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$$
\begin{aligned}
f(x)= & \frac{1}{4}\left(\sin ^{2} x-\cos ^{2} x+\sqrt{3}\right)- \\
& \frac{\sqrt{3}}{2} \sin ^{2}\left(x-\frac{\pi}{4}\right)(x \in \mathbf{R}) .
\end{aligned}
$$
Find (1) the smallest positive period of the function $f(x)$;
(2) the intervals where the function $f(x)$ is monotonically increasing. | (1) Notice,
$$
\begin{array}{l}
f(x)=\frac{1}{4}(\sqrt{3}-\cos 2 x)-\frac{\sqrt{3}}{4}\left[1-\cos \left(2 x-\frac{\pi}{2}\right)\right] \\
=\frac{\sqrt{3}}{4} \sin 2 x-\frac{1}{4} \cos 2 x \\
=\frac{1}{2} \sin \left(2 x-\frac{\pi}{6}\right) .
\end{array}
$$
Thus, $T=\frac{2 \pi}{2}=\pi$.
(2) From $2 k \pi-\frac{\pi}{... | \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,533 |
II. (20 points) Let a line $l$ that does not pass through the origin $O$ intersect the circle $x^{2}+y^{2}=1$ at two distinct points $P$ and $Q$. If the slope of line $PQ$ is the geometric mean of the slopes of lines $OP$ and $OQ$, find the range of the area $S$ of $\triangle POQ$. | Let $l_{P Q}: y=k x+b(k \neq 0, b \neq 0)$, substituting into $x^{2}+y^{2}=1$, we get
$$
\left(k^{2}+1\right) x^{2}+2 k b x+b^{2}-1=0 .
$$
From $\Delta=4 k^{2} b^{2}-4\left(k^{2}+1\right)\left(b^{2}-1\right)>0$
$$
\Rightarrow b^{2}<k^{2}+1 \text {. }
$$
Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)\left... | \left(0, \frac{1}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,534 |
Three. (20 points) As shown in Figure 1, $AB$ is the diameter of the semicircle $\odot O$, $C$ is the midpoint of the semicircle arc, $P$ is a point on the extension of $AB$, $PD$ is tangent to the semicircle $\odot O$ at point $D$, and the angle bisector of $\angle APD$ intersects $AC$ and $BC$ at points $E$ and $F$ r... | Proof 1 As shown in Figure 4, let the line $PE$ intersect the semicircle $\odot O$ at points $M$ and $N$, and connect $DE$, $DF$, and $DB$.
Since $\angle APM = \angle DPM$, we have
$\overparen{AM} - \overparen{BN} = \overparen{MC} + \overparen{CD} - \overparen{ND}$.
Also, $\overparen{AM} + \overparen{MC} = \overparen{C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,535 |
Four. (30 points) Given:
Curve $C_{1}: f(x)=\frac{1}{2}\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)$,
Curve $C_{2}: g(x)=\frac{1}{2}\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)$.
The line $x=a$ intersects curves $C_{1}$ and $C_{2}$ at points $A$ and $B$, respectively. The tangent line to curve $C_{1}$ at point $A$ is $l_... | (1) Notice,
$$
\begin{array}{c}
f^{\prime}(x)=\frac{1}{2}\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)=g(x), \\
g^{\prime}(x)=\frac{1}{2}\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)=f(x) .
\end{array}
$$
Then $k_{l_{1}}=\frac{1}{2}\left(\mathrm{e}^{a}-\mathrm{e}^{-a}\right)$,
$$
k_{l 2}=\frac{1}{2}\left(\mathrm{e}^{a}+\m... | \left(-\infty, \frac{1}{2} \ln (\sqrt{5}-2)\right) | Calculus | proof | Yes | Yes | cn_contest | false | 726,536 |
Five. (30 points) Let $P_{0}, P_{1} \cdots, P_{n}\left(n \in \mathbf{N}_{+}\right)$ be $n+1$ points on a plane, and the distance between any two points is no less than 1. Prove:
$$
\sum_{k=1}^{n} \frac{1}{\left(P_{0} P_{k}+1\right)^{4}}<\frac{7}{2} .
$$ | Let the point set
$$
S_{k}=\left\{P_{m}|k<| P_{0} P_{m} \mid \leqslant k+1\right\}(k=1,2, \cdots) \text {. }
$$
Let $\max _{1<m \leqslant\{}\left\{\left|P_{0} P_{m}\right|\right\}=M$.
Then when $k \geqslant[M]$, $\left|S_{k}\right|=0$, where $[M]$ denotes the greatest integer not exceeding the real number $M$.
If $P_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,537 |
2. $\int_{1}^{2} \sqrt{4-x^{2}} \mathrm{~d} x=$ | 2. $\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$.
By the geometric meaning of definite integrals, we have
$$
\begin{array}{l}
\int_{1}^{2} \sqrt{4-x^{2}} \mathrm{~d} x=\frac{\pi}{6} \times 2^{2}-\frac{1}{2} \times 1 \times \sqrt{3} \\
=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}
\end{array}
$$ | \frac{2 \pi}{3}-\frac{\sqrt{3}}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,539 |
3. In $\triangle A B C$, it is known that the lengths of the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and $S_{\triangle A B C}=a^{2}-(b-c)^{2}$. Then $\tan \frac{A}{2}=$ $\qquad$ | 3. $\frac{1}{4}$.
Notice,
$$
\begin{array}{l}
a^{2}-(b-c)^{2}=a^{2}-\left(b^{2}-2 b c+c^{2}\right) \\
=-\left(b^{2}+c^{2}-a^{2}\right)+2 b c \\
=-2 b c \cos A+2 b c . \\
\text { Hence } \frac{1}{2} b c \sin A=-2 b c \cos A+2 b c \\
\Rightarrow 4-4 \cos A=\sin A \\
\Rightarrow 8 \sin ^{2} \frac{A}{2}=2 \sin \frac{A}{2}... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,540 |
Property 5 (Archimedes' Broken Chord Theorem) Let $M$ be the midpoint of the arc $\overparen{B C}$, $A$ be another point on the circumference of the circle containing the arc, and $A$ is on the same side of the chord $B C$ as $M$, satisfying $A B>A C$. Draw $M D \perp A B$ at point $D$, then $B D=D A+A C$. | Property 5 Proof As shown in Figure 1, take point $E$ on $BA$ such that $BE=AC$.
It is easy to see that $\triangle MBE \cong \triangle MCA$.
Thus, $ME=MA$.
Also, $MD \perp EA$, so $ED=DA$.
Therefore, $BD=BE+ED=AC+DA$.
Corollary 2 Let $M$ be the midpoint of the arc $\overparen{BC}$, $A$ be any point on the circumference... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,541 |
4. There are three sets of cards in red, yellow, and blue, each set containing five cards, marked with the letters $A, B, C, D, E$. If five cards are drawn from these 15 cards, with the requirement that the letters are all different and all three colors are included, then the number of different ways to draw the cards ... | 4. 150.
Divide into two categories: $3, 1,1$ and $2,2,1$.
Calculate respectively:
$$
\frac{C_{3}^{1} C_{5}^{3} C_{2}^{1} C_{2}^{1} C_{1}^{1}}{A_{2}^{2}}+\frac{C_{3}^{1} C_{5}^{2} C_{2}^{1} C_{3}^{2} C_{1}^{1}}{A_{2}^{2}}=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,542 |
5. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies $a_{7}=a_{6} + 2 a_{5}$. If there exist two terms $a_{m} 、 a_{n}$ such that $\sqrt{a_{m} a_{n}}=4 a_{1}$, then the minimum value of $\frac{1}{m}+\frac{4}{n}$ is $\qquad$ | 5. $\frac{3}{2}$.
From $a_{7}=a_{6}+2 a_{5}$, we know $q^{2}=q+2$.
Also, since $q>0$, then $q=2$.
Since $\sqrt{a_{m} a_{n}}=4 a_{1}$, we have
$16 a_{1}^{2}=a_{m} a_{n}=a_{1} q^{m-1} a_{1} q^{n-1}$.
Thus, $2^{m+n-2}=2^{4} \Rightarrow m+n=6$.
Therefore, $\frac{1}{m}+\frac{4}{n}=\frac{1}{6}\left(\frac{1}{m}+\frac{4}{n}\r... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,543 |
6. Given the function
$$
f(x)=\frac{1}{3} x^{3}-(a-1) x^{2}+b^{2} x,
$$
where, $a \in\{1,2,3,4\}, b \in\{1,2,3\}$.
Then the probability that the function $f(x)$ is increasing on $\mathbf{R}$ is | 6. $\frac{3}{4}$.
Since $f(x)$ is an increasing function on $\mathbf{R}$, we have
$$
f^{\prime}(x)=x^{2}-2(a-1) x+b^{2} \geqslant 0
$$
always holds, which means
$$
\Delta=4(a-1)^{2}-4 b^{2} \leqslant 0 .
$$
Solving this, we get $|a-1| \leqslant|b|$.
$$
\begin{array}{l}
\text { Therefore, }(a, b) \\
=(1,1),(2,1),(1,2... | \frac{3}{4} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,544 |
7. Let the ellipse $C_{1}: \frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ intersect the parabola $C_{2}$ : $y^{2}=8 x$ at a point $P\left(x_{0}, y_{0}\right)$. Define
$$
f(x)=\left\{\begin{array}{ll}
2 \sqrt{2 x} ; & 0<x \leq x_{0} \\
-2 \sqrt{2 x} ; & x>x_{0} .
\end{array}\right.
$$
If the line $y=a$ intersects $y=f(x)$ at poi... | 7. $\left(\frac{20}{3}, 8\right)$.
From the problem, it is easy to get $x_{0}=\frac{4}{3}$.
Since $N(2,0)$ is exactly the common focus of the ellipse $C_{1}$ and the parabola $C_{2}$, by the definition of a parabola, we have
$$
|N A|=x_{A}+2 \text {. }
$$
Since point $B$ is on the ellipse, then
$$
\frac{x_{B}^{2}}{16... | \left(\frac{20}{3}, 8\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,545 |
8. Given a moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
2 x+y \leqslant 2, \\
x \geqslant 0, \\
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1 .
\end{array}\right.
$$
Then the area of the figure formed by the moving point $P(x, y)$ is | 8. 2 .
From equation (1), we have
$$
\begin{array}{l}
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y \\
\Rightarrow \ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
\end{array}
$$
It is easy to see that $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$ is a strictly increasing function.
$$
\begin{a... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,546 |
9. (14 points) In $\triangle A B C$, it is known that the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively, and $a^{2}$, $b^{2}$, $c^{2}$ form an arithmetic sequence.
(1) Find the range of $B$;
(2) If the equation about $B$
$$
\sqrt{3} \cos B+\sin B=m
$$
has exactly one solut... | 9. (1) From the problem, we know $2 b^{2}=a^{2}+c^{2}$.
Then $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$ $=\frac{a^{2}+c^{2}}{4 a c} \geqslant \frac{2 a c}{4 a c}=\frac{1}{2}$.
Also, $0<\angle B<\pi$, so $\angle B \in\left(0, \frac{\pi}{3}\right]$.
(2) Note that,
$\sqrt{3} \cos B+\sin B=2 \sin \left(B+\frac{\pi}{3}\right... | 2 \text{ or } \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,547 |
10. (14 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}, \\
a_{n+1}=2 a_{n}-a_{n-1}(n \geqslant 2) ;
\end{array}
$$
The sequence $\left\{b_{n}\right\}$ satisfies
$$
b_{1} \neq \frac{1}{4}, 3 b_{n}-b_{n-1}=n(n \geqslant 2) \text {. }
$$
Let the sum o... | 10. (1) From $a_{n+1}=2 a_{n}-a_{n-1}(n \geqslant 2)$, we know $a_{n+1}-a_{n}=a_{n}-a_{n-1}(n \geqslant 2)$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence.
Also, since $a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}$, we have
$$
\therefore a_{n}=\frac{2 n-1}{4} \text {. }
$$
From $3 b_{n}-b_{n-1}=n$, we know $b_{n... | \frac{1}{4} n^{2}-\left(\frac{1}{3}\right)^{n}+1 | Algebra | proof | Yes | Yes | cn_contest | false | 726,548 |
11. (14 points) As shown in Figure 1, let $S-ABCD$ be a pyramid with a height of 3, where the base $ABCD$ is a square with side length 2, and the projection of vertex $S$ on the base is the center of the square $ABCD$. $K$ is the midpoint of edge $SC$, and a plane passing through $AK$ intersects segments $SB$ and $SD$ ... | 11. (1) Let the angle between $A K$ and plane $S B C$ be $\theta$.
Since $S C=\sqrt{3^{2}+(\sqrt{2})^{2}}=\sqrt{11}$, therefore, $C K=\frac{\sqrt{11}}{2}$,
and $\cos \angle S C A=\frac{\sqrt{2}}{\sqrt{11}}=\frac{\sqrt{22}}{11}$.
Then $A K^{2}=A C^{2}+C K^{2}-2 A C \cdot C K \cos \angle S C A=\frac{27}{4}$.
Thus, $A K=... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,549 |
12. (14 points) Given the function
$$
\begin{array}{l}
f(x)=\ln x-a x(a>0), \\
g(x)=f(x)+f^{\prime}(x) .
\end{array}
$$
(1) If the maximum value of the function $f(x)$ is -4 when $1 \leqslant x \leqslant \mathrm{e}$, find the expression for the function $f(x)$;
(2) Determine the range of values for $a$ such that the fu... | 12. (1) Notice that, $f^{\prime}(x)=\frac{1}{x}-a$.
Thus, $f(x)$ is monotonically increasing in $\left(0, \frac{1}{a}\right)$, and $f(x)$ is monotonically decreasing in $\left(\frac{1}{a},+\infty\right)$.
Therefore, when $x=\frac{1}{a}$, $f(x)$ reaches its maximum value.
(i) When $01$,
$f(x)_{\max }=f(1)=-4$.
Solving ... | f(x)=\ln x-4 x, \left[\frac{1}{4},+\infty\right) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,550 |
13. (15 points) Given an ellipse $E: \frac{x^{2}}{8}+\frac{y^{2}}{b^{2}}=1$ with its foci on the $x$-axis, which contains a circle $C: x^{2}+y^{2}=\frac{8}{3}$. A tangent line $l$ of circle $C$ intersects ellipse $E$ at points $A$ and $B$, and satisfies $\overrightarrow{O A} \perp \overrightarrow{O B}$ (where $O$ is th... | 13. (1) As shown in Figure 2, let the tangent line $l$ of circle $C$ intersect the ellipse $E$ at two points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$.
When the slope of line $l$ exists, let the equation of $l$ be
$$
y=k x+m \text {. }
$$
Substituting into the equation of the ellipse, we get
$$
\le... | \frac{4 \sqrt{6}}{3} \leqslant|A B| \leqslant 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,551 |
Example 1 In $\triangle ABC$, it is known that $AB > AC$, the bisector of an exterior angle of $\angle A$ intersects the circumcircle of $\triangle ABC$ at point $E$, and a perpendicular line $EF$ is drawn from $E$ to $AB$ at point $F$. Prove:
$$
2 AF = AB - AC.
$$
(1989, National High School Mathematics Competition) | Proof As shown in Figure 2, let $K$ be a point on the extension of $CA$. Connect $EB$ and $EC$. Since $AE$ bisects the exterior angle $\angle BAC$, we have $\angle EBC = \angle EAK = \angle EAB = \angle ECB$.
Therefore, $EC = EB$, which means $E$ is the midpoint of the arc $\overparen{BAC}$.
By Property 5, we know $BF ... | 2AF = AB - AC | Geometry | proof | Yes | Yes | cn_contest | false | 726,552 |
14. (15 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$.
(1) Find the minimum value of $f(n)$;
(2) When $n=2 \times 10^{k}-1\left(k \in \mathbf{N}_{+}\right)$, find $f(n)$;
(3) Does there exist a positive integer $n$ such that
$$
f(n)=201... | 14. (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, we know that $f(n) \neq 1$.
Assume $f(n)=2$. Then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e.,
$$
3 n^{2}+n+1=10^{k}+1 \text {. }
$$
Clearly, when $k=1$, $n$ does not exist.
Therefore, $k$ is ... | 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,553 |
1. Given $\sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a}$. Then $\sqrt{4 x+x^{2}}=$
(A) $a-\frac{1}{a}$
(B) $\frac{1}{a}-a$
(C) $a+\frac{1}{a}$
(D) cannot be determined | -1. B.
$$
\begin{array}{l}
\text { Given } \sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a} \Rightarrow x=\frac{1}{a}+a-2 \\
\Rightarrow x+2=\frac{1}{a}+a .
\end{array}
$$
Then $\sqrt{4 x+x^{2}}=\sqrt{(x+2)^{2}-4}$
$$
\begin{array}{l}
=\sqrt{\left(\frac{1}{a}+a\right)^{2}-4}=\sqrt{\left(\frac{1}{a}-a\right)^{2}} \\
=\left|\frac{1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,554 |
2. Given $a^{2}+b^{2}=1, b^{2}+c^{2}=c^{2}+a^{2}=2$. Then the minimum value of $a b+b c+a c$ is ( ).
(A) $\sqrt{3}-\frac{1}{2}$
(B) $\frac{1}{2}-\sqrt{3}$
(C) $-\frac{1}{2}-\sqrt{3}$
(D) $\frac{1}{2}+\sqrt{3}$ | 2. B.
$$
\begin{array}{l}
\text { Given } a^{2}+b^{2}=1, b^{2}+c^{2}=2, a^{2}+c^{2}=2 \\
\Rightarrow(a, b, c)=\left( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{6}}{2}\right) .
\end{array}
$$
Then $a b+b c+c a$
$$
\begin{array}{l}
\geqslant \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}+\frac{\sqrt{2... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,555 |
3. Given point $D$ on side $A B$ of $\triangle A B C$, $\angle A$ $=40^{\circ}, \angle B=50^{\circ}, \angle B C D=30^{\circ}, B C=a, A C$ $=b$. Then $C D=(\quad$.
(A) $\frac{\sqrt{2}}{2} \sqrt{a b}$
(B) $\frac{1}{2} \sqrt{a^{2}+b^{2}}$
(C) $\frac{1}{3}(a+b)$
(D) $\frac{1}{4}(2 a+b)$ | 3. B.
Obviously, $\angle A C B=90^{\circ}$.
As shown in Figure 5, let the midpoint of $A B$ be $M$, and connect $C M$.
$$
\begin{array}{c}
\text { Since } M A=M C, \\
\angle A=40^{\circ}, \\
\text { we know } \angle C M D \\
=2 \times 40^{\circ}=80^{\circ} . \\
\text { Also, } \angle C D M \\
=\angle B+\angle B C D=50... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,556 |
4. The system of inequalities about $x$
$$
\left\{\begin{array}{l}
x^{2}-x-2>0, \\
2 x^{2}+(2 k+5) x+5 k<0
\end{array}\right.
$$
has a unique integer solution -2. Then the range of the real number $k$ is ( ).
(A) $[-3,2)$
(B) $\left(2, \frac{5}{2}\right)$
(C) $[-2,2]$
(D) $\left[-3, \frac{5}{2}\right]$ | 4. A.
From $x^{2}-x-2>0 \Rightarrow x>2$ or $x-2>-\frac{2}{5}$.
Thus, $-\frac{5}{2}<x<-k$.
Since -2 is the only integer solution that satisfies inequality (2), then $-2 \leqslant -k \leqslant 3$.
Therefore, $-3 \leqslant k<2$. | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 726,557 |
5. As shown in Figure 1, hexagon $A B C D E F$ is composed of five unit squares. A line that can bisect the area of this hexagon is called a "good line". Then the number of good lines is ( ) lines.
(A) 1
(B) 2
(C) 3
(D) infinitely many | 5. D.
Let the midpoint of $CD$ be $H$, and the centers of rectangle $ABCH$ and rectangle $EFHD$ be $O_{1}$ and $O_{2}$, respectively. Then $O_{1}O_{2}$ is a good line, and any line passing through the midpoint $M$ of segment $\mathrm{O}_{1} \mathrm{O}_{2}$ (intersecting segment $DE$) is also a good line. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,558 |
6. In a graduation photo at a school, 100 students and teachers from two graduating classes of the ninth grade are to be arranged on steps in a trapezoidal formation (with more people in the front rows than in the back rows, and the number of rows being greater than or equal to 3). The photographer requires that the nu... | 6. B.
Let the last row have $k$ people, with a total of $n$ rows. Then the number of people in each row from back to front are $k, k+1, \cdots, k+(n-1)$.
According to the problem,
$$
k n+\frac{n(n-1)}{2}=100,
$$
which simplifies to $n[2 k+(n-1)]=200$.
Since $k$ and $n$ are both positive integers, and $n \geqslant 3$,... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,559 |
1. The positive integer $n=$ $\qquad$ that makes $2^{n}+256$ a perfect square. | When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$.
If it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a positive integer). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, hence $k=1$.
Thus, $n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,560 |
3. As shown in Figure 3, in trapezoid $ABCD$, $DC \parallel AB$, $\frac{DC}{AB} = \frac{1}{3}$, $MN$ is the midline, $EF \parallel AB$ and passes through the intersection of $AC$ and $BD$, points $E$ and $F$ are on $AD$ and $BC$ respectively. Then the ratio of the areas of trapezoids $EFCD$, $MNEF$, and $ABNM$ is $\qqu... | 3. 5: 7:20.
It is easy to prove that trapezoid $E F C D \backsim$ trapezoid $A B N M$, and trapezoid $M N C D \backsim$ trapezoid $A B F E$.
Let $D C=1$. Then $A B=3, M N=\frac{1}{2}(1+3)=2$.
Assume the area of trapezoid $E F C D$ is 1. Then the area of trapezoid $A B N M$ is 4. Let the area of trapezoid $M N E F$ be ... | 5: 7: 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,562 |
Example 1 Find all $n$ such that there exist $a$ and $b$ satisfying
$$
S(a)=S(b)=S(a+b)=n .
$$ | From property (1), we know
$$
a \equiv b \equiv a+b \equiv n(\bmod 9) \text {. }
$$
Thus, $2 n=n(\bmod 9)$.
Therefore, $n$ is a multiple of 9.
When $n=9 k$, take $a=b=10^{k}-1$, at this time,
$$
S\left(10^{k}-1\right)=S\left(2\left(10^{k}-1\right)\right)=9 k \text {, }
$$
satisfying the problem's requirements. | n=9k | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,563 |
5. Given that $a, b (a \leqslant b)$ are positive real numbers, for a positive integer $n$, let $x_{n}$ denote the value of $S([a n+b])$, where $[x]$ represents the greatest integer not exceeding the real number $x$. Prove: $\left\{x_{n}\right\}$ $(n \geqslant 1)$ contains a constant subsequence. | For an integer $k$, let $n_{k}=\left[\frac{10^{k}+a-b}{a}\right]$. Then
$$
\begin{array}{l}
10^{k}=a\left(\frac{10^{k}+a-b}{a}-1\right)+b<a n_{k}+b \\
=a\left[\frac{10^{k}+a-b}{a}\right]+b \leqslant 10^{k}+b .
\end{array}
$$
Thus, $10^{k} \leqslant\left[a n_{k}+b\right] \leqslant 10^{k}+b$.
If $k$ is sufficiently larg... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,566 |
6. Prove: For any $N$, there exists $n>N$, such that $S\left(3^{n}\right) \geqslant S\left(3^{n+1}\right)$. | If there exists $N$, such that for any $n>N$, we have
$$
S\left(3^{n+1}\right)-S\left(3^{n}\right)>0 \text {. }
$$
By property (1), we know $S\left(3^{n+1}\right)-S\left(3^{n}\right) \geqslant 9$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=N+1}^{n}\left(S\left(3^{k+1}\right)-S\left(3^{k}\right)\right) \geqslant 9(n-N... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,567 |
Example 1 In the Cartesian coordinate system, a 100-gon $P$ is given, satisfying:
(1) The coordinates of the vertices of $P$ are all integers;
(2) The sides of $P$ are all parallel to the coordinate axes;
(3) The side lengths of $P$ are all odd.
Prove: The area of $P$ is odd. | Note that, in this problem, we only need to determine the parity of the area, not the specific area.
We call the sides parallel to the $x$-axis "horizontal sides" and the sides parallel to the $y$-axis "vertical sides."
Since each horizontal side of the polygon must connect to vertical sides at both ends, and each ve... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,568 |
Example 3 The King of the Spider Republic (the Spider King) plans to build $n$ cities and $n-1$ roads in his territory, so that each road connects two cities without passing through any other cities, and any two cities can reach each other through these roads. He also requires that the shortest distances between cities... | (1) When $n=6$, the Spider King's plan can be realized; the distribution of 6 points and the design of each segment's length are shown in Figure 2.
Obviously, the distances between any two of the 6 cities, $1,2, \cdots, 15$, all appear $\left(15=\mathrm{C}_{6}^{2}\right)$.
(2) When $n=2013$, the Spider King's plan can... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,570 |
Example 1 Let $A=\{x \mid x \geqslant 10, x \in \mathbf{N}\}, B \subseteq A$, and the elements in $B$ satisfy:
(i) The digits of any element are all different;
(ii) The sum of any two digits of any element is not equal to 9.
(1) Find the number of two-digit and three-digit numbers in $B$;
(2) Does there exist a five-di... | (1) For two-digit numbers, the digit in the tens place can be $1,2, \cdots, 9$; the digit in the units place, since it cannot be the same as the digit in the tens place and the sum of the two digits cannot be 9, has 8 possible choices for each digit in the tens place.
Therefore, the total number of two-digit numbers th... | 4012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,572 |
Example 2 For a four-digit number, at most two of its digits are different. How many such four-digit numbers are there?
保留源文本的换行和格式，所以翻译结果如下：
Example 2 For a four-digit number, at most two of its digits are different.
Ask: How many such four-digit numbers are there? | Solution: Clearly, there are exactly 9 four-digit numbers where all four digits are the same.
Below, we consider four-digit numbers with exactly two different digits in three steps.
(1) First, consider the thousands place, which has 9 possible choices: $1,2, \cdots, 9$.
(2) Next, consider the hundreds, tens, and units... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,573 |
Example 3 Given $A \cup B \cup C=\{1,2, \cdots, 6\}$, and $A \cap B=\{1,2\},\{1,2,3,4\} \subseteq B \cup C$.
Then the number of $(A, B, C)$ that satisfy the conditions is $\qquad$ groups (different orders of $A, B, C$ are considered different groups). | As shown in Figure 1, for $1$ and $2$, they can belong to regions I and II, which gives $2^{2}$ possibilities; for $3$ and $4$, they can belong to $B \cup C$ except for regions I and II, i.e., regions III, IV, VI, and VII, which gives $4^{2}$ possibilities; for $5$ and $6$, they can belong to $A \cup B$ except for regi... | 1600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,574 |
Example 3 (1) Find the minimum value of the sum of the digits of integers of the form $3 n^{2}+n+1\left(n \in \mathbf{Z}_{+}\right)$;
(2) Does there exist a number of this form whose sum of digits is 1999? | (1) Since
$$
3 n^{2}+n+1=n(3 n+1)+1
$$
is an odd number, the last digit must be odd.
If the sum of the digits is 1, then the number is 1, but $n$ is a positive integer, $3 n^{2}+n+1 \geqslant 5$, which is a contradiction.
If the sum of the digits is 2 and it is an odd number, then the first and last digits must both ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,575 |
Example 4 Given two sets of real numbers
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{50}\right\} \text {. }
$$
If the mapping $f$ from $A$ to $B$ makes each element in $B$ have a preimage, and
$$
f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant f\lef... | Let's assume $b_{1}<b_{2}<\cdots<b_{50}$.
Since each element in set $B$ has a preimage, let the set of preimages of $b_{i}$ be $A_{i}(i=1,2, \cdots, 50)$, with the number of elements being $x_{i}$. Then
$$
x_{1}+x_{2}+\cdots+x_{50}=100 \text {. }
$$
Thus, the problem is transformed into finding the number of positive ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,576 |
Example 5 If 6 college graduates apply to three employers, and each employer hires at least one of them, then the number of different hiring scenarios is $\qquad$ kinds. | The number of ways to hire 3 people is $\mathrm{A}_{6}^{3}=120$; the number of ways to hire 4 people is $\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{1} \mathrm{C}_{3}^{1}}{2!} \times \mathrm{A}_{3}^{3}=540$; the number of ways to hire 5 people is
$$
\frac{C_{6}^{2} C_{4}^{2} C_{2}^{1}}{2!} \times A_{3}^{3}+\frac{C_{6}^{3}... | 2100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,577 |
Example 6 In a $6 \times 6$ grid, three identical red cars and three identical black cars are parked, with one car in each row and each column, and each car occupies one cell. The number of ways to park the cars is ( ).
(A) 720
(B) 20
(C) 518400
(D) 14400 | Assume first that the three red cars are distinct and the three black cars are also distinct. The first car can obviously be placed in any of the 36 squares, giving 36 ways. The second car, which cannot be in the same row or column as the first car, has 25 ways to be placed.
Similarly, the third, fourth, fifth, and si... | 14400 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,578 |
Example 7: There are 4 red cards, 3 blue cards, 2 yellow cards, and 1 white card. Cards of the same color are indistinguishable. Questions:
(1) How many ways are there to arrange these 10 cards in a row from left to right?
(2) How many ways are there to arrange the cards so that the first 3 cards from the left are of t... | (1) $\frac{10!}{4!\times 3!\times 2!\times 1!}=12600$ ways.
(2) For the left 3 cards being red, there are
$$
\frac{7!}{1!\times 3!\times 2!\times 1!}=420 \text { (ways); }
$$
For the left 3 cards being blue, there are
$$
\frac{7!}{4!\times 2!\times 1!}=105 \text { (ways). }
$$
Thus, there are $420+105=525$ ways in to... | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,579 |
1. The sequence $\left\{a_{n}\right\}$ has 11 terms, satisfying
$$
a_{1}=0, a_{11}=4,
$$
and $\left|a_{k+1}-a_{k}\right|=1(k=1,2, \cdots, 10)$. The number of different sequences is ( ).
(A) 100
(B) 120
(C) 140
(D) 160 | According to the problem, $a_{k+1}-a_{k}=1$ or -1.
In $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{11}-a_{10}$, suppose there are $x$ ones, then there are $10-x$ negative ones.
According to the problem, $4=x-(10-x) \Rightarrow x=7$.
Therefore, the number of sequences that meet the condition is $\mathrm{C}_{10}^{7}=120$. Hence... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,580 |
2. If 9 students are to be assigned to three different factories for social practice activities, with three students at each factory, then the number of different allocation schemes is ( ).
(A) $\mathrm{C}_{9}^{3} \mathrm{C}_{6}^{3} \mathrm{C}_{3}^{3}$
(B) $3 \mathrm{C}_{9}^{3} \mathrm{C}_{6}^{3} \mathrm{C}_{3}^{3}$
(C... | First, select three students from nine to go to the first factory, then select three students from the remaining six to go to the second factory, and the remaining three students will go to the third factory, there are $\mathrm{C}_{9}^{3} \mathrm{C}_{6}^{3} \mathrm{C}_{3}^{3}$ ways. | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,581 |
3. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1, 2, \cdots, 14$, such that the following conditions are satisfied:
$$
a_{2}-a_{1} \geqslant 3 \text { and } a_{3}-a_{2} \geqslant 3 \text {. }
$$
Then the number of different ways to choose such numbers is $\qquad$ kinds. | Let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}$,
$$
a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4} \text {. }
$$
Then $x_{1}+x_{2}+x_{3}+x_{4}=14$.
Thus, the problem is transformed into finding the number of integer solutions to the equation under the conditions
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,582 |
4. Let $A$ and $B$ be two sets, and $(A, B)$ is called a "pair". When $A \neq B$, $(A, B)$ and $(B, A)$ are considered different pairs. Then the number of different pairs satisfying the condition $A \cup B=\{1,2,3,4\}$ is $\qquad$ | Prompt: Following Example 3, we know there are $3^{4}=81$ pairs.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Prompt: Following Example 3, we know there are $3^{4}=81$ pairs. | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,583 |
5.2011 is a four-digit number whose sum of digits is 4. Then the total number of four-digit numbers whose sum of digits is 4 is $\qquad$. | In fact, the number of four-digit numbers $\overline{x_{1} x_{2} x_{3} x_{4}}$ is the number of integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=4
$$
satisfying the conditions $x_{1} \geqslant 1, x_{2}, x_{3}, x_{4} \geqslant 0$. It is easy to see that there are $\mathrm{C}_{6}^{3}=20$ such s... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,584 |
6. There are 8 English letters $K, Z, A, I, G, A, K$, and $\mathrm{U}$, each written on 8 cards. Ask:
(1) How many ways are there to arrange these cards in a row?
(2) How many ways are there to arrange 7 of these cards in a row? | (1) $\frac{8!}{2!\times 2!}=10080$ ways.
(2) If the letter taken away is $K$ or $A$, then there are
$$
2 \times \frac{7!}{2!}=5040 \text { (ways); }
$$
If the letter taken away is $Z$, $I$, $G$, or $U$, then there are
$$
4 \times \frac{7!}{2!\times 2!}=5040 \text { (ways). }
$$
Therefore, there are $5040+5040=10080$ ... | 10080 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,585 |
Example 1 Given that $p$ is a prime number greater than 3. Find
$$
\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)
$$
the value. | Let $\omega=\mathrm{e}^{\frac{2 \pi i}{p}}$. Then
$$
\begin{array}{l}
\omega^{p}=1, \omega^{-\frac{p}{2}}=-1, 2 \cos \frac{2 k \pi}{p}=\omega^{k}+\omega^{-k} . \\
\text { Hence } \prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=\prod_{k=1}^{p-1}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{p-1} \omega... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,587 |
Example 2 Find
$$
I=\sin \frac{\pi}{n} \cdot \sin \frac{2 \pi}{n} \cdots \cdots \sin \frac{(n-1) \pi}{n}
$$
the value of ( $n$ is a natural number greater than 1). | Let $\omega=\cos \frac{\pi}{n}+\mathrm{i} \sin \frac{\pi}{n}$. Then $\omega^{2 n}=1$.
Thus, $1, \omega^{2}, \omega^{4}, \cdots, \omega^{2(n-1)}$ are all roots of $x^{2 n}-1=0$.
Therefore, $x^{2 n}-1$
$$
\begin{array}{l}
=\left(x^{2}-1\right)\left[x^{2(n-1)}+x^{2(n-2)}+\cdots+x^{2}+1\right] \\
=\left(x^{2}-1\right)\left... | \frac{n}{2^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,588 |
Let $S(n)$ denote the sum of the digits of the integer $n$. Then for any integer $m (m>1)$, there does not exist a positive integer $n$ such that
$$
n-S(n)=10^{m}-10 \text {. }
$$ | Prove that for some integer $m(m>1)$, assuming there exists a positive integer $n$ such that equation (1) holds. Let
$$
n=10^{k} a_{k}+10^{k-1} a_{k-1}+\cdots+10 a_{1}+a_{0} \text {, }
$$
where $a_{k}, a_{k-1}, \cdots, a_{0}$ are non-negative integers no greater than 9, and $a_{k} \geqslant 1$.
Thus, $S(n)=a_{k}+a_{k-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,589 |
1. $4 \sqrt{3+2 \sqrt{2}}-\sqrt{41+24 \sqrt{2}}=(\quad)$.
(A) $\sqrt{2}-1$
(B) 1
(C) $\sqrt{2}$
(D) 2 | $\begin{array}{l}\text {-1. B. } \\ 4 \sqrt{3+2 \sqrt{2}}-\sqrt{41+24 \sqrt{2}} \\ =4(\sqrt{2}+1)-(3+4 \sqrt{2})=1\end{array}$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,590 |
2. The sum of all real numbers $m$ that satisfy $(2-m)^{m^{2}-m-2}=1$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. A.
When $2-m=1$, i.e., $m=1$, it satisfies the condition.
When $2-m=-1$, i.e., $m=3$,
$$
(2-m)^{m^{2}-m-2}=(-1)^{4}=1 \text{, }
$$
it satisfies the condition.
When $2-m \neq \pm 1$, i.e., $m \neq 1$ and $m \neq 3$, by the condition, $m^{2}-m-2=0$, and $2-m \neq 0$.
Solving gives $m=-1$.
Therefore, the sum is $1+3+... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,591 |
3. Given that $AB$ is the diameter of $\odot O$, $C$ is a point on $\odot O$, $\angle CAB=15^{\circ}$, and the angle bisector of $\angle ACB$ intersects $\odot O$ at point $D$. If $CD=\sqrt{3}$, then $AB=(\quad)$.
(A) 2
(B) $\sqrt{6}$
(C) $2 \sqrt{2}$
(D) 3 | 3. A.
As shown in Figure 3, connect $O C$, and draw $O M \perp C D$ at point $M$.
Then $\angle O C M$
$$
=45^{\circ}-15^{\circ}=30^{\circ} \text {. }
$$
Thus, $C M=\frac{\sqrt{3}}{2} O C$.
Therefore, $A B=2 O C$
$$
=\frac{4}{\sqrt{3}} C M=\frac{2}{\sqrt{3}} C D=2 \text {. }
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,592 |
4. Indeterminate equation
$$
3 x^{2}+7 x y-2 x-5 y-17=0
$$
all positive integer solutions $(x, y)$ have $(\quad)$ groups.
(A) 1
(B) 2
(C) 3
(D) 4 | 4. B.
From the given conditions, we have
$$
y=\frac{-3 x^{2}+2 x+17}{7 x-5} \text {. }
$$
Since $x$ and $y$ are positive integers, it follows that
$$
-3 x^{2}+2 x+17 \geqslant 7 x-5 \text {. }
$$
Therefore, $3 x^{2}+5 x \leqslant 22$.
Thus, $x=1$ or 2.
When $x=1$, $y=8$; when $x=2$, $y=1$.
In conclusion, $(x, y)=(1,... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,593 |
5. In rectangle $ABCD$, the side lengths are $AD=3, AB=2, E$ is the midpoint of $AB$, $F$ is on segment $BC$, and $\frac{BF}{FC}=\frac{1}{2}, AF$ intersects $DE$ and $DB$ at points $M$ and $N$. Then $MN=(\quad$.
(A) $\frac{3 \sqrt{5}}{7}$
(B) $\frac{5 \sqrt{5}}{14}$
(C) $\frac{9 \sqrt{5}}{28}$
(D) $\frac{11 \sqrt{5}}{2... | 5. C.
It is easy to know that $\triangle B N F \backsim \triangle D N A$
$$
\Rightarrow \frac{F N}{A N}=\frac{B N}{N D}=\frac{B F}{A D}=\frac{1}{3} \text {. }
$$
Thus, $F N=\frac{1}{3} A N=\frac{1}{4} A F$.
As shown in Figure 4, extend $D E$ and $C B$ to intersect at point $G$.
Then $\triangle A M D \backsim \triangl... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,594 |
6. Let $n$ be a positive integer, and call $n$ a "good number" if the number of prime numbers not exceeding $n$ equals the number of composite numbers not exceeding $n$. Then the sum of all good numbers is ( ).
(A) 33
(B) 34
(C) 2013
(D) 2014 | 6. B.
Since 1 is neither a prime number nor a composite number, a good number must be an odd number.
Let the number of prime numbers not exceeding $n$ be $a_{n}$, and the number of composite numbers be $b_{n}$. When $n \leqslant 15$, only consider the case where $n$ is odd (as shown in Table 1).
Table 1
\begin{tabula... | 34 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,595 |
1. Given real numbers $x, y, z$ satisfy
$$
x+y=4,|z+1|=x y+2 y-9 \text {. }
$$
then $x+2 y+3 z=$ $\qquad$ | $=, 1.4$.
From $x+y=4$, we get $x=4-y$. Then
$$
\begin{array}{l}
|z+1|=x y+2 y-9 \\
=6 y-y^{2}-9=-(y-3)^{2} \\
\Rightarrow z=-1, y=3 \Rightarrow x=1 \\
\Rightarrow x+2 y+3 z=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,596 |
Example 5 Proof: There exist 50 different positive integers $n_{i}$ $(1 \leqslant i \leqslant 50)$, such that for each $i$, $n_{i}+S\left(n_{i}\right)$ is the same positive integer. | Prove by replacing 50 with $k$, to prove a more general case.
When $k=1$, it is obviously true.
Assume when $k-1$, $m_{1}m_{k-1}$.
Let $n_{i}=10^{l+1}+m_{i}(i<k)$, the newly constructed number
$$
\begin{array}{l}
n_{k}=m+10^{l+1}-10 . \\
\text { Then } n_{1}+S\left(n_{1}\right) \\
=\left(10^{l+1}+m_{1}\right)+\left(1+S... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,597 |
2. A cube is painted red on all its faces, then cut into $n^{3}(n>2)$ identical smaller cubes. If the number of smaller cubes with only one face painted red is the same as the number of smaller cubes with no faces painted red, then $n=$ $\qquad$ . | 2.8.
The total number of small cubes with only one face painted red is $6(n-2)^{2}$, and the total number of small cubes with no faces painted red is $(n-2)^{3}$. According to the problem,
$$
6(n-2)^{2}=(n-2)^{3} \Rightarrow n=8 \text {. }
$$ | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,598 |
3. In $\triangle A B C$, it is known that $\angle A=60^{\circ}, \angle C=$ $75^{\circ}, A B=10$, points $D, E, F$ are on sides $A B, B C, C A$ respectively. Then the minimum perimeter of $\triangle D E F$ is $\qquad$ | $3.5 \sqrt{6}$.
As shown in Figure 5, construct the symmetric points $P$ and $Q$ of point $E$ with respect to $AB$ and $AC$, and connect $AE$, $AP$, $AQ$, $DP$, $FQ$, and $PQ$.
Then $\angle PAQ = 120^{\circ}$, and $AP = AQ = AE$.
Construct $AH \perp BC$ at point $H$.
Then the perimeter of $\triangle DEF$
$$
\begin{arr... | 5 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,599 |
4. If real numbers $x, y, z$ satisfy
$$
x^{2}+y^{2}+z^{2}-(x y+y z+z x)=8 \text {, }
$$
let $A$ denote the maximum value of $|x-y|, |y-z|, |z-x|$, then the maximum value of $A$ is $\qquad$. | 4. $\frac{4 \sqrt{6}}{3}$.
From the problem, we know
$$
(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=16 \text {. }
$$
Let's assume $A=|x-y|$. Then
$$
\begin{array}{l}
A^{2}=(x-y)^{2}=[(y-z)+(z-x)]^{2} \\
\leqslant 2\left[(y-z)^{2}+(z-x)^{2}\right] \\
=2\left[16-(x-y)^{2}\right] \\
=2\left(16-A^{2}\right) .
\end{array}
$$
Solving t... | \frac{4 \sqrt{6}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,600 |
One, (20 points) Given real numbers $a, b, c, d$ satisfy $2a^2 + 3c^2 = 2b^2 + 3d^2 = (ad - bc)^2 = 6$. Find the value of $\left(a^2 + \dot{b}^2\right)\left(c^2 + d^2\right)$. | Let $m=a^{2}+b^{2}, n=c^{2}+d^{2}$. Then
$$
\begin{array}{l}
2 m+3 n=2 a^{2}+2 b^{2}+3 c^{2}+3 d^{2}=12 . \\
\text { By }(2 m+3 n)^{2}=(2 m-3 n)^{2}+24 m n \geqslant 24 m n \\
\Rightarrow 12^{2} \geqslant 24 m n \\
\Rightarrow m n \leqslant 6 . \\
\text { Also } m n=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
=... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,601 |
II. (25 points) As shown in Figure 1, point $C$ is on the circle $\odot O$ with diameter $AB$. Tangents to $\odot O$ are drawn through points $B$ and $C$, intersecting at point $P$. Connect $AC$. If $OP = \frac{9}{2} AC$, find the value of $\frac{PB}{AC}$. | II. Connect $O C$ and $B C$.
Since $P C$ and $P B$ are tangents to $\odot O$, we have $\angle P O C = \angle P O B$.
Also, $\angle C O B = 2 \angle O A C$, so $\angle P O B = \angle O A C$.
Therefore, $O P \parallel A C$.
From the problem, we know $\angle A C B = \angle O B P = 90^{\circ}$, $\angle P O B = \angle O A C... | 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,602 |
Three. (25 points) Given that $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
If positive integers $a$, $b$, and $m$ satisfy the equation
$$
(a t+m)(b t+m)=31 m
$$
find the value of $a b$. | Since $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
$t$ is an irrational number, and $t^{2}=1-t$.
From the problem, we have
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=31 m \\
\Rightarrow a b(1-t)+m(a+b) t+m^{2}=31 m \\
\Rightarrow[m(a+b)-a b] t+\left(a b+m^{2}-31 m\right)=0 .
\end{array}
$$
Since $a, b,... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,603 |
One, (20 points) Given $t=\sqrt{2}-1$. If positive integers $a$, $b$, and $m$ satisfy
$$
(a t+m)(b t+m)=17 m
$$
find the value of $a b$. | Given that $t=\sqrt{2}-1$, we have
$$
t^{2}=3-2 \sqrt{2} \text {. }
$$
From the problem, we know
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=17 m \\
\Rightarrow a b(3-2 \sqrt{2})+m(a+b)(\sqrt{2}-1)+m^{2}=17 m \\
\Rightarrow \sqrt{2}[m(a+b)-2 a b]+ \\
\quad\left[3 a b-m(a+b)+m^{2}-17 m\right]=0 .
\end{array}
$$
Since... | 72 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,604 |
II. (25 points) As shown in Figure 2, in $\triangle ABC$, it is given that $AB > AC$, and $O$, $I$ are the circumcenter and incenter of $\triangle ABC$, respectively, and it satisfies $AB - AC = 2OI$.
Prove:
(1) $OI \parallel BC$;
(2) $S_{\triangle AOC} - S_{\triangle AOB} = 2S_{\triangle AOI}$. | (1) As shown in Figure 6, draw $O M \perp B C$ at point $M$, and $I N \perp B C$ at point $N$.
Let $B C=a, A C=b, A B=c$.
It is easy to see that $C M=\frac{1}{2} a, C N=\frac{1}{2}(a+b-c)$.
Therefore, $M N=C M-C N=\frac{1}{2}(c-b)=O I$.
Since $M N$ is the perpendicular segment between the two parallel lines $O M$ and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,605 |
Three. (25 points) If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-... | Three, since $a, b, c$ have cyclic symmetry, without loss of generality, assume $0 < a < b < c$, then
$$
c-a>b>0, c-b>a>0 \text {. }
$$
Thus $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1+\frac{(c-b)^{2}-a^{2}}{2 b c}>1$,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1+\frac{(c-a)^{2}-b^{2}}{2 c a}>1, \\
\frac{a^{2}+b^{2}-c... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,606 |
1. The function $f(x)$ satisfies $f(x+3)=-\frac{1}{f(x)}$ for any real number $x$.
If $f(0)=2$, then $f(2013)=(\quad)$.
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) 2
(D) 2013 | -1. A.
From the problem, we know
$$
f(x+6)=-\frac{1}{f(x+3)}=f(x) \text {. }
$$
Thus, $f(x)$ is a periodic function with a period of 6. Therefore, $f(2013)=f(3)=-\frac{1}{f(0)}=-\frac{1}{2}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,607 |
Example 6 Proof: Among any 79 consecutive numbers, there must be at least one number whose digit sum is a multiple of 13.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The instruction itself is not part of the transl... | Prove that among the first 40 numbers, at least four numbers end with 0, thus, there must exist a number whose tens digit is less than or equal to 6, let this number be $y$.
By property (3) we know
$$
y, y+1, \cdots, y+9, y+19, y+29, y+39
$$
These 13 numbers have digit sums that are exactly 13 consecutive integers, so... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 726,608 |
2. Let the arithmetic sequence $\left\{a_{n}\right\}$ and the geometric sequence $\left\{b_{n}\right\}$ satisfy $0b_{3}$, (3) $a_{6}>b_{6}$, (4) $a_{5}<b_{6}$, the number of correct statements is $($ ) .
(A) 0
(B) 1
(C) 2
(D) 3 | 2. B.
Notice,
$$
a_{3}=\frac{a_{1}+a_{5}}{2}=\frac{b_{1}+b_{5}}{2}>\sqrt{b_{1} b_{5}}=b_{3},
$$
thus, conclusion (2) is correct, and conclusion (1) is incorrect.
$$
\text { Let } a_{n}=1+20(n-1) \text {. }
$$
If we take $b_{n}=3^{n-1}$, we know $a_{6}b_{6}$, thus, conclusion (4) is incorrect. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,609 |
3. A number $x$ is randomly chosen from $[0,10]$, and a number $y$ is randomly chosen from $[0,6]$. The probability that
$$
|x-5|+|y-3| \leqslant 4
$$
is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) $\frac{3}{4}$ | 3. C.
As shown in Figure 2, the graph of $|x-5|+|y-3| \leqslant 4$ is a square $B N E M$ and its interior, centered at $G(5,3)$. Suppose $B M$ intersects $R Q$ at point $A$. It is easy to see that $A(4,6)$. Similarly, $F(6,6)$.
Thus, the required probability is
$$
p=\frac{S_{\text {hexagon } A B C D E F}}{S_{\text {re... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,610 |
4. When the coordinates $x, y$ of a point $(x, y)$ on the plane are both rational numbers, the point is called a "rational point". Let $r$ be a given positive real number. Then the number of rational points on the circle $(x-1)^{2}+(y-\sqrt{2})^{2}=r^{2}$ is ( ).
(A) at most one
(B) at most two
(C) at most four
(D) can... | 4. B.
Let $(a, b),(c, d)$ be two rational points on this circle.
$$
\begin{array}{l}
\text { Then }(a-1)^{2}+(b-\sqrt{2})^{2} \\
=(c-1)^{2}+(d-\sqrt{2})^{2} \\
\Rightarrow 2 \sqrt{2}(b-d)=a^{2}+b^{2}-c^{2}-d^{2}+2(c-a) .
\end{array}
$$
Clearly, the right side of the equation is a rational number.
Therefore, the left ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,611 |
6. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=30^{\circ}$, $A C=2$, $M$ is the midpoint of $A B$, and $\triangle A C M$ is folded along $C M$ such that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the volume $V$ of the tetrahedron $A-B C M$ is ( ).
(A) $\frac{\sqrt{2}}{3}$
(B... | 6. D.
As shown in Figure 3, draw $B D \perp C M$, and let it intersect the extension of $C M$ at point $D$. Draw $A F \perp C M$ at point $F$, and draw $E F / / B D$ such that $E F = B D$. Connect $B E$.
Clearly, $A F = F E = B D = \sqrt{3}$, and $E B = D F = 2$.
Thus, $A E^{2} = A B^{2} - E B^{2} = 8 - 4 = 4$.
The he... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,613 |
7. Given the function $f(x)=\frac{3+x}{1+x}$. Let
$$
\begin{array}{l}
f(1)+f(2)+f(4)+\cdots+f(1024)=m, \\
f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{1024}\right)=n .
\end{array}
$$
Then $m+n=$ . $\qquad$ | Ni, 7.42.
From $f(x)=1+\frac{2}{1+x}$, we know $f\left(\frac{1}{x}\right)=1+\frac{2 x}{1+x}$.
Therefore, $f(x)+f\left(\frac{1}{x}\right)=4$.
Also, $f(1)=2$, so, $m+n=4 \times 10+2=42$. | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,614 |
8. Given that $\mathrm{i}$ is the imaginary unit. If
$$
z=1+\mathrm{i}+\cdots+\mathrm{i}^{2013},
$$
denote the complex conjugate of $z$ as $\bar{z}$, then $z \cdot \bar{z}=$ $\qquad$ | 8. 2 .
Let $a_{n}=\mathrm{i}^{n}$. Then $a_{n+4}=a_{n}$, and
$$
1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}=0 \text {. }
$$
Thus $z=1+\mathrm{i} \Rightarrow \bar{z}=1-\mathrm{i}$
$$
\Rightarrow z \cdot \bar{z}=(1+\mathrm{i})(1-\mathrm{i})=1-\mathrm{i}^{2}=2 .
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,615 |
9. Given real numbers $x, y$ satisfy $x^{2}+\frac{y^{2}}{16}=1$. Then the maximum value of $x \sqrt{2+y^{2}}$ is $\qquad$ | 9. $\frac{9}{4}$.
From the problem, we know $16 x^{2}+y^{2}=16$.
Let $s=x \cdot \sqrt{2+y^{2}}$. Then
$$
\begin{array}{l}
s^{2}=x^{2}\left(2+y^{2}\right)=\frac{1}{16} \times 16 x^{2}\left(2+y^{2}\right) \\
\leqslant \frac{1}{16}\left(\frac{16 x^{2}+2+y^{2}}{2}\right)^{2}=\frac{81}{16} .
\end{array}
$$
Therefore, $s \... | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,616 |
10. For the curve $C: x^{4}+y^{2}=1$, consider the following statements:
(1) The curve $C$ is symmetric with respect to the origin;
(2) The curve $C$ is symmetric with respect to the line $y=x$;
(3) The area enclosed by the curve $C$ is less than $\pi$;
(4) The area enclosed by the curve $C$ is greater than $\pi$.
The... | 10. (1), (4).
Notice that, by substituting $-x$ for $x$ and $-y$ for $y$, the equation of curve $C$ remains unchanged, so statement (1) is correct.
By substituting $x$ for $y$ and $y$ for $x$, the equation of curve $C$ changes, so statement (2) is incorrect.
Since $1=x^{4}+y^{2}<x^{2}+y^{2}(0<|x|<1)$, the points on ... | (1), (4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,617 |
11. Let $n$ be a positive integer less than 100, and satisfies $\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n$ is an integer. Then the sum of all positive integers $n$ that meet the condition is $\qquad$ | 11. 635 .
Notice that,
$$
\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n=\frac{5 n^{2}+3 n-5}{15}
$$
is an integer, so, $15 \mid \left(5 n^{2}+3 n-5\right)$.
Thus, $5 \mid n$, and $3 \mid \left(n^{2}-1\right)$.
Therefore, $n=15 k+5$ or $15 k+10$.
Hence, the sum of all positive integers $n$ that satisfy the condition i... | 635 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,618 |
Example 7 Let $k$ be a positive integer. Then among any consecutive $2 \times 10^{k}$ positive integers, there must exist a positive integer whose sum of digits is a multiple of $k$.
| Proof: Let the first term of the continuous $2 \times 10^{k}$ positive integers be $x$,
$$
x \equiv a\left(\bmod 10^{k}\right)\left(0<a \leqslant 10^{k}\right) .
$$
Therefore, the segment of $10^{k}$ consecutive natural numbers starting from $b=x-a+10^{k}$ is all within the original $2 \times 10^{k}$ consecutive posit... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,619 |
12. Given the function $f(x)=\frac{a}{x}-x$. If for any $x \in(0,1)$, $f(x) f(1-x) \geqslant 1$ always holds, then the range of the real number $a$ is $\qquad$ . | 12. $a \in\left(-\infty,-\frac{1}{4}\right] \cup[1,+\infty)$.
Let $y=1-x$. Then $y \in(0,1)$. Therefore,
$$
\begin{array}{l}
f(x) f(1-x)=f(x) f(y) \\
=\left(\frac{a}{x}-x\right)\left(\frac{a}{y}-y\right) \\
=\frac{x^{2} y^{2}-a\left(x^{2}+y^{2}\right)+a^{2}}{x y} \\
=\frac{(x y)^{2}-a\left[(x+y)^{2}-2 x y\right]+a^{2}... | a \in\left(-\infty,-\frac{1}{4}\right] \cup[1,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,620 |
13. Given the function
$$
f(x)=\sin ^{2} \omega x+\sqrt{3} \sin \omega x \cdot \sin \left(\omega x+\frac{\pi}{2}\right)
$$
has the smallest positive period of $\frac{\pi}{2}$, where $\omega>0$. Find the maximum and minimum values of $f(x)$ on $\left[\frac{\pi}{8}, \frac{\pi}{4}\right]$. | Three, 13. From the problem, we have
$$
\begin{array}{l}
f(x)=\frac{1-\cos 2 \omega x}{2}+\frac{\sqrt{3}}{2} \sin 2 \omega x \\
=\sin \left(2 \omega x-\frac{\pi}{6}\right)+\frac{1}{2}
\end{array}
$$
Also, $T=\frac{2 \pi}{2 \omega}=\frac{\pi}{2}$, so $\omega=2$. Therefore,
$$
f(x)=\sin \left(4 x-\frac{\pi}{6}\right)+\f... | 1 \leqslant f(x) \leqslant \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,621 |
14. Given the function $f(x)=\frac{x^{3}+3 x}{3 x^{2}+1}$, the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=2, x_{n+1}=f\left(x_{n}\right)\left(n \in \mathbf{N}_{+}\right)$.
Let $b_{n}=\log _{3}\left(\frac{x_{n+1}-1}{x_{n+1}+1}\right)\left(n \in \mathbf{N}_{+}\right)$.
(1) Prove that $\left\{b_{n}\right\}$ is a geo... | 14. (1) From the problem, we have
$$
\begin{array}{l}
\frac{x_{n+1}-1}{x_{n+1}+1}=\frac{f\left(x_{n}\right)-1}{f\left(x_{n}\right)+1} \\
=\frac{x_{n}^{3}-3 x_{n}^{2}+3 x_{n}-1}{x_{n}^{3}+3 x_{n}^{2}+3 x_{n}+1}=\left(\frac{x_{n}-1}{x_{n}+1}\right)^{3} .
\end{array}
$$
Thus, $\log _{3} \frac{x_{n+1}-1}{x_{n+1}+1}=3 \log... | T_{n}=\frac{3^{n+1}(2 n-1)+3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,622 |
15. Given point $B(0,1), P, Q$ are any two points on the ellipse $\frac{x^{2}}{4}+y^{2}=1$ different from point $B$, and $B P \perp B Q$.
(1) If the projection of point $B$ on the line segment $P Q$ is $M$, find the equation of the trajectory of $M$;
(2) Find the range of the x-intercept of the perpendicular bisector $... | 15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then the equation of $P Q$ is $y=k x+m$.
Combining with the ellipse equation, we get
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-4=0 \\
\Rightarrow x_{1}+x_{2}=\frac{-8 k m}{4 k^{2}+1}, x_{1} x_{2}=\frac{4 m^{2}-4}{4 k^{2}+1} .
\... | -\frac{9}{20} \leqslant b \leqslant \frac{9}{20} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,623 |
1. For $\triangle A B C$, extend side $B C$ to point $D$, such that $C D=C B$, extend side $C A$ to point $E$, such that $A E=2 A C$. If $A D=B E$, prove: $\triangle A B C$ is a right triangle. | 1. As shown in Figure 1, extend $A C$ to point $F$, such that $A C = C F$.
Then $\triangle A C D \cong \triangle F C B$
$\Rightarrow B F = A D = B E$.
Also, $A E = 2 A C = A F$, thus $\angle B A C = 90^{\circ}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,625 |
3. Let $n$ be a positive integer. Prove:
(1) There exists a set $S$ consisting of $6 n$ positive integers, such that: the least common multiple of any two elements in $S$ is no greater than $32 n^{2}$;
(2) For any set $T$ consisting of $6 n$ positive integers, there exist two elements whose least common multiple is gre... | 3. (1) Construct the set
$$
\begin{array}{l}
S=\{1,2,3, \cdots, 4 n, 4 n+2,4 n+4, \\
\quad 4 n+6, \cdots, 8 n\}, \\
|S|=6 n .
\end{array}
$$
Below, we explain that $S$ meets the requirements.
(i) If $a, b \in\{1,2, \cdots, 4 n\}$, then
$$
[a, b] \leqslant(4 n)^{2}=16 n^{2} \text {; }
$$
(ii) If $a \in\{1,2, \cdots, 4 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,627 |
4. Find all positive integers $a, b$ such that: there exist three consecutive integers for which the polynomial $P(n)=\frac{n^{5}+a}{b}$ takes integer values.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. Let three consecutive integers be $x-1, x, x+1$ such that
$$
\begin{array}{l}
(x-1)^{5}+a \equiv 0(\bmod b), \\
x^{5}+a \equiv 0(\bmod b), \\
(x+1)^{5}+a \equiv 0(\bmod b) .
\end{array}
$$
Then $A=(x+1)^{5}-(x-1)^{5}$
$$
\begin{array}{l}
=10 x^{4}+20 x^{2}+2 \equiv 0(\bmod b), \\
B=(x+1)^{5}-x^{5} \\
=5 x^{4}+10 x^... | (a, b) = (k, 1), (11k-10, 11), (11k-1, 11) \left(k \in \mathbf{N}_{+}\right) | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 726,628 |
5. Given that $\odot O$ is the circumcircle of $\triangle A B C$, $\odot I$ is tangent to $A C$, $B C$, and internally tangent to $\odot O$ at point $P$. A line parallel to $A B$ is tangent to $\odot I$ at point $Q$ (inside $\triangle A B C$). Prove: $\angle A C P=\angle Q C B$. | 5. As shown in Figure 3, let $AC$ and $BC$ be tangent to $\odot I$ at points $E$ and $F$, respectively, and $PC$ intersect $\odot I$ at point $D$. Connect $PE$, $PQ$, and $PF$ and extend them to intersect $\odot O$ at points $K$, $M$, and $L$, respectively.
Since $\odot O$ and $\odot I$ are homothetic with respect to ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,629 |
Example 8 Suppose $N$ consecutive positive integers satisfy the following conditions: the sum of the digits of the 1st number is divisible by 1, the sum of the digits of the 2nd number is divisible by 2, $\cdots$. The sum of the digits of the $N$th number is divisible by $N$. Find the maximum possible value of $N$.
| Let the $N$ numbers be $a_{1}, a_{2}, \cdots, a_{N}$.
If $N \geqslant 22$, then among $a_{2}, a_{3}, \cdots, a_{21}$, these 20 numbers, at least two numbers have a units digit of 9, and among these two numbers, at least one has a tens digit that is not 9, let this number be $a_{i}$.
Thus, $i \leqslant 21$.
If $i$ is ev... | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,630 |
6. Snow White and the seven dwarfs live in a small house in the forest. Over 16 consecutive days, some dwarfs go mining each day, while the rest pick fruits in the forest, and no dwarf does both jobs on the same day. For any two different days, at least three dwarfs have done both jobs on these two days. It is known th... | 6. If a dwarf $X$ works the same on days $D_{1}, D_{2}, D_{3}$, then these three days are called "monotonic" for dwarf $X$.
First, prove a lemma.
Lemma: It is impossible to have three dwarfs $X_{1}, X_{2}, X_{3}$ such that days $D_{1}, D_{2}, D_{3}$ are monotonic for all of them.
Proof by contradiction.
Assume the conc... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,631 |
1. Given an acute triangle $\triangle A B C$ satisfying $\angle A<\angle B$, $\angle A<\angle C$, $P$ is a moving point on side $B C$, and $D, E$ are points on sides $A B, A C$ respectively, such that $B P=P D, C P=P E$. Prove: As point $P$ moves along side $B C$, the circumcircle of $\triangle A D E$ passes through a ... | 1. It is only necessary to prove: the fixed point is the orthocenter $H$ of $\triangle ABC$.
As shown in Figure 1, let the projections of points $C$ and $B$ on sides $AB$ and $AC$ be $X$ and $Y$, respectively, and let $M$ be the midpoint of $BC$. Then
$$
M B = M X = M Y = M C \text{. }
$$
Without loss of generality, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,632 |
2. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for each pair of real numbers $(x, y)$, we have
$$
f\left(x+y^{2}\right)=f(x)+|y f(y)| .
$$ | 2. First, prove that $f$ is non-decreasing.
Let any real numbers $r, s$ satisfy $r \geqslant s, t=\sqrt{r-s}$. Then $f(r)=f\left(s+t^{2}\right)=f(s)+|t f(t)| \geqslant f(s)$.
In the original equation, let $x=0$. Then
$$
f\left(y^{2}\right)-f(0)=|y f(y)| \text {. }
$$
For any real number $a$, define $g(a)=f(a)-f(0)$, ... | f(x)=c x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,633 |
3. Do there exist integers $a, b, c > 2010$, satisfying the equation $a^{3}+2 b^{3}+4 c^{3}=6 a b c+1 ?$ | 3. There exist integers $a, b, c$.
Obviously, $\left(a_{1}, b_{1}, c_{1}\right)=(1,1,1)$ satisfies the original equation.
For $n \geqslant 1$, define
$$
\begin{array}{l}
\left(a_{n+1}, b_{n+1}, c_{n+1}\right) \\
=\left(a_{n}+2 c_{n}+2 b_{n}, b_{n}+a_{n}+2 c_{n}, c_{n}+b_{n}+a_{n}\right) .
\end{array}
$$
Then
$$
a_{n+... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,634 |
5. Find all positive integers $n$ such that the ternary polynomial
$$
\begin{array}{l}
P_{n}(x, y, z) \\
=(x-y)^{2 n}(y-z)^{2 n}+(y-z)^{2 n}(z-x)^{2 n}+ \\
(z-x)^{2 n}(x-y)^{2 n}
\end{array}
$$
divides the ternary polynomial
$$
\begin{array}{l}
Q_{n}(x, y, z) \\
=\left[(x-y)^{2 n}+(y-z)^{2 n}+(z-x)^{2 n}\right]^{2 n}... | 5. The only positive integer that satisfies the condition is $n=1$, and it is easy to verify that $Q_{1}=4 P_{1}$.
Below is the proof: For $n \geqslant 2, P_{n} \times Q_{n}$.
Assume $P_{n} \mid Q_{n}$, then set
$$
Q_{n}(x, y, z)=R_{n}(x, y, z) P_{n}(x, y, z) .
$$
Define $p_{n}(x)=P_{n}(x, 0,-1)$,
$$
q_{n}(x)=Q_{n}(x,... | n=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,636 |
6. Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $P$, and the projections of $P$ on $AB$ and $CD$ are $E$ and $F$ respectively, the line segments $BF$ and $CE$ intersect at point $Q$. Prove:
$$
PQ \perp EF \text{. }
$$ | 6. As shown in Figure 2, let the projections of point $P$ on $EF$, $EC$, and $FB$ be $G$, $X$, and $Y$ respectively. Then $E$, $P$, $Y$, $B$ and $F$, $P$, $X$, $C$ are each sets of four concyclic points.
$$
\begin{array}{l}
\text { Hence } \angle E Y F=90^{\circ}+\angle E Y P \\
=90^{\circ}+\angle E B P=90^{\circ}+\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,637 |
7. Find all positive integers $n(n \geqslant 2)$, such that the following
conclusion holds: if $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ is a sequence of positive integers, and satisfies
$$
a_{1}+a_{2}+\cdots+a_{n}=2 n-1,
$$
then there exist some consecutive terms (at least two terms) in the sequence, whose arithmet... | 7. For all integers $n \geqslant 4$ to satisfy the condition, while $n=2,3$ do not satisfy the condition, the sequences $(1,2)$ and $(2,1,2)$ are counterexamples.
Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be any sequence of positive integers with a sum of $2 n-1$, and define
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k}-2 k ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,638 |
1. Let $f(x)$ be a decreasing function on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
f(x+2013)=2013 f(x) .
$$
Then a function that satisfies this condition is $\qquad$ | 1. $-2013^{\frac{x}{2013}}$ (the answer is not unique).
Consider the function $f(x)=-a^{x}(a>1)$.
By the problem, we have
$$
\begin{array}{l}
-a^{x+2013}=-2013 a^{x} \\
\Rightarrow a^{2013}=2013 \Rightarrow a=2013^{\frac{1}{2013}} .
\end{array}
$$
Thus, $f(x)=-2013^{\frac{x}{2013}}$ meets the condition.
Moreover, for... | -2013^{\frac{x}{2013}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,640 |
Example 9 Proof: Among any 36 consecutive numbers less than 10000, there must be a number that is divisible by the sum of its digits (i.e., a Niven number). | Prove that in any 36 consecutive numbers, there must be two that are multiples of 18. Therefore, if one of them has a digit sum of 9 or 18, by property (1), it is a Niven number.
On the other hand, the only number with a digit sum of 36 is 9999, which is not a multiple of 18. Thus, by combining the property, we only n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,641 |
2. Let the volume of the regular triangular prism $A B C-A^{\prime} B^{\prime} C^{\prime}$ be $V$, and points $P, Q$ lie on edges $A A^{\prime}, C C^{\prime}$ respectively, such that $A P = C^{\prime} Q$. Then the volume of the tetrahedron $B P Q B^{\prime}$ is $\qquad$ | $\begin{array}{l}\text { 2. } \frac{1}{3} V . \\ V_{\text {tetrahedron } B P Q B^{\prime}}=V_{\text {triangular prism } P-B B^{\prime} Q}=V_{\text {triangular prism } A-B B^{\prime} Q} \\ =V_{\text {triangular prism } A-B B^{\prime} C}=\frac{1}{3} V .\end{array}$ | \frac{1}{3} V | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,642 |
3. Let $n$ be a given positive integer, and the set
$$
M=\left\{\frac{1}{2^{n}}, \frac{1}{2^{n+1}}, \cdots, \frac{1}{2^{2 n}}\right\} \text {. }
$$
Let the subsets of $M$ be denoted as $M_{1}, M_{2}, \cdots, M_{i}$. For $1 \leqslant i \leqslant t$, let $S\left(M_{i}\right)$ represent the sum of all elements in the set... | $3.2-\frac{1}{2^{n}}$.
For any element $a$ in set $M$, since there are $2^{n}$ subsets containing $a$, each element appears $2^{n}$ times in the "sum".
$$
\begin{array}{l}
\text { Therefore, } S\left(M_{1}\right)+S\left(M_{2}\right)+\cdots+S\left(M_{t}\right) \\
=2^{n} S(M)=2^{n}\left(\frac{1}{2^{n-1}}-\frac{1}{2^{2 n}... | 2-\frac{1}{2^{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,643 |
4. Let $a, b \in \mathbf{R}$. Then the minimum value of $M=\frac{\left(a^{2}+a b+b^{2}\right)^{3}}{a^{2} b^{2}(a+b)^{2}}$ is $\qquad$ . | 4. $\frac{27}{4}$.
Let $a=b=1$, we get
$$
M=\frac{\left(a^{2}+a b+b^{2}\right)^{3}}{a^{2} b^{2}(a+b)^{2}}=\frac{27}{4} \text {. }
$$
Now we prove: $M \geqslant \frac{27}{4}$.
Let $x=a^{2}+b^{2}, y=a b$. Then $x \geqslant 0$, and $x \geqslant 2 y$. Thus, the inequality becomes
$$
4(x+y)^{3} \geqslant 27 y^{2}(x+2 y) .... | \frac{27}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,644 |
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