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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
16. Let the set $M=\{n \mid n$ ! can be expressed as the product of $n-3$ consecutive positive integers, and $n>4\}$. Prove: $M$ is a finite set, and find all elements of $M$. | 16. Let $n \in M$, then
$$
n!=1 \times 2 \times \cdots \times n=m(m+1) \cdots(m+n-4) \text {. }
$$
If $m \leqslant 4$, then
$$
m(m+1) \cdots(m+n-4) \leqslant 4 \times 5 \times \cdots \times n<n!.
$$
Therefore, it must be that $m \geqslant 5$.
Since $m(m+1) \cdots(m+n-5) \geqslant 5 \times 6 \times \cdots \times n$, w... | M=\{6,7,23\} | Number Theory | proof | Yes | Yes | cn_contest | false | 718,860 |
1. Let $x_{1}, x_{2}, \cdots, x_{n}$ be numbers that can take one of the values $-3, 0, 1$, and
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=279, \\
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=-585 .
\end{array}
$$
Then the value of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ is ( ).
(A) 2005
(B) 2006
(C) 2007
(D) 200... | - 1.C.
Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ numbers, where $-3$ appears $b$ times, $1$ appears $b$ times, and $0$ appears $c$ times. Then
$$
\left\{\begin{array} { l }
{ 9 a + b = 2 7 9 , } \\
{ - 2 7 a + b = - 5 8 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=24, \\
b=63 .
\end{array}\right.\right.
$$
T... | 2007 | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,861 |
2. Given $n(n \geqslant 8)$ positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{n}=a_{1} q^{n-1}$, where $q$ is a positive number not equal to 1. Then the relationship between $a_{1}+a_{8}$ and $a_{4}+a_{5}$ is ( ).
(A) greater than
(B) equal to
(C) less than
(D) cannot be determined | 2. A.
$$
\begin{array}{l}
\left(a_{1}+a_{8}\right)-\left(a_{4}+a_{5}\right) \\
=\left(a_{1}+a_{1} q^{7}\right)-\left(a_{1} q^{3}+a_{1} q^{4}\right) \\
=a_{1}\left(1+q^{7}-q^{3}-q^{4}\right) \\
=a_{1}\left(1-q^{3}\right)\left(1-q^{4}\right) \\
=a_{1}(1-q)^{2}(1+q)\left(1+q^{2}\right)\left(1+q+q^{2}\right) .
\end{array}
... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,862 |
3. Divide the sides of the equilateral $\triangle A B C$ into four equal parts, as shown in Figure 1. Then the number of pairs of congruent equilateral triangles in the figure is ( ).
(A) 100
(B) 121
(C) 144
(D) 169 | 3. C.
Let the side length of the equilateral triangle be 4. From Figure 1, we can see:
(1) There are 16 equilateral triangles with a side length of 1, all of which are congruent, and the number of pairs is $15+14+\cdots+1=120$;
(2) There are 7 equilateral triangles with a side length of 2, all of which are congruent, ... | 144 | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,863 |
Example 8 Given that $x, y, z$ are positive real numbers, and $x+y+z=1$. Prove:
$$
\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}} \geqslant 0 \text {. }
$$
Analysis: This problem is slightly different from the previous two examples, but the numerator can factor out $3x-1$, so there is no... | Prove: Introduce a parameter $\lambda$, set
$$
\frac{3 x^{2}-x}{1+x^{2}} \geqslant \lambda(3 x-1) \text {, }
$$
i.e., $(3 x-1)\left(\lambda-\frac{x}{1+x^{2}}\right) \leqslant 0$.
(1) When $x=\frac{1}{3}$, $\lambda \in \mathbf{R}$.
(2) When $\frac{1}{3}<x<1$, $\lambda \leqslant \frac{x}{1+x^{2}}$, i.e., $\lambda \leqsl... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,864 |
4. In Rt $\triangle A B C$, $\angle C=90^{\circ}, A C=3$, $B C=4$, points $I$ and $O$ are the incenter and circumcenter of $\triangle A B C$, respectively. Then $\tan \angle I O A=(\quad)$.
(A) 1
(B) $\frac{\sqrt{3}}{3}$
(C) $\sqrt{3}$
(D) 2 | 4.D.
As shown in Figure 2, let $\odot I$ be tangent to sides $A C$, $B C$, and $A B$ at points $D$, $E$, and $F$, respectively. Connect $I D$, $I E$, and $I F$. Then $I D \perp A C$, $I E \perp B C$, and $I F \perp A B$.
In the right triangle $\triangle A B C$, by the Pythagorean theorem, we have
$$
A B=\sqrt{A C^{2}... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,865 |
6. Let $\frac{1}{3} \leqslant x \leqslant 5$. Then the maximum value of $3 \sqrt{5-x}+2 \sqrt{3 x-1}$ is ( ).
(A) $2 \sqrt{14}$
(B) $\frac{5 \sqrt{14}}{2}$
(C) $5 \sqrt{2}$
(D) $7 \sqrt{2}$ | 6.D.
Let $\sqrt{5-x}=a, \sqrt{3 x-1}=b$. Squaring both equations and eliminating $x$ yields
$$
3 a^{2}+b^{2}=14 \text {. }
$$
Now let $y=3 \sqrt{5-x}+2 \sqrt{3 x-1}$, then
$$
y=3 a+2 b \text {. }
$$
From equation (2), we get $a=\frac{y-2 b}{3}$, substituting this into equation (1) and simplifying yields
$$
7 b^{2}-4... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,867 |
1. Given $a+b+c=2, a^{2}+b^{2}+c^{2}=2$. Try to compare the size of the following three expressions:
$$
\begin{array}{l}
a(1-a)^{2}-b(1-b)^{2}-c(1-c)^{2} \\
\text { (fill in “>”, “= ” or “<”). }
\end{array}
$$ | ニ、1. $=,=$.
From the given, we have $\left\{\begin{array}{l}a+b=2-c, \\ a^{2}+b^{2}=2-c^{2}\end{array}\right.$.
(1) ${ }^{2}$ - (2) yields $2 a b=2\left(1-2 c+c^{2}\right)$.
Transforming the above equation, we get $a b=(1-c)^{2}$.
Multiplying both sides by $c$ gives $a b c=c(1-c)^{2}$.
Similarly, $a b c=b(1-b)^{2}, a ... | a(1-a)^{2} = b(1-b)^{2} = c(1-c)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,868 |
2. Given a square $A B C D$ with side length $1, P$ is a point on $A C$ different from $A$ and $C$, point $E$ is on $C D$, and $P E$ $\perp P B$. If $A P=x$, the area of pentagon $A B P E D$ is $y$, then the relationship between $y$ and $x$ is $\qquad$ . | 2. $y=-\frac{1}{2} x^{2}+\sqrt{2} x\left(0<x<\frac{\sqrt{2}}{2}\right)$.
As shown in Figure 4, draw $PG \perp BC$ and $PH \perp CD$, with $G$ and $H$ being the feet of the perpendiculars. Then quadrilateral $PGCH$ is a rectangle.
Since $CP$ is the angle bisector of $\angle BCD$, we have $PG = PH$.
Therefore, quadril... | y=-\frac{1}{2} x^{2}+\sqrt{2} x\left(0<x<\frac{\sqrt{2}}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,869 |
3. The irrational root of the fractional equation $\frac{1}{x-5}+\frac{2}{x-4}+\frac{3}{x-3}+\frac{4}{x-2}$ $=4$ is | 3. $\frac{7 \pm \sqrt{5}}{2}$.
Transform the original equation to
$$
\begin{array}{l}
\left(1-\frac{1}{x-5}\right)+\left(1-\frac{2}{x-4}\right)+\left(1-\frac{3}{x-3}\right)+\left(1-\frac{4}{x-2}\right)=0 \\
\Leftrightarrow \frac{x-6}{x-5}+\frac{x-6}{x-4}+\frac{x-6}{x-3}+\frac{x-6}{x-2}=0 \\
\Leftrightarrow(x-6)\left[\... | \frac{7 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,870 |
4. Given that the three sides of $\triangle A B C$ are $a, b, c$, and $D, E$ are the midpoints of sides $A B, A C$ respectively. Line $D E$ intersects the circumcircle of $\triangle A B C$ at points $F, G$. Then $D F-E G=$ $\qquad$ . | 4. $\pm \frac{b^{2}-c^{2}}{2 a}$.
When line $D E$ intersects the arc $\overparen{A B}$ opposite $\angle C$ at point $F$, and intersects the arc $\overparen{A C}$ opposite $\angle B$ at point $G$, as shown in Figure 5. By the Midline Theorem of a triangle, we have
$$
D E=\frac{1}{2} B C=\frac{1}{2} a .
$$
By the Inter... | \pm \frac{b^{2}-c^{2}}{2 a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,871 |
One, (20 points) The vertex of the quadratic function $y=x^{2}-4 x+3$ is $P$. A line $A B$ passing through the point $\left(\frac{3}{2},-\frac{7}{4}\right)$ is tangent to the parabola, and the line $A B$ passes through the first, third, and fourth quadrants, intersecting the $x$-axis and $y$-axis at points $A$ and $B$,... | 1. As shown in Figure 7, let the equation of the line $AB$ be $y = kx + b$.
Substituting $x = \frac{3}{2}$ and $y = -\frac{7}{4}$, we get
$$
b = -\frac{7}{4} - \frac{3}{2}k.
$$
Therefore, $y = kx - \frac{3}{2}k - \frac{7}{4}$.
From the system of equations
$$
\left\{\begin{array}{l}
y = kx - \... | \frac{\sqrt{2}}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,872 |
II. (25 points) Let $D$ be the midpoint of the base $BC$ of isosceles $\triangle ABC$, and $E$, $F$ be points on $AC$ and its extension, respectively. Given that $\angle EDF=90^{\circ}$, $DE=a$, $DF=b$, $AD=c$. Try to express the length of $BC$ in terms of $a$, $b$, and $c$.
---
The translation preserves the original... | II. As shown in Figure 8, draw $D G \perp A C$ at $G$. By the properties of isosceles triangles, we know that $A D \perp B C$ and $A D$ bisects $\angle B A C$.
In the right triangle $\triangle E D F$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
E F=\sqrt{D E^{2}+D F^{2}} \\
=\sqrt{a^{2}+b^{2}} .
\end{array... | \frac{2 a b c}{\sqrt{a^{2} c^{2}+b^{2} c^{2}-a^{2} b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,873 |
Three, (25 points) Given that $m$ and $n$ ($m>n$) are positive integers, and the last three digits of $3^{m}$ and $3^{n}$ are the same. Find the minimum value of $m-n$.
保留源文本的换行和格式，翻译结果如下：
```
Three, (25 points) Given that $m$ and $n$ ($m>n$) are positive integers, and the last three digits of $3^{m}$ and $3^{n}$ are... | Three, from the given information, $3^{m}-3^{n}$ is a multiple of 1000, i.e.,
$$
3^{m}-3^{n}=3^{n}\left(3^{m-n}-1\right)
$$
is a multiple of 1000.
Also, $\left(3^{n}, 1000\right)=1$, hence $3^{m-n}-1$ is a multiple of 1000.
Let $s=m-n$, then $3^{s}-1$ is a multiple of 1000.
We only need to find the smallest $s$ such t... | 100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,874 |
Example 9 Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=3$. Prove:
$$
\begin{array}{l}
\frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}}+\frac{(b+3)^{2}}{2 b^{2}+(3-b)^{2}}+ \\
\frac{(c+3)^{2}}{2 c^{2}+(3-c)^{2}} \leqslant 8 .
\end{array}
$$
Analysis: Directly plotting the graph of the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{... | Prove: Introduce parameters $\lambda_{1}$ and $\lambda_{2}$, set
$$
\frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant \lambda_{1} a+\lambda_{2}(0<a<3).
$$
(2) When $\frac{3}{2}<x<3$, $f^{\prime}(x)<0$.
The graph of the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{2}}$ is monotonically increasing in the interval $\left(0... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,875 |
1. Let $a, b, c$ be three distinct positive integers. If
$$
\begin{array}{l}
\{a+b, b+c, c+a\} \\
=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\}\left(n \in \mathbf{N}_{+}\right),
\end{array}
$$
then the minimum value of $a^{2}+b^{2}+c^{2}$ is ( ).
(A) 2007
(B) 1949
(C) 1297
(D) 1000 | - 1.C.
Assume $a>b>c$, then $a+b>c+a>b+c$.
Since $(a+b)+(b+c)+(c+a)=2(a+b+c)$ is an even number, $n^{2} 、(n+1)^{2} 、(n+2)^{2}$ must be two odd and one even. Therefore, $n$ must be an odd number.
Also, since $b+c>1$, $n$ must be an odd number no less than 3.
If $n=3$, then $\{a+b, b+c, c+a\}=\left\{3^{2}, 4^{2}, 5^{2}\... | 1297 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,876 |
2. Given $\alpha, \beta, \gamma$ are three unequal acute angles. If $\tan \alpha=\frac{\sin \beta \cdot \sin \gamma}{\cos \beta-\cos \gamma}$, then $\tan \beta$ equals ( ).
(A) $\frac{\sin \gamma \cdot \sin \alpha}{\cos \gamma-\cos \alpha}$
(B) $\frac{\sin \gamma \cdot \sin \alpha}{\cos \gamma+\cos \alpha}$
(C) $\frac{... | 2. B.
From the given, we know that $\tan \alpha>0$,
$\sin \beta \cdot \sin \gamma>0$, then
$$
\cos \beta-\cos \gamma>0.
$$
Therefore, we can construct a right-angled $\triangle ABC$ as shown in Figure 4, such that $\angle C$
$$
\begin{array}{c}
=90^{\circ}, \angle A=\alpha, \\
AC=\cos \beta-\cos \gamma, \\
BC=\sin \b... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,877 |
3. As shown in Figure $1, P$ is any point on the right branch of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>$ $0, b>0 ＼mathrm{~ right ~ branch ~}$, a line passing through point $P$ intersects the two asymptotes of the hyperbola at points $P_{1}$ and $P_{2}$, respectively, and point $P$ internally divide... | 3. A.
Let point $P$ divide $P_{1} P_{2}$ in the ratio $\lambda(\lambda>0), P(x, y)$, $P_{1}\left(x_{1}, \frac{b}{a} x_{1}\right), P_{2}\left(x_{2},-\frac{b}{a} x_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x=\frac{x_{1}+\lambda x_{2}}{1+\lambda}, \\
y=\frac{\frac{b}{a} x_{1}-\lambda \cdot \frac{b}{a} x_{2}}{1+\lambda... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,878 |
4. From the 8 vertices of a cube, any 4 points $M, N, P, Q$ that do not lie on the same plane are chosen to form a dihedral angle $M-P Q-N$. Then the number of such dihedral angles of different sizes is ( ).
(A) 28
(B) 27
(C) 9
(D) 8 | 4. D.
Consider the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, which can be divided into the following three categories based on the edge of the dihedral angle:
(1) Taking the edge of the cube as the edge of the dihedral angle (let's take $A B$ as the edge of the dihedral angle), there are 2 essentially different dihedral... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,879 |
5. Two vectors $O A$ and $O B$ on a plane satisfy $|O A|=a, |O B|=b$, and $a^{2}+b^{2}=4, O A \cdot O B$ $=0$. If the vector $O C=\lambda O A+\mu O B(\lambda, \mu \in \mathbf{R})$, and $\left(\lambda-\frac{1}{2}\right)^{2} a^{2}+\left(\mu-\frac{1}{2}\right)^{2} b^{2}=1$, then the maximum value of $\mid O C$ is ( ).
(A)... | 5.C.
Since $|O A|=a,|O B|=b$, and $a^{2}+b^{2}=4, O A \perp O B$, therefore, points $O, A, B$ lie on a circle with the midpoint $M$ of $A B$ as the center and 1 as the radius.
Also, $O M=\frac{1}{2}(O A+O B), O C=\lambda O A+\mu O B$, then $M C=O C-O M=\left(\lambda-\frac{1}{2}\right) O A+\left(\mu-\frac{1}{2}\right) ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,880 |
6. Let $m, n$ be positive integers. If there exists a positive integer $k$ such that $\log _{2} m, \log _{2} n$ and $\log _{2} k$ can be the lengths of the three sides of a triangle, then $k$ is called a “good number”. It is known that there are exactly 100 good numbers $k$. Then the maximum possible value of $m n$ is ... | 6.C.
Assume $m \geqslant n$. From the problem, we have
$$
\begin{array}{l}
\log _{2} m-\log _{2} n\frac{m}{n}, m n-101 \leqslant \frac{m}{n},
\end{array}
$$
which implies $\frac{100}{1-\frac{1}{n^{2}}}<m n \leqslant \frac{101}{1-\frac{1}{n^{2}}}$.
It is easy to see that when $n=2$, $\frac{101}{1-\frac{1}{n^{2}}}$ rea... | 134 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,881 |
Sure, here is the translated text:
```
II. Fill-in-the-Blanks (9 points each, total 54 points)
1. Let the three interior angles of $\triangle A B C$ be $\angle A, \angle B, \angle C$ with the corresponding side lengths $a, b, c$. If $a < b < c$, and
$$
\left\{\begin{array}{l}
\frac{b}{a}=\frac{\left|b^{2}+c^{2}-a^{2}\... | II. 1.1:2:4.
From equation (1) and the Law of Sines and Cosines, we get
$$
\frac{\sin B}{\sin A}=2|\cos A| \Rightarrow \sin B=|\sin 2 A| \text {. }
$$
Since \(0<\angle A<\angle B<\frac{\pi}{2}\), we have \(\angle B=2 \angle A\).
Similarly, from equation (2), we get \(\angle C=2 \angle B\).
Combining \(\angle A+\angle ... | 1:2:4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,882 |
2. For non-empty sets $A$ and $B$, define the operation:
$A \vee B=\{x \mid x \in A \cup B$, and $x \notin A \cap B\}$.
Given two intervals $M=(a, b)$ and $P=(c, d)$, where $a+b<c+d$, $ab=cd<0$. Then $M \vee P=$ $\qquad$ | 2. $(a, c) \cup(b, d)$.
Let $a b=c d=t<0$, then $a<0<b, c<0<d$.
$$
\begin{array}{l}
\text { Let } f(x)=x^{2}-(a+b) x+t, \\
g(x)=x^{2}-(c+d) x+t .
\end{array}
$$
Thus, $a$ and $b$ are the roots of the equation $f(x)=0$, and $c$ and $d$ are the roots of the equation $g(x)=0$.
$$
\begin{array}{l}
\text { Also, } f(c)=c^... | (a, c) \cup (b, d) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,883 |
3. Given the function $f(x)=f_{1}(x)=|1-2 x|$, $x \in[0,1]$. When $n \geqslant 2$, $f_{n}(x)=f\left(f_{n-1}(x)\right)$, $x \in[0,1]$. Then the number of real solutions to the equation $f_{2007}(x)=\frac{1}{2} x$ is . $\qquad$ | $3.2^{2007}$
First, $f_{1}(x)=|1-2 x|, x \in[0,1]$, the range is $[0,1]$. Next, consider the zeros and the variation of the graph of $f_{1}(x)=|1-2 x|$.
It is easy to know that $f_{1}(x)$ has 1 zero at $x=\frac{1}{2}$, and the graph is a broken line composed of two line segments, from 1 to 0, then from 0 to 1.
For $f... | 2^{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,884 |
4. Two boxes are each filled with black and white balls, and the total number of balls in the two boxes is 25. Each time, a ball is randomly drawn from each box. If the probability of drawing two black balls is $\frac{27}{50}$, then, the probability of drawing two white balls is | 4. $\frac{1}{25}$.
Let the events of drawing a black ball from the two boxes be $A$ and $B$, respectively, then
$$
P(A \cdot B)=P(A) \cdot P(B)=\frac{27}{50} \text {. }
$$
On the other hand, suppose one of the boxes contains $n$ balls, of which $a$ are black, and the other box contains $b$ black balls. Then
$$
P(A \c... | \frac{1}{25} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,885 |
Example 1 Figure 1 is a shape composed of 4 unit squares, with each vertex of the unit squares called a grid point. By taking any three grid points as vertices, how many different isosceles right triangles can be formed?
Regarding this problem, the answer given in [1] is 24, [2] supplements it to 30, and [3] provides ... | Solution: Through the above classification, mark the number of isosceles right triangles with each grid point as the right-angle vertex (Figure 2). For example: $B_{3+1}$ indicates that there are 3 upright and 1 slanted isosceles right triangles with $B$ as the right-angle vertex; $E_{1+2}$ indicates that there are 1 u... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,886 |
5. For the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, the right focus is $F$, and the right directrix is $l$. Points $P_{1}, P_{2}, \cdots, P_{24}$ are 24 points arranged in a counterclockwise order on the ellipse, where $P_{1}$ is the right vertex of the ellipse, and
$$
\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdo... | $5.6 \sqrt{5}$.
In the ellipse, $F(\sqrt{5}, 0), e=\frac{\sqrt{5}}{3}, l: x=\frac{9}{\sqrt{5}}$.
It is easy to know that for each $i(i=1,2, \cdots, 12)$, points $P_{i} 、 F 、 P_{i+12}$ are collinear.
Let $l_{P_{i} P_{i+12}}: y=k_{i}(x-\sqrt{5})$, substituting into $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, we get $\left(4+9 k... | 6 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,887 |
6. Given a positive integer $n(n \geqslant 5)$, construct an inverted triangular number table as follows (see Figure 2): In the first row, write the numbers $1,2, \cdots, n$ in sequence. Below each pair of adjacent numbers in the row above, write their sum to form the next row (which has one fewer number than the row a... | $$
6.2^{n-2}(n+1) \text {. }
$$
Without loss of generality, let the first row be $a_{1}, a_{2}, \cdots, a_{n}$ forming an arithmetic sequence with a common difference of $d$. Then the second row is $a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n-1}+a_{n}$, also forming an arithmetic sequence with a common difference of $2d$; ... | 2^{n-2}(n+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,888 |
Three. (20 points) Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies $f\left(f(x)-x^{2}+x\right)=f(x)-x^{2}+x$.
(1) When $f(2)=3$, find $f(1)$; when $f(0)=a$, find $f(a)$.
(2) If there is exactly one real number $x_{0}$ such that $f\left(x_{0}\right)=x_{0}$, find the analytical expression of the function $... | (1) Let $x=2$, we get
$$
f\left(f(2)-2^{2}+2\right)=f(2)-2^{2}+2 \text {. }
$$
Since $f(2)=3$, then
$$
f\left(3-2^{2}+2\right)=3-2^{2}+2 \Rightarrow f(1)=1 \text {. }
$$
Let $x=0$, we get $f(f(0))=f(0)$.
Since $f(0)=a$, then $f(a)=a$.
(2) For any $x \in \mathbf{R}$, we have
$$
f\left(f(x)-x^{2}+x\right)=f(x)-x^{2}+x ... | f(x)=x^{2}-x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,889 |
Four, (20 points) Given $a>0, b>0$, and $a+b<$
1. Prove:
$$
\frac{(a-1)^{2}+b(2 a-b)}{(b-1)^{2}+a(2 b-a)} \geqslant \min \left\{\frac{a}{b}, \frac{b}{a}\right\} .
$$ | Given $a>0, b>0$, and $a+b<1$.
\begin{array}{l}
(a-1)^{2}+b(2 a-b)>0 .
\end{array}
Similarly, $(b-1)^{2}+a(2 b-a)>0$.
\begin{array}{l}
\text { Hence } \frac{(a-1)^{2}+b(2 a-b)}{(b-1)^{2}+a(2 b-a)}>0 \text {. }
\end{array}
(1) If $a \geqslant b$, then $\min \left\{\frac{a}{b}, \frac{b}{a}\right\}=\frac{b}{a}$. Therefore... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,890 |
Five. (20 points) On a piece of paper, there are a line $A B$ and a point $M$ outside the line, and the distance from point $M$ to line $A B$ is $a$. Fold the paper so that point $M$ coincides exactly with a point $M^{\prime}$ on line $A B$. Each such fold leaves a straight line crease on the paper. When $M^{\prime}$ t... | As shown in Figure 5, draw $M H \perp A B$ at $H$, and establish a Cartesian coordinate system with the midpoint $O$ of segment $M H$ as the origin and the line $M H$ as the $y$-axis, then $M\left(0, \frac{a}{2}\right)$.
When point $M^{\prime}$ coincides with $H$, the crease line is the $x$-axis.
When point $M^{\prim... | x^{2} \geqslant 2 a y | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,891 |
One, (50 points) As shown in Figure $3, P$ is a point outside $\odot O$, and $P A, P B$ are two tangents to $\odot O$, touching at points $A, B$ respectively. The secant line $P C D$ of $\odot O$ intersects $A B$ at point $E$, and $E F \perp P A$, with the foot of the perpendicular being $F$. Prove: $F E$ bisects $\ang... | First, prove $\frac{P C}{P D}=\frac{C E}{E D}$.
As shown in Figure 6, connect $A C$, $C B$, $B D$, and $D A$. Since
$$
P A^{2}=P C \cdot P D,
$$
we have $\frac{P C}{P D}=\frac{P C^{2}}{P A^{2}}$.
Also, $\triangle A C E \backsim \triangle D B E$,
$\triangle D A E \backsim \triangle B C E$,
thus $\frac{C E}{B E}=\frac{A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,892 |
II. (50 points) Given an infinite sequence of positive numbers $\left\{a_{n}\right\}$ satisfying:
(1) There exists $m \in \mathbf{R}$, such that $a_{i} \leqslant m(i=1,2, \cdots)$;
(2) For any positive integers $i, j(i \neq j)$, we have $\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$.
Prove: $m \geqslant 1$.
保留了原文的... | For $n \geqslant 4$, let $k_{1}, k_{2}, \cdots, k_{n}$ be a permutation of $1,2, \cdots, n$, and satisfy
$$
0<a_{k_{1}}<a_{k_{2}}<\cdots<a_{k_{n}} \leqslant m .
$$
From condition (2), we have
$$
a_{k_{i}}-a_{k_{i-1}} \geqslant \frac{1}{k_{i}+k_{i-1}}(i=2,3, \cdots, n) \text {. }
$$
Thus, for any $n \geqslant 4$, we h... | m \geqslant 1 | Inequalities | proof | Yes | Yes | cn_contest | false | 718,893 |
Three. (50 points) In a $9 \times 9$ grid, 46 cells are colored red. Prove: there exists a $2 \times 2$ block of 4 cells, in which at least 3 cells are colored red. | Three, first, consider a $9 \times 2$ grid.
As shown in Figure 7, let the first row have $x_{1}$ red squares, and the second row have $x_{2}$ red squares.
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline$a_{1}$ & $a_{2}$ & $a_{3}$ & $a_{4}$ & $a_{5}$ & $a_{6}$ & $a_{7}$ & $a_{8}$ & $a_{9}$ \\
\hline$b_{1}$ & $b_{2}$ & $b_{3... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,894 |
Example 1: Prove that for all $n\left(n \in \mathbf{N}_{+}\right)$, we have
$$
\sum_{k=1}^{n} \frac{1}{k \sqrt{k}}<3 .
$$ | Explanation: The left side of the inequality is the sum of the first $n$ terms of the sequence $\left\{\frac{1}{n \sqrt{n}}\right\}$, and the right side is a constant, so directly using mathematical induction is not feasible.
Notice that $\sum_{k=1}^{n} \frac{1}{k \sqrt{k}}=1+\sum_{k=2}^{n} \frac{1}{k \sqrt{k}}$, and
$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,895 |
3. There are 5 rooms $A, B, C, D, E$ arranged in a circular pattern, with the number of people living in them being $17, 9, 14, 16, 4$ respectively. Now, adjustments are to be made so that the number of people in each room is the same, and it is stipulated that people can only move to the adjacent left or right room. H... | (Tip: Let the number of people moved from room $A$ to room $B$ be $x_{B}$, and so on, then we have
$$
\begin{array}{l}
9+x_{B}-x_{C}=14+x_{C}-x_{D}=16+x_{D}-x_{E} \\
=4+x_{E}-x_{A}=17+x_{A}-x_{B} \\
=\frac{1}{5}(17+9+14+16+4)=12 .
\end{array}
$$
This is transformed into finding
$$
\begin{array}{l}
y=\left|x_{B}-5\righ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 718,896 |
2. The sequence of real numbers $a_{0}, a_{1}, \cdots$ is defined as: $a_{0}=-1$, for all positive integers $n$, $\sum_{k=0}^{n} \frac{a_{n-k}}{k+1}=0$. Prove: For all positive integers $n$, $a_{n}>0$. | 2. Prove by mathematical induction.
When $n=1$, we get $a_{1}=\frac{1}{2}>0$.
Assume that when $n \geqslant 1$, we have $a_{i}>0(i=1,2, \cdots, n)$.
From $\sum_{k=0}^{n} \frac{a_{k}}{n-k+1}=0, \sum_{k=0}^{n+1} \frac{a_{k}}{n-k+2}=0$, we get
$$
\begin{array}{l}
0=(n+2) \sum_{k=0}^{n+1} \frac{a_{k}}{n-k+2}-(n+1) \sum_{k... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,897 |
3. If the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3$, $3 z_{1}-2 z_{2}=\frac{3}{2}-\mathrm{i}$, then $z_{1} z_{2}=$ $\qquad$ . | $3.6\left(-\frac{5}{13}+\frac{12}{13} i\right)$.
Let $3 z_{1}=6(\cos \alpha+i \sin \alpha)$,
$2 z_{2}=6(\cos \beta+i \sin \beta)$.
Then $6(\cos \alpha-\cos \beta)=\frac{3}{2}, 6(\sin \alpha-\sin \beta)=-1$.
Thus $\tan \frac{\alpha+\beta}{2}=\frac{3}{2}, \cos (\alpha+\beta)=-\frac{5}{13}, \sin (\alpha+\beta)=\frac{12}{1... | 6\left(-\frac{5}{13}+\frac{12}{13} i\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,898 |
4. All positive integer pairs $(x, y)$ that satisfy $1!+2!+\cdots+x!=y^{2}$ are $\qquad$ | $$
\text { 4. }(1,1),(3,3) \text {. }
$$
Since when $x \geqslant 5$, we have
$$
\begin{array}{l}
1!+2!+\cdots+x! \equiv 3(\bmod 5), \\
y^{2} \equiv 0,1,4(\bmod 5),
\end{array}
$$
thus, $x<5$.
By direct calculation, we know $(x, y)=(1,1),(3,3)$. | (1,1),(3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,899 |
5. The equation of the largest circle above the parabola $y=a x^{2}(a>0)$, which is tangent to the parabola at its vertex, is $\qquad$ | 5. $x^{2}+\left(y-\frac{1}{2 a}\right)^{2}=\frac{1}{4 a^{2}}$.
Let the required circle be $x^{2}+(y-r)^{2}=r^{2}(r>0)$, and let $P(x, y)$ be any point on the parabola $y=a x^{2}$. Then
$$
\begin{array}{l}
\sqrt{x^{2}+(y-r)^{2}} \geqslant r \text { holds for all } y \geqslant 0 \\
\Leftrightarrow y^{2}+\left(\frac{1}{a... | x^{2}+\left(y-\frac{1}{2 a}\right)^{2}=\frac{1}{4 a^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,900 |
6. The line $y=2 x$ intersects the curve
$$
(8 \sin \theta+\cos \theta+1) y=2(2 \sin \theta-\cos \theta+3) x^{2}
$$
to form a chord of maximum length of $\qquad$ . | $6.8 \sqrt{5}$.
Obviously, the origin is an intersection point.
Substituting $y=2 x$ into the curve equation, we get
$$
(2 x-8) \sin \theta-(x+1) \cos \theta=1-3 x
$$
For $\theta$ to have a solution, then
$$
(x+1)^{2}+(2 x-8)^{2} \geqslant(3 x-1)^{2} \text {. }
$$
Solving this, we get $-8 \leqslant x \leqslant 2,|x|_... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,901 |
Three. (20 points) Does there exist a cube in space such that the distances from its 8 vertices to a certain plane are exactly 0, 1, 2, 3, 4, 5, 6, 7? If it exists, indicate the position of the cube and the corresponding plane; if not, please explain the reason. | Three, let the coordinates of the 8 vertices of a unit cube be
$$
\begin{array}{l}
(0,0,0),(0,0,1),(0,1,0),(0,1,1), \\
(1,0,0),(1,0,1),(1,1,0),(1,1,1) .
\end{array}
$$
If we consider them as three-digit binary numbers, they are precisely $0,1,2,3,4,5,6,7$.
Let $(x y z)_{2}$ be a three-digit binary number, then
$$
(x y... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,902 |
Four. (20 points) Let $x, y, z > 0$, and $k \geqslant 1, k \in \mathbf{N}$. Prove:
$$
\begin{aligned}
R= & \frac{y^{k}-x^{k}}{z^{k-1}+z^{k-2} x+\cdots+x^{k-1}}+ \\
& \frac{z^{k}-y^{k}}{x^{k-1}+x^{k-2} y+\cdots+y^{k-1}}+ \\
& \frac{x^{k}-z^{k}}{y^{k-1}+y^{k-2} z+\cdots+z^{k-1}} \geqslant 0 .
\end{aligned}
$$ | Let
$$
\begin{aligned}
S= & \frac{y^{k}-z^{k}}{z^{k-1}+z^{k-2} x+\cdots+x^{k-1}}+\frac{z^{k}-x^{k}}{x^{k-1}+x^{k-2} y+\cdots+y^{k-1}}+ \\
& \frac{x^{k}-y^{k}}{y^{k-1}+y^{k-2} z+\cdots+z^{k-1}} .
\end{aligned}
$$
Then \( R-S \)
$$
\begin{array}{l}
=\frac{z^{k}-x^{k}}{z^{k-1}+z^{k-2} x+\cdots+x^{k-1}}+ \\
\frac{x^{k}-y^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,903 |
One, (50 points) Try to find all real-coefficient polynomials $f(x)$, such that for all real numbers $a, b, c$ satisfying $a b + b c + c a = 0$, the following holds:
$$
f(a-b) + f(b-c) + f(c-a) = 2 f(a+b+c).
$$ | Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$.
Since for all $a, b, c \in \mathbf{R}$ satisfying $a b+b c+c a=0$, we have
$$
\begin{array}{l}
f(a-b)+f(b-c)+f(c-a) \\
=2 f(a+b+c),
\end{array}
$$
Therefore, when $a=b=c=0$, we have
$$
3 f(0)=2 f(0), f(0)=a_{0}=0 \text {; }
$$
When $a=x, b=c=0$, we have
$$
f(x)+f(-... | f(x)=A x^{4}+B x^{2}(A, B \in \mathbf{R}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,904 |
II. (50 points) As shown in Figure 3, $BC$ is the diameter of the circumcircle $\odot O$ of $\triangle CEF$, $A$ is a point on the arc $\overparen{EF}$, and $D$ is the midpoint of the arc $\overparen{AB}$. A line through point $O$ parallel to $DA$ intersects $AC$ at point $J$. If $J$ is the incenter of $\triangle CEF$,... | $$
\begin{array}{l}
AF, FJ, FO. \\
\text {Let } \angle AOD = \angle BOD \\
= \alpha, \angle ACF = \beta.
\end{array}
$$
Since $D$ is the midpoint of $\overparen{AB}$, we know that
$$
\begin{array}{l}
\angle BOD = \angle ACB, \\
OD \parallel AC.
\end{array}
$$
Also, $AD \parallel OJ$, so quadrilateral $ADOJ$ is a para... | 30^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,905 |
Three, (50 points) Find all positive integer solutions $(x, y, z, t)$ of the equation $2^{y}+2^{x} \times 5^{t}-5^{x}=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $(x, y, z, t)$ be a positive integer solution of the equation. Then $2^{y} \equiv 1(\bmod 5)$.
Since $2^{4} \equiv 1(\bmod 5)$, we have $4 \mid y$.
Taking the equation modulo 4, we get
$$
2^{x}=2(\bmod 4).
$$
Thus, $z=1$.
Let $y=4 r$, then $5^{x}+1=2^{4 r}+2 \times 5^{t}$, which simplifies to
$$
5^{x}-2 \times 5^... | (2,4,1,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,906 |
Five. (20 points) Prove: For any point $A$ on the parabola $y^{2}=2 p x$ $(p>0)$, there always exist two points $B$ and $C$ on the parabola such that $\triangle A B C$ is an equilateral triangle. | Given $\triangle A B C$ with side length $a$, the angle between the positive direction of the $x$-axis and $A B$ is $\theta$, and $A\left(x_{0}, y_{0}\right)$. Then,
$$
\begin{array}{l}
B\left(x_{0}+a \cos \theta, y_{0}+a \sin \theta\right), \\
C\left(x_{0}+a \cos \left(60^{\circ}-\theta\right),\right. \\
\left.y_{0}+a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,907 |
3. The sequence $c_{0}, c_{1}, \cdots$ is defined as: $c_{0}=1, c_{1}=0$, and for non-negative integers $n, c_{n+2}=c_{n+1}+c_{n}$. Consider the set $S$ of ordered pairs $(x, y)$, where $(x, y)$ satisfies: there exists a finite set $J$ of positive integers, such that
$$
x=\sum_{j \in J} c_{j}, y=\sum_{j \in J} c_{j-1} ... | 3. Let $\varphi=\frac{1+\sqrt{5}}{2}, \psi=\frac{1-\sqrt{5}}{2}$ be the two roots of the quadratic equation $t^{2}-t-1=0$. Then we have $\varphi \psi=-1, \varphi+\psi=1, 1+\psi=\psi^{2}$. By the method of characteristic roots, for $n \geqslant 0$, we have
$c_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n-1}-\psi^{n-1}\right)$.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,908 |
Given $x, y$ are positive numbers satisfying $x+y=1$. Prove: $\frac{2}{1+x^{2}}+\frac{1}{1-y^{2}}>2$.
---
The translation is provided as requested, maintaining the original formatting and line breaks. | Proof: Since $\frac{1+x^{2}}{2}+1-y^{2}>0$, and
$$
\begin{array}{l}
\left(\frac{1+x^{2}}{2}+1-y^{2}\right)\left(\frac{2}{1+x^{2}}+\frac{1}{1-y^{2}}\right) \\
=2+\frac{1+x^{2}}{2\left(1-y^{2}\right)}+\frac{2\left(1-y^{2}\right)}{1+x^{2}} \\
=4+\left[\sqrt{\frac{1+x^{2}}{2\left(1-y^{2}\right)}}-\sqrt{\frac{2\left(1-y^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,909 |
Initially 210 As shown in Figure 1, from a point $P$ outside a circle, two secants $P A B$ and $P C D$ are drawn, intersecting the circle at points $A, B,$ and $C, D$ respectively. $A D$ and $B C$ intersect at point $Q$. A secant $P E F$ passing through point $Q$ intersects the circle at points $E$ and $F$. Parallel li... | Proof: Let $B N$ and $D M$ intersect $E F$ at points $R$ and $R^{\prime}$, respectively, and connect $D R$.
Since $D N / / P F$, we have,
$$
\angle N R P=\angle D N B=\angle D A B \text{. }
$$
Therefore, points $A, B, R, Q$ are concyclic.
Thus, $P Q \cdot P R=P A \cdot P B=P C \cdot P D$.
Hence, points $C, D, R, Q$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,910 |
Given 209 Let $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$, and $\tan \alpha + \tan \beta + \tan \gamma = 1$. Prove: $\sin 2\alpha + \sin 2\beta + \sin 2\gamma \leq \frac{9}{5}$. | Prove: For the function
$$
f(x)=(1+x^{2})(2-x)(x \in(0,1))
$$
Differentiating, we get
$$
f^{\prime}(x)=2 x(2-x)-x^{2}-1=(1-x)(3 x-1).
$$
By $f^{\prime}(x)=0$, we get $x=\frac{1}{3}$.
When $0<x<\frac{1}{3}, f^{\prime}(x)>0, f(x)$ is an increasing function.
Therefore, when $x \in(0,1)$, we have
$$
(1+x^{2})(2-x) \geqs... | \sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma \leqslant \frac{9}{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 718,911 |
Given a prime $p(p>3)$, define
$$
S=\left\{n \mid \sum_{k=1}^{p-1} k^{n} \equiv 0\left(\bmod p^{3}\right), n \in \mathbf{N}\right\} .
$$ | Proof: $S$ is an infinite set.
Proof: Let $S_{n}=\sum_{k=1}^{p-1} k^{n}$.
Define the set
$$
T=\{r \mid r=(m p+3)(p-1)+3, m \in \mathbf{N}\},
$$
For any $r \in T$,
$$
r=(m p+3)(p-1)+3(m \in \mathbf{N}) .
$$
By the binomial theorem, we have
$$
\begin{array}{l}
2 S_{r}=\sum_{k=1}^{p-1}\left[k^{r}+(p-k)^{r}\right] \\
=\s... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,912 |
4. For positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, prove:
$$
\sum_{i<j} \frac{a_{i} a_{j}}{a_{i}+a_{j}} \leqslant \frac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)} \sum_{i<j} a_{i} a_{j} .
$$ | 4. Let $S=\sum_{i=1}^{n} a_{i}$.
Since $\sum_{i<j}\left(a_{i}+a_{j}\right)=(n-1) S$, then
$$
\begin{array}{l}
\sum_{i<j} \frac{a_{i} a_{j}}{a_{i}+a_{j}}=\sum_{i<j} \frac{1}{4}\left[a_{i}+a_{j}-\frac{\left(a_{i}-a_{j}\right)^{2}}{a_{i}+a_{j}}\right] \\
\quad=\frac{n-1}{4} S-\frac{1}{4} \sum_{i<j} \frac{\left(a_{i}-a_{j... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,913 |
5. Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Prove:
$$
\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leqslant 3 .
$$ | 5. Proof 1: Note that $\sqrt{a}+\sqrt{b}>\sqrt{a+b}>\sqrt{c}$,
$$
\begin{array}{l}
\sqrt{b}+\sqrt{c}>\sqrt{a}, \sqrt{c}+\sqrt{a}>\sqrt{b}. \\
\text { Let } x=\sqrt{b}+\sqrt{c}-\sqrt{a}, y=\sqrt{c}+\sqrt{a}-\sqrt{b}, \\
z=\sqrt{a}+\sqrt{b}-\sqrt{c}.
\end{array}
$$
Then $x, y, z>0$, and we have
$$
\begin{array}{l}
b+c-a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,914 |
2. Given $x \in(0,1)$, let $y \in(0,1)$, and the $n$-th digit after the decimal point of $y$ is the $2^{n}$-th digit after the decimal point of $x$. Prove: if $x$ is a rational number, then $y$ is also a rational number. | 2. Since $x$ is a rational number, $x$ is periodic from some digit after the decimal point. Let the period length be $d$, and let $d=2^{u} v$, where $v$ is an odd number, then there exists a positive integer $w$, such that $2^{w} \equiv 1(\bmod v)$. In particular, we can take $w=\varphi(v)$, where $\varphi$ is the Eule... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,915 |
3. The sequence $f(1), f(2), \cdots$ is defined as
$$
f(n)=\frac{1}{n}\left(\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\cdots+\left[\frac{n}{n}\right]\right) \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding $x$: Prove:
(1) There are infinitely many $n$ such that
$$
f(n+1)>f(n) \text {; }
$$
... | 3. Let $g(n)=n f(n) (n \geqslant 1)$, then $g(0)=0$.
For $k=1,2, \cdots, n$, if $k$ is not a divisor of $n$, then
$$
\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=0 ;
$$
if $k$ is a divisor of $n$, then
$$
\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=1 \text {. }
$$
For $n \geqslant 1$, let $d(n)$ be th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,916 |
5. Find all integer solutions to the equation $\frac{x^{7}-1}{x-1}=y^{5}-1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. No integer solutions.
Lemma If $x$ is an integer, and $p$ is a prime factor of $\frac{x^{7}-1}{x-1}$, then either $p \equiv 1(\bmod 7)$, or $p=7$.
Proof of the lemma: Since $p \left\lvert\, \frac{x^{7}-1}{x-1}\right.$, therefore, $p \mid\left(x^{7}-1\right)$.
Thus, $(x, p)=1$.
By Fermat's Little Theorem, we have $x... | No integer solutions. | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,917 |
7. Prove: For every positive integer $d$, there exists an integer $m$, such that $d \mid\left(2^{m}+m\right)$. | 7. Prove by mathematical induction on $d$: For every positive integer $N$, there exist positive integers $b_{0}, b_{1}, \cdots, b_{d-1}$, such that for each $i(i=0,1, \cdots, d-1)$, we have $b_{i}>N$, and
$2^{b_{i}}+b_{i}=i(\bmod d)$.
$m=b_{0}$ is the original proposition.
When $d=1$, the conclusion is obviously true.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,919 |
Example 2 Given a positive integer $n(n \geqslant 2)$, let positive integers $a_{i}(i=1,2, \cdots, n)$ satisfy $a_{1}<a_{2}<\cdots<a_{n}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$. Prove that for any real number $x$, we have
$$
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}{2} \time... | Explanation: Based on the known condition $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$ and the structure of the inequality to be proven, it is not difficult to think of first using the Cauchy-Schwarz inequality for magnification, obtaining
$$
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \sum_{i=1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,921 |
2. The distance from Mr. Wang's home to his workplace is $3500 \mathrm{~m}$, and Mr. Wang leaves home on foot to go to work every morning at $7: 30$, arriving at his workplace between $8: 10$ (inclusive) and $8: 20$ (inclusive). If Mr. Wang's walking speed is set as $x \mathrm{~m} / \mathrm{min}$, then the range of Mr.... | 2. A.
According to the problem, the time Mr. Wang spends on his way to work is between $40 \mathrm{~min}$ and $50 \mathrm{~min}$, so $\frac{3500}{50} \leqslant x \leqslant \frac{3500}{40}$, which means $70 \leqslant x \leqslant 87.5$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,922 |
3. As shown in Figure $2, A B$ is the diameter of a semicircle, chords $A D$ and $B C$ intersect at point $P$, it is known that $\angle D P B=$ $60^{\circ}, D$ is the midpoint of $\overparen{B C}$. Then $\tan \angle A D C$ equals ( ).
(A) $\frac{1}{2}$
(B) 2
(C) $\sqrt{3}$
(D) $\frac{\sqrt{3}}{3}$ | 3. D.
Connect $A C$ (figure omitted). Let $\angle D A C=\angle D A B=\alpha, \angle A B C=\beta$. Then $\alpha+\beta=60^{\circ}, 2 \alpha+\beta=90^{\circ}$. Solving these, we get $\alpha=30^{\circ}$. Therefore, $\tan \angle D A C=\frac{\sqrt{3}}{3}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,923 |
4. The graph of the parabola $y=x^{2}+x+p(p \neq 0)$ intersects the $x$-axis at one point with the $x$-coordinate $p$. Then, the coordinates of the vertex of the parabola are ( ).
(A) $(0,-2)$
(B) $\left(\frac{1}{2},-\frac{9}{4}\right)$
(C) $\left(-\frac{1}{2}, \frac{9}{4}\right)$
(D) $\left(-\frac{1}{2},-\frac{9}{4}\r... | 4.D.
From the problem, we get $p^{2}+p+p=0$.
Solving, we get $p_{1}=-2, p_{2}=0$ (discard).
When $p=-2$, the parabola is $y=x^{2}+x-2$.
Therefore, the vertex coordinates are $\left(-\frac{1}{2},-\frac{9}{4}\right)$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,924 |
5. As shown in Figure 3, in $\triangle A B C$, $A B$ $=A C, \angle A=36^{\circ}, C D$ is the angle bisector. Then the ratio of the area of $\triangle D B C$ to the area of $\triangle A B C$ is ( ).
(A) $\frac{\sqrt{5}-2}{2}$
(B) $\frac{\sqrt{5}-2}{3}$
(C) $\frac{3-\sqrt{5}}{2}$
(D) $\frac{3-\sqrt{5}}{3}$ | 5.C.
$$
\begin{array}{l}
\text { It is easy to prove } A D=D C=B C \Rightarrow \triangle C B D \subset \triangle A B C \\
\Rightarrow \frac{B C}{A B}=\frac{B D}{B C} \Rightarrow \frac{B C}{A B}=\frac{A B-B C}{B C}=\frac{A B}{B C}-1 \\
\Rightarrow \frac{B C}{A B}=\frac{\sqrt{5}-1}{2} \\
\Rightarrow \frac{S_{\triangle D ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,925 |
6. The line $l: y=p x$ (where $p$ is an integer not equal to 0) intersects the line $y=x+10$ at a point that is an integer point (both the x-coordinate and y-coordinate are integers). Therefore, the number of lines $l$ that satisfy this condition is $(\quad)$.
(A)6
(B) 7
(C) 8
(D) infinitely many | 6.B.
Solve the system of equations $\left\{\begin{array}{l}y=p x, \\ y=x+10,\end{array}\right.$ to get $x=\frac{10}{p-1}$.
Since $x$ and $p$ are integers, we have
$$
p-1= \pm 10, \pm 5, \pm 2, \pm 1 \text {. }
$$
Thus, $p=11,-9,6,-4,3,-1,2,0$ for a total of 8 values, and $p=0$ is discarded. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,926 |
7. Place three consecutive positive integers $a, b, c$ in any order (different orders are considered different groups) into
$$
\square x^{2}+\square x+\square=0
$$
as the quadratic coefficient, linear coefficient, and constant term of a quadratic equation. The number of groups of $a, b, c$ that make the resulting equa... | 7.C.
Let the three consecutive positive integers be $n-1, n, n+1$ (where $n$ is an integer greater than 1). When the coefficient of the linear term is $n-1$ or $n$, $\Delta$ is less than zero, and the equation has no real roots; when the coefficient of the linear term is $n+1$, $\Delta=(n+1)^{2}-4 n(n-1)=-3(n-1)^{2}+4... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,927 |
8. A uniform cube has the numbers $1,1,2,3,3,5$ marked on its six faces as shown in Figure 4. When this cube is rolled once, the number on the top face is recorded as the x-coordinate of a point in the Cartesian coordinate system, and the number on the bottom face is recorded as the y-coordinate of that point. Accordin... | 8.A.
Each throw may result in 6 points with coordinates (two of which are overlapping): $(1,1),(1,1),(2,3),(3,2),(3,5)$, $(5,3)$. By plotting the points and calculation, it can be found that the line determined by any two of the three points $(1,1)(2,3)$, $(3,5)$ passes through point $P(4,7)$. Therefore, the probabili... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,928 |
9. If $a$ is a perfect square, then the smallest perfect square greater than $a$ is $\qquad$ .
| The smallest perfect square greater than $a$ is
$$
(\sqrt{a}+1)^{2}=a+2 \sqrt{a}+1 .
$$ | a+2\sqrt{a}+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,929 |
11. Given in acute $\triangle A B C$, $\angle A=$ $50^{\circ}, A B>B C$. Then the range of $\angle B$ is $\qquad$ . | $11.40^{\circ}<\angle B<80^{\circ}$.
As shown in Figure 12, when $BC$ is the shortest, $\angle ABC=40^{\circ}$. Now, with point $B$ as the center and $AB$ as the radius, draw a circle intersecting line $AC$ at point $C_{1}$. When $BC_{1}=AB$, $\angle ABC_{1}=80^{\circ}$. Therefore, $40^{\circ}<\angle B<80^{\circ}$. | 40^{\circ}<\angle B<80^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,931 |
Example 3 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and
$$
a_{n+1}=\frac{1+2 a_{n}}{1+a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: $a_{n}<\frac{1+\sqrt{5}}{2}$. | Explanation: Since the right side of the inequality is a constant, it is difficult to directly apply mathematical induction to achieve the transition from $k$ to $k+1$. Therefore, we can prove a strengthened version of the original inequality:
$$
0 < a_n < \frac{1+\sqrt{5}}{2} \quad (1)
$$
On the other hand, we have
$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,932 |
12. Let the side length of the equilateral $\triangle ABC$ be $a$. If $\triangle ABC$ is rotated $60^{\circ}$ about its center (the center of the circumcircle of the equilateral triangle), the resulting triangle is $\triangle A^{\prime} B^{\prime} C^{\prime}$. Then the distance between points $A$ and $B^{\prime}$ is $\... | 12. $\frac{\sqrt{3}}{3} a$ or $\frac{2 \sqrt{3}}{3} a$.
When $\triangle A B C$ is rotated $60^{\circ}$ clockwise, as shown in Figure 13
(a), connect $O A$ and $O B^{\prime}$. Since $\angle A O B^{\prime}=60^{\circ}$, $\triangle O A B^{\prime}$ is an equilateral triangle. Therefore, $A B^{\prime}=\frac{\sqrt{3}}{3} a$.... | \frac{\sqrt{3}}{3} a \text{ or } \frac{2 \sqrt{3}}{3} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,933 |
13. As shown in Figure 6, a right trapezoid $A O C D$ is placed in the Cartesian coordinate system. It is known that $A D=3, A O=8, O C=5$. If point $P$ is inside the trapezoid, and
$$
\begin{array}{l}
S_{\triangle P A D}=S_{\triangle P O C}, \\
S_{\triangle P A O}=S_{\triangle P C D},
\end{array}
$$
then, the coordin... | 13. $\left(\frac{17}{8}, 3\right)$.
$S_{\text {trapezoid } A O C D}=\frac{(3+5) \times 8}{2}=32$.
As shown in Figure 14, draw $P E$ $\perp y$-axis at point $E$.
Since $S_{\triangle P R D}=S_{\triangle P Q C}$, we have $3 A E=5 O E$, that is,
$$
3(8-O E)=5 O E \text {. }
$$
Solving this, we get $O E=3$.
Thus, $S_{\tri... | \left(\frac{17}{8}, 3\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,934 |
14. Given that the weights of $A$, $B$, $C$, and $D$ are all integers in kilograms, where $A$ is the lightest, followed by $B$, $C$, and $D$, the weights of each pair of them are as follows (unit: kg):
$45,49,54,55,60,64$.
Then the weight of $D$ is $\qquad$ kg. | 14.35.
Since $A+B=45, A+C=49, B+D=60, C+D$ $=64$, therefore, $C-B=4$.
Then $B+C=B+(B+4)=2B+4$ is an even number.
Among $54 \text{~kg}$ and $55 \text{~kg}$, only 54 is an even number, so $B=25$. Then $D=35$. | 35 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 718,935 |
15. (12 points) Given $b-a=\frac{1}{8}, 2 a^{2}+a=$ $\frac{1}{4}$. Find the value of $\frac{b}{a}-a$. | Three, 15. Multiply the equation $b-a=\frac{1}{8}$ by 2 and subtract the equation $2 a^{2}+a=\frac{1}{4}$, to get $2 b-2 a^{2}=3 a$.
Given that $a \neq 0$.
Divide both sides by $2 a$, to get $\frac{b}{a}-a=\frac{3}{2}$. | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,936 |
16. (12 points) There are $a$ matchsticks of the same length. When placed as shown in Figure 7, they can form $m$ squares; when placed as shown in Figure 8, they can form $2 n$ squares.
(1) Express $m$ in terms of $n$;
(2) When these $a$ matchsticks can also be arranged as shown in Figure 9, find the minimum value of $... | 16. (1) The total number of matchsticks in Figure 7 is $3 m+1$, and the total number of matchsticks in Figure 8 is $5 n+2$.
Since the total number of matchsticks is the same, we have $3 m+1=5 n+2$.
Solving for $m$ gives $m=\frac{5 n+1}{3}$.
(2) Suppose there are $3 p$ squares in Figure 9, then the total number of match... | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,937 |
17. (12 points) As shown in Figure 10, given that the diameter of a circle is equal to the height of equilateral $\triangle ABC$, the circle is tangent to sides $AB$ and $BC$ at points $D$ and $E$, respectively, and intersects side $AC$ at points $F$ and $G$. Find the measure of $\angle DEF$. | 17. As shown in Figure 15, draw a perpendicular from point $E$ to $BC$ intersecting the circle at point $H$ and $AC$ at point $O$. Connect $AH$ and $DH$, and draw $AM \perp BC$, with the foot of the perpendicular at $M$.
Since $E$ is the point of tangency, $EH$ must pass through the center of the circle, i.e., $EH$ is... | 105^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,938 |
18. (14 points) Given the parabola $C_{1}$:
$$
y=a x^{2}-2 a m x+a m^{2}+2 m+1(a>0, m>0)
$$
with vertex $A$, and the vertex $B$ of parabola $C_{2}$ is on the $y$-axis, and parabolas $C_{1}$ and $C_{2}$ are centrally symmetric about $P(1,3)$.
(1) When $a=1$, find the equation of parabola $C_{2}$ and the value of $m$;
(... | 18. (1) When $a=1$, we have
$$
\begin{array}{l}
y=a x^{2}-2 a m x+a m^{2}+2 m+1 \\
=(x-m)^{2}+2 m+1,
\end{array}
$$
Therefore, the vertex $A(m, 2 m+1)$.
Also, $P(1,3)$, and let the equation of line $A B$ be $y=k x+b$.
Substituting the coordinates of points $A$ and $P$ gives
$$
\left\{\begin{array}{l}
2 m+1=k m+b, \\
3... | a_{1}=\frac{1}{19}, a_{2}=\frac{1}{49} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,939 |
$1.2007^{2007}$ 's last digit is ( ).
(A) 1
(B) 3
(C) 7
(D) 9 | $-1 . B$.
When $k=1,2, \cdots$, the last digit of $2007^{k}$ cycles through $7,9,3,1$. Since 2007 is a number of the form $4 n+3$, the last digit of $2007^{200}$ is 3. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,940 |
2. Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$, the result is ( ).
(A) $\sqrt{3}$
(B) $\sqrt{2}$
(C) $\sqrt{3}+\sqrt{2}$
(D) $\sqrt{3}-\sqrt{2}$ | 2.B.
Since $2+\sqrt{3}=\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)^{2}, 2-\sqrt{3}=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)^{2}$, then
$$
\begin{array}{l}
\text { Original expression }=\frac{2+\sqrt{3}}{\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}} \\
=\sqrt{2}\left(\frac{2... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,941 |
3. Given that $a$, $b$, and $c$ are positive numbers, the quadratic equation $a x^{2}+b x+c=0$ has two equal real roots. Then the root situation of the equation $(a+1) x^{2}+(b+2) x+(c+1)=0$ is $(\quad)$.
(A) No real roots
(B) Two equal real roots
(C) Two unequal real roots
(D) The root situation is uncertain | 3. D.
Let the equal root of $a x^{2}+b x+c=0$ be $r$. Then
$$
a x^{2}+b x+c=a(x-r)^{2} \text {. }
$$
When $r=-1$,
$$
\begin{array}{l}
(a+1) x^{2}+(b+2) x+(c+1) \\
=\left(a x^{2}+b x+c\right)+(x+1)^{2}=(a+1)(x+1)^{2},
\end{array}
$$
it can be known that $x=-1$ is also the equal root of $(a+1) x^{2}+(b+2) x+(c+1)=0$;
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,942 |
4. If the three vertices of a right-angled triangle are all taken from the vertices of a certain regular dodecagon, then the number of such right-angled triangles is $(\quad)$.
(A) 36
(B) 60
(C) 96
(D) 120 | 4. B.
Considering on the circumcircle $\odot O$ of this dodecagon, any triangle formed by any three vertices of the dodecagon has $\odot O$ as its circumcircle. If it is a right triangle, then one of its sides must be the diameter of $\odot O$. Among the diagonals of the regular dodecagon, there are 6 diameters. For e... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,944 |
6. For a given unit square, if its two diagonals and the lines connecting the midpoints of each pair of sides are drawn, figure 1 is obtained. Then, the number of pairs of similar triangles in figure 1 is ( ).
(A) 44
(B) 552
(C) 946
(D) 1892 | 6.C.
In Figure 1, all triangles are isosceles right triangles. First, let's count the number of triangles:
Classified by the length of the hypotenuse: there are 4 triangles with a hypotenuse length of $\sqrt{2}$; 8 triangles with a hypotenuse length of 1; 16 triangles with a hypotenuse length of $\frac{\sqrt{2}}{2}$;... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,946 |
1. The sum of all real numbers $x$ that satisfy the equation $\sqrt{3 x-4}+\sqrt[3]{5-3 x}=1$ is $\qquad$ . | II. $1 . \frac{22}{3}$.
Let $\sqrt{3 x-4}=t$, then $t \geqslant 0$.
The original equation becomes $1-t=\sqrt[3]{1-t^{2}}$, that is,
$t(t-1)(t-3)=0$.
Therefore, $t=0,1,3$.
From $t=0$, we get $x=\frac{4}{3}$; from $t=1$, we get $x=\frac{5}{3}$; from $t=3$, we get $x=\frac{13}{3}$. After verification, all are solutions to... | \frac{22}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,947 |
2. The number of triangles with integer side lengths and a perimeter of 20 is $\qquad$ . | 2.8.
Let the three sides of a triangle be $a, b, c$, and $a \geqslant b \geqslant c$, $a+b+c=20$, then $a \geqslant 7$.
Also, from $b+c>a$, we get $2 a<a+b+c=20 \Rightarrow a<10$. Therefore, $7 \leqslant a \leqslant 9$. We can list
$$
\begin{array}{l}
(a, b, c)=(9,9,2),(9,8,3),(9,7,4),(9,6,5), \\
(8,8,4),(8,7,5),(8,6,... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,948 |
4. Arrange the four numbers $1, 2, 3, 4$ to form a four-digit number, such that this number is a multiple of 11. Then the number of such four-digit numbers is $\qquad$.
| 4.8.
Since $1+4=2+3$, we can place 1 and 4 in the even positions, and 2 and 3 in the odd positions, which gives us four arrangements; placing 2 and 3 in the even positions, and 1 and 4 in the odd positions also gives us four arrangements. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,950 |
One. (20 points) Find all integers $a$ such that the quadratic equation in $x$
$$
x^{2}-\sqrt{5 a^{2}-26 a-8} x-\left(a^{2}-4 a+9\right)=0
$$
has both roots as integers. | Let the two roots of the equation be $x_{1}$ and $x_{2}$, then,
$\sqrt{5 a^{2}-26 a-8}=x_{1}+x_{2}=$ integer,
which means equation (1) is a quadratic equation with integer coefficients, and its roots are integers, so its discriminant $\Delta$ must be a perfect square. Let
$\left(5 a^{2}-26 a-8\right)+4\left(a^{2}-4 a+9... | a=6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,951 |
II. (25 points) As shown in Figure 3, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $P$. A line through point $P$ intersects $AD$ at point $E$ and $BC$ at point $F$. If $PE=PF$, and $AP+AE=CP+CF$, prove: Quadrilateral $ABCD$ is a parallelogram. | As shown in Figure 5, take points \( M \) and \( N \) on the extensions of \( PA \) and \( PC \) respectively, such that \( AM = AE \) and \( CN = CF \). Then, \( PM = PN \).
Figure 5
Thus, quadrilateral \( EMFN \) is a parallelogram, and \( \angle AME = \angle CNF \),
and \( \triangle AME \) and \( \triangle CNF \) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,952 |
Three. (25 points) If a number $a$ can be expressed as the sum of the squares of two natural numbers (allowing the same), then $a$ is called a "good number". Determine how many good numbers there are among the first 200 positive integers $1,2, \cdots, 200$.
Translate the above text into English, please retain the orig... | Three, the squares not exceeding 200 are $0^{2}, 1^{2}, \cdots, 14^{2}$.
Obviously, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form of $k^{2}+$ $0^{2}$, and there are 14 such numbers.
Each pair of numbers in $1^{2}, 2^{2}, \cdots, 10^{2}$ (which can be the same) has a sum not exceedi... | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,953 |
Example 5 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=a_{n}+\frac{1}{a_{n}}$.
Prove: $14<a_{100}<18$. | Explanation: Prove the strengthened inequality using the method of magnification and reduction
$$
\sqrt{2 n-1}2) \text {. }
$$
When $k>1$,
$$
a_{k}^{2}=\left(a_{k-1}+\frac{1}{a_{k-1}}\right)^{2}>a_{k-1}^{2}+2,
$$
i.e., $a_{k}^{2}-a_{k-1}^{2}>2$.
Then $a_{n}^{2}-a_{1}^{2}=\sum_{k=2}^{n}\left(a_{k}^{2}-a_{k-1}^{2}\righ... | 14<a_{100}<18 | Algebra | proof | Yes | Yes | cn_contest | false | 718,954 |
2. Given the function $f(x)=\pi \sin \frac{x}{4}$. If there exist real numbers $x_{1}, x_{2}$, such that for any real number $x$, we have
$$
f\left(x_{1}\right) \leqslant f(x) \leqslant f\left(x_{2}\right),
$$
then the minimum value of $\left|x_{1}-x_{2}\right|$ is ( ).
(A) $8 \pi$
(B) $4 \pi$
(C) $2 \pi$
(D) $\pi$ | 2.B.
Obviously, $f\left(x_{1}\right)$ and $f\left(x_{2}\right)$ are the minimum and maximum values of $f(x)$, respectively, i.e., $\sin \frac{x_{1}}{4}=-1, \sin \frac{x_{2}}{4}=1$. Therefore, $x_{1}=8 k_{1} \pi-2 \pi, x_{2}=8 k_{2} \pi+2 \pi$.
Thus, $\left|x_{1}-x_{2}\right| \geqslant 4 \pi$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,956 |
3. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}, \cdots .
\end{array}
$$
Let $S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}$. Then the integer closest to $S_{2006}$ is $(\qua... | 3.C.
Since $a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}=\frac{n}{2}$, we have
$$
\begin{array}{l}
S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\fr... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,957 |
4. $n$ is a positive integer, and it is defined that: $n!=1 \times 2 \times \cdots \times$ $n$. Then the remainder when $1!\times 1+2!\times 2+\cdots+59!\times 59$ is divided by 2006 is ( ).
(A) 1
(B) 5
(C) 401
(D) 2005 | 4.D.
Notice that $2006=2 \times 17 \times 59$, so 2006160 !.
$$
\begin{array}{l}
\text { Also } 1! \times 1 + 2! \times 2 + \cdots + 59! \times 59 \\
=1! \times (2-1) + 2! \times (3-1) + \cdots + 59! \times (60-1) \\
=(2!-1!) + (3!-2!) + \cdots + (60!-59!) \\
=60!-1 .
\end{array}
$$
Therefore, the required remainder ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,958 |
5. If two integers $x, y$ satisfy the equation
$$
(2 x+9 y)^{2006}+(4 x-y)^{2006}=7777^{m} \text {, (1) }
$$
then the array $(x, y)$ is called an integer solution of equation (1). The number of integer solutions of equation (1) is ( ).
(A) 0
(B) 1
(C) 2
(D) 2006 | 5.A.
Assume there exist integers $x, y$ satisfying equation (1).
If $7 \mid (2x + 9y)$, then from equation (1) we know $7 \mid (4x - y)$. At this point, the left side of equation (1) can be divided by $7^{200}$, but the right side of equation (1) can only be divided by $7^m$.
Therefore, $7 \nmid (2x + 9y)$, $7 \nmid (... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,959 |
1. If $|a|=1,|b|=2, \boldsymbol{c}=a+b$, and $c \perp$ $a$, then the angle between vector $a$ and $b$ is degrees. | Given $c \perp a \Rightarrow c \cdot a=0 \Rightarrow(a+b) \cdot a=0$
$$
\Rightarrow|\boldsymbol{a}|^{2}+|\boldsymbol{a}| \cdot|\boldsymbol{b}| \cos \alpha=0 \Rightarrow \cos \alpha=-\frac{1}{2} \text {. }
$$
Therefore, the angle $\alpha$ between vector $a$ and $b$ is $120^{\circ}$. | 120^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,960 |
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$. Then
$$
f(1)+f(2)+\cdots+f(2006)=
$$
$\qquad$ | 2.0.
Given that the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$, we know that $f(x)=f(1-x)$.
Also, since $f(x)$ is an odd function defined on $\mathbf{R}$, it follows that $f(1-x)=-f(x-1)$.
Therefore, $f(x)+f(x-1)=0$.
Thus, $f(1)+f(2)+\cdots+f(2006)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,961 |
3. In $\triangle A B C$, it is known that
$$
\frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A},
$$
and $\cos (A-B)+\cos C=1-\cos 2 C$.
Then $\frac{a+c}{b}=$ | $$
\begin{array}{l}
\text { 3. } \frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}} . \\
\text { From } \cos (A-B)+\cos C=1-\cos 2 C \\
\Rightarrow \cos (A-B)-\cos (A+B)=2 \sin ^{2} C \\
\Rightarrow 2 \sin A \cdot \sin B=2 \sin ^{2} C \Rightarrow a b=c^{2} . \\
\text { From } \frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A} \R... | \frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,962 |
4. Let the increasing sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=6, n \in \mathbf{N}_{+}$, and when $n \geqslant 2$, $a_{n}+a_{n-1}=\frac{9}{a_{n}-a_{n-1}}+8$. Then $a_{x}=$ $\qquad$ | 4.29.
$$
\begin{array}{l}
\text { Given } a_{k}+a_{k-1}=\frac{9}{a_{k}-a_{k-1}}+8 \\
\Rightarrow a_{k}^{2}-a_{k-1}^{2}=8\left(a_{k}-a_{k-1}\right)+9 .
\end{array}
$$
Taking $k=2,3, \cdots, n$, summing up we get
$$
a_{n}^{2}-a_{1}^{2}=8\left(a_{n}-a_{1}\right)+9(n-1) \text {. }
$$
Thus, $a_{10}^{2}-6^{2}=8\left(a_{10}... | 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,963 |
5. Given $\triangle A B C$ with three sides $A B=\sqrt{34}, B C$ $=5 \sqrt{10}, C A=2 \sqrt{26}$. Then the area of $\triangle A B C$ is $\qquad$ | 5.10.
As shown in Figure 4, with $BC$ as the hypotenuse, construct a right triangle $\triangle BCD$ on one side of $\triangle ABC$ such that $\angle BDC = 90^\circ$, $BD = 5$, and $CD = 15$.
Then construct a rectangle $DEA'F$ such that $DE = 2$ and $DF = 5$. Thus, $BE = 3$ and $CF = 10$. At this point,
$$
\begin{arra... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,964 |
Example 6 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n+1}=\frac{a_{n}}{2}+\frac{1}{a_{n}} \text {. }
$$
Prove: $1<a_{n}<\frac{3}{2}+\frac{1}{n}$. | Notice that the recursive function of the sequence $\left\{a_{n}\right\}$ is $f(x)=\frac{x}{2}+\frac{1}{x}$, and by solving $f(x)=x$, we obtain the fixed points as $\pm \sqrt{2}$. Therefore, we can first find the general term formula and then use it to prove the inequality. Alternatively, we can directly prove the stre... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,965 |
Three, (10 points) In $\triangle A B C$, $r$ is the inradius, and the radii of the excircles opposite to sides $B C$, $A C$, and $A B$ are denoted as $r_{a}$, $r_{b}$, and $r_{c}$, respectively. Prove: $\frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{1}{r}$.
Note: A circle that is tangent to one side of a triang... | For simplicity, the solution to this problem uses $\triangle A B C$ to also represent its area.
As shown in Figure 5, connect
$A O_{a} 、 B O_{a} 、 C O_{a}$, then
$\triangle A O_{\mathrm{a}} B$
$=\frac{1}{2} r_{a} c$,
$\triangle A O_{a} C$
$=\frac{1}{2} r_{a} b$,
$\triangle B O_{a} C$
$=\frac{1}{2} r_{a} a$.
And $\tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,966 |
Four, (15 points) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers, for real numbers $b_{1}, b_{2}, \cdots, b_{n}$ and $c_{1}, c_{2}, \cdots, c_{n}$, it always holds that $a_{i} c_{i}>b_{i}^{2}(i=1,2, \cdots, n)$. Prove:
$$
\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(c_{1}+c_{2}+\cdots+... | Four, Construct $n$ quadratic functions
$$
f_{i}(x)=a_{i} x^{2}+2 b_{i} x+c_{i}(i=1,2, \cdots, n) \text {. }
$$
Since for each $i$, we have $a_{i}>0, a_{i} c_{i}>b_{i}^{2}$, i.e., $\Delta_{i}=$ $\left(2 b_{i}\right)^{2}-4 a_{i} c_{i} \leqslant 0$, so, $f_{i}(x) \geqslant 0$, that is
$$
f_{i}(x)=a_{i} x^{2}+2 b_{i} x+c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,967 |
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