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 |
|---|---|---|---|---|---|---|---|---|---|
4. Given that $x$ is a
four-digit number, the sum of its digits is $y$. If the value of $\frac{x}{y}$ is minimized, then $x=$ $\qquad$ | 4.1099.
Let $x=\overline{a_{1} a_{2} a_{3} a_{4}}$, where $a_{1} \in \mathbf{N}_{+}, a_{2}, a_{3}, a_{4} \in \mathbf{N}$. Therefore,
$$
\begin{array}{l}
\frac{x}{y}=\frac{1000 a_{1}+100 a_{2}+10 a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} . \\
=1+9 \times \frac{111 a_{1}+11 a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}+a_{4}} .
\end{array... | 1099 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,986 |
5. As shown in Figure 3, there are 16 points, and the distance between any two adjacent points, whether left and right or up and down, is equal to 1. If these points are used as the vertices of triangles, then, a total of right-angled triangles can be obtained. | 5.200.
As shown in Figure 6, the number of right-angled triangles with $A$ as the right-angle vertex is $C_{3}^{1} \cdot C_{3}^{1}=9$;
the number of right-angled triangles with $B$ as the right-angle vertex is
$$
C_{3}^{1} \cdot C_{3}^{1}+C_{2}^{1}+1=12 \text {; }
$$
the number of right-angled triangles with $C$ as t... | 200 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,987 |
6. Given point $A(2,2), P(x, y)$, and $x, y$ satisfy
$$
\left\{\begin{array}{l}
0<x, y \leqslant 2, \\
x+y \geqslant 2, \\
\frac{1}{x}+\frac{1}{y} \geqslant 2 .
\end{array}\right.
$$
Then the range of $|P A|$ is $\qquad$ | 6. $\left[\frac{\sqrt{7}}{2}, 2\right)$.
Since $x, y > 0$, simplifying $\frac{1}{x} + \frac{1}{y} \geqslant 2$ yields $\left(x - \frac{1}{2}\right)\left(y - \frac{1}{2}\right) \leqslant \frac{1}{4}$.
In the Cartesian coordinate system $x O y$, the plane region under the constraint conditions is the shaded part shown ... | \left[\frac{\sqrt{7}}{2}, 2\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,988 |
Three. (20 points) As shown in Figure 4, given that the edge length of the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ is $1, \odot O_{1}$ is the incircle of the square $A B C D$, $\odot O_{2}$ is the circumcircle of the square $A D D_{1} A_{1}$, and $P, Q$ are points on $\odot O_{1}$ and $\odot O_{2}$, respectively. Fin... | Three, establish the spatial rectangular coordinate system as shown in Figure 8.
In the $x 0 y$ plane,
the equation of $\odot O_{1}$ is
$$
\begin{array}{c}
\left(x-\frac{1}{2}\right)^{2}+ \\
\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}
\end{array}
$$
In the $x O_{z}$ plane, the equation of $\odot \mathrm{O}_{2}$ is
$$
... | \frac{\sqrt{3}-\sqrt{2}}{2} \leqslant|P Q| \leqslant \frac{\sqrt{3}+\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,989 |
Four. (20 points) Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and it satisfies $a_{2}=2, S_{n}=\frac{n\left(1+a_{n}\right)}{2}\left(n \in \mathbf{N}_{+}\right)$.
(1) Find the general term of the sequence $\left\{a_{n}\right\}$.
(2) If $b_{n}=a_{n}^{\frac{i}{a_{n}+1}}$, find... | (1) From $S_{1}=\frac{1+a_{1}}{2}=a_{1}$, we know $a_{1}=1$.
When $n>1$,
$$
a_{n}=S_{n}-S_{n-1}=\frac{n\left(1+a_{n}\right)}{2}-\frac{(n-1)\left(1+a_{n-1}\right)}{2} \text {. }
$$
Simplifying, we get $(n-2) a_{n}-(n-1) a_{n-1}+1=0$.
Substituting $n+1$ for $n$ gives
$$
(n-1) a_{n+1}-n a_{n}+1=0 \text {. }
$$
Subtracti... | b_{2}=b_{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,990 |
Five. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0$ ) with the left focus at $F$, a line $\ell$ passing through $F$ intersects the ellipse $C$ at points $A$ and $B$. $P$ is any point on the left directrix, and the direction vectors of lines $P A$, $P F$, and $P B$ are $(1, t)$, $... | (1) Let $l: x=m y-c$, and combine it with the equation of the ellipse $C$, eliminating $x$ and simplifying, we get
$$
\left(a^{2}+b^{2} m^{2}\right) y^{2}-2 b^{2} c m y-b^{4}=0 .
$$
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then
$$
y_{1}+y_{2}=\frac{2 b^{2} c m}{a^{2}+b^{2} m^{2}}, y_{1} y_{2}=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,991 |
一、(50 points) $\odot O$ is the incircle of $\triangle ABC$, $A'$, $B'$, and $C'$ are the points of tangency on sides $BC$, $CA$, and $AB$ respectively. It is known that the Euler line $l$ of $\triangle A'B'C'$ is parallel to $BC$. Prove that $l$ must pass through the circumcenter of $\triangle ABC$. | As shown in Figure 9, ray $AO$ intersects $BC$ and $B'C'$ at $E$ and $F$ respectively. Connect $OA'$, $OC'$, $A'F$, and let $A'F$ intersect $l$ at $G$. Construct the perpendicular bisector of side $BC$ to intersect $l$ at $Q$ and $BC$ at $D$, and connect $QB$. It is known that $BD = \frac{1}{2} BC$, $AO$ is the perpend... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,992 |
Example 1 Let $a$ be the least common multiple of $x$ and $y$, and $A$ be a common multiple of $x$ and $y$. Prove: $a \mid A$.
Analysis: Suppose $a$ does not divide $A$. Then, by the division algorithm of $A$ modulo $a$, we find that the remainder $r$ is also a common multiple of $x$ and $y$, leading to a contradictio... | Proof: Let $A=p a+r(p, r \in \mathbf{Z}, 0 \leqslant r<a)$.
Since $x|A, x| a$, it follows that $x \mid r$.
Similarly, $y \mid r$.
Thus, $r$ is a common multiple of $x$ and $y$.
If $r \neq 0$, then $0<r<a$, which contradicts $[x, y]=a$.
Therefore, $r=0$, i.e., $A=p a(p \in \mathbf{Z})$. Hence, $a \mid A$.
Note: To prove... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,994 |
Example 2: Prove that for each integer $x$, the value of the polynomial $\frac{1}{5} x^{5}+\frac{1}{3} x^{3}+\frac{7}{15} x$ is an integer (such a polynomial is called an integer-valued polynomial).
Analysis:
$$
\text { Let } \begin{aligned}
y & =\frac{1}{5} x^{5}+\frac{1}{3} x^{3}+\frac{7}{15} x \\
& =\frac{3 x^{5}+5... | Proof: First, prove $31\left(2 x^{3}+x\right)$.
When $x=3 k$, $3 \mid\left(2 x^{3}+x\right)$;
When $x=3 k+1$,
$$
2 x^{3}+x \equiv 2 \times 1^{3}+1=3 \equiv 0(\bmod 3) ;
$$
When $x=3 k-1$,
$$
2 x^{3}+x \equiv 2 \times(-1)^{3}-1=-3 \equiv 0(\bmod 3) ;
$$
Therefore, for any integer $x$, $3 \mid\left(2 x^{3}+x\right)$.
T... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,995 |
1. If $p, p+2$ are both primes greater than 3, prove: $6 \mid (p+1)$.
(5th Canadian Mathematics Competition) | (Tip: Consider the representation with remainder modulo 6, where numbers of the form $6n-3$, $6n-2$, $6n$, $6n+2$ are composite. If $p$ is a prime greater than 3, then
$p=6n+1$ or $6n-1$.
If $p=6n+1$, then,
$$
p+2=6n+3=3(2n+1)
$$
is composite. This is a contradiction, so $p=6n-1$. Therefore, $61(p+1)$. ) | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,996 |
1. The total number of integer solutions to the equation $5 x^{2}-6 x y+7 y^{2}=130$ is ( ) groups.
(A) 0
(B) 1
(C) 2
(D) 3 | -.1.C.
Given the equation can be rearranged as
$$
5 x^{2}-6 x y+\left(7 y^{2}-130\right)=0 \text {. }
$$
From $\Delta>0$, we get $y^{2} \leqslant 25$.
Since $y$ is an integer, therefore, $-5 \leqslant y \leqslant 5$.
Upon inspection, $(3,5),(-3,-5)$ satisfy the conditions. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,997 |
2. The function $f(x)$ is defined on $\mathbf{R}$, and satisfies:
(1) $f(x)$ is an even function;
(2) $f(x-1)$ is an odd function.
Then $f(2009)$ is ( ).
(A) 0
(B) 2008
(C) 2009
(D) -2008 | 2. A.
Since $f(x-1)$ is an odd function, we have $f(-1)=0$.
Also, since $f(x)$ is an even function, we get
$$
\begin{array}{l}
f(1)=0, f(-x)=f(x), \\
f(-x-1)=-f(x-1) .
\end{array}
$$
Then $f(x+4)=f(-x-4)=-f(x+2)$
$$
=-f(-x-2)=f(x) \text {. }
$$
Therefore, the period of $f(x)$ is $4$, and $f(2009)=f(1)=0$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,998 |
3. Given $x^{2}+y^{2}=25$. Then the maximum value of the function
$$
z=\sqrt{8 y-6 x+50}+\sqrt{8 y+6 x+50}
$$
is ( ).
(A) $5 \sqrt{10}$
(B) $6 \sqrt{10}$
(C) $7 \sqrt{10}$
(D) $8 \sqrt{10}$ | 3. B.
Since $x^{2}+y^{2}=25$, we have
$$
\begin{array}{l}
z=\sqrt{8 y-6 x+50}+\sqrt{8 y+6 x+50} \\
=\sqrt{(x-3)^{2}+(y+4)^{2}}+\sqrt{(x+3)^{2}+(y+4)^{2}}
\end{array}
$$
The geometric meaning is the sum of the distances from a point on the circle $x^{2}+y^{2}=25$ to points $A(3,-4)$ and $B(-3,-4)$. The maximum value i... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,999 |
4. The directrix of the parabola $y=-\frac{1}{8} x^{2}$ intersects the $y$-axis at point $A$. A line through $A$ intersects the parabola at points $M$ and $N$. Point $B$ is on the axis of symmetry of the parabola, and $\left(\overrightarrow{B M}+\frac{\overrightarrow{M N}}{2}\right) \perp \overrightarrow{M N}$. Then th... | 4.D.
Notice the point $A(0,2)$. Draw a line $M N$ through $A$, with its equation set as $y=k x+2$. Substituting into the parabola equation yields
$$
x^{2}+8 k x+16=0 \text {. }
$$
And $\Delta>0$, then $k^{2}>1, x_{1}+x_{2}=-8 k$.
Let point $B(0, b)$, and the midpoint of $M N$ be
$$
\begin{array}{l}
C\left(-4 k,-4 k^{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,000 |
5. Given $f(x)=\sum_{i=1}^{2009}|x+i|+\sum_{i=1}^{2009}|x-i|$ $(x \in \mathbf{R})$, and $f\left(a^{2}-3 a+2\right)=f(a-1)$. Then the number of values of $a$ is ( ).
(A) 2
$(B) 3$
(C) 4
(D) infinitely many | 5.D.
From the problem, we know that $f(x)$ is an even function.
Considering when $-1 \leqslant x \leqslant 1$, it always holds that
$$
\begin{array}{l}
f(x)=2(1+2+\cdots+2009) \\
=2010 \times 2009 .
\end{array}
$$
Therefore, when $-1 \leqslant a^{2}-3 a+2 \leqslant 1$ and $-1 \leqslant a-1 \leqslant 1$, it always hol... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,001 |
6. Let $a, b, c > 0$, and $a+b+c=1$. Then the maximum value of the real number $\lambda$ such that $a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant 1$ always holds is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\sqrt{3}$
(D) $2 \sqrt{3}$ | 6.D.
Let $a, b, c > 0$, and $a+b+c=1$. Then
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+\lambda \sqrt{a b c} \leqslant 1 \\
\Leftrightarrow \lambda \leqslant 2\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{a c}{b}}+\sqrt{\frac{b c}{a}}\right). \\
\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{a c}{b}}+\sqrt{\frac{b c}{a}}\right)^{2}=\frac{... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 721,002 |
1. Let $m, n \in \mathbf{N}$, and $m>n$, sets $A=\{1,2, \cdots, m\}, B=\{1,2, \cdots, n\}$,
and $C \subset A$. Then the number of $C$ such that $B \cap C \neq \varnothing$ is
$\qquad$ . | $=.1 .2^{m-n}\left(2^{n}-1\right)$
Set $A=B \bigcup\{n+1, n+2, \cdots, m\}$.
Since $B \cap C \neq \varnothing$, and $C \subset A$, thus, the elements of $C$ partly come from the non-empty subsets of $B$, and $B$ has $2^{n}-1$ non-empty subsets; the other part comes from the set $\{n+1, n+2, \cdots, m\}$, and the number... | 2^{m-n}\left(2^{n}-1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,003 |
2. Given $f(x)=x^{2}+2 x+1$, there exists a real number $t$ such that when $x \in[1, m]$, $f(x+t) \leqslant x$ always holds, then the maximum value of $m$ is $\qquad$ . | 2.4.
Translate the graph of $f(x)$ to the right by $-t$ units, and from the graphical analysis, the maximum value of $m$ is the larger of the x-coordinates of the two intersection points of $\left\{\begin{array}{l}y=x, \\ y=f(x+t)\end{array}\right.$
From $f(1+t)=1$, we get $t=-1, t=-3$.
Then, from $f(x-3)=x$, we have
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,004 |
3. In a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $X$ and $Y$ are the centers of the squares $A A_{1} B B_{1}$ and $B B_{1} C_{1} C$, respectively, and $Z$ is on the diagonal $B D$ such that $D Z=3 Z B$. Then the area of the section through $X$, $Y$, and $Z$ is $\qquad$ | 3. $\frac{\sqrt{2}}{2} a^{2}$.
According to the problem, the section through $X, Y, Z$ is a rectangle, with a length equal to the height of the cube $a$, and a width equal to the length of $XY$, which is $\frac{\sqrt{2}}{2} a$. Therefore, the area of the section through $X, Y, Z$ is $\frac{\sqrt{2}}{2} a^{2}$. | \frac{\sqrt{2}}{2} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,005 |
4. Connect the right focus $F$ of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ with a moving point $A$ on the ellipse, and construct a square $F A B C (F, A, B, C$ in clockwise order). When point $A$ moves around the ellipse once, the trajectory equation of the moving point $C$ is $\qquad$ | 4. $\frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9}=1$.
Let $A(3 \cos \alpha, 2 \sin \alpha)(\alpha \in[0,2 \pi])$,
$$
\begin{aligned}
C(x, y) 、 F(\sqrt{5}, 0), \\
\overrightarrow{F A}=(3 \cos \alpha-\sqrt{5}, 2 \sin \alpha) . \\
\text { Then } \overrightarrow{F C}=(2 \sin \alpha,-3 \cos \alpha+\sqrt{5}) . \\
\t... | \frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,006 |
2. Let $n$ be a positive integer. If
$$
A(n)=1986^{n}+1987^{n}+1988^{n}+1989^{n}
$$
is not divisible by 5, find $n$. | (Tip: Only consider the unit digit $G(A(n))$ of $A(n)$. When $n=4 k\left(k \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{4}+7^{4}+8^{4}+9^{4}\right) \\
=G(6+1+6+1)=4,
\end{array}
$$
at this time, 5 does not divide $A(n)$;
$$
\begin{array}{l}
\text { When } ... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,007 |
5. Given $x, y, z \in \mathbf{R}_{+}$, and $\sqrt{x^{2}+y^{2}}+z=1$. Then the maximum value of $x y+2 x z$ is $\qquad$ . | 5. $\frac{\sqrt{3}}{3}$.
From $x^{2}+y^{2}=(1-z)^{2}$, we know
$$
x^{2}=(1-z+y)(1-z-y) \text {. }
$$
Therefore, $x^{2}(y+2 z)^{2}$
$$
\begin{array}{l}
=\frac{1}{3}(1-z+y)(3-3 z-3 y)(y+2 z)(y+2 z) \\
\leqslant \frac{1}{3}\left[\frac{(1-z+y)+(3-3 z-3 y)+(y+2 z)+(y+2 z)}{4}\right]^{4} \\
=\frac{1}{3} .
\end{array}
$$
T... | \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,008 |
6. Given six points in the plane (the lines connecting these points are neither parallel nor perpendicular to each other, nor coincident), through each point draw a perpendicular to the line connecting any two of the remaining five points. Then the maximum number of intersection points of these perpendiculars is $\qqua... | 6.1370 .
(1) Six points form $C_{\hbar}^{2}=15$ lines.
(2) Any three points form a triangle, totaling $\mathrm{C}_{0}^{3}=$ 20.
(3) Any five points form $\mathrm{C}_{5}^{2}=10$ lines; from one point, draw perpendiculars to these 10 lines, resulting in 60 perpendiculars. These perpendiculars can have at most $\mathrm{C}... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,009 |
$$
\begin{array}{l}
\text { Three. (20 points) (1) The quadratic function } \\
f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)
\end{array}
$$
satisfies
(i) For $x \in \mathbf{R}$, there is $4 x \leqslant f(x) \leqslant \frac{1}{2}(x+2)^{2}$
always holds;
(ii) $f(-4+2 \sqrt{3})=0$.
Find $f(x)$.
(2) Let $f_{1}(x)=... | Three, (1) Let $x=2$, we get $8 \leqslant f(2) \leqslant 8$.
Thus, $f(2)=8$,
$4 a+2 b+c=8 \Rightarrow b=\frac{8-4 a-c}{2}$.
Also, $4 x \leqslant f(x)$, so $a x^{2}+(b-4) x+c \geqslant 0$
always holds. Then
$$
\begin{array}{l}
\Delta=(b-4)^{2}-4 a c \\
=\left(\frac{8-4 a-c}{2}-4\right)^{2}-4 a c=\left(2 a-\frac{c}{2}\ri... | f_{2009}(0)=\frac{3^{2010}+3}{3^{2010}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,010 |
Four. (20 points) From a point $M$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two tangents are drawn to the circle with the minor axis as its diameter, with points of tangency $A$ and $B$. The line $AB$ intersects the $x$-axis and $y$-axis at points $P$ and $Q$, respectively. Find the minimum value of $|PQ|$.
... | Let point $M(3 \cos \alpha, 2 \sin \alpha)$, and the points of tangency be $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. The equations of the tangent lines $M A$ and $M B$ are
$$
\begin{array}{l}
x_{1} \cdot 3 \cos \alpha + y_{1} \cdot 2 \sin \alpha = 4, \\
x_{2} \cdot 3 \cos \alpha + y_{2} \cdot 2 \si... | \frac{10}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,011 |
Five. (20 points) Given $a, b > 0$. Find the maximum values of
$$
\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{2 b+a}}, \sqrt{\frac{a}{a+2 b}}+\sqrt{\frac{b}{b+2 a}}
$$ | $$
\begin{array}{l}
\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{2 b+a}} \\
\leqslant \sqrt{2\left(\frac{a}{2 a+b}+\frac{b}{2 b+a}\right)} .
\end{array}
$$
Let $a+b=1$.
Then the right side of equation (1) $=\sqrt{2\left(\frac{a}{1+a}+\frac{b}{1+b}\right)}$
$$
=\sqrt{4-2\left(\frac{1}{1+a}+\frac{1}{1+b}\right)} \text {. }
$$
... | \frac{2 \sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,012 |
一、(50 points) As shown in Figure 1, given an isosceles $\triangle ABC$, $CA=CB$, a circle is drawn through its centroid $G$ and vertex $A$, tangent to $BG$ at point $G$. Extend $CG$ to intersect this circle at point $D$, and $AB$ intersects this circle at point $E$. Prove:
$$
BE \cdot BA = GC \cdot GD \text{. }
$$ | In the isosceles $\triangle ABC$, since $CA = CB$, we have $BG = AG$.
By the secant-tangent theorem, we know $BG^2 = BE \cdot BA$, which means
$$
AG^2 = BE \cdot BA.
$$
Extend $AG$ to point $F$ such that $AG = GF$, then quadrilateral $BF CG$ is a parallelogram. Therefore, $\angle FCG = \angle BGD$.
By the inscribed an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,013 |
II. (50 points) On an infinite grid paper, some cells are colored red, and the rest are colored blue. In every $2 \times 3$ rectangle of six cells, there are exactly two red cells. How many red cells are there in a $9 \times 11$ rectangle of 99 cells? | II. 33 Red Squares.
As shown in Figure 3, take any red square $K_{0}$ as the center of a $3 \times 3$ square; it is not allowed to color $K$ red. If $K$ is colored red, then in the $2 \times 3$ rectangles $A F H D$, $A B S T$, and $M N C D$, there will be two red squares each. To ensure that the shape $B C G E$ contain... | 33 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,014 |
Three. (50 points) Let $a_{1}, a_{2}, \cdots, a_{2008}$ be given positive integers, and $a_{1}<a_{2}<\cdots<a_{2008}<10^{100}$. Prove: the set $\left\{a_{1}, a_{2}, \cdots, a_{2008}\right\}$ contains two non-empty and disjoint subsets $A$ and $B$, such that
(1) $A$ and $B$ have the same number of elements;
(2) the sum ... | Three, let the non-empty set $E \subseteq \{a_1, a_2, \cdots, a_{200}\}$.
Construct the mapping $f: E \rightarrow (E^0, E^1, E^2)$, where the number of non-empty subsets (pre-images) in the set $\{a_1, a_2, \cdots, a_{208}\}$ is $2^{208}-1$. Here, $E^i (i=0,1,2)$ represents the sum of the $i$-th powers of the elements ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,015 |
As shown in Figure 2, in the acute triangle $\triangle ABC$, $AC > BC$, and $CD$ is the median on side $AB$. Points $E$, $F$, and $G$ are points on segments $CA$, $CB$, and $CD$ respectively, distinct from the endpoints of the segments, and points $C$, $E$, $G$, and $F$ are concyclic. Let the circumcenter and circumrad... | Proof: Let $\angle O_{1} D G=\alpha, \angle O_{2} D G=\beta$.
By the cosine rule, we have
$$
\begin{array}{l}
O_{1} G^{2}=D G^{2}+D O_{1}^{2}-2 D G \cdot D O_{1} \cos \alpha, \\
O_{2} G^{2}=D G^{2}+D O_{2}^{2}-2 D G \cdot D O_{2} \cos \beta .
\end{array}
$$
Thus, to prove $G O_{1}^{2}+G O_{2}^{2}=r_{1}^{2}+r_{2}^{2}$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,016 |
Let $k$ be an odd number. Can the quartic polynomial with integer coefficients $x^{4}+8 x^{2}+2008 x+2 k$ be expressed as the product of two polynomials with integer coefficients, both of degree greater than 0? Please explain your reasoning. | Solution: The answer is negative.
The reason is as follows: Suppose $x^{4}+8 x^{2}+2008 x+2 k$ can be factored into the product of two integer-coefficient polynomials with degrees greater than 0.
We will discuss this in two cases.
(1) If there exist four integers $a, b, c, d$, such that
$$
\begin{array}{l}
x^{4}+8 x^{2... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,017 |
3. There are 10 positive integers arranged from smallest to largest: $1, 4, 8$, $10, 16, 19, 21, 25, 30, 43$. How many groups of consecutive numbers have a sum that is divisible by 11? | Let 10 positive integers be $a_{1}, a_{2}$, $\cdots, a_{10}$, and let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$$
a_{i+1}+a_{i+2}+\cdots+a_{j}=S_{j}-S_{i} .
$$
The sequence $a_{n}(n=1,2, \cdots, 10)$ has remainders modulo 11 of $1,4,-3,-1,5,-3,-1,3,-3,-1$; the sum of the first $n$ terms $S_{n}$ has remainders modulo 11 ... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,018 |
Given $a, b, c$ are positive numbers satisfying $abc=1$. Prove:
$$
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{1+a+b+c} \geqslant 1 .
$$ | Proof: Since among $1-a, 1-b, 1-c$ there must be two that are either both non-negative or both non-positive, we can assume without loss of generality that
$(1-a)(1-b) \geqslant 0$.
From $(a+b)(1+a b)$
$=b(1+a)^{2}+a(1-b)^{2} \geqslant b(1+a)^{2}$,
we get $\frac{1}{(1+a)^{2}} \geqslant \frac{b}{(a+b)(1+a b)}$.
Similarl... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,019 |
As shown in Figure 3, the convex quadrilateral $ABCD$ is inscribed in $\odot O$. The lines containing the two pairs of opposite sides intersect at points $E$ and $F$, respectively. The diagonals $AC$ and $BD$ intersect at point $G$. Draw $GH \perp EF$ at point $H$, and the chord $MN$ of $\odot O$ passes through point $... | Proof: As shown in Figure 3. Let the intersection point of ray $OC$ and $\odot O$ be $I$. On ray $GI$, take point $H'$ such that
$$
OG \cdot GH' = AG \cdot GC = BG \cdot GD = MG \cdot GN.
$$
As shown in Figure 5, connect the corresponding line segments.
By the intersecting chords theorem, it is easy to prove that $A, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,020 |
6. Prove: There are no integers $x, y$, such that $x^{2}+y^{2}=1995$.
| (Suppose there are integers $x, y$, such that
$$
x^{2}+y^{2}=1995 \text {. }
$$
When $x=2 m, y=2 n$,
$$
1995=x^{2}+y^{2}=4\left(m^{2}+n^{2}\right) \text {, }
$$
so, 4|1995, contradiction;
When $x=2 m-1, y=2 n$,
$$
1995=x^{2}+y^{2}=1+4\left(m^{2}-m+n^{2}\right) \text {, }
$$
so, 4|1994, contradiction;
When $x=2 m, y=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,022 |
7. If $p$ and $q$ are both prime numbers, and $7p + q$ and $pq + 11$ are also prime numbers, find the value of $p^q + q^p$.
(1997, Jingzhou City, Hubei Province, Junior High School Mathematics Competition) | (Tip: Since $7 p+q$ is a prime number, and $7 p+q>7$, therefore, $7 p+q$ is odd. Thus, $p, q$ do not share the same parity.
If $p$ is even, then $p=2$. According to the problem, $14+$ $q, 2 q+11$ are both prime numbers. Therefore,
$$
14+q \neq \equiv 0(\bmod 3), 2 q+11 \not \equiv 0(\bmod 3),
$$
which means $q \neq 1... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,023 |
Example 1 Given $a, b, c, d > 0$, and $a+b+c+d=1$. Prove: $\sum \frac{1}{1+a^{3}} \leqslant \frac{256}{65}$ (Note “$\sum$” represents cyclic summation, the same applies hereinafter $)^{[1]}$. | Proof: Let $f(x)=\frac{1}{1+x^{3}}(0<x<1)$. Then
$$
\begin{array}{l}
f\left(\frac{1}{4}\right)=\frac{64}{65}, f^{\prime}(x)=-\frac{3 x^{2}}{\left(1+x^{3}\right)^{2}}, \\
f^{\prime}\left(\frac{1}{4}\right)=-\frac{768}{4225} .
\end{array}
$$
Thus, the equation of the tangent line to $f(x)$ at $x=\frac{1}{4}$ is
$$
g(x)=... | \frac{256}{65} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,024 |
Example 2 Given $x_{i}>0(i=1,2,3,4,5)$, and $\sum_{i=1}^{5} \frac{1}{1+x_{i}}=1$. Prove: $\sum_{i=1}^{3} \frac{x_{i}}{4+x_{i}^{2}} \leqslant 1$.
(2003, China Western Mathematical Olympiad) | Analysis: The condition does not meet the requirements.
Let $\frac{1}{1+x_{i}}=a_{i}(i=1,2,3,4,5)$. Then the original condition
becomes $\sum_{i=1}^{5} a_{i}=1$, which now meets the requirements.
Proof: Let $\frac{1}{1+x_{i}}=a_{i}(i=1,2,3,4,5)$. Then
$a_{i}>0, \sum_{i=1}^{5} a_{i}=1$,
$\frac{x_{i}}{4+x_{i}^{2}}=\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,025 |
Example 3 Given $a, b, c > 0$, and $a^{4} + b^{4} + c^{4} = 3$. Prove: $\sum \frac{1}{4 - ab} \leqslant 1$.
(2005, Moldova Selection Competition) | Analysis: From $a^{4}, b^{4}, c^{4}$ to $a b, b c, c a$, we think of the inequality $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Proof: Notice
$$
\begin{array}{l}
3=a^{4}+b^{4}+c^{4}=\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+\left(c^{2}\right)^{2} \\
\geqslant a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} .
\end{array}
$$
Let... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,026 |
Example 4 Let $x, y, z > 0, x^{4} + y^{4} + z^{4} = 1$. Find the minimum value of $\sum \frac{x^{3}}{1 - x^{8}}$.
(2000, Jiangsu Province High School Mathematics Competition) | Analysis: Let $x^{4}=a, y^{4}=b, z^{4}=c$, and the expression $\sum \frac{x^{3}}{1-x^{8}}=\sum \frac{\left(x^{4}\right)^{\frac{3}{4}}}{1-\left(x^{4}\right)^{2}}$ must be transformed into an expression in terms of $a, b, c$.
Solution: Let $x^{4}=a, y^{4}=b, z^{4}=c$. Then $a, b, c > 0, a+b+c=1$,
$$
\sum \frac{x^{3}}{1-x... | \frac{9 \sqrt[4]{3}}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,027 |
Example 3 Let integers $a, b, a-b$ not be multiples of 3. Prove: $a^{3}+b^{3}$ is a multiple of 9.
Analysis: From the condition, $a, b$ can only be in the form of $3 n \pm 1$. Since $a-b$ is not a multiple of 3, $a, b$ cannot both be in the form of $3 n+1$ or $3 n-1$, thus, $a, b$ can only be in the forms of $3 n+1, 3... | Proof: According to the problem, without loss of generality, let
$$
a=3 m+1, b=3 n-1 \text {. }
$$
Then $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$$
\begin{aligned}
= & 3(m+n)\left[(3 m+1)^{2}+(3 n-1)^{2}-\right. \\
& (3 m+1)(3 n-1)] \\
= & 3(m+n)\left(9 m^{2}+6 m+1+ \\
& 9 n^{2}-6 n+1-9 m n+3 m-3 n+1\right) \\
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,028 |
Example 5 Given $x, y, z > 0$. Prove:
$$
\frac{x^{3}}{y(x+y)^{2}}+\frac{y^{3}}{z(y+z)^{2}}+\frac{z^{3}}{x(z+x)^{2}} \geqslant \frac{3}{4} \text {. }
$$
Analysis: Since each term on the left side of the inequality can only contain one variable, the left side must be transformed, simultaneously producing the conditions ... | Prove: The original inequality can be transformed into
$$
\frac{\left(\frac{x}{y}\right)^{3}}{\left(\frac{x}{y}+1\right)^{2}}+\frac{\left(\frac{y}{z}\right)^{3}}{\left(\frac{y}{z}+1\right)^{2}}+\frac{\left(\frac{z}{x}\right)^{3}}{\left(\frac{z}{x}+1\right)^{2}} \geqslant \frac{3}{4} .
$$
Let $\frac{x}{y}=a, \frac{y}{z... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,029 |
Example 6 Given $a, b, c, d > 0$. Prove:
$$
\sum \frac{a^{3}}{(a+b)(a+c)(a+d)} \geqslant \frac{1}{2} \text {. }
$$
Analysis: From the product in the denominator
$$
(a+b)(a+c)(a+d)
$$
we think of the AM-GM inequality
$$
\sqrt[3]{x y z} \leqslant \frac{x+y+z}{3} .
$$ | Proof: Since the inequality is a homogeneous inequality, we can assume \(a+b+c+d=1\). Notice that
\[
\begin{array}{l}
(a+b)(a+c)(a+d) \\
\leqslant\left[\frac{(a+b)+(a+c)+(a+d)}{3}\right]^{3} \\
=\left(a+\frac{b+c+d}{3}\right)^{3}=\left(a+\frac{1-a}{3}\right)^{3} \\
=\frac{1}{27}(2 a+1)^{3} .
\end{array}
\]
\[
\begin{ar... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,030 |
Example 7 Given $a, b, c > 0$. Prove:
$$
\sum \frac{a}{2 \sqrt{b^{2}-b c+c^{2}}+3 a} \leqslant \frac{3}{5} \text {. }
$$
Analysis: When $b=c$,
$$
\sqrt{b^{2}-b c+c^{2}}=\sqrt{b^{2}}=b=\frac{b+c}{2} \text {, }
$$
From this, we conjecture that $2 \sqrt{b^{2}-b c+c^{2}} \geqslant b+c$. | Proof: Notice
$$
\begin{array}{l}
2 \sqrt{b^{2}-b c+c^{2}}=\sqrt{4\left(b^{2}+c^{2}\right)-4 b c} \\
1 \geqslant \sqrt{\left(b^{2}+c^{2}\right)+3 \times 2 b c-4 b c}=\sqrt{(b+c)^{2}} .
\end{array}
$$
Then $2 \sqrt{b^{2}-b c+c^{2}} \geqslant b+c$
$$
\Rightarrow \sum \frac{a}{2 \sqrt{b^{2}-b c+c^{2}}+3 a} \leqslant \sum... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,031 |
Example 8 Given $x, y, z>-\frac{1}{3}$, and $x+y+z$ =3. Prove: $\sum \frac{1}{\sqrt{3 x+1}+\sqrt{3 y+1}} \geqslant \frac{3}{4}$.
Analysis: $x, y$ are respectively in two radicals $\sqrt{3 x+1}$, $\sqrt{3 y+1}$, which leads to the inequality
$$
\sqrt{a}+\sqrt{b} \leqslant \sqrt{2(a+b)},
$$
This can make $x, y$ in the ... | Proof: By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sqrt{3 x+1}+\sqrt{3 y+1} \\
\leqslant \sqrt{\left(1^{2}+1^{2}\right)(3 x+1+3 y+1)} \\
=\sqrt{2} \cdot \sqrt{3(x+y)+2} \\
=\sqrt{2} \cdot \sqrt{3(3-z)+2} \\
=\sqrt{2} \cdot \sqrt{11-3 z} .
\end{array}
$$
Thus,
$$
\sum \frac{1}{\sqrt{3 x+1}+\sqrt{3 y+1}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,032 |
Example 9 Let positive numbers $x, y, z$ satisfy $x y z \geqslant 1$. Prove:
$$
\sum \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant 0 \text {. }
$$
(46th IMO) | Analysis: The degrees of $x, y, z$ are different. According to the condition $x y z \geqslant 1$, we can multiply $x^{2}, y^{2}, z^{2}$ by $x y z$ to make the degrees of $x, y, z$ all 5.
$$
\begin{array}{l}
\text { Proof: } \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} \cdot x y z}{x^{5}+\left(y^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,033 |
Example 10 Given that $a, b, c$ are the side lengths of a triangle, prove:
$$
\sum \frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}} \geqslant \frac{3(\sqrt{a}+\sqrt{b}+\sqrt{c})}{a+b+c} \text {. }
$$ | Analysis: The inequality has a side without a constant, and both sides must be multiplied by $\frac{a+b+c}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$ to make the right side a constant. This leads to the consideration of the difference of squares formula from the smaller part $(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{a}+\sqrt{b}+\sqrt{c})$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,034 |
Example 1 Let $a_{n}$ be the number of natural numbers $N$ such that the sum of the digits of $N$ is $n$ and each digit can only be $1$, $3$, or $4$. Prove that $a_{2n}$ ($n=1,2, \cdots$) is a perfect square.
(1991, National High School Mathematics Competition) | Proof: Note that the first digit of $n$ can only be 1, 3, or 4. If we remove the first digit, then we have the recurrence relation
$$
a_{n}=a_{n-1}+a_{n-3}+a_{n-4} \text {. }
$$
It is easy to see that $a_{1}=1, a_{2}=1, a_{3}=2, a_{4}=4$. The characteristic equation of the sequence $\left\{a_{n}\right\}$ is $x^{4}-x^{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,036 |
As shown in Figure 1, let point $P$ outside the line segment $A C B$ form angles $\alpha, \beta$ with the segments $A C, C B$ respectively. Then
$$
\begin{array}{l}
\frac{\sin (\alpha+\beta)}{P C} \\
=\frac{\sin \alpha}{P B}+\frac{\sin \beta}{P A} .
\end{array}
$$ | Proof: Since $S_{\triangle P A B}=S_{\triangle P M C}+S_{\triangle P C B}$, therefore,
$$
\begin{array}{l}
\frac{1}{2} P A \cdot P B \sin (\alpha+\beta) \\
=\frac{1}{2} P A \cdot P C \sin \alpha+\frac{1}{2} P C \cdot P B \sin \beta .
\end{array}
$$
Dividing both sides by $\frac{1}{2} P A \cdot P B \cdot P C$, we get t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,038 |
Example 1 Given that $\triangle A B C$ is an inscribed equilateral triangle, $P$ is any point on $\overparen{B C}$, and $N$ is the intersection of $P A$ and $B C$. Prove: $\frac{1}{P B}+\frac{1}{P C}=\frac{1}{P N}$. | Proof: As shown in Figure 2, since $\alpha=\beta=60^{\circ}$, applying the Angle Subtended Theorem from point $P$ to points $B$, $N$, and $C$ yields
$$
\begin{array}{l}
\frac{\sin 120^{\circ}}{P N} . \\
=\frac{\sin 60^{\circ}}{P B}+\frac{\sin 60^{\circ}}{P C} .
\end{array}
$$
Since $\sin 60^{\circ}=\sin 120^{\circ}$, ... | \frac{1}{P B}+\frac{1}{P C}=\frac{1}{P N} | Geometry | proof | Yes | Yes | cn_contest | false | 721,040 |
Example 2 Given that the extensions of two pairs of opposite sides of quadrilateral $ABCD$ intersect at points $K$ and $L$, a line is drawn through $K$ and $L$, and the diagonals $AC$ and $BD$ intersect line $KL$ at points $G$ and $F$ respectively. Prove that $\frac{1}{KF}, \frac{1}{KL}, \frac{1}{KG}$ form an arithmeti... | Given:
$$
\begin{array}{l}
K A=a, \\
K B=b, \\
K C=c, \\
K D=d, \\
K F=f, \\
K C=g, K L=l .
\end{array}
$$
From the perspective of point $K$, applying the Angle Subtended Theorem to points $A, B, L, D, C, L, D, B, F$, and $A, C, G$ respectively, we get:
$$
\begin{array}{l}
\frac{\sin (\alpha+\beta)}{b}=\frac{\sin \al... | \frac{1}{K F}, \frac{1}{K L}, \frac{1}{K G} \text{ form an arithmetic sequence} | Geometry | proof | Yes | Yes | cn_contest | false | 721,041 |
Example 3 In $\triangle A B C$, the angle bisector of $\angle A$ is $A D$. Prove: $A D^{2}=A B \cdot A C-B D \cdot D C$. | Prove: As shown in Figure 4, taking point $A$ as the viewpoint and applying the Angle Subtended Theorem to points $B, D, C$, we get
$$
\begin{array}{l}
\frac{\sin 2 \alpha}{A D} \\
=\frac{\sin \alpha}{b}+\frac{\sin \alpha}{c} .
\end{array}
$$
Thus, $A D=\frac{2 b c \cos \alpha}{b+c}$.
In $\triangle B D A$ and $\triang... | A D^{2}=A B \cdot A C-B D \cdot D C | Geometry | proof | Yes | Yes | cn_contest | false | 721,042 |
Example 4 In a convex quadrilateral $ABCD$, diagonals $AC$ and $BD$ are perpendicular to each other. Through the intersection point $O$ of $AC$ and $BD$, draw any two lines intersecting $AD$, $BC$, $AB$, and $CD$ at points $E$, $F$, $G$, and $H$ respectively. $GF$ and $EH$ intersect $BD$ at points $I$ and $J$. Prove:
$... | Proof: As shown in Figure 5, from the given conditions, we have
$$
\begin{array}{l}
\angle A O G \\
=\angle C O H \\
=90^{\circ}-\alpha, \\
\angle C O F \\
=\angle A O E \\
=90^{\circ}-\beta .
\end{array}
$$
Taking $O$ as the viewpoint, applying the Angle Subtended Theorem to points $A, G, B$, we get
$$
\frac{\sin 90^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,043 |
Example 5 Given in $\square A B C D$, $E, F$ are points on sides $A B, A D$ respectively, and $E F$ intersects diagonal $A C$ at point $P$. If $\frac{A E}{E B}=\frac{a}{b}, \frac{A F}{F D}=\frac{m}{n}\left(a, b, m, n \in \mathbf{R}_{+}\right)$, then the value of $\frac{A P}{P C}$ is $(\quad$.
(A) $\frac{a m}{a m+b m}$
... | Solution: As shown in Figure 6,
connect $B D$ and $A C$
intersecting at point $O$.
Taking $A$ as the viewpoint, apply the Angle Subtended Theorem to points $E$,
$P$, $F$ and $B$, $O$, $D$ respectively, we get
$$
\begin{array}{l}
\frac{\sin (\alpha+\beta)}{A P}=\frac{\sin \alpha}{A E}+\frac{\sin \beta}{A F}, \... | \frac{a m}{a m+a n+b m} | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,044 |
Example 6 Given $B P: P Q: Q C=1: 2: 1, C G: A G$ $=1: 2$. Then $B E: E F: F G=(\quad$.
(A) $12: 17: 7$
(B) 11:16:6
(C) $10: 15: 6$
(D)9:14:5
(15th Wuyang Cup Junior High School Mathematics Competition (Grade 9)) | Solution: As shown in Figure 7, let
$$
\begin{array}{l}
\angle C B G=a, \angle G B A \\
=\beta, B P=Q C=x, \\
P Q=2 x, B A=y .
\end{array}
$$
From the perspective of point $B$,
applying the Angle Bisector Theorem to points $P, E, A$, $Q, F, A$, and $C, G, A$ respectively, we get
$$
\begin{array}{l}
\frac{\sin (\alpha+... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,045 |
Let the notation $|X|$ represent the number of elements in the set $X$. Suppose $|X|=n\left(n \in \mathbf{N}_{+}\right)$, and all subsets of $X$ are $\left\{A_{1}, A_{2}, \cdots, A_{m}\right\}\left(m=2^{n}\right)$. It is known that the sum of the number of elements in the intersections of these subsets is
$$
S=\sum_{i=... | Consider the following $m \times m$ matrix
$$
\left(\begin{array}{cccc}
\left|A_{1} \cap A_{1}\right| & \left|A_{1} \cap A_{2}\right| & \cdots & \left|A_{1} \cap A_{m}\right| \\
\left|A_{2} \cap A_{1}\right| & \left|A_{2} \cap A_{2}\right| & \cdots & \left|A_{2} \cap A_{m}\right| \\
\vdots & \vdots & \vdots & \vdots \\... | \frac{n}{2}\left(4^{n-1}-2^{n-1}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,046 |
1. If
$$
\frac{4^{5}+4^{5}+4^{5}+4^{5}}{3^{5}+3^{5}+3^{5}} \times \frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5}}{2^{5}+2^{5}}=2^{n},
$$
then, $n=$ . $\qquad$ | $-1.12$
The left side of the original equation can be transformed into
$$
\frac{4 \times 4^{5}}{3 \times 3^{5}} \times \frac{6 \times 6^{5}}{2 \times 2^{5}}=\frac{4^{6}}{3^{6}} \times \frac{6^{6}}{2^{6}}=4^{6}=2^{12} .
$$
Thus, $2^{12}=2^{n}$. Therefore, $n=12$. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,048 |
Example 5 Let the integer $a$ divided by 7 leave a remainder of 3, and the integer $b$ divided by 7 leave a remainder of 5. If $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7.
(1994, Tianjin City Junior High School Mathematics Competition) | Solution: Let $a=7 m+3, b=7 n+5$. Then
$$
\begin{array}{l}
a^{2}-4 b=(7 m+3)^{2}-4(7 n+5) \\
=49 m^{2}+42 m+9-28 n-20 \\
=7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{array}
$$
Therefore, the remainder when $a^{2}-4 b$ is divided by 7 is 3. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,049 |
2. The City Mathematics Invitational Competition sets up three types of medals: gold, silver, and bronze. The organizing committee puts these medals in five boxes, with each box containing only one type of medal. The number of medals in each box is 3, 6, 9, 14, and 18, respectively. It is known that the silver medals a... | 2.12, 14, 24.
Based on "the number of bronze medals is twice the number of gold medals," the sum of bronze and gold medals should be a multiple of 3.
The sum of bronze and gold medals should also be the sum of four numbers from the known 3, 6, 9, 14, 18, due to the fact that "there is only one box of silver medals."
... | 12, 14, 24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,050 |
3. Given that $p$ is a prime number, and the equation
$$
x^{2}+p x-444 p=0
$$
has two integer roots. Then $p=$ $\qquad$ | 3.37.
According to
$$
x^{2}=p(444-x)
$$
we know that $p(444-x)$ is a perfect square.
Since $p$ is a prime number, it follows that $p\left|x^{2} \Rightarrow p\right| x$.
Let $x=n p(n \in \mathbf{Z})$ and substitute into equation (1) to get
$$
(n p)^{2}=p(444-n p) .
$$
Since $p \neq 0$, we have $n^{2} p=444-n p$, whic... | 37 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,051 |
4. Given real numbers $a, b, c$ simultaneously satisfy $a-7b+8c=4$ and $8a+4b-c=7$. Then, $a^{2}-b^{2}+c^{2}=$ $\qquad$ . | 4.1.
According to the conditions $a+8c=4+7b, 8a-c=7-4b$. Squaring both sides of the two equations and then adding them yields
$$
\begin{array}{l}
(a+8c)^{2}+(8a-c)^{2} \\
=(7+4b)^{2}+(7-4b)^{2} .
\end{array}
$$
Simplifying and organizing, we get $65\left(a^{2}+c^{2}\right)=65\left(1+b^{2}\right)$.
Therefore, $a^{2}-b... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,052 |
5. In trapezoid $A B C D$, $A B / / C D$, the base angles $\angle D A B=36^{\circ}, \angle C B A=54^{\circ}, M$ and $N$ are the midpoints of sides $A B$ and $C D$, respectively. If the lower base $A B$ is exactly 2008 units longer than the upper base $C D$, then the line segment $M N=$ $\qquad$ | 5.1004.
As shown in Figure 4, draw $N S / / A D$ and $N T / / C B$ through point $N$.
Then, from $\square A S N D$ and $\square B T N C$, we get
$D N=A S, N C=T B$,
and
$$
\begin{array}{l}
\angle N S T=\angle D A B=36^{\circ}, \\
\angle N T S=\angle C B A=54^{\circ} . \\
\text { Hence, } \angle S N T=180^{\circ}-(\a... | 1004 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,053 |
6. Let $a$, $b$, $c$ be non-zero real numbers, and satisfy $a^{2}-b^{2}=b c$ and $b^{2}-c^{2}=c a$.
Prove: $a^{2}-c^{2}=a b$. | 6. Adding the known equations yields
$$
a^{2}-c^{2}=c(a+b) \text {. }
$$
From the condition $a^{2}-b^{2}=b c$ we get $a^{2}=b(b+c)$, which means
$$
b+c=\frac{a^{2}}{b} .
$$
From the condition $b^{2}-c^{2}=c a$ we get
$$
(b+c)(b-c)=c a \text {. }
$$
Using equation (2) we have $\frac{a^{2}}{b}(b-c)=a c$, which means
$... | a^{2}-c^{2}=a b | Algebra | proof | Yes | Yes | cn_contest | false | 721,054 |
7. In an acute triangle $\triangle A B C$, $A D \perp B C$ at point $D$, $B E \perp A C$ at point $E$, and $A D$ intersects $B E$ at point $H$. $M$ and $N$ are the midpoints of sides $A B$ and $C H$, respectively. Prove: $M N \perp D E$. | 7. As shown in Figure 5, connect $M D$ and $M E$.
In the right triangle $\triangle A B C$, $D M$ is the median of the hypotenuse $A B$, therefore,
$$
M D=\frac{1}{2} A B .
$$
In the right triangle $\triangle A B E$, similarly, we get
$$
M E=\frac{1}{2} A B .
$$
Thus, $M D=M E$.
By the same method, connect $N D$ and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,055 |
1. Calculate: $1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+$ $\cdots+\frac{1}{20}(1+2+\cdots+20)$.
| $\begin{array}{l}\text { 1. Original expression }=1+\frac{1}{2} \times \frac{2 \times 3}{2}+\frac{1}{3} \times \frac{3 \times 4}{2}+ \\ \cdots+\frac{1}{20} \times \frac{20 \times 21}{2} \\ =\frac{1}{2}(2+3+4+\cdots+21) \\ =\frac{1}{2} \times \frac{23 \times 20}{2}=115 .\end{array}$ | 115 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,058 |
2. Let real numbers $x, y$ satisfy
$$
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1 \text {. }
$$
Find the value of $x+y$. | 2. First, we have $x+y=\sqrt{y^{2}+1}-\sqrt{x^{2}+1}$. Squaring both sides and simplifying, we get
$$
\sqrt{x^{2}+1} \cdot \sqrt{y^{2}+1}=1-x y \text {. }
$$
Squaring both sides again, we get
$$
x^{2} y^{2}+x^{2}+y^{2}+1=x^{2} y^{2}-2 x y+1 \text {, }
$$
which simplifies to $(x+y)^{2}=0$.
Therefore, $x+y=0$. | x+y=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,059 |
Example 6: Prove that when a positive integer is divided by a smaller positive integer, the dividend is always greater than twice the remainder.
Analysis: Let $a>b(a, b \in \mathbf{N})$, and let $a=b q+r$ $(0 \leqslant r < b)$. We need to prove $a > 2r$. To go from the condition to the goal, we need to "eliminate" $b$... | Proof: Let $a>b(a, b \in \mathbf{N})$, and let $a=b q+r$ $(0 \leqslant r < b)$. If $r \geqslant b/2$, then $a-b=b(q-1)+r \geqslant b/2$. If $r < b/2$, then $b-r < b/2$. Since $a-b=(b-r)q+r$, we have $a-b \leqslant (b/2)q+r$. If $q \geqslant 2$, then $(b/2)q \geqslant b$, so $a-b \geqslant b/2$. If $q=1$, then $a-b=b-r+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,060 |
3. As shown in Figure 3, in the concave quadrilateral $ABCD$, its three interior angles $\angle A, \angle B, \angle C$ are all $45^{\circ}$. $E, F, G, H$ are the midpoints of sides $AB, BC, CD, DA$ respectively. Prove: Quadrilateral $EFGH$ is a square. | 3. To prove that quadrilateral $E F G H$ is a square, it is only necessary to prove:
(1) Quadrilateral $E F G H$ is a parallelogram;
(2) $E H=H G$;
(3) $E H \perp H G$
(1) As shown in Figure 7,
Connect $A C, B D$,
extend $B D$ to intersect $A C$
at point $K$, and extend
$C D$ to intersect $A B$ at point
$L$. Then, by
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,061 |
4. Let real numbers $x, y, z$ simultaneously satisfy
$$
\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}
$$
Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-1)^{2}$. | 4. Transform to get
$$
\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}
$$
Multiplying the above three equations yields
$$
\begin{array}{l}
(x-2)(y-2)(z-2) \\
=-6(x-2)(y-2)(z-2) . \\
(x+1)^{2}(y+1)^{2}(z+1)^{2} .
\end{array}... | 6027 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,062 |
5. For a sequence of numbers in order, define the following operation rules:
Rule $A$: For three adjacent numbers $a, b, c$, their order changes to $c, b, a$, which is called one "transformation";
Rule $B$: For four adjacent numbers $a, b, c, d$, their order changes to $d, c, b, a$, which is called one "transformation... | 5. (1) Rule $A$ cannot achieve the goal.
For any number under rule $A$, its position's parity does not change. If the goal could be achieved, then “1” would move from an odd position to an even position, which is impossible.
(2) Rule $B$ can achieve the goal.
For any five adjacent elements 12345, the operations are as... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,063 |
6. If an integer can be expressed in the form $x^{2}+2 x y+2 y^{2}$ (where $x, y$ are integers), then the number is called a "good number".
(1) Determine whether 29 is a good number;
(2) List the good numbers among $80, 81, \cdots, 100$;
(3) If $m$ and $n$ are both good numbers, prove that $m n$ is also a good number. | 6. (1) Since $x^{2}+2 x y+2 y^{2}=(x+y)^{2}+y^{2}$, therefore, a good number can be expressed as the sum of two perfect squares. Given that $29=5^{2}+2^{2}$, we know that 29 is a good number.
(2) The perfect squares within 100 are as follows:
$$
0,1,4,9,16,25,36,49,64,81,100 \text {. }
$$
The good numbers within the r... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,064 |
7. In an equilateral triangle $\triangle A B C$ with side length $2 \sqrt{7}$, $M$ is the midpoint of side $B C$, and $P$ is a point on $A C$.
(1) For what value of $P C$ is $B P+P M$ minimized?
(2) Find the minimum value of $B P+P M$. | 7. As shown in Figure 8, connect $A M$, and flip $\triangle A B C$ along the axis $A C$ by $180^{\circ}$ to get $\triangle A D C$. Let $N$ be the symmetric point of $M$ (about $A C$). At this point, we always have $P M = P N$. Since $B$ and $N$ are fixed points, $B P + P N \geqslant B N$, with equality holding if and o... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,065 |
8. Find the integer pairs $(x, y)$ that satisfy the condition $x^{4}+x^{3}+x^{2}+x=y^{2}+y$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 8. Multiplying both sides of the conditional equation by 4, we have
$$
4 x^{4}+4 x^{3}+4 x^{2}+4 x=4 y^{2}+4 y \text {. }
$$
By completing the square, we get
$$
4 x^{4}+4 x^{3}+4 x^{2}+4 x+1=(2 y+1)^{2} \text {. }
$$
Notice that
$$
\left(2 x^{2}+x\right)^{2}=4 x^{4}+4 x^{3}+x^{2}
$$
and $\left(2 x^{2}+x+1\right)^{2}... | (-1,0),(-1,-1),(0,0),(0,-1),(2,-6),(2,5) | Other | math-word-problem | Yes | Yes | cn_contest | false | 721,066 |
Example 7 There is a sequence of numbers: $1,3,4,7,11,18, \cdots$, starting from the third number, each number is the sum of the two preceding numbers.
(1) What is the remainder when the 1991st number is divided by 6?
(2) Group the above sequence as follows:
$(1),(3,4),(7,11,18), \cdots$,
where the $n$-th group has ex... | Solution: Let the $n$-th number of the sequence be $a_{n}$. Then
$$
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
$$
Through experimentation, it is known that the remainders of $a_{1}, a_{2}, \cdots, a_{26}$ when divided by 6 are
$$
\begin{array}{l}
1,3,4,1,5,0,5,5,4,3,1,4,5, \\
3,2,5,1,0,1,1,2,3,5,2,1,3 .
\end{array}
$$
No... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,071 |
```
Prove:
\[
\begin{array}{l}
\tan 3 - \tan 2 - \tan 1 + \tan \frac{1}{2} - \tan \frac{1}{3} - \tan \frac{1}{6} \\
= \tan 3 \cdot \tan 2 \cdot \tan 1 + \tan \frac{1}{2} \cdot \tan \frac{1}{3} \cdot \tan \frac{1}{6}
\end{array}
\]
``` | $=、 1 、 2 、 3$ radians are not equal to $\frac{\pi}{2}+n \pi(n \in$ Z). Therefore, $\tan 1 、 \tan 2 、 \tan 3$ are all defined, and
$\tan 1 \cdot \tan 2 \neq 1 \Rightarrow 1-\tan 1 \cdot \tan 2 \neq 0$.
From $1+2=3$, we get $\tan (1+2)=\tan 3$, that is
$\frac{\tan 1+\tan 2}{1-\tan 1 \cdot \tan 2}=\tan 3$.
Thus, $\tan 1+... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,073 |
Three, (15 points) As shown in Figure 3, $AB$ is a chord of a given circle, dividing the circle into two parts. $M$ and $N$ are the midpoints of the two arcs, respectively. With point $B$ as the center of rotation, the segment $AMB$ is rotated clockwise by an angle to form the segment $A_{1}MB$. The midpoint of $AA_{1}... | Three, as shown in Figure 4, take the midpoint $C$ of $AB$ and the midpoint $C_1$ of $A_1B$. It is easy to see that $A_1B = AB$, thus $A_1C_1 = AC$.
Connect $MC_1$ and $NC$.
Then $MC_1 \perp A_1B$,
$NC \perp AB$.
When not rotated, $C_1$ and $C$ are the same point, and $MN$ is a diameter perpendicular to $AB$. By the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,074 |
Four. (15 points) The function $f(x)$ defined on the interval $[0,1]$ satisfies $f(0)=f(1)=0$, and for any $x_{1}, x_{2} \in [0,1]$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant f\left(x_{1}\right)+f\left(x_{2}\right) \text {. }
$$
(1) Prove: For any $x \in [0,1]$, we have
$$
f(x) \geqslant 0 \text {; }
$$
... | (1) Take $x_{1}=x_{2}=x \in[0,1]$, then
$$
f\left(\frac{2 x}{2}\right) \leqslant f(x)+f(x),
$$
i.e., $f(x) \leqslant 2 f(x)$.
Thus, $f(x) \geqslant 0$.
Therefore, for any $x \in[0,1]$, $f(x) \geqslant 0$.
(2) From $f(0)=f(1)=0$, we get
$$
f\left(\frac{0+1}{2}\right) \leqslant f(0)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{1... | 0 | Algebra | proof | Yes | Yes | cn_contest | false | 721,075 |
Five. (15 points) There are $m(m>2)$ middle school students in Beijing who submitted a total of $n$ different suggestions for the “Beijing Olympic Games”. It is known that among any two students, there is at least one suggestion that is the same, and at least one suggestion that is different. Prove: The number of stude... | Let $A$ be the set of different suggestions proposed by $m$ middle school students, and $A_{i} (i=1,2, \cdots, m)$ represents the set of suggestions proposed by the $i$-th student. Let $|X|$ denote the number of elements in set $X$, and $\bar{X}$ denote the complement of $X$. Then, $|A|=n$. Since any two students' sugg... | m \leqslant 2^{n-1} | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,076 |
One, (25 points) As shown in Figure 1,
$\odot O$ is the incircle of trapezoid $ABCD$, with points of tangency
$$
E, F, G, H, AB \parallel
$$
$CD$. Draw $BP \parallel AD$ intersecting the extension of $DC$ at point $P$, and the extension of $AO$ intersects $CP$ at point $Q$. If $AE = BE$, prove: $\angle CBQ = \angle PBQ... | Connect $O E$. By the problem, $O E$ is the perpendicular bisector of $A B$. Thus, trapezoid $A B C D$ is an isosceles trapezoid, so $A D=$
$B C$. And quadrilateral $A B P D$ is a parallelogram, hence $A D=B C=B P$.
Let $D H=x, A H=y$. Then $x+y=A D \Rightarrow B P=x+y$.
Therefore, $C P=2 y-2 x=2(y-x)$.
Due to the tang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,077 |
II. (25 points) Given that $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$. Prove:
$$
\begin{array}{l}
\frac{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}{\sqrt{3}} \\
\geqslant \sqrt[6]{\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2}} .
\end{array}
$$
(Supplied b... | In $\triangle A B C$, note that $\tan \frac{A}{2} \cdot \tan \frac{B}{2}+\tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{C}{2} \cdot \tan \frac{A}{2}=1$. Therefore,
$$
\left(\frac{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}{\sqrt{3}}\right)^{2}
$$
$$
=\frac{\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,078 |
Four. (25 points) Prove: (1) There exist infinitely many positive integers $n$ such that the largest prime factor of $n^{2}+1$ is less than $n$;
(2) There exist infinitely many positive integers $n$ such that $n^{2}+1$ divides $n!$.
(Zhang Lei provided the problem) | (1) Let $n=2 k^{2}\left(k \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
n^{2}+1=4 k^{4}+1 \\
=\left(2 k^{2}+1-2 k\right)\left(2 k^{2}+1+2 k\right) .
\end{array}
$$
Since $2 k^{2}+1-2 k<n$, it is sufficient to prove that there are infinitely many positive integers $k$ such that $2 k^{2}+1+2 k$ is composite and i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,080 |
Five. (25 points) As shown in Figure 3, given $\square A B C D$, the circle $\odot O_{1}$ passing through points $A$, $B$, and $C$ intersects $A D$ and $B D$ at points $E$ and $F$, respectively. The circle $\odot O_{2}$ passing through points $C$, $D$, and $F$ intersects $A D$ at point $G$. Let the radii of $\odot O_{1... | Five, as shown in Figure 5, connect $C A$, $C F$, and $C G$. Then
$$
\begin{array}{l}
\angle A C F \\
=\angle A B F \\
=\angle C D F .
\end{array}
$$
Therefore, $A C$ is
a tangent to $\odot \mathrm{O}_{2}$.
Similarly, $C C$ is a tangent to $\odot O_{1}$.
By the power of a point theorem, we get
$$
E G \cdot A G=C G^{2}... | \frac{E G}{A D}=\frac{R_{2}^{2}}{R_{1}^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 721,081 |
Example 8 A positive integer, if it can be expressed as the difference of squares of two positive integers, is called a "wise number". Arrange all the wise numbers in ascending order. Find the wise number at the 2009th position. | Solution: Let $n$ be any positive integer.
When $n=2 k+1\left(k \in \mathbf{N}_{+}\right)$, we have $n=2 k+1=(k+1)^{2}-k^{2}$.
Thus, every odd number greater than 1 is a wise number.
When $n=4 k\left(k \in \mathbf{N}_{+}, k \geqslant 2\right)$, we have
$$
n=4 k=(k+1)^{2}-(k-1)^{2} \text {. }
$$
Thus, every number grea... | 2681 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,082 |
Six. (25 points) Let $n$ be a positive integer, and the integer $a$ be a root of the equation $x^{4}+3 a x^{2}+2 a x-2 \times 3^{n}=0$. Find all pairs $(n, a)$ that satisfy the condition. (Provided by Lu Xiushuang) | Six, since $a$ is a root of the equation, then
$$
a^{4}+3 a^{3}+2 a^{2}=2 \times 3^{n} \text {, }
$$
which means $a^{2}(a+2)(a+1)=2 \times 3^{n}$
From this, we get $a \neq 0,-1,-2$.
If $a$ is even, then the left side of the equation is a multiple of 4, while the right side is not, leading to a contradiction. Therefore... | (n, a)=(2,-3),(1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,083 |
Seven, (25 points) Let $a$, $b$, and $c$ be the lengths of the three sides of a right triangle, where $c$ is the length of the hypotenuse. Find the maximum value of $k$ such that $\frac{a^{3}+b^{3}+c^{3}}{a b c} \geqslant k$ holds. (Provided by Li Tiehan) | $$
\text { VII. } k=\left(\frac{a^{3}+b^{3}+c^{3}}{a b c}\right)_{\min } \text {. }
$$
Let $y=\frac{a^{3}+b^{3}+c^{3}}{a b c}$, and set
$$
a=c \cos \theta, b=c \sin \theta, \theta \in\left(0, \frac{\pi}{2}\right) .
$$
Then $y=\frac{\sin ^{3} \theta+\cos ^{3} \theta+1}{\sin \theta \cdot \cos \theta}$
$$
\begin{array}{... | 2+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,084 |
2. Calculate:
(1) $\sqrt{627953481}+\sqrt{672935481}=$ $\qquad$
(2) $\sqrt{\sqrt{254817369}-\sqrt{152843769}}=$
$\qquad$ . | 2. (1) $51000(2) 60$ | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,087 |
Example 9 Find the smallest positive integer $n$, such that the last three digits of $n^{3}$ are 888. | Solution: Since the unit digit of $n^{3}$ is 8, then the unit digit of $n$ is 2. Let $n=10 k+2$. Then
$$
\begin{array}{l}
n^{3}=(10 k+2)^{3}=1000 k^{3}+600 k^{2}+120 k+8 \\
=1000 k^{3}+100 k(6 k+1)+20 k+8 .
\end{array}
$$
According to the problem, we know that the last three digits of $100\left(6 k^{2}+k\right)+20 k+8... | 192 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,093 |
Sure, here is the translated text:
```
II. (20 points) Find all positive integers $n$ such that
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{5}\right]=69 \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
``` | $$
\begin{array}{l}
\frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}-4 \\
<\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{5}\right] \\
\leqslant \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5} . \\
\text { Therefore, } \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}-4 \\
<69 \l... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,095 |
Three. (20 points) As shown in Figure 2, the distance from the center O to a chord of circle $\odot O$ is $O H=2008$. In the two segments, two squares $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ are constructed, where points $A, D, A_{1}, B_{1}$ are on the arc, and points $B, C, C_{1}, D_{1}$ are on the chord. Find the dif... | Three, as shown in Figure 5, let the side lengths of the squares $A_{1} B_{1} C_{1} D_{1}$ and $A B C D$ be $a$ and $b$, respectively. Extend $O H$ to intersect $A_{1} B_{1}$ at point $M$, and draw $O N \perp C D$ at point $N$.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
O M^{2}+M A_{1}^{2}=R^{2} \\
=O N^{2... | 3212.8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,096 |
Four. (20 points) It is known that in the Cartesian coordinate system, there is a disk (including the circumference) $G$ with a center at point $A(2008,525)$ and a radius of 1. The graph of the parabola $y^{2}=2 p x$ intersects the disk $G$. Find the range of values for $p$ (accurate to 0.001). | Let the coordinates of point $P(x, y)$ on the boundary of the disk $G$ be
$$
\left\{\begin{array}{l}
x=2008+\cos \theta, \\
y=525+\sin \theta
\end{array} \quad(0 \leqslant \theta<2 \pi) .\right.
$$
Substituting into the parabola equation $y^{2}=2 p x$, we get
$$
(525+\sin \theta)^{2}=2 p(2008+\cos \theta) \text {. }
$... | 68.369 \leqslant p \leqslant 68.895 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,097 |
一、(20 points) As
shown in Figure 3, the
shaded part is a
closed figure within
the semicircle $\odot 0$, it is known that chord $M N$
is parallel to chord $P Q$, and
the distance between the parallel lines $M N$ and $P Q$ is $2 \mathrm{~m}, M N=$
$6 \mathrm{~m}, P Q=8 \mathrm{~m}$. Find the length of $\overparen{M P}$ (... | - 1. As shown in Figure 6, draw a radius $O T$ perpendicular to the parallel chords $M N$ and $P Q$, with the feet of the perpendiculars being $C$ and $D$, respectively. Connect $O P$ and $O M$. Let the distance from the center $O$ to $P Q$ be $s$, and the radius of $\odot O$ be $R$. Then, in the right triangles $\tria... | 2.265 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,098 |
II. (20 points) As shown in Figure 4, the side length of the equilateral $\triangle ABC$ is $1, D$ and $E$ are points on sides $AB$ and $AC$ respectively. When the triangle is folded along $DE$, vertex $A$ exactly lands on point $F$ on side $BC$, with $BD=x$ and $BF=y$. Find the relationship between $x$ and $y$, and sk... | II. Connect $D F$.
In $\triangle B D F$, by the cosine rule we have
$$
D F^{2}=B D^{2}+B F^{2}-2 B D \cdot B F \cos 60^{\circ} \text {, }
$$
which is $(1-x)^{2}=y^{2}+x^{2}-2 y \cdot x \cdot \frac{1}{2}$.
Thus, $y^{2}-x y+2 x-1=0(0 \leqslant y \leqslant 1)$.
Using a graphing calculator, we can obtain its approximate g... | x \in [0, 4 - 2\sqrt{3}] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,099 |
Three. (20 points) For a set $S\left(S \subseteq \mathbf{N}_{+}\right)$, if for any $x \in S, \dot{x}$ cannot divide the sum of elements of any non-empty subset of $S \backslash\{x\}$, then $S$ is called a "good set" ($S \backslash\{x\}$ represents the set $S$ after removing the element $x$).
(1) If $\{3,4, n\}$ is a g... | Three, (1) Obviously, $n>4$.
If $n=5$, then $4 \mid (3+5)$; if $n=6$, then $3 \mid 6$; if $n=7$, then $7 \mid (3+4)$; if $n=8$, then $4 \mid 8$; if $n=9$, then $3 \mid 9$.
When $n=10$, it is verified that $\{3,4,10\}$ is a good set.
Therefore, the minimum value of $n$ is $n_{0}=10$.
(2) If $\{3,4,10, m\}$ is a good set... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,100 |
1. Given in $\triangle A B C$, $\angle A=90^{\circ}, M$ is the midpoint of side $B C$, and point $N$ on the extension of $B C$ satisfies $A M \perp A N$. The incircle of $\triangle A B C$ touches sides $A B$ and $A C$ at points $E$ and $F$, respectively. $E F$ intersects $A N$ and $B C$ at points $P$ and $Q$. Then $P N... | -1.B.
As shown in Figure 5.
From $A M \perp A N$, we know
$$
\begin{aligned}
& \angle A N M=90^{\circ}-\angle A M N=90^{\circ}-\angle A M C \\
& =90^{\circ}-\left(180^{\circ}-2 \angle A C M\right) \\
& =2\left(\angle A C M-45^{\circ}\right) \\
= & 2(\angle A C M-\angle A F E) \\
=2( & \angle A C M-\angle C F Q)=2 \angl... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,101 |
2. "Beyond the city of Gusu lies the Cold Mountain Temple, where the sound of the bell reaches the guest boats at midnight." On New Year's Eve, the abbot of Cold Mountain Temple strikes the bell 108 times, praying for peace in the world. It is known that there are two boats on the river outside the temple. When the fir... | 2. D.
Let the upstream boat passenger hear the 108th bell at $t \mathrm{~s}$, at this time, the downstream boat passenger has only heard $x$ times. Then
$$
3 t=300(t-107 \times 9) \text {. }
$$
Solving for $t$ gives $t=107 \times \frac{100}{11}(\mathrm{~s})$.
Thus, the distance from the downstream boat to Hanshan Tem... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,102 |
3. Reverse the order of the four digits of a four-digit number, and add the resulting number to the original number. If the sum obtained is divisible by 35, then this four-digit number is called a "good number". Then, among all four-digit numbers, the number of good numbers is ( ).
(A) 234
(B) 252
(C) 270
(D) 369 | 3. A.
Let the four-digit number be $\overline{a_{1} a_{2} a_{3} a_{4}}$. Then
$$
\begin{array}{l}
A=\overline{a_{1} a_{2} a_{3} a_{4}}+\overline{a_{4} a_{3} a_{2} a_{1}} \\
=1001\left(a_{1}+a_{4}\right)+110\left(a_{2}+a_{3}\right) .
\end{array}
$$
Since $1001=7 \times 143$, we have
$$
\begin{array}{l}
7 \mid A \Leftr... | 234 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,103 |
Example 10 Given Theorem: "If three prime numbers $a, b, c$ greater than 3 satisfy the equation $2a + 5b = c$, then $a + b + c$ is a multiple of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
(1997, National Junior High School Mathematics League) | Solution: Let $a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}$. Then
$$
\begin{array}{l}
a+b+c=3(a+2 b) \\
=3\left(3 k_{1}+r_{1}+6 k_{2}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) .
\end{array}
$$
Since $a, b$ are both primes greater than 3, we have
$$
r_{1} r_{2} \neq 0.
$$
If $r_{1} \neq r_{2}$, t... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false | 721,104 |
4. Polynomial
$$
\begin{array}{l}
|x+1|+|x-2|+|x+3|+\cdots+ \\
|x+2007|+|x-2008|+|x+2009|
\end{array}
$$
The minimum value of the above expression is ( ).
(A) 2019044
(B) 2017035
(C) 2009
(D) 0 | 4. A.
$$
\begin{array}{l}
| x+ 1|+| x-2|+| x+3 |+\cdots+ \\
|x+2007|+|x-2008|+|x+2009| \\
=|x+1|+(|x-2|+|x+3|)+\cdots+ \\
(|x-2008|+|x+2009|) \\
\geqslant|x+1|+|(x-2)-(x+3)|+\cdots+ \\
|(x-2008)-(x+2009)| \\
=|x+1|+2+3+\cdots+2009 \\
\geqslant 0+2+3+\cdots+2009 \\
= \frac{2009 \times 2010}{2}-1 \\
= 2019044 .
\end{arr... | 2019044 | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,105 |
5. For $k$ positive real numbers $a_{1}, a_{2}, \cdots, a_{k}$, the quantity $\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}}{k}}$ is called the quadratic mean of these $k$ numbers. Let $A_{n}$ denote the quadratic mean of all numbers in $1,2, \cdots, 2009$ that are divisible by $n$. Then the order of $A_{2} 、 A_{3} ... | 5.D.
$$
\begin{array}{l}
A_{2}=\sqrt{\frac{2^{2}+4^{2}+\cdots+2008^{2}}{\frac{2008}{2}}} \\
=2 \sqrt{\frac{1^{2}+2^{2}+\cdots+1004^{2}}{1004}} \\
=2 \sqrt{\frac{1005 \times 2009}{6}}=\sqrt{\frac{2010 \times 4018}{6}} \text {, } \\
A_{3}=\sqrt{\frac{3^{2}+6^{2}+\cdots+2007^{2}}{\frac{2007}{3}}} \\
=3 \sqrt{\frac{1^{2}+2... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,106 |
6. As shown in Figure 1, two skaters, Jia and Yi, are at points $A$ and $B$ on a circular skating rink, with $OB=20 \, \text{m}$, $OA=15 \, \text{m}$, and $OA \perp OB$. Yi starts from point $B$ and skates along the circumference of the circle $\odot O$ in a clockwise direction at a speed of $5 \, \text{m/s}$. At the s... | 6.C.
As shown in Figure 6, rays $AO$ and $BO$ intersect
$\odot O$ at points $A'$ and $B'$, respectively.
Points $C$ and $D$ are taken sequentially on $\overparen{B A'}$.
$AC$ and $AD$ intersect $\odot O$ at points $C'$ and $D'$, respectively.
Perpendiculars are drawn from point $O$ to chords $CC'$ and $DD'$, with feet... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,107 |
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