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Example 6 Let $a, b, c, d$ be positive real numbers, $a < b$ and $c < d$. There is a triangle with side lengths $\sqrt{a^{2}+c^{2}}, \sqrt{b^{2}+d^{2}}, \sqrt{(b-a)^{2}+(d-c)^{2}}$. Then the area of this triangle is $\qquad$
(12th Five Sheep Cup Junior High School Mathematics Competition) | Solution: As shown in Figure 3, construct rectangle $ABCD$ such that $AB = b - a$, $AD = c$. Extend $DA$ to point $E$ such that $DE = d$, and extend $DC$ to point $F$ such that $DF = b$. Connect $EF$ and $FB$. Then,
$$
\begin{array}{l}
BF = \sqrt{a^2 + c^2}, \quad EF = \sqrt{b^2 + d^2}, \\
BE = \sqrt{(b-a)^2 + (d-c)^2}... | \frac{1}{2}(bc - ad) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,539 |
6. Given that $x, y, z$ are three non-negative real numbers, satisfying
$$
3 x+2 y+z=5, x+y-z=2 \text{. }
$$
If $S=2 x+y-z$, then the sum of the maximum and minimum values of $S$ is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 6. A.
From $\left\{\begin{array}{l}3 x+2 y+z=5 \\ x+y-z=2, \\ 2 x+y-z=S,\end{array}\right.$ we get $\left\{\begin{array}{l}x=S-2 \geqslant 0, \\ y=5-\frac{4}{3} S \geqslant 0, \\ z=1-\frac{1}{3} S \geqslant 0 .\end{array}\right.$ Therefore, $2 \leqslant S \leqslant 3$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,541 |
1. Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ . | ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,542 |
2. In a convex quadrilateral $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$. If $S_{\triangle O A D}=4, S_{\triangle O B C}=9$, then the minimum value of the area of the convex quadrilateral $A B C D$ is | 2. 25 .
Let $S_{\triangle A O B}=x, S_{\triangle C O D}=y$. Then
$$
\frac{4}{x}=\frac{O D}{O B}=\frac{y}{9} \Rightarrow x y=36 \text {. }
$$
Thus, $x+y \geqslant 2 \sqrt{x y}=12$.
Equality holds if and only if $x=y=6$.
Therefore, $S_{\text {quadrilateral } A B C D}=4+9+x+y \geqslant 25$.
Equality holds if and only if... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,543 |
3. Given real numbers $x, y$ satisfy $x^{2}-2 x-4 y=5$, let $t=x-2 y$. Then the range of $t$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $t \leqslant \frac{9}{2}$.
From $t=x-2 y \Rightarrow y=\frac{1}{2}(x-t)$. Then $x^{2}-4 x+2 t-5=0$.
Since $x$ is a real number,
$$
\Delta=36-8 t \geqslant 0 \Rightarrow t \leqslant \frac{9}{2} \text {. }
$$ | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,544 |
4. As shown in Figure 3, given that $\triangle A B C$ is inscribed in $\odot O$, and $A B=A C$, diameter $A D$ intersects $B C$ at point $E, F$ is the midpoint of $O E$. If $B D / / C F, B C=2 \sqrt{5}$, then the length of line segment $C D$ is $\qquad$ | 4. $\sqrt{6}$.
Since $A B=A C, A D$ is the diameter, so $C E=B E=\sqrt{5}$.
Also, $B D / / C F$, which gives $\triangle B D E \cong \triangle C F E$. Therefore, $D E=E F$. Since $F$ is the midpoint of $O E$, then $O F=F E=E D$.
Let $O F=x$. Then $O A=O D=3 x, A E=5 x$. By the intersecting chords theorem, we have $C E ... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,545 |
Three, (20 points) Given that $a, b, c, d$ are different real numbers, and $a, c$ are the roots of the equation $x^{2} + a x - b = 0$, $b, d$ are the roots of the equation $x^{2} + c x + d = 0$. Find the values of $a, b, c, d$.
---
Please note that the translation retains the original formatting and structure of the ... | $$
\begin{array}{l}
a+c=-a, a c=-b ; \\
b+d=-c, b d=d .
\end{array}
$$
(1) If $d \neq 0$, from $b d=d$, we know $b=1$.
From $a+c=-a$, we know $c=-2 a$.
From $a c=-b$, we know $-2 a^{2}=-1$, thus $a= \pm \frac{\sqrt{2}}{2}$.
When $a= \pm \frac{\sqrt{2}}{2}$, $c=\mp \sqrt{2}$, we get
$$
\begin{array}{l}
d=-c-b=\sqrt{2}-... | (a, b, c, d) = \left(\frac{\sqrt{2}}{2}, 1, -\sqrt{2}, \sqrt{2} - 1\right), \left(-\frac{\sqrt{2}}{2}, 1, \sqrt{2}, -\sqrt{2} - 1\right), (1, 2, -2, 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,546 |
Four. (25 points) As shown in Figure 4, quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is inscribed in a circle, and the incenters of $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{2} A_{3} A_{4}$, and $\triangle A_{3} A_{4} A_{1}$ are $I_{1}$, $I_{2}$, and $I_{3}$, respectively. Prove:
(1) $A_{2}$, $I_{1}$, $I_{2}$, and $A_{3}... | (1) As shown in Figure 6, connect $I_{1} A_{1}$, $I_{1} A_{2}$, $I_{1} A_{3}$, $I_{2} A_{2}$, and $I_{2} A_{3}$. Since $I_{1}$ is the incenter of $\triangle A_{1} A_{2} A_{3}$, we have:
$$
\begin{array}{l}
\angle I_{1} A_{1} A_{2}=\angle I_{1} A_{1} A_{3}=\frac{1}{2} \angle A_{2} A_{1} A_{3}, \\
\angle I_{1} A_{2} A_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,547 |
Five. (25 points) As shown in Figure 5, place 3 identical coins sequentially into a $4 \times 4$ square grid (each square can only hold 1 coin). Find the probability that any two of the 3 coins are not in the same row or column.
Place the above text into the requested format:
Five. (25 points) As shown in Figure 5, p... | First, calculate the total number of ways $N$.
The first coin can be placed in 16 cells, which gives 16 ways; the second coin can be placed in the remaining 15 cells, which gives 15 ways; the third coin can be placed in the remaining 14 cells, which gives 14 ways. Therefore, the total number of ways is
$$
N=16 \times 1... | \frac{6}{35} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,548 |
1. Define the set operation:
$$
A \otimes B=\{z \mid z=x y, x \in A, y \in B\} \text {. }
$$
Let $A=\{2,0\}, B=\{0,8\}$. Then the sum of all elements in the set $A \otimes B$ is ( ).
(A) 16
(B) 18
(C) 20
(D) 22 | -1. A.
The elements of the set $A \otimes B$ are
$$
\begin{array}{l}
z_{1}=2 \times 0=0, z_{2}=2 \times 8=16, \\
z_{3}=0 \times 0=0, z_{4}=0 \times 8=0 .
\end{array}
$$
Therefore, the sum of all elements in the set $A \otimes B$ is 16. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,549 |
Example 7 Given that $a, b, c, d, e$ are all real numbers, and
$$
a+b+c+d+e=4, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=4 \text {. }
$$
Find the maximum value of the real number $a$. | For the arrays $b, c, d, e$, we have
$$
\begin{aligned}
\bar{x} & =\frac{b+c+d+e}{4}=\frac{4-a}{4}, \\
S^{2} & =\frac{1}{4}\left[\left(b^{2}+c^{2}+d^{2}+e^{2}\right)-4 \bar{x}^{2}\right] \\
& =\frac{1}{4}\left[4-a^{2}-4\left(\frac{4-a}{4}\right)^{2}\right] \geqslant 0 .
\end{aligned}
$$
Simplifying, we get $a(5a-8) \l... | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,550 |
2. Given $\left\{a_{n}\right\}$ is a geometric sequence, $a_{2}=2, a_{5}=\frac{1}{4}$. Then the range of $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}\left(n \in \mathbf{N}_{+}\right)$ is ( ).
(A) $[12,16)$
(B) $[8,16)$
(C) $\left[8, \frac{32}{3}\right)$
(D) $\left[\frac{16}{3}, \frac{32}{3}\right)$ | 2. C.
Let the common ratio of $\left\{a_{n}\right\}$ be $q$. Then $q^{3}=\frac{a_{5}}{a_{2}}=\frac{1}{8}$. Therefore, $q=\frac{1}{2}$. Hence, the sequence $\left\{a_{n} a_{n+1}\right\}$ is a geometric sequence with the first term $a_{1} a_{2}=8$ and the common ratio $q^{2}=\frac{1}{4}$. Thus,
$$
a_{1} a_{2}+a_{2} a_{3... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,551 |
3.5 Names of volunteers randomly enter three different Olympic venues to participate in reception work. Then the probability that each venue has at least one volunteer is ( ).
(A) $\frac{3}{5}$
(B) $\frac{1}{15}$
(C) $\frac{5}{8}$
(D) $\frac{50}{81}$ | 3. D.
The number of ways for 5 volunteers to randomly enter three different Olympic venues is $3^{5}=243$. The scenario where each venue has at least one volunteer can be divided into two categories:
The first category, 3 people go to one venue, and the remaining two venues each have 1 person, the number of ways for t... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 721,552 |
4. Given that $a, b$ are non-zero and non-collinear vectors, let condition $M: \boldsymbol{b} \perp (\boldsymbol{a}-\boldsymbol{b})$;
condition $N$: for all $x \in \mathbf{R}$, the inequality $|\boldsymbol{a}-x \boldsymbol{b}| \geqslant$ $|\boldsymbol{a}-\boldsymbol{b}|$ always holds.
Then $M$ is ( ) of $N$.
(A) a nece... | 4. C.
Let $\overrightarrow{O A}=\boldsymbol{a}, \overrightarrow{O B}=\boldsymbol{b}$. Then $x \boldsymbol{b}$ represents any vector collinear with $\overrightarrow{O B}$, $|\boldsymbol{a}-x \boldsymbol{b}|$ represents the distance $A C$ from point $A$ to any point $C$ on the line $O B$, and $|\boldsymbol{a}-\boldsymbo... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,553 |
5. Let the function $f(x)$ be defined on $\mathbf{R}$, and consider the following three statements:
(1) The graph of a function satisfying $f(x+2)+f(2-x)=4$ is symmetric about the point $(2,2)$;
(2) The graph of a function satisfying $f(x+2)=f(2-x)$ is symmetric about the line $x=2$;
(3) The graphs of the functions $f(... | 5. D.
Substitute $x-2$ for $x$ in $f(x+2)+f(2-x)=4$, we get $f(x)+f(4-x)=4$.
If the point $(x, y)$ is on the graph of $y=f(x)$, then $4-y=f(4-x)$, which means the point $(x, y)$'s symmetric point about $(2,2)$, $(4-x, 4-y)$, is also on the graph of $y=f(x)$. Conversely, this is also true, so proposition (1) is a true... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,554 |
6. The line segment connecting two points on a sphere is called a chord of the sphere.
For a sphere with a radius of 4, the lengths of two chords \( AB \) and \( CD \) are \( 2 \sqrt{7} \) and \( 4 \sqrt{3} \) respectively. \( M \) and \( N \) are the midpoints of \( AB \) and \( CD \) respectively, and the endpoints ... | 6. A.
Assume $AB$ and $CD$ intersect at point $N$, then $AB$ and $CD$ are coplanar, so points $A, B, C, D$ are concyclic. Clearly, in this circle, the length of chord $AB$ passing through point $N$ is greater than or equal to the length of chord $CD$ with $N$ as its midpoint, so proposition (2) is incorrect.
It is ea... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,555 |
7. Let $a=\sin \left(\sin 2008^{\circ}\right), b=\sin \left(\cos 2008^{\circ}\right)$, $c=\cos \left(\sin 2008^{\circ}\right), d=\cos \left(\cos 2008^{\circ}\right)$. Then the size relationship of $a, b, c, d$ is ( ).
(A) $a<b<c<d$
(B) $b<a<d<c$
(C) $c<d<b<a$
(D) $d<c<a<b$ | 7. B.
Since $2008^{\circ}=5 \times 360^{\circ}+180^{\circ}+28^{\circ}$, therefore,
$$
\begin{array}{l}
a=\sin \left(-\sin 28^{\circ}\right)=-\sin \left(\sin 28^{\circ}\right)0, \\
d=\cos \left(-\cos 28^{\circ}\right)=\cos \left(\cos 28^{\circ}\right)>0 .
\end{array}
$$
$$
\text { Also } \sin 28^{\circ}<\cos 28^{\circ}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,556 |
8. Let the function $f(x)=x^{3}+3 x^{2}+6 x+14$, and $f(a)=1, f(b)=19$. Then $a+b=(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -2 | 8. D.
$$
\begin{array}{l}
\text { Given } f(x)=x^{3}+3 x^{2}+6 x+14 \\
=(x+1)^{3}+3(x+1)+10,
\end{array}
$$
Let $g(y)=y^{3}+3 y$. Then $g(y)$ is an odd function and strictly increasing.
And $f(a)=(a+1)^{3}+3(a+1)+10=1$,
$$
f(b)=(b+1)^{3}+3(b+1)+10=19 \text {, }
$$
Then $g(a+1)=-9, g(b+1)=9$,
$$
g(-b-1)=-9 \text {. }
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,557 |
10. Given the set $\Omega=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 2008\right\}$. If points $P(x, y)$ and $P^{\prime}\left(x^{\prime}, y^{\prime}\right)$ satisfy $x \leqslant x^{\prime}$ and $y \geqslant y^{\prime}$, then point $P$ is said to dominate $P^{\prime}$. If a point $Q$ in the set $\Omega$ satisfies: there do... | 10. $\left\{(x, y) \mid x^{2}+y^{2}=2008, x \leqslant 0\right.$ and $\left.y \geqslant 0\right\}$.
$P$ is better than $P^{\prime}$, meaning $P$ is located to the left and above $P^{\prime}$. “There does not exist any other point in $\Omega$ that is better than $Q$”, means “there is no point in $\Omega$ to the left and ... | \left\{(x, y) \mid x^{2}+y^{2}=2008, x \leqslant 0\right. \text{ and } \left.y \geqslant 0\right\} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,559 |
11. In the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, the coefficient of $x^{150}$ is $\qquad$ (answer with a number). | 11.7651 .
By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$
that do not exceed 100. Clearly, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural nu... | 7651 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,560 |
Example 8 Find the maximum value of the function $y=\sqrt{5-2 x}+\sqrt{3+2 x}$.
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。
This part is not translated as it contains instructions for the translation task itself. Here is the requested translation above. | Solution: From the given condition, we know $y>0$.
Notice that the variance of the two numbers $\sqrt{5-2 x}$ and $\sqrt{3+2 x}$ is
$$
\begin{array}{l}
S^{2}=\frac{1}{2}\left[(\sqrt{5-2 x})^{2}+(\sqrt{3+2 x})^{2}-2\left(\frac{y}{2}\right)^{2}\right] \\
\geqslant 0 .
\end{array}
$$
Solving this, we get $16-y^{2} \geqsl... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,561 |
12. The base of a hexagonal prism is a regular hexagon, and its lateral edges are perpendicular to the base. It is known that all vertices of the hexagonal prism lie on the same sphere, and the volume of the hexagonal prism is $\frac{9}{8}$, with the perimeter of the base being 3. Then the volume of this sphere is $\qq... | 12. $\frac{4}{3} \pi$.
Since the base circumference is 3, the base side length is $\frac{1}{2}$, and the base area is $S=\frac{3}{8} \sqrt{3}$.
Given the volume is $\frac{9}{8}$, the height is $\sqrt{3}$. The diameter of the sphere is $\sqrt{1^{2}+(\sqrt{3})^{2}}=2$, and the volume of the sphere $V=\frac{4}{3} \pi R^... | \frac{4}{3} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,562 |
13. In a $4 \times 4$ chessboard, 8 small squares are to be colored black such that each row and each column has exactly two black squares. Then there are $\qquad$ different ways to do this (answer with a number). | 13. 90 .
The first row can be colored with 2 black cells in $\mathrm{C}_{4}^{2}$ ways. After the first row is colored, there are the following three scenarios:
(1) The black cells in the second row are in the same columns as those in the first row, in which case, the remaining rows have only one way to be colored;
(2)... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,563 |
14. An extracurricular activity group at a school designed a tree planting plan on graph paper for a certain desert as follows: The $k$-th tree is planted at point $P_{k}\left(x_{k}, y_{k}\right)$, where $x_{1}=1, y_{1}=1$. When $k \geqslant 2$,
$$
\left\{\begin{array}{l}
x_{k}=x_{k-1}+1-5\left[\frac{k-1}{5}\right]+5\l... | 14. $(3,402)$.
Let $f(k)=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right]$. Then
$$
\begin{array}{l}
f(k+5)=\left[\frac{k+5-1}{5}\right]-\left[\frac{k+5-2}{5}\right] \\
=\left[1+\frac{k-1}{5}\right]-\left[1+\frac{k-2}{5}\right] \\
=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right]=f(k) .
\end{array}
$$
There... | (3,402) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,564 |
15. (14 points) Let real numbers $a, b \in [\alpha, \beta]$, where $\alpha, \beta$ are positive real numbers. Prove that: $\frac{b}{a} + \frac{a}{b} \leq \frac{\beta}{\alpha} + \frac{\alpha}{\beta}$, and the equality holds if and only if $a = \alpha, b = \beta$ or $a = \beta, b = \alpha$. | Three, 15. By symmetry, without loss of generality, let $a \leqslant b$, and let $\frac{a}{b}=t$.
Since $\alpha \leqslant a \leqslant b \leqslant \beta$, we have
$\frac{\alpha}{\beta} \leqslant t=\frac{a}{b} \leqslant \frac{\beta}{\alpha}$.
Let $f(t)=t+\frac{1}{t}\left(\frac{\alpha}{\beta} \leqslant t \leqslant \frac{\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,565 |
16. (14 points) Player A and Player B are having a table tennis match, using a best-of-five format (i.e., the first to win three games wins the championship). For each game, the probability of Player A winning is $\frac{2}{3}$, and the probability of Player B winning is $\frac{1}{3}$. If the event of "Player B winning ... | 16. If one side wins with a score of 3:1 or 3:0, the remaining game is played but does not affect the match result. Therefore, the problem is transformed into: finding the probability that "Party B wins at least three out of five games."
The probability of Party B winning all five games is $\left(\frac{1}{3}\right)^{5}... | \frac{17}{81} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,566 |
17. (16 points) Given the function $f(x)=\ln (1+x)-x$ on the interval $[0, n]\left(n \in \mathbf{N}_{+}\right)$, the minimum value is $b_{n}$. Let $a_{n}=\ln (1+n)-b_{n}, p_{k}=\frac{a_{1} a_{3} \cdots a_{2 k-1}}{a_{2} a_{4} \cdots a_{2 k}}\left(k \in \mathbf{N}_{+}\right)$. Prove: $p_{1}+p_{2}+\cdots+p_{n}<\sqrt{2 a_{... | 17. It is known that the domain of the function is $(-1,+\infty)$.
Also, $f^{\prime}(x)=\frac{1}{1+x}-1=-\frac{x}{1+x}$.
When $x \in[0, n]$, $f^{\prime}(x)<0$, which means $f(x)$ is a decreasing function on $[0, n]\left(n \in \mathbf{N}_{+}\right)$, hence
$$
\begin{array}{l}
b_{n}=f(n)=\ln (1+n)-n, \\
a_{n}=\ln (1+n)-... | \sqrt{2 a_{n}+1}-1 | Calculus | proof | Yes | Yes | cn_contest | false | 721,567 |
18. (18 points) Through a point $P$ on the line $l: 5 x-7 y-70=0$, draw the tangents $P M$ and $P N$ to the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, with the points of tangency being $M$ and $N$, respectively. Connect $M N$.
(1) When point $P$ moves along the line $l$, prove that the line $M N$ always passes throu... | 18. Let \( P\left(x_{0}, y_{0}\right) \), \( M\left(x_{1}, y_{1}\right) \), and \( N\left(x_{2}, y_{2}\right) \). Then the equations of the tangent lines to the ellipse at points \( M \) and \( N \) are
$$
\frac{x_{1} x}{25}+\frac{y_{1} y}{9}=1, \frac{x_{2} x}{25}+\frac{y_{2} y}{9}=1 \text {. }
$$
Since both tangent l... | \left(\frac{25}{14},-\frac{9}{10}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 721,568 |
1. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$. If $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{2}}$ where $(a, b, c$ are positive integers). Then $a+b+c=(\quad)$.
(A) 8
(B) 32
(C) 48
(D) 50 | -、1. D.
From the given condition,
$(\sin x-\cos x)^{2}=\left(\frac{\pi}{4}\right)^{2}$.
Therefore, $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Hence, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$ $=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}}$.
Thus, $a+b+c=32+16+2=50$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,569 |
2. Given a regular triangular prism with a base edge length of 1, and the diagonals of the two non-adjacent side faces are perpendicular to each other. Then the length of its lateral edge is ( ).
(A) $\sqrt{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 2
(D) $\frac{\sqrt{3}}{2}$ | 2. B.
Let $A B C-A_{1} B_{1} C_{1}$ be a regular triangular prism, with the side edge length being $a$, and the side diagonals $A B_{1}$ and $B C_{1}$ being perpendicular to each other. Then
$$
\begin{array}{l}
\overrightarrow{B_{1} A} \cdot \overrightarrow{B C_{1}}=0 \\
\Rightarrow\left(\overrightarrow{B_{1} B}+\over... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,570 |
Example 9 Given that $a, b$ are positive numbers, and the parabolas $y=x^{2}+a x+2 b$ and $y=x^{2}+2 b x+a$ both have common points with the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
$(2000$, National Junior High School Mathematics League) | Solution: From the given conditions, we have
$$
a^{2}-8 b \geqslant 0,4 b^{2}-4 a \geqslant 0 \text {. }
$$
Thus, $a^{4} \geqslant 64 b^{2} \geqslant 64 a$, which means $a \geqslant 4$.
Furthermore, $b^{2} \geqslant a \geqslant 4$. Therefore, $a^{2}+b^{2} \geqslant 20$.
Moreover, when $a=4, b=2$, the parabolas $y=x^{2... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,572 |
4. The number of real solutions to the equation $\left|x^{2}-3 x+2\right|+\left|x^{2}+2 x-3\right|=11$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 4. C.
The original equation is $|x-1|(|x-2|+|x+3|)=11$.
By discussing four cases: $x \leqslant -3$, $-3 < x \leqslant 2$, $2 < x \leqslant 1$, and $x > 1$, we know that there are 2 real solutions that satisfy the equation. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,573 |
5. Given the function $f(x)$ satisfies for all real numbers $x$, $y$,
$$
f(x)+f(2 x+y)+5 x y=f(3 x-y)+2 x^{2}+1 \text {. }
$$
Then the value of $f(10)$ is ( ).
(A) -49
(B) -1
(C) 0
(D) 25 | 5. A.
Let $x=10, y=5$, we get $f(10)+f(25)+250=f(25)+200+1$. Therefore, $f(10)=-49$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,574 |
6. Given a real number $a$ such that there is only one real number $x$ satisfying the inequality $\left|x^{2}+2 a x+3 a\right| \leqslant 2$. Then the number of all real numbers $a$ that satisfy the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) infinitely many | 6. B.
To make the inequality $\left|x^{2}+2 a x+3 a\right| \leqslant 2$ have only one solution, the graph of the parabola $f(x)=x^{2}+2 a x+3 a$ must be tangent to the line $y=2$.
Therefore, the equation $x^{2}+2 a x+3 a=2$, i.e., $x^{2}+2 a x+$ $3 a-2=0$ must have a discriminant
$$
\Delta=4 a^{2}-4(3 a-2)=0 .
$$
So... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 721,575 |
7. If the point with the abscissa $\frac{3}{2} a$ on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is farther from the right focus than from the left directrix, then the range of the acute angle between the two asymptotes of the hyperbola is $\qquad$. | II. 7. $\left(0^{\circ}, 60^{\circ}\right)$.
For a point on the hyperbola with the abscissa $\frac{3}{2} a$, the distance to the right focus is $e\left|\frac{3}{2} a-\frac{a^{2}}{c}\right|$, and the distance to the left directrix is $\left|\frac{3}{2} a-\left(-\frac{a^{2}}{c}\right)\right|$.
From the condition, we know... | \left(0^{\circ}, 60^{\circ}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,576 |
8. Let the polynomial be
$$
\begin{aligned}
P(x)= & x^{15}-2008 x^{14}+2008 x^{13}-2008 x^{12}+ \\
& 2008 x^{11}-\cdots+2008 x^{3}-2008 x^{2}+2008 x .
\end{aligned}
$$
Then \( P(2007) = \) | 8. 2007 .
$$
\begin{array}{c}
P(x)=(x-2007)\left(x^{14}-x^{13}+x^{12}- \\
x^{11}+\cdots+x^{2}-x\right)+x .
\end{array}
$$
Therefore, $P(2007)=2007$. | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,577 |
9. Real numbers $x, y$ satisfy $\left\{\begin{array}{l}x+\sin y=2008, \\ x+2008 \cos y=2007\end{array}\right.$ $\left(0 \leqslant y \leqslant \frac{\pi}{2}\right)$. Then $x+y=$ $\qquad$ | 9. $2007+\frac{\pi}{2}$.
Subtracting the two equations gives $\sin y=1+2008 \cos y$. Given $0 \leqslant y \leqslant \frac{\pi}{2}$, we know $1+2008 \cos y \geqslant 1$. Therefore, it can only be that $\sin y=1, \cos y=0$. Hence, $y=\frac{\pi}{2}$. Consequently, $x=2007$.
Therefore, $x+y=2007+\frac{\pi}{2}$. | 2007+\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,578 |
10. In the Cartesian coordinate system, let point $A(0,4)$, $B(3,8)$. If point $P(x, 0)$ makes $\angle A P B$ maximum, then $x=$ $\qquad$ | 10. $5 \sqrt{2}-3$.
Let $\angle A P B=\theta$.
It is easy to know that when the circumcircle of $\triangle A B P$ is tangent to the $x$-axis, $\angle A P B$ is maximized.
Extend $B A$ to intersect the $x$-axis at point $Q$. Then $Q(-3,0)$. By the power of a point theorem, we have $Q P^{2}=Q A \cdot Q B=5 \times 10$. T... | 5 \sqrt{2}-3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,579 |
11. Prime numbers $p$, $q$, $r$ satisfy $p+q=r$, and $(r-p) \cdot$ $(q-p)-27 p$ is a perfect square. Then all the triples $(p, q, r)=$ $\qquad$ | 11. $(2,29,31)$.
From the problem, we know $p<q<r$, combined with $p+q=r$, we know $p=$ 2. Therefore, $q=r-2$. Thus,
$$
(r-p)(q-p)-27 p=(q-1)^{2}-55 \text {. }
$$
Let $(q-1)^{2}-55=n^{2}$ (where $n$ is a positive integer). Then
$$
(q-1-n)(q-1+n)=55 \text {. }
$$
Solving this, we get $(q, n)=(29,27)$ or $(9,3)$ (disc... | (2,29,31) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,580 |
12. For a positive integer $n \leqslant 500$, it has the property: when an element $m$ is randomly selected from the set $\{1,2, \cdots, 500\}$, the probability that $m \mid n$ is $\frac{1}{100}$. Then the maximum value of $n$ is $\qquad$ | 12. 81 .
From the problem, we know that $n$ has exactly 5 positive divisors. Let the prime factorization of $n$ be $n=p_{1}^{\alpha_{1}} p_{1}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$.
Then the number of positive divisors of $n$ is
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)=5 ... | 81 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,581 |
13. Given the parabola $C: y=a x^{2}(a>0)$, the line $y=x+2$ intersects the parabola at points $A$ and $B$, and $M$ is the midpoint of segment $AB$. A vertical line through $M$ intersects the parabola $C$ at point $N$.
(1) Prove: The tangent line $l$ to the parabola $C$ at point $N$ is parallel to $AB$.
(2) Does there ... | 13. (1) From $\left\{\begin{array}{l}y=x+2, \\ y=a x^{2}\end{array}\right.$ we get $a x^{2}-x-2=0$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then $x_{1}+x_{2}=\frac{1}{a}, x_{1} x_{2}=-\frac{2}{a}$.
Thus, $x_{N}=x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{1}{2 a}, y_{N}=a x_{N}^{2}=\frac{1}{4 a}$.
From ... | a=\frac{7}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,582 |
Example 10 When $a, b$ take what real numbers, the equation
$$
x^{2}+2(1+a) x+\left(3 a^{2}+4 a b+4 b^{2}+2\right)=0
$$
has real roots?
(1987, National Junior High School Mathematics League) | Solution: When $\Delta \geqslant 0$, the equation has real roots
$$
\begin{array}{l}
\Leftrightarrow[2(1+a)]^{2}-4\left(3 a^{2}+4 a b+4 b^{2}+2\right) \geqslant 0 \\
\Leftrightarrow 2 a^{2}+4 a b+4 b^{2}-2 a+1 \leqslant 0 \\
\Leftrightarrow(a+2 b)^{2}+(a-1)^{2} \leqslant 0 . \\
\text { Also, }(a+2 b)^{2}+(a-1)^{2} \geq... | a=1, b=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,583 |
14. As shown in Figure 1, given that the circumradius $R$ of the acute triangle $\triangle ABC$ is $1$, $\angle BAC = 60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively. The line segment $OH$ intersects the extension of $BC$ at point $P$. Find:
(1) The area of the concave ... | 14. (1) As shown in Figure 2, connect $A H$, and draw $O D \perp B C$ at point D.
Since $O$ is the
circumcenter of
$\triangle A B C$ and $\angle B A C$ $=60^{\circ}$, we have
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
O D=O C \cos 60^{\circ}=\frac{1}{2} .
\end{array}
$$
By the property of ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,584 |
15. Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$. | 15. When $a=b=c=1$, we can get $k \geqslant 2$.
Below is the proof: the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,585 |
16. Let $x_{1}, x_{2}, \cdots$ be distinct positive real numbers. Prove: $x_{1}, x_{2}, \cdots$ is a geometric sequence if and only if for all integers $n(n \geqslant 2)$, we have
$$
\frac{x_{1}}{x_{2}} \sum_{k=1}^{n-1} \frac{x_{n}^{2}}{x_{k} x_{k+1}}=\frac{x_{n}^{2}-x_{1}^{2}}{x_{2}^{2}-x_{1}^{2}} .
$$ | 16. Necessity. If $x_{1}, x_{2}, \cdots$ is a geometric sequence, set $x_{k}=a r^{k-1}$. Then
$$
\begin{array}{l}
\frac{x_{1}}{x_{2}} \sum_{k=1}^{n-1} \frac{x_{n}^{2}}{x_{k} x_{k+1}}=\frac{r^{2(n-1)}}{r} \sum_{k=1}^{n-1} \frac{1}{r^{2 k-1}}=\sum_{k=0}^{n-2} r^{2 k} \\
=\frac{r^{2(n-1)}-1}{r^{2}-1}=\frac{x_{n}^{2}-x_{1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,586 |
1. Add a number to $\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\frac{1}{210}+\frac{1}{336}$ so that the sum of the seven numbers equals $\frac{1}{4}$. What is the number to be added?
(A) $\frac{1}{56}$
(B) $\frac{1}{84}$
(C) $\frac{1}{112}$
(D) $\frac{1}{224}$ | -1. C.
Add this number
$$
\begin{aligned}
= & \frac{1}{4}-\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\frac{1}{210}+\frac{1}{336}\right) \\
= & \frac{1}{4}-\left(\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\right. \\
& \left.\frac{1}{4 \times 5 \times 6}+\frac... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,587 |
2. The graph of the quadratic function $y=a x^{2}+$ $b x+c$ is shown in Figure 1. Which of the following conclusions is correct? ( ).
(A) $3|a|+|c|>2|b|$
(B) $3|a|+|c|=2|b|$
(C) $3|a|+|c|<2|b|$
(D) $3|a|+|c| \leqslant 2|b|$ | 2. C.
From the graph, we know $a0, c0$, i.e.,
$$
a+b+c>0 \text {. }
$$
Since $-\frac{b}{2 a}>1$, we have,
$$
2 a+b>0 \text {. }
$$
From equations (1) and (2), we get $3 a+2 b+c>0$.
Then $3|a|+|c|-2|b|=-3 a-c-2 b$
$$
=-(3 a+2 b+c)<0 \text {. }
$$
Therefore, $3|a|+|c|<2|b|$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,588 |
3. As shown in Figure $2, \odot O$'s diameter $A B$ intersects with the line of chord $C D$ (not a diameter) at a point $P$ outside the circle, $P A=A B=4$. Then the range of $P D$ is ( ).
(A) $4<P D<8$
(B) $4 \sqrt{2}<P D<8$
(C) $4<P D<4 \sqrt{2}$
(D) $4 \sqrt{2} \leqslant P D \leqslant 8$ | 3. B.
As shown in Figure 5, connect $O D$. In $\triangle P O D$, we have $P O - O D \angle P E C = \angle E D C$.
Therefore, $P D > P E$, which means $P D > 4 \sqrt{2}$.
In conclusion, $4 \sqrt{2} < P D < 8$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,589 |
4. The root situation of the equation $x^{3}+x^{2}+x-2=0$ is ( ).
(A) There is one positive real root
(B) There are two positive real roots
(C) There is one negative real root
(D) There are two negative real roots | 4. A.
Obviously, $x \neq 0$.
Thus, $\frac{x^{3}+x^{2}+x-2}{x}=0$, which means $x^{2}+x+1=\frac{2}{x}$. Let $y_{1}=x^{2}+x+1$, $y_{2}=\frac{2}{x}$. As shown in Figure 6, plot the graphs of the two functions. From the graph, we can see that the two function graphs intersect at only one point and it is in the first quadr... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,590 |
5. If prime numbers $p, q$ satisfy $p^{2}-5 q-36=0$, then the median of the sample $p, q, 9, 16$ is $(\quad$.
(A) 16
(B) 13
(C) 11
(D) 13.5 | 5. D.
From $p^{2}-5 q-36=0$ we get $(p+6)(p-6)=5 q=q 5=1 \times 5 q=5 q \times 1$.
$$
\left\{\begin{array} { l }
{ p + 6 = 1 , } \\
{ p - 6 = 5 q ; }
\end{array} \left\{\begin{array}{l}
p+6=5 q, \\
p-6=1 .
\end{array}\right.\right.
$$
Solving, we get $(p, q)=(-1,-7)$ (discard), (11,17), $\left(-5,-\frac{11}{5}\right... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,591 |
6. There is a triangular activity site, with side $A B=20, \angle B=60^{\circ}, \angle C=75^{\circ}, A D$ is a lime line, and $A D \perp B C$ at point $D$. Xiao Ming starts from point $D$, passes through a point on side $A B$ and a point on side $A C$, and finally returns to $D$. Then the shortest distance Xiao Ming wa... | 6. C.
As shown in Figure 7, construct the symmetric point $D_{1}$ of point $D$ with respect to $AB$, and the symmetric point $D_{2}$ of point $D$ with respect to $AC$. Connect $D_{1}D_{2}$, which intersects $AB$ and $AC$ at points $E$ and $F$, respectively. Then $DE + EF + FD$ is the shortest path that Xiao Ming can t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,592 |
1. Given
$$
x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}=1 \text {. }
$$
then the value of $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}$ is | $$
\text { II, 1. }-1 \text {. }
$$
Notice that
$$
\begin{array}{l}
x \sqrt{x^{2}+3 x+18}+x \sqrt{x^{2}-6 x+18} \\
=\frac{x^{2}\left(x^{2}+3 x+18\right)-x^{2}\left(x^{2}-6 x+18\right)}{x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}} \\
=\frac{x^{2} \cdot 9 x}{1}=9 x^{3} .
\end{array}
$$
Subtracting the above equation fr... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,593 |
Example 1 In an equilateral $\triangle ABC$, there is a point $P$ inside, and its distances to the three vertices $A, B, C$ are $1, \sqrt{2}, \sqrt{3}$, respectively. Find the degree of $\angle APB$.
Analysis: The three segments $PA, PB, PC$ form a "Y" shape, and their lengths are exactly the three sides of a right-an... | Solution: As shown in Figure 1, construct an equilateral $\triangle P B D$ with $P B$ as a side, such that point $D$ is on the opposite side of $A$ with respect to $B C$, and connect $C D$.
Since $\angle A B C=\angle P B D=60^{\circ}$, we have
$$
\begin{array}{l}
\angle A B C-\angle P B C=\angle P B D-\angle P B C \\
\... | 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,594 |
Example 2 When a point $P$ inside a non-obtuse $\triangle ABC$ makes the value of $P A+P B+P C$ the smallest, find the degree measures of $\angle A P B$, $\angle B P C$, and $\angle C P A$.
Analysis: Let $Q$ be any point inside $\triangle ABC$, and $Q A$, $Q B$, $Q C$ form a "Y" shape. According to the shortest distan... | Solution: Without loss of generality, let $\angle B A C \leqslant 60^{\circ}$.
As shown in Figure 2, construct an equilateral
$\triangle A Q D$ with $A Q$ as a side, such that point
$D$ is on the opposite side of $B$ with respect to $A C$,
and construct an equilateral $\triangle A C E$ (where $E$ is a fixed point) outs... | \angle A P B=\angle B P C=\angle C P A=120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,595 |
Example 5 Let $a, b, c > 0, a + b + c = 1$. Prove:
$$
\sum \frac{1}{bc + a + \frac{1}{a}} \leqslant \frac{27}{31} \text{. }
$$
(2008, Serbian Mathematical Olympiad)
Analysis: Transform the fraction into $\frac{a}{abc + a^2 + 1}$. If we treat $abc$ as a constant, then this expression is only related to $a$. When $a$ inc... | Proof: Without loss of generality, let $a \geqslant b \geqslant c$. Then,
$$
a b c+a^{2}+1 \geqslant a b c+b^{2}+1 \geqslant a b c+c^{2}+1 \text {. }
$$
Thus, $\frac{a}{a b c+a^{2}+1}-\frac{b}{a b c+b^{2}+1}$
$$
=\frac{a b c(a-b)+(1-a b)(a-b)}{\left(a b c+a^{2}+1\right)\left(a b c+b^{2}+1\right)} \geqslant 0 .
$$
Sim... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,596 |
Example 6 Let $a, b, c > 0$,
$$
\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}=1 \text {. }
$$
Prove: $a+b+c \geqslant a b+b c+c a$.
(2007, Romanian Mathematical Olympiad)
Analysis: Note that the conclusion is a relationship between $a, b, c$ and $ab, bc, ca$. Transform the fraction $\frac{1}{a+b+1}$ into $\frac{c}{a... | Prove: Notice
$$
\sum \frac{1}{a+b+1}=1 \Leftrightarrow \sum \frac{c}{a c+b c+c}=1 .
$$
Without loss of generality, assume $a \geqslant b \geqslant c$. Then
$$
\begin{array}{l}
a c+b c+c \leqslant a b+b c+b \leqslant a b+a c+a, \\
\frac{c}{a c+b c+c} \leqslant \frac{b}{a b+b c+b} \leqslant \frac{a}{a b+a c+a} \text {.... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,597 |
Example 7 Given a real number $c \in\left(\frac{1}{2}, 1\right)$. Find the smallest constant $M$, such that for any integer $n \geqslant 2$ and real numbers $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, as long as they satisfy
$$
\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{n} a_{k},
$$
we always hav... | Proof: From the given conditions, we have
$$
\sum_{k=1}^{m}(c n-k) a_{k}=\sum_{k=m+1}^{n}(k-c n) a_{k} \text {. }
$$
Since $\frac{1}{2}c n \geqslant m \geqslant 1$.
Notice that $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{m}$,
$$
c n-1 \geqslant c n-2 \geqslant \cdots \geqslant c n-m \text {. }
$$
By Chebyshe... | \frac{1}{1-c} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,598 |
Example 8 Given positive integers $r, s, t$, satisfying $1<r<s<t$. For all positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the condition
$$
\frac{x_{j}}{x_{j+1}} \leqslant 1+\frac{s-t}{j+t}(j=1,2, \cdots, n)
$$
find
$$
M=\frac{\sum_{j=1}^{n} j(j+1) \cdots(j+s-1) x_{j}}{\sum_{j=1}^{n}(j+r) \cdots(j+s-1... | Solution: Let $x_{j}=(j+s)(j+s+1) \cdots(j+t-1)$ $(j=1,2, \cdots, n)$. Then
$$
M=\frac{\sum_{j=1}^{n} j(j+1) \cdots(j+t-1)}{\sum_{j=1}^{n}(j+r) \cdots(j+t-1)}.
$$
We will prove that for positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the given conditions, we have
$$
M \geqslant \frac{\sum_{j=1}^{n} j(j+... | \frac{t-r+1}{t+1} \cdot \frac{n(n+1) \cdots(n+t)}{(n+r) \cdots(n+t)-r(r+1) \cdots t} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,599 |
Example 1 In $\triangle ABC$, $\angle B \neq \angle C$, the incircle $\odot I$ of $\triangle ABC$ touches $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Let $P$ be the intersection of $AD$ and $\odot I$ different from point $D$. Draw a perpendicular from point $P$ to $AD$ intersecting $EF$ at point $Q$, and let... | Proof: As shown in Figure 1, let the radius of $\odot I$ be $R$, and establish a Cartesian coordinate system with $I$ as the origin and the line $ID$ as the $y$-axis.
Then the equation of $\odot I$ is $x^{2}+y^{2}=R^{2}$.
Let point $A\left(x_{0}, y_{0}\right)$. Then the equation of the chord of contact $EF$ is
$x_{0} x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,600 |
Example 2 Given that $Q$ is a point on the circle with diameter $AB$ $(Q \neq A, B)$, and the projection of $Q$ on $AB$ is $H$. The circle with center $Q$ and radius $QH$ intersects the circle with diameter $AB$ at points $C$ and $D$. Prove that $CD$ bisects the segment $QH$.
(2006, Turkish National Team Selection Exam... | Proof: As shown in Figure 2, let the center of the circle with diameter $AB$ be $O$, and the radius be $R$. Establish a Cartesian coordinate system with $O$ as the origin and the line $AB$ as the $x$-axis. Then the equation of $\odot O$ is
$$
x^{2}+y^{2}=R^{2} \text {. }
$$
Let point $Q\left(x_{0}, y_{0}\right)$. Then... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,601 |
Example 3 In isosceles $\triangle A B C$, $A B=A C, D$ is the midpoint of side $A C$, $E$ is the projection of point $D$ on side $B C$, $F$ is the midpoint of $D E$. Prove: $B F \perp A E$ if and only if $\triangle A B C$ is an equilateral triangle.
(2005, Slovenia Mathematical Olympiad) | Proof: As shown in Figure 3, let $O$ be the midpoint of $BC$, and establish a Cartesian coordinate system with $O$ as the origin and the line $AO$ as the $y$-axis. Without loss of generality, let the points be $B(-1,0)$ and $C(1,0)$.
Sufficiency.
Since $\triangle ABC$ is an equilateral triangle, the point $A(0, \sqrt{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,602 |
Example 4 Let the circumcircle of square $ABCD$ be circle $\Gamma$, $M$ be a point on the arc $\overparen{CD}$ not containing point $A$, $AM$ intersects $BD$ and $CD$ at points $P$ and $R$, $BM$ intersects $AC$ and $DC$ at points $Q$ and $S$. Prove: $PS \perp QR$.
(2006, French National Team Selection Exam) | Proof: As shown in Figure 4, let the radius of circle $\Gamma$ be 1, with the center at $O$. Establish a Cartesian coordinate system with $O$ as the origin and line $AC$ as the $x$-axis. Then the equation of circle $\Gamma$ is $x^{2}+y^{2}=1$.
Let point $M\left(x_{0}, y_{0}\right)$.
Then $x_{0}^{2}+y_{0}^{2}=1$.
$$
l_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,603 |
Example 5 Given that $A A_{1}$, $B B_{1}$, and $C C_{1}$ are the altitudes of the acute-angled $\triangle A B C$. Prove that the feet of the perpendiculars from $C_{1}$ to the segments $A C$, $A A_{1}$, $B B_{1}$, and $B C$ lie on the same straight line.
(2005, Belarus Mathematical Olympiad) | Proof: As shown in Figure 5, let the perpendicular feet from point $C_{1}$ to line segments $A C$, $A A_{1}$, $B B_{1}$, and $B C$ be $B_{2}$, $N$, $M$, and $A_{2}$, respectively. Establish a Cartesian coordinate system with $C_{1}$ as the origin and the line $A B$ as the $x$-axis. Let the coordinates of points $A$, $B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,604 |
2. Given trapezoid $A B C D$ satisfies $A B / / C D$, there is a point $E$ on the extension of $C B$, and a point $F$ on line segment $A D$, such that $\angle D A E=\angle C B F$. Let the lines $C D$ and $A B$ intersect $E F$ at points $I$ and $J$, respectively, and let the midpoint of line segment $E F$ be $K$, and $K... | 2. As shown in Figure 1.
$$
\begin{array}{l}
\text { Since } \angle E B F=180^{\circ}-\angle C B F \\
=180^{\circ}-\angle E A F,
\end{array}
$$
Therefore, quadrilateral $A E B F$ is a cyclic quadrilateral.
Thus, $A J \cdot J B=F J \cdot J E$.
Hence, the necessary and sufficient condition for point $I$ to be on the cir... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,605 |
Example 3 In $\triangle A B C$, $\angle C=90^{\circ}, B C=2$, $P$ is a point inside $\triangle A B C$ such that the minimum value of $P A+P B+P C$ is $2 \sqrt{7}$. Find the degree measure of $\angle A B C$.
Analysis: The key to this problem is to find the line segment when $P A+P B+P C$ reaches its minimum value. From... | Solution: As shown in Figure 3, construct an equilateral $\triangle ACD$ outside $\triangle ABC$ with $AC$ as a side, and connect $BD$. Then the length of $BD$ is the minimum value of $PA + PB + PC$, i.e., $BD = 2\sqrt{7}$. Draw $DE \perp BC$ intersecting the extension of $BC$ at point $E$.
Let $AC = x$. Then $CD = x$.... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,606 |
3. Let $P, Q$ be two points inside the convex quadrilateral $ABCD$, and satisfy that quadrilaterals $PQDA$ and $QPBC$ are cyclic. If there exists a point $E$ on the line segment $PQ$ such that $\angle PAE = \angle QDE, \angle PBE = \angle QCE$, prove: quadrilateral $ABCD$ is cyclic. | 3. Proof 1: As shown in Figure 2.
Let $F$ be a point on line $AD$ such that
$EF \parallel PA$.
If point $F$ is between $A$ and $D$, from the fact that quadrilateral $PQDA$ is a cyclic quadrilateral, we have
$$
\angle EFD = \angle PAD = 180^{\circ} - \angle EQD.
$$
Thus, quadrilateral $EFDQ$ is a cyclic quadrilateral.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,607 |
4. Given that $B E$ and $C F$ are altitudes of acute $\triangle A B C$, and two circles passing through points $A$ and $F$ are tangent to line $B C$ at points $P$ and $Q$, respectively, with point $B$ between $C$ and $Q$. Prove: The intersection of $P E$ and $Q F$ lies on the circumcircle of $\triangle A E F$. | 4. Since $B P^{2}=B F \cdot B A=B Q^{2}$, therefore, $B P=B Q$.
As shown in Figure 3, let $B E$ and $C F$ intersect at point $H$. Then $H$ is the orthocenter of $\triangle A B C$. Connect $A H$ with $B C$ intersecting at point $D$. Then $A D \perp B C$.
Thus, quadrilaterals $C D F A$ and $C D H E$ are both cyclic quad... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,608 |
5. Given integers $k, n$ satisfying $0 \leqslant k \leqslant n-2$. Consider a set $L$ of $n$ lines in the plane such that no two lines are parallel and no three lines are concurrent. Let the set of intersection points of the lines in $L$ be $I$, and let $O$ be a point not on any line in $L$. If $X \in I$, and the open ... | 5. For each point $P \in I$, define its order as the number of lines in $L$ that intersect the open line segment $OP$.
By definition, if the order of point $P$ is at most $k$, then $P$ is red.
First, we prove: There is at least one point $X \in I$ whose order is 0.
In fact, the lines in $L$ divide the plane into sever... | \frac{1}{2}(k+1)(k+2) | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,609 |
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+d^{2}+(1-a)^{2}+(a-b)^{2}+ \\
(b-c)^{2}+(c-d)^{2}+(d-1)^{2}
\end{array}
$$
When the value of the above algebraic expression is minimized, what is the value of $a+b+c+d$? Prove your conclusion. | 1. Arrange in descending powers of $a$,
$$
\text { original expression }=3 a^{2}-(2+2 b) a+\cdots \text {. }
$$
Therefore, when the value of the original expression is minimized,
$$
a=\frac{2+2 b}{2 \times 3}=\frac{1+b}{3} \text {. }
$$
Similarly, when the value of the original expression is minimized, we have
$$
d=\... | \frac{6}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,611 |
2. Given that $\triangle A B C$ is inscribed in $\odot O$ with radius 1, and $A B \cdot A C=A B-A C=1$. Try to find the measures of the three interior angles of $\triangle A B C$, and prove your conclusion. | 2. From $A B+A C=\sqrt{(A B-A C)^{2}+4 A B \cdot A C}=\sqrt{5}$, we solve to get $A B=\frac{\sqrt{5}+1}{2}, A C=\frac{\sqrt{5}-1}{2}$.
Take a point $D$ on $O A$ such that $O D=A C=\frac{\sqrt{5}-1}{2}$, then $A D=1-O D=\frac{3-\sqrt{5}}{2}$.
In $\triangle O A C$ and $\triangle C A D$, we have
$$
\angle O A C=\angle C A... | \angle A B C=18^{\circ}, \angle A C B=54^{\circ}, \angle B A C=108^{\circ} \text{ or } \angle A B C=18^{\circ}, \angle A C B=126^{\circ}, \angle B A C=36^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,612 |
3. Does there exist a positive integer such that its cube plus 101 is exactly a perfect square? Prove your conclusion. | 3. If there exists a positive integer $x$ that satisfies the conditions of the problem, then $x^{3}+101=y^{2}\left(y \in \mathbf{N}_{+}\right)$.
If $x$ is even, then $y$ is odd.
Let $x=2 n, y=2 m+1(n, m \in \mathbf{N})$. Then $2 n^{3}+25=m^{2}+m$,
the left side of this equation is odd, while the right side is even, whi... | 95 | Number Theory | proof | Yes | Yes | cn_contest | false | 721,613 |
4. Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$
is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n$ is | 4.39.
Notice that
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}}=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{(k+1)^{2} k-k^{2}(k+1)} \\
=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{k^{2}+k}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} .
\end{array}
$$
Then the original expression $=1-\frac{1}{\sqrt{n+1}}$.
Given $\frac{19}{20}<1... | 39 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,615 |
5. Given the equation in $x$
$$
x^{4}-6 x^{3}+a x^{2}+6 x+b=0
$$
the left side can be divided by $x-1$, and the remainder when divided by $x+2$ is 72. Then all the solutions of this equation (in ascending order) are $\qquad$ | 5. $-1,1,2,4$.
From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
1-6+a+6+b=0, \\
(-2)^{4}-6(-2)^{3}+a(-2)^{2}+6(-2)+b=72
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a=7, \\
b=-8 .
\end{array}\right.
\end{array}
$$
Thus, the original equation is
$$
x^{4}-6 x^{3}+7 x^{2}+6 x-8... | -1,1,2,4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,616 |
Example 4 As shown in Figure $4, D$ is a point outside $\triangle A B C, A B=A C$ $=B D+C D, \angle A B D=$ $60^{\circ}$. Find the degree measure of $\angle A C D$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution 1: As shown in Figure 4, extend $BD$ to point $E$ such that $DE = CD$, and connect $AD$, $AE$.
From $AB = BD + CD = BD + DE = BE$,
$$
\angle ABE = 60^{\circ},
$$
we know that $\triangle ABE$ is an equilateral triangle. Therefore,
$\angle E = 60^{\circ}, AE = AB$.
Thus, $AC = AE$.
So, $\triangle ACD \cong \tri... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,617 |
6. If the difference between two positive integers is 21, and the sum of their greatest common divisor and least common multiple is 1463, then these two positive integers are $\qquad$ | 6. 112, 91.
Let these two positive integers be $x, y(x>y)$, their greatest common divisor be $d$, and $x=d x_{1}, y=d y_{1}$. Then their least common multiple is $d x_{1} y_{1}$, and
$$
\left\{\begin{array}{l}
d\left(x_{1}-y_{1}\right)=21, \\
d\left(1+x_{1} y_{1}\right)=1463 .
\end{array}\right.
$$
Since $(1463,21)=7... | 112, 91 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,618 |
7. Let $AB$ and $CD$ be two perpendicular diameters of $\odot O$, $E$ be a point on segment $OB$, and $OE=\frac{1}{n} BE$. The extension of line segment $CE$ intersects $\odot O$ at point $F$, and line segment $AF$ intersects $DO$ at point $G$. Then $\frac{DG}{GC}=$ $\qquad$ | 7. $\frac{1}{n+1}$.
Let the radius of the circle be $n+1$. From the given information,
$$
O E=1, C E=\sqrt{1^{2}+(n+1)^{2}} \text {. }
$$
Connect $D F$. Then $\angle C F D=90^{\circ}$.
Thus, $\triangle E O C \backsim \triangle D F C \Rightarrow \frac{E C}{D C}=\frac{O C}{C F}$.
So, $C F=\frac{2(n+1)^{2}}{\sqrt{(n+1)^... | \frac{1}{n+1} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,619 |
1. The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$ | 1.1008015.
By
$$
\begin{array}{l}
1-\sqrt{\frac{n}{n+2009}}>\frac{1}{1005} \Rightarrow \sqrt{\frac{n}{n+2009}}<\frac{1004}{1005} \\
\Rightarrow \frac{n}{n+2009}<\frac{1004^2}{1005^2} \Rightarrow n<\frac{1004^2}{1005^2-1004^2} \cdot 2009=1008014.25
\end{array}
$$
Thus, the number of positive integer values of $n$ for w... | 1008015 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,620 |
2. Simplify $\sqrt[3]{10+7 \sqrt{2}}+\sqrt[3]{10-7 \sqrt{2}}$ to get the result $\qquad$ . | 2. $2 \sqrt[3]{4}$
Let the original expression $=x>0$. Then
$$
\begin{array}{l}
x^{3}=20+3 \sqrt[3]{100-49 \times 2} \cdot x \\
\Rightarrow x^{3}-3 \sqrt[3]{2} \cdot x-20=0 . \\
\text { Let } x=\sqrt[3]{4} y>0 \text {. Then } \\
4 y^{3}-6 y-20=0 \\
\Rightarrow(y-2)\left(2 y^{2}+4 y+5\right)=0 .
\end{array}
$$
Solving... | 2 \sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,621 |
8. Given that $E$ is a point on the extension of side $C D$ of the cyclic quadrilateral $A B C D$, and $I$ is the incenter of $\triangle A B C$. If $\angle A B C=70^{\circ}, \angle A C B=60^{\circ}, D E=D A$, then the degree measure of $\angle D E I$ is $\qquad$ | 8. $25^{\circ}$.
From the given information, we have
$$
\begin{array}{l}
\angle B A C=50^{\circ}, \angle E D A=\angle A B C=70^{\circ}, \\
\angle D E A=\angle D A E=55^{\circ}, \\
\angle C I A=90^{\circ}+\frac{1}{2} \angle A B C=180^{\circ}-\angle C E A .
\end{array}
$$
Therefore, points $C$, $I$, $A$, and $E$ are co... | 25^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,622 |
9. Factory A bought a machine, which produces a certain product every day. After several days, Factory B also bought the same machine and started producing the same product at the same speed.
When the total number of products produced by Factory A was 6 times that of Factory B, both factories had produced for an integ... | 9.5 9th May.
Let when the total number of products produced by Factory A is 6 times that of Factory B, Factory A has produced for $x$ days, and Factory B has produced for $\frac{x}{6}$ days. After another $y$ days, the total number of products produced by Factory A is 4 times that of Factory B. Then
$$
x+2 y=4\left(\f... | 9th May | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,623 |
10. Let $n$ be a positive integer. If in the $2 n+1$ consecutive natural numbers including 2009, the sum of the squares of the first $n+1$ numbers is equal to the sum of the squares of the last $n$ numbers, then the value of $n$ is $\qquad$. | 10. 31 .
Let $m, m+1, \cdots, m+2 n$ be $2 n+1$ consecutive natural numbers that satisfy the above conditions. Then
$$
\begin{array}{l}
m^{2}+(m+1)^{2}+\cdots+(m+n)^{2} \\
=(m+n+1)^{2}+(m+n+2)^{2}+\cdots+(m+2 n)^{2} . \\
\text { Let } S_{k}=1^{2}+2^{2}+\cdots+k^{2} \text {. Then } \\
S_{m+n}-S_{m-1}=S_{m+2 n}-S_{m+n} ... | 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,624 |
1. Find all integer pairs $(x, y)$ that satisfy the equation $x^{2}-2 x y+126 y^{2}=2009$. (Supplied by Zhang Pengcheng) | 1. Let the integer pair $(x, y)$ satisfy the equation
$$
x^{2}-2 x y+126 y^{2}-2009=0 \text {, }
$$
considering it as a quadratic equation in $x$, its discriminant
$$
\Delta=500\left(4^{2}-y^{2}\right)+36
$$
should be a perfect square.
If $y^{2}>4^{2}$, then $\Delta<0$.
If $y^{2}<4^{2}$, then $y^{2}$ can take $0,1^{2... | (1,4),(7,4),(-1,-4),(-7,-4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,625 |
2. In a convex pentagon $A B C D E$, it is known that $A B = D E$, $B C = E A$, $A B \neq E A$, and points $B, C, D, E$ are concyclic. Prove that the necessary and sufficient condition for points $A, B, C, D$ to be concyclic is $A C = A D$.
(Xiong Bin, problem contributor) | 2. Necessity.
If points $A, B, C, D$ are concyclic, then from $AB = DE$, $BC = EA$, we get
$\angle BAC = \angle EDA$, $\angle ACB = \angle DAE$.
Therefore, $\angle ABC = \angle DEA \Rightarrow AC = AD$.
Sufficiency.
Let the circle on which points $B, C, D, E$ lie be $\odot O$. If $AC = AD$, then the center $O$ lies on... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,626 |
3. Let $x, y, z \in \mathbf{R}_{+}, \sqrt{a}=x(y-z)^{2}, \sqrt{b}=$ $y(z-x)^{2}, \sqrt{c}=z(x-y)^{2}$. Prove:
$$
a^{2}+b^{2}+c^{2} \geqslant 2(a b+b c+c a) \text {. }
$$
(Tang Lihua) | 3. First prove: $\sqrt{a}, \sqrt{b}, \sqrt{c}$ cannot form the three sides of a triangle.
$$
\begin{array}{l}
\text { By } \sqrt{b}+\sqrt{c}-\sqrt{a}=-(y+z)(z-x)(x-y), \\
\sqrt{c}+\sqrt{a}-\sqrt{b}=-(z+x)(x-y)(y-z), \\
\sqrt{a}+\sqrt{b}-\sqrt{c}=-(x+y)(y-z)(z-x),
\end{array}
$$
we know that $(\sqrt{b}+\sqrt{c}-\sqrt{a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,627 |
Example 5 In $\triangle A B C$, $A B=A C, \angle A=80^{\circ}$, $P$ is a point inside $\triangle A B C$, and $\angle P B C=10^{\circ}, \angle P C A=30^{\circ}$. Find the degree measure of $\angle P A C$. | Solution: As shown in Figure 6, construct the symmetric point $P^{\prime}$ of point $P$ with respect to $A C$, and connect $P^{\prime} A$, $P^{\prime} B$, $P^{\prime} C$, and $P^{\prime} P$.
Since $\angle P C A=30^{\circ}$, by symmetry we have
$$
\begin{array}{l}
\angle P C P^{\prime} \\
=2 \angle P C A=60^{\circ}, \\... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,628 |
4. Given twelve red points on a circle. Find the minimum value of $n$, such that there exist $n$ triangles with red points as vertices, so that every chord with red points as endpoints is a side of one of these triangles.
（Supplied by Tao Pingsheng） | 4. Let the set of red points be $A=\left\{A_{1}, A_{2}, \cdots, A_{12}\right\}$.
There are 11 chords passing through point $A_{1}$, and any triangle containing vertex $A_{1}$ contains exactly two chords passing through point $A_{1}$. Therefore, these 11 chords passing through point $A_{1}$ must be distributed among at... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,629 |
5. Let the set of all permutations $X=(x_{1}, x_{2}, \cdots, x_{9})$ of $1,2, \cdots, 9$ be $A$. For any $X \in A$, let
\[
\begin{array}{l}
f(X)=x_{1}+2 x_{2}+\cdots+9 x_{9}, \\
M=\{f(X) \mid X \in A\} .
\end{array}
\]
Find $|M|$ (where $|M|$ denotes the number of elements in the set $M$).
(Xiong Bin) | 5. Generally prove: when $n \geqslant 4$, for all permutations $X_{n}=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of the first $n$ positive integers $1,2, \cdots, n$, if
$$
\begin{array}{l}
f\left(X_{n}\right)=x_{1}+2 x_{2}+\cdots+n x_{n}, \\
M_{n}=\{f(X)|X \in A|,
\end{array}
$$
$$
\text { then }\left|M_{n}\right|=\frac... | 121 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,630 |
6. As shown in Figure 1, given
$\odot O, \odot I$ are the circumcircle and incircle of $\triangle A B C$ respectively. Prove: Through any point $D$ on $\odot O$, a $\triangle D E F$ can be constructed such that $\odot O, \odot I$ are the circumcircle and incircle of $\triangle D E F$ respectively. (Provided by Tao Ping... | 6. As shown in Figure 5, let $O I = d$, and $R$, $r$ be the radii of the circumcircle and incircle of $\triangle A B C$, respectively. Extend $A I$ to intersect $\odot O$ at point $K$. Then
$$
\begin{array}{l}
K I = K B \\
= 2 R \sin \frac{A}{2}, \\
A I = \frac{r}{\sin \frac{A}{2}} .
\end{array}
$$
Extend $O I$ to int... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,631 |
7. Let $f(x, y, z)=\sum \frac{x(2 y-z)}{1+x+3 y}(x, y, z \geqslant 0)$, and $x+y+z=1$. Find the maximum and minimum values of $f(x, y, z)$.
(Li Shenghong) | 7. First, prove: $f \leqslant \frac{1}{7}$, and the equality holds if and only if $x=y=z=\frac{1}{3}$.
Notice that
$$
f=\sum \frac{x(x+3 y-1)}{1+x+3 y}=1-2 \sum \frac{x}{1+x+3 y}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sum \frac{x}{1+x+3 y} \geqslant \frac{\left(\sum x\right)^{2}}{\sum x(1+x+... | f_{\text {max }}=\frac{1}{7}, f_{\text {min }}=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,632 |
8. In an $8 \times 8$ grid, what is the minimum number of small squares that need to be removed so that it is impossible to cut out a complete "T-shaped" pentomino from the remaining grid?
（Sun Wenxian, problem contributor）
Will the above translation meet your needs? | 8. At least 14 small squares must be removed from Figure 6.
Figure 6
Figure 7
As shown in Figure 7, an $8 \times 8$ chessboard is divided into five regions. The central region must have at least two small squares removed to prevent a T-shaped pentomino from being placed. The two crossed positions are not equivalent; o... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,633 |
One. (25 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n}=\sqrt{x_{n-1}^{2}+x_{n-1}}+x_{n-1}(n \geqslant 2) \text {. }
$$
Find the general term formula for the sequence $\left\{x_{n}\right\}$.
(Zhang Lei, provided) | One can easily prove that $x_{n}$ is positive.
Notice that
$$
\begin{array}{l}
\frac{1}{x_{n}}=\frac{1}{\sqrt{x_{n-1}^{2}+x_{n-1}}+x_{n-1}} \\
=\frac{\sqrt{x_{n-1}^{2}+x_{n-1}}-x_{n-1}}{x_{n-1}}=\sqrt{1+\frac{1}{x_{n-1}}}-1,
\end{array}
$$
which means
$$
\begin{array}{l}
\frac{1}{x_{n}}+1=\sqrt{1+\frac{1}{x_{n-1}}}=\c... | x_{n}=\frac{1}{2^{2^{1-n}}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,634 |
II. (25 points) As shown in Figure 1, in the acute triangle $\triangle ABC$, it is known that $AB > AC$, $\cos B + \cos C = 1$, and points $E$ and $F$ are on the extensions of $AB$ and $AC$, respectively, such that $\angle ABF = \angle ACE = 90^{\circ}$.
(1) Prove that $BE + CF = EF$;
(2) Let the angle bisector of $\an... | (1) Since $\angle A B F=\angle A C E=90^{\circ}$, therefore, $E, B, C, F$ are concyclic. Thus,
$$
\angle C F E=\angle A B C, \angle B E F=\angle A C B \text {. }
$$
Hence $\cos \angle C F E+\cos \angle B E F$
$$
=\cos \angle A B C+\cos \angle A C B=1 \text {, }
$$
which means $\frac{C F}{E F}+\frac{B E}{E F}=1$.
Ther... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,635 |
Four. (25 points) A captain and three sailors together received 2,009 coins of the same denomination. The four agreed to distribute the coins according to the following rules: Sailor 1, Sailor 2, and Sailor 3 each write down a positive integer, denoted as \( b_{1}, b_{2}, b_{3} \), satisfying \( b_{1} \geqslant b_{2} \... | Four, the maximum value is 673.
First, the captain can ensure getting no less than 673 gold coins.
In fact, when the captain divides the gold coins into 3 piles with numbers $671, 670, 668$,
(1) If $b_{1} \geqslant 671$, then the captain can get no less than $671+2=673$ coins;
(2) If $b_{1} < 671$, then the captain ca... | 673 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,636 |
Five. (25 points) As shown in Figure 2, in the given sector $A O B$, the central angle is acute.
Take a point $C$ on the arc $\overparen{A B}$, different from $A$ and
$B$, and take a point $P$ on the line segment
$O C$, connect
$A P$, and draw a line
$B Q / / A P$ through point $B$ to intersect the ray
$O C$ at point $... | Five, connect $A B$ intersecting $O C$ at point $M$.
Since $B Q / / A P$, then quadrilateral $A P B Q$ is a trapezoid.
Therefore, $S_{\triangle A Q M}=S_{\triangle B P M}$.
Thus, $S_{O A Q P B}=S_{O A M P B}+S_{\triangle \Lambda Q M}$
$=S_{\text {OAMPB }}+S_{\triangle B M P}=S_{\triangle O A B}$
is a constant, meaning ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,637 |
Example 6 In isosceles $\triangle A B C$, $\angle B=\angle C=40^{\circ}$, extend $A B$ to point $D$, such that $A D=B C$. Find the degree measure of $\angle B C D$. | Solution 1: As shown in Figure 7, construct an equilateral $\triangle ABE$ outside $\triangle ABC$ with $AB$ as a side, and connect $CE$. It is easy to see that
$$
AB=AC, \angle BAC=100^{\circ}.
$$
Since $\triangle ABE$ is an equilateral triangle, then
$$
\begin{array}{l}
AB=BE=AE, \\
\angle BAE=60^{\circ}, \\
\angle ... | 10^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,639 |
Seven, (25 points) Let $[m]$ denote the greatest integer not exceeding the real number $m$. Suppose $x$ and $y$ are positive real numbers, and for all positive integers $n$, we have $[x[n y]]=n-1$. Prove that $x y=1$, and $y$ is an irrational number greater than 1.
(Liu Kangning, problem contributor) | $$
\begin{array}{l}
\text { VII. For } x[n y]-11$, or $x y1$, it is obviously true;
when $0<y<1$, take $n=\left[\frac{1}{1-y}\right]+1$, then $\frac{1}{1-y}<n \leqslant \frac{1}{1-y}+1$.
Solving gives $\frac{n-2}{n-1} \leqslant y<\frac{n-1}{n}$, that is,
$$
\frac{n^{2}-2 n}{n-1} \leqslant n y<n-1 \text {. }
$$
Therefo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,640 |
Eight, (25 points) Find the smallest positive integer that is divisible by 209 and whose sum of digits equals 209.
(Zhang Lei provided the question) | Eight, the smallest number is $2 \times 10^{24} + 2 \times 10^{23} - 10^{15} - 1$. Prove: Since
$$
209 = 11 \times 19, 209 = 9 \times 23 + 2 \text {, }
$$
the number is at least 24 digits long and is divisible by 11 and 19.
(1) If the number is a 24-digit number, let the $i$-th digit from right to left be $a_{i} (1 \l... | 2 \times 10^{24} + 2 \times 10^{23} - 10^{15} - 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,641 |
Three. (25 points) Given 26 distinct positive integers, among any six of these numbers, there are at least two numbers such that one number divides the other. Prove that there must exist six numbers, one of which is divisible by the other five.
(Zhang Tongjun, problem contributor) | Three, arrange 26 numbers in ascending order. Number the smallest number as 1, and the principle for numbering subsequent numbers is: if none of the numbers before it can divide it, number this number as 1; if some of the numbers before it can divide it, let the largest number among those that can divide it be numbered... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,642 |
1. Given that $x$ is an irrational number, and $(x+1)(x+3)$ is a rational number. Under the above assumptions, there are the following four conclusions:
(1) $x^{2}$ is a rational number;
(2) $(x-1)(x-3)$ is an irrational number;
(3) $(x+1)^{2}$ is a rational number;
(4) $(x-1)^{2}$ is an irrational number.
Among them, ... | - 1. C.
Take the irrational number $x=\sqrt{3}-2$, at this time,
$$
(x+1)(x+3)=(\sqrt{3}-1)(\sqrt{3}+1)=2
$$
is a rational number.
While $x^{2}=(\sqrt{3}-2)^{2}=7-4 \sqrt{3}$ is an irrational number, hence conclusion (1) is incorrect;
By $(x+1)^{2}=(\sqrt{3}-1)^{2}=4-2 \sqrt{3}$ is an irrational number, hence conclu... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,643 |
2. Given the equation in terms of $x$
$$
\frac{5}{2} x-a=\frac{8}{5} x+142 \text {, }
$$
when $a$ is some positive integer, the solution of the equation is a positive integer. Then the smallest positive integer value of $a$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 2. A.
From the original equation, we get $a=\frac{9}{10} x-142$.
Since $a$ is a positive integer, we have,
$$
\frac{9}{10} x>142 \Rightarrow x>157 \frac{7}{9} \text {. }
$$
Also, $x$ is a positive integer, to make $\frac{9}{10} x$ an integer, $x$ must be a multiple of 10, and to make $a$ as small as possible, we shou... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,644 |
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