problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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5. Find the smallest positive integer $n$ such that: If each vertex of a regular $n$-gon is arbitrarily colored with one of the three colors red, yellow, or blue, then there must exist four vertices of the same color that form the vertices of an isosceles trapezoid (a convex quadrilateral with one pair of parallel side... | 5. The smallest value of $n$ sought is 17.
First, prove: When $n=17$, the conclusion holds.
Proof by contradiction.
Assume there exists a method to three-color the vertices of a regular 17-gon such that no four vertices of the same color form the vertices of an isosceles trapezoid.
Since $\left[\frac{17-1}{3}\right]+... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,531 |
6. Determine all triples \((p, q, n)\) that simultaneously satisfy
$$
q^{n+2} \equiv 3^{n+2}\left(\bmod p^{n}\right), p^{n+2} \equiv 3^{n+2}\left(\bmod q^{n}\right)
$$
where \(p\) and \(q\) are odd primes, and \(n\) is an integer greater than 1.
(Proposed by Yonggao Chen) | 6. It is evident that $(3,3, n) (n=2,3, \cdots)$ are all arrays that meet the requirements.
Assume $(p, q, n)$ is another array that meets the requirements, then $p \neq q, p \neq 3, q \neq 3$. Without loss of generality, let $q > p \geqslant 5$.
If $n=2$, then $q^{2} \mid \left(p^{4}-3^{4}\right)$, which means $q^{2}... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,532 |
1. Given the function $y=x^{2}+\frac{1}{\sqrt{-x}}$, point $P(x, y)$ lies on the graph of this function. Then, point $P(\ddot{x}, y)$ should be in the ( ) of the Cartesian plane.
(A) first quadrant
(B) second quadrant
(C) third quadrant
(D) fourth quadrant | -.1.B.
Since $-x>0$, then, $x<0$, thus, we know that point $P(x, y)$ should be in the second quadrant. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,533 |
2. A box contains $m$ red balls, 10 white balls, and $n$ black balls. Each ball is identical except for its color. If a ball is randomly drawn from the box, the probability of drawing a white ball is the same as the probability of not drawing a white ball. What is the relationship between $m$ and $n$? ( ( ).
(A) $m+n=1... | 2. A.
The total number of balls in the box is $m+n+10$, the probability of drawing a white ball is $P=\frac{10}{m+n+10}$; the probability of not drawing a white ball is $P^{\prime}=\frac{m+n}{m+n+10}$.
According to the problem, $\frac{10}{m+n+10}=\frac{m+n}{m+n+10}$. Therefore, $m+n=10$. | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,534 |
3. Our province stipulates: The junior high school mathematics competition is held on the last Sunday of November each year. Therefore, the date of the junior high school mathematics competition in 2008 is ( ).
(A) November 26
(B) November 27
(C) November 29
(D) November 30 | 3. D.
November 25, 2007, was a Sunday, 2008 is a leap year, from November 25, 2007, to November 25, 2008, there are 366 days, 366 divided by 7 leaves a remainder of 2, meaning November 25, 2008, is a Tuesday, so November 30, 2008, is the last Sunday of November. | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 719,535 |
4. In the Cartesian coordinate system, there are points $A(-2,2)$, $B(3,2)$, and $C$ is a point on the coordinate axes. If $\triangle ABC$ is a right-angled triangle, then the number of points $C$ that satisfy the condition is ( ) .
(A) 1
(B) 2
(C) 4
(D) 6 | 4.D.
(1) If $\angle A C B=90^{\circ}$, at this time, $A C^{2}+C B^{2}=A B^{2}$.
(i) As shown in Figure 7, point $C(a, 0)$ is on the $x$-axis, we have
$$
\begin{array}{l}
(a+2)^{2}+ \\
2^{2}+(3-a)^{2}+2^{2} \\
= 5^{2} .
\end{array}
$$
Solving this, we get $a=2$ or $a=-1$.
Thus, we have points $C_{1}(2,0)$ and $C_{2}(-1... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,536 |
5. As shown in Figure 1, on the sides $BC$ and $CA$ of the equilateral $\triangle ABC$, there are points $E$ and $F$ respectively, and it satisfies
$$
\begin{array}{l}
B E=C F=a, E C=F A \\
=b(a>b) . \text { When } B F \text { bisects }
\end{array}
$$
$A E$, then the value of $\frac{a}{b}$ is ( ).
(A) $\frac{\sqrt{5}-... | 5.C.
As shown in Figure 8, draw $ED \parallel BF$ intersecting $AC$ at point $D$. Clearly, $AF = FD = b$.
Therefore, $DC = a - b$.
Notice that $\frac{DC}{DF} = \frac{EC}{BE} = \frac{b}{a}$, which means $\frac{a - b}{b} = \frac{b}{a}$. Thus, $\left(\frac{a}{b}\right)^2 - \left(\frac{a}{b}\right) - 1 = 0$. Solving this,... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,537 |
6. A unit ordered 22 boxes of fast food from a fast food store, spending a total of 140 yuan. There are three types of boxed meals: A, B, and C, with unit prices of 8 yuan, 5 yuan, and 3 yuan, respectively. Therefore, the number of different possible ordering schemes is ( . ) types.
(A) 1
(B) 2
(C) 3
(D) 4 | 6. C.
Let the unit order $x$, $y$, and $z$ boxes of meals A, B, and C, respectively. Then,
$$
\left\{\begin{array}{l}
x+y+z=22 \\
8 x+5 y+3 z=140 .
\end{array}\right.
$$
(1) $\times 8$ - (2) gives
$$
3 y+5 z=36 \Rightarrow 5 z=36-3 y \leqslant 36 \text {. }
$$
From this, we know $z \leqslant 7$, and $z$ is a multiple... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,538 |
7. Given $a>0, b>0$ and
$$
\sqrt{a}(\sqrt{a}+4 \sqrt{b})=3 \sqrt{b}(\sqrt{a}+2 \sqrt{b}) \text {. }
$$
Then the value of $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}$ is $(\quad)$.
(A) 1
(B) 2
(C) $\frac{19}{11}$.
(D) $\sqrt{2}$ | 7. B.
From $\sqrt{a}(\sqrt{a}+4 \sqrt{b})=3 \sqrt{b}(\sqrt{a}+2 \sqrt{b})$, we get $a+\sqrt{a b}-6 b=0$, which is
$$
(\sqrt{a}+3 \sqrt{b})(\sqrt{a}-2 \sqrt{b})=0 \text {. }
$$
Since $\sqrt{a}+3 \sqrt{b} \neq 0$, we have $\sqrt{a}-2 \sqrt{b}=0$, which means $a=4 b$.
Therefore, $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,539 |
8. As shown in Figure 2, in trapezoid $A B C D$, $\angle D=90^{\circ}, M$ is the midpoint of $A B$. If $C M=$ $6.5, B C+C D+D A=$ 17, then the area of trapezoid $A B C D$ is ( ).
(A) 20
(B) 30
(C) 40
(D) 50 | 8. B.
As shown in Figure 9, extend $C M$ to intersect the extension of $D A$ at point $E$, then $\triangle B C M \cong \triangle A E M$. Therefore, $C E = 2 C M = 13$, and $A E = B C$. Let the area of trapezoid $A B C D$ be $S$. Then
$$
\left\{\begin{array}{l}
D E^{2} + C D^{2} = C E^{2}, \\
\frac{1}{2} D E \cdot C D ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,540 |
Example 4 Given that $C$ is the midpoint of line segment $AB$, the circle $\odot O_{1}$ passing through points $A$ and $C$ intersects the circle $\odot O_{2}$ passing through points $B$ and $C$ at points $C$ and $D$. $P$ is the midpoint of the arc $\overparen{AD}$ on $\odot O_{1}$ (excluding point $C$), and $Q$ is the ... | Explanation: As shown in the figure,
5, let $A D$ and $P C$ intersect
at point $E$, and $B D$ and $Q C$
intersect at point $F$.
In $\odot O_{1}$,
Since $\overparen{P A}=\overparen{P D}$, we know
$$
\begin{array}{c}
\angle P D E \\
=\angle P A D=\angle P C D .
\end{array}
$$
Also, $\angle D P E=\angle C P D$, so,
$\tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,541 |
9. As shown in Figure 3, in rhombus $A B C D$, $\angle A=100^{\circ}$, $M$ and $N$ are the midpoints of sides $A B$ and $B C$ respectively, and $M P \perp C D$ at point $P$. Then the degree measure of $\angle N P C$ is $\qquad$ | $=9.50^{\circ}$.
Since quadrilateral $ABCD$ is a rhombus, therefore,
$$
AB=BC, \angle B=180^{\circ}-\angle A=80^{\circ} \text {. }
$$
Also, $M$ and $N$ are the midpoints of sides $AB$ and $BC$, respectively, so $BM=BN$,
$$
\angle BMN=\angle BNM=\frac{1}{2}\left(180^{\circ}-80^{\circ}\right)=50^{\circ} \text {. }
$$
L... | 50^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,542 |
10. As shown in Figure 4, in $\triangle A B C$, $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=3, C D=2$, then $S_{\triangle A B C}$ $=$ $\qquad$ . | 10.15 .
Solution 1: Let $A B=c, A C=b$. Then
$$
\begin{aligned}
A D^{2} & =c^{2}-9=b^{2}-4, b=\sqrt{c^{2}-5}, \\
A D & =\sqrt{c^{2}-9} .
\end{aligned}
$$
As shown in Figure 11, draw $C E \perp A B$ at point $E$, $\triangle A E C$ is an isosceles right triangle. Thus,
$$
\begin{array}{l}
E C=\frac{\sqrt{2}}{2} A C=\fr... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,543 |
11. The linear function $y=-\frac{\sqrt{3}}{3} x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. A square $A B C D$ is constructed with segment $A B$ as one side in the first quadrant (as shown in Figure 5). In the second quadrant, there is a point $P\left(a, \frac{1}{2}\right)$, which sati... | 11. $\frac{\sqrt{3}}{2}-8$.
Let $x=0, y=0$, we get $A(\sqrt{3}, 0) 、 B(0,1)$. Then $O A=\sqrt{3}, O B=1, A B=2, S_{\text {square } A B C D}=4$.
Therefore, $S_{\triangle A B P}=4$.
As shown in Figure 12, connect
$P O$. Then
$$
\begin{array}{l}
S_{\triangle A O P} \\
=\frac{1}{2} \times \frac{1}{2} \sqrt{3} \\
=\frac{... | \frac{\sqrt{3}}{2}-8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,544 |
12. If the real number $a$ satisfies
$$
a^{3}+a^{2}-3 a+2=\frac{3}{a}-\frac{1}{a^{2}}-\frac{1}{a^{3}} \text {, }
$$
then $a+\frac{1}{a}=$ | 12.2 or -3.
Let $a+\frac{1}{a}=b$. Then $a^{2}+\frac{1}{a^{2}}=b^{2}-2$.
$$
\begin{array}{l}
\text { Also } a^{3}+\frac{1}{a^{3}}=\left(a+\frac{1}{a}\right)\left(a^{2}+\frac{1}{a^{2}}\right)-\left(a+\frac{1}{a}\right) \\
=b\left(b^{2}-2\right)-b=b^{3}-3 b .
\end{array}
$$
According to the problem,
$$
\begin{array}{l}... | 2 \text{ or } -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,545 |
14. In a dormitory of a school, there are several students, one of whom serves as the dorm leader. During New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator. Each dormitory administrator also gives a gre... | 14. Let there be $x$ students in this dormitory, and $y$ administrators in the dormitory building $\left(x, y \in \mathbf{N}_{+}\right)$. According to the problem, we have
$$
x(x-1)+x y+y=51 \text{. }
$$
Simplifying, we get $x^{2}+(y-1) x+y-51=0$.
Thus, $\Delta=(y-1)^{2}-4(y-51)$
$$
=y^{2}-6 y+205=(y-3)^{2}+196 \text{... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,547 |
15. If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive integers, and $a_{1}<a_{2}<\cdots<a_{n} \leqslant 2007$, to ensure that there always exist four distinct numbers $a_{i} 、 a_{j} 、 a_{k} 、 a_{l}$, such that $a_{i}+a_{j}=a_{k}+a_{l}=a_{n}$, what is the minimum value of $n$? And explain the reason. | 15. Let $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005$, $a_{1004}=2007$, a total of 1004 odd numbers. Clearly, the sum of any two numbers is not equal to 2007.
If we add the number 2006 to the above 1004 numbers, i.e., $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005, a_{1004}=$ $2006, a_{1000}=2007$, at this point, there exists o... | 1006 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,548 |
In acute $\triangle ABC$, $BD$ and $CE$ are altitudes on sides $AC$ and $AB$, respectively. A circle is drawn with $AB$ as its diameter, intersecting $CE$ at point $M$. Point $N$ is taken on $BD$ such that $AN = AM$. Prove: $AN \perp CN$.
(Bi Jikun, problem contributor) | Proof 1: As shown in Figure 1, connect $DM$.
Since $AB$ is the diameter and $BD \perp AC$, points $A$, $B$, $M$, and $D$ are concyclic. Therefore,
$$
\begin{array}{l}
\angle ABD = \angle AMD. \\
\text{Also, } \angle ACE = 90^{\circ} - \angle CAE \\
= \angle ABD = \angle AMD,
\end{array}
$$
Thus, $\triangle ADM \sim \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,549 |
II. (25 points) Let the three sides of $\triangle ABC$ be $a, b, c$, and $a+b+c=3$. Find the minimum value of
$$
f(a, b, c)=a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c
$$
(Supplied by Jia Yinghong) | $$
\begin{array}{l}
=\left\{f(a, b, c)=a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c\right. \\
=(a+b+c)^{2}-2(a b+b c+c a)+\frac{4}{3} a b c \\
=9-2\left(a b+b c+c a-\frac{2}{3} a b c\right) .
\end{array}
$$
Since $a, b, c$ are the side lengths of $\triangle ABC$, and $a+b+c=3$, we have $0<a, b, c<\frac{3}{2}$. Therefore,
$$
\b... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,550 |
Three. (25 points) In the sequence $\left\{a_{n}\right\}$, $a_{0}=2007$, $a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}(n \in \mathbf{N})$. Prove that when $0 \leqslant n \leqslant 1004$, we have $\left[a_{n}\right]=2007-n$ (where, $[x]$ denotes the greatest integer not exceeding $x$).
(Harbin Normal University Affiliated High Sch... | Three, first consider the general problem.
Let $a_{0} \in \mathbf{N}_{+}, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}$. Prove:
$$
\left[a_{n}\right]=a_{0}-n\left(0 \leqslant n \leqslant \frac{1}{2}\left(a_{0}+2\right)\right) \text {. }
$$
For any positive integer $n$, by the recursive formula, we know $a_{n}>0$.
Since $a_{n}-a_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,551 |
Example 5 As shown in Figure 6, in $\triangle ABC$, $O$ is the circumcenter, the three altitudes $AD, BE, CF$ intersect at point $H$, and lines $ED, FD$ intersect $AB, AC$ at points $M, N$ respectively. Prove:
(1) $OB \perp DF, OC \perp DE$;
(2) $OH \perp MN$.
(2001, National High School Mathematics Competition) | Explanation: (1) Since points $A, C, D, F$ are concyclic, then
$$
\begin{array}{l}
\angle B D F=\angle B A C . \\
\text { Also, } \angle O B C=\frac{1}{2}\left(180^{\circ}-\angle B O C\right) \\
=90^{\circ}-\angle B A C,
\end{array}
$$
which means $\angle O B D=90^{\circ}-\angle B D F$.
Therefore, $O B \perp D F$.
Sim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,552 |
Four. (25 points) Each point on the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$... | Obviously, $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arcs $\overparen{A B}$ (including $A$ but not $B$), $\overparen{B C}$ (including $B$ but not $C$), and $\overparen{C A}$ (including $C$ but not $A$) with colors $1, 2,$ and $3$ respectively, and color all oth... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,553 |
Five. (25 points) Let $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$. Find the maximum value of
$$
A=\frac{\left(1-\sqrt{\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}\right)^{2}}{\cot \alpha+\cot \beta}
$$
(Supplied by Northwest Polytechnical University High School) | Five, from $\cot \alpha+\cot \beta$
$$
\begin{aligned}
&=\frac{1-\tan ^{2} \frac{\alpha}{2}}{2 \tan \frac{\alpha}{2}}+\frac{1-\tan ^{2} \frac{\beta}{2}}{2 \tan \frac{\beta}{2}} \\
&=\frac{\left(\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}\right)\left(1-\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}\right)}{2 \tan \frac... | 3-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,554 |
Six. (25 points) Given
$$
f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x .
$$
(1) Solve the equation: $f(x)=0$;
(2) Find the number of subsets of the set
$$
M=\left\{n \mid f\left(n^{2}-214 n-1998\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
(Li Tiehan, problem contributor) | (1) For any $0 < x_1 < x_2$, we have $\frac{x_1 + 1}{x_2 + 1} > \frac{x_1}{x_2}$, thus $\lg \frac{x_1 + 1}{x_2 + 1} > \lg \frac{x_1}{x_2}$. Therefore,
$$
\begin{array}{l}
f(x_1) - f(x_2) > \lg \frac{x_1}{x_2} - \log \frac{x_1}{x_2} \\
= \lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9}.
\end{array}
$$
Since $0 <... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,555 |
Seven, (25 points) Let $n$ be a positive integer, $a=[\sqrt{n}]$ (where $[x]$ denotes the greatest integer not exceeding $x$). Find the maximum value of $n$ that satisfies the following conditions:
(1) $n$ is not a perfect square;
(2) $a^{3} \mid n^{2}$.
(Zhang Tongjun
Zhu Yachun, problem contributor) | Seven, from (1) we get $a<\sqrt{n}<a+1$, then $a^{2}<n<a^{2}+2 a+1$,
that is
$a^{2}+1 \leqslant n \leqslant a^{2}+2 a$.
Let $n=a^{2}+t(t \in\{1,2, \cdots, 2 a\})$.
From (2) we have
$a^{3}\left|\left(a^{4}+2 a^{2} t+t^{2}\right) \Rightarrow a^{2}\right| t^{2} \Rightarrow a \mid t$.
Furthermore, $a^{3} \mid \left(a^{4}+2... | 24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,556 |
Eight. (25 points) Let the inradius of $\triangle ABC$ be 1, and the side lengths $BC=a$, $CA=b$, $AB=c$. If $a$, $b$, and $c$ are all integers, prove that $\triangle ABC$ is a right triangle. | Let $\triangle A B C$ have an incircle that touches sides $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively. Let $A E = A F = x$, $B F = B D = y$, and $C D = C E = z$. Then,
$$
x = \frac{b + c - a}{2}, \quad y = \frac{c + a - b}{2}, \quad z = \frac{a + b - c}{2}.
$$
Since $a$, $b$, and $c$ are all integ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,557 |
1. Given that $f(x)$ is an even function, and
$$
f(2+x)=f(2-x) \text {, }
$$
when $0 \leqslant x \leqslant 2$, $f(x)=3^{-x}$. If $n \in \mathbf{N}_{+}, a_{n}=$ $f(n)$, then $a_{200}$ equals $(\quad)$.
(A) 27
(B) 3
(C) $\frac{1}{3}$
(D) $\frac{1}{27}$ | -1.C.
Since $f(2+x)=f(2-x)$, we have
$$
\begin{array}{l}
f(4+x)=f(2+(2+x)) \\
=f(2-(2+x))=f(-x) .
\end{array}
$$
Also, since $f(x)$ is an even function, $f(-x)=f(x)$, thus
$$
f(4+x)=f(x) \text {. }
$$
Therefore, we know that $f(x)$ is a periodic function with a period of 4. Hence,
$$
\begin{array}{l}
a_{2007}=f(2007)... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,558 |
2. Choose two elements from the set $\{1,2, \cdots, 11\}$ as $m$ and $n$ in the ellipse equation $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1$. Then the number of ellipses that can be formed within the rectangular region
$$
B=\{(x, y)|| x|<11,| y \mid<9\}
$$
is ( ).
(A) 43
(B) 72
(C) 86
(D) 9 | 2. B.
According to the problem, $m$ is a positive integer not greater than 10, and $n$ is a positive integer not greater than 8. First, determine $n$, which has 8 possibilities; for each determined $n$, $m$ has $10-1=9$ possibilities. Therefore, the number of ellipses that satisfy the conditions is $8 \times 9=72$. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,559 |
4. Let $[x]$ denote the greatest integer not exceeding $x$, and $\mathbf{Z}$ denote the set of integers. The solution set of the equation $[x]^{3}-[x]^{2}=x^{3}-x^{2}$ is $M$. Then, $(\quad)$.
(A) $M \subset \mathbf{Z}$
(B) $\mathbf{Z} \subset M$
(C) $M=\mathbf{Z}$
(D) $M$ and $\mathbf{Z}$ are not subsets of each other | 4.C.
Obviously, $\mathbf{Z} \subset M$.
Let $x \in M$, and let $x=t+\alpha$ (where $t \in \mathbf{Z}, 0 \leqslant \alpha < 1$).
Then $3 t^{2}+(3 \alpha-2) t+\alpha^{2}-\alpha>0$.
Thus, $\alpha=0$.
When $t=0$, $\alpha^{2}-\alpha=0$, so $\alpha=0$, which means $x \in \mathbf{Z}$, hence $M \subset \mathbf{Z}$.
Therefore,... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,561 |
5. Given the inequality $(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant 9$ holds for any positive real numbers $x, y$. Then the minimum value of the positive real number $a$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 5.B.
$$
(x+y)\left(\frac{1}{x}+\frac{a}{y}\right)=1+a+\frac{y}{x}+\frac{a x}{y} \text {. }
$$
When $\frac{y}{x}=2$, $1+a+\frac{y}{x}+\frac{a x}{y}=3+\frac{3}{2} a$.
From $3+\frac{3}{2} a \geqslant 9$, we get $a \geqslant 4$.
$$
\text { Also, } 1+4+\frac{y}{x}+\frac{4 x}{y} \geqslant 5+2 \sqrt{\frac{y}{x} \cdot \frac{4... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 719,562 |
Example 6 As shown in Figure 7, quadrilateral $ABCD$ is circumscribed around $\odot O$, with the points of tangency on $AB$, $BC$, $CD$, and $DA$ being $E$, $F$, $G$, and $H$, respectively. The line $HE$ intersects $FG$ at point $P$. Prove:
$OP \perp AC$. | Explanation 1: As shown in Figure 7, connect $PA$ and $PC$. In $\triangle AEP$, point $H$ is on $PE$. By Stewart's Theorem, we have
$$
\begin{array}{l}
AE^{2} \cdot PH + AP^{2} \cdot EH = AH^{2} \cdot EP + EH \cdot PH \cdot EP. \\
\text{Also, } AE = AH, PE = EH + HP, \text{ then} \\
AE^{2} \cdot PH + AP^{2} \cdot EH \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,563 |
6. Students $A$ and $B$ each have 3 cards. They play a game by tossing a fair coin. When the coin lands heads up, $A$ wins a card from $B$, otherwise, $B$ wins a card from $A$. If one person wins all the cards, the game ends. What is the probability that the game ends exactly after 5 coin tosses? ( )
(A) $\frac{1}{16}$... | 6.D.
Toss a coin 5 times. If $A$ wins, then there is exactly 1 tail in the first 3 tosses, and the remaining 4 tosses are all heads; if $B$ wins, then there is exactly 1 head in the first 3 tosses, and the remaining 4 tosses are all tails.
Thus, the required probability is $P=\frac{3+3}{2^{5}}=\frac{3}{16}$. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,564 |
7. In $\triangle A B C$, the sides opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$ respectively, and $\tan B=\frac{\sqrt{3} a c}{a^{2}+c^{2}-b^{2}}$. Then the size of $\angle B$ is or $\qquad$ | II.7. $\frac{\pi}{3}, \frac{2 \pi}{3}$.
By the cosine rule, $b^{2}=a^{2}+c^{2}-2 a c \cos B$. Then $\tan B=\frac{\sqrt{3} a c}{a^{2}+c^{2}-b^{2}}=\frac{\sqrt{3} a c}{2 a c \cos B}=\frac{\sqrt{3}}{2 \cos B}$, i.e., $\sin B=\frac{\sqrt{3}}{2}$.
Therefore, $\angle B$ is $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$. | \frac{\pi}{3}, \frac{2 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,565 |
8. When tossing a coin once, the probability of getting heads or tails is $\frac{1}{2}$. When the $n$-th toss results in heads, let $a_{n}=1$; when the $n$-th toss results in tails, let $a_{n}=-1$. Also, let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$, then the probability that $S_{n}=2$ is $\qquad$. | 8. When $n$ is even, it is $\frac{\mathrm{C}_{n}^{\frac{n+2}{2}}}{2^{n}}$, and when $n$ is odd, it is 0.
Given that $S_{n}=2$, which means tossing $n$ times, with exactly $\frac{n+2}{2}$ times showing heads and $\frac{n-2}{2}$ times showing tails, we can infer that the integer $n$ must be even. Therefore, the required... | \frac{\mathrm{C}_{n}^{\frac{n+2}{2}}}{2^{n}} \text{ when } n \text{ is even, and } 0 \text{ when } n \text{ is odd} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,566 |
9. As shown in Figure 2, in the polyhedron ABCDEF, it is known that quadrilateral $ABCD$ is a square with side length 3, $EF$ $/ / AB, EF=\frac{3}{2}$. If the volume of the polyhedron is $\frac{15}{2}$, then the distance between $EF$ and $AC$ is | 9.2 .
Take the midpoints of $A B$ and $C D$ as $M$ and $N$, and connect $F M$, $F N$, and $M N$.
Since $E F / / A M$ and $E F = \frac{3}{2} = A M$, we know that the polyhedron $A D E - M N F$ is a triangular prism.
Let the distance between $E F$ and $A C$ be $h$. From
$$
V_{A D E-M N F} + V_{F-B C M M} = \frac{15}{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,567 |
10. Given five points $A, B, C, D, E$ on a plane, no three of which are collinear. By connecting these points with 4 lines, each point is an endpoint of at least one line segment. The number of different connection methods is $\qquad$ kinds. | 10.135.
In Figure 5, 4 connection methods meet the requirements of the problem (the diagram only represents the connection form between points and lines, without considering the position of the points).
Case (1), based on the choice of the central point, there are 5 connection methods;
Case (2), it can be considered... | 135 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,568 |
11. Given $\tan \alpha+\log _{2}(2 \tan \alpha-6)=8$, $\tan \beta+2^{\tan \beta-1}=5$. Then $\tan \alpha+\tan \beta$ equals $\qquad$ . | 11.8 .
Let $t=\log _{2}(2 \tan \alpha-6)$. Then $t+2^{t-1}=5$. Also, $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, and $\tan \beta+2^{\tan \beta-1}=5$, so $\tan \beta=t$.
Therefore, $\tan \alpha+\tan \beta=\tan \alpha+t=8$.
Hence, $\tan \alpha+\tan \beta=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,569 |
12. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with the left vertex $A$ and the right focus $F$. Let $P$ be any point on the hyperbola in the first quadrant. If $\angle P F A=2 \angle F A P$ always holds, then the eccentricity $e$ of the hyperbola is | 12.2.
From the problem, we can take a point $P$ on the hyperbola such that $P F$ is perpendicular to the $x$-axis, giving $P(c, y)$.
Then $\frac{c^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Hence, $y^{2}=\frac{b^{4}}{a^{2}}$.
Since $y>0$, we have $y=\frac{b^{2}}{a}=\frac{c^{2}-a^{2}}{a}$.
Given that $\angle P F A=2 \angle F ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,570 |
13. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1} a_{n}+a_{n+1}-2 a_{n}=0 \text {. }
$$
For any positive integer $n$, we have
$\sum_{i=1}^{n} a_{i}\left(a_{i}-1\right)<M$ (where $M$ is a constant and an integer).
Find the minimum value of $M$. | Three, 13. From the problem, for $n \in \mathbf{N}_{+}, a_{n} \neq 0$, and
\[
\frac{1}{a_{n+1}}=\frac{1}{2}+\frac{1}{2 a_{n}},
\]
which means
\[
\frac{1}{a_{n+1}}-1=\frac{1}{2}\left(\frac{1}{a_{n}}-1\right).
\]
Given $a_{1}=2$, we have $\frac{1}{a_{1}}-1=-\frac{1}{2}$.
Therefore, the sequence $\left\{\frac{1}{a_{n}}-1\... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,571 |
15. Given the sequence
$$
b_{n}=\frac{1}{3 \sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right](n=0,1, \cdots) \text {. }
$$
(1) For what values of $n$ is $b_{n}$ an integer?
(2) If $n$ is odd and $2^{-\frac{2 n}{3}} b_{n}$ is an integer, what is $n$? | 15. (1) From
$$
\begin{array}{l}
b_{n}=\frac{2}{3 \sqrt{3}}\left[n \sqrt{3}+\mathrm{C}_{n}^{3}(\sqrt{3})^{3}+\mathrm{C}_{n}^{6}(\sqrt{3})^{5}+\cdots\right] \\
=2\left(\frac{n}{3}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5} \cdot 3+\cdots\right),
\end{array}
$$
$b_{n}$ is an integer if and only if $31 n$.
(2) First, $b_{n}$ ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,572 |
一.(50 points) As shown in Figure 3, in trapezoid $ABCD$, $AD // BC, BC=BD=$ $1, AB=AC, CD<1$, and $\angle BAC + \angle BDC = 180^{\circ}$. Find the length of $CD$. | As shown in Figure 6, construct the symmetric point $E$ of point $D$ with respect to $BC$, and connect $AE$, $BE$, and $CE$.
Let $AE$ intersect $BC$ at point $F$. Since $AD \parallel BC$, the distances from points $A$ and $E$ to $BC$ are equal. Therefore,
$$
\begin{array}{l}
AF = EF. \\
\text{Let } CD = CE = x, \\
AF ... | \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,573 |
Example 7 Given that $\odot O$ passes through the vertices $A$ and $C$ of $\triangle ABC$, and intersects $AB$ and $BC$ at points $K$ and $N$ respectively. The circumcircles of $\triangle ABC$ and $\triangle BKN$ intersect at points $B$ and $M$. Prove:
$OM \perp BM$. | Explanation: As shown in Figure 9, let the circumcenters of $\triangle ABC$ and $\triangle BKN$ be $O_1$ and $O_2$, respectively.
From the given conditions, we know that $O$, $O_1$, and $O_2$ are not collinear (otherwise, $B$ and $M$ would coincide, and $AB = BC$. If the common tangent through point $B$ is $l$, then $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,574 |
Three, (50 points) As shown in Figure 4, in a clockwise direction, starting from 1, take 1 step to 2, then take 2 steps to 3, and finally take 3 steps to 4. For a natural number $n$ greater than 1, can the numbers from 1 to $n$ be arranged on a circle so that starting from 1, take one step to $a_{2}$, then take $a_{2}$... | When $n=2 k$, it is possible.
Figure 7 is one arrangement that meets the requirements.
When $n=2 k+1$, there is no arrangement that meets the requirements.
Proof by contradiction.
Assume there is an arrangement that meets the requirements, then each number has both a value (the number itself) and a position (on the ci... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,576 |
14. Let the vertex of the parabola be $P$
$$
y=a x^{2}-3 x+c \quad(a \neq 0)
$$
intersect the positive $x$-axis at points $A$ and $B$, and the positive $y$-axis at point $C$. The circle $\odot D$ passes through points $A$, $B$, and $C$, and is tangent to the $y$-axis. Prove that $P A \perp D A$. | 14. Let \( A\left(x_{1}, 0\right) \), \( B\left(x_{2}, 0\right) \), \( C(0, c) \), then \( D\left(\frac{3}{2 a}, c\right) \).
By Vieta's formulas, we have \( \frac{x_{1}+x_{2}}{2}=\frac{3}{2 a} \).
The tangent from the origin \( O \) to \( \odot D \) is \( O C \), so \( O A \cdot O B=O C^{2} \),
which means \( \frac{c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,577 |
1. Given real numbers $a, b$ satisfy $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{4}{a^{2}+b^{2}}$. Then the value of $\left(\frac{b}{a}\right)^{2008}-\left(\frac{a}{b}\right)^{2009}$ is ( ).
(A) -2
(B) 0
(C) 2
(D) 0 or 2 | -、1.D.
Given the equation, after eliminating the denominator and rearranging, we get $\left(a^{2}-b^{2}\right)^{2}=0$. Therefore, $a=-b$ or $a=b$.
When $a=-b$, the original expression $=1+1=2$;
When $a=b$, the original expression $=1-1=0$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,578 |
2. As shown in Figure $1, M_{1}$, $M_{2}$, $M_{3}$, $M_{4}$ are points on the sides $AB$, $BC$, $CD$, $DA$ of quadrilateral $ABCD$, respectively, and $AM_{1}: M_{1}B = AM_{4}: M_{4}D = CM_{2}: M_{2}B = CM_{3}: M_{3}D = 1:2$. $P$ is a point inside quadrilateral $ABCD$, and the sum of the areas of quadrilaterals $PM_{1}A... | 2. B.
As shown in Figure 5, connect $P A$,
$P B, P C, P D$.
Given $A M_{1}: M_{1} B=$
$1: 2$, it is easy to see that
$$
A B=3 A M_{1} .
$$
Thus, $S_{\triangle P A B}=3 S_{\triangle P A M_{1}}$.
Similarly, $S_{\triangle P B C}=3 S_{\triangle P C M_{2}}, S_{\triangle P C D}=3 S_{\triangle P C M_{3}}$,
$$
S_{\triangle P... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,579 |
3. Given that when $-2<x \leqslant 1$, the function value of the linear function $y=$ $(2 k-3) x+|k|$ is always positive. Then the range of $k$ is ( ).
(A) $1<k<2$
(B) $1 \leqslant k<2$
(C) $1 \leqslant k<2$ and $k \neq 1.5$
(D) $1<k \leqslant 2$ and $k \neq 1.5$ | 3. D.
From the problem, we have
$$
\left\{\begin{array}{l}
2 k-3 \neq 0, \\
(2 k-3) \times(-2)+|k| \geqslant 0, \\
(2 k-3) \times 1+|k|>0 .
\end{array}\right.
$$
From equation (1), we get $k \neq 1.5$.
When $k \geqslant 0$, from equation (2) we solve to get $k \leqslant 2$; from equation (3) we solve to get $k>1$. Th... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,580 |
4. Given a parabola $y=a x^{2}+b x+c$ opening upwards passing through the points $(4,-6)$ and $(-2,0)$, intersecting the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. Then the minimum value of the area $S$ of $\triangle A B C$ is ( ).
(A) $10+4 \sqrt{6}$
(B) $10-4 \sqrt{6}$
(C) $10+4 \sqrt{5}$
(D) $10-... | 4. A.
From the problem, we have
$$
\left\{\begin{array}{l}
16 a+4 b+c=-6, \\
4 a-2 b+c=0 .
\end{array}\right.
$$
(1) - (2) gives $12 a+6 b=-6$, which simplifies to
$$
b=-2 a-1 \text {. }
$$
Substituting equation (3) into equation (2) gives $c=-8 a-2$.
Thus, $y=a x^{2}-(2 a+1) x-8 a-2$
$$
=(a x-4 a-1)(x+2) \text {. }
... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,581 |
5. There are 15 identical chocolate bars, each of which can be divided into at most two smaller pieces (which may not be equal). If these 15 chocolate bars can be evenly distributed among $n$ students, for the following 5 values: $7, 11, 13, 16, 20$, the number of values that $n$ can take is ( ).
(A) 5
(B) 4
(C) 3
(D) ... | 5.A.
Consider each piece of chocolate as 1 unit.
When $n=7$, 4 pieces of chocolate are each divided into $\frac{1}{7}$ and $\frac{6}{7}$, 4 pieces of chocolate are each divided into $\frac{2}{7}$ and $\frac{5}{7}$, 4 pieces of chocolate are each divided into $\frac{3}{7}$ and $\frac{4}{7}$, and another 3 pieces are no... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,582 |
6. As shown in Figure 2, $C D$ is the altitude on the hypotenuse $A B$ of Rt $\triangle A B C$, $I_{1} 、 I_{2} 、 I_{3} 、$ are the incenter of $\triangle A B C 、 \triangle A C D$ 、 $\triangle B C D$ respectively, $A C = 20, B C = 15$. Then the area of $\triangle I_{1} I_{2} I_{3}$ is ( ).
(A) 4
(B) 4.5
(C) 5
(D) 5.5 | 6.C.
As shown in Figure 6, connect $A I_{1}$,
$B I_{1}$, $I_{2} D$, $I_{3} D$. Draw $I_{1} G \perp A B$ at point $G$,
$I_{2} E \perp A B$ at point $E$, and $I_{3} F \perp A B$
at point $F$. In the right triangle $\triangle A B C$, it is easy to see that
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{20^{2}+15^{2}}=25 \text {.... | 5 | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,583 |
1. If two prime numbers $p, q$ satisfy $3 p^{2}+5 q=517$, then $p+q=$ $\qquad$ | II. 1.15 or 103.
If $p$ and $q$ are both odd, then $3p^2 + 5q$ is even, which contradicts the given condition. Therefore, at least one of $p$ or $q$ must be even.
Since $p$ and $q$ are both prime numbers, $p=2$ or $q=2$.
When $p=2$,
$3 \times 2^2 + 5q = 517 \Rightarrow q = 101$ (prime).
In this case, $p + q = 2 + 101 =... | 103 \text{ or } 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,584 |
Example 8 Given $\triangle A B C, \triangle P A B$ and $\triangle Q A C$ are two triangles outside $\triangle A B C$, satisfying $A P=A B$, $A Q=A C$ and $\angle B A P=\angle C A Q$, the line segments $B Q$ and $C P$ intersect at point $R$. Let the circumcenter of $\triangle B C R$ be $O$. Prove that $A O \perp P Q$.
(... | Explanation: As shown in Figure 10, connect $A R$, $O B$, and $O C$. Let the radius of $\odot O$ be $r$.
Then $P R \cdot P C$
$$
=P O^{2}-r^{2} \text {, }
$$
$$
\begin{aligned}
& Q R \cdot Q B \\
= & Q O^{2}-r^{2} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { (1)-(2) gives } \\
P O^{2}-Q O^{2} \\
=P R \cdot P C-Q R \c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,585 |
2. Given that $H$ is the orthocenter of $\triangle A B C$ and $B H=\sqrt{3} A C$. Then the degree measure of $\angle A B C$ is $\qquad$ . | $2.30^{\circ}$ or $150^{\circ}$.
Obviously, $\angle A B C$ cannot be a right angle. Otherwise, $B$ would be the orthocenter. Let $A C=a$. Then $B H=\sqrt{3} a$.
If $\angle A B C$ is acute, let $A H$ intersect $B C$ at point $D$, and $B H$ intersect $A C$ at point $E$. Then $A D \perp B C, B H \perp A C$.
It is easy to... | 30^{\circ} \text{ or } 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,586 |
3. Given that in the right triangle $\triangle A B C$, one side length is 5, and the other two side lengths are exactly the two roots of the equation $x^{2}-7 x+4 k-4=0$. Then the value of $k$ is $\qquad$ . | 3.4 or $\frac{160}{49}$.
Let the other two sides be $m, n (m \geqslant n)$.
By the relationship between roots and coefficients, we have
$$
\left\{\begin{array}{l}
m+n=7, \\
m n=4 k-4 .
\end{array}\right.
$$
When $m>5$, we have
$$
n^{2}+5^{2}=m^{2} \Rightarrow(m+n)(m-n)=25 \text {. }
$$
Then $m-n=\frac{25}{7}$.
Solvi... | 4 \text{ or } \frac{160}{49} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,587 |
One, (20 points) Given that $n$ and $k$ are positive integers, and satisfy the inequality
$$
\frac{1}{7}<\frac{n-k}{n+k}<\frac{63}{439} .
$$
If for a given positive integer $n$, there is only one positive integer $k$ that makes the inequality true. Find the maximum and minimum values of all positive integers $n$ that ... | From the known inequality, we have
$$
\frac{188}{251} n < k < \frac{189}{251} n.
$$
When $n=252, 253, 254$, substituting into equation (1) respectively, we get
$$
\begin{array}{l}
188 \frac{188}{251} < k < 189, \\
189 \frac{125}{251} < k < 189 \frac{3}{4}, \\
190 \frac{62}{251} < k < 190 \frac{1}{2},
\end{array}
$$
no... | 2008 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,589 |
II. (25 points) As shown in Figure 3, the incircle $\odot O$ of quadrilateral $ABCD$ touches sides $AB$, $BC$, $CD$, and $DA$ at points $E$, $F$, $G$, and $H$, respectively. The extensions of $AB$ and $DC$ intersect at point $M$, and $N$ is a point on $BC$ such that $CN = BF$. The extension of $MN$ intersects the arc $... | II. As shown in the figure,
9, connect $O M$,
$O E$, $O F$, and extend
$F O$ to intersect the extension of $M N$
at point $Q$. Construct
the incircle $\odot I$ of
$\triangle M B C$,
which touches $M B$,
$M C$, and $B C$ at points $R$, $S$, and $T$, respectively. Connect $I R$, $I N$, and $I T$.
Since $\odot O$ is the i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,590 |
Three, (25 points) HG TV has an entertainment show, which includes a program called "A Red Afterglow." The venue and rules of this program are as follows: a sheet of tempered glass exactly covers a rectangular area composed of unit square cells, measuring $6 \times 8$. Each square cell contains one red and one green li... | Three, there exists a path that can light up all the red lights in the squares.
As shown in Figure 10,
Let $A$ be the starting square of this path. In the $6 \times 8$ grid, there are some squares with green lights (excluding the starting point), and let one of these green-lit squares be $B$. Player $M$ starts from sq... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,591 |
2. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}>1, a_{n+1}=a_{n}+\frac{1}{a_{n}^{2}} \text {. }
$$
Then the sequence $\left\{a_{n}\right\}()$.
(A) is bounded and decreasing
(B) is bounded and increasing
(C) is unbounded and increasing
(D) is unbounded and decreasing | 2.C.
From $a_{n+1}=a_{n}+\frac{1}{a_{n}^{2}}>a_{n}$, we know that $\left\{a_{n}\right\}$ is increasing.
Notice that
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}^{2}} \\
\Leftrightarrow a_{n+1}^{3}=\left(a_{n}+\frac{1}{a_{n}^{2}}\right)^{3}>a_{n}^{3}+3 a_{n}^{3} \cdot \frac{1}{a_{n}^{2}}=a_{n}^{3}+3 \\
\Rightarrow ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,593 |
3. In the complex plane, the points $Z$ that satisfy
$$
|z-1| \sqrt{5}+|z-1-2 \mathrm{i}|=2|z|
$$
form ( ).
(A) a circle
(B) an ellipse
(C) a circular arc
(D) a hyperbola | 3.C.
Let $A(0,0), B(1,0), C(1,2)$, and the point corresponding to $z$ be $Z$. From $|z-1| \sqrt{5}+|z-1-2 \mathrm{i}|=2|z|$, we can derive
$$
Z B \cdot A C + Z C \cdot A B = Z A \cdot B C.
$$
By the converse of Ptolemy's theorem, the locus of $Z$ is the arc $\overparen{B C}$ of the circumcircle of $\triangle A B C$ t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,594 |
4. Let $x_{1}, x_{2}, x_{3} \in \mathbf{Z}, 0 \leqslant x_{1} \leqslant 2, 1 \leqslant x_{2} \leqslant$ $3, 1 \leqslant x_{3} \leqslant 4$. Then the number of solutions to $x_{1}+x_{2}+x_{3}=7$ is ( ).
(A) 6
(B) 7
(C) 8
(D) 9 | 4. A.
Consider the coefficient of $x^{7}$ in the generating function $\left(1+x+x^{2}\right)\left(x+x^{2}+\right.$ $\left.x^{3}\right)\left(x+x^{2}+x^{3}+x^{4}\right)$
$$
\Leftrightarrow x^{2}\left(1+x+x^{2}\right)^{2}\left(1+x+x^{2}+x^{3}\right) \text { of } x^{7}
$$
coefficient
$\Leftrightarrow\left(1+x+x^{2}\right... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,595 |
1. In quadrilateral $A B C D$, the perpendicular bisectors of $A B$ and $C D$ intersect at point $P$, and the perpendicular bisectors of $A D$ and $B C$ intersect at point $Q$. $M$ and $N$ are the midpoints of $A C$ and $B D$, respectively. Prove: $P Q \perp M N$. | (Tip: By the median length formula, we have
$$
\begin{aligned}
P M^{2} & =\frac{1}{2}\left(P A^{2}+P C^{2}\right)-\frac{1}{4} A C^{2}, \\
P N^{2} & =\frac{1}{2}\left(P B^{2}+P D^{2}\right)-\frac{1}{4} B D^{2}, \\
Q M^{2} & =\frac{1}{2}\left(Q A^{2}+Q C^{2}\right)-\frac{1}{4} A C^{2}, \\
Q N^{2} & =\frac{1}{2}\left(Q B^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,596 |
5. Let the base edge length of the regular tetrahedron $V-ABC$ be 4, and the side edge length be 8. The section through $A$ intersecting the side edges $VB$ and $VC$ is $\triangle AED$. Then the minimum perimeter of the section $\triangle AED$ is ( ).
(A) $12 \frac{1}{5}$
(B) 11
(C) 12
(D) $11 \frac{1}{5}$ | 5. B.
From the edge $V A$, the surface of the tetrahedron is unfolded as shown in Figure 2. The perimeter of $\triangle A D E$ is minimized if and only if $D$ and $E$ lie on the line segment $A A_{1}$.
Let $\angle B V C=\alpha, \angle A V A_{1}=3 \alpha$. Then $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$.
In $\tri... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,597 |
6. If real numbers $x, y$ satisfy
$$
4 x+3 y-2 x\left[\frac{x^{2}+y^{2}}{x^{2}}\right]=0,
$$
then the value of $\frac{y}{x}$ is $(\quad$.
(A) $\frac{2}{3}$ or 2
(B) $-\frac{2}{3}$ or 2
(C) 2
(D) None of the above | 6.B.
Obviously, $4 x+3 y-2 x\left[\frac{x^{2}+y^{2}}{x^{2}}\right]=0$ $\Leftrightarrow 2+\frac{3 y}{2 x}=\left[1+\frac{y^{2}}{x^{2}}\right]$. Then $\frac{3 y}{2 x} \in \mathbf{Z}$. Let $\frac{3 y}{2 x}=t$. Thus, $2+t=\left[1+\frac{4 t^{2}}{9}\right]$ $\Rightarrow \frac{4 t^{2}}{9}<2+t \leqslant 1+\frac{4 t^{2}}{9}$
$\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,598 |
1. $a, b$ are constants. If the parabola $C$:
$$
y=\left(t^{2}+t+1\right) x^{2}-2(a+t)^{2} x+t^{2}+3 a t+b
$$
passes through the fixed point $P(1,0)$ for any real number $t$, find the value of $t$ when the chord intercepted by the parabola $C$ on the $x$-axis is the longest. | $=1.2$.
Substituting $P(1,0)$ into the equation of the parabola $C$ yields
$$
\left(t^{2}+t+1\right)-2(a+t)^{2}+\left(t^{2}+3 a t+b\right)=0 \text {, }
$$
which simplifies to $t(1-a)+\left(1-2 a^{2}+b\right)=0$,
holding for all $t$.
Thus, $1-a=0$, and $1-2 a^{2}+b=0$.
Solving these, we get $a=1, b=1$.
Substituting int... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,599 |
2. To make $p=x^{4}+6 x^{3}+11 x^{2}+3 x+32$ a perfect square of an integer, then the integer $x$ has $\qquad$ solutions. | 2.0 .
From the problem, we know that $p \equiv x^{4}-x^{2}+2(\bmod 3)$.
Since $x^{2} \equiv 0,1(\bmod 3)$, thus,
$$
p \equiv x^{2}\left(x^{2}-1\right)+2 \equiv 2(\bmod 3) \text {. }
$$
Therefore, $p$ is not a perfect square. | 2.0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,600 |
$$
\begin{array}{l}
\text { 3. } 1+\cos ^{2}(2 x+3 y-1) \\
=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}
\end{array}
$$
Then the minimum value of $xy$ is | 3. $\frac{1}{25}$.
On one hand, $1 \leqslant 1+\cos ^{2}(2 x+3 y-1) \leqslant 2$.
On the other hand,
$$
\begin{array}{l}
\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1} \\
=\frac{(x-y)^{2}+2(x-y)+2}{x-y+1} \\
=(x-y+1)+\frac{1}{x-y+1} \geqslant 2 .
\end{array}
$$
Obviously, $x-y+1$ is a positive number.
Therefore, the original ... | \frac{1}{25} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,601 |
4. In the expansion of $(\sqrt[5]{3}+\sqrt[3]{5})^{100}$, there are $\qquad$ terms that are rational numbers. | 4.7.
The general term of the binomial theorem expansion is
$$
\mathrm{C}_{100}^{r} 3^{\frac{1}{5}(100-r)} 5^{\frac{r}{3}}(0 \leqslant r \leqslant 100),
$$
it is a rational number if and only if
$$
3|r, 5|(100-r) \Rightarrow 15 \mid r \text {. }
$$
It is easy to see that there are 7 such $r$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,602 |
5. Given $x, y \in \mathbf{R}$. Then
$$
\cos (x+y)+2 \cos x+2 \cos y
$$
the minimum value is $\qquad$ $\therefore$. | $$
\begin{array}{l}
\text { 5. }-3 \text {. } \\
\cos (x+y)+2 \cos x+2 \cos y+3 \\
=\left(1+\cos \frac{x+y}{2}\right)^{2}\left(1+\cos \frac{x-y}{2}\right)+ \\
\quad\left(1-\cos \frac{x+y}{2}\right)^{2}\left(1-\cos \frac{x-y}{2}\right) \\
\geqslant 0 .
\end{array}
$$
It is clear that equality can be achieved.
Therefore... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,603 |
6. The solution to the equation $\sqrt{3+\sqrt{3+\sqrt{3+x}}}=x$ is
$\qquad$ . | 6. $\frac{1+\sqrt{13}}{2}$.
Obviously, $x>0$.
First, prove: If $x>\sqrt{3+x}$, then the equation has no solution.
In fact, in this case, we have
$$
\begin{array}{l}
\sqrt{3+\sqrt{3+\sqrt{3+x}}}<\sqrt{3+\sqrt{3+x}} \\
<\sqrt{3+x}<x .
\end{array}
$$
Similarly, when $x<\sqrt{3+x}$, the equation also has no solution.
The... | \frac{1+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,604 |
$$
\begin{array}{l}
\text { Four, (20 points) Let } x_{2}=1, x_{3}=2, \\
x_{n}=(n-1)\left(x_{n-1}+x_{n-2}\right)(n \geqslant 4) .
\end{array}
$$
Find the general term formula for $\left\{x_{n}\right\}$. | Let's make the substitution $x_{k}=k!y_{k}$, and substitute it into the given recurrence relation to get $k y_{k}=(k-1) y_{k-1}+y_{k-2}$, that is,
$$
y_{k}-y_{k-1}=-\frac{1}{k}\left(y_{k-1}-y_{k-2}\right) \text {. }
$$
Taking $k=4,5, \cdots, m$ in equation (1) and multiplying the corresponding equations.
Noting that $... | x_{n}=n!\sum_{m=2}^{n} \frac{(-1)^{m}}{m!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,606 |
Example 1 If the inequality about $x$
$$
a x^{2}-|x+1|+2 a<0
$$
has a solution set of $\varnothing$, find the range of real number $a$. | Explanation: First, convert the condition to "for any $x \in \mathbf{R}$,
there is $a x^{2}-|x+1|+2 a \geqslant 0$, then separate the parameter $a$, transforming it into a function problem.
By the problem, for any $x \in \mathbf{R}$, we have
$$
a x^{2}-|x+1|+2 a \geqslant 0 \text {, }
$$
which means $a \geqslant f(x)... | \left[\frac{\sqrt{3}+1}{4},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,608 |
Example 1 Given the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=-1, \\
a_{n+2} a_{n+1}=4 a_{n+1} a_{n}-6 a_{n+1} a_{n-1}+9 a_{n} a_{n-1}-6 a_{n}^{2} .
\end{array}
$$
Find $a_{n}$. | Analysis: This recursive sequence is a nonlinear homogeneous recursive sequence and cannot be directly solved using the characteristic root method. Note that the recursive formula is a quadratic homogeneous form, which can be transformed into a linear homogeneous form through factorization.
Solution: Since
$$
\begin{ar... | a_{n}=\frac{5}{4}+\frac{1}{12}(-3)^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,609 |
Given 222 As shown in Figure $4, A D$ is the angle bisector of $\triangle A B C$, $\angle A D C=60^{\circ}$, point $M$ is on $A D$, satisfying $D M = D B$, the rays $B M, C M$ intersect $A C, A B$ at points $E, F$. Prove: $D F \perp E F$. | Prove: As shown in Figure 4, construct the bisector of the exterior angle $\angle C A K$ of $\angle B A C$ to intersect the ray $B C$ at point $G$.
Since $A D$ and $A G$ are the internal and external angle bisectors of $\triangle A B C$, we have
$$
\frac{B D}{D C}=\frac{A B}{A C}=\frac{B G}{G C}.
$$
Also, since $A D$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,610 |
Example 3 Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=1$, $b_{0}=0$, and
$$
\left\{\begin{array}{l}
a_{n+1}=7 a_{n}+6 b_{n}-3, \\
b_{n+1}=8 a_{n}+7 b_{n}-4
\end{array}\right.
$$
$(n=0,1,2, \cdots)$. Prove: $a_{n}(n=0,1,2, \cdots)$ is a perfect square.
(2000, National High School M... | Analysis: This is a binary non-homogeneous linear recursive sequence. We can first reduce the binary to unary, then use the difference method to transform the non-homogeneous linear recursive sequence into a homogeneous linear recursive sequence, aiming to convert it into a form suitable for the characteristic root met... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,611 |
Example 6 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a_{2}=1$, and $a_{n+2}=3 a_{n+1}+18 a_{n}+2^{n}\left(n \in \mathbf{N}_{+}\right)$.
Find $a_{n}$. | Analysis: This is a non-linear recursive sequence, and the method of making a difference step by step cannot be used to reduce it to a linear recursive sequence. Instead, the method of undetermined coefficients can be used to transform it into a linear recursive sequence.
Solution: Let
$a_{n+2}-A \cdot 2^{n+2}$
$=3\lef... | a_{n}=\frac{6^{n}}{12}-\frac{(-3)^{n}}{5}-\frac{2^{n}}{20} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,612 |
2. In $\triangle A B C$ with an area of 1, $P$ is the midpoint of side $B C$, and point $Q$ is on side $A C$ such that $A Q=2 Q C$. Connecting $A P$ and $B Q$ intersects at point $R$. Then the area of $\triangle A B R$ is $\qquad$ | 2. $\frac{2}{5}$.
As shown in Figure 4, draw $P M / / A C$, intersecting $B Q$ at point $M$. Given $P M=\frac{1}{2} C Q=\frac{1}{4} A Q$, we have
$$
\begin{array}{l}
\frac{P R}{R A}=\frac{P M}{A Q}=\frac{1}{4} \\
\frac{A R}{A P}=\frac{4}{5}
\end{array}
$$
Therefore, $S_{\triangle A B R}=\frac{4}{5} S_{\triangle A B P... | \frac{2}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,615 |
3. In $\triangle A B C$, $\angle C=90^{\circ}$, and the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively. If the sum of the squares of the roots of the equation $c\left(x^{2}+1\right)-2 \sqrt{2} b x-a\left(x^{2}-1\right)=0$ is 10, then the value of $\frac{b}{a}$ is . $\qquad$ | 3. $\sqrt{15}$.
Given the equation transformed as
$$
(c-a) x^{2}-2 \sqrt{2} b x+(c+a)=0 \text {. }
$$
From $\Delta=8 b^{2}-4\left(c^{2}-a^{2}\right)=4 b^{2}>0$, we know the equation has two real roots $x_{1} 、 x_{2}$, and
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}=\left(\frac{2 \sqrt{2} b}{c-a}\right)^{2}-2 \cdot \frac{... | \sqrt{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,616 |
4. The numbers $x_{1}, x_{2}, \cdots, x_{100}$ satisfy the following conditions: For $k=1,2, \cdots, 100, x_{k}$ is less than the sum of the other 99 numbers by $k$. Then the value of $x_{25}$ is $\qquad$ . | 4. $\frac{650}{49}$.
Let $S=x_{1}+x_{2}+\cdots+x_{100}$, then
$$
x_{k}=\left(S-x_{k}\right)-k(k=1,2, \cdots, 100) \text {, }
$$
i.e., $k+2 x_{k}=S$.
Summing over $k$ gives
$$
\begin{array}{l}
(1+2+\cdots+100)+2 S=100 S \\
\Rightarrow S=\frac{2525}{49} \\
\Rightarrow x_{25}=\frac{S-25}{2}=\frac{650}{49} .
\end{array}
... | \frac{650}{49} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,617 |
Example 2 If the inequality about $x$
$$
\frac{x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7}{x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7}<0
$$
has a solution set that is the union of some intervals, and the sum of the lengths of these intervals is no less than 4, find the range of real numbers $a$.
(2001, Shanghai High... | Notice the relationship between the coefficients of the quadratic terms in $x$ in the numerator and denominator on the left side of the inequality:
$$
\left(a^{2}+4 a-5\right)-\left(2 a^{2}+2\right)=-a^{2}+4 a-7 \text {. }
$$
Consider the equations in $x$:
$$
\begin{array}{l}
x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7... | a \in(-\infty, 1] \cup[3,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,618 |
5. Given real numbers $a, b, c$, and $b \neq 0$. If real numbers $x_{1},$ $x_{2}, y_{1}, y_{2}$ satisfy $x_{1}^{2}+a x_{2}^{2}=b, x_{2} y_{1}-x_{1} y_{2}=a$, $x_{1} y_{1}+a x_{2} y_{2}=c$, then the value of $y_{1}^{2}+a y_{2}^{2}$ is $\qquad$ | 5. $\frac{c^{2}+a^{3}}{b}$.
From the given, we have $\left(x_{2} y_{1}-x_{1} y_{2}\right)^{2}=a^{2}$, that is
$$
\begin{array}{l}
2 x_{1} x_{2} y_{1} y_{2}+a^{2}=x_{2}^{2} y_{1}^{2}+x_{1}^{2} y_{2}^{2} . \\
\text { Hence } b\left(y_{1}^{2}+a y_{2}^{2}\right)=\left(x_{1}^{2}+a x_{2}^{2}\right)\left(y_{1}^{2}+a y_{2}^{2... | \frac{c^{2}+a^{3}}{b} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,619 |
6. As shown in Figure 1, let $P$ be a point inside the convex quadrilateral $A B C D$. Draw perpendiculars from $P$ to $A B$, $B C$, $C D$, and $D A$, with the feet of the perpendiculars being $E$, $F$, $G$, and $H$, respectively. Given that $A H=3$, $H D=4$, $D G=1$, $G C=5$, $C F=6$, $F B=4$, and $B E-A E=1$. Then th... | 6.34.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
P A^{2}-A H^{2}=P D^{2}-D H^{2}, \\
P D^{2}-D G^{2}=P C^{2}-C G^{2}, \\
P C^{2}-C F^{2}=P B^{2}-B F^{2}, \\
P B^{2}-B E^{2}=P A^{2}-A E^{2} .
\end{array}
$$
Adding the above four equations, we get
$$
\begin{array}{l}
A H^{2}+D G^{2}+C F^{2}+B E^{2} \\
=A E... | 34 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,620 |
7. As shown in Figure $2, \triangle A B C$ has an area of 1, points $D, G, E,$ and $F$ are on sides $A B, A C, B C$ respectively, $B D < D A$, $D G \parallel B C, D E \parallel A C, G F \parallel A B$. Then the maximum possible area of trapezoid $D E F G$ is $\qquad$ | 7. $\frac{1}{3}$.
Let $\frac{A D}{A B}=x$, then $\frac{B D}{B A}=1-x=\frac{C G}{C A}$.
Thus, by the similarity of triangles, we have
$$
\frac{S_{\triangle D C}}{S_{\triangle A B C}}=x^{2}, \frac{S_{\triangle B D E}}{S_{\triangle B C}}=(1-x)^{2}=\frac{S_{\triangle C F C}}{S_{\triangle A B C}} \text {. }
$$
Since $S_{\... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,621 |
8. Positive integers $x$ not exceeding 1000, such that the sum of the digits of both $x$ and $x+1$ are odd. Then the number of positive integers $x$ that satisfy the condition is $\qquad$. | 8.46.
Obviously, $x \neq 1000$.
Let $x=\overline{a b c}$, where $a, b, c \in\{0,1, \cdots, 9\}$, and not all are zero. $S(x)=a+b+c$ is the sum of the digits of $x$.
(1) If $c \neq 9$, then
$$
\begin{array}{l}
S(x)=a+b+c, \\
S(x+1)=a+b+c+1 ;
\end{array}
$$
(2) If $c=9, b \neq 9$, then
$$
S(x)=a+b+9, S(x+1)=a+b+1 \text ... | 46 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,622 |
9. Given that $k$ is a positive integer not exceeding 50, such that for any positive integer $n, 2 \times 3^{6 n}+k \times 2^{3 n+1}-1$ is always divisible by 7. Then the number of such positive integers $k$ is $\qquad$. | 9.7.
$$
\begin{array}{l}
2 \times 3^{6 n}+k \times 2^{3 n+1}-1 \\
=2 \times 27^{2 n}+2 k \times 8^{n}-1 \\
\equiv 2 \times(-1)^{2 n}+2 k-1 \\
\equiv 2 k+1(\bmod 7) .
\end{array}
$$
But $2 \times 3^{6 n}+k \times 2^{3 n+1}-1 \equiv 0(\bmod 7)$, then
$$
2 k+1 \equiv 0(\bmod 7) \text {, }
$$
which means $2 k+1=7 m$ (whe... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,623 |
10. All prime numbers $p$ that make $\frac{p(p+1)+2}{2}$ a perfect square are $\qquad$ . | 10.2 or 5.
Let $\frac{p(p+1)+2}{2}=k^{2}\left(k \in \mathbf{Z}_{+}\right)$. Then $p(p+1)=2 k^{2}-2=2(k+1)(k-1)$.
(1) When $p=2$, $3=k^{2}-1, k=2$;
(2) When $p \neq 2$, $p \mid(k+1)$ or $p \mid(k-1)$. If $p \mid(k+1)$, then $p+1 \geqslant 2(k-1)$. Thus, $k+2 \geqslant p+1 \geqslant 2(k-1) \Rightarrow k \leqslant 4$.
Wh... | 2 \text{ or } 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,624 |
II. (20 points) As shown in Figure 3, in the right triangle $\triangle ABC$, $\angle C=90^{\circ}, BC = 2, AC = x$, point $F$ is on side $AB$, points $G$ and $H$ are on side $BC$, and quadrilateral $EFGH$ is a square with side length $y$, and $AE = AC$.
(1) Find the function expression of $y$ in terms of $x$.
(2) For w... | (1) As shown in Figure 5, extend $FE$ to intersect $AC$ at point $D$. Clearly, $DF \parallel BC$, then
$\mathrm{Rt} \triangle ADF \sim$
Rt $\triangle ACB$.
Noting that $AE = AC = x$, we have
$DE = \sqrt{x^2 - (x-y)^2} = \sqrt{2xy - y^2}$.
Thus, $\frac{x-y}{x} = \frac{\sqrt{2xy - y^2} + y}{2}$
$$
\Rightarrow 2x - 2y - x... | \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,625 |
Three, (20 points) Find all possible values of the positive integer $n$ such that for such $n$, there exist real numbers $a$ and $b$ for which the function $f(x)=\frac{1}{n} x^{2}+a x+b$ is an integer for any integer $x$.
---
Please note that the translation retains the original formatting and structure of the text, ... | Three, let $f(x)=\frac{1}{n} x^{2}+a x+b$ be such that for any integer $x, f(x)$ is an integer. Then,
$$
\begin{array}{l}
g(x)=f(x+1)-f(x) \\
=\left[\frac{1}{n}(x+1)^{2}+a(x+1)+b\right]- \\
\quad\left(\frac{1}{n} x^{2}+a x+b\right) \\
=\frac{2}{n} x+\frac{1}{n}+a
\end{array}
$$
is also an integer. Furthermore,
$$
g(x+... | n=1 \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,626 |
Four, (20 points) In a box, there are a total of 88 small balls in three colors: red, yellow, and black. It is known that by randomly taking out 24 balls, it can be guaranteed that at least 10 of the balls are of the same color. Under the condition that this is true, regardless of how the balls of each color are distri... | First, prove that taking out only 43 balls is not enough.
In fact, when there are 42 red balls, 41 yellow balls, and 5 black balls in the box, if you take any 24 balls, then the number of red and yellow balls is at least $24-5=19$, so there must be at least 10 red or yellow balls; but taking 19 red balls, 19 yellow bal... | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,627 |
Example 3 Try to find all positive integers $k$, such that for any positive integers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a$, $b$, and $c$.
(2002, Girls' Mathematical Olympiad) | Explanation: First, from the inequality relationship between $a b+b c+c a$ and $a^{2}+b^{2}+c^{2}$, we derive $k>5$. Then, by constructing an example, we find that $k \leqslant 6$. Therefore, since $k \in \mathbf{Z}_{+}$, we conclude that $k=6$. Finally, we provide the proof.
Notice that for any positive real numbers ... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,629 |
4. If $f(x)=(k-1) x^{2}+2 k x+2007$ is an even function on $\mathbf{R}$, then on $(-\infty, 2007)$, $f(x)$ ( ).
(A) is an increasing function
(B) is a decreasing function
(C) first decreases then increases
(D) first increases then decreases | 4.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,632 |
5. In $\triangle A B \mathrm{C}$, if
$$
a b \cos C+b c \cos A+c a \cos B=c^{2},
$$
where the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, then the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{2} a b$
(B) $\frac{1}{2} b c$
(C) $\frac{1}{2} c a$
(D) $\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\... | 5.A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,633 |
3. The first term of an arithmetic sequence is a non-zero real number $a$, and for each positive integer $n$, the sum of the first $n$ terms of the sequence is equal to $a n^{2}$. Then the common difference of this sequence is $\qquad$
Translate the text above into English, please keep the original text's line breaks ... | 3. $2 a$ | 2 a | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,636 |
4. Use a set of student's
triangular rulers to form quadrilateral $A B C D$ as shown in Figure 3, where $\angle C B D=\angle C D B=$ $45^{\circ}, \angle B A D=2 \angle B D A$ $=60^{\circ}$, and let the angle formed by diagonal $C A$ and side $C B$ be $\theta$. Then $\tan \theta=$ $\qquad$ | 4. $\frac{\sqrt{3}-1}{2}$ | \frac{\sqrt{3}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,637 |
5. On a plane, 2007 non-coincident lines $l_{1}, l_{2}, \cdots, l_{2007}$ are drawn, always following the rule of alternating perpendicular and parallel $\left(l_{2} \perp l_{1}, l_{3} / / l_{2}, l_{4} \perp l_{3}, l_{5} / /\right.$ $\left.l_{4}, \cdots\right)$. These 2007 non-coincident lines have a total of intersect... | 5.
$$
1007012
$$ | 1007012 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,638 |
Three, (15 points) Given the linear function $f(x)=a x+b$
for any $x, y \in[0,1]$ it satisfies
$|f(x)+f(y)-x y| \leqslant \frac{1}{4}$.
Determine such $f(x)$.
| For a linear function $f(x)=a x+b$, let
$$
F(x, y)=a x+b+a y+b-x y .
$$
According to the given, $F(0,0)=2 b \geqslant-\frac{1}{4}$, which means
$$
b \geqslant-\frac{1}{8} \text {. }
$$
$$
F(0,1)=a+2 b \leqslant \frac{1}{4} \text {. }
$$
Also, $F(1,1)=2 a+2 b-1 \geqslant-\frac{1}{4}$, which means $a+b \geqslant \frac{... | f(x)=\frac{1}{2} x-\frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,639 |
Example 4 Given a positive integer $n(n \geqslant 2)$. Find the largest real number $\lambda$, such that the inequality
$$
a_{n}^{2} \geqslant \lambda\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n},
$$
holds for any positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfying $a_{1}<a_{2}<\cdots<a_{n}$.
(2003, Girls' ... | Explanation: First, derive the range of $\lambda$ for a special sequence $\left\{a_{n}\right\}$, then provide a proof for any sequence $\left\{a_{n}\right\}$.
Let $a_{i}=i(i=1,2, \cdots, n)$.
From equation (1), it is easy to find that $\lambda \leqslant \frac{2(n-2)}{n-1}$.
Next, we prove: For any positive integers $a_... | \frac{2(n-2)}{n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,640 |
Four, (10 points) As shown in Figure 4, for the cyclic quadrilateral $ABCD$:
one pair of opposite sides
$AB, DC$ are extended to meet at point $P$,
the other pair of opposite sides
$AD, BC$ are extended to meet at point
$Q$, from $P, Q$ respectively draw the tangents $PE, QF$ to the circle,
where $E, F$ are the points ... | To prove: $P Q^{2}=P E^{2}+P F^{2}$ is sufficient.
As shown in Figure 5, construct an auxiliary circle passing through points $Q$, $D$, and $C$, intersecting $P Q$ at point $G$, and connect $C G$.
Since $D$, $C$, $G$, and $Q$ are concyclic, we have,
$$
\begin{array}{l}
\angle P G C=\angle Q D C \\
=\angle A B C .
\en... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,641 |
Five. (15 points) Given a sequence of real numbers $x_{0}, x_{1}, x_{2}, \cdots$, $x_{n}, \cdots$ defined by the recurrence relation
$$
x_{0}=5, x_{n}=x_{n-1}+\frac{1}{x_{n-1}}(n=1,2, \cdots)
$$
Prove that: $45<x_{1000}<45.1$. | It is not hard to see that all terms of the sequence are positive and increasing, i.e.,
$$
x_{0}45$.
$x_{n-1}^{2}=\left(x_{n-2}+\frac{1}{x_{n-2}}\right)^{2}=x_{n-2}^{2}+\frac{1}{x_{n-2}^{2}}+2$,
$x_{n-2}^{2}=\left(x_{n-3}+\frac{1}{x_{n-3}}\right)^{2}=x_{n-3}^{2}+\frac{1}{x_{n-3}^{2}}+2$,
Furthermore, when $n=100$, from... | 45<x_{1000}<45.1 | Algebra | proof | Yes | Yes | cn_contest | false | 719,642 |
One, (15 points) Given $T=\{1,2, \cdots, 8\}$. For $A \subseteq T, A \neq \varnothing$, define $S(A)$ as the sum of all elements in $A$. Question: How many non-empty subsets $A$ of $T$ are there such that $S(A)$ is a multiple of 3 but not a multiple of 5?
(Chen Yonggao provided) | For $\varnothing$, define $S(\varnothing)=0$. Let
$$
T_{0}=\{3,6\}, T_{1}=\{1,4,7\}, T_{2}=\{2,5,8\} \text {. }
$$
For $A \subseteq \dot{T}$, let
$$
A_{0}=A \cap T_{0}, A_{1}=A \cap T_{1}, A_{2}=A \cap T_{2} .
$$
Then $S(A)=S\left(A_{0}\right)+S\left(A_{1}\right)+S\left(A_{2}\right)$
$$
\equiv\left|A_{1}\right|-\left... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,643 |
II. (15 points) As shown in Figure 1, circles $\odot O_{1}$ and $\odot O_{2}$ intersect at points $C$ and $D$. A line through $D$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $B$, respectively. Point $P$ lies on the arc $\overparen{A D}$ of $\odot O_{1}$, and $P D$ intersects the extension of line segme... | Let $\triangle A B C$ have a circumcircle $\odot O$ with radius $R$, and let the tangent from point $N$ to $\odot O$ be $N X$. Then,
$$
N O^{2}=N X^{2}+R^{2}=N C \cdot N B+R^{2} \text {. }
$$
Similarly, $M O^{2}=M C \cdot M A+R^{2}$.
Since $A 、 C 、 D 、 P$ are concyclic, we have,
$$
M C \cdot M A=M D \cdot M P \text {.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,644 |
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