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9. Given two points $M(-5,0)$ and $N(5,0)$. If there exists a point $P$ on a line such that $|P M|-|P N|=6$, then the line is called a "harmonious line". Given the following lines:
(1) $y=x-1$;
(2) $y=2$;
(3) $y=\frac{5}{3} x$;
(4) $y=2 x+1$,
which of these lines are harmonious lines? $\qquad$ | 9. (1),(2).
This problem is transformed into finding the intersection of a line with the right branch of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$. By analyzing the positional relationship between the lines in the options and the asymptote $y=\frac{4}{3} x$ of the hyperbola, it is easy to see that (1) and (2)... | (1),(2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,146 |
10. Let $M=(5+\sqrt{24})^{2 n}\left(n \in \mathbf{N}_{+}\right), N$ be the fractional part of $M$. Then the value of $M(1-N)$ is $\qquad$ | 10.1.
Since $(5+\sqrt{24})^{2 n}+(5-\sqrt{24})^{2 n}$ is a positive integer, and $0<(5-\sqrt{24})^{2 n}<1$, therefore,
$$
\begin{array}{l}
N=1-(5-\sqrt{24})^{2 n} \\
M(1-N)=(5+\sqrt{24})^{2 n}(5-\sqrt{24})^{2 n}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,147 |
11. As shown in Figure 1, given
$\odot C:(x+1)^{2}+y^{2}=$
8, a fixed point $A(1,0)$, and $M$ is
a moving point on the circle. Points $P$ and $N$
are on $A M$ and $M C$, respectively, and
satisfy $\overrightarrow{A M}=2 \overrightarrow{A P}, \overrightarrow{N P}$.
$\overline{A M}=0$, the trajectory of point $N$ is curv... | 11. $[3-2 \sqrt{2}, 1)$.
From the problem, we know that $P$ is the midpoint of $A M$, and $N P \perp A M$, so $N A = N M$.
From $N C + N A = N C + N M = C M = 2 \sqrt{2}$, we get that the locus of point $N$ is an ellipse with foci at $C(-1,0)$ and $A(1,0)$ (as shown in Figure 3).
From $c=1, a=\sqrt{2}$, and $b=1$, we... | 3-2 \sqrt{2} \leqslant \lambda<1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,148 |
12. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ satisfying:
$$
S_{n}+a_{n}=\frac{n-1}{n(n+1)}(n=1,2, \cdots) \text {. }
$$
Then the general term $a_{n}=$ | 12. $\frac{1}{2^{n}}-\frac{1}{n(n+1)}$.
Notice
$$
\begin{array}{l}
a_{n+1}=S_{n+1}-S_{n} \\
=\frac{n}{(n+1)(n+2)}-a_{n+1}-\frac{n-1}{n(n+1)}+a_{n},
\end{array}
$$
i.e., $2 a_{n+1}$
$$
\begin{array}{l}
=\frac{n+2-2}{(n+1)(n+2)}-\frac{1}{n+1}+\frac{1}{n(n+1)}+a_{n} \\
=\frac{-2}{(n+1)(n+2)}+a_{n}+\frac{1}{n(n+1)} .
\en... | a_{n}=\frac{1}{2^{n}}-\frac{1}{n(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,149 |
13. Given the function $f(x)=|\sin x|$ intersects the line $y=k x(k>0)$ at exactly three points, and the maximum value of the x-coordinates of these points is $\alpha$. Prove:
$$
\frac{\cos \alpha}{\sin \alpha+\sin 3 \alpha}=\frac{1+\alpha^{2}}{4 \alpha} .
$$ | Three, 13. As shown in Figure $4, f(x)$'s graph intersects with the line $y=k x$ $(k>0)$ at three points, then within $x \in\left(\pi, \frac{3 \pi}{2}\right)$ it is tangent, with the tangent point being $A(\alpha,-\sin \alpha)\left(\alpha \in\left(\pi, \frac{3 \pi}{2}\right)\right)$.
Given $f^{\prime}(x)=-\cos x\left(x... | \frac{1+\alpha^{2}}{4 \alpha} | Calculus | proof | Yes | Yes | cn_contest | false | 724,150 |
14. Let $\angle A O B=\theta\left(\theta\right.$ be a constant and $\left.0<\theta<\frac{\pi}{2}\right)$, and let moving points $P$ and $Q$ be on rays $O A$ and $O B$ respectively, such that the area of $\triangle P O Q$ is always 36. Let the centroid of $\triangle P O Q$ be $G$, and point $M$ be on ray $O G$ such that... | 14. (1) Taking $O$ as the origin and the bisector of $\angle A O B$ as the $x$-axis, establish a rectangular coordinate system $x O y$. Let
$$
\begin{array}{l}
P\left(a \cos \frac{\theta}{2}, a \sin \frac{\theta}{2}\right), Q\left(b \cos \frac{\theta}{2},-b \sin \frac{\theta}{2}\right), \\
G\left(x_{G}, y_{C}\right).
\... | |O G|_{\min }=4 \sqrt{\cot \frac{\theta}{2}} \text{ and } \frac{x^{2}}{36 \cot \frac{\theta}{2}}-\frac{y^{2}}{36 \tan \frac{\theta}{2}}=1(x>0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,151 |
15. Let $\{a_{n}\}$ be a sequence of positive terms with the first term $a_{1}$, and let the sum of the first $n$ terms be $S_{n}$. Let $q$ be a non-zero constant. It is known that for any positive integers $n, m$, when $n > m$, $S_{n} - S_{m} = q^{m} S_{n-m}$ always holds.
(1) Prove that the sequence $\{a_{n}\}$ is a ... | 15. (1) When $n>m$, $S_{n}-S_{m}=q^{m} S_{n-m}$ always holds. Therefore, let $m=n-1$. Then $S_{n}-S_{n-1}=q^{n-1} S_{1}$, i.e., $a_{n}=a_{1} q^{n-1}$.
Thus, the sequence $\left\{a_{n}\right\}$ is a geometric sequence.
(2) If positive integers $m, k, h$ form an arithmetic sequence, then $m+h=2 k$.
When $q=1$, $a_{n}=a_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,152 |
16. As shown in Figure $2, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$. | 16. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), \( C(0, c) \), and assume \( b > c \). The line \( l_{P B} \) is given by \( y - b = \frac{y_{0} - b}{x_{0}} x \), which can be rewritten as
\[
\left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0.
\]
The distance from the circle's center \((1, 0)\) to \( P B \) is ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,153 |
1. If $\frac{1}{t}-|t|=1$, then the value of $\frac{1}{t}+|t|$ is ( ).
(A) $\sqrt{5}$
(B) $\pm \sqrt{5}$
(C) $\sqrt{3}$
(D) $\pm \sqrt{3}$ | - 1. A.
Given that $t>0$, then $\frac{1}{t}+|t|>0$.
Therefore, $\frac{1}{t}+|t|=\sqrt{\left(\frac{1}{t}-|t|\right)^{2}+4}=\sqrt{5}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,154 |
2. The number of intersections of the graph of the function $y=x|x|-\left(4 \cos 30^{\circ}\right) x+2$ with the $x$-axis is ( ).
(A) 4
(B) 3
(C) 2
(D) 0 | 2. B.
When $x>0$,
$$
y=x^{2}-2 \sqrt{3} x+2=(x-\sqrt{3})^{2}-1
$$
the graph intersects the $x$-axis at two points;
When $x<0$,
$$
y=-x^{2}-2 \sqrt{3} x+2=-(x+\sqrt{3})^{2}+5
$$
the graph intersects the $x$-axis at one point.
There are three intersections in total. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,155 |
Example 1 Find all non-square numbers $n \in \mathbf{N}_{+}$, such that $[\sqrt{n}]^{3} \mid n^{2}$. (1)
(3rd Northern Mathematical Olympiad Invitational Competition) | 【Analysis】Obviously, $[\sqrt{n}]=1$, i.e., $n=2,3$ satisfy the condition.
Below, let $[\sqrt{n}] \geqslant 2$. Let $A=[\sqrt{n}]$, and $n$ is a non-perfect square. Therefore,
$$
\begin{array}{l}
A<\sqrt{n}<A+1, \\
A^{2}<n<A^{2}+2 A+1 .
\end{array}
$$
Thus, $A^{2}+1 \leqslant n \leqslant A^{2}+2 A$.
Let $n=A^{2}+k(k \... | n=2,3,8,24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,156 |
3. As shown in Figure 1, on the hypotenuse $AB$ of isosceles right $\triangle ABC$, take two points $M$ and $N$ (not coinciding with points $A$ and $B$) such that $\angle MCN=45^{\circ}$. Let $AM=a, MN=b, NB=c$. Then the shape of the triangle with side lengths $a, b, c$ is ( ).
(A) Acute triangle
(B) Obtuse triangle
(C... | 3. C.
As shown in Figure 8, rotate $\triangle A C M$ counterclockwise around point $C$ by $90^{\circ}$, so that $A C$ coincides with $B C$, resulting in $\triangle B C M^{\prime}$. Connect $N M^{\prime}$. Then
$$
\begin{array}{l}
\triangle N C M^{\prime} \cong \triangle N C M \\
\Rightarrow N M^{\prime}=N M=b . \\
\tex... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,157 |
4. As shown in Figure 2, in a rectangular box, $A B=5$, $B C=4, C C_{1}=3$. The shortest distance an ant travels from point $A_{1}$ along the surface of the box to point $C$ is ( ).
(A) $\sqrt{74}$
(B) $\sqrt{78}$
(C) $\sqrt{80}$
(D) $\sqrt{90}$ | 4. A.
From the plane development of the cuboid, the shorter routes are the three cases shown in Figure 9:
In Figure 9(a),
$$
A_{1} C=\sqrt{5^{2}+(3+4)^{2}}=\sqrt{74} \text {; }
$$
In Figure 9(b),
$$
A_{1} C=\sqrt{3^{2}+(5+4)^{2}}=\sqrt{90} \text {; }
$$
In Figure 9(c),
$$
A_{1} C=\sqrt{4^{2}+(3+5)^{2}}=\sqrt{80} \te... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,158 |
5. As shown in Figure 3, in quadrilateral $A B C D$, $\angle B A D=$ $90^{\circ}, A B=B C=\tan 60^{\circ}$, $A C=3, A D=\frac{3}{2}$. Then the length of $C D$ is ( ).
(A) $2 \sqrt{2}$
(B) 2
(C) $\frac{3 \sqrt{3}}{2}$
(D) $\frac{3 \sqrt{2}}{2}$ | 5. C.
As shown in Figure 10, construct
$$
B E \perp A C, D F \perp A C,
$$
with the feet of the perpendiculars at points $E$ and $F$. Then
$$
\begin{array}{c}
A E=E C=\frac{3}{2} \\
\Rightarrow \cos \angle B A E \\
=\frac{A E}{A B}=\frac{\sqrt{3}}{2} \\
\Rightarrow \angle B A C=30^{\circ} \Rightarrow \angle D A F=60^... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,159 |
6. Given the function of $x$, $y=m x^{2}-2 x+1$ $(0 \leqslant x \leqslant 2)$. Which of the following statements is correct? ( ).
(A) When $m=0$, there is no minimum value
(B) When $\frac{1}{2} \leqslant m<1$, $y_{\text {min }}=1$
(C) When $m<0$, $y_{\text {max }}=1-\frac{1}{m}$
(D) When $m \geqslant 1$, $y_{\text {max... | 6. D.
When $m=0$, $y$ has a minimum value of -3; when $\frac{1}{2} \leqslant m<1$, from the graph and the monotonicity of the function, we know $y_{\text {min }}=1-\frac{1}{m}$;
When $m<0$, from the graph and the monotonicity of the function, we know $y_{\max }=1$;
When $m \geqslant 1$, $y_{\max }=4 m-3$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,160 |
1. Let $x$ be a positive integer, and $x<50$. Then the number of $x$ such that $x^{3}+11$ is divisible by 12 is $\qquad$. | $$
\begin{array}{l}
x^{3}+11=(x-1)\left(x^{2}+x+1\right)+12 \\
=(x-1) x(x+1)+(x-1)+12 .
\end{array}
$$
Notice that $6 \mid(x-1) x(x+1)$, so $6 \mid(x-1)$;
also, $2 \mid \left(x^{2}+x+1\right)$, hence $12 \mid (x-1)$.
Since $x<50$, we have
$$
x-1=0,12,24,36,48 \text {. }
$$
Thus, $x=1,13,25,37,49$, a total of 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,161 |
2. Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation
$$
x^{2}-2009 x+2011=0
$$
real numbers $m$ and $n$ satisfy
$$
\begin{array}{l}
2009 m x_{1}+2009 n x_{2}=2009, \\
2010 m x_{1}+2010 n x_{2}=2010 .
\end{array}
$$
Then $2011 m x_{1}+2011 n x_{2}=$ $\qquad$ | 2. -2009 .
From the given, we know
$$
\begin{array}{l}
=2010 \times 2009-2011 \times 2009=-2009 . \\
\end{array}
$$ | -2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,162 |
3. As shown in Figure 4, it is known that quadrilateral $ABCD$ is inscribed in $\odot O$, $AB$ is the diameter, $AD=DC$, and $BA$, $CD$ are extended to intersect at point $E$, $BF \perp EC$ intersects the extension of $EC$ at point $F$. If $AE=AO, BC=6$, then the length of $CF$ is $\qquad$. | 3. $\frac{3 \sqrt{2}}{2}$.
As shown in Figure 11, connect $A C$, $B D$, and $O D$.
Since $A B$ is the diameter of $\odot O$, we have
$$
\begin{array}{l}
\angle B C A \\
=\angle B D A \\
=\angle B F C \\
=90^{\circ} .
\end{array}
$$
Also, since quadrilateral $A B C D$ is an inscribed quadrilateral of $\odot O$, we ge... | \frac{3 \sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,163 |
4. As shown in Figure 5, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and $G$ is the midpoint of $C D$. A line through $G$ intersects $A C$ and $B C$ at points $P$ and $Q$, respectively. Then the value of $\frac{C A}{C P}+\frac{C B}{C Q}$ is $\qquad$. | 4. 4 .
As shown in Figure 12, draw $A E / / P Q$ intersecting $C D$ at point $E$, and $B F / / P Q$ intersecting the extension of $C D$ at point $F$. Then $A E / / B F$.
Since $A D = B D$, therefore, $E D = F D$.
$$
\begin{array}{c}
\text { Also, } \frac{C A}{C P} = \frac{C E}{C G}, \\
\frac{C B}{C O} = \frac{C F}{C ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,164 |
One, (20 points) As shown in Figure 6, in trapezoid $ABCD$, $AD$ $/ / BC, \angle BAD=90^{\circ}$, the radius of the incircle of trapezoid $ABCD$ is $R$, and the diagonals $AC$ and $BD$ intersect at point $M$. Prove:
$$
S_{\triangle CDM}=R^{2}
$$ | As shown in Figure 13, let $O$ be the center of the inscribed circle, and $E, F$ be the points of tangency of $AD, BC$ with $\odot O$. Connect $EF$ (through $O$), $CO, DO$. It is easy to prove:
$$
\begin{aligned}
& \angle COD=90^{\circ} \\
\Rightarrow & \angle FCO=\angle DOE \\
\Rightarrow & \triangle OCF \backsim \tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,165 |
Example 2 Let $n \in \mathbf{N}, n>4$. Prove:
$$
(n-1) \left\lvert\,\left[\frac{(n-1)!}{n}\right]\right. \text {. }
$$
(2002, Australian Mathematical Olympiad) | 【Analysis】If $n>4$ is not a prime number, then there exist $p, q \geqslant 2$, such that $n=p q$.
When $p<q$, we have
$$
(n-1)!=1 \times 2 \times \cdots \times p \times \cdots \times q \times \cdots \times(n-1) \text {. }
$$
Clearly, $\frac{(n-1)!}{p q} \in \mathbf{N}_{+}$, and
$$
(n-1) \left\lvert\, \frac{(n-1)!}{p q... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,167 |
Three. (25 points) Let positive integers $a, b, c$ be pairwise coprime, and $a^{3}+b^{3}+c^{3}$ can be simultaneously divided by $a^{2} b, b^{2} c, c^{2} a$.
(1) Prove: $a^{2} b^{2} c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right)$;
(2) If $a \geqslant b \geqslant c$, find the values of $a, b, c$. | Three, (1) From
$$
a^{2} b\left|\left(a^{3}+b^{3}+c^{3}\right) \Rightarrow a^{2}\right|\left(a^{3}+b^{3}+c^{3}\right) \text {. }
$$
Similarly, $b^{2}$ and $c^{2}$ also divide $a^{3}+b^{3}+c^{3}$.
Since $a$, $b$, and $c$ are pairwise coprime, we have
$$
a^{2} b^{2} c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right) \text {. }
$... | a=b=c=1 \text{ or } a=3, b=2, c=1 | Number Theory | proof | Yes | Yes | cn_contest | false | 724,168 |
1. Given that the set $M$ is a subset of $\{1,2, \cdots, 2011\}$, and the sum of any four elements in $M$ cannot be divisible by 3. Then $|M|_{\text {max }}=$ $\qquad$ | - 1. 672.
Consider the set $A=\{3,6,9, \cdots, 2010\}$,
$$
\begin{array}{l}
B=\{1,4,7, \cdots, 2011\}, \\
C=\{2,5,8, \cdots, 2009\} .
\end{array}
$$
If $M \cap A \neq \varnothing$, then $|M \cap B|<3,|M \cap C|<3,|M \cap A|<4$. Therefore, $|M|<10$.
Now assume $M \cap A=\varnothing$.
If $|M \cap B| \geqslant 2$, then ... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,169 |
2. If $n \in \mathbf{N}, n \geqslant 2, a_{i} \in\{0,1, \cdots, 9\}$
$$
\begin{array}{l}
(i=1,2, \cdots, n), a_{1} a_{2} \neq 0 \text {, and } \\
\sqrt{a_{1} a_{2} \cdots a_{n}}-\sqrt{a_{2} a_{3} \cdots a_{n}}=a_{1}, \\
\end{array}
$$
then $n=$ $\qquad$, where $\overline{a_{1} a_{2} \cdots a_{n}}$ is the $n$-digit num... | 2. 2 .
Let $x=\overline{a_{2} a_{3} \cdots a_{n}} \in \mathbf{N}_{+}$. Then $\overline{a_{1} a_{2} \cdots a_{n}}=10^{n-1} a_{1}+x$.
By the given condition, $\sqrt{10^{n-1} a_{1}+x}-\sqrt{x}=a_{1}$. Therefore, $10^{n-1} a_{1}+x=x+a_{1}^{2}+2 a_{1} \sqrt{x}$.
Thus, $10^{n-1}-2 \sqrt{x}=a_{1}$.
If $n \geqslant 3$, then
$... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,170 |
3. In $\triangle A B C$, $I$ is the incenter of $\triangle A B C$. If $A C+A I=B C, A B+B I=A C$, then $\angle B=$ $\qquad$ | 3. $\frac{2 \pi}{7}$.
As shown in Figure 1, take a point $D$ on the extension of $C A$ such that $A D=A I$. Then
$$
\begin{array}{l}
C D=C A+A D \\
=C A+A I=B C .
\end{array}
$$
By the given conditions,
$$
\begin{array}{l}
\angle C D I=\angle C B I \\
=\angle A B I .
\end{array}
$$
Since $A D=A I$, we have
$$
\angle... | \frac{2 \pi}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,171 |
4. For any positive integer $n$, let $a_{n}$ be the smallest positive integer such that $n \mid a_{n}$!. If $\frac{a_{n}}{n}=\frac{2}{5}$, then $n=$ $\qquad$ . | 4.25.
From $\frac{a_{n}}{n}=\frac{2}{5} \Rightarrow a_{n}=\frac{2 n}{5} \Rightarrow 51 n$.
Let $n=5 k\left(k \in \mathbf{N}_{+}\right)$. If $k>5$, then $5 k \mid k!$.
Thus, $a_{\mathrm{n}} \leqslant k<\frac{2 n}{5}$, a contradiction.
Clearly, when $k=2,3,4$, $a_{n} \neq \frac{2 n}{5}$.
Also, $a_{25}=10=\frac{2}{5} \ti... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,172 |
5. On a straight line, three points $A$, $B$, and $C$ are arranged in sequence, and $A B=6, A C=24, D$ is a point outside the line, and $D A$ $\perp A B$. When $\angle B D C$ takes the maximum value, $A D=$ $\qquad$ . | 5. 12 .
Let $\angle B D C=\theta\left(\theta<90^{\circ}\right), \triangle B C D$'s circumcircle $\odot O$ has a radius of $R$. Then $\sin \theta=\frac{B C}{2 R}$.
When $R$ decreases, $\theta$ increases.
Therefore, when $\odot O$ is tangent to $A D$ at point $D$, $\theta$ is maximized.
At this time, $A D^{2}=A B \cdot ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,173 |
6. Given the function $f: Z \rightarrow R$ satisfies for any integers $x, y$ that
$$
f(x+y) f(x-y)=f(x)+f(y) .
$$
Then $f(2011)=$ $\qquad$ . | 6.0 or 2 or -1.
From equation (1) we get
$$
\begin{array}{l}
f(x)+f(-y)=f(x+y) f(x-y)=f(x)+f(y) \\
\Rightarrow f(y)=f(-y) .
\end{array}
$$
In equation (1), let $x=y$, we get
$$
f(2 x) f(0)=2 f(x) \text {. }
$$
In equation (2), let $x=0$, we get $f(0)=0$ or 2.
If $f(0)=0$, then from equation (2) we get
$$
f(x)=0(\for... | 0 \text{ or } 2 \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,174 |
7. Given that on the side $C A$ of $\angle A C B$ there are 2011 points $A_{1}, A_{2}, \cdots, A_{2011}$, and on the side $C B$ there are 2011 points $B_{1}, B_{2}, \cdots, B_{2011}$, satisfying
$$
\begin{array}{l}
A_{1} A_{2}=A_{2} A_{3}=\cdots=A_{2010} A_{2011}, \\
B_{1} B_{2}=B_{2} B_{3}=\cdots=B_{2010} B_{2011} .
\... | 7.2010 .
$$
\begin{array}{l}
\text { Let } \angle A C B=\alpha, C A_{1}=a, C B_{1}=b, \\
A_{i} A_{i+1}=s, B_{i} B_{i+1}=t(i=1,2, \cdots, 2010), \\
S_{\text {trapezoid } A_{i} A_{i+1} B_{i+2} B_{i}}=S_{i}(i=1,2, \cdots, 2009) . \\
\text { Then } S_{i}=\frac{1}{2}\{[a+(i+1) s][b+(i+1) t]- \\
\quad[a+(i-1) s][b+(i-1) t]\}... | 2010 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,175 |
8. Given positive integers $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!+c! \text {. }
$$
then $c\left(a^{5}+b^{5}+c^{2}\right)+3=$ $\qquad$ | 8.2011 .
$$
\begin{array}{l}
\text { Given }(a!)(b!)=a!+b!+c! \\
\Rightarrow(a!-1)(b!-1)=c!+1 \text {. }
\end{array}
$$
Assume without loss of generality that $a \geqslant b$. Clearly, $c > a$. Then $(a!-1) \mid(c!+1)$.
Also, $c!+1=\frac{c!}{a!}(a!-1)+\frac{c!}{a!}+1$, so $(a!-1) \left\lvert\,\left(\frac{c!}{a!}+1\rig... | 2011 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,176 |
9. (16 points) Given that the three vertices of the equilateral triangle $\triangle ABC$ lie on the parabola $y=x^{2}$. Try to find the equation of the locus of the center of the equilateral triangle $\triangle ABC$.
| 9. Let $l_{A B}: y=k x+b$.
(1)
By symmetry, we first assume $k \geqslant 0$.
Combining equation (1) with $y=x^{2}$, we get
$$
x^{2}-k x-b=0.
$$
Then $x_{A}+x_{B}=k, x_{A} x_{B}=-b, \Delta=k^{2}+4 b>0$.
Thus, the midpoint $M\left(\frac{k}{2}, \frac{k^{2}}{2}+b\right)$ of $A B$, and
$$
|A B|=\sqrt{1+k^{2}}\left|x_{A}-x_... | x = \frac{k}{2} \cdot \frac{3-k^{2}}{1-3 k^{2}}, \quad y = \frac{9 k^{6}+18 k^{4}+33 k^{2}+8}{4\left(1-3 k^{2}\right)^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,177 |
Example 3 The partition of the set $H=\{1,2, \cdots, 9\}$ is to represent $H$ as the union of pairwise disjoint subsets. For $n \in H$ and partition $P$, the number of elements in the subset containing $n$ is denoted by $P(n)$. For example, if
$$
P:\{1,4,5\} \cup\{2\} \cup\{3,6,7,8,9\} \text {, }
$$
then $P(6)=5$. | Proof: For any two partitions $P_{1}$ and $P_{2}$ of $H$, there exist two different elements $m, n$ in $H$ such that
$$
\begin{array}{l}
P_{1}(m)=P_{1}(n), \\
P_{2}(m)=P_{2}(n).
\end{array}
$$
(2008-2009 Hungarian Mathematical Olympiad)
**Analysis:** Since the partitions $P_{1}$ and $P_{2}$ are arbitrary, directly pro... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,179 |
Example 4: Two people, A and B, play a number guessing game. A selects a positive integer $a$ and tells B that $a \leqslant 2006$. Each time B tells A a positive integer $b$, A then tells B whether $a+b$ is a prime number. Prove: B can guess the number $a$ chosen by A in fewer than 2006 questions. ${ }^{[2]}$
(2006, Au... | Proof. Suppose there exist $i, k \in \mathbf{N}_{+}$, such that
$$
1 \leqslant i<i+k \leqslant 2006 \text {, and } S_{i}=S_{i+k} \text {. }
$$
Then for any $u=i+1, i+2, \cdots, i+2005$,
we have $f(u)=f(u+k)$.
If $k=1$, note that
$$
\begin{array}{l}
f(2003)=f(2011)=1, \\
f(2004)=f(2012)=0 .
\end{array}
$$
Thus, $2003,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,180 |
Example 5 Proof: There does not exist a positive integer $n(n>1)$ such that $n^{2} \mid\left(3^{n}+1\right)$. | Proof Assume there exists a positive integer $n(n>1)$, such that $n^{2} \mid \left(3^{n}+1\right)$.
Obviously, $3 \times n$.
If $2 \mid n$, then
$$
3^{n}+1 \equiv(-1)^{n}+1=2(\bmod 4) \text {. }
$$
But $4 \mid \left(3^{n}+1\right)$, contradiction.
So $2 \nmid n$.
If $n$ is a prime, then by Fermat's Little Theorem we h... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,181 |
Example 6 Given positive integers $a>b>1$, and the equation
$$
\frac{a^{x}-1}{a-1}=\frac{b^{y}-1}{b-1}(x>1, y>1)
$$
has at least two different positive integer solutions $(x, y)$. Prove: $a$ and $b$ are coprime. [3] | Proof: Assume $a$ and $b$ are not coprime, and let $p$ be a prime common divisor of them. Denote $V_{p}(n)$ as the highest power of $p$ in the prime factorization of $n$.
Let $\left(x_{i}, y_{i}\right)(i=1,2)$ be two different solutions satisfying the problem, and assume without loss of generality that $x_{1}>x_{2}>1$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,182 |
Example 7 Prove: There does not exist a triangle whose sides, area, and internal angles (in degrees) are all rational numbers.
(14th Turkish Mathematical Olympiad) | Assume there exists a triangle satisfying the conditions. Then, by the cosine theorem and area formula, the sine and cosine values of the internal angles of the triangle are all rational numbers. Take the smallest internal angle $\theta \in (0, 60^\circ]$.
Clearly, $\theta \neq 60^\circ$. Otherwise, the triangle would... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,183 |
In $\triangle A B C$, $\angle A C B=90^{\circ}$. A circle is drawn with $B$ as the center and $B C$ as the radius. Point $D$ is on side $A C$, and line $D E$ is tangent to $\odot B$ at point $E$. A line perpendicular to $A B$ through point $C$ intersects $B E$ at point $F, A F$ intersects $D E$ at point $G$, and $A H$ ... | From the Law of Sines, we have
$$
\begin{array}{l}
\frac{R}{B F}=\frac{\sin \angle B F C}{\sin \angle B C F}=\frac{\cos (\beta+\gamma)}{\sin \alpha}, \\
\frac{B F}{F G}=\frac{\sin \angle B G F}{\sin \angle F B G}=\frac{\sin (\theta+\gamma)}{\sin \beta} .
\end{array}
$$
(1) $\times$ (2) gives
$$
\frac{R}{F G}=\frac{\sin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,184 |
1. For any set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ consisting of four different positive integers, let $S_{A}=a_{1}+a_{2}+a_{3}+a_{4}$. Let $n_{A}$ be the number of pairs $(i, j)$ such that $a_{i}+a_{j}(1 \leqslant i<j \leqslant 4)$ divides $S_{1}$. Find all sets $A$ consisting of four different positive inte... | For a set of four positive integers
$$
A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\} \text {, }
$$
without loss of generality, assume $a_{1} < a_{2} < a_{3} < a_{4}$.
Let $S_{1}=a_{1}+a_{2}+a_{3}+a_{4}$, and $u=\operatorname{gcd}\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$.
Then, $u \mid S_{1}$.
Since $a_{1}+a_{2} \mid S_{1... | A=\{a, 5a, 7a, 11a\} \text{ or } A=\{a, 11a, 19a, 29a\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,185 |
3. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a real-valued function defined on the set of real numbers, satisfying for all real numbers $x, y$,
$$
f(x+y) \leqslant y f(x)+f(f(x)) .
$$
Prove: For all real numbers $x \leqslant 0$, we have $f(x)=0$. | 3. Let $y=t-x$. Then equation (1) can be written as
$$
f(t) \leqslant t f(x)-x f(x)+f(f(x)) \text {. }
$$
In equation (2), let $t=f(a), x=b$ and $t=f(b), x=a$. We get
$$
\begin{array}{l}
f(f(a))-f(f(b)) \leqslant f(a) f(b)-b f(b), \\
f(f(b))-f(f(a)) \leqslant f(a) f(b)-a f(a) .
\end{array}
$$
Adding the above two equ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,187 |
4. Given an integer $n>0$. There is a balance scale and $n$ weights with masses $2^{0}, 2^{1}, \cdots, 2^{n-1}$.
Now, through $n$ steps, all the weights are to be placed on the balance one by one, ensuring that during the process, the weight on the left side never exceeds the weight on the right side. In each step, on... | 4. (Based on the solution by Yao Bowen)
The number of different methods in the operation process is
$(2 n-1)!!=1 \times 3 \times \cdots \times(2 n-1)$.
We will use mathematical induction on $n$.
When $n=1$, there is only one weight, which can only be placed on the left side of the balance, so there is only 1 method.
... | (2 n-1)!! | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,188 |
Example 3 Given that the longest diagonal of a regular octagon is equal to $a$, and the shortest diagonal is equal to $b$. Then the area of this regular octagon is ( ). ${ }^{[3]}$
(A) $a^{2}+b^{2}$
(B) $a^{2}-b^{2}$
(C) $a+b$
(D) $a b$
(2009, Beijing Middle School Mathematics Competition (Grade 8)) | Solve As shown in Figure 3, perform an equal-area transformation on the regular octagon. Then
$$
\begin{array}{l}
A E=M N=P Q=a, \\
P N=Q M=F D=b .
\end{array}
$$
Then $S_{\text {octagon } A C D E F C H}$
$$
=S_{\text {quadrilateral MNPQ }}=a b \text {. }
$$
Therefore, the answer is D. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,189 |
5. Let $f$ be a function defined on the set of integers with positive integer values. It is known that for any two integers $m, n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove: For all integers $m, n$, if $f(m) \leqslant f(n)$, then $f(n)$ is divisible by $f(m)$. | 5. Let integers $x, y$ such that $f(x)<f(y)$.
Let $m=x, n=y$. Then
$$
f(x-y) \mid | f(x)-f(y) |,
$$
i.e., $f(x-y) \mid (f(y)-f(x))$.
Thus, $f(x-y)<f(y)-f(x)<f(y)$.
Therefore, the difference $d=f(x)-f(x-y)$ satisfies $-f(y)<-f(x-y)<d<f(x)<f(y)$.
Let $m=x, n=x-y$. Then $f(y) \mid d$. Hence, $d=0$.
Thus, $f(x)=f(x-y)$.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,190 |
6. Let the circumcircle of acute triangle $\triangle ABC$ be circle $\Gamma$, and let $l$ be a tangent line to circle $\Gamma$. Denote the reflections of the tangent line $l$ across the lines $BC$, $CA$, and $AB$ as $l_{a}$, $l_{b}$, and $l_{c}$, respectively. Prove that the circumcircle of the triangle formed by the l... | 6. (Based on the solution by Chen Lin)
As shown in Figure 1, let the circles symmetric to circle $\Gamma$ with respect to $BC$, $CA$, and $AB$ be $\Gamma_{a}$, $\Gamma_{b}$, and $\Gamma_{c}$, respectively. Let the points symmetric to point $P$ with respect to $BC$, $CA$, and $AB$ be $P_{a}$, $P_{b}$, and $P_{c}$, resp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,191 |
1. At a concert, there are 20 singers who will perform. For each singer, there is a set of other singers (possibly an empty set) such that he wishes to perform later than all the singers in this set. Question: Is there a way to have exactly 2,010 ways to order the singers so that all their wishes are satisfied? | 1. Such examples exist.
A ranking of singers that satisfies everyone's wishes is called "good".
If for a set of wishes of $k$ singers there exist $N$ good rankings, then $N$ is called "achievable by $k$ singers" (or simply "$k$-achievable").
Next, we prove that 2010 is "20-achievable".
First, we prove a lemma.
Lemma ... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,192 |
2. On some planets, there are $2^{N}(N \geqslant 4)$ countries, each with a flag composed of $N$ unit squares forming a rectangle of width $N$ and height 1. Each unit square is either yellow or blue. No two countries have the same flag. If $N$ flags can be arranged to form an $N \times N$ square such that all $N$ unit ... | 2. Note: In the problem-solving process, the diagonals of the squares involved are all main diagonals.
Let $M_{N}$ be the smallest positive integer that satisfies the condition.
First, prove: $M_{N}>2^{N-2}$.
Consider $2^{N-2}$ flags, where the first unit square of each flag is yellow, and the second unit square is blu... | 2^{N-2}+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,193 |
3.2500 chess kings are placed on a $100 \times 100$ chessboard, such that:
（1）no king can capture another king (i.e., no two kings are placed on unit squares that share a vertex);
（2）each row and each column contains exactly 25 kings.
Find the number of ways to place the kings (placements that can be obtained from each... | 3. Assume there is a placement that satisfies the conditions.
Divide the chessboard into 2500 $2 \times 2$ squares, and call these $2 \times 2$ squares "blocks." Each block cannot have more than 1 king (otherwise, two kings can attack each other).
By the pigeonhole principle, each block has exactly one king.
Use the l... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,194 |
5. There are $n(n \geqslant 4)$ players participating in a tennis tournament, where any two players play exactly one match, and there are no ties. If among four players, one player loses to the other three, and between these four players, each player wins one match and loses one match, then this group of four is called... | 5. For any tournament $T$, let
$$
S(T)=\sum_{i=1}^{n}\left(w_{i}-l_{i}\right)^{3} \text {. }
$$
First, prove that for a tournament $T$ with 4 participants, the conclusion holds.
In fact, let $A=\left(w_{1}, w_{2}, w_{3}, w_{4}\right)$ represent the number of wins for these 4 participants, and assume without loss of g... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,195 |
6. Given a positive integer $k$ and two other positive integers $b, w (b > w > 1)$, there are two pearl strings, one with $b$ black pearls and the other with $w$ white pearls, the number of pearls on the string is called the string length.
A person cuts these pearl strings according to the following steps, each step s... | 6. The state after the $i$-th step is denoted by $A_{i}$, with the initial state being $A_{0}$. The transition from $A_{i-1}$ to $A_{i}$ represents the $i$-th step. A pearl string of length $m$ is referred to as an $m$-string, with black and white pearl strings of length $m$ being denoted as $m$-$b$-string and $m$-$w$-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,196 |
1. (50 points) Given an acute triangle $\triangle ABC$, draw a perpendicular from point $A$ to $BC$ intersecting the circle $\odot O_{1}$ with diameter $BC$ at points $D$ and $E$; draw a perpendicular from point $B$ to $CA$ intersecting the circle $\odot O_{2}$ with diameter $CA$ at points $F$ and $G$. Prove that point... | 1. Proof 1 As shown in Figure 1, let $A E$ intersect $B C$ and $B G$ intersect $C A$, and $A E$ intersect $B G$ at points $A_{1}$, $B_{1}$, and $H$ respectively. Then $H$ is the orthocenter of $\triangle A B C$, and points $A_{1}$, $B_{1}$ lie on $\odot O_{2}$ and $\odot O_{1}$ respectively.
Since $B C$ and $C A$ are ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,198 |
2. (50 points) Let $d(n)$ be the number of positive divisors of the positive integer $n$. Define the sequence $\left\{a_{n}\right\}$ as follows:
$$
a_{1}=A, a_{n+1}=d\left(\left[\frac{3}{2} a_{n}\right]\right)+2011 \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Prove: For any... | 2. For positive integer $n$, it is easy to see that
$$
n-1, n-2, \cdots,\left[\frac{n}{2}\right]+1
$$
these $\left[\frac{n-1}{2}\right]$ numbers cannot be positive divisors of $n$, i.e.,
$$
d(n) \leqslant n-\left[\frac{n-1}{2}\right] \leqslant \frac{n}{2}+1 .
$$
Then $a_{n+1} \leqslant \frac{\left[\frac{3}{2} a_{n}\r... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,199 |
Example 4 As shown in Figure $4, M, N, P$ are the midpoints of sides $A B, B C, C A$ of $\triangle A B C$ respectively, and $B P$ intersects $M N$ and $A N$ at points $E$ and $F$.
(1) Prove:
$$
B F=2 F P \text {; }
$$
(2) If $S_{\triangle A B C}=$
$S$, find the area of $\triangle N E F$ | Proof (1) omitted.
(2) From $A M=M B, C N=N B$, we get
$M N / / A C, M N=\frac{1}{2} A C$.
Also, from $A P=P C$, we get $M E=E N$.
Thus, $E N=\frac{1}{2} M N, N F=\frac{1}{3} A N$.
Therefore, $S_{\triangle N E F}=\frac{N E}{M N} \cdot \frac{N F}{A N} S_{\triangle M M N}$
$=\frac{1}{2 \times 3} \cdot \frac{1}{2} S_{\tri... | \frac{1}{24} S | Geometry | proof | Yes | Yes | cn_contest | false | 724,200 |
3. (50 points) Given that $p$ is a prime number, the fractional part of $\sqrt{p}$ is $x$, and the fractional part of $\frac{1}{x}$ is $\frac{\sqrt{p}-31}{75}$. Find all prime numbers $p$ that satisfy the condition. | 3. Let $p=k^{2}+r$, where $k, r$ are integers, and satisfy $0 \leqslant r \leqslant 2 k$.
Since $\frac{\sqrt{p}-31}{75}$ is the fractional part of $\frac{1}{x}$, we have $0 \leqslant \frac{\sqrt{p}-31}{75} < 1$. Let
$\frac{1}{x}=\frac{1}{\sqrt{p}-k}=N+\frac{\sqrt{p}-31}{75}(N \geqslant 1)$.
Then $\frac{\sqrt{p}+k}{r}=N... | 2011 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,201 |
4. (50 points) On an $n$-row $n$-column chessboard, place $n^{2}-1(n \geqslant 3)$ chess pieces. The pieces are numbered as follows:
\[
\begin{array}{l}
(1,1), \cdots,(1, n),(2,1), \cdots,(2, n), \\
\cdots,(n, 1), \cdots,(n, n-1) .
\end{array}
\]
If the piece numbered $(i, j)$ is exactly in the $i$-th row and $j$-th c... | 4. Cannot.
Re-number the chess pieces in lexicographical order, i.e., $(i, j)$ is numbered as $(i-1) n+j$. The squares on the chessboard are also numbered in lexicographical order, with the $i$-th row and $j$-th column being $(i-1) n+j$. The standard state is when the $k$-th chess piece is in the $k$-th square.
Recor... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 724,202 |
5. (50 points) Let $O$ be the circumcenter of acute $\triangle ABC$. The extensions of $AO$, $BO$, and $CO$ intersect $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. If $\triangle ABC \sim \triangle DEF$, prove that $\triangle ABC$ is an equilateral triangle. | 5. First, prove that $O$ is the orthocenter of $\triangle D E F$.
As shown in Figure 2, let $H$ be the orthocenter of $\triangle D E F$, and $D H, E H, F H$ intersect $E F, F D, D E$ at points $L, M, N$ respectively.
$$
\begin{array}{l}
\text { By } \angle E H F=180^{\circ}-\angle E D F \\
=180^{\circ}-\angle E A F,
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,203 |
$\begin{array}{l}\text { 6. (50 points) For any } x, y, z \in \mathbf{R}, \text { prove: } \\ -\frac{3}{2}\left(x^{2}+y^{2}+2 z^{2}\right) \leqslant 3 x y+y z+z x \\ \leqslant \frac{3+\sqrt{13}}{4}\left(x^{2}+y^{2}+2 z^{2}\right) .\end{array}$ | 6. When $x=y=z=0$, the inequality obviously holds. When $x, y, z$ are not all 0, transform the inequality into
$$
-\frac{3}{2} \leqslant \frac{3 x y+y z+z x}{x^{2}+y^{2}+2 z^{2}} \leqslant \frac{3+\sqrt{13}}{4} \text {. }
$$
Let $F(x, y, z)=\frac{3 x y+y z+z x}{x^{2}+y^{2}+2 z^{2}}$.
Below, we find the range of $F(x, ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,204 |
7. (50 points) Among any nine pairwise distinct positive integers not exceeding 9000, there must exist four numbers \(a, b, c, d\) such that
$$
4+d \leqslant a+b+c \leqslant 4 d \text {. }
$$ | 7. Use proof by contradiction.
Assume the proposition is not true, i.e., there exists
$$
1 \leqslant A < B < C < D < E < F < G < H < I,
$$
such that
$$
4 + A = 1 + 2 + (A + 1) \leqslant B + C + D.
$$
From $D \geqslant 4C - (A + B) + 1$, we get
$$
\begin{array}{l}
D \geqslant 4C - (A + B) + 1 \\
= 2C + (C - A) + (C - ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,205 |
8. (50 points) A scientist stored the design blueprint of a time machine on a computer, setting the file opening password as a permutation of $\{1,2, \cdots, 64\}$. They also designed a program that, when eight positive integers between 1 and 64 are input each time, the computer will indicate the order (from left to ri... | 8. Prepare $n^{2}(n=8)$ cards, the front side of which are numbered $1,2, \cdots, n^{2}$, and the back side corresponds to the position of the number in the password (counting from the left). Of course, the operator does not know the numbers on the back side in advance.
First, divide the $n^{2}$ cards into $n$ groups ... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,206 |
1. Given the set $M=\{2,0,11\}$. If $A \varsubsetneqq M$, and $A$ contains at least one even number, then the number of sets $A$ that satisfy the condition is $\qquad$ . | 1. 5.
The sets $A$ that satisfy the conditions are $\{2\}, \{0\}, \{2,0\}, \{2,11\}, \{0,11\}$. There are 5 in total. | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,207 |
2. Let $a, b$ be positive real numbers,
$$
A=\frac{a+b}{2}, B=\frac{2}{\frac{1}{a}+\frac{1}{b}} \text {. }
$$
If $A+B=a-b$, then $\frac{a}{b}=$ | 2. $3+2 \sqrt{3}$.
From the problem, simplifying we get
$$
\begin{array}{l}
a^{2}-6 a b-3 b^{2}=0 \\
\Rightarrow\left(\frac{a}{b}\right)^{2}-6 \cdot \frac{a}{b}-3=0 .
\end{array}
$$
Solving, we get $\frac{a}{b}=3+2 \sqrt{3}$ (negative root is discarded). | 3+2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,208 |
3. Satisfy
$$
\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}+\frac{1-\sin \theta-\cos \theta}{1-\sin \theta+\cos \theta}=2
$$
The maximum negative angle $\theta$ in radians is | 3. $-\frac{\pi}{2}$.
Given the equation can be transformed into
$$
\frac{(1-\sin \theta+\cos \theta)^{2}+(1-\sin \theta-\cos \theta)^{2}}{(1-\sin \theta)^{2}-\cos ^{2} \theta}=2 \text {. }
$$
Simplifying yields $\cos ^{2} \theta=0 \Rightarrow \cos \theta=0$.
Thus, the largest negative angle $\theta$ that satisfies th... | -\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,209 |
4. Let the line $l$ with a slope of $\frac{\sqrt{2}}{2}$ intersect the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at two distinct points $P$ and $Q$. If the projections of points $P$ and $Q$ on the $x$-axis are exactly the two foci of the ellipse, then the eccentricity of the ellipse is | 4. $\frac{\sqrt{2}}{2}$.
Solution 1 Let the semi-focal distance of the ellipse be $c$.
As shown in Figure 5, it is easy to know that point $P\left(c, \frac{\sqrt{2}}{2} c\right)$. Then
$$
\begin{aligned}
& \frac{c^{2}}{a^{2}}+\frac{\left(\frac{\sqrt{2}}{2} c\right)^{2}}{b^{2}}=1 \\
\Rightarrow & \frac{c^{2}}{a^{2}}+\f... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,210 |
Example 5 In $\triangle ABC$, $AB=BC>AC$, $AH$ and $AM$ are the altitude and median from vertex $A$ to side $BC$, respectively, and $\frac{S_{\triangle MHH}}{S_{\triangle ABC}}=\frac{3}{8}$. Determine the value of $\cos \angle BAC$. [s]
$(2009$, Beijing High School Mathematics Competition Preliminary (High $(-))$ | Solve as shown in Figure 5, let $A B=A C$ $=a$, and $D$ be the midpoint of $A C$.
$$
\begin{array}{l}
\text { From } \frac{S_{\triangle M M H}}{S_{\triangle M B C}}=\frac{3}{8}, \\
B M=M C,
\end{array}
$$
we get
$$
\begin{array}{l}
H C=\frac{1}{2} a-\frac{3}{8} a=\frac{1}{8} a \\
\cos \angle B A C=\cos C \\
=\frac{H C... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,211 |
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
保留源文本的换行和格式，直接输出翻译结果如下：
5. In the array of numbers shown in Figure 1, the three nu... | 5.18.
From the problem, we have
$$
\begin{array}{l}
a_{11}+a_{13}=2 a_{12}, a_{21}+a_{23}=2 a_{22}, \\
a_{31}+a_{33}=2 a_{32}, a_{12}+a_{32}=2 a_{22} .
\end{array}
$$
Thus, the sum of all nine numbers is $9 a_{n}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,212 |
6. As shown in Figure 2, the coordinates of the four vertices of rectangle $O A B C$ are $(0,0), (2 \pi, 0), (2 \pi, 2), (0,2)$, respectively. Let the region (shaded in the figure) enclosed by side $B C$ and the graph of the function $y=1+\cos x(0 \leqslant x \leqslant 2 \pi)$ be $\Omega$. If a point $M$ is randomly th... | 6. $\frac{1}{2}$.
Solution 1 Note that, the area of the rectangle $S=4 \pi$, the area of region $\Omega$ is
$$
\begin{array}{l}
S^{\prime}=4 \pi-\int_{0}^{2 \pi}(1+\cos x) \mathrm{d} x \\
=4 \pi-\left.(x+\sin x)\right|_{0} ^{2 \pi}=2 \pi .
\end{array}
$$
Therefore, the required probability $P=\frac{S^{\prime}}{S}=\fr... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 724,213 |
7. Let the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{1}{p}, & x=\frac{q}{p} ; \\
0, & x \neq \frac{q}{p},
\end{array}\right.
$$
where $p$ and $q$ are coprime, and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$ and $f(x)>\frac{1}{5}$ is $\qquad$ . | 7.5.
Obviously, $x=\frac{q}{p}$ (otherwise, $f(x)=0$). At this point, from $f(x)=\frac{1}{p}>\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$. When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$.
Therefore, there are 5 values of $x$ that satisfy the condition. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,214 |
8. Given that $p$ and $q$ are both prime numbers, and $7p+q$, $2q+11$ are also prime numbers. Then $p^{q}+q^{p}=$ $\qquad$ . | 8. 17 .
Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be an odd number.
Therefore, one of $p$ or $q$ must be even (which can only be 2).
Clearly, $q \neq 2$ (otherwise, $2 q+11=15$, which is not a prime number). Thus, $p=2$.
At this point, $14+q$ and $2 q+11$ are both prime numbers.
If $q=3 k+1\left(k \... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,215 |
9. The total length of all curve segments formed by a moving point on the surface of a regular quadrilateral pyramid $P-ABCD$ with lateral edge length and base edge length both being 4, and at a distance of 3 from the vertex $P$, is
保留源文本的换行和格式，直接输出翻译结果如下：
9. The total length of all curve segments formed by a moving ... | 9. $6 \pi$.
On the side faces of a regular quadrilateral pyramid, the moving point forms four arcs with a radius of 3 and a central angle of $\frac{\pi}{3}$, the sum of whose lengths is
$$
l_{1}=4 \times \frac{\pi}{3} \times 3=4 \pi \text {. }
$$
It is also known that the height of the regular quadrilateral pyramid i... | 6 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,216 |
10. The requirements for deciphering codes in modern society are getting higher and higher. In cryptography, the content that can be seen directly is called plaintext, and the content obtained after processing the plaintext in a certain way is called ciphertext. There is a kind of ciphertext that maps the 26 English le... | 10. love.
From the known transformation formula, we have
$$
x=\left\{\begin{array}{ll}
2 y-1, & 1 \leqslant y \leqslant 13, y \in \mathbf{N} ; \\
2 y-26, & 14 \leqslant y \leqslant 26, y \in \mathbf{N} .
\end{array}\right.
$$
According to the above transformation formula, we have
$$
\begin{array}{l}
s \rightarrow 19 ... | love | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,217 |
One. (20 points) Let the function
$$
f(x)=\cos x \cdot \cos (x-\theta)-\frac{1}{2} \cos \theta
$$
where, $x \in \mathbf{R}, 0<\theta<\pi$. It is known that when $x=\frac{\pi}{3}$, $f(x)$ achieves its maximum value.
(1) Find the value of $\theta$;
(2) Let $g(x)=2 f\left(\frac{3}{2} x\right)$, find the minimum value of ... | (1) Notice
$$
\begin{array}{l}
f(x)=\cos x(\cos x \cdot \cos \theta+\sin x \cdot \sin \theta)-\frac{1}{2} \cos \theta \\
=\frac{1+\cos 2 x}{2} \cos \theta+\frac{1}{2} \sin 2 x \cdot \sin \theta-\frac{1}{2} \cos \theta \\
=\frac{1}{2} \cos (2 x-\theta) .
\end{array}
$$
By $f(x)_{\max }=f\left(\frac{\pi}{3}\right)=\frac... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,218 |
II. (20 points) Let $P$ be a moving point on the line $y=x-2$. Draw tangents from $P$ to the parabola $y=\frac{1}{2} x^{2}$, with points of tangency being $A$ and $B$.
(1) Prove that the line $AB$ passes through a fixed point;
(2) Find the minimum value of the area $S$ of $\triangle PAB$, and the coordinates of point $... | Let $P\left(x_{0}, x_{0}-2\right) 、 A\left(x_{1}, \frac{1}{2} x_{1}^{2}\right), B\left(x_{2}, \frac{1}{2} x_{2}^{2}\right)$, and the equation of the chord of tangents gives the line $l_{A B}$:
$$
\begin{array}{l}
\frac{\left(x_{0}-2\right)+y}{2}=\frac{1}{2} x_{0} x \\
\Rightarrow y=x_{0}(x-1)+2 .
\end{array}
$$
Thus, ... | 3 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,219 |
Three. (20 points) As shown in Figure 3, in the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $E$, and $\angle ABD=60^{\circ}, AE=AD$. Extend $AB$ and $DC$ to meet at point $F$. Prove: $B$ is the circumcenter of $\triangle CEF$.
保留源文本的换行和格式，直接输出翻译结果如下：
Three. (20 points) As shown in Figu... | Three, from $A E=A D, A 、 B 、 C 、 D$ being concyclic, we know
$$
\angle B C E=\angle A D E=\angle A E D \text {. }
$$
Therefore, $B C=B E$.
$$
\begin{aligned}
\text { Also, } & \angle B C F=\angle B A D \\
& =180^{\circ}-(\angle A B D+\angle A D B) \\
& =120^{\circ}-\angle A D E, \\
& \angle B F C=\angle A B D-\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,220 |
Four. (30 points) As shown in Figure 4, $M$ is the midpoint of the median $AD$ of $\triangle ABC$, and a line through $M$ intersects sides $AB$ and $AC$ at points $P$ and $Q$ respectively.
Let $\overrightarrow{AP} = x \overrightarrow{AB}, \overrightarrow{AQ} = y \overrightarrow{AC}$, and denote $y = f(x)$.
(1) Find the... | (1) Since the line passing through point $M$ intersects sides $A B$ and $A C$, we have $x>0, y>0$.
Thus, $\overrightarrow{A B}=\frac{1}{x} \overrightarrow{A P}, \overrightarrow{A C}=\frac{1}{y} \overrightarrow{A Q}$.
Therefore, $\overrightarrow{A M}=\frac{1}{2} \overrightarrow{A D}=\frac{1}{4}(\overrightarrow{A B}+\ove... | \left(-\infty,-\frac{2}{3}\right] \cup\left[0, \frac{1}{6}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,221 |
Example 6 Given that $\odot O$ is the incircle of $\triangle ABC$, $D$, $E$, and $N$ are the points of tangency. Connect $NO$ and extend it to intersect $DE$ at point $K$, connect $AK$ and extend it to intersect $BC$ at point $M$. Prove: $N$ is the midpoint of $BC$. ${ }^{[6]}$
(Sixth Northern Mathematical Olympiad Inv... | Notice
$$
\begin{array}{l}
\frac{D K}{K E}=\frac{S_{\triangle A D K}}{S_{\triangle A K E}} \\
=\frac{A D \cdot A K \sin \angle D A K}{A E \cdot A K \sin \angle K A E}=\frac{\sin \angle D A K}{\sin \angle K A E} . \\
\text { On the other hand, } \frac{D K}{K E}=\frac{S_{\triangle D O K}}{S_{\triangle E O K}}=\frac{\sin ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,222 |
Five. (30 points) Let $n \geqslant 4$ be a positive integer. Prove:
$$
0<\frac{(-1)^{[\sqrt{4}]}}{4}+\frac{(-1)^{[\sqrt{5}]}}{5}+\cdots+\frac{(-1)^{[\sqrt{n}]}}{n}<1 \text {, }
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$. | Five, first prove a lemma.
Lemma Let $a_{k}=\sum_{i=k^{2}}^{(k+1)^{2}-1} \frac{1}{t}\left(k \in \mathbf{N}_{+}\right)$. Then $a_{k}>a_{k+1}$.
Proof The expression for $a_{k}$ has $2 k+1$ terms, consider the sum of the first $k$ terms and the sum of the last $k+1$ terms.
Since $\sum_{i=k^{2}}^{k^{2}+k-1} \frac{1}{t}>\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,223 |
1.220 students take a percentage-based test (scores are counted in integers), and no more than three students have the same score. Then the minimum number of scores that exactly three students have the same score is ( ).
(A) 17
(B) 18
(C) 19
(D) 20 | -,1.B.
Among the scores from $0 \sim 100$, each score is obtained by at least two students, thus totaling 202 students. The remaining 18 students must have distinct scores. Therefore, at least 18 scores are obtained by exactly three students. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,224 |
2. Given that $[x]$ represents the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then the number of positive numbers $x$ that satisfy
$$
20\{x\}+1.1[x]=2011
$$
is ( ).
(A) 17
(B) 18
(C) 19
(D) 20 | 2. B.
Since $0 \leqslant\{x\}<1$, we have
$$
\begin{array}{l}
0 \leqslant 20\{x\}<20, \\
0 \leqslant 2011-1.1[x]<20, \\
1810<[x]<1829 .
\end{array}
$$
Therefore, $[x]$ can take $1810+k(k=1,2, \cdots, 18)$, and accordingly $\{x\}$ takes $1-\frac{11 k}{200}$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,225 |
3. Given that $a$ and $b$ are constants, the solution sets of the inequalities
$$
a x+1>x+b \text { and } 3(x-1)<a-b x
$$
are the same. Then
$$
(a+1)^{2}-(b+1)^{2}=(\quad) \text {. }
$$
(A) 0
(B) 3
(C) 4
(D) 8 | 3. A.
The two inequalities are transformed into
$$
(1-a) x<1-b,(b+3) x<a+3 \text {. }
$$
Since the solution sets of the two inequalities are the same, we have
$$
\begin{array}{l}
(1-a)(a+3)=(1-b)(b+3) \\
\Rightarrow(a+1)^{2}=(b+1)^{2} . \\
\text { Therefore, }(a+1)^{2}-(b+1)^{2}=0 .
\end{array}
$$ | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,226 |
4.8 Eight people each receive a different message at the same time. They inform each other of all the messages they know by phone, with each call taking exactly 3 minutes. To ensure that everyone knows all the messages, at least ( ) minutes are needed.
(A)6
(B) 9
(C) 12
(D) 15 | 4. B.
It is easy to prove that 3 or 6 minutes is impossible.
The following constructs a scenario that takes 9 minutes.
Use $1,2, \cdots, 8$ to represent these eight people and the messages they initially know. In the first phone call, 1 and 2, 3 and 4, 5 and 6, 7 and 8 exchange messages, so each person knows two; in t... | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 724,227 |
5. Given that the angle bisector $A D$ of acute triangle $\triangle A B C$ intersects with the altitude $B E$ at point $M$, and $\triangle C D E$ is an equilateral triangle. Then $S_{\triangle D E M}: S_{\triangle A B M}=(\quad$.
(A) $\sqrt{2}: 2$
(B) $1: 2$
(C) $\sqrt{3}: 3$
(D) $1: 4$ | 5. D.
Since $B E \perp A C$ and $\triangle C D E$ is an equilateral triangle, we have
$$
\angle B E C=90^{\circ}, C D=D E.
$$
Therefore, $B D=D E=D C$.
Also, since $A D$ bisects $\angle B A C$, then $A B=A C$.
Thus, $\triangle A B C$ is an equilateral triangle.
Hence, $M$ is the centroid of the equilateral $\triangle... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,228 |
6. As shown in Figure 1, the radius of $\odot O$ is $6, M$ is a point outside $\odot O$, and $O M=12$. A line through $M$ intersects $\odot O$ at points $A, B$. The points $A, B$ are symmetric to points $C, D$ with respect to $O M$. $A D$ and $B C$ intersect at point $P$. Then the length of $O P$ is ( )
(A) 4
(B) 3.5
(... | 6. C.
As shown in Figure 4, extend $M O$ to intersect $\odot O$ at point $N$, and connect $O B$ and $O D$.
By symmetry, point $P$ lies on line $M N$.
Since $\angle B A D=\frac{1}{2} \angle B O D=\angle B O N$, points $A$, $B$, $O$, and $P$ are concyclic.
Thus, $O M \cdot P M=M A \cdot M B=O M^{2}-O N^{2}$
$\Rightarr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,229 |
1. A three-digit number is 29 times the sum of its digits. Then this three-digit number is $\qquad$ | Let the three-digit number be $\overline{a b c}$. Then
$$
\begin{array}{l}
100 a+10 b+c=29(a+b+c) \\
\Rightarrow 71 a=19 b+28 c \\
\Rightarrow b=4 a-2 c+\frac{5}{19}(2 c-a) .
\end{array}
$$
Since $a$, $b$, and $c$ are positive integers, 19 divides $(2 c - a)$.
And $-9 \leqslant 2 c-a \leqslant 17$, then
$$
2 c-a=0 \Ri... | 261 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,230 |
Example 7 As shown in Figure 7, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3$, the distance from point $P$ to line $A N$ is $P D=12, A D=$
30, and a line is drawn through $P$ intersecting
$A N$ and $A M$ at points
$B$ and $C$ respectively. Find the minimum
value of the area of... | Solve As shown in Figure 7, it can be proven: when the line moves to $B_{0} P=P C_{0}$, the area of $\triangle A B C$ is minimized.
Draw $C_{0} Q / / A N$, intersecting $C B$ at point $Q$.
Thus, $\triangle P C_{0} Q \cong \triangle P B_{0} B$.
Also, $S_{\triangle P C_{0} C} \geqslant S_{\triangle P C_{0} Q}=S_{\triangl... | 624 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,233 |
4. As shown in Figure 2, given that $\odot O$ is the incircle of rhombus $A B C D$ with side length 16, points $E$ and $F$ are on sides $A B$ and $B C$ respectively, and $E F$ is tangent to $\odot O$ at point $M$. If $B E=4, B F=13$, then the length of $E F$ is | 4. 10. 5 .
As shown in Figure 6, let
$\odot O$ be tangent to $AB$ at
point $N$, and connect $AC$,
$BD$, $OE$, $OF$,
$ON$.
From the given conditions, we have
$$
\begin{aligned}
& \angle COF \\
= & 180^{\circ}-\angle OFC-\angle OCF \\
= & 90^{\circ}-\frac{1}{2} \angle EFC+\frac{1}{2} \angle EBF \\
= & \frac{1}{2}(\angle... | 10.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,234 |
(1) (20 points) There is a type of notebook originally priced at 6 yuan per book. Store A uses the following promotional method: for each purchase of 1 to 8 books, a 10% discount is applied; for 9 to 16 books, a 15% discount is applied; for 17 to 25 books, a 20% discount is applied; and for more than 25 books, a 25% di... | (1) Since Class $A$ needs 8 books, and the cost per book at Store A and Store B is 5.4 yuan and 5.2 yuan respectively, it is cheaper to buy from Store B.
Since Class $B$ needs 20 books, and the cost per book at Store A and Store B is 4.8 yuan and 4.9 yuan respectively, it is cheaper to buy from Store A.
(2) From the p... | y=\left\{\begin{array}{ll}
4.9 x, & 11 \leqslant x \leqslant 16 ; \\
4.8 x, & 17 \leqslant x \leqslant 20 ; \\
4.5 x, & 21 \leqslant x \leqslant 40 .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,235 |
II. (25 points) As shown in Figure 3, given the right triangle $\triangle ABC$ with the incircle $\odot I$ touching the two legs $AC$ and $BC$ at points $D$ and $E$, respectively, and $AI$, $BI$ intersecting line $DE$ at points $F$ and $G$. Prove:
$$
AB^{2}=2 FG^{2} \text {. }
$$ | As shown in Figure 7, connect $A G$, $B F$, $D I$, and $E I$.
Obviously, quadrilateral $CDIE$ is a square. Therefore,
$$
\begin{array}{l}
\angle A D F = \angle A I B \\
= 135^{\circ}, \\
\angle F A D = \angle I A B .
\end{array}
$$
Thus, $\triangle A D F \sim \triangle A I B$.
Hence, $\frac{A D}{A I} = \frac{A F}{A B... | A B^{2} = 2 F G^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 724,236 |
Three, (25 points) Find all integer triples $(a, b, c)$
such that
$$
a^{3}+b^{3}+c^{3}-3 a b c=2011(a \geqslant b \geqslant c) .
$$ | $$
\begin{array}{l}
(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\
=2 \times 2011 .
\end{array}
$$
Since $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ is always even, we have:
$$
\left\{\begin{array}{l}
a+b+c=1, \\
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=4022 ;
\end{array}\right.
$$
or $\left\{\begin{array}{l}a+b+c=2011, \\ (a-b)^{2}+... | (20,10,-29),(671,670,670) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,237 |
2. If the real number $m$ satisfies
$$
\sin ^{2} \theta+3 m \cos \theta-6 m-4<0
$$
for any $\theta \in\left[0, \frac{\pi}{3}\right]$, then the range of $m$ is $\qquad$ | $m>-\frac{13}{18}$ | m>-\frac{13}{18} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,239 |
3. Let $p, q \in \mathbf{N}$, and $1 \leqslant p<q \leqslant n$, where $n$ is a natural number not less than 3. Then the sum $S$ of all fractions of the form $\frac{p}{q}$ is $\qquad$ | 3. $\frac{1}{4} n(n-1)$.
Notice that
$$
S=\sum_{q=2}^{n} \sum_{p=1}^{q-1} \frac{p}{q}=\sum_{q=2}^{n} \frac{q-1}{2}=\frac{1}{4} n(n-1) .
$$ | \frac{1}{4} n(n-1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,240 |
4. In $\triangle A B C$, it is known that the three interior angles $\angle A, \angle B, \angle C$ form an arithmetic sequence, with their opposite sides being $a, b, c$ respectively, and $c-a$ equals the height $h$ from vertex $A$ to side $AC$. Then $\sin \frac{C-A}{2}=$ | 4. $\frac{1}{2}$.
Notice that
$$
\begin{array}{l}
h=c-a=\frac{h}{\sin A}-\frac{h}{\sin C} \\
\Rightarrow \sin C-\sin A=\sin C \cdot \sin A .
\end{array}
$$
Then $2 \cos \frac{C+A}{2} \cdot \sin \frac{C-A}{2}$
$$
=\frac{1}{2}[\cos (C-A)-\cos (C+A)] \text {. }
$$
Since $\angle A, \angle B, \angle C$ form an arithmetic... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,241 |
5. From the set $\{1,2, \cdots, 10\}$, any two non-adjacent numbers are taken and multiplied. Then the sum of all such products is equal to
| 5. 990 .
Take any two numbers, multiply them, and then find their sum:
$$
\begin{array}{l}
S_{1}=\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \\
=\frac{1}{2} \sum_{k=1}^{10} k(55-k)=\frac{1}{2} \sum_{k=1}^{10}\left(55 k-k^{2}\right) \\
=\frac{1}{2}(55 \times 55-385)=1320,
\end{array}
$$
Among them, t... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,242 |
6. Three equal cylinders are pairwise tangent, and their axes are mutually perpendicular. If the radius of the base of each cylinder is equal to $r$, then the radius of the smallest sphere that is tangent to all three cylindrical surfaces is $\qquad$ . | 6. $\sqrt{2} r-r$.
Construct a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $2 r$, where edge $A A_{1}$ lies on the axis of the first cylinder, edge $D C$ lies on the axis of the second cylinder, and edge $B_{1} C_{1}$ lies on the axis of the third cylinder. Since $A A_{1}, D C, B_{1} C_{1}$ are mutually pe... | \sqrt{2} r - r | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,243 |
Example 8 (1) As shown in Figure 8, within the square $ABCD$, there are two moving circles $\odot O_{1}$ and $\odot O_{2}$ that are externally tangent to each other. $\odot O_{1}$ is tangent to sides $AB$ and $AD$, and $\odot O_{2}$ is tangent to sides $BC$ and $CD$. If the side length of the square $ABCD$ is $1$, and ... | (1) (i) From the given conditions, points $O_{1}$ and $O_{2}$ are both on the diagonal $AC$ of the square.
By symmetry,
$$
\begin{array}{l}
\sqrt{2} r_{1} + r_{1} + r_{2} + \sqrt{2} r_{2} = AC = \sqrt{2} \\
\Rightarrow r_{1} + r_{2} = 2 - \sqrt{2}.
\end{array}
$$
(ii) From $r_{2} = 2 - \sqrt{2} - r_{1} \leqslant \frac{... | \left(\frac{37}{8} - \frac{5 \sqrt{3}}{2}\right) \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,244 |
7. Given positive integers $a_{1}, a_{2}, \cdots, a_{18}$ satisfying
$$
\begin{array}{l}
a_{1}<a_{2}<\cdots<a_{18}, \\
a_{1}+a_{2}+\cdots+a_{18}=2011 .
\end{array}
$$
Then the maximum value of $a_{9}$ is | 7.193.
To maximize $a_{9}$, $a_{1}, a_{2}, \cdots, a_{8}$ should be as small as possible, and $a_{10}, a_{11}, \cdots, a_{18}$ should be as close to $a_{9}$ as possible. Therefore, we take $a_{1}, a_{2}, \cdots, a_{8}$ to be $1, 2, \cdots, 8$, respectively, with their sum being 36.
Let $a_{9}=n$. Then
$$
\begin{array}... | 193 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,245 |
8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4\left(n \in \mathbf{N}_{+}\right)$, let $d_{n}=\left(a_{n}, a_{n+1}\right)$. Then the maximum value of $d_{n}$ is $\qquad$ | 8.433.
Given $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$.
Then $d_{n} \mid\left[-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right)\right]$,
and $\square$
$$
\begin{array}{l}
d_{n} \mid\left(3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}... | 433 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,246 |
9. (16 points) Let $a, b, c \in \mathbf{R}_{+}$, and $1+a+b+c=4abc$. Prove:
$$
\sum \frac{1}{1+a+b} \leqslant 1,
$$
where, " $\sum$ " denotes the cyclic sum. | $$
\begin{array}{l}
1+a+b+c=4 a b c \\
\Rightarrow \sum(1+2 a)(1+2 b)=\prod(1+2 a) \\
\Rightarrow \sum \frac{1}{1+2 a}=1,
\end{array}
$$
where, " $\Pi $" is the cyclic symmetric product.
Thus, it suffices to prove
$$
\begin{array}{l}
\frac{1}{(1+a)+b}+\frac{1}{(1+b)+c}+\frac{1}{(1+c)+a} \\
\leqslant \frac{1}{(1+a)+a}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,247 |
10. (20 points) Let
$$
\frac{1}{2}<a_{1}<\frac{2}{3}, a_{n+1}=a_{n}\left(2-a_{n+1}\right) \text {. }
$$
Prove: $n+\frac{1}{2}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<n+2$. | 10. It is known that $a_{n+1}=\frac{2 a_{n}}{a_{n}+1}$.
Then $a_{n+1}-1=\frac{a_{n}-1}{a_{n}+1}$.
Dividing the two equations yields $\frac{a_{n+1}-1}{a_{n+1}}=\frac{1}{2} \cdot \frac{a_{n}-1}{a_{n}}$.
Thus, $\left\{\frac{a_{n}-1}{a_{n}}\right\}$ is a geometric sequence.
Therefore, $\frac{a_{n}-1}{a_{n}}=\frac{a_{1}-1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,248 |
11. (20 points) Given the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Find the maximum perimeter of the inscribed parallelogram $A B C D$ in the ellipse. | 11. The perimeter of $\square A B C D$ is a bounded continuous number in the closed interval of the positions of points $A, B, C, D$, and it must have a maximum value. Taking a set of points $A, B, C, D$ that maximize this perimeter, the ellipse with foci at $A, C$ and passing through $B, D$ is tangent to the ellipse $... | 4 \sqrt{a^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,249 |
One. (40 points) From a point $P$ outside the circle $\Gamma$, draw two tangents $PA$ and $PB$ to the circle $\Gamma$, with points of tangency at $A$ and $B$ respectively. Let $C$ and $D$ be two points on the minor arc $\overparen{AB}$. Let $O_{1}$, $O_{2}$, $O_{3}$, and $O_{4}$ be the circumcenters of $\triangle PAC$,... | Let the circumcenter of $\triangle PAB$ be $O$. Then
$$
\begin{array}{l}
\angle O_{1} O P=\angle O_{4} O P \\
=90^{\circ}-\frac{1}{2} \angle A P B=\angle P A B, \\
\angle O_{1} P O=\angle O_{1} P A+\angle A P O \\
=\left(\angle A C P-90^{\circ}\right)+\left(90^{\circ}-\angle P A B\right) \\
=\angle A C P-\angle P A B .... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,250 |
$$
\begin{array}{l}
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, \\
1 \leqslant a_{1}<a_{2}<\cdots<a_{n}(n \geqslant 5),
\end{array}
$$
For any $1 \leqslant i \leqslant j \leqslant n$, there is always $\frac{a_{j}}{a_{i}}$ or $a_{i} a_{j}$ that belongs to the set $A$. Prove: $a_{1}, a_{2}, \cdots, a_{n}$ form a geome... | Given $a_{n} a_{n} \notin A$, then $\frac{a_{n}}{a_{n}} \in A$. Hence $a_{1}=1$.
Since $a_{n} a_{i} \notin A(i=2,3, \cdots, n-1)$, it follows that $\frac{a_{n}}{a_{i}} \in A$,
i.e., $1=a_{1}1$.
We can assume $a_{j}=c^{j-1}(j=2,3, \cdots, n-2)$,
$$
\begin{array}{l}
a_{n-1}=a_{3} a_{n-3}=c^{n-2}, \\
a_{n}=a_{2} a_{n-1}=c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,251 |
Three, (50 points) Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum a^{3}+3 a b c \geqslant \sum a b \sqrt{2 a^{2}+2 b^{2}} .
$$ | $$
\begin{array}{l}
\left(\sum a b \sqrt{2 a^{2}+2 b^{2}}\right)^{2} \\
\leqslant\left(\sum a b\right) \sum a b\left(2 a^{2}+2 b^{2}\right) .
\end{array}
$$
Notice that
$$
\begin{array}{l}
\left(\sum a^{3}+3 a b c\right)^{2} \geqslant 2\left(\sum a b\right) \sum a b\left(a^{2}+b^{2}\right) \\
\Leftrightarrow\left(\sum... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,252 |
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