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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 4: Arrange 5 white stones and 10 black stones in a horizontal row, such that the right neighbor of each white stone must be a black stone. How many arrangements are there?
(1996, Japan Mathematical Olympiad Preliminary) | Let the number of black stones between the $i$-th white stone and the $(i+1)$-th white stone be $x_{i+1}(i=1,2,3,4)$, with $x_{1}$ black stones at the far left and $x_{6}$ black stones at the far right, then
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=10,
$$
where $x_{1} \geqslant 0, x_{k} \geqslant 1, k=2,3,4,5,6$.
$$
\be... | 252 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,477 |
Example 5 In coin tossing, if Z represents heads and F represents tails, then the sequence of coin tosses is represented by a string composed of Z and F. We can count the number of occurrences of heads followed by tails (ZF), heads followed by heads (ZZ)...... For example, the sequence ZZFFZZZZFZZFFFF is the result of ... | Solution: The sequences that meet the requirements have the following two possible forms:
(1) Starting with F: $F \cdots F Z \cdots Z F \cdots F Z \cdots Z$;
(2) Starting with Z: $\mathrm{Z} \cdots \mathrm{ZF} \cdots \mathrm{FZ} \cdots \mathrm{ZF} \cdots \mathrm{F}$.
Since the sequence is required to have exactly 3 $\m... | 560 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,478 |
Example 5 As shown in Figure 2, in trapezoid $ABCD$, there is $AD // BC, AB \perp BC, AD=8 \text{~cm}, BC=16 \text{~cm}$, $AB=6 \text{~cm}$. Moving points $M, N$ start from points $B, C$ simultaneously, moving along $BC, CD$ in the directions of $BC, CD$, with the speeds of points $M, N$ being $2 \text{~cm} / \text{s},... | (1) Suppose when points $M$ and $N$ have moved for $x \, \text{s}$, $M N \parallel B D$. Then $B M = 2x$, $C M = 16 - 2x$, $C N = x$.
From the given information, we have $B D = 10$.
Draw $D H \perp B C$ at $H$, then $B H = A D = 8$.
Thus, $C H = 8$, $C D = B D = 10$.
Therefore, $D N = 10 - x$.
When $M N \parallel B D$,... | \frac{40}{9} \, \text{s}, \frac{93}{5} \, \text{cm}^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,479 |
Example 6: Taking three points from $2 n$ equally divided points on a circle, how many acute triangles can be formed? | Given that $\triangle A B C$ is an acute triangle, let there be $x_{1}$ points between points $A$ and $B$, $x_{2}$ points between points $B$ and $C$, and $x_{3}$ points between points $C$ and $A$. Then,
$$
x_{1}+x_{2}+x_{3}=2 n-3 \text {, }
$$
where $1 \leqslant x_{k} \leqslant n-2, k=1,2,3$.
Let $x_{i}^{\prime}=x_{i}... | \frac{n(n-1)(n-2)}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,480 |
Example 7 In a live military exercise, the Red side has set up 20 posts along a straight line. To test 5 new types of weapons, it is planned to equip 5 posts with these new weapons, with the requirement that the first and last posts are not equipped with new weapons, and among every 5 adjacent posts, at least one post ... | Let the 20 positions be ordered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}$,
$$
k=1,2,3,4,5 \text {. }
$$
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$
Then we have
... | 69600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,481 |
Example 1 In a family photo album, there are 10 photos. Each photo has 3 men, the person standing on the left is the son of the person in the middle, and the person on the right is the middle person's brother. It is known that the 10 people in the middle of the 10 photos are all different. How many different people are... | Solution: As shown in Figure 1, there are 16 people, where the horizontal lines represent brothers, and the vertical and diagonal lines represent father-son relationships. The 10 photos taken by them are as follows:
$$
\begin{array}{l}
\quad\{3,1,2\},\{16, \\
2,1\},\{5,3,4\},\{15, \\
4,3\},\{7,5,6\},\{9, \\
6,5\},\{11,... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,482 |
Question 4 Let the function $f(x)=\frac{1}{k-x}$, and denote
$$
\underbrace{f(f(\cdots)}_{\text {j times }}
$$
as $f_{j}(x)$. Given that $f_{n}(x)=x$, and
$$
f_{i}(x) \neq f_{j}(x)(i \neq j, 1 \leqslant i, j \leqslant n) \text {. }
$$
Prove: $\prod_{i=1}^{n} f_{i}(x)=-1$. | Proof: Consider the recursive sequence $g_{j}=k g_{j-1}-g_{j-2}$, $g_{0}=0, g_{1}=1$. We have
$$
\begin{array}{l}
f_{1}(x)=\frac{1}{k-x}=\frac{x g_{0}-g_{1}}{x g_{1}-g_{2}}, \\
f_{2}(x)=\frac{x g_{1}-g_{2}}{x g_{2}-g_{3}}, \\
\cdots \cdots \\
f_{n}(x)=\frac{x g_{n-1}-g_{n}}{x g_{n}-g_{n+1}} .
\end{array}
$$
Multiplyin... | -1 | Algebra | proof | Yes | Yes | cn_contest | false | 716,484 |
Given $\triangle A B C$, point $X$ is a moving point on line $B C$, and point $C$ is between points $B$ and $X$. Also, the incircles of $\triangle A B X$ and $\triangle A C X$ intersect at two distinct points $P$ and $Q$. Prove: $P Q$ passes through a fixed point independent of point $X$.
(45th IMO Shortlist Problem) | Proof: As shown in Figure 1, let the incircles of $\triangle ABX$ and $\triangle ACX$ touch $BX$ at $D$ and $F$, respectively, and touch $AX$ at $E$ and $G$, respectively. Let $M$ and $N$ be the midpoints of $DF$ and $EG$, respectively.
Then $MD^2 = MF^2$, $NE^2 = NG^2$.
Hence, $M$ and $N$ both lie on the radical axis... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,485 |
1. If $a, b$ are real numbers, then which of the following statements is correct? ( ).
(A) $a>b \Rightarrow a^{2}>b^{2}$
(B) $a \neq b \Rightarrow a^{2} \neq b^{2}$
(C) $|a|>b \Rightarrow a^{2}>b^{2}$
(D) $a>|b| \Rightarrow a^{2}>b^{2}$ | - 1.D.
When $a=-1, b=-2$, options (A) and (C) are incorrect; when $a=-1, b=1$, option (B) is incorrect. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,486 |
2. Given $a+b+c=3, a^{2}+b^{2}+c^{2}=3$. Then the value of $a^{2005}+b^{2005}+c^{2005}$ is ( ).
(A) 0
(B) 3
(C) $2^{2005}$
(D) $3 \times 2^{2005}$ | 2.B.
$$
\begin{array}{l}
\text { Given } 0=3\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\
=(a-b)^{2}+(b-c)^{2}+(c-a)^{2},
\end{array}
$$
we know $a=b=c=1$.
Thus, $a^{2005}+b^{2005}+c^{2005}=3$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,487 |
3. There is a kind of football made by sewing together several pieces of black and white cowhide, with black pieces being regular pentagons and white pieces being regular hexagons (as shown in Figure 1). If a well-sewn football of this type has 12 black pieces, then the number of white pieces is ( ).
(A) 16
(B) 18
(C) ... | 3.C.
From the problem, we know that each white patch is adjacent to 3 black patches and 3 white patches, while each black patch is only adjacent to 5 white patches. Therefore, the number of white patches is
$$
\frac{12 \times 5}{3}=20 \text { pieces. }
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,488 |
4. In the right triangle $ABC$, the hypotenuse $AB=5$, and the lengths of the two legs $BC$ and $AC$ are the roots of the quadratic equation
$$
x^{2}-(2 m-1) x+4(m-1)=0 \text {. }
$$
Then the value of $m$ is ( ).
(A) 4
(B) -1
(C) 4 or -1
(D) -4 or 1 | 4.A.
From the given, we have
$$
(2 m-1)^{2}-2 \times 4(m-1)=25 \text {. }
$$
Solving, we get $m_{1}=4, m_{2}=-1$ (discard). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,489 |
5. As shown in Figure 2, the line $x$ $=1$ is the axis of symmetry of the graph of the quadratic function
$$
y=a x^{2}+b x+c
$$
Then we have ( ).
(A) $a+b+c>0$
(B) $b>a+c$
(C) $c>2 b$
(D) $a b c<0$ | 5.C.
Given $a>0, -\frac{b}{2a}=1$, we have $b<0$, $f(1)=a+b+c=b-a=b+\frac{b}{2}>2b$, hence the answer is (C). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,491 |
6. In the Cartesian coordinate system, a point whose both coordinates are integers is called an integer point. Let $k$ be an integer. When the intersection point of the lines $y=x-3$ and $y=k x+k$ is an integer point, the number of possible values for $k$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 6.C.
From $\left\{\begin{array}{l}y=x-3, \\ y=k x+k,\end{array}\right.$ we solve to get $x=\frac{k+3}{1-k}=-1+\frac{4}{1-k}$. Therefore, $k=-3,-1,0,2,3,5$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,492 |
1. Given $x$ is a non-zero real number, and $x^{\frac{1}{2}}+x^{-\frac{1}{2}}=a$. Then $\frac{x^{2}+1}{x}=$ $\qquad$ . | $$
=1 \cdot a^{2}-2 \text {. }
$$
Let $x^{\frac{1}{2}}+x^{-\frac{1}{2}}=a$ and square both sides to get $x+x^{-1}+2=a^{2}$.
$$
\text { Then } \frac{x^{2}+1}{x}=x+x^{-1}=a^{2}-2 \text {. }
$$ | a^2 - 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,493 |
2. Given that $a$ is a real number, and makes the quadratic equation in $x$, $x^{2}+a^{2} x+a=0$, have real roots. Then the maximum value that $x$ can take is $\qquad$ | 2. $\frac{\sqrt[3]{2}}{2}$.
Given that $a$ is a real number, when $a \neq 0$, the quadratic equation in $a$, $a^{2} x+a+x^{2}=0$, has real roots. Therefore, $\Delta=1-4 x^{3} \geqslant 0$. Hence, $x \leqslant \frac{\sqrt[3]{2}}{2}$. When $a=0$, $x=0$. | \frac{\sqrt[3]{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,494 |
3. $P$ is a point on the extension of the diameter $A B$ of $\odot O$, $P C$ is tangent to $\odot O$ at point $C$, and the angle bisector of $\angle A P C$ intersects $A C$ at point $Q$. Then $\angle P Q C=$ $\qquad$ | $3.45^{\circ}$.
As shown in Figure 4, connect $O C$. Since $P C$ is tangent to $\odot O$ at point $C$, then $O C$ $\perp P C$.
Since $O A=$ $O C$, therefore,
$$
\begin{array}{l}
\angle O A C=\angle O C A=\frac{1}{2} \angle P O C . \\
\text { Also } \angle A P Q=\angle C P Q=\frac{1}{2} \angle A P C \text {, so } \\
\a... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,495 |
4. For a positive integer $n$, if there exist positive integers $a$ and $b$ such that $n=a+b+a b$, then $n$ is called a "good number". For example, $3=1+1+1 \times 1$, so 3 is a good number. Then, among the 20 positive integers from $1 \sim 20$, the number of good numbers is $\qquad$ . | 4.12.
$n+1=a+b+a b+1=(a+1)(b+1)$ is a composite number, so the required $n$ is the number obtained by subtracting 1 from the composite numbers between 2 and 21. The composite numbers between 2 and 21 are $4,6,8,9,10,12,14,15,16,18,20,21$, a total of 12. Therefore, the required $n$ has 12 values: $3,5,7,8,9,11,13,14,15,... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,496 |
Three. (20 points) Let $A$ and $B$ be points on the parabola $y=2 x^{2}+4 x-2$, with the origin being the midpoint of segment $AB$. Find the coordinates of points $A$ and $B$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, as shown in Figure 5, let $A(a, b)$. Since the origin is the midpoint of $A B$, $A$ and $B$ are symmetric with respect to the origin, i.e.,
$$
B(-a,-b) \text {. }
$$
Also, $A$ and $B$ are points on the parabola. Substituting them into the equation of the parabola, we get
$$
\left\{\begin{array}{l}
b=2 a^{2}+4 a... | A(1,4), B(-1,-4) \text{ or } A(-1,-4), B(1,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,497 |
Four, (25 points) As shown in the figure,
$3, AB$ is the diameter of $\odot O$, $AB=d$, a tangent line to $\odot O$ is drawn through point $A$ and a point $C$ is taken on it such that $AC = AB$, connect $OC$ to intersect $\odot O$ at point $D$, the extension of $BD$ intersects $AC$ at $E$. Find the length of $AE$.
保留源... | $$
\begin{array}{l}
\angle 1=\angle 2 \\
=\angle 3=\angle 4 .
\end{array}
$$
Therefore, $\triangle C D E \backsim$
$\triangle C A D$, so
$$
\frac{C D}{D E}=\frac{C A}{A D} .
$$
Also, since $\triangle A D E \backsim \triangle B D A$, we have
$$
\frac{A E}{D E}=\frac{A B}{D A} \text {. }
$$
From equations (1), (2), an... | \frac{\sqrt{5}-1}{2} d | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,498 |
Five. (25 points) Let $x=a+b-c, y=a+c-$ $b, z=b+c-a$, where $a, b, c$ are undetermined prime numbers. If $x^{2}=y, \sqrt{z}-\sqrt{y}=2$, find all possible values of $abc$.
---
Translate the above text into English, preserving the original text's line breaks and format, and output the translation directly. | Given three equations, we solve them simultaneously to get
$$
(a, \dot{b}, c)=\left(\frac{1}{2}(x+y), \frac{1}{2}(x+z), \frac{1}{2}(y+z)\right) \text {. }
$$
Since $y=x^{2}$, we have
$$
\begin{array}{l}
a=\frac{1}{2}\left(x+x^{2}\right), \\
b=\frac{1}{2}(x+z), \\
c=\frac{1}{2}\left(x^{2}+z\right) .
\end{array}
$$
Fro... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,499 |
1. In $\triangle A B C$, if $a^{2}+b^{2}=6 c^{2}$, then the value of $(\cot A+\cot B) \tan C$ is ( ).
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $\frac{1}{7}$
(D) $\frac{2}{7}$ | -1.B.
In $\triangle A B C$, we have
$$
\begin{array}{l}
(\cot A+\cot B) \tan C \\
=\frac{\sin (A+B)}{\sin A \cdot \sin B} \cdot \frac{\sin C}{\cos C}=\frac{\sin ^{2} C}{\sin A \cdot \sin B} \cdot \frac{1}{\cos C} \\
=\frac{c^{2}}{a b} \cdot \frac{2 a b}{a^{2}+b^{2}-c^{2}}=\frac{2 c^{2}}{a^{2}+b^{2}-c^{2}} \\
=\frac{2 c... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,500 |
Example 7 As shown in Figure 5, the perimeter of square $ABCD$ is $40 \mathrm{~m}$, two people, Jia and Yi, start from $A$ and $B$ respectively at the same time, and walk along the edges of the square. Jia walks counterclockwise at a speed of $35 \mathrm{~m}$ per minute, and Yi walks clockwise at a speed of $30 \mathrm... | Explanation: After departure, every 2 minutes, both individuals arrive at a vertex simultaneously, with their respective positions as follows:
Time $(\mathrm{min}) \quad 0 \quad 2 \quad 4 \quad 6 \quad \cdots$
Position of A $\quad A \rightarrow D \rightarrow C \rightarrow B \quad \cdots$
Position of B $\quad B \rightar... | (6,10) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,501 |
2. Given that $f(x)$ is a non-zero function defined on $\mathbf{R}$. If for any $a, b \in \mathbf{R}$, it satisfies $f(a b)=a f(b)+b f(a)$, then the function $f(x)(\quad)$.
(A) is an odd function
(B) is an even function
(C) is both an odd and an even function
(D) is neither an odd function nor an even function | 2.A.
Since $f(a b)=a f(b)+b f(a)$, we have $f(-x)=f(-1 \times x)=(-1) f(x)+x f(-1)$.
Also, since $f(-1)=f(-1 \times 1)=-f(1)+f(-1)$, we get $f(1)=0$.
And $f(1)=f((-1) \times(-1))=-2 f(-1)$, we get $f(-1)=0$.
Therefore, $f(-x)=-f(x)$.
Since $f(x)$ is not identically zero on $\mathbf{R}$, $f(x)$ is an odd function. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,502 |
3. Let the sequence be composed of ordered pairs of positive integers $(x, y)$: $(1,1),(1,2),(2,1),(1,3),(2,2)$, $(3,1),(1,4),(2,3),(3,2),(4,1), \cdots$, arranged in ascending order of $x+y$, and when $x+y$ is equal, in ascending order of $x$. Then the position of the ordered pair $(m, n)(m, n$ both being positive inte... | 3. D.
Group by the value of $x+y$.
When $x+y=2$, it is the first group;
When $x+y=3$, it is the second group;
.....
When $x+y=m+n$, it is the $(m+n-1)$-th group.
Therefore, the first $(m+n-2)$ groups of the sequence have a total of ordered pairs
$$
\frac{(m+n-2)(m+n-1)}{2} \text { (pairs). }
$$
For the ordered pair $... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,503 |
4. As shown in Figure 1, in the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$, respectively. The sine value of the plane angle formed by the dihedral angle between plane $C E B_{1}$ and plane $D_{1} F B_{1}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\... | 4.C.
As shown in Figure 4, extend $CE$, $D_{1}F$, and $DA$. In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, since $E$ and $F$ are the midpoints of $AB$ and $AA_{1}$ respectively, it follows that $CE$, $D_{1}F$, and $DA$ intersect at a single point $G$. Connect $B_{1}G$. Given that the edge length of the cube $ABCD-A_{1}B_{1}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,504 |
5. Place five different files $A, B, C, D, E$ into a row of seven drawers numbered $1, 2, 3, 4, 5, 6, 7$, with each drawer containing at most one file. If files $A$ and $B$ must be placed in adjacent drawers, and files $C$ and $D$ must also be placed in adjacent drawers, then the number of different ways to place the f... | 5.C.
Place files $A$ and $B$ in adjacent drawers, denoted as “$AB$”, place files $C$ and $D$ in adjacent drawers, denoted as “$CD$”, and place file $E$ in a drawer, denoted as “$E$”. Thus, “$AB$”, “$CD$”, “$E$”, and the two empty drawers can be considered as five elements, and the number of all permutations of these f... | 240 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,505 |
6. Let the set $M=\left\{a \left\lvert\, a=\frac{x+y}{t}\right., 2^{x}+2^{y}=\right.$ $2^{t}$, where $x, y, t, a$ are all integers $\}$. Then the sum of all elements in the set $M$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8 | 6. D.
Assume $x \leqslant y$, then $2^{x}=2^{x}+2^{y} \leqslant 2^{y}+2^{y}=2^{y+1}$.
Thus, $t \leqslant y+1$.
From $2^{x}>0$, we get $2^{t}=2^{x}+2^{y}>2^{y}$. Therefore, $t>y$.
So, $y<t \leqslant y+1$.
Given that $x, y, t$ are all integers, then $t=y+1$. Hence, $2^{y+1}=$ $2^{x}+2^{y}$, which means $2^{x}=2^{y}$. Th... | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,506 |
7. Given a fixed point $A(4, \sqrt{7})$. If a moving point $P$ is on the parabola $y^{2}=4 x$, and the projection of point $P$ on the $y$-axis is point $M$, then the maximum value of $|P A|-|P M|$ is $\qquad$. | $$
\begin{array}{l}
|P M|=|P N|-|M N|=|P F|-1 \downarrow \\
\text { Then }|P A|-|P M|=|P A|-(|P F|-1) \\
=(|P A|-|P F|)+1 \leqslant|A F|+1=4+1=5 .
\end{array}
$$ | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,507 |
8. Given that the function $f(x)$ is a decreasing function defined on $(-\infty, 3]$, and for $x \in \mathbf{R}$,
$$
f\left(a^{2}-\sin x\right) \leqslant f\left(a+1+\cos ^{2} x\right)
$$
always holds. Then the range of real number $a$ is $\qquad$ | 8. $\left[-\sqrt{2}, \frac{1-\sqrt{10}}{2}\right]$.
From the given, we have $a+1+\cos ^{2} x \leqslant a^{2}-\sin x \leqslant 3$ for $x \in \mathbf{R}$, which means
$$
\left\{\begin{array}{l}
a^{2} \leqslant 3+\sin x, \\
a^{2}-a \geqslant 1+\cos ^{2} x+\sin x
\end{array}\right.
$$
for $x \in \mathbf{R}$.
If $a^{2} \l... | \left[-\sqrt{2}, \frac{1-\sqrt{10}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,508 |
9. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n}+a_{n+1}=1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, then the value of $S_{2003}-2 S_{2004}+S_{2000}$ is $\qquad$ | 9.3.
According to the problem, when $n$ is even, we have $a_{1}+a_{2}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n}{2}$ pairs, so $S_{n}=\frac{n}{2}$. Therefore, $S_{2004}=1002$.
When $n$ is odd, we have $a_{2}+a_{3}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n-1}{2}$ pairs, so
$$... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,509 |
10. As shown in Figure 2, the side length of rhombus $A B C D$ is 1, and $\angle A B C=120^{\circ}$. If
$E$ is any point on the extension of $B C$,
$A E$
intersects $C D$ at point $F$, then
the angle between vector $B F$ and $E D$
is
$\qquad$ | 10.120.
As shown in Figure 5, establish a Cartesian coordinate system. Then
$$
\begin{array}{l}
A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \\
B(0,0), \\
C(1,0), \\
D\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) .
\end{array}
$$
Let $E(a, 0)(a>1)$. Then
$l_{c p}: y=-\sqrt{3}(x-1)$,
$$
l_{\text {AE }}: y=-\frac{\sq... | 120 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,510 |
11. As shown in Figure 3(a), the eight vertices of a cube are assigned values $a, b, c, d, e, f, g, h$, respectively. Then, the arithmetic mean of the values of the three adjacent vertices of each vertex, denoted as $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}, \bar{g}, \bar{h}$, is placed at the corresponding... | 11.20.
According to the problem, we have
$$
\begin{array}{l}
\bar{a}=\frac{b+d+e}{3}, \bar{b}=\frac{a+c+f}{3}, \bar{c}=\frac{b+d+g}{3}, \\
\bar{d}=\frac{a+c+h}{3}, e=\frac{a+f+h}{3}, \bar{f}=\frac{b+e+g}{3}, \\
\bar{g}=\frac{c+f+h}{3}, \bar{h}=\frac{d+e+g}{3} .
\end{array}
$$
Then, $a=(\bar{b}+\bar{d}+\bar{e})-2 \bar... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,511 |
Example 8 A coach can travel from Nanjing to Shanghai on the Shanghai-Nanjing Expressway in $4 \mathrm{~h}$. It is known that a coach departs from Nanjing to Shanghai every half hour, and at the same time, a coach also departs from Shanghai to Nanjing. If a passenger takes a bus from Nanjing to Shanghai, how many coach... | Explanation: This problem can be easily solved by just drawing a schematic diagram (Figure 6).
The traveler encounters 8 trains (including the one arriving at Nanjing station) in the first 2 hours (i.e., traveling half the distance). 2 hours after departure, they encounter exactly 1 train that departed from Shanghai t... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,512 |
12. Given the quadratic function $f(x)$ satisfies $f(-1)=0$, and $x \leqslant f(x) \leqslant \frac{1}{2}\left(x^{2}+1\right)$ for all real numbers $x$. Then, the analytical expression of the function $f(x)$ is $\qquad$ 1. | 12. $\frac{1}{4}(x+1)^{2}$.
Let $f(x)=a x^{2}+b x+c(a \neq 0)$.
From the given, when $x=1$, we have $f(1)=1$, that is,
$$
a+b+c=1 \text {. }
$$
Also, $f(-1)=0$, that is,
$$
a-b+c=0 \text {. }
$$
From equations (1) and (2), we get $a+c=b=\frac{1}{2}$.
Also, $f(x) \geqslant x$, that is, $a x^{2}+(b-1) x+c \geqslant 0$... | \frac{1}{4}(x+1)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,513 |
13. Given the function $f(x)=\left|1-\frac{1}{x}\right|$.
(1) Does there exist real numbers $a, b (a<b)$ such that the domain and range of the function $f(x)$ are both $[a, b]$? If they exist, find the values of $a, b$; if not, explain the reason.
(2) If there exist real numbers $a, b (a<b)$ such that the domain of the... | Three, 13. (1) There do not exist real numbers $a, b (a < b)$ such that $f(x) = mx$ for all $x \in (a, b)$, where $m > 0, a > 0$.
Given:
$$
f(x)=\left\{\begin{array}{l}
1-\frac{1}{x}, x \geqslant 1 ; \\
\frac{1}{x}-1, 0 < x < 1.
\end{array}\right.
$$
By analogy with the solution to (1), we know that when $a, b \in (0... | 0 < m < \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,514 |
14. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, its major axis is $A_{1} A$, and $P$ is a moving point on the ellipse different from points $A_{1}$ and $A$. The lines $P A$ and $P A_{1}$ intersect the same directrix $l$ at points $M$ and $M_{1}$, respectively. Try to prove: the circle with dia... | 14. Given, we can set $A_{1}(-a, 0), A(a, 0)$, and the equation of a directrix $l$ as $x=\frac{a^{2}}{c}$. Let the coordinates of a moving point $P$ on the ellipse be $\left(x_{0}, y_{0}\right)$ with $y_{0} \neq 0$. Then,
$$
l_{P A}: y=\frac{y_{0}}{x_{0}-a}(x-a),
$$
Solving with $x=\frac{a^{2}}{c}$, we get $M\left(\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,515 |
15. If $P$ is a $2n$-digit positive integer composed of the digits $1,2,3,4,5,6,7, 8,9$, and simultaneously satisfies the following two conditions:
(1) The digits $1,2, \cdots, n$ each appear twice in $P$;
(2) Between each pair of the same digit $i(i=1,2, \cdots, n)$, there are exactly $i$ digits.
In this case, we cal... | 15. From the definition of a good number, we know that $n \leqslant 9$.
For a good number $P$, considering the positions of the digits from left to right, if the first occurrence of the digit $i (i=1,2, \cdots, n)$ is denoted as $a_{i}$, then according to the problem, the second occurrence of the digit $i (i=1,2, \cdo... | n=3, 4, 7, 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,516 |
1. For the set $\left\{x, x^{2}-x\right\}$, the range of values for $x$ is ( ).
(A) $(-\infty,+\infty)$
(B) $(-\infty, 0) \cup(0,+\infty)$
(C) $(-\infty, 2) \cup(2,+\infty)$
(D) $(-\infty, 0) \cup(0,2) \cup(2,+\infty)$ | -.1.D.
By the mutual distinctness of the elements of the set, we know that $x$ should satisfy $x \neq x^{2}-x$.
In fact, if $x=x^{2}-x$, then $x=0$ or $x=2$. Therefore, the range of values for $x$ should be all real numbers except 0 and 2. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,517 |
2. If $z^{2}+z+1=0$, then the value of $z^{2005}$ is ( ).
(A) 1
(B) -1
(C) $\frac{1}{2} \pm \frac{\sqrt{3}}{2}$ i
(D) $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$ | $$
\begin{array}{l}
\text { Since } z^{3}-1=(z-1)\left(z^{2}+z+1\right)=0, z^{3}=1 \text {, therefore, } \\
z^{2005}=\left(z^{3}\right)^{668} \cdot z=z . \\
\text { Then } z=\frac{-1 \pm \sqrt{4-1} \mathrm{i}}{2}=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i} .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,518 |
3. The smallest positive period of the function $y=\cos ^{4} x+\sin ^{2} x$ is ( ).
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{2}$
(C) $\pi$
(D) $2 \pi$ | 3. B.
$$
\begin{array}{l}
\text { Since } y=\cos ^{4} x+\sin ^{2} x \\
=\cos ^{2} x\left(1-\sin ^{2} x\right)+\sin ^{2} x=1-\sin ^{2} x \cdot \cos ^{2} x \\
=1-\frac{1}{4} \sin ^{2} 2 x=\frac{7}{8}+\frac{1}{8} \cos 4 x,
\end{array}
$$
Therefore, the smallest positive period of the function $y=\cos ^{4} x+\sin ^{2} x$ ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,519 |
4. Toss a six-sided die with 1, 2, $3,4,5,6$ dots on each face randomly, the mathematical expectation of the number of dots that appear (i.e., the face that is up) is ( ).
(A) 3
(B) 3.5
(C) 4
(D) 4.5 | 4.B.
The probability of rolling a 1, 2, 3, 4, 5, or 6 is $\frac{1}{6}$ each, so the mathematical expectation of the number of points that appear is
$$
\frac{1}{6}(1+2+3+4+5+6)=\frac{21}{6}=3.5 .
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,520 |
6. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of ( ).
(A) 0
(B) 2
(C) 3
(D) 5 | 6.D.
$$
\begin{array}{l}
6^{11}+C_{11}^{1} 6^{10}+\mathrm{C}_{11}^{2} 6^{6}+\cdots+C_{116}^{10} 6-1 \\
=(6+1)^{11}-2=7^{11}-2 . \\
7^{11}-2=7^{11}+1-3 \\
=(7+1)\left(7^{10}-7^{9}+\cdots-7+1\right)-3 \\
=8\left(7^{10}-7^{9}+\cdots-7+1\right)-3 \\
=8\left(7^{10}-7^{9}+\cdots-7\right)+5 .
\end{array}
$$
Therefore, the re... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,522 |
1. A person helps sail a sailboat downstream along the river, and encounters a speedboat coming upstream. He asks the speedboat driver: "Is there a ship coming behind you?" The speedboat driver answers: "I passed a ship half an hour ago." The sailboat continues to sail for another half an hour and meets the ship coming... | (Answer: B.)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The instruction provided seems to be a meta-instruction about how to handle the translation, not part of the text to be translated. The actual translation of ... | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 716,523 |
7. The solution set of the inequality $\left|\frac{1}{\log _{\frac{1}{3}} x+1}\right|>\frac{1}{3}$ is ( ).
(A) $\left(\frac{1}{9}, 81\right)$
(B) $\left(\frac{1}{3}, 3\right) \cup(3,27)$
(C) $\left(\frac{1}{3}, 27\right)$
(D) $\left(\frac{1}{9}, 3\right) \cup(3,81)$ | 7.D.
The original inequality is
$$
\left\{\begin{array}{l}
\log _{\frac{1}{3}} x+1 \neq 0, \\
\left|\log _{\frac{1}{3}} x+1\right| > 0, \text{ and } x \neq 3.
\end{array}\right.
$$
From equation (2), we get $-3 < \log _{\frac{1}{3}} x+1 < 3$, then
$$
\left(\frac{1}{3}\right)^{2} < x < \left(\frac{1}{3}\right)^{-4} \te... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,524 |
8. When $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, the function $f(x)$ satisfies
$2 f(-\sin x)+3 f(\sin x)=\sin 2 x$.
Then $f(x)$ is ( ).
(A) odd function
(B) neither odd nor even function
(C) even function
(D) both odd and even function | 8. A.
Let $t=\sin x$, then $\sin 2 x=2 t \sqrt{1-t^{2}}$, so
$$
2 f(-t)+3 f(t)=2 t \sqrt{1-t^{2}}.
$$
Thus, $2 f(t)+3 f(-t)=-2 t \sqrt{1-t^{2}}$.
(1) + (2) gives
$$
5(f(-t)+f(t))=0 \quad (-1 \leqslant t \leqslant 1).
$$
Therefore, $f(-t)=-f(t)$.
Hence, $f(x)$ is an odd function on $[-1,1]$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,525 |
9. Point $P$ is on the right branch of the hyperbola $x^{2}-y^{2}=a^{2}$, $A_{1}$ and $A_{2}$ are the left and right vertices of the hyperbola, respectively, and $\angle A_{2} P A_{1}=2 \angle P A_{1} A_{2}$. Then $\angle P A_{1} A_{2}=$ ( ).
(A) $30^{\circ}$
(B) $27.5^{\circ}$
(C) $25^{\circ}$
(D) $22.5^{\circ}$ | 9.D.
As shown in Figure 5, draw $PD \perp A_{1}A_{2}$, intersecting the extension of $A_{1}A_{2}$ at point $D$. Let $P(x, y)$, then $y^{2}=x^{2}-a^{2}=(x-a)(x+a)$, which means $PD^{2}=A_{2}D \cdot A_{1}D$.
Therefore, $\mathrm{Rt} \triangle PDA_{2} \sim \mathrm{Rt} \triangle A_{1}DP$. Thus, $\angle DPA_{2}=\angle PA_{... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,526 |
10. As shown in Figure 2, in the regular tetrahedron $A B C D$, $A E=\frac{1}{4} A B$, $C F=\frac{1}{4} C D$. Then the angle between the lines $D E$ and $B F$ is ( ).
(A) $\arccos \frac{4}{13}$
(B) $\arccos \frac{3}{13}$
(C) $\pi-\arccos \frac{4}{13}$
(D) $\pi-\arccos \frac{3}{13}$ | 10.A.
Let the edge length of a regular tetrahedron be $a$, then
$$
\begin{array}{l}
A E=C F=\frac{1}{4} a, \\
D E^{2}=a^{2}+\left(\frac{1}{4} a\right)^{2}-\frac{1}{2} a^{2} \cos 60^{\circ}=\frac{13}{16} a^{2} .
\end{array}
$$
Thus, $D E=\frac{\sqrt{13}}{4} a$.
Similarly, $B F=\frac{\sqrt{13}}{4} a$.
As shown in Figur... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,527 |
11. Given the function
$$
f(x)=\left\{\begin{array}{ll}
-x-1, & -1 \leqslant x < 1 \\
x+1, & x \geqslant 1
\end{array}\right.
$$
the solution set of $f(x) > -1$ is
$\qquad$ | $$
=.11 \cdot\left[-1,-\frac{1}{2}\right) \cup(0,1] \text {. }
$$
(1) When $-1 \leqslant x-1$.
Solving for $x-1$.
Solving for $x<\frac{3}{2}$. Then $0<x \leqslant 1$.
In summary, the solution set is $\left[-1,-\frac{1}{2}\right) \cup(0,1]$. | \left[-1,-\frac{1}{2}\right) \cup(0,1] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,528 |
13. Let two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in a plane be perpendicular to each other, and $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$. Also, $k$ and $t(t \geqslant 0)$ are two real numbers that are not both zero. If the vectors
$$
\boldsymbol{x}=\boldsymbol{a}+(3-t) \boldsymbol{b} \text { and } \boldsymbol{y... | 13.1.
Given $\boldsymbol{a} \cdot \boldsymbol{b}=0$.
$$
\begin{array}{l}
\boldsymbol{x} \cdot \boldsymbol{y}=[\boldsymbol{a}+(3-t) \boldsymbol{b}] \cdot\left(-k \boldsymbol{a}+t^{2} \boldsymbol{b}\right) \\
=-k \boldsymbol{a}^{2}+\left[-k(3-t)+t^{2}\right] \boldsymbol{a} \cdot \boldsymbol{b}+t^{2}(3-t) \boldsymbol{b}^... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,530 |
14. In a promotional sales event with prizes, there are $n$ $(n \geqslant 4)$ people who win the third prize. The third prize consists of four different items, and each winner of the third prize can freely choose one of the four items. The probability that one of the items is not chosen is $\qquad$ | $14 \cdot 4\left(\frac{3}{4}\right)^{n}-12\left(\frac{1}{2}\right)^{n}+\frac{3}{4^{n-1}}$.
Since each person has 4 different choices for the prize, the total number of different possible choices is $4^{n}$.
One prize is not chosen, meaning that three prizes are chosen by someone.
（1）Only one prize is chosen, clearly, t... | 4\left(\frac{3}{4}\right)^{n}-12\left(\frac{1}{2}\right)^{n}+\frac{3}{4^{n-1}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,531 |
15. (12 points) A person takes a loan of $s$ yuan from the bank to buy a house, with an annual interest rate of $p$. The person returns the principal and interest to the bank every two years, and pays off the loan in ten years (requiring the payment amount to the bank to be the same each time). What is the total amount... | Three, 15. Let the principal owed to the bank at the $i$-th payment be $s_{i}$ yuan, and the principal returned in the payment be $a_{i}$ yuan, with the interest being
$$
s_{i}\left[(1+p)^{2}-1\right] \text { (yuan). }
$$
According to the problem, the payment amount to the bank each time is the same, so we have
$$
\be... | \frac{5 s\left[(1+p)^{2}-1\right](1+p)^{10}}{(1+p)^{10}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,532 |
16. (12 points) As shown in Figure 3, the oblique triangular prism $A B C-A_{1} B_{1} C_{1}$ has a side face $A A_{1} C_{1} C$ which is a rhombus with an area of $\frac{\sqrt{3}}{2}$, and $\angle A C C_{1}$ is an acute angle. The side face $A B B_{1} A_{1} \perp$ side face $A A_{1} C_{1} C$, and $A_{1} B=A B=A C=1$.
(1... | 16. Quadrilateral $A A_{1} C_{1} C$ is a rhombus, so,
$$
A A_{1}=A_{1} C_{1}=C_{1} C=C A=1 \text {. }
$$
Thus, we know $\triangle A A_{1} B$ is an equilateral triangle.
Let $D$ be the midpoint of $A A_{1}$, then $B D \perp A A_{1}$.
Also, the side face $A B B_{1} A_{1} \perp$ side face $A A_{1} C_{1} C$, so, $B D \per... | \frac{\sqrt{15}}{5} | Geometry | proof | Yes | Yes | cn_contest | false | 716,533 |
2. A model car racer remotely controls a car, which first moves forward $1 \mathrm{~m}$, then rotates counterclockwise by an angle $\alpha\left(0^{\circ}<\alpha<180^{\circ}\right)$, which is called one operation. If after five operations, the car returns to the starting point, then the angle $\alpha$ is ( ).
(A) $72^{\... | (Tip: Use the theorem of the sum of interior and exterior angles. Answer: D. ) | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,534 |
17. (12 points) Let $f(x)=x^{2}+b x+c$ (where $b, c$ are constants), and the equation $f(x)=x$ has two real roots $x_{1}$, $x_{2}$, satisfying $x_{1}>0, x_{2}-x_{1}>1$.
(1) Prove that $b^{2}>2(b+2 c)$;
(2) If $0<t<x_{1}$, compare the size of $f(t)$ and $x_{1}$. | 17. (1) The equation $f(x)=x$, i.e., $x^{2}+(b-1) x+c=0$, has two real roots $x_{1}$ and $x_{2}$. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-(b-1), x_{1} x_{2}=c . \\
\text { Then }\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2} \\
=[-(b-1)]^{2}-... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,535 |
18. (15 points) As shown in Figure 4, draw two perpendicular chords $OA$ and $OB$ from the origin $O$ to the parabola $y^{2}=2 p x(p>0)$, and then draw the angle bisector of $\angle AOB$ intersecting $AB$ at point $C$. Find the equation of the locus of point $C$. | 18. Let \( A\left(2 p t_{1}^{2}, 2 p t_{1}\right) 、 B\left(2 p t_{2}^{2}, 2 p t_{2}\right) \), then
\[
k_{O A}=\frac{2 p t_{1}}{2 p t_{1}^{2}}=\frac{1}{t_{1}}, k_{O B}=\frac{2 p t_{2}}{2 p t_{2}^{2}}=\frac{1}{t_{2}} \text {. }
\]
Since \( O A \perp O B \), we have \( k_{O A} \cdot k_{O B}=-1 \), i.e., \(\frac{1}{t_{1}... | x^{3}+3 x y^{2}-2 p\left(x^{2}-y^{2}\right)=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,536 |
19. (15 points) There are 800 points on a circle, labeled $1, 2, \cdots, 800$ in a clockwise direction. They divide the circle into 800 gaps. Arbitrarily choose one point and color it red, then color the remaining points red according to the following rule: if the $k$-th point has been colored red, then move clockwise ... | 19. Generally, for a circle with $n$ points, we denote the maximum number of points that can be colored red according to the given rules as $f(n)$.
If the circle has $2 n$ points, and the first point to be colored red is labeled $i$.
(1) If $i=2 k (k \geqslant 1)$ is an even number, then all the points that are colore... | 25 | Number Theory | proof | Yes | Yes | cn_contest | false | 716,537 |
1. Given that functions $h(x)$ and $g(x)$ are both increasing functions defined on the interval $(0,+\infty)$, and let the function $f(x)=$ $h(x) g(x)$. Then the function $f(x)$ on the interval $(0,+\infty)$ ( ).
(A) must be an increasing function
(B) may be an increasing function, or may be a decreasing function, one ... | - 1.D.
Example: $h(x)=-\frac{1}{x}, g(x)=\ln x$, which is an increasing function on $(0,+\infty)$. But $f(x)=h(x) g(x)=-\frac{\ln x}{x}$, its derivative is $f^{\prime}(x)=\frac{\ln x-1}{x^{2}}$.
$$
\begin{array}{l}
\text { On }(0, \mathrm{e}) \text {, } f^{\prime}(x)0 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,538 |
2. Given $\sin 2 x=\frac{\sin \theta+\cos \theta}{2}$, $\cos ^{2} x=\sin \theta \cdot \cos \theta$.
Then, the value of $\cos 2 x$ is ( ).
(A) $\frac{-1 \pm \sqrt{33}}{8}$
(B) $\frac{-1+\sqrt{33}}{8}$
(C) $\frac{-1-\sqrt{33}}{8}$
(D) 0 | 2. C.
Notice
$$
\begin{array}{l}
\sin ^{2} 2 x=\left(\frac{\sin \theta+\cos \theta}{2}\right)^{2}=\frac{1+2 \sin \theta \cdot \cos \theta}{4} \\
=\frac{1+2 \cos ^{2} x}{4},
\end{array}
$$
then $1-\cos ^{2} 2 x=\frac{2+\cos 2 x}{4}$.
Thus, $\cos 2 x=\frac{-1 \pm \sqrt{33}}{8}$.
$$
\begin{array}{l}
\text { Also, } \cos... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,539 |
4. From 6 pairs of different shoes, the probability of picking 4 shoes such that at least 2 of them form a pair is ( ).
(A) $\frac{6}{11}$
(B) $\frac{17}{33}$
(C) $\frac{19}{33}$
(D) $\frac{7}{165}$ | 4.B.
From 6 pairs of different shoes, if 4 are chosen at random, the probability that no 2 shoes form a pair is $\frac{\mathrm{C}_{6}^{4} \cdot 2^{4}}{\mathrm{C}_{12}^{4}}=\frac{16}{33}$. Therefore, the probability that at least 2 shoes form a pair is $1-\frac{16}{33}=\frac{17}{33}$. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,541 |
5. As shown in Figure 1, given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0), A A^{\prime}$ is a chord passing through the right focus $F$ and perpendicular to the real axis $x$, $M$ is the right vertex of the hyperbola, and $\angle A M A^{\prime}=\alpha$. Then, ( ).
(A) $\alpha$ could be $90^{\... | 5.D.
Point $F\left(\sqrt{a^{2}+b^{2}}, 0\right)$, then $A\left(\sqrt{a^{2}+b^{2}}, \frac{b^{2}}{a}\right)$.
Thus, $\tan \frac{a}{2}=\frac{A F}{M F}=\frac{\frac{b^{2}}{a}}{\sqrt{a^{2}+b^{2}}-a}=\frac{\sqrt{a^{2}+b^{2}}+a}{a}$ $>2>\sqrt{3}$.
Therefore, $60^{\circ}<\frac{\alpha}{2}<90^{\circ}$, which means $120^{\circ}<... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,542 |
6. If $x^{2}+y^{2}=169$, then the function
$$
\begin{aligned}
f(x, y)= & \sqrt{24 y-10 x+338}+ \\
& \sqrt{24 y+10 x+338}
\end{aligned}
$$
has the maximum value of ( ).
(A) $10 \sqrt{26}$
(B) $13 \sqrt{2}$
(C) $5 \sqrt{26}$
(D) $26 \sqrt{2}$ | 6.A.
From $x^{2}+y^{2}=169$, we know
$[f(x, y)]^{2}=48 y+676+2 \sqrt{(24 y+338)^{2}-(10 x)^{2}}$.
When $y$ is larger and $|x|$ is smaller, $[f(x, y)]^{2}$ is larger.
Therefore, when $y=13, x=0$, $f(x, y)=10 \sqrt{26}$ is the maximum value. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,543 |
Example 1 As shown in Figure 1, $\triangle ABC$ is an inscribed triangle in a circle and is a scalene triangle. Through points $A, B, C$, draw the tangents to the circle, intersecting lines $BC, CA, AB$ at $A', B', C'$ respectively. Prove that points $A', B', C'$ are collinear. | Let $B C=a, C A=b, A B=c$. It is easy to know that $\frac{A C^{\prime}}{C^{\prime} B}=\frac{S_{\triangle C^{\prime} C}}{S_{\triangle C C^{\prime} B}}$.
It is also easy to prove that $\triangle A C^{\prime} C \backsim \triangle C C^{\{\prime} B$. Therefore, $\frac{S_{\triangle C^{\prime} C}}{S_{\triangle C C^{\prime} B}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,545 |
Example 2 As shown in Figure 2, with one side $BC$ of the acute $\triangle ABC$ as the diameter, construct $\odot O$. Draw two tangents from point $A$ to $\odot O$, with the points of tangency being $M$ and $N$. Point $H$ is the orthocenter of $\triangle ABC$. Prove that points $M, H, N$ are collinear. | Ray $A H$ intersects $B C$ at $D$, obviously $A D$ is the altitude. Let the intersection of $A B$ and $\odot O$ be $E$, it is easy to know that $C, H, E$ are collinear.
Connect $O M, O N, D M, D N, M H, H N$. It is easy to see that
$$
\angle A M O=\angle A N O=\angle A D O=90^{\circ}.
$$
Thus, $A, M, O, D, N$ are conc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,546 |
Example 4 In $\triangle A B C$, $\angle A B C=70^{\circ}, \angle A C B$ $=30^{\circ}, P, Q$ are two points inside the triangle, such that $\angle Q B C=$ $\angle Q C B=10^{\circ}, \angle P B Q=\angle P C B=20^{\circ}$. Prove: $A$, $P$, $Q$ are collinear. ${ }^{[2]}$ | Proof 1: As shown in Figure 4, connect $A Q$ and $P Q$. Since
$$
\begin{array}{l}
\angle B A C \\
=80^{\circ}, \\
\angle B Q C \\
=160^{\circ}, \\
B Q=Q C,
\end{array}
$$
Therefore, point $Q$ is the circumcenter of $\triangle A B C$.
Also, $\angle A B Q=60^{\circ}$, so $\triangle A B Q$ is an equilateral triangle.
Thu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,547 |
5. In Rt $\triangle A B C$, $\angle C=90^{\circ}, A C=3$, $B C=4$. Then the value of $\cos (A-B)$ is ( ).
(A) $\frac{3}{5}$
(B) $\frac{4}{5}$
(C) $\frac{24}{25}$
(D) $\frac{7}{25}$ | 5.C.
In $\triangle A B C$, construct $\angle B A D=\angle B$ (with $D$ on side $B C$), then $B D=A D$.
Let $A D=x$, then $C D=4-x$.
In the right triangle $\triangle A C D$, by the Pythagorean theorem we have
$$
3^{2}+(4-x)^{2}=x^{2} .
$$
Solving for $x$ gives $x=\frac{25}{8}$.
Thus, $\cos (A-B)$
$$
=\cos \angle C A D... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,548 |
6. Given that quadrilateral $A B C D$ is a trapezoid, and $A D / /$ $B C, A D < B C$. Also, $\odot O$ is tangent to $A B$, $A D$, and $C D$ at points $E$, $F$, and $G$ respectively, with the center $O$ on $B C$. Then the relationship between $A B +$ $C D$ and $B C$ is ( ).
(A) greater than
(B) equal to
(C) less than
(D... | 6. B.
As shown in Figure 2, let the radius of $\odot O$ be $r$. Connect $O E, O F, O G,$ and $O A, O D$. Then $O E \perp A B, O F \perp A D, O G \perp C D$. Therefore,
$$
O E=O F=O G=r, O F
$$
is the height of trapezoid $A B C D$.
$S_{\triangle O A D}+S_{\triangle O C D}$, we get
$$
\frac{1}{2}(A D+B C)_{r}=\frac{1}{... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,549 |
1. Given that quadrilateral $A B C D$ is inscribed in $\odot O$, and $A D$ $/ / B C$. Then the shape of quadrilateral $A B C D$ is $\qquad$ . | 2.1. Isosceles Trapezoid or Rectangle.
When $A D \neq B C$, by $A D / / B C$ we get $A B=C D$, at this time, quadrilateral $A B C D$ is an isosceles trapezoid.
When $A D=B C$, then quadrilateral $A B C D$ is a parallelogram. Therefore, $\angle A=\angle C$. Also, $\angle A+\angle C=180^{\circ}$, so, $\angle A=$ $90^{\c... | Isosceles Trapezoid or Rectangle | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,550 |
2. If $\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}=$
1 , then $a b c=$ | 2.1 .
From the given, we have
$$
\begin{array}{l}
\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\left(\frac{c}{c a+c+1}-1\right)=0 \\
\Rightarrow \frac{a}{a b+a+1}+\frac{b}{b c+b+1}-\frac{c a+1}{c a+c+1}=0 \\
\Rightarrow\left(\frac{a}{a b+a+1}-\frac{c a}{c a+c+1}\right)+\left(\frac{b}{b c+b+1}-\frac{1}{c a+c+1}\right)=0 \\
\Rig... | a b c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,551 |
3. Let $a, b, c$ be positive numbers, and
$$
a^{2}+b^{2}+c^{2}-2 a b-2 a c-2 b c<0 \text {. }
$$
Then with $\sqrt{a}, \sqrt{b}, \sqrt{c}$ as sides \qquad a triangle can be formed (fill in “can” or “cannot”).
| 3. Can.
Assume $a \geqslant b \geqslant c$. Because
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-2 a b-2 a c-2 b c \\
= {\left[\left(a^{2}+2 a b+b^{2}\right)-2(a c+b c)+c^{2}\right]-4 a b } \\
=(a+b-c)^{2}-(2 \sqrt{a b})^{2} \\
=(a+b-c+2 \sqrt{a b})(a+b-c-2 \sqrt{a b}) \\
= {\left[(\sqrt{a}+\sqrt{b})^{2}-(\sqrt{c})^{2}\righ... | Can | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,552 |
4. In the Cartesian coordinate system, let point $A(4,-5)$, $B(8,-3)$, $C(m, 0)$, $D(0, n)$. When the perimeter of quadrilateral $A B C D$ is minimized, the value of $\frac{m}{n}$ is $\qquad$. | 4. $-\frac{3}{2}$.
As shown in Figure 3, the length of $AB$ is determined. To minimize the perimeter of quadrilateral $ABCD$, we need to minimize $AD + DC + CB$. Construct the point $A'$, the reflection of point $A$ over the $y$-axis, and the point $B'$, the reflection of point $B$ over the $x$-axis. Connect $A'B'$, w... | -\frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,553 |
One, (20 points) Find all integer solutions to the equation $a^{3} b-a b^{3}+2 a^{2}+2 b^{2}+4=0$.
| One, transform the original equation to
$$
2^{2}+\left(a^{2}+b^{2}\right) \times 2+\left(a^{3} b-a b^{3}\right)=0 \text {. }
$$
Let $2=x$, then
$$
x^{2}+\left(a^{2}+b^{2}\right) x+\left(a^{3} b-a b^{3}\right)=0 \text {. }
$$
Considering the above equation as a quadratic equation in $x$, by the quadratic formula we ge... | (a, b) = (-3, 1), (3, -1), (-3, 2), (3, -2), (1, 3), (-1, -3), (2, 3), (-2, -3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,554 |
II. (25 points) In $\triangle A B C$, $D$ is the midpoint of $B C$, $G$ is a moving point on $A D$ (excluding $A, D$), the extensions of $B G, C G$ intersect $A C, A B$ at $E, F$ respectively. Find the maximum value of $\frac{S_{\triangle B G F}+S_{\triangle C G E}}{S_{\triangle A B C}}$. | As shown in Figure 4, it is evident that
$$
\frac{B D}{C D}=\frac{S_{\triangle A E D}}{S_{\triangle A C D}}=\frac{S_{\triangle B C D}}{S_{\triangle C G D}}=\frac{S_{\triangle A C B}}{S_{\triangle A C C}},
$$
$$
\frac{C E}{E A}=\frac{S_{\triangle B C E}}{S_{\triangle B A E}}=\frac{S_{\triangle C C E}}{S_{\triangle A C E... | 6-4 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,555 |
Three, (25 points) Find the equation of the parabola obtained by rotating the parabola $y=x^{2}-2 x-3$ around point $A(3,0)$ by $90^{\circ}$ (Note: $x$ is a quadratic function of $y$). | Three, as shown in Figure 5, $A(3,0)$ is the intersection point of the parabola with the positive half of the $x$-axis. Since
$$
\begin{array}{l}
y=x^{2}-2 x-3 \\
=(x-1)^{2}-4,
\end{array}
$$
the vertex of the original parabola is $P(1,-4)$, and the intersection point of the axis of symmetry with the $x$-axis is $B(1,... | x=y^{2}-4 y+3 \text { or } x=-y^{2}-4 y+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,556 |
1. Given a function $f(x)$ defined on $\mathbf{R}$ whose graph is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, and satisfies $f(x)=-f\left(x+\frac{3}{2}\right), f(-1)=1, f(0)=-2$. Then, the value of $f(1)+f(2)+\cdots+f(2006)$ is ( ).
(A) 1
(B) 2
(C) -1
(D) -2 | $-1 . B$.
From the graph of the function $f(x)$ being centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, we know that $f(x)=-f\left(-x-\frac{3}{2}\right)$.
Also, $f(x)=-f\left(x+\frac{3}{2}\right)$, so $f\left(-x-\frac{3}{2}\right)=f\left(x+\frac{3}{2}\right)$.
Therefore, $f(x+3)=f\left(\left(x+\frac{3... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,557 |
Example 5 As shown in Figure $5, \triangle A B C$ is tangent to $\odot O$, with the points of tangency on sides $B C, C A, A B$ being $D, E, F$ respectively. Ray $D O$ intersects $E F$ at point $A^{\prime}$, and similarly define points $B^{\prime}$ and $C^{\prime}$. Prove that $A A^{\prime}, B B^{\prime}, C C^{\prime}$... | Proof: Since $D A^{\prime} 、 E B^{\prime} 、 F C^{\prime}$ are concurrent at point $O$, by Ceva's Theorem, we have
$$
\begin{array}{l}
\frac{F A^{\prime}}{A^{\prime} E} \cdot \frac{E C^{\prime}}{C^{\prime} D} \cdot \frac{D B^{\prime}}{B^{\prime} F} \\
=1 .
\end{array}
$$
Since $D 、 E 、$. $F$ are all points of tangency,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,558 |
2. Given $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P$ is any point on the right branch of the hyperbola. If the minimum value of $\frac{\left|P F_{1}\right|^{2}}{\left|P F_{2}\right|}$ is $8 a$, then the range of the eccentricity $e$ ... | 2. A.
$$
\begin{array}{l}
\frac{\left|P F_{1}\right|^{2}}{\left|P F_{2}\right|}=\frac{\left(2 a+\left|P F_{2}\right|\right)^{2}}{\left|P F_{2}\right|}=\frac{4 a^{2}}{\left|P F_{2}\right|}+\left|P F_{2}\right|+4 a \\
\geqslant 2 \sqrt{4 a^{2}}+4 a=8 a,
\end{array}
$$
When and only when $\frac{4 a^{2}}{\left|P F_{2}\rig... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,559 |
3. If for any rectangular prism $A$, there exists a rectangular prism $B$ of the same height as $A$, such that the ratio of the lateral surface area of $B$ to that of $A$ and the ratio of the volume of $B$ to that of $A$ are both equal to $\lambda$, then the range of values for $\lambda$ is ( ).
(A) $\lambda>0$
(B) $0<... | 3.D.
Let the length and width of the base of rectangular prism $A$ be $a_{1}$ and $a_{2}$, and the length and width of the base of rectangular prism $B$ be $b_{1}$ and $b_{2}$. From the given conditions, we have
$$
\lambda=\frac{b_{1}+b_{2}}{a_{1}+a_{2}}=\frac{b_{1} b_{2}}{a_{1} a_{2}},
$$
which implies $b_{1}+b_{2}=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,560 |
4. Let $S_{n}=1+\frac{1}{1+\frac{1}{3}}+\frac{1}{1+\frac{1}{3}+\frac{1}{6}}+\cdots+$ $\frac{1}{1+\frac{1}{3}+\frac{1}{6}+\cdots+\frac{1}{k_{n}}}$, where $k_{n}=\frac{n(n+1)}{2}$ $\left(n \in \mathbf{N}_{+}\right)$, and let $T_{0}$ be the largest integer $T$ that satisfies the inequality $S_{2006}>T$. Among the followin... | 4.C.
$$
\begin{array}{l}
\text { Given } \frac{1}{a_{n}}=1+\frac{1}{3}+\frac{1}{6}+\cdots+\frac{2}{n(n+1)} \\
=2\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{n(n+1)}\right] \\
=2\left(1-\frac{1}{n+1}\right)=\frac{2 n}{n+1},
\end{array}
$$
we know that $a_{n}=\frac{n+1}{2 n}=\frac{1}{2}\left(1+\frac{1... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,561 |
5. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. If none of $x, y, z$ is more than twice another, then the minimum value of the product $xyz$ is ( ).
(A) $\frac{1}{127}$
(B) $\frac{4}{125}$
(C) $\frac{1}{36}$
(D) $\frac{1}{32}$ | 5.D.
Let $m=x_{0} y_{0} z_{0}$ be the minimum value of the product, and assume $x_{0} \leqslant y_{0} \leqslant z_{0}$. Suppose $x_{1}=x_{0}-t, z_{1}=z_{0}+t$, where $t>0$, and sufficiently small so that $z_{1} \leqslant 2 x_{1}$.
Thus, $x_{1}, y_{0}, z_{1}$ still satisfy the original constraints, therefore,
$x_{1} y_... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,562 |
6. Given the set $S=\{1,2,3,4,5,6\}$, a one-to-one mapping $f: S \rightarrow S$ satisfies the condition: for any $x \in S$, $f(f(f(x)))=x$. Then the number of mappings $f$ that satisfy the condition is ( ).
(A) 81
(B) 80
(C) 40
(D) 27 | 6.A.
It is clear that there does not exist $x \in S$ such that $f(x) \neq x$ and $f(f(x)) = x$. Otherwise, $f(f(f(x))) = f(x) \neq x$, which contradicts the given condition. Therefore, for any $x \in S$, either $f(x) = x$, or $f(x) = x_{1}$, $f(x_{1}) = x_{2}$, $f(x_{2}) = x$, and $x, x_{1}, x_{2}$ are distinct. Hence... | 81 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,563 |
1. As shown in Figure $1, O, A, B$ are three points on a plane, vectors $\boldsymbol{O A}$ $=a, O B=b$. On the plane $A O B$, $P$ is any point on the perpendicular bisector of line segment $A B$, vector $O P=p$, and $|a|=3$, $|b|=2$. Then $p \cdot(a-b)=$ | ii. $1 \cdot \frac{5}{2}$.
Let the perpendicular bisector of $AB$ intersect $AB$ at $C$, and connect $OC$. Then $OC=\frac{1}{2}(a+b)$. Therefore,
$$
PC=OC-OP=\frac{1}{2}(a+b)-p.
$$
Since $AB \perp PC$, i.e., $AB \cdot PC=0$, we have,
$(b-a) \cdot\left[\frac{1}{2}(a+b)-p\right]=0$.
Thus, $p \cdot(a-b)=\frac{1}{2}\left(... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,564 |
2. Two Go teams, Team A and Team B, each have 5 players who compete in a relay match according to a predetermined order. The No. 1 players of both teams start the match, and the loser is eliminated; then the No. 2 player of the losing team competes against the winner, and the loser is eliminated again. This continues u... | 2. $\frac{5}{18}$.
Since a total of 9 players from both teams have been eliminated, the match must have lasted 9 rounds, and Player 5 of Team A must have defeated Player 5 of Team B, with their order being fixed. Additionally, the number of ways the 5 players from Team A can appear is $\mathrm{C}_{10}^{5}$, and the nu... | \frac{5}{18} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,565 |
3. Let $m$ and $n$ be positive integers. If the equation
$$
m+\cos x=n(m+\sin x)
$$
has real solutions, then the conditions that $m$ and $n$ should satisfy are | 3. $\left\{\begin{array}{l}m=1, \\ n \in \mathbf{N}_{+}\end{array}\right.$ or $\left\{\begin{array}{l}m \in \mathbf{N}_{+}, \\ n=1\end{array}\right.$ or $\left\{\begin{array}{l}m=2, \\ n=2 .\end{array}\right.$
The original equation can be transformed into
$$
\cos (x+\theta)=\frac{m(n-1)}{\sqrt{n^{2}+1}},
$$
where $\t... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,566 |
4. As shown in Figure 2, the radius of hemisphere $O$ is $R$, and the rectangular prism $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ is inscribed in the hemisphere $O$ with one of its faces $A B C D$ on the base of the hemisphere $O$. Then the maximum value of the sum of all the edges of the rectangular prism is | 4.12R.
Let $A B=a, A D=b, A A_{1}=c$, and connect $O A_{1}$.
In the right triangle $\triangle A_{1} A O$, we have $\frac{a^{2}+b^{2}}{4}+c^{2}=R^{2}$.
Thus, $\frac{5}{4} a^{2}+\frac{5}{4} b^{2}+8 c^{2}$
$=\left(a^{2}+b^{2}\right)+\left(\frac{b^{2}}{4}+4 c^{2}\right)+\left(4 c^{2}+\frac{a^{2}}{4}\right)$
$\geqslant 2 a... | 12R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,567 |
5. If there exists a permutation $a_{1}$, $a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$, such that $k+a_{k}(k=1,2, \cdots, n)$ are all perfect squares, then $n$ is called a "middle number". Then, in the set $\{15,17,2000\}$, the number of elements that are middle numbers is $\qquad$ . | 5.3.
(1) 15 is a median number. Because in the arrangement of Table 1, $k+a_{k}$ $(k=1,2, \cdots, 15)$ are all perfect squares:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline$a_{k}$ & 15 & 14 & 13 & 12 & 11 & 10 & 9 & 8 & ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,568 |
Example 6 As shown in Figure 6, construct $\triangle A B F$ and $\triangle A C E$ outside $\triangle A B C$ with sides $A B$ and $A C$ respectively, such that $\triangle A B F \backsim \triangle A C E$, and $\angle A B F=90^{\circ}$. Prove that $B E$, $C F$, and the altitude $A H$ from $A$ to side $B C$ are concurrent. | Prove: Applying the trigonometric form of Ceva's theorem to $\triangle A B C$ with respect to points $E$ and $F$ respectively, we have
$$
\begin{array}{l}
\frac{\sin \angle A B E}{\sin \angle E B C} \cdot \frac{\sin \angle B C E}{\sin \angle E C A} \cdot \frac{\sin \angle C A E}{\sin \angle E A B}=1, \\
\frac{\sin \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,569 |
6. Let $P$ be a fixed point on the circle $x^{2}+y^{2}=1$, and points $Q$ and $R$ are inside or on the circumference of the circle, and $\triangle P Q R$ is an equilateral triangle with side length $\frac{2}{\sqrt{3}}$. Then the minimum value of $|O Q|^{2}+|O R|^{2}$ is | 6. $\frac{2}{3}$.
Without loss of generality, let $P(-1,0)$, then the equation of the circle with $P$ as the center and $\frac{2}{\sqrt{3}}$ as the radius is
$$
(x+1)^{2}+y^{2}=\frac{4}{3} \text {. }
$$
It is easy to see that the two intersection points of $\odot P$ and the circle $x^{2}+y^{2}=1$ are $A\left(-\frac{1... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,570 |
Three. (20 points) Let the function $f(x)=a^{x}+3 a(a>$ 0 , and $a \neq 1)$ have the inverse function $y=f^{-1}(x)$, and the graph of the function $y=g(x)$ is symmetric to the graph of $y=f^{-1}(x)$ about the point $(a, 0)$.
(1) Find the analytical expression of the function $y=g(x)$.
(2) Does there exist a real number... | (1) From $f(x)=a^{x}+3 a$, we get $f^{-1}(x)=\log _{a}(x-3 a)(x>3 a)$.
Also, the graph of $y=g(x)$ is symmetric to the graph of $y=f^{-1}(x)$ about the point $(a, 0)$, thus we have
$$
g(x)=-\log _{a}(-x-a)(x3 a),
$$
which means $-1 \leqslant \log _{a}\left(x^{2}-4 a x+3 a^{2}\right) \leqslant 1$.
From $3 a0$, we get $... | 0<a \leqslant \frac{9-\sqrt{57}}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,571 |
Four, (20 points) Let $k(k \geqslant 3)$ positive integers $a_{1}, a_{2}$, $\cdots, a_{k}$ form a geometric sequence in order. Prove that when $k \geqslant a_{1}$, the common ratio $q$ of this geometric sequence must be a positive integer. | Assume $q$ is not a positive integer, let's say $q=\frac{r}{t}$, where $(t, r)=1$, and $t \geqslant 2$.
Therefore, $\left(t^{k-1}, r^{k-1}\right)=1(k \geqslant 3)$.
By $a_{k}=a_{1} q^{k-1}=\frac{a_{1} r^{k-1}}{t^{k-1}}$, and since $a_{k} \in \mathbf{N}_{+}$, it follows that,
$t^{k-1} \mid a_{1}$.
Let $a_{1}=m t^{k-1}\l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,572 |
Five. (20 points) Given the parabola $y^{2}=4 x$ with focus $F$, draw two perpendicular chords $A B$ and $C D$ through $F$. Let the midpoints of $A B$ and $C D$ be $M$ and $N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point;
(2) Construct circles with diameters $A B$ and $C D$ respectively, ... | (1) It is known that $\boldsymbol{F}(1,0)$.
Let $l_{A B}: y=k(x-1)$, substituting into $y^{2}=4 x$, we get
$k^{2} x^{2}-2\left(k^{2}+2\right) x+k^{2}=0$.
It is easy to see that, $x_{M}=\frac{x_{A}+x_{B}}{2}=\frac{k^{2}+2}{k^{2}}$,
$y_{M}=k\left(x_{M}-1\right)=\frac{2}{k}$.
Thus, $M\left(\frac{k^{2}+2}{k^{2}}, \frac{2}{... | \left(x-\frac{3}{2}\right)^{2}+y^{2}=\frac{9}{4}(y \neq 0) | Geometry | proof | Yes | Yes | cn_contest | false | 716,573 |
One. (50 points) As shown in Figure 3, let $P$ be a point inside $\triangle ABC$ such that
$$
\begin{array}{l}
\angle APB - \angle ACB \\
= \angle APC - \angle ABC.
\end{array}
$$
Let $I_{1}$ and $I_{2}$ be the incenters of $\triangle APB$ and $\triangle APC$, respectively. Prove that the lines $AP$, $BI_{1}$, and $CI... | Extend $A P$ to intersect the circumcircle of $\triangle A B C$ at point $K$, and connect $K B, K C$. Then
$$
\begin{array}{l}
\angle A P B-\angle A C B \\
=\angle A P B-\angle A K B \\
=\angle P B K, \\
\angle A P C-\angle A B C \\
=\angle A P C-\angle A K C \\
=\angle P C K .
\end{array}
$$
Therefore, $\angle P B K$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,574 |
Given a positive number $\lambda$ and a positive integer $n(n \geqslant$ 2). Find the smallest positive number $M(\lambda)$, such that for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ we have
$$
M(\lambda)\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\lambda \prod_{k=1}^{n} x_{... | Let $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, then we have $M(\lambda) \geqslant \frac{n}{n^{n}}+\frac{\lambda}{n^{n}}=\frac{1}{n^{n}}(n+\lambda)$.
Let $x_{1}=1, x_{2}=x_{3}=\cdots=x_{n}=0$, then we have $M(\lambda) \geqslant 1$.
Therefore, $M(\lambda) \geqslant \max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\}$.
We disc... | \max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,575 |
Three. (50 points) A $98 \times$ 98 chessboard is displayed on a computer screen, colored in the usual way (i.e., with two colors alternating). A person can drag the mouse to select a rectangle with borders along the chessboard lines, then click the mouse, and all the colors within this frame will change (i.e., white t... | Three, we prove that for an $n \times n$ chessboard, if $n$ is odd, then at least $n-1$ mouse clicks are required to turn the entire chessboard into one color; if $n$ is even, then at least $n$ mouse clicks are required to turn the entire chessboard into one color.
Consider the $4(n-1)$ small squares along the border ... | 98 | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,576 |
As 171 As shown in Figure 2, given that $\odot O_{1}, \odot O_{2}, \odot O_{3}$ intersect at point $G$, and the three circles intersect each other in pairs at points $D, E, F$. A line through point $D$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $B$ respectively, and $A D=B D$. Connect $A E$ and extend... | Proof: As shown in Figure 3, connect $C F$, $B F$, $G D$, $G E$, and $G F$. Since
$$
\begin{array}{l}
\angle C F G \\
=\angle A E G, \\
\angle B F G \\
=\angle A D G,
\end{array}
$$
Therefore, $\angle C F G +$
$$
\begin{array}{l}
\angle B F G = \angle A E G + \\
\angle A D G = 180^{\circ}.
\end{array}
$$
Thus, points... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,577 |
Initially 172 As shown in Figure 4, on the three sides of the equilateral $\triangle ABC$, six points $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ are taken in sequence, such that $A_{1} A_{2}=A_{2} B_{1}=$ $B_{1} B_{2}=B_{2} C_{1}=C_{1} C_{2}=C_{2} A_{1}$. Prove:
$$
\triangle A B_{2} C_{1} \cong \triangle B C_{2} A_{1} ... | Proof: As shown in Figure 5, connect $B_{1} C_{1}$ and $B_{2} C_{2}$ to get the intersection point $O$, then connect $A O$, $B_{1} C_{2}$, $A_{2} C_{1}$, and $A_{2} C_{2}$.
Since $\angle B_{1} B_{2} C_{1} = \angle A C_{1} B_{2} + 60^{\circ}$,
$\angle C_{2} C_{1} B_{2} = \angle A B_{2} C_{1} + 60^{\circ}$,
$B_{1} B_{2} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,578 |
Let $F$ be the set of real polynomials $f(x)$ satisfying
(1) the degree of $f(x)$ is less than or equal to 3;
(2) for any $x \in [0,1]$, $|f(x)| \leqslant 1$.
Find $\max _{f \in F} f(2)$. | Solution: Let $f(x)=a x^{3}+b x^{2}+c x+d$. Denote
$$
\left.\begin{array}{l}
A=f(1)=a+b+c+d, \\
B=f\left(\frac{3}{4}\right)=\frac{27}{64} a+\frac{9}{16} b+\frac{3}{4} c+d, \\
C=f\left(\frac{1}{4}\right)=\frac{1}{64} a+\frac{1}{16} b+\frac{1}{4} c+d, \\
D=f(0)=d .
\end{array}\right\}
$$
By condition (2), we know $A, B,... | 99 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,579 |
Example 7 As shown in Figure 7, six small circles are inside a large circle and are all internally tangent to the large circle. Each pair of adjacent small circles are externally tangent to each other. The points of tangency between the six small circles and the large circle are sequentially $A_{1}$, $A_{2}$, $A_{3}$, ... | Proof: As shown in Figure 7, connect $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, $A_{4} A_{5}$, $A_{5} A_{6}$, $A_{6} A_{1}$, $A_{2} A_{4}$, $A_{4} A_{6}$, $A_{6} A_{2}$, $O A_{1}$, $O A_{2}$, $O_{1} O_{2}$. Thus, point $O_{1}$ is on $O A_{1}$, and point $O_{2}$ is on $O A_{2}$. Let the centers and radii of the six sm... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,580 |
Let $n$ be a positive integer. Find the non-negative integer $m$ that satisfies $\sum_{k=0}^{m}\left[\frac{n-\log _{2}(2 k+1)}{2}\right]=0$, where $[x]$ denotes the greatest integer less than or equal to $x$.
---
The translation preserves the original text's formatting and structure. | Solution: Let the positive integer $k_{0}$ satisfy
$$
\begin{array}{l}
\frac{n-\log _{2}\left(2 k_{0}+1\right)}{2} \geqslant 0, \\
\frac{n-\log _{2}\left[2\left(k_{0}+1\right)+1\right]}{2}1$ when, $p(n)>0$, so it is necessary to make
$$
\sum_{k=2^{n-1}}^{m}\left[\frac{n-\log _{2}(2 k+1)}{2}\right]=-\frac{2^{n-1}+(-1)^{... | m=\frac{2^{n+1}+(-1)^{n}-3}{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,581 |
Example 8 In rhombus $A B C D$, $\angle A=60^{\circ}, E$ is a point on the minor arc $\overparen{B D}$ of the circumcircle of $\triangle A B D$, and line $D E$ intersects $A B$ at point $F$. Prove that $A D$, $B E$, and $C F$ are concurrent. | Proof: As shown in Figure 8, let $\angle DFC = x$, and denote $\angle BDF = \alpha, \angle CDF = \beta$. Therefore, $\alpha + \beta = 60^\circ$, $\angle DBE = \beta = \angle BFD$. Applying the trigonometric form of Ceva's Theorem to $\triangle DBF$ and point $C$, we have
$$
\begin{aligned}
1= & \frac{\sin \angle DBC}{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,582 |
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