problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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3. Let $a, b \in \mathbf{N}_{+}$, and satisfy
$$
56 \leqslant a+b \leqslant 59,0.9<\frac{a}{b}<0.91 \text {. }
$$
Then $b^{2}-a^{2}$ equals ( ).
(A) 171
(B) 177
(C) 180
(D) 182 | 3. B.
From the problem, we get $0.9 b+b56$.
Solving, we get $29<b<32 \Rightarrow b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we solve to get $27<a<28$, there is no such positive integer $a$;
When $b=31$, from $0.9 b<a<0.91 b$, we solve to get $27<a<29 \Rightarrow a=28$.
Thus, $b^{2}-a^{2}=177$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,645 |
4. As shown in Figure 1, in $\triangle A B C$, it is given that $A B>A C$, points $D$ and $E$ are on $A B$ and $A C$ respectively, and $B D=C E$. Taking the midpoints $M$ and $N$ of $B E$ and $C D$ respectively, line $M N$ intersects $A B$ and $A C$ at points $P$ and $Q$. Then ( ).
(A) $A P>A Q$
(B) $A P<A Q$
(C) $A P=... | 4. C.
As shown in Figure 3, take the midpoint $K$ of $BC$. Then $KM$ and $KN$ are the midlines of $\triangle BCE$ and $\triangle BCD$, respectively. Hence,
$$
\begin{array}{c}
K N=\frac{1}{2} B D \\
=\frac{1}{2} C E=K M \\
\Rightarrow \angle K M N=\angle K N M .
\end{array}
$$
But from $KM \parallel AC$ and $KN \para... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,646 |
5. The graph of a certain linear function is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through the point $(-1,-25)$. Then, on the line segment $AB$ (including $A$ and $B$), the number of points with both integer coordina... | 5. B.
Let $y=\frac{5}{4} x+b$.
From $-25=\frac{5}{4} \times(-1)+b$, we get $b=-\frac{95}{4}$.
Therefore, $y=\frac{5}{4} x-\frac{95}{4}$.
Thus, $A(19,0)$ and $B\left(0,-\frac{95}{4}\right)$.
From $y=\frac{5(x-19)}{4}, 0 \leqslant x \leqslant 19$, taking $x=3,7$,
$11,15,19$, $y$ is an integer.
Therefore, on the line seg... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,647 |
6. In $\triangle A B C$, it is known that $B C=a, C A=b, A B$ $=c$, and the inradius is $r$. If $r=\frac{1}{2}(a+b-c)$, then $\triangle A B C$ is ( ).
(A) Acute triangle
(B) Right triangle
(C) Isosceles triangle
(D) Equilateral triangle | 6. B.
Let the incircle $\odot O$ touch $AC$ and $BC$ at points $D$ and $E$, respectively. Then $CD = CE = \frac{a+b-c}{2} = r$.
Also, $OD \perp CD, OE \perp CE$
$\Rightarrow CD \perp CE \Rightarrow \angle C = 90^{\circ}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,648 |
1. Simplify
$$
\sqrt{1+2 \sin \alpha \cdot \cos \alpha}+\sqrt{1-2 \sin \alpha \cdot \cos \alpha}
$$
$\left(0^{\circ}<\alpha \leqslant 90^{\circ}\right)$ The result is $\qquad$ . | II. 1.2 $2 \cos \alpha$ or $2 \sin \alpha$.
It is easy to know that $1+2 \sin \alpha \cdot \cos \alpha$
$$
\begin{array}{l}
=\sin ^{2} \alpha+2 \sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha \\
=(\sin \alpha+\cos \alpha)^{2} .
\end{array}
$$
Similarly, $1-2 \sin \alpha \cdot \cos \alpha=(\sin \alpha-\cos \alpha)^{2}$.... | 2 \cos \alpha \text{ or } 2 \sin \alpha | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,649 |
Example 1 Given $0 \leqslant a \leqslant b \leqslant c \leqslant d \leqslant e$, $a+b+c+d+e=1$.
Prove: $a d+c d+c b+b e+e a \leqslant \frac{1}{5}$.
Analysis: Careful observation reveals that each letter appears twice. When combining like terms containing the same letter, it is found to be the sum of two sets of number... | Prove: Notice
$$
\begin{array}{l}
2(a d+c d+c b+b e+e a) \\
=a(d+e)+b(c+e)+c(b+d)+ \\
\quad d(a+c)+e(a+b) .
\end{array}
$$
Since $a \leqslant b \leqslant c \leqslant d \leqslant e$, we have
$$
d+e \geqslant c+e \geqslant b+d \geqslant a+c \geqslant a+b \text {. }
$$
By Chebyshev's inequality,
$$
\begin{array}{l}
a(d+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,650 |
2. If for any real number $a$, the equation about $x$
$$
x^{2}-2 a x-a+2 b=0
$$
has real roots, then the range of the real number $b$ is $\qquad$ | 2. $b \leqslant -\frac{1}{8}$.
According to the problem, we have
$$
\Delta=4 a^{2}-4(-a+2 b)=4 a^{2}+4 a-8 b \geqslant 0
$$
for all real numbers $a$, which means
$$
8 b \leqslant 4 a^{2}+4 a=(2 a+1)^{2}-1
$$
for all real numbers $a$.
Clearly, when $a=-\frac{1}{2}$, $(2 a+1)^{2}-1$ reaches its minimum value -1.
Thus,... | b \leqslant -\frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,651 |
3. Given that the three vertices $A$, $B$, and $C$ of the right triangle $\triangle ABC$ are all on the parabola $y=x^{2}$, and the hypotenuse $AB$ is parallel to the $x$-axis. Then the height $h$ from the hypotenuse is $\qquad$. | 3. 1 .
Let point $A\left(a, a^{2}\right)$ and $C\left(c, c^{2}\right)(|c|<a)$. Since $|AB|=|AC|$, we have $a^{2}-c^{2}=1$.
Therefore, the altitude from $C$ to the hypotenuse $AB$ is $a^{2}-c^{2}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,652 |
4. As shown in Figure 2, in $\triangle A B C$, it is known that $A B=9, B C$ $=8, C A=7, A D$ is the angle bisector, and a circle is drawn with $A D$ as a chord,
touching $B C$ and intersecting $A B$ and $A C$ at points $M$ and $N$, respectively. Then $M N=$
$\qquad$ | 4. 6 .
As shown in Figure 4, connect $D M$.
$$
\begin{array}{l}
\text { By } \angle B D M \\
=\angle B A D \\
=\angle C A D \\
=\angle D M N,
\end{array}
$$
we get $M N \parallel B C$.
Thus, $\triangle A M N \sim \triangle A B C$.
It is easy to know $B D=\frac{9}{9+7} \times 8=\frac{9}{2}, B M \cdot B A=B D^{2}$
$\Ri... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,653 |
One, (20 points) Given $\triangle A B C$ with side lengths $a, b, c$, let $l=\frac{1}{2}(a+b+c)$. If $l^{2}=2 a b$, prove: $l<2 a$.
| \begin{array}{l}\text { I. From } a+c>b \Rightarrow a+b+c>2 b \\ \Rightarrow 2 l=a+b+c>2 b \Rightarrow 2 b<2 l . \\ \text { Also } l^{2}=2 a b \Rightarrow l^{2}<2 a l \Rightarrow l<2 a .\end{array} | l<2a | Inequalities | proof | Yes | Yes | cn_contest | false | 721,654 |
II. (25 points) In $\triangle ABC$, it is known that $AC > BC$, $\angle C = 60^{\circ}$. Three equilateral triangles are constructed outward on sides $AB$, $BC$, and $CA$, namely $\triangle ABC'$, $\triangle BCA'$, and $\triangle CAB'$. What is the relationship between $S_{\triangle ABC}$ and $S_{\triangle ABC'}$, $S_{... | II. Conclusion: $S_{\triangle A B C}=S_{\triangle B C A^{\prime}}+S_{\triangle C A B^{\prime}}-S_{\triangle A B C^{\prime}}$.
As shown in Figure 5, take $C D=C B$ on $A C$, and connect $D B, D C^{\prime}, D B^{\prime}$.
It is easy to see that
$\angle D B C=60^{\circ}$.
Therefore, $\angle C^{\prime} B D$
$=\angle A B C$... | S_{\triangle A B C}=S_{\triangle B C A^{\prime}}+S_{\triangle C A B^{\prime}}-S_{\triangle A B C^{\prime}} | Geometry | proof | Yes | Yes | cn_contest | false | 721,655 |
Three. (25 points) Prove: There exist infinitely many pairs of positive integers $(m, n)$ satisfying the equation
$$
m^{2}+25 n^{2}=10 m n+7(m+n) .
$$ | Three, the original equation can be transformed into
$$
m^{2}-(10 n+7) m+25 n^{2}-7 n=0 \text {. }
$$
Therefore, $\Delta=168 n+49$ is a perfect square.
Let $168 n+49=49(12 k+1)^{2}\left(k \in \mathbf{N}_{+}\right)$.
Thus, $n=42 k^{2}+7 k$.
Then $m=\frac{10 n+7 \pm \sqrt{\Delta}}{2}$
$$
\begin{array}{l}
=\frac{10\left(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,656 |
1. In Rt $\triangle A B C$, it is known that $\overrightarrow{A B}=(2,3), \overrightarrow{A C}$ $=(1, k)$. Then the value of $k$ is | $-1, -\frac{2}{3}, \frac{11}{3}, \frac{3 \pm \sqrt{13}}{2}$.
Discuss in three cases.
(1) When $\angle A=\frac{\pi}{2}$, from $\overrightarrow{A B} \cdot \overrightarrow{A C}=0$, we get
$$
2 \times 1+3 k=0 \Rightarrow k=-\frac{2}{3} \text {. }
$$
(2) When $\angle B=\frac{\pi}{2}$, from the given, we have $\overrightarro... | k=-\frac{2}{3}, \frac{11}{3}, \frac{3 \pm \sqrt{13}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,657 |
2. If the function $f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x$ has a range of $[n, M]$ on the interval $[-k, k](k>0)$, then $M+n$ $=$ $\qquad$ | 2. 2 .
Notice that
$$
f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x=1+\frac{2^{x}-1}{2^{x}+1}+\sin x \text {. }
$$
Let $g(x)=f(x)-1$. Then
$$
g(-x)=\frac{2^{-x}-1}{2^{-x}+1}+\sin (-x)=\frac{1-2^{x}}{1+2^{x}}-\sin x=-g(x)
$$
is an odd function.
Let the maximum value of $g(x)$ be $g\left(x_{0}\right)$. By the given information,... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,658 |
3. The 29th Summer Olympic Games were held in Beijing on August 8, 2008, forming a memorable number 20080808. The number of different positive divisors of 20080808 divided by 8 is $\qquad$ | 3. 8 .
From $20080808=2^{3} \times 11 \times 17 \times 31 \times 433$, we know that the number of different positive divisors of 20080808 is
$$
N=(3+1)(1+1)^{4}=8^{2} \text { (divisors). }
$$
Dividing by 8 gives 8. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,659 |
Example 2 Let $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n), \sum_{i=1}^{n} x_{i}$ $=1$. Prove:
$$
\sum_{i=1}^{n} \sqrt{\frac{1}{x_{i}}-1} \geqslant(n-1) \sum_{i=1}^{n} \frac{1}{\sqrt{\frac{1}{x_{i}}-1}} .
$$
Analysis: The difficulty in this problem lies in both sides being summations, and it is hard to combine them. Ho... | Proof: Without loss of generality, let $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$.
For $i<j$,
$$
\begin{array}{l}
\sqrt{\frac{1}{x_{i}}-1}-\sqrt{\frac{1}{x_{j}}-1} \\
=\frac{x_{j}-x_{i}}{x_{i} x_{j}\left(\sqrt{\frac{1}{x_{i}}-1}+\sqrt{\frac{1}{x_{j}}-1}\right)} \leqslant 0 .
\end{array}
$$
Then $\sqrt{\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,661 |
5. In $\triangle A B C$, suppose
$$
\frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A},
$$
and $\cos (A-B)+\cos C=1-\cos 2 C$. Then the value of $\frac{a+c}{b}$ is | 5. $\frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}}$.
The first equation, derived from the Law of Sines, is
$$
\begin{array}{l}
\frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A}=\frac{b}{b-a} \\
\Rightarrow b^{2}-a^{2}=a b \Rightarrow\left(\frac{a}{b}\right)^{2}+\left(\frac{a}{b}\right)-1=0 .
\end{array}
$$
Solving the equ... | \frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,662 |
6. If the parabola $y=-\frac{x^{2}}{4}+4$ intersects the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{3^{2}}$ $=1$ at four distinct points, then the value of the positive number $a$ is | 6. $a>\sqrt{7}+1$.
Combining the equations $y=-\frac{x^{2}}{4}+4$ and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{3^{2}}=1$, and eliminating $x$, we obtain the quadratic equation in $y$:
$$
a^{2} y^{2}-36 y+\left(144-9 a^{2}\right)=0.
$$
Let $f(y)=a^{2} y^{2}-36 y+\left(144-9 a^{2}\right)$, which is an upward-opening parabola.... | a>\sqrt{7}+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,663 |
7. Given $f(x)=x^{3}-3 x^{2}+3 x$ has a range of $[a, b]$ on the interval $[a, b](b>a)$. Then the interval $[a, b]$ that satisfies the condition is $\qquad$ | 7. $[0,1],[0,2],[1,2]$.
From $f^{\prime}(x)=3 x^{2}-6 x+3=3(x-1)^{2}$, we know that except for $x=1$, $f^{\prime}(x)>0$.
Therefore, $f(x)$ is an increasing function on $(-\infty,+\infty)$.
According to the problem, the function takes the minimum value $a$ at $x=a$ and the maximum value $b$ at $x=b$, then
$$
f(a)=a, f(... | [0,1],[0,2],[1,2] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 721,664 |
8. Given $a=\frac{11 \cdots 1}{2009 \uparrow}, b=\frac{100 \cdots 05}{2008 \uparrow}$. Then $10^{2009}-\sqrt{a b+1}=$ $\qquad$ | 8. $\underbrace{66 \cdots 6}_{20091}$.
From the given, we have
$$
\begin{array}{l}
a=\frac{11 \cdots 1}{2009 \uparrow}=\sum_{i=0}^{2008} 10^{i}=\frac{10^{2009}-1}{9}, \\
b=\underset{2008 \uparrow}{100 \cdots 05}=10^{2009}+5 .
\end{array}
$$
Then $a b+1=\frac{10^{2009}-1}{9}\left(10^{2009}+5\right)+1$
$$
\begin{array}... | 66 \cdots 6_{20091} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,665 |
9. (14 points) (1) Prove that the sum of the distances from each vertex of an equilateral triangle to any tangent of its circumcircle is a constant;
(2) Conjecture whether the spatial proposition “the sum of the distances from each vertex of a regular tetrahedron to any tangent plane of its circumsphere is a constant” ... | (1) As shown in Figure 4, let the center of the circumcircle of the equilateral triangle $\triangle ABC$ be $O$, with radius $R$, and let $l$ be any tangent line. Connect $CO$ to intersect $AB$ at point $D$. Then $D$ is the midpoint of $AB$, and $CD = 3OD$.
In general, let the notation $d_{P}$ represent the distance f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,666 |
10. (15 points) Given that $P$ is a point on the hyperbola $E: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and points $A$ and $B$ are on the two asymptotes of the hyperbola, with $A$ in the first quadrant and $B$ in the fourth quadrant. When $P$, $A$, and $B$ are collinear, find the range of the area of $\... | 10. As shown in Figure 6, given the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
$$
the points on the first and fourth quadrants of the two asymptotes are
$A(a m, b m)$,
$$
B(a n,-b n)(m>0, n>0) \text {. }
$$
Then the equation of the line $A B$ is
$$
(a m-a n)(y-b m)=(b m+b n)(x-a m) \text {, }
$$
which si... | [a b,+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,667 |
11. (15 points) Given that $a, b, c$ are real numbers, and for any integer $n > 1$,
$$
\left(a^{n}-b^{n}\right)\left(b^{n}-c^{n}\right)\left(c^{n}-a^{n}\right) \neq 0.
$$
(1) If $\frac{b-c}{b^{2}-c^{2}}, \frac{c-a}{c^{2}-a^{2}}, \frac{a-b}{a^{2}-b^{2}}$ form an arithmetic sequence, prove that $a^{2}, b^{2}, c^{2}$ also... | 11. From the given, we know that $a \pm b, b \pm c, c \pm a$ are all not 0.
(1) Given that $\frac{b-c}{b^{2}-c^{2}}, \frac{c-a}{c^{2}-a^{2}}, \frac{a-b}{a^{2}-b^{2}}$ form an arithmetic sequence, we have
$$
\begin{array}{l}
\frac{c-a}{c^{2}-a^{2}}-\frac{b-c}{b^{2}-c^{2}}=\frac{a-b}{a^{2}-b^{2}}-\frac{c-a}{c^{2}-a^{2}} ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,668 |
一、(50 points) As shown in Figure 1, let the circumradius of the acute triangle $\triangle ABC$ be $R$. Inside $\triangle ABC$, take a concentric circle $\Gamma$ with the circumcircle, with radius $r (R > r > 0)$. From any point $P$ on circle $\Gamma$, draw $PD \perp BC$ at point $D$, $PE \perp AC$ at point $E$, and $PF... | (1) As shown in Figure 7, connect $A P$ intersecting the circumcircle at point $A_{1}$, connect $B P$ intersecting the circumcircle at point $B_{1}$, and draw the diameter $M N$ through $P$.
By the intersecting chords theorem, we have
$$
\begin{array}{l}
P B \cdot P B_{1} \\
\quad=P M \cdot P N=(R+r)(R-r)=R^{2}-r^{2} .... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,669 |
II. (50 points) Given a function $f(x)$ defined on $\mathbf{R}$ with $f(1)=2$, and for any $x, y \in \mathbf{R}$, we have
$$
f(x+y)=f(x) f(y) \text {. }
$$
Prove that for any rational number $\alpha > 1$, and real number $\beta \in (0,2)$, we have
$$
\sum_{i=1}^{n} \frac{1}{f(\alpha+i)-\beta}<\frac{1}{f(\alpha)-\beta}... | Second, we first prove: for rational number $x$, we have $f(x)=2^{x}$. In fact, from $f(x+y)=f(x) f(y)$, it is easy to get
$$
f\left(\sum_{i=1}^{n} x_{i}\right)=\prod_{i=1}^{n} f\left(x_{i}\right) \text {. }
$$
For positive integers $p$ and $q$, we have
$$
\begin{array}{l}
f(p)=f\left(\sum_{i=1}^{p} 1\right)=\prod_{i=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,670 |
Four. (50 points) As shown in Figure 2, in the cyclic isosceles trapezoid \(ABCD\), it is known that \(AB > CD\), the diagonals \(AC\) and \(BD\) intersect at point \(E\), and all line segments in Figure 2 are positive integers. \(\angle AED = 2 \angle ABD = 120^\circ\), and the radius of the circle \(R = \frac{7 \sqrt... | (1) In $\triangle A B D$, $\angle A B D=60^{\circ}$, and the circumradius is $R=\frac{7 \sqrt{3}}{3}$. By the Law of Sines, we have
$$
A D=2 R \sin \angle A B D=2 \times \frac{7 \sqrt{3}}{3} \sin 60^{\circ}=7 \text {. }
$$
In $\triangle A E D$, given $\angle A E D=120^{\circ}, A D=7$, we know $A D$ is the longest side... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,671 |
Example 3 Let $x_{i}>0(i=1,2, \cdots, n), k \geqslant 1$. Prove:
$$
\sum_{i=1}^{n} \frac{1}{1+x_{i}} \cdot \sum_{i=1}^{n} x_{i} \leqslant \sum_{i=1}^{n} \frac{1}{x_{i}^{k}} \cdot \sum_{i=1}^{n} \frac{x_{i}^{k+1}}{1+x_{i}} \text {. }
$$
(2006, Girls' Mathematical Olympiad)
Analysis: This problem is quite obviously solve... | Proof: Without loss of generality, let $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$.
From the conditions, we easily get
$$
\begin{array}{l}
\frac{1}{x_{1}^{k}} \leqslant \frac{1}{x_{2}^{k}} \leqslant \cdots \leqslant \frac{1}{x_{n}^{k}}, \\
\frac{x_{1}^{k}}{1+x_{1}} \geqslant \frac{x_{2}^{k}}{1+x_{2}} \geqs... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,672 |
Three. (50 points) For any given positive integer $n$ $(n \geqslant 2)$, the sequence $\left\{b_{k}\right\}$ satisfies $b_{1}=1$, and
$$
\frac{b_{k+1}}{b_{k}}=\frac{k-n}{k+1}(k=1,2, \cdots, n-1) \text {. }
$$
(1) Find $b_{1}+b_{2}+\cdots+b_{n}$;
(2) Let $a_{n}=\frac{1}{b_{1}+b_{2}+\cdots+b_{n}}(n=1,2, \cdots)$, prove t... | (1) From $\frac{b_{k+1}}{b_{k}}=\frac{k-n}{k+1}$, we have
$n b_{k}=k b_{k}-(k+1) b_{k}$.
Let $k=1,2, \cdots, n-1$. Adding them up, we get
$$
n \sum_{i=1}^{n} b_{i}=b_{1} \Rightarrow \sum_{i=1}^{n} b_{i}=\frac{b_{1}}{n}=\frac{1}{n} \text {. }
$$
(2) From (1), we have $a_{n}=\left(\sum_{i=1}^{n} b_{i}\right)^{-1}=n(n=1$,... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,673 |
As shown in Figure 3, given that $\triangle ABC$ is an acute triangle, point $A'$ is outside $\triangle ABC$ and within $\angle A$, satisfying
$$
\begin{array}{l}
\angle ABA' + \angle ABC \\
= 180^{\circ}, \\
\angle ACA' + \angle ACB \\
= 180^{\circ}.
\end{array}
$$
Similarly, points $B'$ and $C'$ can be defined.
Prov... | Proof: As shown in Figure 3, let the three interior angles of $\triangle ABC$ be $\angle A, \angle B, \angle C$. From the given information, we have
$$
\begin{array}{l}
\angle ABA^{\prime} + \angle B + \angle ACA^{\prime} + \angle C \\
= 180^{\circ} + 180^{\circ} = 360^{\circ}.
\end{array}
$$
In the convex quadrilater... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,674 |
Initially 258 Draw a line from vertex $C$ of square $A B C D$ outward, intersecting the extensions of $A B$ and $A D$ at points $M$ and $N$ respectively, and let $D M$ and $B N$ intersect at point $H$. Prove: $A H \perp M N$. | Proof: As shown in Figure 4, let $DM$ intersect $BC$ at point $F$, and $BN$ intersect $DC$ at point $F$. Connect $AE$, $AF$, and $EF$.
From $DF \parallel AB$ and $BC \parallel AN$, we get
$$
\begin{array}{l}
\frac{DF}{AB}=\frac{ND}{NA}, \\
\frac{CE}{CB}=\frac{ND}{NA}.
\end{array}
$$
From equations (1) and (2), we hav... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,675 |
258 Test: Can 2008 be expressed as
$$
a_{1}^{x_{1}}+a_{2}^{x_{2}}+\cdots+a_{m}^{x_{m}}-b_{1}^{y_{1}}-b_{2}^{y_{2}}-\cdots-b_{n}^{y_{n}}
$$
? If so, are there infinitely many such representations? Here, $m, n (m-n \geqslant 50)$ are positive integers greater than 100 and less than 170; $a_{1}, a_{2}, \cdots, a_{m}, b_{... | Solution: The representation that satisfies the requirements of the problem does exist, and there are infinitely many. The reason is as follows:
Notice that $7^{2}+11^{2}=170=1^{2}+13^{2}$, thus,
$$
A_{i}=\left(\frac{7}{c_{i}}\right)^{2}+\left(\frac{11}{c_{i}}\right)^{2}-\left(\frac{1}{c_{i}}\right)^{2}-\left(\frac{13}... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,677 |
Example 4 Let $a, b, c > 0, abc = 1$. Prove that for integer $k (k \geqslant 2), \frac{a^{k}}{b+c}+\frac{b^{k}}{c+a}+\frac{c^{k}}{a+b} \geqslant \frac{3}{2}$.
(Fourth China Southeast Mathematical Olympiad)
Analysis: When the variables are sorted, the larger the numerator, the smaller the denominator, and the larger the... | Proof: Without loss of generality, let $a \geqslant b \geqslant c$. Then
$$
\begin{array}{l}
\frac{a^{k}}{b+c} \geqslant \frac{b^{k}}{c+a} \geqslant \frac{c^{k}}{a+b}, \\
b+c \leqslant c+a \leqslant a+b .
\end{array}
$$
By Chebyshev's inequality, we have
$$
\begin{array}{l}
\left(\frac{a^{k}}{b+c}+\frac{b^{k}}{c+a}+\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,678 |
Question As shown in Figure 1, let $A B$ and $C D$ be two diameters of $\odot O$. Draw $P B \perp A B$ through point $B$, and let it intersect the extension of $C D$ at point $P$. Draw line $P E$ through $P$ to intersect $\odot O$ at points $E$ and $F$. Connect $A E$ and $A F$ to intersect $C D$ at points $G$ and $H$ r... | Analysis: This problem is not new; it has been introduced in text [2]. However, few of the summer camp participants can solve it, the reason being the difficulty in connecting equal line segments with a circle. If we utilize a bundle of parallel lines, considering equal line segments on other lines parallel to it, the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,679 |
Example 1 In the pentagram $A B C D E$, the intersection points of the intersecting line segments are labeled as shown in Figure 5. Given that $A Q=Q C, B R=$ $R D, C R=R E, D S=S A$. Prove: $B T=T P=P E$.
| Analysis: In this problem, there are many conditions. Utilizing the knowledge of midlines and doubling the midline, we can find many parallel lines. Projecting the line segments to be proven onto the parallel line bundle, the conclusion becomes evident.
Proof: As shown in Figure 6, connect $A E$ and extend it to inter... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,680 |
3. As shown in Figure 15, in $\triangle ABC$, $D$ and $E$ are two points on side $BC$, and $BD$
$$
= DE = EC, GF \parallel
$$
$BC$, intersecting $AD$ and $AE$
at points $P$ and $Q$, $BF$ intersects
$AD$ and $AE$ at points $M$ and
$K$, $CG$ intersects $AD$ and $AE$
at points $I$ and $N$. Prove:
$$
\frac{1}{DE}, \frac{1}... | (Given $B D=D E=E C, G F // B C$, we get $G P=P Q=Q F$.
Then $P I: I D=P G: C D=\frac{G F}{3}: \frac{2 B C}{3}=Q K: K E$.
Therefore, $I K // D E$. Similarly, $M N // D E$.
Hence $P Q // I K // M N // D E$.
Next, we only need to prove $\frac{P Q}{M N}+1=\frac{2 P Q}{I K}$ and $\frac{D E}{I K}+1=$ $\frac{2 D E}{M N}$. ) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,681 |
One, (50 points) As shown in Figure 1, let quadrilateral $ABCD$ be a given convex quadrilateral, with $BC = DA$, but $BC \neq D$. There are moving points $E$ and $F$ on sides $BC$ and $DA$, respectively. Line $AC$ intersects $BD$ at point $P$, and line $EF$ intersects $BD$ and $AC$ at points $Q$ and $R$, respectively. ... | One, Auxiliary Lines
As shown in Figure 3. It is easy to see that
$$
\begin{array}{l}
O A=O C, \\
O B=O D .
\end{array}
$$
From $B C=A D$
$$
\begin{aligned}
\Rightarrow & \triangle B O C \\
& \cong \triangle D O A \\
\Rightarrow & \angle C B O=\angle A D O, \angle B O C=\angle A O D \\
\Rightarrow & \angle A O C
\end{... | BE = DF | Geometry | proof | Yes | Yes | cn_contest | false | 721,682 |
II. (50 points) Let $a_{1}, a_{2}, \cdots, a_{2009}$ be non-negative real numbers, satisfying:
(1) $a_{1}+a_{2}+\cdots+a_{2009}=2$;
(2) $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{208} a_{2009}+a_{2009} a_{1}=1$.
Find the maximum and minimum values of $S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{2009}^{2}$. | Let's assume $a_{2009}$ is the smallest among these $a_{i}$, and $a_{1}>0$. From the given information, we have
$$
\begin{aligned}
4= & \left(a_{1}+a_{2}+\cdots+a_{2009}\right)^{2} \\
\geqslant & \left(a_{1}+a_{2}+\cdots+a_{2009}\right)^{2}- \\
& \left(a_{1}-a_{2}+a_{3}-\cdots-a_{2008}+a_{2009}\right)^{2} \\
= & 4\left... | S_{\text{max}} = 2, \, S_{\text{min}} = \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,683 |
Three, (50 points) Find the number of all sequences that satisfy the following conditions:
(1) Each term is an integer no less than 2;
(2) The sum of all terms equals a fixed value $m$.
| Three, the sequence that satisfies conditions (1) and (2) is called an "$f(m)$ sequence". Below, we calculate the number of $f(m)$ sequences.
Since for a sequence $\left\{a_{n}\right\}$ with $k$ terms, there is
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{k} \\
=m\left(a_{i} \geqslant 2, i=1,2, \cdots, k\right),
\end{arra... | \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{m-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{m-1}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,684 |
Four, (50 points) Prove: There exist infinitely many rectangular prisms with edge lengths as positive integers, whose volume is exactly equal to the square of the length of the diagonal, and each face of the prism can always be divided into two squares with integer side lengths.
保留源文本的换行和格式，直接输出翻译结果如下：
Four, (50 poin... | Let the edge lengths of a rectangular prism be $x, y, z$. According to the problem, we have $x^{2}+y^{2}+z^{2}=x y z$.
The problem is to prove that the equation $x^{2}+y^{2}+z^{2}=x y z$ has infinitely many positive integer solutions $\left(a_{n}, b_{n}, c_{n}\right)$, and that among $a_{n}, b_{n}, c_{n}$, the product ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,685 |
The number 2592009 can be decomposed into the sum of the squares of five positive integers, the sum of these five positive integers is 99, and one of the numbers is 19. Write out this decomposition. | Solution: From $99-19=80$, we know that the sum of the four numbers other than 19 is 80, and the average is 20. Therefore, we can set
$$
\sum_{i=1}^{4}\left(20+a_{i}\right)^{2}+19^{2}=2009,
$$
where, $\sum_{i=1}^{4} a_{i}=0, a_{i}(i=1,2,3,4)$ are integers.
Notice that
$$
400 \times 4+40 \sum_{i=1}^{4} a_{i}+\sum_{i=1}... | 2009=26^{2}+19^{2}+18^{2}+18^{2}+18^{2} \text{ and } 2009=22^{2}+22^{2}+22^{2}+19^{2}+14^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,686 |
As shown in Figure 1, in quadrilateral $ABCD$, it is given that $\angle C=90^{\circ}, \angle CDB=66^{\circ}, \angle BDA=30^{\circ}$, $\angle DAB=96^{\circ}$. Prove: $AD=2CD$.
保留源文本的换行和格式，直接输出翻译结果如下：
As shown in Figure 1, in quadrilateral $ABCD$, it is given that $\angle C=90^{\circ}, \angle CDB=66^{\circ}, \angle BDA... | Proof: As shown in the figure,
1. First, consider the circumcenter $E$ of $\triangle A B D$.
$$
\begin{array}{l}
\text { By } \angle A E B \\
=2 \angle A D B \\
=60^{\circ},
\end{array}
$$
we know that
$$
\begin{array}{l}
E D=E A \\
=E B=A B .
\end{array}
$$
It is easy to see that
$$
\angle E B C=18^{\circ} \text {. ... | AD=2CD | Geometry | proof | Yes | Yes | cn_contest | false | 721,687 |
259 As shown in Figure 2, $P$ is a point outside circle $\odot O$, and $PAB$, $PCD$, and $PEF$ are three secants of $\odot O$. $A, B, C, D, E,$ and $F$ are the intersection points of the secants with $\odot O$. Secant $PEF$ intersects $AD$ and $BC$ at points $M$ and $N$. Let $\angle APE = \alpha$ and $\angle EPC = \bet... | Proof: (1) As shown in Figure 2, draw a circle with center $O$ passing through point $P$, and intersecting the secant lines $P A B$, $P C D$, and $P E F$ at points $K$, $L$, and $Q$. By the perpendicular diameter theorem, it is easy to prove
$$
\begin{array}{l}
P A = B K, \\
P C = D L, \\
P E = F Q.
\end{array}
$$
By ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,688 |
Given 260 Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1(a, b, c>0)$, $n \in \mathbf{N}_{+}$. Prove:
$$
\frac{a^{n}}{a+2 b+3 c}+\frac{b^{n}}{b+2 c+3 a}+\frac{c^{n}}{c+2 a+3 b} \geqslant \frac{3^{n}}{6} .
$$ | Prove: When $n=1$, make the substitution
$$
\begin{aligned}
a+2 b+3 c & =18 x, b+2 c+3 a=18 y, \\
c+2 a+3 b & =18 z .
\end{aligned}
$$
Solving, we get $a=-5 x+7 y+z, b=x-5 y+7 z$, $c=7 x+y-5 z$.
$$
\begin{array}{l}
\text { Then } \frac{a}{a+2 b+3 c}+\frac{b}{b+2 c+3 a}+\frac{c}{c+2 a+3 b} \\
=\frac{-5 x+7 y+z}{18 x}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,689 |
4. As shown in Figure 16, given that the chord $AB$ of $\odot O$ is trisected by points $C$ and $D$, and $E$ and $F$ are the trisection points of arc $\overparen{AB}$. Connecting $EC$ and $FD$ intersect at point $S$, and connecting $SA$ and $SB$. Prove:
$$
\begin{array}{l}
\angle A S B \\
=\frac{1}{3} \angle A O B
\end... | (提示: Auxiliary lines as shown in Figure 16. It is easy to know that $A B$ // $M N$. It is easy to prove that $M E=A E=E F=F B=F N ; A F \perp S A$.
Also, $A F \perp O E$, so $O E / / S A$.
Therefore, $\angle A S B=\angle E O F=\frac{1}{3} \angle A O B$. ) | \angle A S B = \frac{1}{3} \angle A O B | Geometry | proof | Yes | Yes | cn_contest | false | 721,690 |
Example 1 Let $O$ be the vertex of a parabola, $F$ the focus, and $P Q$ a chord passing through point $F$. Given $|O F|=a,|P Q|=$ $b$. Find the area of $\triangle O P Q$.
(1991, National High School Mathematics Competition) | Explanation: As shown in Figure 1, establish a polar coordinate system with $F$ as the pole and $F x$ as the polar axis. Then the equation of the parabola is
$$
\rho=\frac{2 a}{1-\cos \theta} .
$$
Let point $P\left(\rho_{1}, \theta\right)(0<\theta<\pi)$. Then
$$
\begin{array}{l}
Q\left(\rho_{2}, \theta+\pi\right) . \\... | a \sqrt{a b} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,691 |
Example 2 Let $P_{i}(i=1,2, \cdots, n)$ be $n$ points on an ellipse, $F$ be the left focus of the ellipse, and the line segments $F P_{1}, F P_{2}$, $\cdots, F P_{n}$ divide the angle at $F$ into $n$ equal parts. Prove:
$$
\sum_{i=1}^{n} \frac{1}{\left|F P_{i}\right|}=\frac{n}{e p} .
$$ | Explanation: Establish a polar coordinate system with $F$ as the pole and $F x$ as the polar axis. Let point $P_{1}\left(\rho_{1}, \theta\right)$. Then, we have $P_{2}\left(\rho_{2}, \theta+\frac{2 \pi}{n}\right), P_{3}\left(\rho_{3}, \theta+\frac{4 \pi}{n}\right), \cdots, P_{n}\left(\rho_{n}, \theta+\frac{2(n-1) \pi}{... | \frac{n}{e p} | Geometry | proof | Yes | Yes | cn_contest | false | 721,692 |
Example 3 Given the parabola $C: y^{2}=4 x, F$ is the focus of the parabola $C$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$, and let $\overrightarrow{F B}=\lambda \overrightarrow{A F}$. If $\lambda \in[4,9]$, find the range of the y-intercept of line $l$. | Explanation: The line $l$ is not fixed, and its position relationship satisfies $\overrightarrow{F B}=\lambda \overrightarrow{A F}, \lambda$ varies within the range $[4,9]$. Since $\overrightarrow{F B}$ and $\overrightarrow{A F}$ are in the same direction, $\lambda$ represents the ratio of the lengths of $\overrightarr... | \left[-\frac{4}{3},-\frac{3}{4}\right] \cup\left[\frac{3}{4}, \frac{4}{3}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,693 |
Example 4 As shown in Figure 3, let $F$ be the focus of the parabola $y^{2}=4 x$, and $A, B$ be two points on the parabola different from the origin $O$, satisfying $\overrightarrow{F A} \cdot \overrightarrow{F B}=0$. Extend $A F, B F$ to intersect the parabola at points $C, D$ respectively. Find the minimum value of t... | As shown in Figure 3, with $F$ as the pole and $F x$ as the polar axis, the equation of the parabola is
$$
\rho=\frac{2}{1-\cos \theta} \text {. }
$$
Let point $A\left(\rho_{1}, \theta\right)(0<\theta<\pi)$. Then
$$
\begin{array}{l}
B\left(\rho_{2}, \theta+\frac{\pi}{2}\right) 、 C\left(\rho_{3}, \theta+\pi\right) 、 \\... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,694 |
Example 5: Through the right focus $F$ of the hyperbola $x^{2}-y^{2}=1$, a chord $M N$ is drawn intersecting the right branch, and $P$ is the midpoint of the chord $M N$.
(1) Find the equation of the locus of point $P$;
(2) Draw $P Q \perp M N$ intersecting the $x$-axis at point $Q$, prove that $\frac{|M N|}{|F Q|}=\sq... | (1) Since the chord $M N$ passes through the right focus $F$, we can consider using the polar equation of the hyperbola to solve the problem. Establish a polar coordinate system with the right focus $F(\sqrt{2}, 0)$ as the pole and $F x$ as the polar axis. Then the polar equation of the right branch of the hyperbola is... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,695 |
Example 6 Given the circle $C_{0}: x^{2}+y^{2}=1$ and the ellipse $C_{1}$ : $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. Try to find: Under what conditions on $a$ and $b$ does there exist, for any point $P$ on the ellipse $C_{1}$, a parallelogram with $P$ as a vertex, that is externally tangent to the circle $C_... | Explanation: This problem can start from a special figure, taking $P$ as one of the vertices of the ellipse, finding the conditions that $a$ and $b$ should satisfy under this condition, and then further proving the general case. By plane geometry knowledge, a circumscribed parallelogram of a circle is definitely a rhom... | \frac{1}{a^{2}}+\frac{1}{b^{2}}=1 | Geometry | proof | Yes | Yes | cn_contest | false | 721,696 |
1. Given the length of the major axis of an ellipse $\left|A_{1} A_{2}\right|=6$, and the focal distance $\left|F_{1} F_{2}\right|$ $=4 \sqrt{2}$, a line is drawn through the left focus $F_{1}$ intersecting the ellipse at points $M$ and $N$. Let $\angle F_{2} F_{1} M=\alpha(0 \leqslant \alpha<\pi)$. When $|M N|$ equals... | (Tip: The polar equation of the ellipse is
$$
\rho=\frac{1}{3-2 \sqrt{2} \cos \theta} \text {. }
$$
Let $M\left(\rho_{1}, \alpha\right) 、 N\left(\rho_{2}, \alpha+\pi\right)$. Then
$$
\begin{array}{l}
|M N|=\frac{1}{3-2 \sqrt{2} \cos \alpha}+\frac{1}{3-2 \sqrt{2} \cos (\alpha+\pi)} \\
=\frac{6}{9-8 \cos ^{2} \alpha} .
... | \alpha=\frac{\pi}{6} \text{ or } \frac{5 \pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,697 |
2. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>$ $0)$ with the right focus at $F$, a line passing through $F$ with a slope of $\sqrt{3}$ intersects the hyperbola $C$ at points $A$ and $B$. If $\overrightarrow{A F}=4 \overrightarrow{F B}$, then the eccentricity of the hyperbola $C$ is ( ).
(... | (Tip: Let the equation of the hyperbola be
$$
\rho=\frac{e p}{1-e \cos \theta} \text {. }
$$
It is easy to know that $\rho_{1}=|\overrightarrow{A F}|=\frac{e p}{1-e \cos \frac{\pi}{3}}$,
$$
\rho_{2}=|\overrightarrow{F B}|=\frac{e p}{1-e \cos \left(\frac{\pi}{3}+\pi\right)} .
$$
From $\overrightarrow{A F}=4 \overright... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,698 |
Example 2 As shown in Figure 7, given that $A B$ is the diameter of $\odot O$, $B C$ is the tangent of $\odot O$, and $O C$ is parallel to chord $A D$. A perpendicular line $D E \perp A B$ is drawn through point $D$ at point $E$, and $A C$ intersects $D E$ at point $P$. Question: Are $E P$ and $P D$ equal? Prove your c... | Proof: As shown in Figure 7, extend \( AD \) to intersect the extension of \( BC \) at point \( F \).
Since \( OC \parallel AD \) and \( AO = OB \), it follows that \( BC = CF \).
Also, since \( ED \parallel BF \), it follows that \( EP = PD \). | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,699 |
3. A line $l$ passing through the right focus $F$ of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$, and there are exactly 3 such lines, find $\lambda$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | (The polar equation of the hyperbola is
$$
\rho=\frac{2}{1-\sqrt{3} \cos \theta} \text {. }
$$
Let $A B$ be a chord passing through the right focus and intersecting only the right branch.
Then $|A B|=\rho_{1}+\rho_{2}$
$$
\begin{array}{l}
=\frac{2}{1-\sqrt{3} \cos \theta}+\frac{2}{1-\sqrt{3} \cos (\theta+\pi)} \\
=\fr... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,700 |
5. Given the ellipse $\frac{x^{2}}{24}+\frac{y^{2}}{16}=1$, the line $l: \frac{x}{12}+\frac{y}{8}=$ $1, P$ is a point on $l$, the ray $O P$ intersects the ellipse at point $R$, and point $Q$ is on the ray $O P$ such that $|O Q| \cdot|O P|=|O R|^{2}$. When point $P$ moves along the line $l$, find the equation of the tra... | (In the polar coordinate system, the equations of the line $l$ and the ellipse are
$$
\rho=\frac{24}{2 \cos \theta+3 \sin \theta}, \rho^{2}=\frac{48}{2+\sin ^{2} \theta} .
$$
Let $Q(\rho, \theta) 、 R\left(\rho_{1}, \theta\right) 、 P\left(\rho_{2}, \theta\right)$. Then
$$
\rho=|O Q| \neq 0 \text { and } \rho \rho_{2}=\... | \frac{(x-1)^{2}}{\frac{5}{2}}+\frac{(y-1)^{2}}{\frac{5}{3}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,701 |
Example 1 A deck of cards consists of 52 cards, with 13 cards of each of the four suits: "Diamonds," "Clubs," "Hearts," and "Spades." The cards are numbered $2,3, \cdots, 10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}$. Two cards of the same suit and consecutive numbers are called "straight flush" cards, and $\mat... | Solution: First, generalize the problem as follows:
$$
\begin{array}{l}
\text { Let } n \geqslant 3 \text {, from } \\
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right) 、 B=\left(b_{1}, b_{2}, \cdots, b_{n}\right) \text { 、 } \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right) 、 D=\left(d_{1}, d_{2}, \cdots, d_{n}\right) \\
\end{ar... | 3^{13}-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,702 |
Example 2 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 or 8. How many different ways are there to select such a group of eight points? Explain your reasoning.
(2001, China Mathematical Oly... | Solution: Number the points in sequence as $1, 2, \cdots, 24$, and list a $3 \times 8$ number table (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
\hline 9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
\hline 17 & 20 & 23 & 2 & 5 & 8 & 11 & 14 \\
\hline
\end{tabular}
In th... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,703 |
Example 3 A deck of tri-color cards, totaling 32 cards, includes 10 cards of each color: red, yellow, and blue, numbered $1,2, \cdots, 10$; there are also a big joker and a small joker, each numbered 0. Several cards are drawn from this deck, and their scores are calculated according to the following rule: each card nu... | Solution: This problem is referred to as the "Two Kings Problem." If an additional joker (called the "Middle King") with a value of 0 is added, and each card is still assigned a value as described, it is called the "Three Kings Problem."
Since $2004 < 2^{11}$, adding some cards with values $2^{11}, 2^{12}, \cdots$ doe... | 1006009 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,704 |
Example 1 Non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, and let $S=\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b+\frac{1}{4}(c-a)^{2}}+\sqrt{c+\frac{1}{4}(a-b)^{2}}$. Prove: $\frac{3}{2} \leqslant S \leqslant 2$.
When $a=b=c=\frac{1}{3}$, $S=\sqrt{3}$, so the maximum and minimum values are not conventional extreme va... | Proof: By symmetry, without loss of generality, assume $a \leqslant b \leqslant c\left(0 \leqslant a \leqslant \frac{1}{3}\right)$.
It is easy to see that $\sqrt{a+\frac{1}{4}(b-c)^{2}} \geqslant \sqrt{a}$.
Next, we prove:
$$
\sqrt{b+\frac{1}{4}(c-a)^{2}} \geqslant b+\frac{c-a}{2} \text {. }
$$
In fact,
Equation (1)
$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,705 |
Example 2 Non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Prove:
$$
S=\frac{x^{2} y}{x+z}+\frac{y^{2} z}{y+x}+\frac{z^{2} x}{z+y} \leqslant \frac{1}{4} .
$$
When $x=y=z=\frac{1}{3}$, $S=\frac{1}{6}$ is not the maximum value, so it is an unconventional extremum. | Proof: Given $x+y+z=1$, we know that at least one of $x, y, z$ is not less than $\frac{1}{3}$. Without loss of generality, assume $x \geqslant \frac{1}{3}$. Then
$$
\begin{aligned}
& S=\frac{x y(x+z)}{x+z}-\frac{x y z}{x+z}+\frac{y z(y+x)}{y+x}- \\
& \frac{x y z}{y+x}+\frac{z x(z+y)}{z+y}-\frac{x y z}{z+y} \\
= & x y+y... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,706 |
Example 3 Non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Prove:
$$
S=x^{2} y+y^{2} z+z^{2} x \leqslant \frac{4}{27}.
$$
When $x=y=z=\frac{1}{3}$, $S=\frac{1}{9}$ is not the maximum value, so it is an unconventional extremum. | Prove: Let $S$ be expressed as a function of $x$ and $y$:
$$
\begin{array}{l}
S=x^{2} y+y^{2}(1-x-y)+(1-x-y)^{2} x \\
=x^{3}+(3 y-2) x^{2}+(1-2 y) x+y^{2}(1-y) .
\end{array}
$$
For any given value $y_{0}$ of $y$, $S$ can be considered as a cubic function of the variable $x$.
Differentiating with respect to $x$ gives:
... | \frac{4}{27} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,707 |
Prove: $\mathrm{C}_{n}^{0} \cdot \frac{3^{0}}{3^{0}+1}+\mathrm{C}_{n}^{1} \cdot \frac{3^{1}}{3^{1}+1}+\cdots+\mathrm{C}_{n}^{n} \cdot \frac{3^{n}}{3^{n}+1}$ $\geqslant \frac{3^{n} \times 2^{n}}{3^{n}+2^{n}}$. | Proof: Since $\mathrm{C}_{n}^{k}\left[1+\left(\frac{1}{3}\right)^{k}\right]>0(k=0,1$, $\cdots, n)$, by the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
C_{n}^{0} \cdot \frac{3^{0}}{3^{0}+1}+C_{n}^{1} \cdot \frac{3^{1}}{3^{1}+1}+\cdots+C_{n}^{n} \cdot \frac{3^{n}}{3^{n}+1} \\
=\frac{\left(\mathrm{C}_{n}^{0}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,709 |
Example 3 As shown in Figure 8, in $\triangle ABC$, $AB > BC$. The median and angle bisector from point $B$ intersect $AC$ at points $M$ and $L$, respectively. $MD \parallel AB$ intersects $BL$ at point $D$, $LE \parallel BC$ intersects $BM$ at point $E$ and intersects $MD$ at point $X$. Prove: $ED \perp BL$. (2000, Na... | Analysis: This problem also involves completing the basic figure of a set of parallel lines.
Proof: As shown in Figure 8, extend $M D$ to intersect $B C$ at point $P$.
Since $A B / / M D$ and $A M = M C$, we get $B P = P C$.
Also, since $E L / / B C$, we have $E X = X L$, and
$\angle A B L = \angle C B L$.
Therefore, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,710 |
1. Let $n$ be a positive integer, and $p$ be a prime number. Prove: If integers $a, b, c$ (not necessarily positive) satisfy
$$
a^{n}+p b=b^{n}+p c=c^{n}+p a,
$$
then $a=b=c$. | Proof 1: If $a, b, c$ have two equal, then we immediately get $a=b=c$. If $a, b, c$ are all distinct, then
$$
a^{n}-b^{n}=-p(b-c),
$$
$$
b^{n}-c^{n}=-p(c-a),
$$
$$
c^{n}-a^{n}=-p(a-b).
$$
Thus, $\frac{a^{n}-b^{n}}{a-b} \cdot \frac{b^{n}-c^{n}}{b-c} \cdot \frac{c^{n}-a^{n}}{c-a}=-p^{3}$.
If $n$ is odd, then $a^{n}-b^{n}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,711 |
2. Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ be distinct positive integers. Prove: there exist distinct indices $i, j$, such that $a_{i}+a_{j}$ does not divide any of $3 a_{1}, 3 a_{2}, \cdots, 3 a_{n}$. | 2. Without loss of generality, let $0<a_{1}<a_{2}<\cdots<a_{n}$. Assume for contradiction that $a_{1}+a_{j} \mid 3 a_{i}$ for some $i, j$ with $1 \leq i, j \leq n$ and $i \neq j$.
If $i<n$, then $a_{1}+a_{j} \mid 3 a_{n-1}$. Since $a_{1}+a_{j} \leq a_{1}+a_{n-1}<2 a_{n-1}$, it follows that $a_{1}+a_{j}=a_{n-1}$, which... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,712 |
3. Let the sequence of positive integers $a_{0}, a_{1}, \cdots$ satisfy that the greatest common divisor of any two consecutive terms is greater than the term preceding them, i.e., $\left(a_{i}, a_{i+1}\right)>a_{i-1}$. Prove: For all non-negative integers $n$, $a_{n} \geqslant 2^{n}$. | 3. Since $a_{i} \geqslant\left(a_{i}, a_{i+1}\right)>a_{i-1}$, the sequence is strictly increasing. In particular, $a_{0} \geqslant 1, a_{1} \geqslant 2$.
For each $i \geqslant 1$, we have
$a_{i+1}-a_{i} \geqslant\left(a_{i}, a_{i+1}\right)>a_{i-1}$.
Thus, $a_{i+1} \geqslant a_{i}+a_{i-1}+1$.
Hence $a_{2} \geqslant 4, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,713 |
5. Let $\mathbf{N}_{+}$ be the set of all positive integers. For each $n \in \mathbf{N}_{+}$, let the number of positive divisors of $n$ be $d(n)$. Find all functions $f\left(f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}\right)$ that satisfy the following properties.
(1) For all $x \in \mathbf{N}_{+}$,
$$
d(f(x))=x \te... | 5. There exists a unique function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ satisfying the condition:
$$
f(1)=1, f(n)=\prod_{i=1}^{k} p_{i}^{p_{i}^{p_{i-1}}},
$$
where, $n=\prod_{i=1}^{k} p_{i}^{a_{i}}$ is the prime factorization of $n(n>1)$.
Direct verification shows that the function defined by (1) satisfies th... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,715 |
2. Inputting a positive integer $n$ into a machine will produce a positive integer $\frac{n(n+1)}{2}$. If 5 is input into the machine, and then the number produced is input into the machine again, the final number produced by the machine is $\qquad$ . | 2. 120 .
The first result is $\frac{5(5+1)}{2}=15$;
The second result is $\frac{15(15+1)}{2}=120$. | 120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,717 |
3. $A, B, C$ three people pick watermelons. The sum of the number of watermelons picked by $A$ and $B$ is 6 less than that picked by $C$; the sum of the number of watermelons picked by $B$ and $C$ is 16 more than that picked by $A$; the sum of the number of watermelons picked by $C$ and $A$ is 8 more than that picked b... | 3. 60.
Let the number of watermelons collected by $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Then
\[
\begin{array}{l}
a+b-c=-6, \\
b+c-a=16, \\
c+a-b=8 .
\end{array}
\]
By adding (1) + (2) + (3), we get $a+b+c=18$.
Thus, we have $c=12, b=5, a=1 \Rightarrow a b c=60$. | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,718 |
4. As shown in Figure 1, the center of a semicircle is $O$. A beam of light travels from point $M$ perpendicular to $PA$ towards the semicircle, and reflects at point $C$ on the circumference. The reflection angle $\angle OCB$ equals the incidence angle $\angle MCO$; then the light intersects the circumference again at... | 4. $36^{\circ}$.
As shown in Figure 7, a circle is drawn with $PA$ as its diameter, and the trajectory of the light refraction forms a regular pentagon $ABCDE$. Then
$$
\begin{array}{l}
\angle DCB=108^{\circ} \\
\angle OCM=54^{\circ} \\
\angle COM=36^{\circ}
\end{array}
$$ | 36^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,719 |
5. There are 19 children with ages ranging from $1 \sim 19$ years old standing in a circle. Record the difference in ages between each pair of adjacent children. The maximum possible sum of these 19 differences is . $\qquad$ | 5. 180 .
19 differences are derived from 19 equations, where each integer from $1 \sim 19$ appears twice. To maximize the sum of the differences, the larger numbers 11 and 19 should all be used as minuends, and the smaller numbers $1 \sim 9$ should all be used as subtrahends. The number 10 should be used once as a subt... | 180 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,720 |
Example 4 As shown in Figure $9, \odot O_{1}$ and $\odot O_{2}$ intersect at points $M$ and $N$. Let $l$ be the one of the two common tangents of $\odot O_{1}$ and $\odot O_{2}$ that is closer to $M$. $l$ is tangent to $\odot O_{1}$ at point $A$ and to $\odot O_{2}$ at point $B$. Let the line passing through point $M$ ... | Analysis: This problem requires proving the equality of line segments, which essentially involves using the property of isosceles triangles that the altitude, median, and angle bisector coincide. We can use symmetry to prove perpendicularity and then use the properties of parallel lines to prove the midpoint.
Proof: A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,721 |
6. $\frac{\left(2^{4}+2^{2}+1\right)\left(4^{4}+4^{2}+1\right)\left(6^{4}+6^{2}+1\right)\left(8^{4}+8^{2}+1\right)\left(10^{4}+10^{2}+1\right)}{\left(3^{2}+1\right)\left(5^{4}+5^{2}+1\right)\left(7^{4}+7^{2}+1\right)\left(9^{4}+9^{2}+1\right)\left(11^{4}+11^{2}+1\right)}$ $=$ | 6. $\frac{3}{133}$.
Given $x^{4}+x^{2}+1=\left(x^{2}+1\right)^{2}-x^{2}$ $=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$,
and $(x+1)^{2}-(x+1)+1=x^{2}+x+1$,
then the original expression
$$
\begin{aligned}
= & \frac{1^{2}+1+1}{3^{2}+3+1} \times \frac{3^{2}+3+1}{5^{2}+5+1} \times \frac{5^{2}+5+1}{7^{2}+7+1} \times \\
& ... | \frac{3}{133} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,722 |
7. In a certain exam, the passing score is 65 points. The overall average score of the class is 66 points, and the average score of all students who passed is 71 points, while the average score of all students who failed is 56 points. To reduce the number of failing students, the teacher adds 5 points to each student's... | 7. 24 .
Let the class have $n$ students. If before the bonus points, $x$ students passed the exam, then
$$
71 x+56(n-x)=66 n \Rightarrow x=\frac{2 n}{3},
$$
which means $n$ is a multiple of 3.
Let $y$ be the number of students who passed after the bonus points. Then
$$
75 y+59(n-y)=71 n \Rightarrow y=\frac{3 n}{4},
$... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,723 |
8. As shown in Figure 2, given that $A, B, C, D$ are four points on a plane that are not concyclic,
$$
\begin{array}{l}
\triangle A B D, \triangle A D C \text {, } \\
\triangle B C D, \triangle A B C
\end{array}
$$
have circumcenters
$$
\text { as } E, F, G, H \text {, }
$$
respectively. The line segments $E G, F H$
... | 8.4.
Since points $E$ and $G$ lie on the perpendicular bisector of $BD$, $EG$ is the perpendicular bisector of segment $BD$. Similarly, $FH$ is the perpendicular bisector of segment $AC$. Since $I$ is the intersection of $EG$ and $FH$, it follows that $CI = AI = 4$. | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,724 |
9. The number of different right-angled triangles (congruent triangles are considered the same) with $2009^{12}$ as one of the legs and all three sides as integers is $\qquad$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 9. 612 .
Let $a$, $2009^{12}$, and $c$ be the sides of a right triangle ($c$ being the hypotenuse). Then
$$
(c-a)(c+a)=2009^{24}=41^{24} \times 7^{48}.
$$
Since $c+a > c-a$, and these two numbers have the same parity, they are both odd.
Also, $2009^{24}=41^{24} \times 7^{48}$ has $25 \times 49$ different factors, wh... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,725 |
10. There is a six-digit number, the sum of its digits is divisible by 26. When this six-digit number is increased by 1, the sum of the digits of the resulting number is also divisible by 26. The smallest six-digit number that satisfies the above conditions is $\qquad$ | 10. 898999.
Obviously, this six-digit number will have a carry when 1 is added.
To make the sum of the digits a multiple of 26, at least three positions need to carry over. Therefore, the last three digits of this six-digit number are all 9, and the sum of the digits of the first three is 25.
Thus, the smallest number... | 898999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,726 |
11. On a circle, there is 1 red point and 2009 blue points. Xiao Dan calculates the number of convex polygons where all vertices are blue points, while Xiao Dong calculates the number of convex polygons where one vertex is the red point. The difference between the two numbers they get is $\qquad$ | 11. 2017036.
For every convex $n$-sided polygon $(n \in \mathbf{N}_{+})$ calculated by Xiao Dan, there is a corresponding convex $(n+1)$-sided polygon calculated by Xiao Dong. Xiao Dong's convex $(n+1)$-sided polygon has one more red vertex than Xiao Dan's convex $n$-sided polygon. This correspondence does not hold wh... | 2017036 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,727 |
1. In an international chess tournament with 10 participants, each player must play exactly one game against every other player. After several games, it is found that among any three players, at least two have not yet played against each other. How many games have been played at most by this point? | 2. Suppose there are 5 men and 5 women among the 10 players, and all matches so far have been between men and women, thus satisfying the condition, i.e., 25 matches have been played.
Next, we prove: 25 is indeed the maximum value.
Let $k$ be the number of matches played by the player who has played the most matches, de... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,729 |
2. $P$ is a point inside $\triangle A B C$ such that $\angle P B C=$ $30^{\circ}, \angle P B A=8^{\circ}$, and $\angle P A B=\angle P A C=22^{\circ}$. What is $\angle A P C$ in degrees? | 2. As shown in Figure 8, extend $A C$ to point $Q$ such that $A B = A Q$. Then
$$
\begin{array}{l}
\triangle B A P \cong \triangle Q A P . \\
\text { Also, } \angle A P B \\
= 180^{\circ} - \angle P B A - \\
\angle P A B \\
= 150^{\circ}, \\
\text { then } \angle B P Q \\
= 360^{\circ} - \angle A P B - \angle A P Q = ... | 142^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,730 |
3. Find the smallest positive integer that can be expressed as the sum of the squares of four positive integers and can divide some integer of the form $2^{n}+15\left(n \in \mathbf{N}_{+}\right)$. | 3. The smallest five positive integers that can be expressed as the sum of four positive integer squares are
$$
\begin{array}{l}
4=1+1+1+1, \\
7=4+1+1+1, \\
10=4+4+1+1, \\
12=9+1+1+1, \\
13=4+4+4+1 .
\end{array}
$$
Obviously, since $2^{n}+15$ is odd, the smallest positive integer cannot be 4, 10, or 12.
Also, $2^{n} ... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,731 |
Example 5 As shown in Figure $10, \triangle A B C$ has an incircle $\odot O$ that touches $B C, C A, A B$ at points $D, E, F$, respectively. A line parallel to $B C$ is drawn through $F$ and intersects line $D A$ and $D E$ at points $H$ and $G$, respectively. Prove that $F H = H G$.
---
The translation preserves the ... | Proof: As shown in Figure 10, draw a line through point $A$ parallel to $BC$, intersecting $DF$ and $DG$ at points $P$ and $Q$ respectively.
From $\triangle APE \sim \triangle BDF$, and given $BF=BD$, we get $AP=AF$.
Similarly, $AE=AQ$.
Since $AE=AF$, it follows that $AP=AQ$.
Given that $FG \parallel BC$, therefore, $F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,732 |
1. (40 points) There is a stack of cards numbered $1 \sim 15$. After arranging them in a certain order, the following two operations are performed: Place the top card on the table, then place the second card at the bottom of the stack. These two operations are repeated until all 15 cards are sequentially placed on the ... | 1. Assuming the cards are arranged in descending order from top to bottom as shown from left to right in Table 1. The first card originally becomes the bottom card, so it must be 15. The third card must be 14. And so on.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & $\beta$ & 13 &... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,733 |
2. (40 points) Find the smallest positive integer such that its factors include at least one number ending in each of the digits $0, 1, \cdots, 9$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. If a factor ends with 0, then it must contain the prime factors $2, 5$. Clearly, it has a factor ending with 1, and it also contains the factor $3^{3}$. Therefore, the number is $2 \times 3^{3} \times 5 = 270$. At this point, the factors are
$$
\begin{array}{l}
3,2 \times 3^{3}=54,2 \times 3=6, \\
3^{3}=27,2 \times ... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 721,734 |
3. (40 points) Fill in the squares in Figure 3 with the numbers $1 \sim 6$, so that each row, each column, and both main diagonals contain the numbers 1～6 exactly once.
Figure 3 should be filled as follows: | 3. Label each row from bottom to top as 1 to 6, and each column from left to right as $a$ to $f$. It can be observed that $(f, 1)$ and $(f, 6)$ are either 5 or 6. Therefore, $(f, 4)$ must be 3, and $(f, 3)$ must be 2; in the 2nd row, 2 must be at $(d, 2)$, and 3 must be at $(e, 2)$; in the $b$ column, 3 must be at $(b,... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,735 |
4. (40 points) Figure 4 is a schematic diagram of a commercial center, where the areas marked with the letter $P$ are three parking lots, and the other seven shaded areas are stores. The arrows in Figure 4 indicate that the segments of the street are one-way, with the arrow direction showing the direction of travel (se... | 4. Number the seven stores as shown in Figure 10, and assume the right side is the east side, starting from the easternmost parking lot.
Since Store 4 can only exit from its west exit, and Store 7 can only exit from its north exit, it follows that Store 5 must immediately follow Store 4, and the parking lot on the wes... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,736 |
6. (40 points) Given
$$
a_{n}=\frac{2^{n}}{2^{2 n+1}-2^{n+1}-2^{n}+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: $a_{1}+a_{2}+\cdots+a_{2009}<1$. | 6. Notice that
$$
a_{n}=\frac{2^{n}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}=\frac{1}{2^{n}-1}-\frac{1}{2^{n+1}-1} \text {. }
$$
Thus, $a_{1}+a_{2}+\cdots+a_{2008}$
$$
\begin{array}{l}
=\frac{1}{2-1}-\frac{1}{2^{2}-1}+\frac{1}{2^{2}-1}-\frac{1}{2^{3}-1}+\cdots+ \\
\frac{1}{2^{2009}-1}-\frac{1}{2^{2010}-1} \\
=1-\f... | 1-\frac{1}{2^{2010}-1}<1 | Algebra | proof | Yes | Yes | cn_contest | false | 721,738 |
7. (40 points) Find all partitions of positive integers that satisfy the following conditions:
Divide all positive integers into two categories: "cold numbers" and "hot numbers," such that the sum of any hot number and cold number is a hot number, and their product is a cold number. | 7. First prove: All cold numbers are multiples of some smallest cold number, which can be any integer greater than 1.
Let \( m, n \) be cold numbers, and \( k \) be a hot number. Then \( m+k \) is a hot number, and \( (m+k) n \) is a cold number.
On the other hand, \( k n \) is a cold number, and by
\[
(m+k) n = m n + ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,739 |
8. (40 points) Figure 6 shows that a regular octagon can be cut into one $1 \times 1$ square and four congruent pentagons, which can be reassembled into a square. Try to find the perimeter of each pentagon. | 8. As shown in Figure 12 (a), let the side length of the square be $a$.
Then
$$
A B+B C=C G+B C=a .
$$
It is known that the side length of the regular octagon is 1.
Cut the regular octagon as shown in Figure 12 (b).
Thus, the area of the regular octagon is
$1+1 \times \frac{\sqrt{2}}{2} \times 4+\frac{1}{2} \times \f... | 2+\sqrt{2+2\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,740 |
9. (40 points) In each game of bridge, four players play together. It is known that the number of games played is exactly equal to the number of players, and any two players have played at least one game together. Try to find the maximum number of players.
| 9. There are $n$ players, and each game can produce 6 pairs of players.
From $\mathrm{C}_{n}^{2} \leqslant 6 n$ we get
$$
n \leqslant 13 \text {. }
$$
Next, we prove that $n=13$ is feasible.
As shown in Figure 13, number the 13
players from $0 \sim 12$.
The players in each game are as follows:
$$
\begin{array}{l}
(0,2... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,741 |
10. (40 points) Try to determine: Among the numbers $2008$, $2009$, and $2010$, which can be expressed in the form $x^{3}+y^{3}+z^{3}-3 x y z$ $\left(x, y, z \in \mathbf{N}_{+}\right)$? | 10. When $x=670, y=z=669$, we get 2008; when $x=669, y=z=670$, we get 2009.
2010 cannot be expressed in this form.
The proof below uses proof by contradiction:
If $31 t$ and $9 \times t$, then $t$ cannot be written in the form $x^{3}+y^{3}+z^{3}-3 x y z$.
If not, then because $3 \mid t$, we have
$$
31(x+y+z)
$$
or $3 ... | 2008 \text{ and } 2009 \text{ can be expressed, but } 2010 \text{ cannot.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,742 |
Example 6 Given a trapezoid $A B C D$ with the upper and lower bases satisfying $A B>C D$, points $K 、 L$ are on sides $A B 、 C D$ respectively, and satisfy $\frac{A K}{K B}=\frac{D L}{L C}$. If there exist points $P 、 Q$ on segment $K L$ such that $\angle A P B=\angle B C D, \angle C Q D=\angle A B C$. Prove: $P 、 Q 、... | Analysis: An inverse proposition of a set of parallel lines is:
If a set of parallel lines intercepts corresponding segments on three lines in proportion, then these three lines are concurrent (can be proven by the same method).
This can be used to prove three lines are concurrent.
Proof: Given $A B / / C D, \frac{A K}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,743 |
1. Prove that the equation $a b c=2009(a+b+c)$ has only a finite number of positive integer solutions $(a, b, c)$. (Supplied by Liang Yingde) | 1. It suffices to prove: The positive integer solutions of the original equation satisfying $a \leqslant b \leqslant c$ are only finitely many.
In fact, by $a \leqslant b \leqslant c$, we know
$$
\begin{array}{l}
a b c=2009(a+b+c) \leqslant 6027 c \\
\Rightarrow a b \leqslant 6027 .
\end{array}
$$
Therefore, there are... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,744 |
2. As shown in Figure 1, in $\triangle A B C$, $\angle B A C=90^{\circ}$, point $E$ is on the arc $\overparen{B C}$ (excluding point $A$) of the circumcircle $\Gamma$ of $\triangle A B C$, and $A E>E C$.
Connect $E C$ and extend it to point $F$ such that
$$
\angle E A C=\angle C A F,
$$
Connect $B F$ and let it inter... | 2. Using the method of the same.
As shown in Figure 3, let
the circumcircle $\Gamma$ of $\triangle A E F$ intersect the extension of $A C$ at point $P$, and connect $P E$ and $P F$.
$$
\begin{array}{l}
\text { Since } \angle E A C \\
=\angle F A C,
\end{array}
$$
Therefore, quadrilateral $A E P F$ is a cyclic quadril... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,745 |
3. In the Cartesian coordinate system, let the set of points be
$$
\begin{array}{l}
\left\{P_{1}, P_{2}, \cdots, P_{4 n+1}\right\} \\
=\{(x, y) \mid x, y \text { are integers, }|x| \leqslant n,|y| \leqslant n, x y=0\},
\end{array}
$$
where, $n \in \mathbf{N}_{+}$. Find
$$
\left(P_{1} P_{2}\right)^{2}+\left(P_{2} P_{3}... | 3. First, prove a lemma.
Lemma Let real numbers $s_{1} \geqslant s_{2} \geqslant \cdots \geqslant s_{m}$, and
$$
\begin{array}{l}
f\left(s_{1}, s_{2}, \cdots, s_{m}\right) \\
=\min _{\left(t_{1}, t_{2}, \cdots, t_{m}\right)}\left\{\left|t_{1}-t_{2}\right|^{2}+\left|t_{2}-t_{3}\right|^{2}+\cdots+\right. \\
\left.\left|... | 16n-8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,746 |
4. Let there be $n(n \geqslant 4)$ points $V_{1}, V_{2}, \cdots, V_{n}$ in a plane, with no three points collinear. Some of these points are connected by line segments. Place $n$ chess pieces, labeled $1, 2, \cdots, n$, on these $n$ points, with exactly one piece on each point. Now perform the following operation on th... | 4. The minimum value sought is $n+1$.
First, when the number of line segments is no more than $n$, it is easy to see that one of the following two scenarios must occur:
(1) There exists a point among $V_{1}, V_{2}, \cdots, V_{n}$ that is connected by a line segment to at most one other point;
(2) For each point among ... | n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,747 |
5. Let real numbers $x, y, z$ be greater than or equal to 1. Prove that: $\left(x^{2}-2 x+2\right)\left(y^{2}-2 y+2\right)\left(z^{2}-2 z+2\right)$ $\leqslant(x y z)^{2}-2 x y z+2$.
(Xiong Bin) | 5. Note that $x \geqslant 1, y \geqslant 1$. Then
$$
\begin{array}{l}
\left(x^{2}-2 x+2\right)\left(y^{2}-2 y+2\right)-\left[(x y)^{2}-2 x y+2\right] \\
=(-2 y+2) x^{2}+\left(6 y-2 y^{2}-4\right) x+\left(2 y^{2}-4 y+2\right) \\
=-2(y-1)\left[x^{2}+(y-2) x+1-y\right] \\
=-2(y-1)(x-1)(x+y-1) \leqslant 0 . \\
\text { Henc... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,748 |
6. As shown in Figure 2, circles $\Gamma_{1}$ and $\Gamma_{2}$ are internally tangent at point $S$, the chord $A B$ of circle $\Gamma_{2}$ is tangent to circle $\Gamma_{1}$ at point $C$, and $M$ is the midpoint of arc $\overparen{A B}$ (excluding point $S$). A line $M N \perp A B$ is drawn through point $M$, with the f... | 6. Proof 1: As shown in Figure 4, let the centers of circles $\Gamma_{1}$ and $\Gamma_{2}$ be $O_{1}$ and $O_{2}$, and their radii be $r$ and $R$ respectively.
By the perpendicular chord theorem, the extension of $M N$ passes through $O_{2}$, and $N$ is the midpoint of chord $A B$.
Since circles $\Gamma_{1}$ and $\Gam... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,749 |
7. In a $10 \times 10$ grid, there is a shape composed of $4 n$ $1 \times 1$ small squares, which can be covered by $n$ "田" shaped figures, or by $n$ "世" or "円" shaped figures (which can be rotated). Find the minimum value of the positive integer $n$.
(Zhu Huaiwei, problem contributor) | 7. Let the given shapes be denoted as type $A$ and type $B$.
First, we prove that $n$ is even.
Color the $10 \times 10$ grid as shown in Figure 6.
Regardless of which 4 squares type $A$ covers, the number of black squares must be even, while for type $B$ it is odd. If $n$ is odd, the number of black squares covered b... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,750 |
8. Let $a_{n}=n \sqrt{5}-[n \sqrt{5}]$. Find the maximum and minimum terms in the sequence $a_{1}, a_{2}$, $\cdots, a_{2009}$, where $[x]$ denotes the greatest integer less than or equal to the real number $x$. (Wang Zhixiong) | 8. Let $b_{0}=0, b_{1}=1$,
$$
b_{n}=4 b_{n-2}+b_{n-1}(n \geqslant 2) \text {. }
$$
Then $b_{n}=\frac{(2+\sqrt{5})^{n}-(2-\sqrt{5})^{n}}{2 \sqrt{5}}$.
In particular, $b_{6}=1292, b_{7}=5473$.
For any $k(k=1,2, \cdots, 5473)$, there exist unique integers $x_{k} 、 y_{k}$, such that
$$
1292 k=x_{k}+5473 y_{k}\left(1 \leqs... | a_{1292}, a_{1597} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,751 |
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