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2. Consider a tangent line to the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$, which intersects the two symmetry axes of the ellipse at points $A$ and $B$. Then the minimum length of segment $AB$ is $\qquad$ . | 2. 8 .
Let the point of tangency be \( P(5 \cos \theta, 3 \sin \theta) \). Then the equation of the tangent line to the ellipse at point \( P \) is
$$
\frac{\cos \theta}{5} x + \frac{\sin \theta}{3} y = 1,
$$
which intersects the \( x \)-axis and \( y \)-axis at
$$
A\left(\frac{5}{\cos \theta}, 0\right), B\left(0, \f... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,896 |
3. In rectangle $A B C D$, it is known that $A B=2, B C=3$, $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Rotate $\triangle F A B$ $90^{\circ}$ around $E F$ to $\triangle F A^{\prime} B^{\prime}$. Then the volume of the tetrahedron $A^{\prime} B^{\prime} C D$ is $\qquad$ . | 3. 2 .
It is known that $E F=B C=3$, and the plane of $\triangle F A^{\prime} B^{\prime}$ divides the tetrahedron $A^{\prime} B^{\prime} C D$ into two tetrahedrons of equal volume, $C F A^{\prime} B^{\prime}$ and $D F A^{\prime} B^{\prime}$. Their heights are $C F=D F=1$, and $S_{\triangle F A^{\prime} B^{\prime}}=3$.... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,897 |
4. $\sin 7.5^{\circ}+\cos 7.5^{\circ}=$ | 4. $\frac{\sqrt{4+\sqrt{6}-\sqrt{2}}}{2}$.
Notice,
$$
\begin{array}{l}
\sin 15^{\circ}=\sqrt{\frac{1-\cos 30^{\circ}}{2}} \\
=\sqrt{\frac{8-4 \sqrt{3}}{16}}=\frac{\sqrt{6}-\sqrt{2}}{4} .
\end{array}
$$
Then $\left(\sin 7.5^{\circ}+\cos 7.5^{\circ}\right)^{2}$
$$
\begin{array}{l}
=1+2 \sin 7.5^{\circ} \cdot \cos 7.5^{... | \frac{\sqrt{4+\sqrt{6}-\sqrt{2}}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,898 |
5. Use the digits $1,2, \cdots, 7$ to form a seven-digit number such that it is a multiple of 11. The number of seven-digit numbers that can be formed is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. 576 .
Let $n$ be a seven-digit number satisfying the condition, and let $a$ and $b$ be the sums of the digits in the odd and even positions, respectively. Then $a+b=28$, and $a-b$ is a multiple of 11.
Since $a+b$ and $a-b$ have the same parity, they must both be even. Clearly, $|a-b| \neq 22$, so only $a-b=0$.
Thus... | 576 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,899 |
Example 5 Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$
All real roots. ${ }^{[4]}$
(2011, International Invitational Competition for Young Mathematicians in Cities) | 【Analysis】The most basic method to solve higher-degree equations is to convert them into lower-degree equations for solving, that is, to handle them as linear or quadratic equations. Factorization is the most powerful tool to achieve such a transformation. The preferred method for factoring higher-degree polynomials is... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,900 |
6. If $7n+1$ and $8n+1$ can both be expressed as the sum of three distinct positive integers in a geometric progression, then the smallest positive integer $n$ is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.6.
Notice that when $n=6$,
$$
\begin{array}{l}
7 n+1=43=1+6+6^{2}, \\
8 n+1=49=3^{2}+3 \times 5+5^{2}
\end{array}
$$
are both the sums of three positive integers in geometric progression.
When $n \leqslant 5$, it is easy to see that,
$$
\begin{array}{l}
7 \times 1+1=8, \\
8 \times 2+1=17, \\
8 \times 3+1=25, \\
7 \... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,901 |
7. For any $x, y \in [0,1]$, the function
$$
f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}
$$
has a maximum value of $\qquad$ . | 7. 1.
Since $x, y \in [0,1]$, then $x \leqslant \sqrt{x}, y \leqslant \sqrt{y}$.
Let $x=\sin ^{2} \alpha, y=\sin ^{2} \beta\left(\alpha, \beta \in\left[0, \frac{\pi}{2}\right]\right)$.
Thus, $f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}$
$$
\begin{array}{l}
\leqslant \sqrt{x(1-y)}+\sqrt{y(1-x)} \\
=\sin \alpha \cdot \cos \beta+\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,902 |
8. Let $A$ be a 20-element subset of the set
$$
M=\{1,2, \cdots, 2012\}
$$
such that the difference between any two elements in $A$ is a multiple of 12. Then the number of such subsets $A$ is $\qquad$ | 8. $8 C_{168}^{20}+4 C_{167}^{20}$.
For $x, y \in M$, if $12 \mid (x-y)$, then $x, y$ are called "congruent".
Thus, when $n \in \{1, 2, \cdots, 8\}$, $n$ has 168 congruent numbers. For each such $n$, the 168-element set
$$
T_{n}=\{n+12 k \mid k=0,1, \cdots, 167\}
$$
has any 20-element subset that meets the condition... | 8 C_{168}^{20}+4 C_{167}^{20} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,903 |
Ni.(16 points) The vertex of the parabola is $O$ and the focus is $F$. When the moving point $P$ moves on the parabola, find the maximum value of the distance ratio $\left|\frac{P O}{P F}\right|$. | Let the equation of the parabola be $y^{2}=4 a x(a>0)$. The vertex is $O(0,0)$, and the focus is $F(a, 0)$.
If the coordinates of a moving point on the parabola are $P(x, y)$, then
$$
\begin{array}{l}
\left(\frac{P O}{P F}\right)^{2}=\frac{x^{2}+y^{2}}{(x-a)^{2}+y^{2}} \\
=\frac{x^{2}+4 a x}{(x-a)^{2}+4 a x}=\frac{x^{2... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,904 |
Three. (20 points) Given $\triangle A B C$ is inscribed in $\odot O, I$ is its incenter: lines $A I, B I$ intersect $\odot O$ at points $D, E$ respectively, a line $l_{1} / / A B$ is drawn through point $I$, and a tangent line $l_{c}$ to $\odot O$ is drawn through point $C$. If $l_{c}$ intersects $l_{1}$ at point $F$, ... | Three, as shown in Figure 1, let the line $l_{1}$ intersect $DE$ at point $F_{1}$, and connect $F_{1}C$. By a well-known theorem, we have
$$
\begin{array}{l}
DC = DI, \\
EC = EI.
\end{array}
$$
Thus, $\triangle IDE$ and $\triangle CDE$ are symmetric with respect to the line $DE$, meaning that $DE$ is the perpendicular... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,905 |
Four, (20 points) Question: In how many different ways can the elements of the set $M=\{1,2,3,4,5\}$ be assigned to three (ordered) sets $A$, $B$, and $C$, such that each element is contained in at least one of the sets, the intersection of these three sets is empty, and the intersection of any two of these sets is not... | As shown in Figure 2, consider the seven parts divided by the Venn diagram, represented by $x, u, v, w, a, b, c$ respectively.
Now, fill the elements of $M$ into these parts. According to the problem, $x$ cannot be filled with any number, while $u, v, w$ must be filled with numbers, and the numbers filled in these par... | 1230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,906 |
1. Use the digits $1,2,3$ to form a four-digit number, requiring all three digits to appear, and the same digit should not be adjacent. How many such four-digit numbers are there?
(A) 24
(B) 18
(C) 15
(D) 12 | -、1. B.
From the problem, we know that one of the digits $1, 2, 3$ must be used repeatedly, which gives us three scenarios; placing the two identical numbers in non-adjacent positions in the four-digit number also has three scenarios; placing the remaining two numbers in the remaining two positions of the four-digit nu... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,907 |
2. Let $A_{k}=\left\{x \left\lvert\, x=k t+\frac{1}{k t}\right., \frac{1}{k^{2}} \leqslant t \leqslant 1\right\}$, where $k=2,3, \cdots, 2012$. Then the intersection of all $A_{k}$ is ( ).
(A) $\varnothing$
(B) $\{2\}$
(C) $\left[2, \frac{5}{2}\right]$
(D) $\left[2, \frac{2012^{2}+1}{2012}\right]$ | 2. C.
It is easy to see that $A_{k}=\left[2, k+\frac{1}{k}\right]$, and $k+\frac{1}{k}$ is increasing on $[2,+\infty)$. Therefore, the intersection of all $A_{k}$ is $A_{2}=\left[2, \frac{5}{2}\right]$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,908 |
3. $f(x)$ is a function defined on $(0,1)$, for any $1<x<y<+\infty$, we have
$$
f\left(\frac{1}{x}\right)-f\left(\frac{1}{y}\right)=f\left(\frac{x-y}{1-x y}\right) .
$$
Let $a_{n}=f\left(\frac{1}{n^{2}+5 n+5}\right)\left(n \in \mathbf{N}_{+}\right)$. Then $a_{1}+a_{2}+\cdots+a_{8}=(\quad)$.
(A) $f\left(\frac{1}{2}\rig... | 3. C.
Notice,
$$
\begin{array}{l}
a_{n}=f\left(\frac{1}{n^{2}+5 n+5}\right) \\
=f\left(\frac{(n+2)-(n+3)}{1-(n+2)(n+3)}\right) \\
=f\left(\frac{1}{n+2}\right)-f\left(\frac{1}{n+3}\right) .
\end{array}
$$
Then $a_{1}+a_{2}+\cdots+a_{8}$
$$
\begin{array}{l}
=f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)+f\left(\f... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,909 |
4. Let the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have a focus at $F$. Draw a line $l$ through $F$ perpendicular to the $x$-axis, intersecting the two asymptotes at points $A$ and $B$. Let $P$ be one of the points where $l$ intersects the hyperbola. Let $O$ be the origin. If there exist real nu... | 4. A.
Substitute the coordinates of points $A\left(c, \frac{b c}{a}\right), B\left(c,-\frac{b c}{a}\right)$ into
$$
\overrightarrow{O P}=m \overrightarrow{O A}+n \overrightarrow{O B} \text {, }
$$
to get point $P\left((m+n) c,(m-n) \frac{b c}{a}\right)$.
Substitute the coordinates of point $P$ into the hyperbola equa... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,910 |
Example 6 Solve the system of equations
$$
\left\{\begin{array}{l}
x-5 y+18 \sqrt{2 y}=20, \\
6 \sqrt{2 x}-x-5 y=11 .
\end{array}\right.
$$ | 【Analysis】The basic method for handling irrational equations (systems) is rationalization. First, add and subtract the two equations, then multiply the resulting two equations, and use the difference of squares formula to achieve rationalization. After factoring by grouping, the solution can be completed.
Solving by (2... | \left(\frac{9}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,911 |
5. In $\triangle A B C$, let $B C=a, A C=b, A B=c$.
Then the equation
$$
\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2}
$$
holds if and only if ( ).
(A) $c a=b^{2}$
(B) $a+b=2 c$
(C) $b+c=2 a$
(D) $c+a=2 b$ | 5. D.
$$
\begin{array}{l}
\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2} \\
\Leftrightarrow \frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\frac{1-\cos C}{2} \\
=\frac{1+\cos B}{2} \\
\Leftrightarrow \cos A+\cos C=2-2 \cos B \\
\Leftrightarrow 2 \cos \frac{A+C}{2} \cdot \cos \frac{A-C}{2... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,912 |
6. Let $S=\{(x, y)|-2 \leqslant y \leqslant| x \mid,-2 \leqslant x \leqslant 2\}$.
Then when $(x, y) \in S$, and such that the quadratic equation
$$
t^{2}+(|x|-1) t+|y|-2=0
$$
has one root greater than $1$ and one root less than $1$, the probability is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) 1 | 6. A.
As shown in Figure 2, S is a square
$$
\begin{array}{l}
-2 \leqslant x \leqslant 2, \\
-2 \leqslant y \leqslant 2
\end{array}
$$
The area below the graph of $y=|x|$, is
$$
A_{s}=16-4=12 \text {. }
$$
The equation
$$
t^{2}+(|x|-1) t+|y|-2=0
$$
has one root greater than 1 and one root less than 1
$$
\begin{arra... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,913 |
8. $\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+\cdots+2012}\right)$ $=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8. $\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right... | 8. $\frac{1007}{3018}$.
Given $1-\frac{1}{1+2+\cdots+n}=\frac{(n-1)(n+2)}{n(n+1)}$, then
$$
\begin{array}{l}
\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+\cdots+2012}\right) \\
=\frac{1 \times 4}{2 \times 3} \cdot \frac{2 \times 5}{3 \times 4} \cdots \cdots \frac{2011 \times 20... | \frac{1007}{3018} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,915 |
9. Let the function
$$
\begin{array}{l}
f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}, \\
S_{n}=\sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)\left(n \geqslant 2, n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $S_{n}=$ | 9. $\frac{n-1}{2}$.
When $x_{1}+x_{2}=1$,
$$
f\left(x_{1}\right)+f\left(x_{2}\right)=1+\log _{2} \frac{x_{1} x_{2}}{\left(1-x_{1}\right)\left(1-x_{2}\right)}=1 \text {. }
$$
Then $2 S_{n}=\sum_{i=1}^{n-1}\left[f\left(\frac{i}{n}\right)+f\left(\frac{n-i}{n}\right)\right]=n-1$
$$
\Rightarrow S_{n}=\frac{n-1}{2} \text {... | \frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,916 |
10. The solution set of the inequality
$$
\frac{8}{(x+1)^{3}}+\frac{10}{x+1}-x^{3}-5 x>0
$$
is $\qquad$ . | 10. $(-\infty,-2) \cup(-1,1)$.
The original inequality is transformed into
$$
\left(\frac{2}{x+1}\right)^{3}+5 \cdot \frac{2}{x+1}>x^{3}+5 x \text {. }
$$
Let $f(t)=t^{3}+5 t$. It is easy to see that $f(t)$ is monotonically increasing on $\mathbf{R}$, and when $x \neq-1$, the original inequality becomes
$$
f\left(\fr... | (-\infty,-2) \cup(-1,1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,917 |
11. Given that point $P$ is on the curve $y=\mathrm{e}^{x}$, and point $Q$ is on the curve $y=\ln x$. Then the minimum value of $|P Q|$ is $\qquad$ . | 11. $\sqrt{2}$.
Notice that the curve $y=\mathrm{e}^{x}$ and $y=\ln x$ are symmetric about the line $y=x$.
Therefore, the minimum value of $|P Q|$ is twice the minimum distance from a point on the curve $y=\mathrm{e}^{x}$ to the line $y=x$.
Let $P\left(x, \mathrm{e}^{x}\right)$ be any point on the curve $y=\mathrm{e... | \sqrt{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 725,918 |
12. In the tetrahedron $ABCD$, it is known that $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$. The radius of the sphere that passes through $D$ and is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron is 1, then the radius of the circumscribed... | 12.3.
As shown in Figure 3, draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$ and $DF \perp AC$, with the feet of the perpendiculars being $E$ and $F$ respectively.
Then $HE \perp AB$, $HF \perp AC$, and
$AE = AF = AD \cos 45^{\circ} = \sqrt{6}$.
From ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,919 |
14. Given an ellipse centered at the origin $O$, with foci on the $x$-axis, and an eccentricity of $\frac{\sqrt{3}}{2}$, the ellipse passes through the point $\left(\sqrt{2}, \frac{\sqrt{2}}{2}\right)$. Let a line $l$ that does not pass through the origin $O$ intersect the ellipse at points $P$ and $Q$, and the slopes ... | 14. From the problem, we can set the equation of the ellipse as
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
Notice,
$$
\left\{\begin{array} { l }
{ \frac { c } { a } = \frac { \sqrt { 3 } } { 2 } , } \\
{ \frac { 2 } { a ^ { 2 } } + \frac { 1 } { 2 b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left... | (0,1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,921 |
Example 7 Given that $a, b, c$ are positive numbers, satisfying
$$
\begin{array}{l}
a+b+c=32, \\
\frac{b+c-a}{b c}+\frac{c+a-b}{c a}+\frac{a+b-c}{a b}=\frac{1}{4} .
\end{array}
$$
Prove: A right-angled triangle can be formed with $\sqrt{a}, \sqrt{b}, \sqrt{c}$ as the side lengths. ${ }^{(s)}$
(2009, National Junior Hi... | Proof Note that,
$$
\begin{array}{c}
\left(\frac{b+c-a}{b c}+\frac{c+a-b}{c a}+\frac{a+b-c}{a b}\right)(a+b+c)=8 \\
\Rightarrow \frac{(b+c)^{2}-a^{2}}{b c}+\frac{(c+a)^{2}-b^{2}}{c a}+ \\
\frac{(a+b)^{2}-c^{2}}{a b}=8 \\
\Rightarrow \frac{(b+c)^{2}-a^{2}}{b c}-4+\frac{(c+a)-b^{2}}{c a}-4+ \\
\frac{(a+b)^{2}-c^{2}}{a b}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,922 |
15. As shown in Figure 1, given that $PA$ and $PB$ are two tangents drawn from a point $P$ outside circle $\odot O$, $M$ and $N$ are the midpoints of segments $AP$ and $AB$ respectively, line $MN$ intersects $\odot O$ at points $C$ and $E$, point $N$ is between $M$ and $C$, $PC$ intersects $\odot O$ at point $D$, and t... | 15. As shown in Figure 4, connect $O P, O A, O C, E P$. Clearly, $O, P, N=$ points are collinear, and $O P \perp A B$.
Since $M, N$ are the midpoints of $P A, A B$ respectively, we have
$$
\begin{array}{l}
M N=M P=M A, M N \parallel P B \\
\Rightarrow P M^{2}=A M^{2}: M E \cdot M C \\
\Rightarrow \triangle M P E \sim \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,923 |
16. Let the increasing sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1,4 a_{n+1}=5 a_{n}+\sqrt{9 a_{n}^{2}+16}(n \geqslant 1) \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<2$. | 16. (1) Solution 1 From the problem, we get
$$
a_{2}=\frac{5}{2}, a_{3}=\frac{21}{4}, a_{4}=\frac{85}{8}.
$$
From $5=1+4, 21=1+4+4^{2}$,
$$
85=1+4+4^{2}+4^{3},
$$
we conjecture: $a_{n}=\frac{1+4+4^{2}+\cdots+4^{n-1}}{2^{n-1}}$
$$
=\frac{4^{n}-1}{3 \times 2^{n-1}}=\frac{2}{3}\left(2^{n}-\frac{1}{2^{n}}\right).
$$
In ... | S_{n}<2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,924 |
1. A communication network consists of several terminals. If among any three terminals, at least two are directly connected, then this communication network is called "triple-connected." A communication network that satisfies the following conditions is referred to as a "windmill with $n$ blades."
$n$ pairs of terminal... | 1. Let each vertex in graph $G$ represent a terminal. If two terminals are directly connected, a red line is drawn between the corresponding vertices; otherwise, a blue line is drawn.
From the problem, we know that there are no blue triangles in graph $G$.
Let $|G|=m$.
(1) $n=1$.
A windmill with one blade is a red tria... | f(n)=\left\{\begin{array}{ll}6, & n=1 ; \\ 4 n+1, & n \geq 2 .\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,925 |
2. Given an acute triangle $\triangle A B C$ with altitudes $A D, B E$, and $C F$ intersecting at point $H$, and circle $\odot O$ passing through points $A, H$, and intersecting sides $A B, A C$ at points $Q, P$ different from $A$. If the circumcircle of $\triangle O P Q$ is tangent to side $B C$ at point $R$, prove: $... | 2. As shown in Figure 1, let the circumcircle of $\triangle B Q H$ intersect line $B C$ at point $R_{1}$ (not coinciding with point $B$).
From the fact that $A, P, H, Q$ and $B, Q, H, R_{1}$ are concyclic, we have
$$
\begin{array}{l}
\angle P H Q=180^{\circ}-\angle P A Q, \\
\angle Q H R_{1}=180^{\circ}-\angle Q B R_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,926 |
3. Find the minimum value of the real number $k$ such that for any real numbers $x, y, z$ not all positive, the inequality
$$
\begin{array}{l}
k\left(x^{2}-x+1\right)\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \\
\geqslant(x y z)^{2}-x y z+1
\end{array}
$$
always holds. | 3. First, prove a lemma.
Lemma When $s, t$ are at least one not greater than 0, we have
$$
\frac{4}{3}\left(s^{2}-s+1\right)\left(t^{2}-t+1\right) \geqslant(s t)^{2}-s t+1 .
$$
Proof Equation (1) is equivalent to
$$
\begin{array}{l}
\left(s^{2} t^{2}-4 s^{2} t+4 s^{2}\right)-\left(4 s t^{2}-7 s t+4 s\right)+ \\
\left... | \frac{16}{9} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,927 |
4. For a given positive integer $n$. Try to find the smallest positive integer $d_{n}$ that cannot be expressed in the form $\sum_{i=1}^{n}(-1)^{a_{i}} \times 2^{b_{i}}$, where $a_{i} 、 b_{i}(i=1,2, \cdots, n)$ are all non-negative integers. | 4. For any positive integer $p$, let $t(p)$ denote the smallest value of the positive integer $n$ when $p$ can be expressed in the form given in the problem, and call this representation the "minimal representation" of $p$.
(1) If $p$ is even.
Then in any representation that yields $p$, the number of 0s in $b_{i}(i=1,2... | \frac{2^{2 n+1}+1}{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,928 |
5. Let $P$ be a convex $n\left(n \in \mathbf{N}_{+}, n>7\right)$-sided polygon. It is known that non-intersecting diagonals in $P$ divide $P$ into $n-2$ triangles. If one of these triangles has all three sides as diagonals of $P$, then it is called an "internal triangle". Try to find the number of ways to divide $P$ su... | 5. Label the vertices of $P$ in counterclockwise order as $A_{0}, A_{1}, \cdots, A_{n-1}$.
First, consider the number of partition methods in which two inner triangles are located as shown in Figure 3. Such a partition is said to start from point $A_{0}$.
In Figure 3, the numbers $m_{1}, m_{2}$ represent the number of... | 2^{n-9} n \mathrm{C}_{n-4}^{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,929 |
6. Construct $\triangle P A B$ and $\triangle Q A C$ externally on $\triangle A B C$ such that $A P=A B, A Q=A C, \angle B A P=\angle C A Q$. The line segments $B Q$ and $C P$ intersect at point $R$, and $O$ is the circumcenter of $\triangle B C R$. Prove: $A O \perp P Q$. | 6. Proof 1 It is easy to know, $\triangle A P C \cong \triangle A B Q$.
Then $\angle A P R=\angle A P C=\angle A B Q=\angle A B R$.
Thus, $A, P, B, R$ are concyclic.
Similarly, $A, Q, C, R$ are concyclic.
Let $\angle P A B=2 \alpha$.
As shown in Figure 4, let the midpoint of the arc $\overparen{B C}$ of the circumcirc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,930 |
1. If $\frac{20122012 \cdots 201215}{n \uparrow}$ is divisible by 15, then the minimum value of $n$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | -1.A.
Obviously, 5। $\underset{n \uparrow}{20122012 \cdots 201215}$.
And the sum of the digits of the permutation of $n$ 2012s is $5n$, so the minimum value of $n$ is 3. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,931 |
2. Given that $x$, $y$, $z$ are all non-negative numbers, and satisfy
$$
y+z-1=4-y-2 z=x \text{. }
$$
If $w=2 x^{2}-2 y+z$, then the minimum value of $w$ is ).
(A) -1
(B) $\frac{23}{9}$
(C) $-\frac{1}{2}$
(D) 0 | 2. C.
From the given, we have $y=3 x-2 \geqslant 0, z=-2 x+3 \geqslant 0$. Thus, $\frac{2}{3} \leqslant x \leqslant \frac{3}{2}$.
And $w=2 x^{2}-2 y+z$ $=2 x^{2}-8 x+7=2(x-2)^{2}-1$.
Therefore, when $x=\frac{3}{2}$, $w_{\min }=-\frac{1}{2}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,932 |
Example 8 Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition. ${ }^{(6)}$
(2010, National Junior High School Mathematics Competiti... | 【Analysis】In a right-angled triangle, the three sides satisfy the Pythagorean theorem. Given the condition $c=\frac{1}{3} a b-(a+b)$, one unknown can be eliminated to obtain a quadratic equation in two variables, which is generally difficult to solve. However, the problem of integer solutions to a quadratic equation in... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,933 |
3. In $\triangle A B C$, let $\angle C=90^{\circ}, \angle A=$ 22. $5^{\circ}, A B=4$. Then the area of $\triangle A B C$ is ( ).
(A) 2
(B) 3
(C) $2 \sqrt{2}$
(D) $2 \sqrt{3}$ | 3. C.
As shown in Figure 4, draw the median $C M$ on the hypotenuse $A B$. Then
$$
\begin{array}{l}
C M=B M=2, \\
\angle B M C=2 \angle A \\
=45^{\circ} .
\end{array}
$$
Draw $C H \perp A B$ at point $H$. Then $C H=\sqrt{2}$. Therefore, $S_{\triangle A B C}=\frac{1}{2} C H \cdot A B^{\prime}=2 \sqrt{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,934 |
4. In $\triangle A B C$, let the largest angle $\angle A$ be twice the smallest angle $\angle C$, and $A B=2, A C=3$. Then the perimeter of $\triangle A B C$ is ( ).
(A) $9-\sqrt{13}$
(B) $5 \sqrt{3}-\sqrt{10}$
(C) $4 \sqrt{2}$
(D) $5+\sqrt{10}$ | 4. D.
As shown in Figure 5, extend $CA$ to point $D$ such that $AD = AB$, and connect $BD$. Then
$$
\begin{array}{l}
\angle D = \angle ABD \\
= \frac{1}{2} \angle CAB \\
= \angle C.
\end{array}
$$
Therefore, $\triangle CBD \sim \triangle DAB$
$$
\Rightarrow \frac{BD}{AB} = \frac{CD}{BD} \Rightarrow BC = BD = \sqrt{10... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,935 |
5. If the equation ||$y^{2}-1|-2|=m$ has exactly five distinct real roots, then $m=(\quad$ ).
(A) 0
(B) 1
(C) 2
(D) greater than 2 | 5. B.
When $m=1$, the original equation is $\left|y^{2}-1\right|=3$ or 1.
Therefore, there are five different real roots $\pm 2, \pm \sqrt{2}, 0$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,936 |
6. As shown in Figure 1, in $\triangle A B C$, $A B=A C, C M$
bisects $\angle A C B$, intersecting $A B$
at point $M, A D$
$\perp B C$ at point $D$,
$M E \perp B C$ at point
$E, M F \perp M C$ intersects
$B C$ at point $F$.
Then the value of $C F-4 D E$ is ( ).
(A) positive number
(B) 0
(C) negative numb... | 6. B.
As shown in Figure 6, extend $C A$ and $F M$ to intersect at point $P$. Then $C F = C P$.
Draw $M N \parallel B C$ intersecting $A D$ and $A C$ at points $H$ and $N$ respectively. Then
$$
\begin{array}{l}
M N = \frac{1}{2} C F, D E = M H . \\
\text { Since } A B = A C, A D \perp B C \\
\Rightarrow D E = M H = H... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,937 |
1. Let positive numbers $x, y, z$ satisfy
$$
\frac{1}{x^{3}}=\frac{8}{y^{3}}=\frac{27}{z^{3}}=\frac{k}{(x+y+z)^{3}} \text {. }
$$
Then $k=$ $\qquad$ | $$
\begin{array}{l}
\sqrt[3]{k}=\frac{x+y+z}{x}=\frac{2(x+y+z)}{y} \\
=\frac{3(x+y+z)}{z}=\frac{6(x+y+z)}{x+y+z}=6 .
\end{array}
$$
Therefore, $k=216$. | 216 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,938 |
2. If the equation with respect to $x$
$$
x^{2}+2(m+3) x+m^{2}+3=0
$$
has two real roots $x_{1}$ and $x_{2}$, then the minimum value of $\left|x_{1}-1\right|+\left|x_{2}-1\right|$ is $\qquad$. | 2.6 .
According to the problem, we have
$$
\begin{array}{l}
\Delta=[2(m+3)]^{2}-4\left(m^{2}+3\right) \geqslant 0 \\
\Rightarrow m \geqslant-1 .
\end{array}
$$
Then $x_{1}+x_{2}=-2(m+3)<0$.
When $x=1$, the left side of the equation is greater than 0, thus, $x_{1}$ and $x_{2}$ are on the same side of 1.
$$
\begin{arra... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,939 |
3. Let the line $y=-\frac{1}{2} x+1$ intersect the $x$-axis and $y$-axis at points $B$ and $A$, respectively. Point $C$ is the reflection of point $B$ over the $y$-axis. Construct an isosceles right triangle $\triangle ACD$ with $AC$ as one of the legs in the second quadrant. Draw $DE \perp x$-axis at point $E$. If the... | 3. $-\frac{3}{14}$.
As shown in Figure 7, from the given conditions we have
$$
A(0,1), B(2,0), C(-2,0) \text {. }
$$
It is easy to prove that Rt $\triangle D E C \cong$ Rt $\triangle C O A$.
Thus, $D E=C O=2, E C=O A=1, O E=3$.
It is evident that the line $y=k x-2 k$ passes through point $B$, and this line intersects... | -\frac{3}{14} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,940 |
4. Let $x_{n}$ denote the unit digit of the number $n^{4}$. Then
$$
x_{1}+x_{2}+\cdots+x_{2012}=
$$
$\qquad$ | 4.6640 .
Notice that, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, and the unit digits of $1^{4}, 2^{4}, \cdots, 10^{4}$ are $1,6,1,6,5,6,1,6,1,0$ respectively.
Thus, $x_{1}+x_{2}+\cdots+x_{10}=33$.
Therefore, $x_{1}+x_{2}+\cdots+x_{2012}$
$$
=201 \times 33+(1+6)=6640 \text {. }
$$ | 6640 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,941 |
One. (20 points) As shown in Figure 2, in the isosceles right triangle $\triangle ABC$, $\angle C=90^{\circ}$, points $D$ and $E$ are on side $BC$, and point $F$ is on the extension of $AC$, such that $BE=ED=CF$. Find the tangent value of $\angle CEF + \angle CAD$.
---
The translation preserves the original text's li... | $\begin{array}{l}\text { I. Draw } D G \perp A B \text { at point } G . \\ \text { Let } A C=B C=x, \\ B E=E D=C F=y\left(0<y<\frac{x}{2}\right) . \\ \text { Then } G D=\sqrt{2} y \Rightarrow A G=\sqrt{2}(x-y)=\sqrt{2} C E \text {. } \\ \text { Therefore, } \mathrm{Rt} \triangle E C F \backsim \mathrm{Rt} \triangle A G... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,942 |
II. (25 points) As shown in Figure 3, circles $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $O$. Line $AB$ is tangent to $\odot O_{1}$ and $\odot O_{2}$ at points $B$ and $A$, respectively, and intersects the $x$-axis and $y$-axis at points $M(2 \sqrt{3}, 0)$ and $C(0,2)$.
(1) Find the radius of $\odo... | (1) It is easy to know that $l_{A B}: y=-\frac{\sqrt{3}}{3} x+2$.
Then $\angle A M O_{2}=30^{\circ}$.
It is easy to know that $O$ is the midpoint of $M \mathrm{O}_{2}$.
Therefore, the radius of $\odot \mathrm{O}_{2}$ is $R=2 \sqrt{3}$.
(2) It is easy to see that $\triangle M O B$ is an isosceles triangle, and
$\angle B... | P(0,2) \text{ or } P(-4 \sqrt{3}, 6) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,943 |
Example 1 Positive numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{100}=3, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}>1 .
\end{array}
$$
Prove: $a_{1}, a_{2}, \cdots, a_{100}$ must contain three numbers whose sum is greater than 1. | Proof By symmetry, without loss of generality, assume
$a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{100}$.
It suffices to prove $a_{1}+a_{2}+a_{3}>1$.
Proof by contradiction.
Suppose $a_{1}+a_{2}+a_{3} \leqslant 1$, as shown in Figure 1, construct a square with side length 1, and form a $1 \times 3$ rectangle. T... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,944 |
Three, (25 points) Given that $m, n, p, q$ satisfy $m n p q = 6(m-1)(n-1)(p-1)(q-1)$.
(1) If $m, n, p, q$ are all positive integers, find the values of $m, n, p, q$;
(2) If $m, n, p, q$ are all greater than 1, find the minimum value of $m+n+p+q$. | (1) Let's assume $m \geqslant n \geqslant p \geqslant q$.
Clearly, $q \geqslant 2$.
If $q \geqslant 3$, then
$$
\begin{aligned}
\frac{1}{m} & \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{q} \leqslant \frac{1}{3} \\
\Rightarrow & \frac{1}{6}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\... | (9,4,2,2),(6,5,2,2),(4,3,3,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,945 |
1. In the Cartesian coordinate system, there are eight points:
$$
A_{i}(i, 0), B_{i}(0, i)(i=1,2,3,4),
$$
Divide them into two groups, each containing four points, to form convex quadrilaterals $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Then
the probability is $\qquad$ | $-1 \cdot \frac{5}{12}$.
When and only when two points are taken from each of $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ and $\left\{B_{1}, B_{2}, B_{3}, B_{4}\right\}$ to form a pair, two convex quadrilaterals are formed. There are $\left(\frac{4 \times 3}{2}\right)^{2}=36$ ways to do this.
Consider the case where $... | \frac{5}{12} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,946 |
2. Let $f(x)$ be a continuous even function, and for $x>0$, $f(x)$ is a strictly increasing function. Then the range of $x$ that satisfies
$$
f(x)<f\left(\frac{x+3}{x+4}\right)
$$
is | 2. $\left(\frac{-5-\sqrt{13}}{2},-4\right) \cup\left(-4, \frac{-3-\sqrt{21}}{2}\right) \cup$
$$
\left(\frac{-5+\sqrt{13}}{2}, \frac{-3+\sqrt{21}}{2}\right) \text {. }
$$
Discuss the following three cases.
(1) $\left\{\begin{array}{l}x \geqslant 0, \\ \frac{x+3}{x+4}>0, \\ x0 \\ -x<\frac{x+3}{x+4}\end{array}\right.$
$\... | \left(\frac{-5-\sqrt{13}}{2},-4\right) \cup\left(-4, \frac{-3-\sqrt{21}}{2}\right) \cup \left(\frac{-5+\sqrt{13}}{2}, \frac{-3+\sqrt{21}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,947 |
3. Given three vertices of a cube are
$$
P(4,7,4), Q(7,11,4), R(11,8,9) \text {. }
$$
then the coordinates of the center of the cube are $\qquad$ | 3. $\left(\frac{15}{2}, \frac{15}{2}, \frac{13}{2}\right)$.
Notice that,
$$
\begin{array}{l}
P Q=\sqrt{(4-7)^{2}+(7-11)^{2}+(4-4)^{2}}=5, \\
Q R=\sqrt{(7-11)^{2}+(11-8)^{2}+(4-9)^{2}}=5 \sqrt{2}, \\
R P=\sqrt{(11-4)^{2}+(8-7)^{2}+(9-4)^{2}}=5 \sqrt{3} .
\end{array}
$$
Thus, $P Q: Q R: R P=1: \sqrt{2}: \sqrt{3}$.
The... | \left(\frac{15}{2}, \frac{15}{2}, \frac{13}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,948 |
5. Suppose there are 10 red, 10 yellow, and 10 blue small balls. Now, all of them are to be placed into two bags, A and B, such that each bag contains balls of two colors, and the sum of the squares of the number of balls of two colors in bags A and B are equal. There are $\qquad$ ways to do this. | 5.61.
Let the number of red, yellow, and blue balls in bag A be $x, y, z (1 \leqslant x, y, z \leqslant 9)$. Then the number of balls of corresponding colors in bag B are $10-x, 10-y, 10-z$.
First, assume $x \leqslant y \leqslant z$.
From the problem, we know
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=(10-x)^{2}+(10-y)^{2... | 61 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,950 |
6. In a Cartesian coordinate system, there are 25 non-coincident horizontal and vertical lines, each dyed one of two colors: black or red. Then, the intersection points of black horizontal lines and black vertical lines are dyed black; the intersection points of red horizontal lines and red vertical lines are dyed red;... | 6. 1:6 or 6:1.
Let the number of black and red horizontal lines be $x, y$ respectively, and the number of black and red vertical lines be $u, v$ respectively. Then the number of black, red, yellow, and green points are $x u, y v, x v, y u$ respectively.
From the problem, we have
$$
\left\{\begin{array}{l}
x+y+u+v=25 \... | 1:6 \text{ or } 6:1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,951 |
7. Given two lines with a slope of 1, $l_{1}$ and $l_{2}$, passing through the two foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $l_{1}$ intersects the ellipse at points $A$ and $B$, $l_{2}$ intersects the ellipse at points $C$ and $D$. If quadrilateral $\square A B C D$ satisfies $A C \pe... | 7.2012.
It is known that $\square A B C D$ is symmetric about the origin $O$. As shown in Figure 1, let $\angle A \dot{F}_{1} F_{2}=\alpha$. Then
. $\tan \alpha=1$
. $\Rightarrow \alpha=45^{\circ}$.
Since $A C \perp A B$, we know
$A C \perp A F_{1}$.
Thus, $\triangle A F_{1} O$ is an
isosceles right triangle. Therefo... | 2012 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,952 |
8. Given $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$ has a segment of digits $\overline{2012}$ in its decimal part, where $n$ is the smallest number satisfying the condition. Then $\left[\frac{m}{\sqrt{n}}\right]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 8. 2 .
Let $\frac{m}{n}=\overline{A . B 2012 C}$, where $A, B, C$ are digit strings, and the lengths of $A, B$ are $k, l (k, l \in \mathbf{N})$. Then
$$
\frac{10^{6}(m-n A)-n B}{n}=\overline{0.2012 C} \triangleq \frac{a}{b},
$$
where $(a, b)=1, 1 \leqslant b \leqslant n$.
Assume $\frac{m}{n}=\overline{0.2012 C}$.
$$
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,953 |
9. (16 points) Find all positive integers $n$ and real numbers
$$
x\left(x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right) \text {, }
$$
such that $n \tan x+\sqrt{3}$ and $\cot x+\sqrt{3}$ are both rational numbers. | Given that $n \tan x+\sqrt{3}$ and $\cot x+\sqrt{3}$ are both rational numbers, we have
$$
\begin{array}{l}
(n \tan x+\sqrt{3})+(\cot x+\sqrt{3}) \\
=(n \tan x+\cot x)+2 \sqrt{3} \in \mathbf{Q} \\
(n \tan x+\sqrt{3})(\cot x+\sqrt{3}) \\
=n+(n \tan x+\cot x) \sqrt{3}+3 \in \mathbf{Q}
\end{array}
$$
From equation (1), t... | n=2 \text{ or } 3, \quad x=\arctan \frac{-\sqrt{3}+1}{2}, \arctan \frac{-\sqrt{3}-1}{2}, -\frac{\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,954 |
Example 2 Prove Kantorovich's (Канторович) inequality:
Let $a_{j}>0, \sum_{j=1}^{n} a_{j}=1, 0<\lambda_{1}<\lambda_{2}<\cdots<\lambda_{n}$, where $j=1,2, \cdots, n, n \geqslant 2$. Then
$$
\left(\sum_{j=1}^{n} a_{j} \lambda_{j}\right) \sum_{j=1}^{n} \frac{a_{j}}{\lambda_{j}} \leqslant \frac{\left(\lambda_{1}+\lambda_{... | Proof using geometric construction. First, we prove a lemma.
Lemma: Let $P$ be any point on the diagonal $AC$ of rectangle $ABCD$, and let $PE \perp AB$, $PF \perp BC$, as shown in Figure 2.
Then $S_{\text{quadrilateral } PEHF} \leq \frac{1}{4} S_{\text{rectangle } ABCD}$.
Proof: By
$1=\frac{AP}{AC}+\frac{PC}{AC}=\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,955 |
10. (20 points) Rolling Dice (a uniform cube, with six faces marked with $1,2,3,4,5,6$) Game rules are as follows: First roll 9 dice, take out the dice showing 1 and set them aside; on the second roll, take out the dice showing 1 from the remaining dice; $\cdots \cdots \cdots$, until no dice show 1 or all dice are take... | 10. According to the game rules, if the game ends exactly after 9 rounds, then in the first eight rounds, each time exactly 1 die shows a 1, and the ninth round ends the game regardless of whether it shows a 1 or not. Among these, the probability that exactly 1 die shows a 1 in the $k(k=1,2, \cdots, 8)$-th round, where... | 2012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,956 |
11. (20 points) In the plane of the equilateral $\triangle B_{1} B_{2} B_{3}$, there is any point $P$ (not on the lines of the three altitudes of $\triangle B_{1} B_{2} B_{3}$). The reflections of point $P$ about the lines $B_{2} B_{3}$, $B_{3} B_{1}$, and $B_{1} B_{2}$ are $P_{1}$, $P_{2}$, and $P_{3}$, respectively. ... | 11. Establish the complex plane, and let the complex number corresponding to point $X$ still be denoted as $X$.
Obviously, the circumcenter $S_{j}$ of $\triangle P P_{j} B_{j}$ lies on the line $B_{k} B_{l}$, where $\{j, k, l\}=\{1,2,3\}$.
Let $\frac{S_{1}-B_{2}}{S_{1}-B_{3}}=t_{1}$.
$$
\begin{array}{l}
\frac{S_{1}-B_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,957 |
One. (40 points) In trapezoid $ABCD$, $AD \parallel BC$, $AB = AD + BC > BD$, $M$ is the midpoint of $CD$, and $E$ is a point outside the trapezoid, on the opposite side of $CD$; satisfying $2 \angle AEB = \angle DEC = \angle ABC$, $O$ is the circumcenter of $\triangle BCE$. Prove: $\angle OME = 90^{\circ}$. | As shown in Figure 2, take a point $G$ on the extension of $A D$ such that $D G = B C$.
Then $A B = A D + B C = A G$.
Connect $B G$, $C G$, and $E G$. In the quadrilateral, and
$$
\angle A G B = \frac{1}{2} \angle A B C = \angle A E B.
$$
Therefore, point $E$ lies on the circumcircle of $\triangle A B G$.
Since $A B >... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,958 |
II. (40 points) Given the function
$$
f(x)=3\left(\sin ^{3} x+\cos ^{3} x\right)+m(\sin x+\cos x)^{3}
$$
has a maximum value of 2 in $x \in\left[0, \frac{\pi}{2}\right]$. Find the value of the real number $m$. | $$
\begin{array}{l}
\sin ^{3} x+\cos ^{3} x \\
=(\sin x+\cos x)\left[(\sin x+\cos x)^{2}-3 \sin x \cdot \cos x\right] \\
=(\sin x+\cos x)\left\{(\sin x+\cos x)^{2}-\right. \\
\left.\quad \frac{3}{2}\left[(\sin x+\cos x)^{2}-1\right]\right\} .
\end{array}
$$
Let $t=\sin x+\cos x$
$$
=\sqrt{2} \sin \left(x+\frac{\pi}{4}... | m=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,959 |
Three, (50 points) Let the number of all positive integers satisfying the following conditions be $N$:
(1) less than or equal to 2,012;
(2) the number of 1s in their binary representation is at least 2 more than the number of 0s.
Find the sum of the digits of $N$.
| Three, from $2012=(11111011100)_{2}$, we know that the numbers satisfying the conditions have at most 11 digits in binary representation.
The first digit must be 1, so the number of $d+1$-digit numbers with exactly $k+1$ digits being 1 is $\mathrm{C}_{d}^{k}$, and condition (2) is equivalent to
$$
\begin{array}{l}
k+1... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,960 |
Four. (50 points) Let $n \in \mathbf{N}_{+}, f(n)$ be the number of all integer sequences $\left\{a_{k} \mid k=0,1, \cdots, n\right\}$ that satisfy the following conditions:
$$
\begin{array}{l}
\text { (1) } a_{0}=0, a_{n}=2 n, \text { and } \\
1 \leqslant a_{k+1}-a_{k} \leqslant 3(k=0,1, \cdots, n-1) ;
\end{array}
$$
... | Divide a circle of length $2 \cdot n$ into $2n$ equal parts, and label the points sequentially as $0,1, \cdots, 2n$. Then color the points labeled $a_{i} (i=0,1, \cdots, n-1)$ black, and the other $n$ points white. The sequence given in the problem corresponds one-to-one with the following coloring method:
(1) The poin... | 2012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,961 |
Given $P$ is a point inside $\triangle A B C$, and the rays $A P, B P, C P$ intersect the sides $B C, C A, A B$ at points $D, E, F$ respectively. If $S_{\triangle A B C}=\lambda S_{\triangle O E F}$, find the value of $\frac{P A}{P D}+\frac{P B}{P E}+\frac{P C}{P F}$ (expressed in terms of $\lambda$). | As shown in Figure 2, let $E F$ intersect $A P$ at point $H$. By the area ratio, we have
$$
\begin{array}{l}
\frac{A H}{P H}=\frac{S_{\triangle A E F}}{S_{\triangle P E F}} \\
=\frac{S_{\triangle A E F}}{S_{\triangle A P F}} \cdot \frac{S_{\triangle A P F}}{S_{\triangle P E F}} \\
=\frac{E B}{P B} \cdot \frac{A C}{E C}... | 2\lambda-2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,962 |
Given non-zero real numbers $a$, $b$, $c$, satisfying
$$
(a+b+c)(b+c-a)(c+a-b)(a+b-c)=a^{4} \text{. }
$$
Find the maximum and minimum values of $\frac{b}{c}-\frac{c}{b}$, and indicate the values of $|a|:|b|:|c|$ when $\frac{b}{c}-\frac{c}{b}$ takes its maximum and minimum values. | Consider $a$ as the main variable, obtaining a quartic equation in $a$:
$$
2 a^{4}-2\left(b^{2}+c^{2}\right) a^{2}+\left(b^{2}-c^{2}\right)^{2}=0 .
$$
Then, regard equation (1) as a quadratic equation in the unknown $a^{2}$.
Since the equation has real solutions, we have:
$$
\begin{array}{l}
\Delta=4\left(b^{2}+c^{2}\... | |a|:|b|:|c|=\sqrt{2}: \sqrt{2+\sqrt{2}}: \sqrt{2-\sqrt{2}} \text{ or } |a|:|b|:|c|=\sqrt{2}: \sqrt{2-\sqrt{2}}: \sqrt{2+\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,963 |
Let $x_{0}=\frac{\sqrt{5}-1}{2}$. Prove: For any rational number $\frac{p}{q}(p, q \in \mathbf{Z}$, and $(p, q)=1)$ in the interval $\left[x_{0}-1, x_{0}+1\right]$, we have
$$
\left|\frac{p}{q}-x_{0}\right| \geqslant \frac{1}{(\sqrt{5}+1) q^{2}} .
$$ | Prove that
$$
g(x)=\left(x-x_{0}\right)\left(x+x_{0}+1\right)=x^{2}+x-1 .
$$
It is easy to see that $g\left(\frac{p}{q}\right) \neq 0$.
Thus, $f(p, q)=q^{2} g\left(\frac{p}{q}\right)=p^{2}+p q-q^{2} \neq 0$.
Since $p, q \in \mathbf{Z}$, it follows that $f(p, q) \in \mathbf{Z}$.
Therefore, $|f(p, q)| \geqslant 1$.
For ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,964 |
Let $a, b, c, d$ be positive real numbers. Prove:
$$
\begin{array}{l}
\sqrt{\frac{a^{3}}{a^{3}+15 b c d}}+\sqrt{\frac{b^{3}}{b^{3}+15 c d a}}+ \\
\sqrt{\frac{c^{3}}{c^{3}+15 d a b}}+\sqrt{\frac{d^{3}}{d^{3}+15 a b c}} \geqslant 1 .
\end{array}
$$ | Prove that,
$$
\begin{array}{l}
\left(a^{\frac{15}{8}}+b^{\frac{15}{8}}+c^{\frac{15}{8}}+d^{\frac{15}{8}}\right)^{2} \\
\geqslant\left(a^{\frac{15}{8}}+3 b^{\frac{5}{3}} c^{\frac{5}{8}} d^{\frac{5}{8}}\right)^{2} \\
=a^{\frac{15}{4}}+6 a^{\frac{15}{8}} b^{\frac{5}{8}} c^{\frac{5}{8}} d^{\frac{5}{8}}+9 b^{\frac{5}{4}} c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,965 |
Example 1 Given that $a$ and $b$ are real numbers, and $a^{2} + ab + b^{2} = 3$. If the maximum value of $a^{2} - ab + b^{2}$ is $m$, and the minimum value is $n$, find the value of $m + n$. ${ }^{\text {[2] }}$ | Let $a^{2}-a b+b^{2}=t$.
Combining this with the given equation, we get
$$
a b=\frac{3-t}{2}, a+b= \pm \sqrt{\frac{9-t}{2}} .
$$
Thus, $a$ and $b$ are the two real roots of the quadratic equation in $x$:
$$
x^{2} \pm \sqrt{\frac{9-t}{2}} x+\frac{3-t}{2}=0
$$
Therefore, $\Delta=\left( \pm \sqrt{\frac{9-t}{2}}\right)^{... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,966 |
Example 2 The system of equations in $x, y, z$
$$
\left\{\begin{array}{l}
3 x+2 y+z=a, \\
x y+2 y z+3 z x=6
\end{array}\right.
$$
has real solutions $(x, y, z)$. Find the minimum value of the positive real number $a$. | Solving the system of equations, we get
$$
3 x+2 y=a-z, 3 x \times 2 y=6\left(z^{2}-a z+6\right) \text {. }
$$
It is easy to see that $3 x$ and $2 y$ are the two real roots of the quadratic equation
$$
t^{2}-(a-z) t+6\left(z^{2}-a z+6\right)=0
$$
Thus, $\Delta=[-(a-z)]^{2}-24\left(z^{2}-a z+6\right) \geqslant 0$, whi... | \sqrt{23} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,967 |
9. (16 points) Given the function $f(x)$ satisfies $f(0)=0$, and for any $x \in \mathbf{R}, f(2 x)=\sin x+f(x)$. Prove: $f(1)<1$.
---
The translation retains the original text's line breaks and formatting. | 9. From $f(2 x)=\sin x+f(x)$, we get
$$
\begin{array}{l}
f(1)=\sin \frac{1}{2}+f\left(\frac{1}{2}\right) \\
f\left(\frac{1}{2}\right)=\sin \frac{1}{4}+f\left(\frac{1}{4}\right) \\
\cdots \cdots \\
f\left(\frac{1}{2^{n-1}}\right)=\sin \frac{1}{2^{n}}+f\left(\frac{1}{2^{n}}\right)
\end{array}
$$
Adding the above equatio... | f(1)<1 | Algebra | proof | Yes | Yes | cn_contest | false | 725,968 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, $a_{2}=9$, and for any positive integer $n$,
$$
n a_{n+2}-6(n+1) a_{n+1}+9(n+2) a_{n}=0 \text {. }
$$
Find the general term formula of the sequence $\left\{a_{n}\right\}$. | 10. From the given, we have
$$
\begin{array}{l}
n\left(a_{n+2}-3 a_{n+1}\right)=3(n+2)\left(a_{n+1}-3 a_{n}\right) \\
\Rightarrow \frac{a_{n+2}-3 a_{n+1}}{3(n+2)}=\frac{a_{n+1}-3 a_{n}}{n} \\
\Rightarrow \frac{a_{n+2}-3 a_{n+1}}{3^{n+1}(n+2)(n+1)}=\frac{a_{n+1}-3 a_{n}}{3^{n}(n+1) n} \\
\Rightarrow \frac{a_{n+1}-3 a_{n... | a_{n}=3^{n-2}\left(n^{3}-n+3\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,969 |
11. (20 points) Through a point $P$ on the directrix $l$ of the parabola $C: y^{2}=4 x$, draw two tangent lines to the parabola $C$, touching at points $A$ and $B$ respectively, and let $M$ be the intersection of the directrix $l$ with the $x$-axis.
(1) Let the focus of the parabola be $F$, prove:
$$
|P F|^{2}=|A F||B ... | 11. (1) It is known that the equation of the directrix $l$ of the parabola $C$ is $x=-1$.
Let $P\left(-1, y_{0}\right), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $l_{A B}: y_{0} y=2(x-1)$.
When $y_{0} \neq 0$, the slope of line $A B$ is $k=\frac{2}{y_{0}}$.
It is known that $F(1,0)$.
Then the slope ... | P(-1, \pm 2), \text{ and the equation of line } A B \text{ is } y=x-1 \text{ or } y=-x+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,970 |
One. (40 points) As shown in Figure 1, in $\triangle ABC$, it is given that $AB > AC$, the incircle $\odot I$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively, $M$ is the midpoint of side $BC$, line $EF$ intersects the extensions of $BI$ and $CI$ at points $P$ and $Q$ respectively, and inters... | As shown in Figure 2, connect $I D$, $I E$, $I F$, $M P$, $Q D$, $B Q$, and $C P$. Then $I D \perp B C$, $I E \perp C A$, $I F \perp A B$.
From $A E=A F$, we know
$$
\begin{array}{l}
\angle A E F=\angle A F E=\frac{1}{2}\left(180^{\circ}-\angle B A C\right). \\
\text { Also, } \angle B I Q=\angle C I P \\
=\frac{1}{2}(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,971 |
II. (40 points) Given that $a, b, c$ are non-negative real numbers. Prove:
$\sum \sqrt{a^{2}+a b+b^{2}} \leqslant \sqrt{5 \sum a^{2}+4 \sum a b}$, where, " $\sum$ " denotes the cyclic sum. | $$
\begin{array}{l}
\left(\sum \sqrt{a^{2}+a b+b^{2}}\right)^{2} \\
\leqslant\left(\sum(a+b)\right) \sum \frac{a^{2}+a b+b^{2}}{a+b} .
\end{array}
$$
Thus, it suffices to prove
$$
\begin{array}{l}
2(a+b+c) \sum \frac{a^{2}+a b+b^{2}}{a+b} \leqslant 5 \sum a^{2}+4 \sum a b \\
\Leftrightarrow 2 \sum \frac{a^{2}+a b+b^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,972 |
$$
\text { Three. (50 points) Given }(2 k+1) \times(2 k+1)
$$
$\left(k \in \mathbf{N}_{*}\right)$, each element in the square grid is a real number with an absolute value no greater than 1, and the sum of all elements in the grid is equal to 0. Find the smallest non-negative real number $C$, such that in every such gri... | Three, first, the specialized square grid table
$$
a_{i j}=\left\{\begin{array}{ll}
1, & i, j \leqslant k+1 ; \\
-1, & i, j>k+1 ; \\
-\frac{2 k+1}{2 k(k+1)}, & \text { other cases. }
\end{array}\right.
$$
In this grid, the sum of the elements in the first $k+1$ rows and the first $k+1$ columns is all equal to
$$
k+1-\... | C=k+\frac{1}{2 k+2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,973 |
Four. (50 points) Given a sequence of positive integers $\left\{a_{n}\right\}$ that satisfies for any positive integer $n$,
$$
0<a_{n+1}-a_{n} \leqslant 2013 \text {. }
$$
Prove: There exist infinitely many pairs of positive integers $(p, q)$ $(p \neq q)$, such that $a_{p} \mid a_{q}$. | Assume that there are only finitely many pairs of positive integers $(p, q)$ $(p \neq q)$ satisfying $a_{p} \mid a_{q}$, i.e., there exists a positive integer $N$ such that all $q$ satisfying the condition are less than $N$.
We will prove by mathematical induction: For any positive integer $k \geqslant 1$, there exist... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,974 |
As shown in Figure 2, $P$ and $Q$ are points on the sides $AB$ and $BC$ of the square $ABCD$, respectively, and $AP = CQ$. $\odot O$ is the circle passing through points $C$, $D$, and $Q$. $PE$ is tangent to $\odot O$ at point $E$. Prove: $PE = BQ$.
Prove: $PE = BQ$. | Prove as shown in Figure 2, connect $C P$, intersecting $\odot O$ at point $H$, then connect $B H$, $D H$, $Q H$, and $D Q$.
From the given conditions, we have $B C = C D$, $B P = B Q$,
$\angle P B C = \angle D C Q = 90^{\circ}$.
Since $C$, $D$, $H$, and $Q$ are concyclic,
$\angle H D Q = \angle H C Q$,
$\angle D H Q =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,975 |
Initially 340 Given two points $A$ and $B$ on a straight line $l$, the distance between them is $10000 \mathrm{~cm}$. At points $A$ and $B$, there are two movable barriers, designated as Barrier 1 and Barrier 2, respectively. Assume there is a ping-pong ball between $A$ and $B$, moving along the straight line $l$ at a ... | Let's assume that when the ping-pong ball contacts the 2nd board for the $n$-th time, the position of the 2nd board after it moves quickly is $B_{n}$; when the ping-pong ball contacts the 1st board for the $n$-th time, its position is $A_{n}$, and we set
$$
\begin{array}{l}
B_{0}=B, A_{n} B_{n-1}=x_{n}, A_{n} B_{n}=y_{... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 725,976 |
339 Given a quadrilateral $ABCD$ inscribed in $\odot O$, let $AB=a, BC=b, CD=c, DA=d, \odot O$ have a radius of $R, p=\frac{1}{2}(a+b+c+d)$. Prove:
$$
R^{2}=\frac{\left(a^{2} b^{2} c^{2}+a^{2} b^{2} d^{2}+b^{2} c^{2} d^{2}+c^{2} d^{2} a^{2}\right)+a b d\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{16(p-a)(p-b)(p-c)(p-d)} .
$$ | Prove as shown in Figure 3, connect
$A C$.
By the Law of Sines, we have
$$
A C=2 R \sin B \text {. }
$$
By the Law of Cosines, we have
$$
\begin{array}{l}
a^{2}+b^{2}-2 a b \cos B=A C^{2}, \\
c^{2}+d^{2}-2 c d \cos D=A C^{2} .
\end{array}
$$
Since $\angle B+\angle D=180^{\circ}$, we have
$$
\cos D=-\cos B \text {. }
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,977 |
Example 1 Let the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfy
$$
a_{1}=2, a_{m+n}+a_{m-n}-m+n=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {, }
$$
where $m, n \in \mathbf{N}, m \geqslant n$. Prove:
(1) For all $n \in \mathbf{N}$, we have
$$
a_{n+2}=2 a_{n+1}-a_{n}+2 \text {; }
$$
(2) $\frac{1}{a_{1}}+\frac... | Prove (1) In the given relation,
let $m=n$, we get $a_{0}=0$;
let $n=0$, we get $a_{2 m}=4 a_{m}-2 m$;
let $m=n+2$, we get
$$
a_{2 n+2}+a_{2}-2=\frac{1}{2}\left(a_{2 n+4}+a_{2 n}\right) \text {. }
$$
From equation (1), we get
$$
\begin{array}{l}
a_{2 n+2}=4 a_{n+1}-2(n+1), \\
a_{2}=4 a_{1}-2=6, \\
a_{2 n+4}=4 a_{n+2}-... | 1-\frac{1}{2010}<1 | Algebra | proof | Yes | Yes | cn_contest | false | 725,978 |
Given $x, y, z$ are all real numbers greater than 1, and satisfy $\sqrt{x y}+\sqrt{z}=\sqrt{z(x-1)(y-1)}$. Prove:
$$
\begin{array}{l}
\frac{\sqrt{x}}{3 x+(y-1)(z-1)}+\frac{\sqrt{y}}{3 y+(z-1)(x-1)}+ \\
\frac{\sqrt{z}}{3 z+(x-1)(y-1)} \leqslant \frac{2}{7} .
\end{array}
$$ | Proof: From the given, we can let
$$
\sqrt{x-1}=\tan A, \sqrt{y-1}=\tan B,
$$
where $\angle A, \angle B$ are two interior angles of the acute triangle $\triangle ABC$. Then
$$
x=\sec ^{2} A, y=\sec ^{2} B.
$$
$$
\begin{array}{l}
\text { Then } \sqrt{z}=\frac{\sqrt{x y}}{\sqrt{(x-1)(y-1)}-1} \\
= \frac{\sec A \cdot \se... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,979 |
Example 2 Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{1}=1, a_{n+1}=2 a_{n}+\sqrt{3 a_{n}^{2}+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove:
(1) When $n>1$, $a_{n+1}+a_{n-1}=4 a_{n}$;
(2) $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1+\sqrt{3}}{2}$. | Proof (1) It is known that $\{a_n\}$ is an increasing sequence with all terms being positive.
From $a_{n+1}=2a_n+\sqrt{3a_n^2+1}$, we can simplify to get
$$
a_{n+1}^2 - 4a_n a_{n+1} + a_n^2 = 1.
$$
When $n \geq 2$, we have
$$
a_n^2 - 4a_{n-1} a_n + a_{n-1}^2 = 1.
$$
Subtracting the two equations, we get
$$
\begin{arr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,980 |
Example 3 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=\frac{\sqrt{1+a_{n}^{2}}-1}{a_{n}}(n=1,2, \cdots) \text {, }
$$
$S_{n}$ is the sum of its first $n$ terms. Prove:
$$
S_{n}>\frac{3\left(2^{n}-1\right)}{2^{n+1}} .
$$ | It is known that for any $n \in \mathbf{N}_{+}, a_{n}>0$.
Let $a_{n}=\tan \alpha_{n}\left(n=1,2, \cdots, \alpha_{n} \in\left(0, \frac{\pi}{2}\right)\right)$.
Then $\alpha_{1}=\frac{\pi}{4}$, and
$$
\begin{array}{l}
\tan \alpha_{n+1}=\frac{\sqrt{1+\tan \alpha_{n}^{2}}-1}{\tan \alpha_{n}} \\
=\frac{\sec \alpha_{n}-1}{\ta... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,981 |
Example 4 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2, a_{n+1}=a_{n}^{2}-a_{n}+1\left(n \in \mathbf{N}_{*}\right)$. | Prove: When $n \geqslant 4$,
$$
1-\frac{1}{2^{n+1}}2 \times 3 \times 7 \times 2^{n-4}>2^{n},
$$
i.e., $\sum_{k=1}^{n} \frac{1}{a_{k}}=1-\frac{1}{a_{n+1}-1}>1-\frac{1}{2^{n+1}}$.
Therefore, when $n \geqslant 4$, $1-\frac{1}{2^{n+1}}<\sum_{k=1}^{n} \frac{1}{a_{k}}<1$. | 1-\frac{1}{2^{n+1}}<\sum_{k=1}^{n} \frac{1}{a_{k}}<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,982 |
Example 5 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{2}=2, \frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}(n \geqslant 1) \text {. }
$$
(1) Find the recursive relation between $a_{n+1}$ and $a_{n}$, i.e., $a_{n+1}=f\left(a_{n}\right)$;
(2) Prove: $63<a_{2008}<78$. | (1) Solution: It is obvious that, $a_{n}>0$.
Then from $\frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}$, we get $\frac{a_{n+2} a_{n+1}}{a_{n+1}^{2}+1}=\frac{a_{n+1} a_{n}}{a_{n}^{2}+1}$.
Thus, $\left\{\frac{a_{n+1} a_{n}}{a_{n}^{2}+1}\right\}$ is a constant sequence. Since $\frac{a_{2} a_{1}}{a_{1}^{2}+1}=1$, ... | 63<a_{2008}<78 | Algebra | proof | Yes | Yes | cn_contest | false | 725,983 |
Example 6 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n} a_{n+1}=n+1\left(n \in \mathbf{N}_{+}\right)$. | Prove: $\sum_{k=1}^{n} \frac{1}{a_{k}} \geqslant 2(\sqrt{n+1}-1)$.
Proof It is easy to see that for all $n \in \mathbf{N}_{+}$, we have $a_{n}>0$.
When $n \geqslant 2$, $a_{n} a_{n+1}=n+1, a_{n-1} a_{n}=n$.
Subtracting the two equations gives $\frac{1}{a_{n}}=a_{n+1}-a_{n-1}(n \geqslant 2)$.
$$
\begin{array}{l}
\text {... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,984 |
Example 7 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, a_{n+1}=a_{n}+\frac{a_{n}^{2}}{n^{2}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: For all $n \in \mathbf{N}_{+}$, we have
(1) $a_{n}<\frac{1}{2}-\frac{1}{4 n}$. | Proof (1) It is clear that, $a_{n}>0$.
Thus, $a_{n+1}-a_{n}=\frac{a_{n}^{2}}{n^{2}}>0$
$$
\begin{array}{l}
\Rightarrow a_{n}-\frac{1}{n^{2}} \\
\Rightarrow \frac{1}{a_{n}}-\frac{1}{a_{1}}>-\sum_{k=1}^{n-1} \frac{1}{k^{2}}>-1-\sum_{k=2}^{n} \frac{1}{k(k-1)} \\
=-1-\left(1-\frac{1}{n-1}\right)=-2+\frac{1}{n-1} \\
\Righta... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,985 |
Example 8 Given the function $f(x)=\frac{16 x+7}{4 x+4}$, the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}>0, b_{1}>0, \\
a_{n}=f\left(a_{n-1}\right), b_{n}=f\left(b_{n-1}\right)(n=2,3, \cdots) .
\end{array}
$$
(1) Find the range of $a_{1}$ such that for any positive int... | (1) Sol. From $f(x)=\frac{16 x+7}{4 x+4}=4-\frac{9}{4 x+4}$, we have
$$
\begin{array}{l}
a_{n+1}-a_{n}=f\left(a_{n}\right)-f\left(a_{n-1}\right) \\
=\frac{9}{4 a_{n-1}+4}-\frac{9}{4 a_{n}+4} \\
=\frac{9}{4} \cdot \frac{a_{n}-a_{n-1}}{\left(1+a_{n-1}\right)\left(1+a_{n}\right)}=\cdots \\
=\left(\frac{9}{4}\right)^{n-1} ... | 0<b_{n}-a_{n} \leqslant \frac{1}{8^{n-1}}(n=1,2, \cdots) | Algebra | proof | Yes | Yes | cn_contest | false | 725,986 |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=\frac{1}{4}, \\
a_{n+1}=\frac{(n-1) a_{n}}{n-a_{n}}(n=2,3, \cdots) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: For all $n \in \mathbf{N}_{+}$, $\sum_{k=1}^{n} a_{k}^{2}<\fr... | (1) From
$$
\frac{1}{n a_{n+1}}-\frac{1}{(n-1) a_{n}}=-\left(\frac{1}{n-1}-\frac{1}{n}\right),
$$
we get $a_{n}=\frac{1}{3 n-2}\left(n \in \mathbf{N}_{*}\right)$.
(2) Using $k \geqslant 2$, we have
$$
\begin{array}{l}
a_{k}^{2}=\frac{1}{(3 k-2)^{2}}<\frac{1}{(3 k-4)(3 k-1)} \\
=\frac{1}{3}\left(\frac{1}{3 k-4}-\frac{1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,987 |
Example 3 Let $a b \neq 0$, and the functions
$$
\begin{aligned}
f_{1}(x) & =x^{2}+2 a x+4 b \\
\text { and } f_{2}(x) & =x^{2}+4 a x+2 b
\end{aligned}
$$
have the same minimum value $u$, and the functions
$$
\begin{aligned}
f_{3}(x) & =-x^{2}+2 b x+4 a \\
\text { and } f_{4}(x) & =-x^{2}+4 b x+2 a
\end{aligned}
$$
h... | Notice that,
$$
\begin{array}{l}
f_{1}(x)=(x+a)^{2}+4 b-a^{2} \geqslant 4 b-a^{2}, \\
f_{2}(x)=(x+2 a)^{2}+2 b-4 a^{2} \geqslant 2 b-4 a^{2}, \\
f_{3}(x)=-(x-b)^{2}+4 a+b^{2} \leqslant 4 a+b^{2}, \\
f_{4}(x)=-(x-2 b)^{2}+2 a+4 b^{2} \leqslant 2 a+4 b^{2} .
\end{array}
$$
From $4 b-a^{2}=u=2 b-4 a^{2} \Rightarrow-2 b=3... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,988 |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{4}{a_{n}}\right)\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
$S_{n}$ is the sum of its first $n$ terms. Prove: when $n \geqslant 2$,
$$
2 n-1<S_{n}<2 n-\frac{1}{4} \text {. }
$$ | Hint: By constructing a geometric sequence $\left\{\ln \frac{a_{n}-2}{a_{n}+2}\right\}$, we obtain $\frac{a_{n}-2}{a_{n}+2}=\left(\frac{1}{3}\right)^{2 n-1}$, and through bounding
$$
\begin{array}{l}
4 \times\left(\frac{1}{3}\right)^{2 n-1}<a_{n}-2=\frac{4 \times\left(\frac{1}{3}\right)^{2 n-1}}{1-\left(\frac{1}{3}\rig... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,989 |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\sqrt{2}, a_{n+1}=\sqrt{2-\sqrt{4-a_{n}^{2}}}(n=1,2, \cdots) \text {, }
$$
$S_{n}$ is the sum of its first $n$ terms. Prove: $S_{n}<\frac{13}{4}$. | It is known that $0 < a_{n} < \sqrt{2}$.
Let $a_{n} = 2 \sin \alpha_{n} \left(\alpha_{n} \in \left(0, \frac{\pi}{4}\right], n \in \mathbf{N}_{+}\right)$.
We find that $\alpha_{n+1} = \frac{\alpha_{n}}{2}$.
Thus, $a_{n} = 2 \sin \frac{\pi}{2^{n+1}} < 2 \times \frac{\pi}{2^{n+1}} < \frac{13}{2^{n+2}}$.
Using $\left\{\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,990 |
4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, 2 n a_{n}=(2 n-3) a_{n-1}(n \geqslant 2) .
$$ | Prove: For all $n \in \mathbf{N}$, we have $\sum_{k=1}^{n} a_{k}<1$.
Hint: The given equation can be transformed to
$$
\begin{array}{l}
a_{n-1}=2\left[(n-1) a_{n-1}-n a_{n}\right] . \\
\text { Hence } \sum_{k=1}^{n} a_{k}=1-2(n+1) a_{n+1} .
\end{array}
$$ | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,991 |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=\left(1+\frac{1}{n^{2}+n}\right) a_{n}+\frac{1}{2^{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: (1) $a_{n-1} \leqslant\left(1+\frac{1}{n^{2}+n}+\frac{1}{2^{n}}\right) a_{n}$;
(2) $a_{0}<\mathrm{e}^{2}$. | (1) By induction, we get $a_{n} \geqslant 1\left(n \in \mathrm{N}_{+}\right)$.
$$
\begin{array}{l}
\text { Then } a_{n+1}=\left(1+\frac{1}{n^{2}+n}\right) a_{n}+\frac{1}{2^{n}} \\
\leqslant\left(1+\frac{1}{n^{2}+n}\right) a_{n}+\frac{1}{2^{n}} a_{n} .
\end{array}
$$
(2) By $\ln a_{n+1} \leqslant \ln \left(1+\frac{1}{n^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,992 |
Example 1 Given the sequence $\left\{b_{n}\right\}$ satisfies $b_{1}=1, b_{n}>0$ $(n=2,3, \cdots)$, and its first $n$ terms product is
$$
T_{n}=\left(a^{n-1} b_{n}\right)^{n}(n=1,2, \cdots) \text {. }
$$
(1) Prove: $\left\{b_{n}\right\}$ is a geometric sequence;
(2) Find the sum of the products of all different pairs o... | (1) Proof Note that,
$$
\begin{array}{l}
b_{n+1}=\frac{T_{n+1}}{T_{n}}=\frac{\left(a^{n} b_{n+1}\right)^{n+1}}{\left(a^{n-1} b_{n}\right)^{n}}=a^{2 n}\left(\frac{b_{n+1}}{b_{n}}\right)^{n} b_{n+1} \\
\Rightarrow \frac{b_{n+1}}{b_{n}}=\frac{1}{a^{2}}
\end{array}
$$
$\Rightarrow\left\{b_{n}\right\}$ is a geometric sequen... | \frac{\left(a^{2 n}-1\right)\left(a^{2 n-2}-1\right)}{a^{4 n-6}\left(a^{2}-1\right)\left(a^{4}-1\right)} \text{ for } a \neq \pm 1, \quad \frac{n^{2}-n}{2} \text{ for } a = \pm 1 | Algebra | proof | Yes | Yes | cn_contest | false | 725,993 |
Example 2 Define the function $f(x)=\frac{4^{x}}{4^{x}+2}$ on $\mathbb{R}$. Let $S_{n}=f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right)(n=2,3, \cdots)$.
(1) Find $S_{n}$;
(2) Does there exist a constant $M>0$, such that for any $n \geqslant 2$, we have
$$
\frac{1}{S_{2}}+\frac{1}{S... | (1) Notice,
$$
\begin{array}{l}
f(x)+f(1-x)=\frac{4^{x}}{4^{x}+2}+\frac{4^{1-x}}{4^{1-x}+2} \\
=\frac{4^{x}}{4^{x}+2}+\frac{4}{4+2 \times 4^{x}}=1 .
\end{array}
$$
In equation $\mathrm{E}$, let $x=\frac{1}{n}, \cdots, x=\frac{n-1}{n}$, and add them up to get
$$
S_{n}=\frac{n-1}{2} .
$$
(2) Take $n=2^{m}$. Then
$$
\beg... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,994 |
Example 3 Given that $\left\{a_{n}\right\}$ is a sequence of real numbers, for all $n \in \mathbf{N}$. It satisfies the relation
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=\frac{1}{a_{1} a_{2} \cdots a_{n}} .
$$
(1) Find the relationship between $a_{n}$ and $a_{n+1}(n \geqslant 2)$;
(2) Prove: If $a_{1} ... | (1) Solution From the given condition, we have
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{1} a_{2} \cdots a_{n+1}}-\frac{1}{a_{1} a_{2} \cdots a_{n}} \\
=\left(\frac{1}{a_{n+1}}-1\right) \frac{1}{a_{1} a_{2} \cdots a_{n}} \\
\Rightarrow 1-a_{n+1}=a_{1} a_{2} \cdots a_{n} \\
\Rightarrow 1-a_{n}=a_{1} a_{2} \cdots... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,995 |
Example 4 Given the function
$$
\begin{array}{l}
f(x)=\frac{2 x}{a x+b}, f(1)=1, f\left(\frac{1}{2}\right)=\frac{2}{3} . \\
\text { Let } x_{1}=\frac{1}{2}, x_{n+1}=f\left(x_{n}\right) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{x_{n}\right\}$;
(2) Prove: $x_{1} x_{2} \cdots x_{n+1}>\frac... | (1) Solution: From
$$
\begin{array}{l}
f(1)=1, f\left(\frac{1}{2}\right)=\frac{2}{3} \\
\Rightarrow a=b=1 \Rightarrow f(x)=\frac{2 x}{x+1} .
\end{array}
$$
First, find \( x_{1}=\frac{1}{2}, x_{2}=\frac{2}{3}, x_{3}=\frac{4}{5}, x_{4}=\frac{8}{9} \), conjecture: \( x_{n}=\frac{2^{n-1}}{2^{n-1}+1} \).
Prove by mathemati... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,996 |
Example 5 Toss a fair coin $n$ times, and let $p_{n}$ denote the probability of not getting three consecutive heads.
(1) Find $p_{1}, p_{2}, p_{3}, p_{4}$;
(2) Investigate the recursive formula for the sequence $\left\{p_{n}\right\}$, and provide a proof;
(3) Discuss the monotonicity and limit of the sequence $\left\{p... | (1) Obviously, $p_{1}=p_{2}=1$,
$$
p_{3}=1-\frac{1}{8}=\frac{7}{8}, p_{4}=1-\frac{3}{16}=\frac{13}{16} \text {. }
$$
(2) There are three cases in total.
(i) If the $n$-th toss is a tail, then the probability of not having three consecutive heads in the first $n$ tosses is the same as not having three consecutive heads ... | \lim _{n \rightarrow \infty} p_{n}=0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,997 |
Example 6 As shown in Figure 1, let the points on the curve $y=\frac{1}{x} + 1$ and the points on the $x$-axis form isosceles right triangles $\triangle O B_{1} A_{1}$, $\triangle A_{1} B_{2} A_{2}, \cdots$, with the right-angle vertices on the curve $y=\frac{1}{x}$. Try to find the coordinate expression for point $A_{... | Solve: Since $\triangle A_{n-1} B_{n} A_{n}$ is an isosceles right triangle, if we set $A_{n}\left(x_{n}, 0\right)$, then
$$
B_{n}\left(\frac{x_{n}+x_{n-1}}{2}, \frac{x_{n}-x_{n-1}}{2}\right) \text {. }
$$
Substituting into $y=\frac{1}{x}$, we get
$$
\frac{x_{n}+x_{n-1}}{2} \cdot \frac{x_{n}-x_{n-1}}{2}=1 \Rightarrow ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,998 |
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