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 |
|---|---|---|---|---|---|---|---|---|---|
9. As shown in Figure 3, in $\triangle A B C$, it is known that $D$ is the midpoint of side $B C$, and $E, F$ are the trisection points of side $A B$. $A D$ intersects $C E, C F$ at points $M, N$ respectively. Then $A M: M N: N D=$ $\qquad$ | 9. 5:3:2.
As shown in Figure 6, connect $D F$.
Since $D$ is the midpoint of side $B C$, and $E, F$ are the trisection points of side $A B$, we have
$$
\begin{array}{l}
D F \parallel C E \\
\Rightarrow A M=\frac{1}{2} A D .
\end{array}
$$
Draw a line through point $D$ parallel to $A B$, intersecting $C F$ at point $G$... | 5:3:2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,245 |
Example 5 As shown in Figure 7, the side length of square $ABCD$ is 1, and points $E, F, G$ are on sides $AB, AD, BC$ respectively (they can coincide with the vertices). If $\triangle EFG$ is an equilateral triangle, find the maximum and minimum values of the area of $\triangle EFG$.
(2010, Wuhan Junior High School Mat... | Solve as shown in Figure 7, construct $E K \perp F G$. Then $K$ is the midpoint of side $F G$. Connect $A K$ and $B K$.
$$
\begin{array}{l}
\text { By } \angle E K G=\angle E B G=\angle E K F \\
=\angle E A F=90^{\circ},
\end{array}
$$
we get that $E, K, G, B$ and $E, K, F, A$ are respectively concyclic.
Thus, $\angle... | \frac{\sqrt{3}}{4} \text{ and } 2\sqrt{3} - 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,246 |
10. If there is a square $A B C D$ in a plane and $M$ is any point in the plane, then the minimum value of $\frac{M A+M C}{M B+M D}$ is | 10. $\frac{\sqrt{2}}{2}$.
Let the side length of square $A B C D$ be $a$.
If point $M$ coincides with $A$ or $C$, then
$$
\frac{M A+M C}{M B+M D}=\frac{\sqrt{2} a}{2 a}=\frac{\sqrt{2}}{2} \text {. }
$$
If point $M$ is neither $A$ nor $C$, then $M A \cdot M C>0$.
By $(M B-M D)^{2} \geqslant 0$, we know
$$
M B^{2}+M D^... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,247 |
11. Given the parabola $y=x^{2}+m x+n$ passes through the point $(2,-1)$, and intersects the $x$-axis at points $A(a, 0)$ and $B(b, 0)$. If $P$ is the vertex of the parabola, find the equation of the parabola that minimizes the area of $\triangle P A B$. | Three, 11. Given that the parabola $y=x^{2}+m x+n$ passes through the point $(2,-1)$, we have
$$
4+2 m+n=-1 \Rightarrow 2 m+n=-5 \text {. }
$$
Since the parabola $y=x^{2}+m x+n$ intersects the $x$-axis at points $A(a, 0), B(b, 0)$, we have,
$$
a+b=-m, a b=n \text {. }
$$
Thus, $|A B|=|a-b|=\sqrt{(a+b)^{2}-4 a b}$ $=\... | y=x^{2}-4 x+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,248 |
12. As shown in Figure 4, with the vertices of the equilateral triangle $\triangle ABC$ with side length 1 as the centers and its side length as the radius, draw three equal circles, obtaining intersection points $D, E, F$. Connect $CF$ to intersect $\odot C$ at point $G$. With point $E$ as the center and $EG$ as the r... | 12. According to the problem, $\angle A C G=30^{\circ}$. As shown in Figure 7, connect $E A$, $E C$, and $E M$.
Since the radii of $\odot A$ and $\odot C$ are both 1, $\triangle A C E$ is an equilateral triangle, so $\angle A C E=60^{\circ}$.
Thus, $\angle E C F=90^{\circ}$.
In the right triangle $\triangle E C G$, we... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,249 |
13. Let $p$ and $5 p^{2}-2$ both be prime numbers. Find the value of $p$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 13. Let $m$ be a non-negative integer.
If $p=3 m$, then $p=3,5 p^{2}-2=43$, both are prime numbers;
$$
\begin{array}{l}
\text { If } p=3 m+1 \text {, then } m>0, \\
5 p^{2}-2=5(3 m+1)^{2}-2 \\
=3\left(15 m^{2}+10 m+1\right)
\end{array}
$$
is not a prime number;
$$
\begin{array}{l}
\text { If } p=3 m+2 \text {, then }... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,250 |
14. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
x < a \\
x > -a
\end{array}\right.
$$
the solution set contains exactly two integers. Find the range of real values for $a$.
| 14. From the given, we know $\frac{a+2}{2}<x<a+1$.
From the problem, there exists an integer $k$ satisfying
$$
\begin{array}{l}
\left\{\begin{array}{l}
k \leqslant \frac{a+2}{2}<k+1, \\
k+2<a+1 \leqslant k+3
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2 k-2 \leqslant a<2 k, \\
k+1<a \leqslant k+2 .
\end{a... | 3<a<4 \text{ or } 4<a \leqslant 5 \text{ or } a=6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,251 |
1. In a finite group of boys and girls, call a set of boys a "boy-friendly set" if each girl knows at least one boy in the set; call a set of girls a "girl-friendly set" if each boy knows at least one girl in the set. Prove: the number of boy-friendly sets has the same parity as the number of girl-friendly sets. (The r... | 1. Proof 1 If any boy in the set $X$ of boys does not know any girl in the set $Y$ of girls, then $X$ and $Y$ are called "separated."
Consider the number $S$ of all separated subset pairs $(X, Y)$.
For a given set $X$ of boys, let $Y_{X}$ be the largest set of girls that are separated from $X$. Then, the number of girl... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,252 |
2. Given in non-isosceles $\triangle ABC$, points $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$ respectively. Line $BE$ intersects the circumcircle of $\triangle BCF$ at point $P$ (different from point $B$), line $AD$ intersects the circumcircle of $\triangle ABE$ at point $Q$ (different from point ... | 2. As shown in Figure 1, take point $T$ on ray $G F$ such that $G F \cdot G T = G Q \cdot G D$.
Thus, points $F, D, Q, T$ are concyclic.
Therefore, $\angle F Q G = \angle G T D = \angle C T D$. It is easy to see that $\angle D P B = \angle C M D$.
Let $G E = x, G F = y$.
From $G P \cdot G B = G C \cdot G F$, we get $G ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,253 |
3. Given that each positive integer is colored either red or blue. The function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ satisfies the following two conditions:
(1) If $x \leqslant y$, then $f(x) \leqslant f(y)$;
(2) If positive integers $x, y, z$ (which can be equal) have the same color, and $x+y=z$, then $f(x)+... | 3. For integers $x, y$, let $[x, y]$ denote all integers $t$ satisfying $x \leqslant t \leqslant y$, and call its length $y-x$.
If $\frac{f(x)}{x}=\frac{f(y)}{y}$ holds for any same-colored integers $x, y$, then take $a=\max \left\{\frac{f(r)}{r}, \frac{f(b)}{b}\right\}$, where $r, b$ are integers colored red and blue... | proof | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,254 |
4. Prove: There exist infinitely many positive integers $n$, such that $2^{2^{n}+1}+1$ is divisible by $n$, but $2^{n}+1$ is not divisible by $n$. | 4. (1) First, we explain that $n=57$ is a positive integer that meets the requirements.
On one hand, because
$$
\begin{array}{l}
2^{57}+1 \equiv\left(2^{9}\right)^{6} \times 2^{3}+1 \\
\equiv(-1)^{6} \times 2^{3}+1 \equiv 9(\bmod 19),
\end{array}
$$
thus, $2^{57}+1$ cannot be divisible by 57.
On the other hand, becaus... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,255 |
5. Given an integer $n \geqslant 3$ and $\left[\frac{(n+2)^{2}}{3}\right]$ colors. Each cell of an $n \times n$ grid is colored with one of these colors, and each color is used at least once. Prove: there must exist a $1 \times 3$ or $3 \times 1$ sub-rectangle in the grid, whose three cells are colored with three diffe... | 5. If at least two of the three cells in a $1 \times 3$ or $3 \times 1$ small rectangle are of the same color, then the small rectangle is said to be "dominated" by that color. Clearly, each small rectangle can be dominated by at most one color.
First, we prove two lemmas.
Lemma 1 In a row, if a certain color (let's as... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,256 |
Example 6 Given that $a, b, c, x, y, z$ are all positive real numbers, and $a+x=b+y=c+z=k$. Prove:
$$
a z+b x+c y<k^{2} .
$$ | Prove as shown in Figure 8, construct an equilateral $\triangle ABC$, such that $AB = BC = CA = k$, points $D$, $E$, and $F$ are on sides $AB$, $BC$, and $CA$ respectively, and $AD = a$, $BE = b$, $CF = c$. Then
$S_{\triangle ABC} = \frac{1}{2} k^2 \sin 60^{\circ}$, $S_{\triangle ADF} = \frac{1}{2} a z \sin 60^{\circ}$... | a z + b x + c y < k^2 | Inequalities | proof | Yes | Yes | cn_contest | false | 725,257 |
6. Given that $I$ and $O$ are the incenter and circumcenter of $\triangle ABC$, respectively, circle $\omega_{A}$ passes through points $B$ and $C$, and is tangent to the incircle of $\triangle ABC$. Similarly, define circles $\omega_{B}$ and $\omega_{C}$. Let circles $\omega_{B}$ and $\omega_{C}$ intersect at two dist... | 6. As shown in Figure 2, let the incircle of $\triangle ABC$ be $\Gamma$, and let it touch sides $BC$, $CA$, and $AB$ at points $A_1$, $B_1$, and $C_1$ respectively. Let $\Gamma$ and circle $\omega_A$ be tangent at point $X_A$, and extend $X_A A_1$ to intersect circle $\omega_A$ at point $M_A$.
Since $\Gamma$ and $\om... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,258 |
1. The system of equations
$$
\left\{\begin{array}{l}
y^{x^{2}+7 x+12}=1, \\
x+y=1
\end{array}\right.
$$
has the solution set | $$
\text { - 1. }\{(0,1),(2,-1),(-3,4),(-4,5)\} \text {. }
$$
Notice,
| \{(0,1),(2,-1),(-3,4),(-4,5)\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,259 |
2. In the Cartesian coordinate system, a line segment $A B$ of length 1 moves on the $x$-axis (point $A$ is to the left of $B$), point $P(0,1)$ is connected to $A$ by a line, and point $Q(1,2)$ is connected to $B$ by a line. Then the equation of the trajectory of the intersection point $R$ of lines $P A$ and $Q B$ is $... | 2. $y(x-2)=-2$.
Let $A(a, 0), B(a+1,0)$, the intersection point of line $P A$ and $Q B$ is $R(x, y)$.
When $a=0$, the two lines do not intersect, so $a \neq 0$.
At this time, $l_{P A}: \frac{x}{a}+y=1$,
$$
l_{Q B}: y=-\frac{2}{a}(x-a-1) \text {. }
$$
From equations (1) and (2), we get $x=a+2, y=-\frac{2}{a}$.
Elimina... | y(x-2)=-2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,260 |
3. Given that $M$ is a moving point on the arc of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ in the first quadrant, $M N \perp y$ axis, with the foot of the perpendicular being $N$. When the area of $\triangle O M N$ is maximized, the radius $r=$ $\qquad$ of its inscribed circle | 3. $\frac{\sqrt{2}}{2}$.
Let $M(4 \cos \theta, 3 \sin \theta)\left(0<\theta<\frac{\pi}{2}\right)$. Then
$$
\begin{array}{l}
S_{\triangle M O N}=\frac{1}{2}|O N||N M| \\
=\frac{1}{2} \times 3 \sin \theta \times 4 \cos \theta=3 \sin 2 \theta .
\end{array}
$$
Therefore, when $\theta=\frac{\pi}{4}$, $\left(S_{\triangle M... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,261 |
4. Given that the circumradius of $\triangle A B C$ is $1, \angle A$, $\angle B$, and $\angle C$'s angle bisectors intersect the circumcircle of $\triangle A B C$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Then
$$
\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin ... | 4. 2 .
Connect $B A_{1}$. By the Law of Sines, we have
$$
\begin{array}{l}
A A_{1} \cos \frac{A}{2}=2 \sin \left(B+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Similarly, $B B_{1} \cos \frac{B}{2}=\sin C+\sin A$,
$$
C C_{1} \cos \frac{C}{2}=\sin A+\sin B \text {. }
$$
Therefore, the o... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,262 |
5. Let $f(x)=a \sin [(x+1) \pi]+b \sqrt[3]{x-1}+2$, where $a$ and $b$ are real constants. If $f(\lg 5)=5$, then $f(\lg 20)=$ | 5. -1 .
Obviously,
$$
f(x)=a \sin [(x-1) \pi]+b \sqrt[3]{x-1}+2 .
$$
Let $t=x-1$. Then
$$
f(t+1)=a \sin \pi t+b \sqrt[3]{t}+2=g(t)+2,
$$
where, $g(t)=a \sin \pi t+b \sqrt[3]{t}$ is an odd function.
According to the problem,
$$
\begin{aligned}
5 & =f(\lg 5)=f(1-\lg 2) \\
& =g(-\lg 2)+2=-g(\lg 2)+2 .
\end{aligned}
$$
... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,263 |
6. In the Cartesian coordinate system, $O$ is the origin, points $A(3, a) 、 B(3, b)$ make $\angle A O B=45^{\circ}$, where $a 、 b$ are integers, and $a>b$. Then the number of pairs $(a, b)$ that satisfy the condition is. | 6.6.
Let $\angle A O X=\alpha, \angle B O X=\beta$. Then $\tan \alpha=\frac{a}{3}, \tan \beta=\frac{b}{3}$.
Given $a>b$, we have
$$
\begin{array}{l}
1=\tan 45^{\circ}=\tan (\alpha-\beta) \\
=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{3(a-b)}{9+a b} .
\end{array}
$$
Rearranging gives $(a+3)(b... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,264 |
7. Given the equation of circle $\odot C$ is
$$
x^{2}+y^{2}-4 x-2 y+1=0 \text {, }
$$
the line $y=\left(\tan 10^{\circ}\right) x+\sqrt{2}$ intersects $\odot C$ at points
$A$ and $B$. The sum of the inclination angles of lines $A C$ and $B C$ is
$\qquad$ | 7. $200^{\circ}$.
The equation of $\odot C$ is $(x-2)^{2}+(y-1)^{2}=4$, the distance from the center $(2,1)$ to the line $y=\left(\tan 10^{\circ}\right) x+\sqrt{2}$ is
$$
d=\frac{\left|2 \tan 10^{\circ}-1+\sqrt{2}\right|}{\sqrt{1+\tan ^{2} 10^{\circ}}}<2,
$$
which means the given line intersects $\odot C$ at points $A... | 200^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,265 |
8. A and B, two table tennis players, are in the middle of a match. A must win 2 more games to finally win; B must win 3 more games to finally win. If the probability of A or B winning each game is $\frac{1}{2}$, then the probability of A finally winning is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 8. $\frac{11}{16}$.
Player A has the following six possible ways to win:
(1) A wins, A wins;
(2) A wins, B wins, A wins;
(3) A wins, B wins, B wins, A wins;
(4) B wins, A wins, A wins;
(5) B wins, A wins, B wins, A wins;
(6) B wins, B wins, A wins, A wins.
Their probabilities are $\frac{1}{4}$, $\frac{1}{8}$, $\frac{... | null | Other | math-word-problem | Yes | Yes | cn_contest | false | 725,266 |
9. (14 points) For two real numbers $a, b$, $\min \{a, b\}$ represents the smaller of $a, b$. Find all non-zero real numbers $x$ such that
$$
\min \left\{x+\frac{4}{x}, 4\right\} \geqslant 8 \min \left\{x, \frac{1}{x}\right\} .
$$ | When $x>0$, $x+\frac{4}{x} \geqslant 2 \sqrt{x \cdot \frac{4}{x}}=4$; when $x>0$; \\
$x+\frac{4}{x}, & x>1$; \\
$x, & x \leqslant-1 \text { or } 0<1$; \\
(1) When $x>1$, the original inequality is $4 \geqslant \frac{8}{x} \Rightarrow x \geqslant 2$.
At this time, $x \in[2,+\infty)$.
(2) When $0<x \leqslant 1$, the... | (-\infty, 0) \cup\left(0, \frac{1}{2}\right] \cup[2,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,267 |
Example 9 Given $a, b>0$. Find the area of the triangle with side lengths
$$
\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+4 b^{2}}, \sqrt{4 a^{2}+b^{2}}
$$ | Solve as shown in Figure 9, construct rectangle $ABCD$, such that $AB$
$$
=CD=2b, AD=BC
$$
$=2a$, and $E, F$ are the midpoints of
$CD, AD$ respectively. In this case, in $\triangle DEF$,
$$
\begin{array}{l}
EF=\sqrt{a^{2}+b^{2}}, BF=\sqrt{a^{2}+4b^{2}}, \\
BE=\sqrt{4a^{2}+b^{2}} . \\
\text{Therefore, } S_{\triangle DEF... | \frac{3}{2}ab | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,268 |
10. (14 points) As shown in Figure 1, in $\triangle ABC$, $O$ is the midpoint of side $BC$, points $M$ and $N$ are on sides $AB$ and $AC$ respectively, and $AM=6$, $MB=4$, $AN=$ $4$, $NC=3$, $\angle MON=$ $90^{\circ}$. Find the size of $\angle A$. | 10. Solution 1 As shown in Figure 3, extend $N O$ to point $P$, such that $O P=O N$. Connect $B P, M P$.
Since $B O=O C$, we know
$B P \parallel A C, B P=C N=3$.
Since point $M$ is on the perpendicular bisector of $N P$, therefore,
$M P=M N$.
Let $M N=a$. Then, in $\triangle A M N$ and $\triangle M B P$, by the cosine ... | \arccos \frac{3}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,269 |
11. (16 points) For an integer $k$, define the set
$$
S_{k}=\{n \mid 50 k \leqslant n<50(k+1), n \in \mathbf{Z}\} \text {. }
$$
How many of the 600 sets $S_{0}, S_{1}, \cdots, S_{599}$ do not contain any perfect squares? | 11. Notice,
$$
\begin{array}{l}
(x+1)^{2}-x^{2}=2 x+1 \leqslant 50(x \in \mathbf{N}) \\
\Leftrightarrow x \leqslant 24(x \in \mathbf{N}) .
\end{array}
$$
While $(24+1)^{2}=625 \in S_{12}$, thus, the square numbers in $S_{0}, S_{1}$, $\cdots, S_{12}$ do not exceed $25^{2}$, and each set consists of 50 consecutive non-n... | 439 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,270 |
12. (16 points) Find all positive integers $n$ greater than 1, such that for any positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, the inequality
$$
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \geqslant n\left(x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n} x_{1}\right) \text {. }
$$
holds. | 12. When $n=2$, the inequality is
$$
\begin{array}{l}
\left(x_{1}+x_{2}\right)^{2} \geqslant 2\left(x_{1} x_{2}+x_{2} x_{1}\right) \\
\Leftrightarrow\left(x_{1}-x_{2}\right)^{2} \geqslant 0 .
\end{array}
$$
Thus, $n=2$ satisfies the condition.
When $n=3$, the inequality is
$$
\begin{array}{l}
\left(x_{1}+x_{2}+x_{3}\r... | 2, 3, 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,271 |
$$
\begin{array}{l}
\text { 2. If } f(g(x))=\sin 2 x, \\
g(x)=\tan \frac{x}{2}(0<x<\pi),
\end{array}
$$
then $f\left(\frac{\sqrt{2}}{2}\right)=$ $\qquad$ | 2. $\frac{4 \sqrt{2}}{9}$.
Notice that,
$$
\begin{array}{c}
f(g(x))=\sin 2 x=2 \sin x \cdot \cos x \\
=2 \cdot \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \cdot \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} .
\end{array}
$$
Then $f\left(\frac{\sqrt{2}}{2}\right)=2 \times \frac{\sqrt{2}}{1+\frac{1}{2}... | \frac{4 \sqrt{2}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,273 |
3. If a square pyramid with a base edge length of 2 is inscribed with a sphere of radius $\frac{1}{2}$, then the volume of this square pyramid is .. $\qquad$ | 3. $\frac{16}{9}$.
Let $\theta$ be the dihedral angle between the base and a side face of a regular quadrilateral pyramid. Then $\tan \frac{\theta}{2}=\frac{1}{2}$.
Thus, $\tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}=\frac{4}{3}$. Therefore, the height of the regular quadrilateral pyramid i... | \frac{16}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,274 |
4. In the Cartesian coordinate system, point $A(1,2)$, $B(4,1)$, and a moving point $P(x, y)$ on the circle $x^{2}+y^{2}=25$ form a triangle. Then the maximum value of $S_{\triangle A B P}$ is $\qquad$ | 4. $\frac{1}{2}(7+5 \sqrt{10})$.
The length of segment $AB$ is
$$
\sqrt{(4-1)^{2}+(1-2)^{2}}=\sqrt{10} \text {. }
$$
The equation of the line on which segment $AB$ lies is $x+3 y-7=0$, and its distance from the origin is $\frac{|-7|}{\sqrt{1^{2}+3^{2}}}=\frac{7}{\sqrt{10}}$.
Since points $A$ and $B$ are both inside ... | \frac{1}{2}(7+5 \sqrt{10}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,275 |
5. Divide the positive integers $1,2, \cdots, 7$ into two groups in any way, such that each group has at least one number. Then the probability that the sum of the numbers in the first group is equal to the sum of the numbers in the second group is $\qquad$ | 5. $\frac{4}{63}$.
There are $2^{7}-2=126$ possible scenarios.
Notice that $1+2+\cdots+7=28$.
Therefore, when the sums of the two groups of numbers are equal, the value of the sum is 14. At this point, the first group of numbers must contain at least one of the numbers $5,6,7$. Thus,
when the first group of numbers co... | \frac{4}{63} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,276 |
6. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$. Then the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is | 6.3.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(a_{n}+1\right)}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \\
\Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\sum_{n=0}^{2011}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,277 |
7. The four real roots of the quartic polynomial $f(x)$ form an arithmetic sequence with a common difference of 2. Then, the difference between the largest and smallest roots of $f^{\prime}(x)$ is | 7. $2 \sqrt{5}$.
Let the four real roots of $f(x)$ be
$$
\begin{array}{l}
a-3, a-1, a+1, a+3. \\
\text { Then } f(x)= k(x-a-3)(x-a-1) . \\
(x-a+1)(x-a+3)(k \neq 0) .
\end{array}
$$
Let $x-a=t$. Then
$$
\begin{array}{l}
f(x)=k(t-3)(t-1)(t+1)(t+3) \\
=k\left(t^{4}-10 t^{2}+9\right)=g(t) .
\end{array}
$$
Thus, $g^{\pr... | 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,278 |
8. $[x]$ represents the greatest integer not exceeding the real number $x$. The area of the figure formed by points satisfying $[x]^{2}+[y]^{2}=50$ on the plane is $\qquad$ . | 8. 12 .
First, consider the case in the first quadrant.
When $x>0, y>0$, from $[x]^{2}+[y]^{2}=50$, we get
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l }
{ 7 \leqslant x < 8 , } \\
{ 1 \leqslant y < 2 }
\end{array} \left\{\begin{array} { l }
{ 5 \leqslant x < 6 , } \\
{ 5 \leqslant y < 6 ; }
\end{array} \... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,280 |
9. (14 points) Given a positive sequence $\left\{a_{n}\right\}$ satisfies:
(1) $a_{1}=2012$;
(2) $a_{2} 、 a_{3}$ are integers;
(3) The sequence $\left\{n a_{n}-n^{2}\right\}$ is a geometric sequence with a common ratio no greater than 10.
Find the general term formula of the sequence $\left\{a_{n}\right\}$. | From condition (3), we know that $n a_{n}-n^{2}=c q^{n-1}$.
Thus, $a_{n}=\frac{c q^{n-1}}{n}+n \quad (n=1,2, \cdots)$.
From condition (1), we get $c=2011$. Therefore,
$$
a_{n}=\frac{2011 q^{n-1}}{n}+n \quad (n=1,2, \cdots).
$$
Since $a_{2}=\frac{2011 q}{2}+2$ is an integer, $\frac{2011 q}{2}$ must be an integer.
Assum... | a_{n}=\frac{2011 \times 6^{n-1}}{n}+n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,281 |
10. (14 points) Given $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, and point $P$ is on the curve $C$. If $S_{\triangle P F_{1} F_{2}}=\sqrt{3}$, find $\angle F_{1} P F_{2}$. | 10. Suppose point \( P\left(x_{0}, y_{0}\right) \) is on the right branch of the hyperbola.
From the given conditions, it is easy to get \( F_{1} F_{2}=2 \sqrt{2} \). Then
\[
\left.S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2} \|\right| y_{0}\left|=\frac{1}{2} \times 2 \sqrt{2}\right| y_{0} \right\rvert\,=\s... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,282 |
11. (18 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove:
$$
\sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \geqslant \frac{n^{2}+1}{n} .
$$ | 11. By Cauchy-Schwarz inequality,
$$
\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \geqslant n^{2} \\
\Rightarrow \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \geqslant n^{2} .
\end{array}
$$
Again by Cauchy-Schwarz inequality,
$$
\b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,283 |
12. (18 points) Let $n (n \geqslant 11)$ be a positive integer. The set $A$ consists of the sums of 10 consecutive positive integers not greater than $n$, and the set $B$ consists of the sums of 11 consecutive positive integers not greater than $n$. If the number of elements in $A \cap B$ is 181, find the maximum and m... | 12. Clearly,
\[
\begin{array}{l}
A=\{55+10 k \mid k=0,1, \cdots, n-10\}, \\
B=\{66+11 l \mid l=0,1, \cdots, n-11\} .
\end{array}
\]
To find the number of elements in \( A \cap B \), let
\[
55+10 k=66+11 l \text{. }
\]
Then \( 10 k=(l+1) 11 \).
Let \( k=11 m \). Then \( l=10 m-1 \).
Since \( 0 \leqslant k \leqslant n-... | 2011 \text{ and } 2001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,284 |
1. Given $a>b, a+b=2$. Then the minimum value of $\frac{a^{2}+b^{2}}{a-b}$ is ( ).
(A) $2 \sqrt{2}$
(B) $\sqrt{2}$
(C) 2
(D) 1 | - 1. C.
Let $a=1+t, b=1-t(t>0)$. Then
$$
\frac{a^{2}+b^{2}}{a-b}=\frac{2\left(1+t^{2}\right)}{2 t}=t+\frac{1}{t} \geqslant 2 \text {. }
$$
Equality holds if and only if $t=1$, i.e., $a=2, b=0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,285 |
2. As shown in Figure 1, the hyperbola $y=\frac{k}{x}(k>0, x>0)$ intersects the sides
$C B$ and $B A$ of rectangle $O A B C$
at points $E$ and $F$, respectively, and $A E$
$=\lambda A B$, connecting $E F$.
Then $\frac{S_{\triangle B E F}}{S_{\triangle O E F}}=(\quad)$.
(A) $\frac{\lambda}{1-\lambda}$
(B) $\frac{\lambda... | 2. D.
Let point $B(b, a)$. Then point $E(\lambda b, a), F(b, \lambda a)$.
Thus, $B E=b(1-\lambda), B F=a(1-\lambda)$.
Since point $F$ is on the hyperbola, we have $a b=\frac{k}{\lambda}$.
Therefore, $S_{\triangle O E F}=S_{\text {rectangle } O A B C}-S_{\triangle O A F}-S_{\triangle O C E}-S_{\triangle B E F}$
$$
\beg... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,286 |
3. Satisfy
$$
(2 m+1)^{2}-(2 n+1)^{2}=2016
$$
the number of positive integer pairs $(m, n)$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. D.
Notice,
$$
\begin{array}{l}
(2 m+1)^{2}-(2 n+1)^{2} \\
=4(m+n+1)(m-n) .
\end{array}
$$
Since $m+n+1$ and $m-n$ have opposite parity, we have
$$
\begin{array}{l}
(m-n, m+n+1) \\
=(1,504),(3,168),(7,72), \\
(8,63),(9,56),(21,24) \\
\Rightarrow(m, n) \\
\quad=(252,251),(85,82),(39,32), \\
(35,27),(32,23),(22,1) .
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,287 |
4. As shown in Figure 2, equilateral $\triangle A B C$ is inscribed in $\odot O, P$ is any point on the minor arc $\overparen{B C}$, connect $P A, P B, P C, P A$ intersects $B C$ at point $D$, let the area of quadrilateral $A B P C$ be $4 \sqrt{3}$. Then ( ).
(A) $P D \leqslant 1$
(B) $P D \geqslant 1$
(C) $P D \leqsla... | 4. A.
As shown in Figure 5, $\triangle A B P$ is rotated counterclockwise by $60^{\circ}$ around point $A$ to $\triangle A C E$.
It is easy to see that points $P$, $C$, and $E$ are collinear, and $\triangle A P E$ is an equilateral triangle.
$$
\begin{array}{l}
\text { Also, } S_{\triangle A P E} \\
= S_{\text {quad... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,288 |
5. On a Sunday, ninth-grade student Xiao Hua went to the market to buy vegetables for his mother at a little past 6 o'clock in the morning. Before leaving, he glanced at his watch and noticed that the angle between the hour and minute hands was $\alpha^{\circ}$ (where $\alpha$ is an integer). When he returned, he also ... | 5. B.
Let the angle between the hour hand and the minute hand be $\alpha$ when it is $n$ o'clock and $m$ minutes. Then, when the hour hand is ahead,
$$
\alpha=30 n-6 m+0.5 m=30 n-5.5 m \text {; }
$$
When the minute hand is ahead, $\alpha=5.5 m-30 n$.
Suppose the time of departure is 6 o'clock and $x$ minutes, and the... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,289 |
6. As shown in Figure 3, let $I$ be the incenter of $\triangle A B C$. If $\angle A B C=35^{\circ}$, and $B C$ $=A I+A C$, then $\angle B A C$ $=(\quad)$.
(A) $55^{\circ}$
(B) $60^{\circ}$
(C) $65^{\circ}$
(D) $70^{\circ}$ | 6. D.
As shown in Figure 6, on $BC$, intercept $CD = CA$, and connect $ID$. Then
$\triangle ACI \cong \triangle DCI$.
Thus, $AI = DI$,
$\angle CDI = \angle CAI$.
Since $BC = AI + AC$
$$
= CD + BD \text{, }
$$
Therefore, $BD = AI = DI$.
Hence, $\angle BID = \angle IBD = \angle ABI$.
Thus, $\angle BAC = 2 \angle IDC = ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,291 |
1. Definition:
$$
\begin{array}{l}
f(n)=\sqrt{n(n+1)(n+2)(n+3)+1}, \\
g(n)=(n+1)^{2} .
\end{array}
$$
Given the following conclusions:
(1) For any rational number $n, f(n)$ is a rational number;
(2) $f(1)-g(1)=1$;
(3) If $f(n)-g(n)=2012$, then $n=2012$;
(4) For any rational number $n$, $f(n)>g(n)$.
The correct conclu... | Sure, here is the translation:
```
2. (1)(2)(3).
Notice,
\[
\begin{array}{l}
\sqrt{n(n+1)(n+2)(n+3)+1} \\
=\sqrt{\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1} \\
=\left|n^{2}+3 n+1\right|,
\end{array}
\]
Thus, conclusions (1) and (2) are correct;
When \( n^{2}+3 n+1 > 0 \),
\[
f(n)-g(n)=n,
\]
In this case, \( f(... | (1)(2)(3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,292 |
3. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=$ $\qquad$ . | 3. 1 .
From the given, $\sqrt{2-\sqrt{3}}=\sqrt{x}-\sqrt{y}$.
Then $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{(\sqrt{3}-1)^{2}}}{\sqrt{2}}$ $=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Thus, $x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}$.
S... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,294 |
4. There is a single-function calculator that can only perform the operation of finding the difference between any two integers and then taking the absolute value of the result. The operation process is: input integers $x_{1}, x_{2}$, and it displays $\left|x_{1}-x_{2}\right|$ (for example, input 1, 2, and the output i... | 4. $n-1$ or $n$ or $n+1$.
Let the order of the $n$ input numbers be $x_{1}, x_{2}, \cdots, x_{n}$.
Then $m=|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots\left|-x_{n}\right|$, $m$ must not exceed the maximum number among $x_{1}, x_{2}, \cdots, x_{n}$.
Therefore, $0 \leqslant m \leqslant n$.
It is easy to see that $m... | n-1 \text{ or } n \text{ or } n+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,295 |
One. (20 points) Let positive real numbers $a, b$ satisfy $a+b=ab$. Prove: $\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4} \geqslant \frac{1}{2}$. | $$
\begin{array}{l}
\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4} \\
=\frac{a\left(a^{2}+4\right)+b\left(b^{2}+4\right)}{\left(b^{2}+4\right)\left(a^{2}+4\right)} \\
=\frac{a^{3}+b^{3}+4(a+b)}{a^{2} b^{2}+4\left(a^{2}+b^{2}\right)+16} \\
=\frac{(a+b)^{3}-3 a b(a+b)+4(a+b)}{a^{2} b^{2}+4\left[(a+b)^{2}-2 a b\right]+16} \\
=\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,296 |
$$
\begin{array}{l}
\text { II. (25 points) (1) Given the parabola } \\
y=x^{2}-2 m x+4 m-8
\end{array}
$$
with vertex $A$, construct an inscribed equilateral $\triangle A M N$ (points $M, N$ are on the parabola). Find the area of $\triangle A M N$;
(2) If the parabola $y=x^{2}-2 m x+4 m-8$ intersects the $x$-axis at ... | (1) From the symmetry of the parabola and the equilateral triangle, we know that $M N \perp y$-axis.
As shown in Figure 8, let the axis of symmetry of the parabola intersect $M N$ at point $B$. Then
$$
A B=\sqrt{3} B M .
$$
Let $M(a, b)$ $(m<a)$. Then
$$
\begin{aligned}
& B M=a-m . \\
& \text { Also, } A B=y_{B}-y_{A... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,297 |
2. In a $5 \times 5$ square grid, three chess pieces are placed arbitrarily at the centers of 25 smaller squares, with at most one piece per center. The probability that the centers of these three pieces form the vertices of a triangle is
保留源文本的换行和格式，直接输出翻译结果。 | 2. $\frac{537}{575}$.
Let $A$, $B$, and $C$ represent the centers of three chess pieces. The number of different ways to place $A$, $B$, and $C$ at the centers of 25 small squares is $25 \times 24 \times 23$.
The number of points forming a straight line in the centers of the small squares can be categorized as follow... | \frac{537}{575} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,300 |
Conclusion 3 As shown in Figure 3, let the incircle of $\triangle A B C$ touch sides $B C, C A, A B$ at points $D, E, F$, respectively. Line $A D$ intersects the incircle at point $P$, and lines $P C, P B$ intersect the incircle at points $G, H$, respectively. Then
(1) Lines $A D, B E, C F$ are concurrent;
(2) Lines $A... | Prove (1) By applying the converse of Ceva's Theorem and the Tangent-Secant Theorem, it is proven that $A D$, $B E$, and $C F$ are concurrent at $N$.
(2) Note that, $\frac{H F}{F P}=\frac{B F}{B P}=\frac{H D}{D P}$, $\frac{P E}{E G}=\frac{P D}{D G}$.
Multiplying the above two equations yields $\frac{P E}{E G} \cdot \fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,301 |
3. Given $a, b, c \in \mathbf{R}_{+}$, and $abc=4$. Then the minimum value of the algebraic expression $a^{a+b} b^{3b} c^{c+b}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 64.
It is easy to know that $a^{a+b} b^{3 b} c^{c+b}=a^{a} b^{2 b} c^{c} \cdot 4^{b}=a^{a}(2 b)^{2 b} c^{c}$.
And $a b c=4$ can be transformed into $a \cdot 2 b \cdot c=8$, considering $2 b$ as a whole.
Since the function $f(x)=\ln x$ is increasing on $(0,+\infty)$, for any $a, b \in(0,+\infty)$, we always have $(... | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,302 |
4. Given a regular tetrahedron $P-ABC$ with the center of the base being $O$, a moving plane through $O$ intersects the lateral edges or their extensions of the tetrahedron at points $S, R, Q$. If
$$
\begin{array}{l}
\overrightarrow{P S}=m \overrightarrow{P A}, \overrightarrow{P R}=n \overrightarrow{P B}, \overrightarr... | 4. $\frac{1}{3}$.
Since points $S, R, Q, O$ are coplanar, we have
$$
\overrightarrow{P O}=x \overrightarrow{P S}+y \overrightarrow{P R}+z \overrightarrow{P Q},
$$
and $x+y+z=1$.
Thus, $\overrightarrow{P O}=x m \overrightarrow{P A}+y n \overrightarrow{P B}+z k \overrightarrow{P C}$.
From the fact that the center of th... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,303 |
5. Given vectors
$$
a=(x-y+1, x-y), b=\left(x-y+1,10^{x}\right) \text {. }
$$
Then the number of all integer pairs $(x, y)$ that satisfy $a \cdot b=2012$ is $\qquad$ | 5.0.
From the problem, we have
$$
\begin{array}{l}
(x-y+1)^{2}+10^{x}(x-y)=2012 \\
\Rightarrow(x-y)\left(x-y+2+10^{x}\right)=2011 .
\end{array}
$$
Obviously, $x \neq 0$, otherwise, $y(y-3)=2011$, this equation has no integer solutions.
If $x>0$, then $10^{x}$ is a positive integer.
Thus, $x-y+2+10^{x}>x-y$.
We get th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,304 |
6. The maximum value of the volume $V$ of a tetrahedron inscribed in a sphere with radius $R$ is $\qquad$ . | 6. $\frac{8 \sqrt{3}}{27} R^{3}$.
Let the tetrahedron be $P-ABC$, and the circumradius of $\triangle ABC$ be $r$. Then
$$
S_{\triangle ABC}=2 r^{2} \sin A \cdot \sin B \cdot \sin C \leqslant \frac{3 \sqrt{3}}{4} r^{2} .
$$
The equality holds if and only if $\angle A=\angle B=\angle C=60^{\circ}$.
If the distance from... | \frac{8 \sqrt{3}}{27} R^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,305 |
7. Given an acute angle $\alpha$ satisfies the equation
$$
\begin{array}{l}
\sin \left(2 \alpha-20^{\circ}\right) \cdot \sin \left(2 \alpha-10^{\circ}\right) \\
=\cos \left(\alpha-10^{\circ}\right) \cdot \sin 10^{\circ} .
\end{array}
$$
Then $\alpha=$ . $\qquad$ | 7. $20^{\circ}$.
Given $\alpha \in\left(0^{\circ}, 90^{\circ}\right)$, we know $\cos \left(\alpha-10^{\circ}\right)>0$.
Dividing both sides of the given equation by $\cos \left(\alpha-10^{\circ}\right)$, we get
$2 \sin \left(\alpha-10^{\circ}\right) \cdot \sin \left(2 \alpha-10^{\circ}\right)=\sin 10^{\circ}$.
When $\... | 20^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,306 |
9. (16 points) For a given positive integer $M$, define $f_{1}(M)$ as the square of the sum of the digits of $M$. When $n>1$ and $n \in \mathbf{N}$, $f_{n}\left(f_{n-1}(M)\right)$ represents the $r_{n}$-th power of the sum of the digits of $f_{n-1}(M)$, where, when $n$ is odd, $r_{n}=2$; when $n$ is even, $r_{n}=3$. Fi... | Let the positive integer $M=a_{1} a_{2} \cdots a_{m}$, where $a_{i} \in \mathbf{N}, a_{1}>0 (i=1,2, \cdots, m)$. Then
$$
f_{1}(M)=\left(a_{1}+a_{2}+\cdots+a_{m}\right)^{2} \equiv M^{2}(\bmod 9) .
$$
When $M=3^{2010}$, $f_{1}(M) \equiv M^{2} \equiv 0(\bmod 9)$.
By mathematical induction, it is easy to see that for $n \... | 729 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,308 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2}=5 \text {, }
$$
and $\left(a_{n+1}+a_{n-1}\right)\left(a_{n}^{2}+a_{n} a_{n-1}+a_{n-1}^{2}\right)$
$$
=6\left(a_{n}^{3}-a_{n-1}^{3}\right)\left(n \geqslant 2, n \in \mathbf{N}_{+}\right) \text {. }
$$
Let $c_{n}=\sum_{i=1}^{n}\left[... | 10. From $a_{n}^{2}+a_{n} a_{n-1}+a_{n-1}^{2}>0$, we know
$$
\begin{array}{l}
a_{n+1}+a_{n}=6\left(a_{n}-a_{n-1}\right) \\
\Rightarrow a_{n+1}=5 a_{n}-6 a_{n-1} \\
\Rightarrow a_{n+1}-2 a_{n}=3\left(a_{n}-2 a_{n-1}\right) \\
\Rightarrow a_{n+1}-2 a_{n}=3^{n-1}\left(a_{2}-2 a_{1}\right)=3^{n} \\
\Rightarrow \frac{a_{n+1... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,309 |
11. (20 points) Let the hyperbola $C_{1}: x^{2}-y^{2}=2 m^{2}$, and the parabola $C_{2}$ has its vertex at $N(n, 0)$, with its focus being the left focus $F$ of the hyperbola $C_{1}$. A line $l$ with a slope of 1 passes through $F$ and intersects the parabola $C_{2}$ at points $P$ and $Q$. If there exist real numbers $... | 11. Clearly, $m \neq 0$.
It is easy to see that $F(-2|m|, 0)$.
Since the vertex of the parabola $C_{2}$ is $N(n, 0)$ and the focus is $F(-2|m|, 0)$, its equation is
$$
y^{2}=-4(2|m|+n)(x-n) .
$$
The equation of the line $l$ passing through point $F$ with a slope of 1 is $y=x+2|m|$.
From $\left\{\begin{array}{l}y^{2}=... | \frac{n}{m}= \pm \sqrt{2}-2,2 \mp \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,310 |
One. (40 points) Given $\triangle A B C, \angle A=80^{\circ}, A B=$ $A C$, point $F$ is on side $A C$, and $\angle F B C=10^{\circ}$. Take a point $E$ on segment $B F$ such that the distance from $E$ to side $B C$ is half of $E C$. Connect $A E$, and let point $D$ be the midpoint of side $A E$. $B D$ intersects the ext... | As shown in Figure 1, construct a regular $\triangle PBC$ on the same side of $\triangle ABC$ with $BC$ as the side, and connect $AP$.
Then $\triangle PAB \cong \triangle PAC$
$$
\begin{array}{l}
\Rightarrow \angle APB = \angle APC = 30^{\circ}. \\
\quad \angle ABC = (180^{\circ} - 80^{\circ}) \div 2 = 50^{\circ}, \tex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,311 |
Example 1 Given that $x, y, z$ are all integers. If
$$
11 \mid (7x + 2y - 5z) \text{, }
$$
Prove: $11 \mid (3x - 7y + 12z)$.
(1987, Beijing Junior High School Mathematics Competition) | Prove that,
$$
\begin{array}{l}
4(3 x-7 y+12 z)+3(7 x+2 y-5 z) \\
=11(3 x-2 y+3 z) .
\end{array}
$$
Since 11|11 $(3 x-2 y+3 z)$, and
$$
111(7 x+2 y-5 z) \text {, }
$$
Therefore, 11|4(3x-7y+12z).
Also, since $(11,4)=1$, then
$$
11 \mid(3 x-7 y+12 z) \text {. }
$$ | 11 \mid (3 x-7 y+12 z) | Number Theory | proof | Yes | Yes | cn_contest | false | 725,312 |
Example 2 Let $72 \mid \overline{a 679 b}$. Find the values of $a$ and $b$. $(1980$, Canadian Mathematics Competition) | Note that, $72=8 \times 9, (8,9)=1$.
Therefore, we only need to discuss the values of $a$ and $b$ when both 8 and 9 divide $\bar{a} \overline{7} 9 b$.
By $8|\overline{a 679 b} \Rightarrow 8| \overline{79 b} \Rightarrow b=2$;
By $9|\overline{a 679 b} \Rightarrow 9|(a+6+7+9+b)$
$$
\Rightarrow 9 \text { | }(a+6+7+9+2) \Ri... | a=3, b=2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,313 |
3. Prove: $(3 n+1) 7^{n}-1$ is divisible by 9. | Let $f(n)=(3 n+1) 7^{n}-1$. When $n=1$, $f(1)=27$ is divisible by 9. Assume $9 \mid f(k)(k=1,2, \cdots)$.
Also, $f(k+1)=f(k)+9(2 k+3) \times 7^{k}$, so $9 \mid f(k+1)$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,314 |
2. Given that $P$ is a point inside $\square A B C D$, and through $P$ lines $E F / / B C, G H / / A B$ are drawn, with points $E, F, G, H$ on sides $A B, C D, A D, B C$ respectively. If $S_{\square P E B H}=6, S_{\square P G D F}=4$, then the area of $\triangle P A C$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2. A.
As shown in Figure 1, let
$$
\begin{array}{l}
S_{\square P E A G}=a, \\
S_{\square P F C H}=b .
\end{array}
$$
Then $S_{\square A B C D}$
$$
\begin{array}{l}
=a+b+6+4 \\
=a+b+10, \\
S_{\triangle A B C}=\frac{1}{2} S_{\square A B C D}=\frac{1}{2} a+\frac{1}{2} b+5, \\
S_{\text {quadrilateral } A B C}=S_{\triangl... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,315 |
3. The sum of 10 distinct integers is 2012. Then, the minimum value of the difference between the largest and smallest of these 10 integers is ( ).
(A) 10
(B) 11
(C) 12
(D) 13 | 3. B.
Let these 10 distinct integers be $x_{1}$, $x_{2}, \cdots, x_{10}$, and $x_{1}<x_{2}<\cdots<x_{10}$. Then
$$
\begin{array}{l}
x_{2} \geqslant x_{1}+1, x_{3} \geqslant x_{2}+1 \geqslant x_{1}+2, \\
x_{4} \geqslant x_{1}+3, \cdots, x_{10} \geqslant x_{1}+9, \\
x_{9} \leqslant x_{10}-1, x_{8} \leqslant x_{9}-1 \leq... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,316 |
4. It is known that the three sides of a triangle are the three roots of the equation
$$
x^{3}-9 x^{2}+26 x-24=0
$$
Then this triangle is ( ) triangle.
(A) Right
(B) Obtuse
(C) Acute
(D) Uncertain | 4. B.
$$
\begin{array}{l}
\text { Given } x^{3}-9 x^{2}+26 x-24=0 \\
\Rightarrow\left(x^{3}-2 x^{2}\right)-\left(7 x^{2}-14 x\right)+(12 x-24)=0 \\
\Rightarrow(x-2)\left(x^{2}-7 x+12\right)=0 \\
\Rightarrow(x-2)(x-3)(x-4)=0 \\
\Rightarrow x_{1}=2, x_{2}=3, x_{3}=4 .
\end{array}
$$
Therefore, the lengths of the sides o... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,317 |
5. Given that $a$ is a prime number less than 2012, and 2012 $-a$ is coprime with $a$. Then the number of $a$ that satisfies the condition is ( ).
(A) 1003
(B) 1004
(C) 1005
(D) 1006 | 5. B.
If $a$ is even, then $2012-a$ is also even. In this case, $a$ and $2012-a$ have a common divisor of 2, so they are not coprime. Therefore, $a$ cannot be even and must be odd.
Since $2012=503 \times 4$, and 503 is a prime number, we have
$$
a \neq 503 \times 1 \text { and } 503 \times 3 \text {. }
$$
There are 1... | 1004 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,318 |
6. In $\triangle A B C$, $A B=A C=m, B C=n$, $\angle B A C=160^{\circ}$. Then the relationship between $n^{3}-\sqrt{3} m^{3}$ and $3 m^{2} n$ is $n^{3}-\sqrt{3} m^{3}(\quad) 3 m^{2} n$.
(A) greater than
(B) less than
(C) equal to
(D) uncertain | 6. C.
From $A B=A C$, we get
$$
\angle B=\angle A C B=\frac{180^{\circ}-\angle B A C}{2}=10^{\circ} \text {. }
$$
Construct $\angle B A D=\angle B$ inside $\angle B A C$, with point $D$ on side $B C$, and draw $C E \perp D A$ at point $E$. Then
$$
\begin{array}{l}
A D=B D, \\
\angle D A C=\angle B A C-\angle B A D=15... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,319 |
$\qquad$ 1. A six-digit number $\overline{a b c d e f}$, when multiplied by 4, becomes $\overline{f a b c d e}$. The number of six-digit numbers that satisfy this condition is $\qquad$. | $=1.6$
Let $\overline{a b c d e}=x$. Then
$$
\begin{array}{l}
4(10 x+f)=100000 f+x \\
\Rightarrow x=2564 f .
\end{array}
$$
Since $f$ is a single digit and $x$ is a five-digit number, it is easy to see that,
$$
f=4,5,6,7,8,9 \text {. }
$$
Therefore, the six-digit numbers $\overline{a b c d e f}$ that satisfy the cond... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,320 |
2. In $\triangle A B C$, $\angle A=45^{\circ}, \angle B=30^{\circ}$, $A D$ is the median of $\triangle A B C$. Then $\angle A D C=$ $\qquad$ | $2.45^{\circ}$.
As shown in Figure 2, draw $C E \perp A B$ at point $E$, and connect $D E$. Then
$$
\begin{array}{l}
D E=\frac{1}{2} B C \\
=B D=D C, \\
A E=E C, \\
\angle D C E=60^{\circ} .
\end{array}
$$
Therefore, $\triangle D C E$ is an equilateral triangle
$$
\Rightarrow D E=E C, \angle D E C=60^{\circ} \text {. ... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,321 |
3. If $a-b=2, \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4$, then $a^{5}-b^{5}=$ | 3. 82 .
$$
\begin{aligned}
\text { Given } & \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4 \\
\Rightarrow & a(1-a)^{2}-b(1+b)^{2}=4 a b \\
\Rightarrow & a-2 a^{2}+a^{3}-b-2 b^{2}-b^{3}=4 a b \\
\Rightarrow & (a-b)-2\left(a^{2}+b^{2}\right)+\left(a^{3}-b^{3}\right)=4 a b \\
\Rightarrow & (a-b)-2\left[(a-b)^{2}+2 a b\right]+... | 82 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,322 |
4. $\triangle A B C$ is inscribed in $\odot O, F$ is a point on arc $\overparen{B C}$, $O G \perp B F$ at point $G$, and $O G=\frac{1}{2} A C$. Then the positional relationship between $A F$ and $B C$ is $\qquad$ . | 4. $A F \perp B C$.
As shown in Figure 3, draw the diameter $F M$, and connect $B M, A M$. Then $\angle M A F=90^{\circ}$.
By $O G \perp B F$
$$
\Rightarrow B G=G F \text {. }
$$
In $\triangle F B M$, by $O F=O M$, we get
$$
O G=\frac{1}{2} B M .
$$
Also, $O G=\frac{1}{2} A C$, then
$$
\begin{array}{l}
B M=A C \Righ... | A F \perp B C | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,323 |
One, (20 points) Find the integer part of $\left(\frac{1+\sqrt{5}}{2}\right)^{19}$. | Let $a=\frac{1+\sqrt{5}}{2}, b=\frac{1-\sqrt{5}}{2}$. Then $a+b=1, a^{2}=a+1, b^{2}=b+1$.
Thus $a^{2}+b^{2}=(a+1)+(b+1)$ $=(a+b)+2=3$, $a^{3}+b^{3}=\left(a^{2}+a\right)+\left(b^{2}+b\right)$ $=\left(a^{2}+b^{2}\right)+(a+b)=1+3=4$, $a^{4}+b^{4}=\left(a^{3}+a^{2}\right)+\left(b^{3}+b^{2}\right)$ $=\left(a^{3}+b^{3}\righ... | 9349 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,324 |
II. (25 points) Given that $D$ is a point inside $\triangle A B C$, $E$ is the midpoint of side $A C$, $A B=6, B C=10, \angle B A D=$ $\angle B C D, \angle E D C=\angle A B D$. Find the length of $D E$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | II. As shown in Figure 4, extend \( CD \) to point \( F \) such that \( DF = CD \), and connect \( AF \) and \( BF \).
Then \( AF \parallel DE \), and \( DE = \frac{1}{2} AF \).
Thus, \( \angle AFD = \angle EDC = \angle ABD \)
\(\Rightarrow A, F, B, D\) are concyclic
\(\Rightarrow \angle BFD = \angle BAD = \angle BCD \... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,326 |
Three. (25 points) Find the minimum value of the function
$$
y=\sqrt{x^{4}-3 x^{2}-6 x+13}+\sqrt{x^{4}-x^{2}+1}
$$ | $$
\begin{aligned}
y= & \sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}}+ \\
& \sqrt{(x-0)^{2}+\left(x^{2}-1\right)^{2}} .
\end{aligned}
$$
The above expression represents the sum of the distances from point $P\left(x, x^{2}\right)$ to points $A(3,2)$ and $B(0,1)$. The image of point $P\left(x, x^{2}\right)$ is the parabola ... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,327 |
1. Given that there are 10 squares with a side length of 1 inscribed in square $A B C D$ (as shown in Figure 1). Then the area of square $A B C D$ is $\qquad$ | $-1 . \frac{1521}{58}$.
As shown in Figure 3, let $P B$
$$
=a, Q B=b, A B=
$$
c. Then
$$
\begin{array}{l}
P E=4 b, \\
E F=4 a \text {, } \\
H F=2 b \text {, } \\
H I=2 a \text {, } \\
J I=2 b \text {, } \\
S J=2 a, D R=a, D S=b . \\
\text { Therefore }\left\{\begin{array}{l}
b+2 a+2 b+4 a=c \\
2 b-2 a+4 b+a=c
\end{arra... | \frac{1521}{58} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,328 |
3. As shown in Figure 2, in the equilateral $\triangle ABC$, points $D, E, F$ on sides $BC, CA$, and $AB$ satisfy $BD=2CD, CE=3AE, AF=4BF$. If $S_{\triangle ABC}=36$, then the area of $\triangle LMN$ formed by lines $AD, BE, CF$ is . $\qquad$ | 3. $\frac{529}{52}$.
Consider any $\triangle XYZ$, with points $S$ and $T$ on sides $YZ$ and $XZ$ respectively, such that $YS = uSZ$ and $ZT = vTX$.
Let $YT$ and $XS$ intersect at point $K$, as shown in Figure 4.
By Menelaus' Theorem, we have
$$
\begin{array}{l}
\frac{YS}{SZ} \cdot \frac{ZX}{XT} \cdot \frac{TK}{KY} =... | \frac{529}{52} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,330 |
4. Place 27 balls numbered $1 \sim 27$ into three bowls, Jia, Yi, and Bing, such that the average values of the ball numbers in bowls Jia, Yi, and Bing are $15$, $3$, and $18$, respectively, and each bowl must contain no fewer than 4 balls. Then the maximum value of the smallest ball number in bowl Jia is $\qquad$ | 4. 10 .
Let there be $a$, $b$, and $c$ balls in bowls 甲, 乙, and 丙, respectively. Then,
$$
\begin{aligned}
a+b+c & =27, \\
15 a+3 b+18 c & =\frac{27 \times 28}{2} .
\end{aligned}
$$
From equation (2), we get
$$
5 a+b+6 c=126 \text {. }
$$
If $b \geqslant 6$, then the average value of the ball numbers in bowl 乙 is not... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,331 |
5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for all non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Then the remainder when $a_{2012}$ is divided by 2012 is ـ. $\qquad$ | 5.1.
In equation (1), let $n=0$, we get
$$
a_{2 m}=4 a_{m}+2 m-2-a_{0} \text {. }
$$
In equation (2), let $m=0,1$, we get
$$
a_{0}=1, a_{2}=3 \text {. }
$$
In equation (1), let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right) \\
=\frac{1}{2}\left(4 a_{m}+2 m-3+3\right)=2 a_... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,332 |
6. Given a positive integer $k$ that satisfies for any positive integer $n$, the smallest prime factor of $n^{2}+n-k$ is no less than 11. Then, $k_{\text {min }}$ $=$ . $\qquad$ | 6.43 .
Notice,
$$
\begin{array}{l}
n^{2}+n \equiv 0(\bmod 2), \\
n^{2}+n \equiv 0,2(\bmod 3),
\end{array}
$$
$$
\begin{array}{l}
n^{2}+n \equiv 0,2,1(\bmod 5), \\
n^{2}+n \equiv 0,2,6,5(\bmod 7),
\end{array}
$$
and the smallest prime factor of $n^{2}+n-k$ is not less than 11, then
$$
\begin{aligned}
k & \equiv 1(\bmo... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,333 |
7. Given positive integers $a, b$ satisfy
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \text {. }
$$
Then $|10(a-5)(b-15)|+2=$ | 7.2012.
Notice,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{... | 2012 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,334 |
8. Given a cyclic quadrilateral $A_{1} A_{2} A_{3} A_{4}$ with an inscribed circle $\odot I$ that is tangent to the sides $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, and $A_{4} A_{1}$ at points $B_{1}$, $B_{2}$, $B_{3}$, and $B_{4}$ respectively, then
$$
\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{... | 8. 8 .
As shown in Figure 5, let the radius of $\odot I$ be $r$,
$$
\begin{array}{l}
A_{i} B_{i}=A_{i} B_{i-1} \\
=a_{i},
\end{array}
$$
where $i=1,2,3,4, B_{0}=B_{4}$.
Then $A_{i} A_{i+1}=a_{i}+a_{i+1}$,
$$
B_{i} B_{i+1}=\frac{2 A_{i+1} B_{i} \cdot B_{i} I}{\sqrt{A_{i+1} B_{i}^{2}+B_{i} I^{2}}}=2\left(a_{i+1}^{-2}+r... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,335 |
5. Prove: For all integers $n, n^{2}+2 n+12$ is not a multiple of 121.
(1971, Canadian Mathematical Olympiad) | Assume $n^{2}+2 n+12=121 k$. Then $(n+1)^{2}=11(11 k-1)$ $\Rightarrow 11|(n+1) \Rightarrow 11|(11 k-1)$. Contradiction. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,336 |
9. (16 points) Let the moving point $P$ be on the right branch of the hyperbola
$$
\frac{x^{2}}{16}-\frac{y^{2}}{9}=1
$$
(excluding the vertex), $Q$ and $R$ are the left and right foci of the hyperbola, respectively, $S$ is the excenter of $\triangle PQR$ inside $\angle PQR$, and $D(0,1)$. Find the minimum value of $|... | 9. Let the equation of the hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\).
Let \(P\left(x_{0}, y_{0}\right), Q(-c, 0), R(c, 0), S(x, y)\), and \(P R\) intersects \(Q S\) at point \(T\). Let the complex numbers corresponding to points \(P, Q, R, S, T\) be \(p, q, r, s, t\), respectively.
It is easy to see th... | \sqrt{25.9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,337 |
10. (20 points) How many nine-digit numbers $\overline{a_{1} a_{2} \cdots a_{9}}$ satisfy $a_{1} \neq 0$, all digits are distinct, and
$$
a_{1}+a_{3}+a_{5}+a_{9}=a_{2}+a_{4}+a_{6}+a_{8} .
$$ | 10. Let the missing digit be $a$, and
$$
\sum_{i=\pi} a_{i}=\sum_{i=m} a_{i}=t .
$$
Then $2 t=\sum_{i=1}^{9} a_{i}=45-a$.
Thus, $a \in\{1,3,5,7,9\}$.
If $a=1$, then $t=22$, we have
$$
\begin{array}{l}
22=2+3+8+9=2+4+7+9 \\
=2+5+6+9=2+5+7+8 \\
=3+4+6+9=3+4+7+8 \\
=3+5+6+8=4+5+6+7 \\
=0+5+8+9=0+6+7+9 ;
\end{array}
$$
I... | 120384 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,338 |
One. (40 points) Given $\triangle A B C$ and a point $O$, construct circles passing through point $O$ with the midpoints $A_{1}, B_{1}, C_{1}$ of sides $B C, C A, A B$ of $\triangle A B C$ as centers, respectively. Let the intersections of these circles, other than $O$, be $A_{2}, B_{2}, C_{2}$. Prove:
(1) If $\triangl... | Obviously, the circumcenter of $\triangle ABC$ is the orthocenter of $\triangle A_{1} B_{1} C_{1}$. Points $A_{2}, B_{2}, C_{2}$ are the reflections of point $O$ over $B_{1} C_{1}$, $C_{1} A_{1}$, and $A_{1} B_{1}$, respectively.
(1) $\triangle ABC$ is an acute triangle.
As shown in Figure 6, draw lines through $A_{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,339 |
11. (20 points) Find all positive integer pairs \((n, k)\) that satisfy
$$
(n+1)^{k}-(2 n+1)=n!
$$ | 11. Notice,
$$
\begin{array}{l}
(n+1)^{k}-(2 n+1)=n! \\
\Leftrightarrow(n+1)\left[(n+1)^{k-1}-1\right]=n[(n-1)!+1] .
\end{array}
$$
Let $p$ be a prime factor of $n+1$.
Since $(n+1, n)=1$, we know $p \mid (n-1)!+1$.
For any prime $q (q \leqslant n-1)$, we have $q \mid (n-1)!+1$.
Thus, $p=n+1$.
$$
\text { Equation (1) }... | (n, k)=(1,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,340 |
II. (40 points) Find all real numbers $x$ that satisfy the equation
$$
\left[x^{2}-2 x\right]+[x]=[x]^{2} \text {, }
$$
where $[a]$ denotes the greatest integer not exceeding the real number $a$. | Let $[x]=u, x-[x]=\varepsilon(u \in \mathbf{Z}, 0 \leqslant \varepsilon < 1)$, then
$$
0 \leqslant (2 \varepsilon - 1) u + \varepsilon^2 - 2 \varepsilon < 1.
$$
Therefore, $\varepsilon \in \left(1-u-\sqrt{u^2-u+2}, 1-u-\sqrt{u^2-u+1}\right]$.
In summary, $x=0$ or $u+\varepsilon$, where, when $u \geqslant 2$, $\varepsi... | x=0 \text{ or } u+\varepsilon, \text{ where } \varepsilon \in \left(1-u-\sqrt{u^2-u+2}, 1-u-\sqrt{u^2-u+1}\right] \text{ for } u \geqslant 2 \text{ and } u < 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,341 |
Three, (50 points) Write down all positive integers from 1 to 10000 from left to right, then remove those numbers that are divisible by 5 or 7, and concatenate the remaining numbers to form a new number. Try to find:
(1) The number of digits in the new number;
(2) The remainder when the new number is divided by 11. | Three, (1) Obviously, the remainder of the remaining numbers when divided by 35 is
$$
\begin{array}{l}
1,2,3,4,6,8,9,11,12,13,16,17,18, \\
19,22,23,24,26,27,29,31,32,33,34,
\end{array}
$$
denoted as $a_{i}(i=1,2, \cdots, 24)$ in sequence.
Among the numbers from $1 \sim 9$, 7 numbers remain, denoted as $b_{i}$ $(i=1,2,... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,342 |
Four, (50 points) Given several rectangular boxes, the lengths of their edges are positive integers not greater than an odd positive integer $n$ (allowing the three edge lengths to be the same), and the thickness of the box walls is negligible. Each box has its three pairs of opposite faces painted red, blue, and yello... | Let the maximum number of harmonious boxes be $f(n)$.
In a three-dimensional Cartesian coordinate system, the coordinate planes
$$
\begin{array}{l}
\{(x, y, z) \mid x=0, y \geqslant 0, z \geqslant 0\}, \\
\{(x, y, z) \mid x \geqslant 0, y=0, z \geqslant 0\}, \\
\{(x, y, z) \mid x \geqslant 0, y \geqslant 0, z=0\}
\end... | \frac{3 n^{2}+1}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,343 |
As shown in Figure 2, in trapezoid $A B C D$, $A B$ // $C D$, and $A B>C D$. Points $K$ and $L$ are on sides $A B$ and $C D$ respectively, such that $\frac{A K}{K B}=\frac{D L}{L C}$. If there exist points $P$ and $Q$ on segment $K L$, and $A P$ intersects $D Q$ at point $E$, $C Q$ intersects $B P$ at point $F$, and it... | Prove: As shown in Figure 3, connect $E F$.
From $\frac{A K}{K B}=\frac{D L}{L C}$, we know that $A D$, $K L$, and $B C$ intersect at one point $S$.
By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{A D}{D S} \cdot \frac{S Q}{Q P} \cdot \frac{P E}{E A}=1, \\
\frac{B C}{C S} \cdot \frac{S Q}{Q P} \cdot \frac{P F}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,344 |
In $\square A B C D$, $A B=8$, the circumradius of $\triangle A B D$ is 5. Find the maximum value of the diagonal $A C$.
The original text has been translated into English, preserving the original formatting and line breaks. | As shown in Figure 4, let the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $P$.
Then $BP = DP$, $AC = 2AP$.
Let the circumcenter of $\triangle ABD$ be $O$, and connect $OB$, $OP$.
Since $P$ is the midpoint of $BD$, we know $\angle OPB = 90^\circ$.
Thus, the locus of point $P$ is the circle $\odot ... | 3\sqrt{17} + 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,345 |
Find all real roots of the equation
$$
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right)
$$ | Solve:
$$
\begin{array}{l}
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right) \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}=x^{2}+2 x+4 \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}-3=x^{2}+2 x+1 \\
\Rightarrow \frac{-2\left(x^{3}+1\right)^{2}}{x^{6}+x^{3}+1}=(x+1)^{2} \\
\Rightarrow(x+1)^{2}\le... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,346 |
Example 1 Find all triples of integers $(x, y, z)$ such that
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003^{[1]} \text {. }
$$
$(2003$, Nordic Mathematical Contest) | Notice,
$$
\begin{aligned}
(x+y & +z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \\
& =2 \times 2003,
\end{aligned}
$$
and $(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \equiv 0(\bmod 2)$.
Thus, we have $\left\{\begin{array}{l}x+y+z=1, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=4006 ;\end{array}\right.$
or $\left\{\begin{array}{l}x+y+z=2003, ... | (668,668,667),(668,667,668),(667,668,668) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,347 |
Let $n$ be a positive integer, $[x]$ be the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$.
(1) Find all positive integers $n$ that satisfy
$$
\sum_{k=1}^{2013}\left[\frac{k n}{2013}\right]=2013+n
$$
(2) Find all positive integers $n$ that maximize $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right... | (1) Let $m=2013, d=(m, n), m=d m_{1}, n=d n_{1}$.
For $1 \leqslant k \leqslant m-1$, we have
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=\left[\frac{k n_{1}}{m_{1}}\right]+\left[\frac{(m-k) n_{1}}{m_{1}}\right] \text{. }
$$
When $m_{1} \mid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\rig... | 1006 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,348 |
Example 2 Proof: There do not exist positive integers $x, y$ satisfying $x^{3}+y^{3}=2^{2009}$. [2]
$(2009$, Brazilian Mathematical Olympiad) | $$
\begin{array}{l}
x^{3} \equiv 0,1,-1(\bmod 7) . \\
\text { and } 2^{2009}=\left(2^{3}\right)^{669} \times 2^{2} \\
\equiv 1^{669} \times 4 \equiv 4(\bmod 7),
\end{array}
$$
and $x^{3}+y^{3} \equiv-2,-1,0,1,2(\bmod 7)$. Therefore, there do not exist positive integers $x, y$ satisfying
$$
x^{3}+y^{3}=2^{2009} \text {... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,349 |
Example 3 Proof: The equation $y+y^{2}=x+x^{2}+x^{3}$ has no integer solutions for $x \neq 0$.
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Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
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Example 3 Proof: The equation $y+y^{2}=x+x^{2}+x^{3}$ has no integer so... | Proof: Suppose the equation has an integer solution $x \neq 0$, then
$$
(y-x)(y+x+1)=x^{3} \text {. }
$$
If $y-x$ and $y+x+1$ have a common prime factor $p$, then by equation (1) we know
$$
p|x \Rightarrow p| y \text {. }
$$
But $p|(x+y+1) \Rightarrow p| 1$, which is a contradiction.
Hence $(y-x, y+x+1)=1$.
Since the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,350 |
Example 5 Given $n \in \mathbf{N}, n>1$. Prove: the indeterminate equation
$$
(x+1)^{n}-x^{n}=n y
$$
has no solutions in the set of positive integers. ${ }^{[4]}$
(2010, Croatian National Team Selection Exam) | Solve by taking the smallest prime factor of $n$ (let it be $p$) for discussion.
It is easy to know that $p \mid x$ and $p \mid(x+1)$ cannot hold simultaneously.
Therefore, when $p \mid x$ or $p \mid(x+1)$, the original equation is taken modulo $p$, which leads to a contradiction.
When $p \nmid x$ and $p \nmid(x+1)$, t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,352 |
Example 6 Find all positive integer solutions of the equation $3^{x}-5^{y}=z^{2}$. ${ }^{[5]}$
(2009, Balkan Mathematical Olympiad) | Solving modulo 2 gives $0 \equiv z^{2}(\bmod 2)$, so $2 \mid z$. Solving modulo 4 gives $(-1)^{x}-1=0(\bmod 4)$, so $2 \mid x$. Let $x=2 x_{1}$. Then $\left(3^{x_{1}}+z\right)\left(3^{x_{1}}-z\right)=5^{y}$. Let $\left\{\begin{array}{l}3^{x_{1}}+z=5^{\alpha}, \\ 3^{x_{1}}-z=5^{\beta},\end{array}\right.$ where $\alpha+\... | (x, y, z)=(2,1,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,353 |
Example 7, find all positive integer solutions $(x, y, z)$ of the equation
$$
(3 x+1)(3 y+1)(3 z+1)=34 x y z
$$
(2010, German Mathematical Olympiad) | Let's assume $x \geqslant y \geqslant z$.
If $z>\frac{1}{\sqrt[3]{34}-3}$, then $3 z+1<\sqrt[3]{34} z$.
Similarly, $3 x+1<\sqrt[3]{34} x, 3 y+1<\sqrt[3]{34} y$.
Multiplying the three inequalities, we get
$$
(3 x+1)(3 y+1)(3 z+1)<34 x y z \text {. }
$$
This contradicts the given equation.
Therefore, $z \leqslant \frac{... | (28,5,2),(11,7,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,354 |
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