(m+1) 1}-3=\frac{3 m(m+1)}{2}-1$, so
$$
\sum_{i=1}^{m} a_{m i}=\frac{m}{2}\left(a_{m l}+a_{m m m}\right)=\fr... | \frac{1}{2} m\left(3 m^{2}+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,406 |
5. The maximum value of the algebraic expression $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$ is $\qquad$ . | 5.2.
It is known that $|a| \leqslant \sqrt{2},|b| \leqslant \sqrt{2}$. Let
$$
a=\sqrt{2} \sin \alpha, b=\sqrt{2} \sin \beta, \alpha, \beta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text {. }
$$
Then $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$
$$
\begin{array}{l}
=2(\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \alp... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,407 |
6. Given in the first quadrant of the Cartesian coordinate plane, there is a regular $\triangle ABC$, which lies within the region (including the boundary) enclosed by the curve $y=-x^{2}+4 a x-3 a^{2}$ $(a>0)$ and the $x$-axis, with the base $BC$ on the $x$-axis. Then, its maximum area function $f(a)$ is _. $\qquad$ | 6. $f(a)=\left\{\begin{array}{ll}\sqrt{3} a^{2}, & a \geqslant \sqrt{3} ; \\ \frac{\sqrt{3}}{3} a^{4}, & 0<a<\sqrt{3} \end{array}\right.$
The two intersection points of the parabola $y=a x^{2}-2 a^{2} x+a^{3} (a>0)$ with the $x$-axis are $D(a, 0)$ and $E(3 a, 0)$. The vertex is $F\left(2 a, a^{2}\right)$.
Since the b... | f(a)=\left\{\begin{array}{ll}\sqrt{3} a^{2}, & a \geqslant \sqrt{3} ; \\ \frac{\sqrt{3}}{3} a^{4}, & 0<a<\sqrt{3} \end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,408 |
Three, (15 points) In the Cartesian coordinate plane, $\triangle ABC$ has two vertices $A, B$ with coordinates $A(-1,0)$, $B(1,0)$. Two points $G, M$ in the plane simultaneously satisfy the following conditions:
(1) $\boldsymbol{G A}+\boldsymbol{G B}+\boldsymbol{G C}=0$;
(2) $|M A|=|M B|=|M C|$;
(3) $G M \parallel A B$... | (1) Let $C(x, y)$, $G(x_0, y_0)$, and $M(x_M, y_M)$.
Since $|MA| = |MB|$, point $M$ lies on the perpendicular bisector of segment $AB$.
Given $A(-1,0)$ and $B(1,0)$, we have $x_M = 0$.
Since $GM \parallel AB$, we have $y_M = y_0$.
Also, $GA + GB + GC = 0$, which means
$$
\begin{array}{l}
\left(-1 - x_0, -y_0\right) + \... | PE \cdot PF \in \left(8, \frac{88}{9}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,409 |
Four. (15 points) If $a_{i} \in \mathbf{C}$ (the set of complex numbers) $(1 \leqslant i \leqslant n)$, and for $1 \leqslant k \leqslant n (k \in \mathbf{N})$, we have
$$
a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k}=0 \text{. }
$$
Prove: For all positive integers $k$,
$$
a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k}=0 .
$$ | Let $f(x) = (x - a_1)(x - a_2) \cdots (x - a_n) = x^n + b_1 x^{n-1} + \cdots + b_n$.
It is easy to see that $f(a_1) = f(a_2) = \cdots = f(a_n) = 0$, i.e.,
\[
\begin{array}{l}
a_1^n + b_1 a_1^{n-1} + \cdots + b_n = 0, \\
a_2^n + b_1 a_2^{n-1} + \cdots + b_n = 0, \\
\cdots \cdots \\
a_n^n + b_1 a_n^{n-1} + \cdots + b_n =... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,410 |
Five, (15 points) Find the smallest positive integer $n$ such that $2^{2005}$ । $\left(161^{n}-1\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Five, (15 points) Find the smallest positive integer $n$ such tha... | Let $n=2^{s} q, s \in \mathbf{N}, 2$ does not divide $q$. Then $161^{n}-1=161^{2^{s} q}-1=\left(161^{2^{s}}-1\right) A$.
It is easy to see that 2 does not divide $A$.
Given $2^{2005} \mid\left(161^{n}-1\right)$, we get $2^{2005} \mid\left(161^{2^{3}}-1\right)$.
Thus, we can set $n=2^{x}$.
Since $161=2^{5} \times 5+1$, ... | 2^{2000} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,411 |
Six. (15 points) A square is divided into 4 squares, the number of edges in the division diagram is 12. If a square is divided into 2005 convex polygons, try to find the maximum number of edges in the division diagram.
| Six, according to Euler's theorem, the number of vertices $a$, the number of faces $b$, and the number of edges $e$ of a simple polyhedron have the following relationship: $a+b-e=2$.
From Euler's theorem, it is easy to see that if a convex polygon is divided into $n$ convex polygons, then the number of vertices $a$, t... | 6016 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,412 |
1. Let the decimal part of $\frac{2}{\sqrt{5}-1}$ be denoted as $\alpha$. Then the value of $\alpha(\alpha+1)$ is ( ).
(A) 1
(B) a positive rational number less than 1
(C) a positive irrational number less than 1
(D) a positive irrational number greater than 1 | -1.A.
From $\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}=1+\frac{\sqrt{5}-1}{2}$, and $0<\frac{\sqrt{5}-1}{2}<\frac{3-1}{2}=1$,
we get $\alpha=\frac{\sqrt{5}-1}{2}=\frac{2}{\sqrt{5}+1}, \alpha+1=\frac{2}{\sqrt{5}-1}$.
Thus, $\alpha(\alpha+1)=\frac{2}{\sqrt{5}+1} \times \frac{2}{\sqrt{5}-1}=1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,413 |
4. An airplane flies from Dunedin to Shanghai every hour, and at the same time, an airplane flies from Shanghai to Dunedin every hour, with each flight taking $10 \mathrm{~h}$ to complete. The number of airplanes flying in the opposite direction that a passenger encounters during a trip from Dunedin to Shanghai is $\qq... | (Prompt: Imitate Example 8. Answer: 21.)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,414 |
2. The positional relationship between the quadratic function $y=x^{2}-|k|$ and the direct proportion function $y=k x(k \neq 0)$ is $(\quad)$.
(A) Disjoint
(B) Tangent
(C) Intersecting
(D) Cannot be determined, depends on the value of $k$ | 2. C.
Idea 1: Draw the function graph. When $k \neq 0$, the origin is inside the parabola $y=x^{2}-|k|$, and any line $y=k x$ passing through the origin will intersect the parabola.
Idea 2: Solve the system of equations by eliminating $y$, resulting in a quadratic equation
$$
x^{2}-k x-|k|=0 \text {. }
$$
Since $\Del... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,415 |
3. As shown in Figure 1, in $\triangle A B C$, $D$ and $E$ are points on $A C$ and $A B$ respectively. Connect $D E$, $B D$, and $C E$, and let $O$ be the intersection of $B D$ and $C E$. If we set $m$
$$
=\frac{S_{\triangle A D E}}{S_{\triangle A B C}}, n=\frac{S_{\triangle O D E}}{S_{\triangle O B C}} \text {, }
$$
... | 3. B.
From the formula for the area of a triangle, we have
$$
m=\frac{S_{\triangle A D E}}{S_{\triangle A B C}}=\frac{\frac{1}{2} A D \cdot A E \sin A}{\frac{1}{2} A B \cdot A C \sin A}=\frac{A D \cdot A E}{A B \cdot A C} .
$$
Similarly, $n=\frac{S_{\triangle O D E}}{S_{\triangle O B C}}=\frac{O D \cdot O E}{O B \cdo... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,416 |
4. Two triangles that share a common side are called a pair of adjacent triangles. Then the number of pairs of adjacent triangles in Figure 1 is ( ).
(A) 8
(B) 16
(C) 24
(D) 32 | 4.D.
In Figure 1, there are 6 points, and except for $A$ and $O$ which are not connected, every two points are connected, resulting in a total of 14 line segments. Considering each line segment as a common side and exhausting all points outside the line segment to form triangles, the number of pairs of triangles shari... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,417 |
5. Let $p$ be a prime number greater than 3, then the remainder when $p^{2}$ is divided by 6 is ( ).
(A) 5
(B) 3
(C) 1
(D) Cannot be determined, depends on the value of $p$ | 5.C.
For primes greater than 3, when divided by 3, the remainder is 1 or 2, which can be expressed as $3m+1$ or $3m+2$. Since primes greater than 3 are necessarily odd, when $p=3m+1$, $m$ must be even $(m=2k)$; when $p=3m+2$, $m$ must be odd $(m=2k-1)$.
Therefore, $p$ can be expressed as $p=6k \pm 1$, yielding
$$
p^{2... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,418 |
2. Given a regular $n$-sided polygon inscribed in a circle. If the line connecting a vertex to the center of the circle is exactly on the perpendicular bisector of a certain side, then $n \equiv$ $\qquad$ $(\bmod 2)$ | 2.1.
As shown in Figure 8, if the line connecting vertex $A_{1}$ and the center $O$ perpendicularly bisects side $A_{i} A_{j}$, then $\triangle A_{1} A_{i} A_{j}$ is an isosceles triangle. Since equal chords subtend equal arcs, the number of vertices in $\overparen{A_{1} A_{i}}$ is equal, and their sum is even. Adding... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,421 |
3. Given a sequence of numbers $a_{1}, a_{2}, \cdots, a_{2006}$, where $a_{1}$ $=1$, and the sum of each pair of consecutive terms is 3. Then $a_{1}-a_{2}+$ $a_{3}-a_{4}+\cdots+a_{2003}-a_{2004}+a_{2005}=$ $\qquad$ | 3. -1001 .
Given $a_{1}+a_{2}=a_{2}+a_{3}=\cdots=a_{2004}+a_{2000}=3$, and $a_{1}=1$, we can sequentially deduce that
$$
\begin{array}{l}
a_{1}=a_{3}=\cdots=a_{2005}=1, a_{2}=a_{4}=\cdots=a_{2004}=2 . \\
\text { Then } a_{1}-a_{2}+a_{3}-a_{4}+\cdots-a_{2004}+a_{2000} \\
=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{4}\righ... | -1001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,422 |
I. (20 points) From the diagram 6, it can be seen that the parabola
$$
y=x^{2}+p x+q
$$
has two points $A\left(a, y_{1}\right)$,
$B\left(b, y_{2}\right)$ (where $a<b$)
below the $x$-axis, then the parabola must intersect the $x$-axis at two points $C\left(x_{1}, 0\right)$ and $D\left(x_{2}, 0\right)$
(where $x_{1}<x_{... | (1) Given point $A\left(a, y_{1}\right)$ is on parabola (1) and below the $x$-axis, we have
$$
0>y_{1}=a^{2}+p a+q .
$$
Completing the square and simplifying, we get
$$
0>y_{1}=\left(a+\frac{p}{2}\right)^{2}+\frac{4 q-p^{2}}{4} \geqslant \frac{4 q-p^{2}}{4} \text {. }
$$
Thus, $p^{2}-4 q>0$.
This indicates that the d... | y=x^{2}+2002 x-4008, y=x^{2}-2006 x, y=x^{2}+394 x-2400, y=x^{2}-398 x-1608 | Algebra | proof | Yes | Yes | cn_contest | false | 716,424 |
5. A passenger train always moves at a constant speed. When it passes a bridge that is $450 \mathrm{~m}$ long, it takes $33 \mathrm{~s}$ from the moment the front of the train enters the bridge until the rear of the train leaves the bridge; when it passes through a tunnel that is $760 \mathrm{~m}$ long, the entire trai... | (Tip: (1) Let the speed of the passenger train be $x \mathrm{~m} / \mathrm{s}$, and the length of the passenger train be $y \mathrm{~m}$. According to the problem, we have $450+y=33 x, 760-y=22 x$. Solving these equations yields $x=22, y=276$. Therefore, $t=\frac{a+276}{v+22}(v>0, a>0)$. (2) When $a=324, 12 \leqslant v... | \frac{600}{37}<t \leqslant \frac{300}{17} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,425 |
II. (25 points) In $\square A B C D$, connect the diagonals $A C$ and $B D$. If $\angle B A D$ is an acute angle, and
$$
(A C \cdot B D)^{2}=A B^{4}+B C^{4} \text {, }
$$
find the degree measure of $\angle B A D$. | $$
\begin{array}{l}
\angle B C E=\angle A C D, \\
\angle C B E=\angle A D C .
\end{array}
$$
Then we have $\angle A B E$
$$
\begin{array}{l}
=360^{\circ}-\angle A D C-\angle A B C \\
=2 \angle B A D \\
\angle A C E=\angle A C B+\angle B C E \\
=\angle A C B+\angle A C D=\angle D C B .
\end{array}
$$
Also, $\triangle ... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,426 |
Three, (25 points) Find the smallest positive integer $k$, such that there exists a positive integer $n$, satisfying $10^{n}=29 k+2$.
---
The translation maintains the original text's format and line breaks. | $$
\begin{array}{l}
\text { Given } \frac{1}{29}=0.0334482758620689655172413793 \text { i, } \\
\frac{2}{29}=0.0068965517241379310344827586 \text { 2e, }
\end{array}
$$
we know that when $n=11$, $k$ takes the minimum value 3448275862.
Explanation: By appending the 28-digit repeating cycle of $\frac{1}{29}$ to 34482758... | 3448275862 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,427 |
1. Let real numbers $x, y$ satisfy $y>x>0$, and
$$
\begin{array}{l}
\tan x=-x, \tan y=-y . \\
\text { Then } \frac{\sin (x-y)}{x-y}-\frac{\sin (x+y)}{x+y}+\frac{2 x^{2} \cos x(x \cos x-\sin x)}{2 x+\sin 2 x}
\end{array}
$$
is ( ).
(A) 0
(B) 2
(C) -1
(D) $-\frac{1}{2}$ | $-1 . B$.
From the given, we have
$$
\begin{array}{l}
\frac{\sin (x-y)}{x-y}=\frac{\sin (x-y)}{\tan y-\tan x}=-\cos x \cdot \cos y, \\
\frac{\sin (x+y)}{x+y}=-\cos x \cdot \cos y, \\
\frac{2 x^{2} \cos x(x \cos x-\sin x)}{2 x+\sin 2 x} \\
=\frac{-2 x \sin x(-\sin x-\sin x)}{-2 \tan x+2 \sin x \cdot \cos x} \\
=\frac{2 ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,428 |
2. Let the function $f(x)(f(x) \neq 0)$ have the domain $(0,+\infty)$, and for $x \in \mathbf{R}_{+}, y \in \mathbf{R}$, it always holds that $f\left(x^{y}\right)=$ $y f(x)$. If $a>b>c>1$, and $a, b, c$ form an arithmetic sequence, then the relationship between $f(a) f(c)$ and $(f(b))^{2}$ is ( ).
(A) $f(a) f(c)(f(b))^... | 2. A.
Let $a=b^{r}, c=b^{s}$, then $r, s$ are positive numbers.
Also, $a+c=2b, a>c$, then
$$
b>\sqrt{ac}=\sqrt{b^{r+s}}>b^{\sqrt{rs}} \text{.}
$$
Thus, $rs<1$.
Therefore, $f(a) f(c)=f\left(b^{r}\right) f\left(b^{s}\right)=rs(f(b))^{2}<(f(b))^{2}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,429 |
3. In the online game "Transformation", the protagonist has a $\frac{2}{3}$ probability of transforming (i.e., from "elephant" to "mouse" or from "mouse" to "elephant") after each level. If the probability that the protagonist does not transform after passing $n$ levels is denoted as $P_{n}$, then ( ).
(A) $P_{5}>P_{4}... | 3. C.
From the given information,
$P_{n+1}=\frac{1}{3} P_{n}+\frac{2}{3}\left(1-P_{n}\right)=\frac{2}{3}-\frac{1}{3} P_{n}$.
Thus, $P_{n+2}-P_{n+1}=-\frac{1}{3}\left(P_{n+1}-P_{n}\right)$.
Furthermore, from $P_{0}=1, P_{1}=\frac{1}{3}$, we can obtain
$$
P_{n+1}-P_{n}=-\frac{2}{3}\left(-\frac{1}{3}\right)^{n} \text {. ... | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,430 |
4. Let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. If the inequality $a_{n}^{2}+\frac{S_{n}^{2}}{n^{2}} \geqslant \lambda a_{1}^{2}$ holds for any arithmetic sequence $\left\{a_{n}\right\}$ and any positive integer $n$, then the maximum value of $\lambda$ is
(A) 0
(B) $\frac{1}{5}$... | 4. B.
Let the common difference of $\left\{a_{n}\right\}$ be $d$, then
$$
\begin{array}{l}
a_{n}^{2}+\frac{S_{n}^{2}}{n^{2}} \\
=\left[a_{1}+(n-1) d\right]^{2}+\frac{1}{n^{2}}\left[n a_{1}+\frac{n(n-1)}{2} d\right]^{2} \\
=2 a_{1}^{2}+3(n-1) a_{1} d+\frac{5}{4}(n-1)^{2} d^{2} \\
=\frac{1}{5} a_{1}^{2}+\left[\frac{3}{\... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,431 |
5. From the 12 edges and the 12 face diagonals of a cube, select $n$ lines such that any two of these lines are skew lines. Then the maximum value of $n$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 5.B.
As shown in Figure 2, take $A B_{1}$, $B C_{1}$, $C D_{1}$, and $D A_{1}$. Clearly, they are pairwise skew, so $n \geqslant 4$. Furthermore, the required skew lines intersect the surface of the cube at distinct points, and there can be at most 8 such points, so $n \leqslant 4$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,432 |
6. Let $A$ and $B$ be the common vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. $P$ and $M$ are two moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy $\boldsymbol{A P}+\boldsymb... | 6.A.
Let's take $A(-a, 0)$ and $B(a, 0)$. Suppose $P\left(x_{1}, y_{1}\right)$ and $M\left(x_{2}, y_{2}\right)$. From the given equation, we know that $O$, $P$, and $M$ are collinear. Therefore,
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}} \\
=\frac{... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,433 |
1. Let $P$ and $Q$ be any two points on the curve
$$
y=x^{3}-3 x^{2}+(3-\sqrt{3}) x+\frac{3}{4}
$$
Then the range of the inclination angle $\alpha$ of the line $PQ$ is $\qquad$ | 二、1. $\left[0, \frac{\pi}{2}\right) \cup\left(\frac{2 \pi}{3}, \pi\right)$.
Finding the derivative, we know that the slope $k \geqslant -\sqrt{3}$ of the tangent line at any point on the curve. Therefore, the slope of the line passing through any two different points on the curve is greater than $-\sqrt{3}$. Thus, the ... | \left[0, \frac{\pi}{2}\right) \cup\left(\frac{2 \pi}{3}, \pi\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,434 |
2. Let $a=\sqrt{x^{2}+x y+y^{2}}, b=p \sqrt{x y}$, $c=x+y$. If for any positive numbers $x$ and $y$, a triangle exists with sides $a$, $b$, and $c$, then the range of the real number $p$ is $\qquad$ | 2. $(2-\sqrt{3}, 2+\sqrt{3})$.
Obviously $a<c$, thus,
$a, b, c$ can form a triangle
$$
\Leftrightarrow c-a<b<c+a
$$
$\Leftrightarrow$ for any positive integers $x, y$, it always holds that
$$
\begin{aligned}
& \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}-\sqrt{\frac{x}{y}+1+\frac{y}{x}} \\
< & p \sqrt{\frac{x}{y}}+\sqrt{\fra... | (2-\sqrt{3}, 2+\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,435 |
Example 1 Find the positive integer solutions of the equation $x^{2}+y^{2}=328$. | Obviously, $x \neq y$. Without loss of generality, assume $x>y>0$. Since 328 is an even number, the parity of $x$ and $y$ must be the same. Therefore, $x \pm y$ is even.
Let $x+y=2 u_{1}, x-y=2 v_{1}$, then $u_{1}, v_{1} \in \mathbf{Z}$, and $u_{1}>v_{1}>0$. Thus,
$$
x=u_{1}+v_{1}, y=u_{1}-v_{1}.
$$
Substituting into... | (18,2),(2,18) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,436 |
3. Let $A=\{1,2,3,4,5\}$. Then the number of mappings $f: A \rightarrow A$ that satisfy the condition $f(f(x))$ $=f(x)$ is $\qquad$ (answer with a number) | 3.196 .
It can be generalized to the case where $A$ has $n$ elements. From the condition, if $a \in A$ and $a$ is in the range of $f$, then it must be that $f(a)=a$. Therefore, we can classify $f$ based on the number of elements in its range. The number of $f$ with $k\left(k \in \mathbf{N}_{+}\right)$ elements in its ... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,437 |
4. Given that two large spherical surfaces with a radius of 1 are externally tangent to each other and both are internally tangent to a cylindrical surface with a radius of 1, and a smaller spherical surface is externally tangent to both large spherical surfaces and internally tangent to the cylindrical surface. A plan... | 4. $\frac{4}{5}$.
It is easy to calculate that the radius of the small sphere is $\frac{1}{4}$. Let the line passing through the center of the small sphere and the center of a large sphere intersect the cylinder at points $C$ and $D$. Then, the ellipse formed by the section through $C$ and $D$ has a constant minor axi... | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,438 |
5. Let $S=\{-3,-2,1,2,3,4\}$, and take any two numbers $a, b$ from $S$, where $a \neq b$. Then the maximum value of the minimum value of the function $f(x)=x^{2}-(a+b) x + a b$ is $\qquad$ . | 5. $-\frac{1}{4}$.
It is easy to see that the minimum value of $f(x)$ is
$$
\varphi(a, b)=-\frac{1}{4}(a-b)^{2} \text {. }
$$
Since $a, b \in S$, and $a \neq b$, the minimum value of $|a-b|$ is 1. Therefore, the maximum value of $\varphi(a, b)$ is $-\frac{1}{4}$. | -\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,439 |
6. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=\frac{1}{8}$, the geometric mean of the first $n$ terms is 8. If the geometric mean of the remaining terms after removing one term from the first $n$ terms is $4 \sqrt{2}$, then the removed term is the $\qquad$th term. | 6.13.
Let the common ratio be $q$, the product of the first $n$ terms be $M$, and the $k$-th term be removed, then $M=8^{n}$, and $\frac{M}{a_{k}}=(4 \sqrt{2})^{n-1}$.
Thus, $a_{k}=(\sqrt{2})^{n+5}$, which means $\frac{1}{8} q^{k-1}=(\sqrt{2})^{n+5}$.
Therefore, $q^{k-1}=(\sqrt{2})^{n+11}$.
Also, $M=8^{n}$, i.e., $\le... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,440 |
Three. (20 points) Let the function $f(x)$ satisfy
$$
f\left(2^{x}\right)=x^{2}-2 a x+a^{2}-1 \text {, }
$$
and $f(x)$ has a range of $[-1,0]$ on $\left[2^{a-1}, 2^{a^{2}-2 a+2}\right]$. Find the range of values for $a$. | Let $g(x)=f\left(2^{x}\right)$, then
$f(x)=g\left(\log _{2} x\right)=\left(\log _{2} x\right)^{2}-2 a \log _{2} x+a^{2}-1$.
Thus, the range of $f(x)$ on the interval $\left[2^{a-1}, 2^{a^{2}-2 a+2}\right]$ being $[-1,0]$ is equivalent to the range of $g(x)=x^{2}-2 a x+a^{2}-1$ on the interval $\left[a-1, a^{2}-2 a+2\ri... | \frac{3-\sqrt{5}}{2} \leqslant a \leqslant 1 \text{ or } 2 \leqslant a \leqslant \frac{3+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,441 |
Four. (20 points) Let $a, b, c, d \in \mathbf{R}$. In the set of complex numbers, the roots of the equations $x^{2}+a x+b=0$ and $x^{2}+c x+d=0$ have moduli less than 1. Prove that the roots of the equation
$$
x^{2}+\frac{a+c}{2} x+\frac{b+d}{2}=0
$$
have moduli less than 1. | $$
\text{Lemma: The roots of the equation } x^{2}+a x+b=0 \text{ have modulus less than } 1 \Leftrightarrow\left\{\begin{array}{l}|a| < 2 , } \\
{ 1 + a + b > 0 , } \\
{ 1 - a + b > 0 , } \\
{ - 1 < - \frac { a } { 2 } < 1 }
\end{array} \text { or } \left\{\begin{array}{l}
a^{2}-4 b<0, \\
a=-2 r \cos \theta(0 \leqslant... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,442 |
Five. (20 points) Let $e$ be the eccentricity of the conic section $\Gamma$, $F$ be one of its foci, $l$ be the axis of symmetry passing through the focus, $O$ be the vertex on $l$ closer to $F$, and $M, N$ be two points on $l$ satisfying $\boldsymbol{O F} \cdot O M + O F \cdot O N = (1-e) O M \cdot O N$. Prove that fo... | Five, with $O$ as the origin and $O F$ as the positive direction of the $x$-axis, establish a rectangular coordinate system. Let $|O F|=s$, then the equation of the curve $\Gamma$ is
$$
(x-s)^{2}+y^{2}=e^{2}\left(x+\frac{s}{e}\right)^{2} \text {. }
$$
Also, because $O F \cdot O M+O F \cdot O N=(1-e) O M \cdot O N$, it... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,443 |
One, (50 points) As shown in Figure 1, two circles are externally tangent at point $T, P Q$ is a chord of $\odot O_{1}$, and lines $P T$ and $Q T$ intersect $\odot O_{2}$ at points $R$ and $S$, respectively. Tangents to $\odot O_{1}$ at $P$ and $Q$ intersect $\odot O_{2}$ at $A$, $B$, $D$, and $C$, respectively. Lines ... | As shown in Figure 3, extend $CA$ to point $M$, and connect $TA$, $TF$, $SR$, $SD$, $SC$, and $AD$.
It is easy to see that $SR \parallel PQ$, so $\angle PFA = \angle ASR = \angle PTA$.
Thus, $P$, $F$, $T$, and $A$ are concyclic. Therefore,
\[
\begin{array}{l}
\angle FAD = \angle FAT + \angle TAD = \angle FPT + \angle T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,444 |
Three. (50 points) If a positive integer $n$ has the sum of its digits in base 3 divisible by 3, then $n$ is called a "proper number." Find the sum of all proper numbers in $S=\{1,2, \cdots, 2005\}$. | For $m \in \mathbf{N}$, in ternary (base 3) representation, there are $3^{m+1}$ non-negative integers with at most $m+1$ digits. Let the number of these integers whose digit sums are congruent to 0, 1, and 2 modulo 3 be $a_{m}$, $b_{m}$, and $c_{m}$, respectively. When $m \geqslant 1$, by classifying the numbers whose ... | 671007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,446 |
Example 2 Find all positive integer triples $(x, y, z)$ such that $y$ is a prime, $y \nmid z, 3 \nmid z$, and $x^{3}-y^{3}=z^{2}$.
(1999, Bulgaria Mathematical Olympiad) | From the problem, we have
$$
(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2} \text {. }
$$
Since $y$ is a prime number, and $y \nmid z, 3 \nmid z$, combining with equation (1), we know
$$
\begin{array}{c}
(x, y)=1, (x-y, 3)=1 \text {. Then } \\
\left(x^{2}+x y+y^{2}, x-y\right) \\
=(3 x y, x-y)=1 .
\end{array}
$$
From equati... | (8,7,13) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,447 |
Given the parabola $y=a x^{2}+b x+c(a \neq 0)$ has two points $A_{1}\left(m_{1}, y_{1}\right)$ and $A_{2}\left(m_{2}, y_{2}\right)$ satisfying
$$
a^{2}+\left(y_{1}+y_{2}\right) a+y_{1} y_{2}=0 \text {. }
$$
(1) Prove that the parabola $y=a x^{2}+b x+c$ must intersect the $x$-axis at two distinct points.
(2) Let the poi... | Prove: (1) From $a^{2}+\left(y_{1}+y_{2}\right) a+y_{1} y_{2}=0$, we have $\left(y_{1}+a\right)\left(y_{2}+a\right)=0$.
Thus, $y_{1}=-a$ or $y_{2}=-a$.
Without loss of generality, let $y_{1}=-a$. Since point $A_{1}$ lies on the parabola, we have $-a=a m_{1}^{2}+b m_{1}+c$.
This indicates that the quadratic equation
$$
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,448 |
In a convex quadrilateral $ABCD$, the extensions of sides $BA$ and $CD$ intersect at point $P$, and the diagonals intersect at point $M$. Line $PM$ intersects $AD$ and $BC$ at points $T$ and $S$, respectively. Then,
$$
\frac{2}{PM}=\frac{1}{PT}+\frac{1}{PS}
$$ | Proof: As shown in Figure 3, we have
$$
\begin{array}{l}
\frac{P T}{M T}=\frac{S_{\triangle P N}}{S_{\triangle M D}} \\
=\frac{S_{\triangle P N D}}{S_{\triangle P M}} \cdot \frac{S_{\triangle P M}}{S_{\triangle M W}} \\
=\frac{D B}{M B} \cdot \frac{P C}{D C} \\
=\frac{S_{\triangle B C D}}{S_{\triangle B C M}} \cdot \fr... | \frac{1}{P T}+\frac{1}{P S}=\frac{2}{P M} | Geometry | proof | Yes | Yes | cn_contest | false | 716,449 |
169 As shown in Figure 4, in the convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Draw any two lines through $O$ that intersect sides $AD$, $BC$, $AB$, and $CD$ at points $E$, $F$, $G$, and $H$ respectively. $GF$ and $EH$ intersect $BD$ at points $I$ and $J$ respectively. Prove:
$$
\frac... | Proof: Let $\angle A O B=\theta, \angle G O B=\alpha, \angle F O B=\beta$.
By the Angle Subtended Theorem, we have
$$
\begin{array}{l}
\frac{\sin \theta}{O G}=\frac{\sin \alpha}{O A}+\frac{\sin (\theta-\alpha)}{O B}, \\
\frac{\sin (\pi-\theta)}{O F}=\frac{\sin \beta}{O C}+\frac{\sin (\pi-\theta-\beta)}{O B}, \\
\frac{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,450 |
In $\triangle A B C$, prove:
$$
\begin{array}{l}
\frac{\sin A}{\sqrt{1-\sin B \cdot \sin C}}+\frac{\sin B}{\sqrt{1-\sin C \cdot \sin A}}+ \\
\frac{\sin C}{\sqrt{1-\sin A \cdot \sin B}} \leqslant 3 \sqrt{3} .
\end{array}
$$ | Proof: Law encountered
$$
\begin{array}{l}
\sin B \cdot \sin C=-\frac{1}{2}[\cos (B+C)-\cos (B-C)] \\
\leqslant-\frac{1}{2}[\cos (B+C)-1]=\frac{1}{2}(1+\cos A)
\end{array}
$$
Thus, $1-\sin B \cdot \sin C \geqslant 1-\frac{1}{2}(1+\cos A)$
$$
=\frac{1}{2}(1-\cos A)=\sin ^{2} \frac{A}{2} \text {. }
$$
Therefore, $\sqrt... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,451 |
Example 3 Try to find the integer solutions of the equation $x^{2}-23 x y^{2}+1989 y^{2}=0$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
The above text has been translated into English, preserving the original text's line breaks and format, as requested. | Obviously, $x=y=0$ is a solution to the equation, and if one of the unknowns is 0, then the other unknown is also 0.
Assume $x \neq 0$ and $y \neq 0$. Clearly, $y^{2} \mid x^{2}$, so we can let $x=t y, t \in \mathbf{Z}$, substituting into the equation we get
$$
t^{2}-23 y t+1989=0 \text {. }
$$
By Vieta's formulas, w... | (x, y)=(90,10), (2210,10), (90,-10), (2210,-10) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,452 |
Example 4 Find all integer solutions to the equation $\frac{x+y}{x^{2}-x y+y^{2}}=\frac{3}{7}$.
(12th All-Russian Mathematical Olympiad) | Notice that
$$
\begin{array}{l}
x^{2}-x y+y^{2} \\
=\frac{1}{4}\left[(x+y)^{2}+3(x-y)^{2}\right] .
\end{array}
$$
Let \( x+y=p, x-y=q \) (where \( p, q \in \mathbf{Z} \)), then the original equation becomes \( 28 p=3\left(p^{2}+3 q^{2}\right) \).
From this, we know that \( p>0 \), and \( p \) is a multiple of 3.
Let \... | (x, y)=(5,4),(4,5) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,453 |
Example 5 Does there exist a positive integer $m$ such that the equation
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a b c}=\frac{m}{a+b+c}
$$
has infinitely many positive integer solutions $(a, b, c)$ ?
(43rd IMO Shortlist) | Explanation: Existence.
If $a=b=c=1$, then $m=12$.
$$
\begin{array}{l}
\text { Let } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a b c}-\frac{12}{a+b+c} \\
=\frac{p(a, b, c)}{a b c(a+b+c)},
\end{array}
$$
where $p(a, b, c)=a^{2}(b+c)+b^{2}(c+a)+$ $c^{2}(a+b)+a+b+c-9 a b c$.
Assume $(x, a, b)$ is a solution to
$$
\beg... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,454 |
Example 6 Find all positive integers $a, b, c$ such that $1 < a < b < c$ and $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.
(33rd IMO) | First, estimate
$$
s=\frac{a b c-1}{(a-1)(b-1)(c-1)}\left({ }_{s} \in \mathbf{N}_{+}\right)
$$
the range.
Let $x=a-1, y=b-1, z=c-1$, then $1 \leqslant x$ $1, \\
s<\frac{(x+1)(y+1)(z+1)}{x y z} \\
=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) \\
\leqslant\left(1+\frac{1}{1}\right)\left... | (3,5,15) \text{ and }(2,4,8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,455 |
Example 7 A positive integer $n$ cannot be divisible by $2$ or $3$, and there do not exist non-negative integers $a$, $b$ such that $\left|2^{a}-3^{b}\right|=n$. Find the minimum value of $n$.
(2003, National Training Team Problem) | Explanation: When $n=1$, $\left|2^{1}-3^{1}\right|=1$;
When $n=5$, $12^{2}-3^{2} \mid=5$;
When $n=7$, $\left|2^{1}-3^{2}\right|=7$;
When $n=11$, $\left|2^{4}-3^{3}\right|=11$;
When $n=13$, $\left|2^{4}-3^{1}\right|=13$;
When $n=17$, $12^{6}-3^{4} \mid=17$;
When $n=19$, $12^{3}-3^{3} \mid=19$;
When $n=23$, $12^{5}-3^{2}... | 35 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,456 |
Example 3 In a mountain bike race held in a city, two cyclists, A and B, start from point A to point B at the same time. Cyclist A runs $\frac{1}{3}$ of the time at speeds $v_{1}, v_{2}, v_{3}$ respectively; Cyclist B runs $\frac{1}{3}$ of the distance at speeds $v_{1}, v_{2}, v_{3}$ respectively. Who will reach point ... | Explanation: Let the distance between $A$ and $B$ be $S$. The times taken by car A and car B to complete the journey are $t_{1}$ and $t_{2}$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{t_{1}}{3}\left(v_{1}+v_{2}+v_{3}\right)=S, \\
\frac{S}{3}\left(\frac{1}{v_{1}}+\frac{1}{v_{2}}+\f... | t_{1} \leqslant t_{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,457 |
Example 8 Given positive integers $a, b$, such that $a b+1$ divides $a^{2}+b^{2}$. Prove that $\frac{a^{2}+b^{2}}{a b+1}$ is the square of some positive integer.
(29th IMO) | Explanation: It is only necessary to prove that, when $k$ is not a perfect square, the indeterminate equation about $a$ and $b$
$$
a^{2}+b^{2}=k(a b+1)
$$
has no positive integer solutions $(a, b)$.
Assume that equation (1) has positive integer solutions ($k$ is not a perfect square), then we can select from all posit... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,458 |
1. Let $p$ be a given odd prime, and the positive integer $k$ makes $\sqrt{k^{2}-p k}$ also a positive integer. Then $k=$ $\qquad$
(2004, National High School Mathematics Competition) | (Prompt: Let $\sqrt{k^{2}-p k}=n, n \in \mathbf{N}_{+}$, then $k^{2}-p k-n^{2}$ $=0, k=\frac{p \pm \sqrt{p^{2}+4 n^{2}}}{2}$. Hence $p^{2}+4 n^{2}$ is a perfect square. Let it be $m^{2}, m \in \mathbf{N}_{+}$, it is easy to get $m=\frac{p^{2}+1}{2}, n=\frac{p^{2}-1}{4}$. Therefore, $k=\frac{(p+1)^{2}}{4}$. ) | \frac{(p+1)^{2}}{4} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,459 |
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying $2001 \mid \left(55^{n}+a \times 32^{n}\right)$.
（14th Irish Mathematical Olympiad） | (Given $2001=87 \times 23$, we get $87 \mid\left(55^{n}+a \times\right.$ $\left.32^{n}\right)$, and $231\left(55^{n}+a \times 32^{n}\right)$. Then $0 \equiv 55^{n}+a \times 32^{n} \equiv$ $(-32)^{n}+a \times 32^{n}=32^{n}(a-1)(\bmod 87)$. Hence $a-1=$ $0(\bmod 87)$. Similarly, we get $a+1 \equiv 0(\bmod 23)$. Let $a=87... | 436 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,460 |
3. Find all positive integers $n$ such that the equation
$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$
has positive integer solutions. | (Suppose $x \leqslant y \leqslant z$. Clearly, $z^{2} \mid \left(x^{3}+y^{3}\right)$, then $z^{2} \leqslant x^{3}+y^{3}$. It is easy to see that $z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geqslant n x^{2} y^{2}-(x+y)$. Therefore, $x^{3}+y^{3} \geqslant z^{2} \geqslant\left[n x^{2} y^{2}-(x+y)\right]^{2}$. Thus, we hav... | n=1 \text{ or } 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,461 |
4. Find all integers $a$ such that the equation
$$
x^{2}+a x y+y^{2}=1
$$
has infinitely many integer solutions $(x, y)$. Prove your conclusion. | (It is obvious that $a \neq 0$. Because $(x, y)$ is a solution to equation (1) $\Leftrightarrow$ $(x, -y)$ is a solution to the equation $x^{2} - a x y + y^{2} = 1$. Therefore, we only need to discuss $ay$, treating (1) as a quadratic equation in $y$. By Vieta's formulas, $(x, -a x - y)$ is also a solution to (1), and ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,462 |
5. Prove: The equation $x^{2}+y^{2}+z^{2}+u^{2}=2 x y z u$ has no positive integer solutions. | (Suppose $(x, y, z, u)$ is a positive integer solution to the original equation. By categorizing and discussing, it is easy to see that all four numbers are even. Let $x=2 x_{1}, y=2 y_{1}$, $z=2 z_{1}, u=2 u_{1}$, and substitute to get $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+u_{1}^{2}=$ $8 x_{1} y_{1} z_{1} u_{1}$. It is easy ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,463 |
Example 1 Find the minimum value of $\frac{\sin x}{2}+\frac{2}{\sin x}(x \in(0, \pi))$. | Solution: Since the derivative of the function $f(t)=\frac{t}{2}+\frac{2}{t}(t \in(0,1])$ is
$$
f^{\prime}(t)=\frac{1}{2}-\frac{2}{t^{2}}=\frac{t^{2}-4}{2 t^{2}}<0,
$$
thus, $f(t)$ is strictly decreasing.
Therefore, the minimum value of $f(t)$ is $f(1)=\frac{5}{2}$.
Hence, the minimum value of $\frac{\sin x}{2}+\frac{... | \frac{5}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,464 |
Example 2 Given $x, y, z \in \mathbf{R}_{+}$. Prove:
$$
\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)} \leqslant \frac{1}{5120} .
$$ | Solving for the function
$$
f(t)=\frac{t}{(a t+b)(a t+d)}, t \in \mathbf{R}_{+}, a, b, c, d > 0
$$
Taking the derivative, we get
$$
\begin{array}{l}
f^{\prime}(t)=p[(a t+b)(a t+d)-t(2 a c t+b c+a d)] \\
=p\left(b d-a c t^{2}\right) . \\
\text { where } p=(a t+b)^{-2}(c t+d)^{-2} \text { represents a positive quantity.... | \frac{1}{5120} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,465 |
Example 3 Given that $a, b, c$ are non-negative real numbers, and $a + b + c = 1$. Prove:
$$
\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \geqslant 2 \text {. }
$$ | Proof: Note that, when $a=1, b=c=0$, the inequality to be proved holds. Without loss of generality, assume $a \geqslant b \geqslant c$. Fix $b$, then $c=1-b-a$ is a function of $a$.
Consider the function of $a$
$$
f(a)=\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2},
$$
$f(a)$ is decreasing ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,466 |
Example 4 Given $a \geqslant b \geqslant c \geqslant 0$, and $a+b+c=$ 3. Prove: $a b^{2}+b c^{2}+c a^{2} \leqslant \frac{27}{8}$. | Prove: When $a=b=\frac{3}{2}, c=0$, the equality of the inequality holds. Fix $b$, then $c=3-b-a$.
Consider the function $f(a)=a b^{2}+b c^{2}+c a^{2}$, we have
$f(a)$ is decreasing $\Leftrightarrow f^{\prime}(a) \leqslant 0$
$$
\begin{array}{l}
\Leftrightarrow b^{2}-2 b c+2 a c-a^{2} \leqslant 0 \\
\Leftrightarrow(b-a... | \frac{27}{8} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,467 |
Example 4 As shown in Figure 1, quadrilateral $ABCD$ is a rectangle. Two people, A and B, start from points $A$ and $B$ respectively at the same time, and move counterclockwise along the rectangle. In which circle does B possibly catch up with A for the first time? In which circle does B certainly catch up with A at th... | Explanation: Let $A D=B C=a \mathrm{~m}, A B=C D=b . \mathrm{m}$, and suppose the first time Yi catches up with Jia is after
$$
\frac{2 a+b}{74-65}=\frac{2 a+b}{9}
$$
minutes. The distance Yi has run when he first catches up with Jia is
$$
\frac{2 a+b}{9} \times 74(\mathrm{~m}) \text {. }
$$
At this point, the number... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,468 |
Example 5 Given $x, y, z \in \mathbf{R}_{+} \cup\{0\}$, and $x+y+z=\frac{1}{2}$. Find
$$
\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1}
$$
the maximum value. | Solution: It is easy to guess that when $x=y=z=\frac{1}{6}$, equation (1) takes the maximum value $\frac{3}{5} \sqrt{\frac{3}{2}}$ (at least this is a "candidate" for the maximum value. Another candidate is when $x=\frac{1}{2}, y=z=0$, the function value is $\frac{1}{3} \sqrt{\frac{1}{2}}$. These special values are wor... | \frac{3}{5} \sqrt{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,469 |
Example 6 Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. Find
$$
\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1}
$$
the maximum value. | Solving: The difference between this problem and Example 5 is that the condition $x + y + z = \frac{1}{2}$ is changed to $x + y + z = 1$.
Assume without loss of generality that $x \geqslant \frac{1}{3} \geqslant z$. Fix $y$, then
$$
f(x) = \sum \frac{\sqrt{x}}{4 x + 1} \text{ is strictly decreasing} \Leftrightarrow f'(... | \frac{3 \sqrt{3}}{7} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,470 |
Example 7 Given $x, y, z \in \mathbf{R}_{+}$, and $xyz=1$. Prove:
$$
\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \leqslant \frac{3 \sqrt{2}}{2} .
$$ | Proof: Without loss of generality, let $x \geqslant y \geqslant z$.
When $y \leqslant 2$, fix $x$, then $z=\frac{1}{x y}$. The function
$$
\begin{array}{l}
f(y)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \text { is decreasing } \\
\Leftrightarrow f^{\prime}(y) \leqslant 0 \\
\Leftrightarrow-\frac{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,471 |
Example 8 Given $x, y, z \in \mathbf{R}_{+} \cup\{0\}$, and $x + y + z = 1$. Prove:
$$
\sum\left(\frac{20}{15-9 x^{2}}-9 x^{2}\right) \leqslant \frac{9}{7} .
$$ | Prove: When $x=y=z=\frac{1}{3}$, the equality holds.
$\sum\left(-9 x^{2}\right)$ reaches its maximum value when $x=y=z=\frac{1}{3}$, but $\sum \frac{20}{15-9 x^{2}}$ does not reach its maximum value when $x=y=z=\frac{1}{3}$. First, explain as follows:
Without loss of generality, assume $x \geqslant y \geqslant z$. Fix ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,472 |
Example 9 Given a positive integer $n$. Find the smallest positive number $\lambda$, such that for any $\theta_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2, \cdots, n)$, as long as
$$
\tan \theta_{1} \cdot \tan \theta_{2} \cdots \cdot \tan \theta_{n}=2^{\frac{n}{2}},
$$
then
$$
\cos \theta_{1}+\cos \theta_{2}+\cdots+\co... | When $n=1$,
$$
\cos \theta_{1}=\left(1+\tan ^{2} \theta_{1}\right)^{-\frac{1}{2}}=\frac{\sqrt{3}}{3}, \lambda=\frac{\sqrt{3}}{3} \text {. }
$$
Assume $n \geqslant 2$. Let $x_{i}=\tan ^{2} \theta_{i}(1 \leqslant i \leqslant n)$, then the given condition becomes
$$
x_{1} x_{2} \cdots x_{n}=2^{n} \text {. }
$$
We need t... | \lambda = \begin{cases} \frac{2}{\sqrt{3}}, & \text{if } n = 2 \\ n-1, & \text{if } n \geq 3 \end{cases} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,473 |
Example 1: Team A and Team B each send out 7 members to participate in a Go team tournament according to a pre-arranged order. Both sides start with the No. 1 member competing, the loser is eliminated; the winner then competes with the No. 2 member of the losing side, ... until all members of one side are eliminated, a... | Solution: Since the losing side is completely eliminated, the total number of eliminations is 7. Let's assume side A wins, and let the $i$-th member of side A eliminate $x_{i}$ members of side B. Then the problem is equivalent to finding the number of non-negative integer solutions to the equation
$$
x_{1}+x_{2}+\cdots... | 3432 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,474 |
Example 2 From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions? | Solution: Notice that
$$
a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(14-a_{3}\right)=14 \text {, }
$$
where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-a_{3}$ $\geqslant 0$, i.e.,
$$
\begin{array}{l}
a_{1}-1+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+ \\
\left(1... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,475 |
Example 3 Put 1996 identical balls into 10 different boxes, so that the $i$-th box contains at least $i$ balls. How many different ways are there to do this? | Solution: Let the $i$-th box contain $x_{i}\left(x_{i} \geqslant i, i=1\right.$, $2, \cdots, 10)$ balls, then
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{10}=1996 . \\
\text { Let } x_{i}^{\prime}=x_{i}-i, \text { then } \\
x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{10}^{\prime}=1941\left(x_{i}^{\prime} \geqslant 0\right) .... | \mathrm{C}_{1950}^{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,476 |
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