problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Let any real numbers $a>b>c>d>0$. To make
$$
\begin{array}{l}
\log _{\frac{b}{a}} 2014+\log _{\frac{d}{b}} 2014+\log _{\frac{d}{c}} 2014 \\
\geqslant m \log _{\frac{d}{a}} 2014
\end{array}
$$
always hold, then the minimum value of $m$ is $\qquad$. | 5.9.
$$
\begin{array}{l}
\text { Let } x_{1}=-\log _{2014} \frac{b}{a}, x_{2}=-\log _{2014} \frac{c}{b}, \\
x_{3}=-\log _{2014} \frac{d}{c} .
\end{array}
$$
Since $a>b>c>d>0$, we have
$$
x_{1}>0, x_{2}>0, x_{3}>0 \text {. }
$$
Thus, the given inequality can be transformed into
$$
\begin{array}{l}
\frac{1}{x_{1}}+\fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,170 |
6. Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$. If $f(4)=2014$, then
$$
f(2014)=
$$
$\qquad$ | 6.4024.
Let $g(x)=f(x)-x$, then we have
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Also, from $f(x+3) \leqslant f(x)+3$,
$$
f(x+2) \geqslant f(x)+2 \text {, }
$$
we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
Fr... | 4024 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,171 |
7. If non-negative integers $m, n$ add up with exactly one carry (in decimal), then the ordered pair $(m, n)$ is called "good". The number of all good ordered pairs whose sum is 2014 is $\qquad$ . | 7. 195 .
If the carry is in the units place, then the combination of units and tens is $5+9$, $6+8$, $7+7$, $8+6$, $9+5$, a total of 5 kinds; the hundreds place can only be $0+0$, a total of 1 kind; the thousands place is $0+2$, $1+1$, $2+0$, a total of 3 kinds. In this case, there are $5 \times 1 \times 3=15$ pairs.
... | 195 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,172 |
8. Given non-negative real numbers $u, v, w$ satisfying $u+v+w=2$. Then the range of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is $\qquad$ . | 8. $[0,1]$.
Obviously, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2} \geqslant 0$, and when $u=0, v=0$, $w=0$, the equality holds.
Therefore, its minimum value is 0.
Assume without loss of generality that $u=\max \{u, v, w\}>0$.
Then $v w \leqslant u^{2}<2 u^{2} \Rightarrow v w-2 u^{2}<0$
$\Rightarrow v w\left(v w-2 u^{2}\righ... | [0,1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,173 |
9. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=2 a_{n}+n\left(1+2^{n}\right)(n=1,2, \cdots) \text {. }
$$
Try to find the general term $a_{n}$ of the sequence. | $$
\begin{array}{l}
\frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}+\frac{n}{2^{n+1}}+\frac{n}{2} \\
\Rightarrow \frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{n}{2^{n+1}}+\frac{n}{2} \\
\Rightarrow \sum_{i=1}^{n}\left(\frac{a_{i+1}}{2^{i+1}}-\frac{a_{i}}{2^{i}}\right)=\sum_{i=1}^{n} \frac{i}{2^{i+1}}+\sum_{i=1}^{n} \fr... | a_{n}=2^{n-2}\left(n^{2}-n+6\right)-n-1(n \geqslant 2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,174 |
10. (20 points) As shown in Figure 1, given $A(-1,0), B(1,0)$, $Q$ and $G$ are the circumcenter and centroid of $\triangle ABC$ respectively, and $QG \parallel AB$.
(1) Find the equation of the locus $E$ of point $C$.
(2) Let the two intersection points of the locus $E$ with the $y$-axis be $A_{1}$ and $A_{2}$ ($A_{1}$... | 10. (1) Let $C(x, y)(y \neq 0)$. Then
$$
G\left(\frac{x}{3}, \frac{y}{3}\right), Q\left(0, \frac{y}{3}\right) \text {. }
$$
From $|Q A|=|Q C| \Rightarrow 1+\frac{y^{2}}{9}=x^{2}+\frac{4 y^{2}}{9}$
$$
\Rightarrow x^{2}+\frac{y^{2}}{3}=1 \text {. }
$$
Thus, the equation of the trajectory $E$ is $x^{2}+\frac{y^{2}}{3}=1... | y=-2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,175 |
11. (20 points) Let \( x, y, z \) all be positive real numbers, and
\[
x+y+z=1 \text{. }
\]
Find the minimum value of the function
\[
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
\]
and provide a proof. | 11. Notice that, $\frac{3 x^{2}-x}{1+x^{2}}=\frac{x(3 x-1)}{1+x^{2}}$.
Consider the function $g(t)=\frac{t}{1+t^{2}}$.
It is easy to see that $g(t)$ is an odd function.
Since when $t>0$, $\frac{1}{t}+t$ is decreasing in the interval $(0,1)$, hence $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in the interval $(0,1)$.
T... | 0 | Algebra | proof | Yes | Yes | cn_contest | false | 728,176 |
Example 8 Find the minimum value of the real number $\lambda$ such that for any integer $n(n \geqslant 2)$ and positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=n$, we always have $\sum_{i=1}^{n} \frac{1}{a_{i}}-\lambda \prod_{i=1}^{n} \frac{1}{a_{i}} \leqslant n-\lambda$.
(2010, Chin... | When the positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ are not all equal,
$$
\sum_{i=1}^{n} \frac{1}{a_{i}}>n \text { and } \prod_{i=1}^{n} \frac{1}{a_{i}}>1
$$
are obvious.
Therefore, the $\lambda$ that satisfies the condition must be positive. Hereafter, assume $\lambda>0$.
The proof below requires the conclusion: t... | e | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,177 |
一、(40 points) As shown in Figure 2, in $\triangle ABC$, $AB > AC$, $H$ is the orthocenter of $\triangle ABC$, $M$ is the midpoint of side $BC$, point $S$ is on side $BC$ and satisfies $\angle BHM = \angle CHS$, and point $A$'s projection on line $HS$ is $P$. Prove: The circumcircle of $\triangle MPS$ is tangent to the ... | As shown in Figure 4, connect $A H$ and extend it to intersect the circumcircle of $\triangle A B C$ at point $D$. Draw $D E \parallel B C$, intersecting the circumcircle of $\triangle A B C$ at point $E$.
It is easy to see that points $D$ and $H$ are symmetric with respect to line $B C$.
Thus, $\angle H C B = \angle B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,178 |
Sure, here is the translated text:
```
II. (40 points) Let $k \in \mathbf{Z}_{+}$, define:
$$
A_{1}=1, A_{n+1}=\frac{n A_{n}+2(n+1)^{2 k}}{n+2}(n=1,2, \cdots) \text {. }
$$
Prove: When $n \geqslant 1$, $A_{n}$ is an integer, and $A_{n}$ is odd if and only if $n \equiv 1$ or $2(\bmod 4)$.
``` | $$
\begin{array}{l}
(n+2) A_{n+1}-n A_{n}=2(n+1)^{2 k}, \\
(n+1) A_{n}-(n-1) A_{n-1}=2 n^{2 k}. \\
(n+1) \times(1)+n \times \text { (2) gives } \\
(n+1)(n+2) A_{n+1}-(n-1) n A_{n-1} \\
=2(n+1)^{2 k+1}+2 n^{2 k+1}.
\end{array}
$$
Repeatedly applying the above equation and summing up, we get
$$
A_{n}=\frac{2 S(n)}{n(n+1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,179 |
Three. (50 points) Given $x, y, z \in [1,2]$. Prove:
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{18}{x+y+z} \\
\geqslant 6\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right),
\end{array}
$$
and identify the conditions under which equality holds. | $$
\begin{array}{l}
\text { Three, let } f(x, y, z) \\
= \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{18}{x+y+z}- \\
6\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right) .
\end{array}
$$
Assume without loss of generality that $x=\max \{x, y, z\}$, and let $t=\frac{y+z}{2}$.
First, we prove that $f(x, y, z) \geqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,180 |
Four, (50 points) Prove: There exists a set $S$ consisting of 2014 positive integers, with the following property: if a subset $A$ of $S$ satisfies that for any $a, a' \in A, a \neq a'$, we have $a + a' \notin S$, then $|A| \leq 152$.
---
The translation maintains the original text's formatting and line breaks. | For $1<k<2014$, let
$2014=k q+r(0 \leqslant r<k)$.
For $i=1,2, \cdots, k$, let
$S_{i}=\left\{2^{i-1} m \mid q \leqslant m \leqslant 2 q-1\right\}$.
Then $\left|S_{i}\right|=q$, and for any
$1 \leqslant i<j \leqslant k, q \leqslant m_{1}, m_{2} \leqslant 2 q-1$,
we have $2^{i-1} m_{1}=2^{j-1} m_{2} \Leftrightarrow 2^{j... | 152 | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,181 |
In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}, AC$ $=4, P$ is a point inside the triangle. If the inradii of $\triangle PBC$, $\triangle PAB$, and $\triangle PAC$ are all $r$, find the value of $r$.
---
The translation maintains the original format and line breaks as requested. | First, we can prove: point $P$ is on the bisector of $\angle A C B$. Let the incenters of $\triangle P A B$ and $\triangle P B C$ be $M$ and $N$, respectively. As shown in Figure 2, draw $M D \perp A B$ at point $D$ and $N E \perp B C$ at point $E$.
Given $A C=4$, we know $A B=4 \sqrt{2}$.
Therefore, $B D=2 \sqrt{2}$.... | r=\sqrt{2}+1-\sqrt{6 \sqrt{2}-5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,182 |
For $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, k), m \geqslant 1$, $n \in \mathbf{Z}_{+}$. Prove:
$$
\sum_{i=1}^{k} \sqrt[n]{\frac{x_{i}}{(m-1) x_{i}+\sum_{j=1}^{k} x_{j}}} \leqslant \frac{k}{\sqrt[n]{k+m-1}} .
$$ | Prove: If $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, k)$, $m \geqslant 1$, then
$$
\sum_{i=1}^{k} \frac{x_{i}}{(m-1) x_{i}+\sum_{j=1}^{k} x_{j}} \leqslant \frac{k}{k+m-1} .
$$
When $m=1$, the equality holds in the inequality.
When $m>1$,
$$
\begin{array}{l}
\sum_{i=1}^{k} \frac{x_{i}}{(m-1) x_{i}+\sum_{j=1}^{k} x_{j}} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,183 |
399 There are $n$ coins arranged in a row, with each coin having one side black and the other side white. Initially, the $i$-th coin from right to left is black side up, and the rest of the coins are white side up. Two players, A and B, take turns, with A starting first, according to the following rules:
(1) If there a... | Consider the binary number $\left(a_{1} a_{2} \cdots a_{n}\right)_{2}$, where $a_{i}=\left\{\begin{array}{ll}1, & \text { if the } i \text {th coin from left is white side up; } \\ 0, & \text { if the } i \text {th coin from left is black side up. }\end{array}\right.$ Here, a coin with the white (black) side up is call... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,184 |
For rational numbers $x, y$ satisfying the equation
$$
y^{2}=x^{3}+a x+b(a, b \in \mathbf{Z}),
$$
prove: there exist integers $r, s, t$, such that $s, r$ are coprime, $t, r$ are coprime, and
$$
x=\frac{s}{r^{2}}, y=\frac{t}{r^{3}}.
$$ | Prove that since $x, y$ are rational numbers, we set
$$
x=\frac{s}{p}, y=\frac{t}{q}((s, p)=(t, q)=1, p>0, q>0) \text {. }
$$
Then $t^{2} p^{3}=s^{3} q^{2}+a s p^{2} q^{2}+b p^{3} q^{2}$.
From equation (1), we know $q^{2}\left|t^{2} p^{3} \Rightarrow q^{2}\right| p^{3}$.
Also from equation (1), we know
$$
p^{3} \mid\l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,185 |
Example 2: Twelve acrobats numbered $1, 2, \cdots, 12$ are divided into two groups, $A$ and $B$, each with six people. Let the actors in group $A$ form a circle, and each actor in group $B$ stand on the shoulders of two adjacent actors in the $A$ circle. If the number of each actor in the $B$ circle is equal to the sum... | Let the sum of the elements in groups $A$ and $B$ be denoted as $x$ and $y$ respectively.
Then $y = 2x$.
Therefore, $3x = x + y = 1 + 2 + \cdots + 12 = 78 \Rightarrow x = 26$.
Clearly, $1, 2 \in A, 11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$.
If $d \leq 7$, then
$$
a + b + c + d \leq 4 + 5 + 6 + 7 = 2... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,186 |
Question 2: Ten boxes are arranged in a circle. Starting from a certain box, place $1, 2, \cdots, 10$ marbles in the boxes in a clockwise direction. According to the following rules: either add one marble to each of two adjacent boxes, or remove one marble from each of two adjacent boxes. Is it possible, through a fini... | The number
At the beginning, $\sum_{i=1}^{10} k_{i}=1+2+\cdots+10=55$ is an odd number, and any operation involves adding or removing two marbles, so the total number of marbles after any operation remains odd. However, the desired final result is $\sum^{10} k_{i}=10 \times 2011$, which is an even number, leading to a ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,187 |
Example 1 Let the function $f(x)$ satisfy
$$
f\left(\frac{1}{1-x}\right)=x f(x)+1 \text {. }
$$
Then $f(5)=$ ( ).
(A) $\frac{1}{2}$
(B) $\frac{4}{5}$
(C) $\frac{8}{5}$
(D) $\frac{9}{5}$ | Let $x=5,-\frac{1}{4}, \frac{4}{5}$, we get respectively
$$
\begin{array}{l}
f\left(-\frac{1}{4}\right)=5 f(5)+1 \\
f\left(\frac{4}{5}\right)=-\frac{1}{4} f\left(-\frac{1}{4}\right)+1 \\
f(5)=\frac{4}{5} f\left(\frac{4}{5}\right)+1
\end{array}
$$
Substituting equations (2) and (1) into equation (3) in sequence, we get... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,188 |
Example 2 Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying $f(0) \neq 0$, and for any $x, y \in \mathbf{R}$ we have
$$
f\left((x-y)^{2}\right)=f^{2}(x)-2 x f(y)+y^{2} .
$$
Then $f(2012)=$ $\qquad$ | Let $x=y=0$.
Then $f(0)=f^{2}(0) \Rightarrow f(0)=1$ or 0 (discard 0).
Let $y=x$.
$$
\begin{array}{l}
\text { Then } f(0)=f^{2}(x)-2 x f(x)+x^{2}=(f(x)-x)^{2} \\
\Rightarrow f(x)=x \pm 1 .
\end{array}
$$
If there exists $x_{0}$ such that $f\left(x_{0}\right)=x_{0}-1$, let
$$
\begin{array}{l}
x=x_{0}, y=0 \text {. } \\... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,189 |
2. Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, for any real number $x$ we have $f\left(x^{2}+x\right)+2 f\left(x^{2}-3 x+2\right)=9 x^{2}-15 x$. Then $f(50)=(\quad$.
(A) 72
(B) 73
(C) 144
(D) 146 | Hint: Choose D. Let $x=\frac{\sqrt{201}-1}{2}$ and $x=\frac{3-\sqrt{201}}{2}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,190 |
11. (20 points) Given an ellipse with eccentricity $\frac{1}{2}$, its left focus $F_{1}$ is the intersection of the directrix of the parabola $y^{2}=4 p x(p>0)$ with the $x$-axis, and its right focus $F_{2}$ is also the focus of the parabola. The intersection point of the ellipse and the parabola above the $x$-axis is ... | 11. From the problem, we know $F_{1}(-p, 0), F_{2}(p, 0)$.
Since $e=\frac{1}{2}$, the equation of the ellipse is
$$
\frac{x^{2}}{4}+\frac{y^{2}}{3}=p^{2} \text {. }
$$
From $\left\{\begin{array}{l}\frac{x^{2}}{4}+\frac{y^{2}}{3}=p^{2}, \\ y^{2}=4 p x,\end{array}\right.$, we get
$$
3 x^{2}+16 p x-12 p^{2}=0 \Rightarro... | \frac{\sqrt{6}}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,191 |
One, (40 points) As shown in Figure 1, for the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $N$, and the midpoint of $AC$ is $M$. If $\frac{BC^2}{CD^2} = \frac{BN}{DN}$, prove:
$$
\frac{MN^2}{MC^2} + \frac{DN^2}{DM^2} = 1.
$$ | Let $\angle A C B=\alpha, \angle A C D=\beta$. Then we have
$$
\begin{array}{l}
\frac{S_{\triangle B C N}}{S_{\triangle C D N}}=\frac{B N}{D N}=\frac{\frac{1}{2} B C \cdot C N \sin \alpha}{\frac{1}{2} C D \cdot C N \sin \beta}=\frac{B C \sin \alpha}{C D \sin \beta} \\
\Rightarrow \frac{B C^{2}}{D C^{2}}=\frac{B N}{D N}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,192 |
II. (40 points) In $\triangle A B C$, prove:
$$
\frac{\cos \frac{B}{2} \cdot \cos \frac{C}{2}}{\cos \frac{A}{2}}+\frac{\cos \frac{C}{2} \cdot \cos \frac{A}{2}}{\cos \frac{B}{2}}+\frac{\cos \frac{A}{2} \cdot \cos \frac{B}{2}}{\cos \frac{C}{2}} \geqslant \frac{3 \sqrt{3}}{2},
$$
with equality holding if and only if $\tr... | For $\triangle A B C$, construct its accompanying $\triangle A_{1} B_{1} C_{1}$.
Then $\cos \frac{B}{2}=\frac{B_{1} E}{B_{1} O}, \cos \frac{C}{2}=\frac{C_{1} G}{A_{1} C_{1}}, \cos \frac{A}{2}=\frac{C_{1} G}{B_{1} C_{1}}$.
Thus, $\frac{\cos \frac{B}{2} \cdot \cos \frac{C}{2}}{\cos \frac{A}{2}}=\frac{\frac{B_{1} E}{B_{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,193 |
Three, (50 points) Find the smallest two positive integers $m$, such that $47\left(m^{2}+46 m+713\right)$ is a perfect square.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, note that,
$$
\begin{array}{l}
47\left(m^{2}+46 m+713\right) \\
=47[47(m+15)+m(m-1)+8] .
\end{array}
$$
Thus, the necessary condition for $47\left(m^{2}+46 m+713\right)$ to be a perfect square is
$$
47 \mid[m(m-1)+8] \text {. }
$$
Let $m=47 k+i(k=0,1, \cdots ; i=0,1, \cdots, 46)$.
Then $m(m-1)+8$
$$
\begin{arr... | 539, 731 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,194 |
Four, (50 points) Given that $A$ is a set composed of 2015 different positive integers, and any three different numbers in $A$ form the side lengths of a non-obtuse triangle, at which point the triangle is called a triangle determined by the set $A$. $S(A)$ represents the sum of the perimeters of all triangles determin... | $$
\begin{array}{l}
\text { When the set } A \text { is given, } S(A) \text { is uniquely determined. Therefore, to make } S(A) \text { as small as possible, the elements in set } A \text { must be as small as possible. } \\
\text { Let } A=\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\}, a_{1}47265625=6875^{2} \geqslant... | 11824020 \mathrm{C}_{2014}^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,195 |
3. Given that the domain of the function $f(x)$ is $(-1,1)$, for any $x, y \in(-1,1)$ we have
$$
\begin{array}{l}
f(x)+f(y)=f\left(\frac{x+y}{1+x y}\right) . \\
\text { If } f\left(\frac{a+b}{1+a b}\right)=1, f\left(\frac{a-b}{1-a b}\right)=2, \text { and }|a|<1,
\end{array}
$$
$|b|<1$, find the values of $f(a)$ and $f... | $$
\left\{\begin{array}{l}
f(a)+f(b)=1, \\
f(a)+f(-b)=2 .
\end{array}\right.
$$
The above text translated into English, retaining the original text's line breaks and format, is as follows:
$$
\left\{\begin{array}{l}
f(a)+f(b)=1, \\
f(a)+f(-b)=2 .
\end{array}\right.
$$
Note: The equations themselves are already in a ... | f(a) = \frac{3}{2}, f(b) = -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,196 |
$$
f(x)=x\left(\sqrt{36-x^{2}}+\sqrt{64-x^{2}}\right)
$$
Find the maximum value of the function. | Algebraic solution Using the Cauchy-Schwarz inequality, we get
$$
\begin{aligned}
f(x) & =x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
& \leqslant \sqrt{\left(x^{2}+36-x^{2}\right)\left(64-x^{2}+x^{2}\right)}=48 .
\end{aligned}
$$
Geometric solution Construct $\triangle A B C, A D \perp B C$, and let $A B=6$, $A C=8, A D=x$... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,198 |
$\begin{array}{l}\text { 6. Let } f(x)=x^{4}+a x^{3}+b x^{2}+c x+d \text {. } \\ \text { If } f(1)=1, f(2)=2, f(3)=3 \text {, then } \\ \frac{1}{4}(f(4)+f(0))=\end{array}$ | Let $g(x)=f(x)-x$. Then
$$
g(1)=g(2)=g(3)=0 \text {. }
$$
Therefore, $g(x)=(x-1)(x-2)(x-3)(x-r)$ (where $r$ is a constant). | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,199 |
Example 1 In $\triangle ABC$, $AD \perp BC$, with the foot of the perpendicular $D$ on the segment $BC$, the inradii of $\triangle ABC$, $\triangle ADB$, and $\triangle ADC$ are $k$, $m$, and $n$ respectively. Given that $m^{2}+n^{2}=k^{2}$. Prove: $\triangle ABC$ is a right triangle. | Algebraic Proof Since point $D$ is on segment $B C$, therefore,
\[
\angle B0 \Leftrightarrow y(x+y+z)-z x>0, \\
a^{2}+b^{2}-c^{2}>0 \Leftrightarrow z(x+y+z)-x y>0, \\
x(x+y+z)-y z=0 \Leftrightarrow b^{2}+c^{2}-a^{2}=0 .
\end{array}
\]
Thus, by the Pythagorean theorem, $\angle B A C=90^{\circ}$.
Geometric Proof As show... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,200 |
Example 2 Let $D$ be the midpoint of side $BC$ of $\triangle ABC$, and $AF$ be the internal angle bisector, intersecting side $BC$ at point $F$ and the circumcircle of $\triangle ABC$ at point $E$. Prove: $AD + AE > AB + AC$. | Connect $B E$.
Let $B C=a, C A=b, A B=c$,
$2 p=a+b+c, A D=m_{a}, A F=w_{a}$.
Then $m_{a}=\frac{1}{2} \sqrt{2 b^{2}+2 c^{2}-a^{2}}, w_{a}=\frac{2 \sqrt{b c p(p-a)}}{b+c}$.
Let the equal circle line of $B C$ be $A K=\sqrt{p(p-a)}$.
First, determine: $m_{a} w_{a} \geqslant A K^{2}=p(p-a)$, which means to prove
$$
\begin{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,201 |
Example 3 Let the base length of isosceles $\triangle A B C$ be $a$, and the length of the legs be $b$. If $\angle A=20^{\circ}$, prove: $a^{3}+b^{3}=3 a b^{2}$. | Trigonometric Proof From the triple angle formula, we have
$$
\frac{1}{2}=\sin 30^{\circ}=3 \sin 10^{\circ}-4 \sin ^{3} 10^{\circ} \text {. }
$$
Substituting $\sin 10^{\circ}=\frac{a}{2 b}$ into the above equation and simplifying, we get
$$
a^{3}+b^{3}=3 a b^{2} \text {. }
$$
Geometric Proof As shown in Figure 3, con... | a^{3}+b^{3}=3 a b^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 728,202 |
Example 4 In $\triangle A B C$, it is known that
$$
\begin{array}{l}
\sin \left(\frac{B}{2}+C\right)|\sin A-\sin C|=\sin \frac{B}{2} \cdot \sin A, \\
a=2 b \cos C .
\end{array}
$$
Find the three interior angles of this triangle. | Triangle Solution From the given condition $a=2 b \cos C$, and the projection theorem, we have
$$
\begin{array}{l}
a=b \cos C+c \cos B=2 b \cos C \\
\Leftrightarrow b \cos C=c \cos B .
\end{array}
$$
By the sine rule, we get
$$
\sin B \cdot \cos C-\sin C \cdot \cos B=0 \Rightarrow \sin (B-C)=0 \text {. }
$$
Therefore... | \angle A=108^{\circ}, \angle B=\angle C=36^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,203 |
Example 3 Let $g(k)$ denote the greatest odd divisor of the positive integer $k$
$$
\begin{aligned}
\text { (for example, } g(3) & =3, g(20)=5) . \text { Find } \\
f(n) & =g(1)+g(2)+g(3)+\cdots+g\left(2^{n}\right) .
\end{aligned}
$$ | For odd $k$, we have $g(k)=k$.
For any $k \in \mathbf{Z}_{+}$, we have
$$
g(2 k)=g(k) \text {. }
$$
$$
\begin{array}{l}
\text { Then } f(n)=1+3+5+\cdots+\left(2^{n}-1\right)+ \\
g(2)+g(4)+\cdots+g\left(2^{n}\right) \\
=\frac{\left(1+2^{n}-1\right) 2^{n-1}}{2}+g(1)+g(2)+\cdots+g\left(2^{n-1}\right) \\
=4^{n-1}+f(n-1) .
... | \frac{4^{n}+2}{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,205 |
The second question As shown in Figure 1, in the acute triangle $\triangle ABC$, $\angle BAC \neq 60^{\circ}$. Through points $B$ and $C$, draw the tangents $BD$ and $CE$ to the circumcircle of $\triangle ABC$, and satisfy $BD = CE = BC$. Line $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respe... | Proof 1 As shown in Figure 2, draw $M P / / A G$, intersecting $A F$ at point $P$; draw $N Q / / A F$, intersecting $A G$ at point $Q$.
From $B D=C E=B C, \angle D B C=\angle E C B$, we get
$B C / / F G$
$\Rightarrow \angle D F B=\angle A B C$,
$\angle F D B=\angle D B C=\angle B A C$
$\Rightarrow \triangle D F B \back... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,206 |
Question 4: Let integers $x_{1}, x_{2}, \cdots, x_{2014}$ be pairwise incongruent modulo 2014, and integers $y_{1}, y_{2}, \cdots, y_{2014}$ be pairwise incongruent modulo 2014. Prove: The integers $y_{1}, y_{2}, \cdots, y_{2014}$ can be rearranged as $z_{1}, z_{2}, \cdots, z_{2014}$, such that $x_{1}+z_{1}, x_{2}+z_{2... | Prove that the original proposition is equivalent to:
There exist permutations $r_{1}, r_{2}, \cdots, r_{2014}$ and $z_{1}, z_{2}, \cdots, z_{2014}$ of $x_{1}, x_{2}, \cdots, x_{2014}$ and $y_{1}, y_{2}, \cdots, y_{2014}$ respectively, such that $r_{1}+z_{1}$, $r_{2}+z_{2}, \cdots, r_{2014}+z_{2014}$ are pairwise incon... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,208 |
Question 1 As shown in Figure $1, P A$ and $P B$ are
tangents to $\odot O$, point $C$ is on the minor arc $\overparen{A B}$ (different from points $A, B$), a line $l$ perpendicular to $P C$ is drawn through point $C$, intersecting the bisector of $\angle A O C$ at point $D$, and the bisector of $\angle B O C$ at point ... | 1 Proof As shown in Figure 1, connect $A C$.
Let $\angle A O D=\angle C O D=\alpha, \angle A P C=\beta$.
Since $P A$ is tangent to $\odot O$,
$$
\text { } O A \perp P A, \angle P A C=\frac{1}{2} \angle A O C=\alpha \text {. }
$$
Also, $P C \perp D E$, i.e., $\angle P C D=90^{\circ}$, then in quadrilateral $P A O C$,
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,209 |
Question 2 As shown in Figure 2, given that $M$ is the midpoint of side $BC$ of $\triangle ABC$, $\odot O$ passes through points $A$ and $C$ and is tangent to $AM$, the extension of $BA$ intersects $\odot O$ at point $D$, and line $CD$ intersects $MA$ at point $P$. Prove:
$$
PO \perp BC .{ }^{[2]}
$$ | Proof As shown in Figure 2, extend $AM$ to point $E$ such that $AM = ME$. Connect $OA, OM, OC, BE, CE$.
Since $M$ is the midpoint of $BC$ and $AM = ME$, quadrilateral $ABEC$ is a parallelogram.
$$
\begin{array}{l}
\Rightarrow AD \parallel EC \Rightarrow \frac{AE}{PA} = \frac{CD}{PD} \\
\Rightarrow AE = \frac{CD}{PD} \c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,210 |
Question 3 As shown in Figure 3, given that the circumcircle of acute triangle $\triangle ABC$ is $\odot O, AD$ is the diameter of $\odot O$, and the lines passing through points $B$ and $C$ and perpendicular to $BC$ intersect the extensions of $CA$ and $BA$ at points $E$ and $F$ respectively. Prove:
$$
\angle ADF = \a... | Proof: As shown in Figure 3, connect $O C$, $B D$, and $C D$.
Let the internal angles of $\triangle A B C$ be $\angle A, \angle B, \angle C$, and the radius of $\odot O$ be $R$.
Since $E B \perp C B$, we have
$$
E C=\frac{B C}{\cos C}=\frac{2 R \sin A}{\cos C},
$$
$E B=B C \tan C=2 R \sin A \cdot \tan C$.
In $\triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,211 |
Given $\triangle A B C$ with orthocenter $H$, circumradius and inradius are $R$ and $r$ respectively. Prove: equation (1) holds. | Proof of the lemma first.
Lemma: Let the distances from the circumcenter $O$ of $\triangle ABC$ to the sides $a, b, c$ be $d_{a}, d_{b}, d_{c}$, respectively. Then $d_{a}+d_{b}+d_{c}=R+r$.
Proof (1) If $\triangle ABC$ is an acute triangle, as shown in Figure 1, let the projections of point $O$ on the three sides be p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,212 |
2. Given the circumcircle $\Gamma$ of $\triangle A B C$, the midpoints of sides $A B$ and $A C$ are $M$ and $N$ respectively, and the midpoint of the arc $\overparen{B C}$ of circle $\Gamma$ not containing point $A$ is $T$. The circumcircles of $\triangle A M T$ and $\triangle A N T$ intersect the perpendicular bisecto... | 2. As shown in Figure 1, let the center of circle $\Gamma$ be $O$. Then $O$ is the intersection of $M Y$ and $N X$.
Let the perpendicular bisector of segment $A T$ be $l$. Then line $l$ passes through point $O$. Denote the reflection transformation about line $l$ as $r$.
Since $A T$ is the angle bisector of $\angle B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,213 |
3. In $\triangle A B C$, it is known that $D$ and $E$ are points on sides $C B$ and $C A$, respectively, and $A D$ and $B E$ bisect $\angle C A B$ and $\angle C B A$, respectively. A rhombus is inscribed in quadrilateral $A E D B$ (the four vertices of the rhombus lie on the four sides of quadrilateral $A E D B$), and ... | 3. As shown in Figure 2, let the four vertices of the rhombus be $K, L, M, N$ on the sides $A E, E D, D B, B A$ respectively.
Let $d(X, Y Z)$ denote the distance from point $X$ to line $Y Z$. Then,
$$
d(D, A B)=d(D, A C), \quad d(E, A B)=d(E, B C),
$$
and $d(D, B C)=d(E, A C)=0$.
This indicates that,
$$
\begin{array}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,214 |
4. In $\triangle A B C$, it is known that $\angle B>\angle C, P$ and $Q$ are two different points on the line $A C$, satisfying $\angle P B A=\angle Q B A=\angle A C B$, and point $A$ is between $P$ and $C$. Assume there exists a point $D$ on segment $B Q$ such that $P D=P B$, and ray $A D$ intersects the circumcircle ... | 4. As shown in Figure 3.
Let $\angle A C B=\alpha$.
Since $\alpha<\angle C B A$, we know $\alpha<90^{\circ}$.
Because $\angle P B A=\alpha$, it follows that line $P B$ is tangent to circle $\Gamma$ at point $B$.
Thus, $P A \cdot P C=P B^{2}=P D^{2}$.
From $\frac{P A}{P D}=\frac{P D}{P C}$, we have
$\triangle P A D \bac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,215 |
Example 4 The function $f$ is defined on the set of ordered pairs of positive integers, and satisfies
$$
\begin{array}{c}
f(x, x)=x, f(x, y)=f(y, x), \\
(x+y) f(x, y)=y f(x, x+y) .
\end{array}
$$
Calculate $f(14,52)$. | Since $f(x, x+y)=\frac{x+y}{y} f(x, y)$, we have,
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac{26}{19} f(14,14+24)=\frac{13}{6} f(14,24) \\
= & \frac{13}{6} f(14,14+10)=\frac{26}{5} f(14,10) \\
= & \frac{26}{5} f(10,14)=\frac{26}{5} f(10,10+4) \\
= & \frac{91}{5} f(10,4)=\frac{91}{5} f(... | 364 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,216 |
5. In a convex hexagon $A B C D E F$, it is known that
$$
A B=D E, B C=E F, C D=F A \text {, }
$$
and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$.
Prove: The diagonals $A D$, $B E$, and $C F$ intersect at one point. | 5. As shown in Figure 4.
Let $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$.
Without loss of generality, assume $\theta \geqslant 0$.
Let $A B=D E=x, C D=F A=y, B C=E F=z$.
Consider points $P, Q, R$ such that quadrilaterals $C D E P$, $E F A Q$, and $A B C R$ are all parallelograms.
$$
\begin{array}{l}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,217 |
1. If positive numbers $a, b$ satisfy
$$
2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b) \text {, }
$$
then $\frac{1}{a}+\frac{1}{b}=$ $\qquad$ . | $-, 1.108$
Let $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$. Then
$$
\begin{array}{l}
a=2^{k-2}, b=3^{k-3}, a+b=6^{k} . \\
\text { Therefore } \frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}=\frac{6^{k}}{2^{k-2} \times 3^{k-3}} \\
=2^{2} \times 3^{3}=108 .
\end{array}
$$ | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,218 |
2. Let the set $\left\{\left.\frac{3}{a}+b \right\rvert\, 1 \leqslant a \leqslant b \leqslant 2\right\}$ have its maximum and minimum elements as $M$ and $m$, respectively. Then the value of $M-m$ is $\qquad$. | $2.5-2 \sqrt{3}$.
From $1 \leqslant a \leqslant b \leqslant 2$, we know
$$
\frac{3}{a}+b \leqslant \frac{3}{1}+2=5 \text {. }
$$
When $a=1, b=2$, the maximum value $M=5$ is obtained.
$$
\text { Also, } \frac{3}{a}+b \geqslant \frac{3}{a}+a \geqslant 2 \sqrt{\frac{3}{a} \cdot a}=2 \sqrt{3} \text {, when } a=b
$$
$=\sq... | 5-2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,219 |
3. If the function $f(x)=x^{2}+a|x-1|$ is monotonically increasing on $[0,+\infty)$, then the range of the real number $a$ is $\qquad$ . | 3. $[-2,0]$.
On the interval $[1,+\infty)$, $f(x)=x^{2}+a x-a$ is monotonically increasing, which is equivalent to $-\frac{a}{2} \leqslant 1 \Rightarrow a \geqslant-2$.
On the interval $[0,1]$, $f(x)=x^{2}-a x+a$ is monotonically increasing, which is equivalent to $\frac{a}{2} \leqslant 0 \Rightarrow a \leqslant 0$.
... | [-2,0] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,220 |
4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, a_{n+1}=\frac{2(n+2)}{n+1} a_{n}\left(n \in \mathbf{Z}_{+}\right) . \\
\text { then } \frac{a_{2014}}{a_{1}+a_{2}+\cdots+a_{2013}}=\text {. }
\end{array}
$$ | 4. $\frac{2015}{2013}$.
From the problem, we know
$$
\begin{array}{l}
a_{n}=\frac{2(n+1)}{n} a_{n-1}=\frac{2(n+1)}{n} \cdot \frac{2 n}{n-1} a_{n-2}=\cdots \\
=\frac{2(n+1)}{n} \cdot \frac{2 n}{n-1} \cdots \cdots \frac{2 \times 3}{2} a_{1} \\
=2^{n-1}(n+1) .
\end{array}
$$
Let the sum of the first $n$ terms of the seq... | \frac{2015}{2013} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,221 |
5. In a regular quadrilateral pyramid $P-ABCD$, it is known that the side faces are equilateral triangles with a side length of 1, and $M, N$ are the midpoints of edges $AB, BC$ respectively. Then the distance between the skew lines $MN$ and $PC$ is $\qquad$. | 5. $\frac{\sqrt{2}}{4}$.
As shown in Figure 2, let the diagonals $A C$ and $B D$ of the base intersect at point $O$. Draw a perpendicular from point $C$ to line $M N$, intersecting $M N$ at point $H$. Since $P O$ is the perpendicular to the base, $P O \perp C H$.
Also, $A C \perp C H$, thus $C H$ is perpendicular to t... | \frac{\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,222 |
6. Let the two foci of the ellipse $\Gamma$ be $F_{1}$ and $F_{2}$, and a line passing through point $F_{1}$ intersects the ellipse $\Gamma$ at points $P$ and $Q$. If $\left|P F_{2}\right|=\left|F_{1} F_{2}\right|$, and $3\left|P F_{1}\right|=4\left|Q F_{1}\right|$, then the ratio of the minor axis to the major axis of... | 6. $\frac{2 \sqrt{6}}{7}$.
Assume $\left|P F_{1}\right|=4,\left|Q F_{1}\right|=3$.
Let the lengths of the major axis and minor axis of the ellipse $\Gamma$ be $2 a$ and $2 b$, respectively, and the focal distance be $2 c$.
Then $\left|P F_{2}\right|=\left|F_{1} F_{2}\right|=2 c$, and by the definition of an ellipse, $... | \frac{2 \sqrt{6}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,223 |
7. Let the incircle radius of equilateral $\triangle A B C$ be 2, and the incenter be
I. If point $P$ satisfies $P I=1$, then the maximum value of the ratio of the areas of $\triangle A P B$ to $\triangle A P C$ is $\qquad$. | 7. $\frac{3+\sqrt{5}}{2}$.
As shown in Figure 4, given $P I=1$, we know that point $P$ lies on the unit circle $\odot I$.
Let $\angle B A P=\alpha$. On $\odot I$, take a point $P_{0}$ such that $\alpha$ reaches its maximum value $\alpha_{0}$. At this time, point $P_{0}$ should fall within $\angle I A C$, and it is th... | \frac{3+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,224 |
8. Let $A, B, C, D$ be four points in space that are not coplanar. With a probability of $\frac{1}{2}$, connect an edge between each pair of points, and whether any two pairs of points are connected is independent. Then the probability that points $A$ and $B$ can be connected by (a single edge or a sequence of edges fo... | 8. $\frac{3}{4}$.
There are 2 possibilities for whether each pair of points is connected by an edge, resulting in a total of $2^{6}=$ 64 scenarios. Consider the number of scenarios where points $A$ and $B$ can be connected by a polyline.
(1) There is an edge $A B$: There are $2^{5}=32$ scenarios.
(2) There is no edge ... | \frac{3}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,225 |
9. (16 points) In the plane rectangular coordinate system $x O y$, $P$ is a moving point not on the $x$-axis, and it satisfies that two tangent lines to the parabola $y^{2}=4 x$ can be drawn through point $P$. The line connecting the two points of tangency $l_{P}$ is perpendicular to $P O$, and the intersection points ... | (1) Let point $P(a, b)(b \neq 0)$. It is easy to know that $a \neq 0$.
As shown in Figure 5, let the points of tangency be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $l_{P A}: y y_{1}=2\left(x+x_{1}\right)$,
$l_{P B}: y y_{2}=2\left(x+x_{2}\right)$.
Since point $P(a, b)$ satisfies equations (1) and ... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,226 |
Example: Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x, y \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$.
(1) Prove: $g(x)$ is a periodic function;
(2) If $f(998)=1002$, find the value of $f(2000)$. | (1) Proof: From $g(x)=f(x)-x$, we get
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Substituting into the inequality in the problem, we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
From equation (1), we get
$$
\begin{... | 2004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,227 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the positive integer $m$ such that
$$
\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}=\frac{1}{100} .
$$ | 10. From the problem, we know that for any positive integer $n$,
$$
a_{n+1} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),
$$
and $\tan a_{n+1}=\sec a_{n}$.
Since $\sec a_{n}>0$, then $a_{n+1} \in\left(0, \frac{\pi}{2}\right)$.
From equation (1), we get $\tan ^{2} a_{n+1}=\sec ^{2} a_{n}=1+\tan ^{2} a_{n}$.
Thus, $\ta... | 3333 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,228 |
11. (20 points) Determine all complex numbers $\alpha$ such that for any complex numbers $z_{1} , z_{2}\left(\left|z_{1}\right| , \left|z_{2}\right|<1, z_{1} \neq z_{2}\right)$, we have
$$
\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}} \neq\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}} .
$$ | 11. Let $f_{\alpha}(z)=(z+\alpha)^{2}+\alpha \bar{z}$. Then
$$
\begin{array}{l}
f_{\alpha}\left(z_{1}\right)-f_{\alpha}\left(z_{2}\right)=\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}}-\left(z_{2}+\alpha\right)^{2}-\alpha \overline{z_{2}} \\
\quad=\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\... | \{\alpha|\alpha \in \mathbf{C},| \alpha \mid \geqslant 2\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,229 |
II. (40 points) As shown in Figure 1, in the acute triangle $\triangle ABC$, $\angle BAC \neq 60^{\circ}$. Through points $B$ and $C$, draw the tangents $BD$ and $CE$ to the circumcircle of $\triangle ABC$, such that $BD = CE = BC$. Line $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respectivel... | Proof 1 As shown in Figure 6, let the two tangents $B D$ and $C E$ intersect at point $K$. Then $B K=C K$.
Figure 6
Combining $B D=C E$, we know $D E \parallel B C$.
Construct the angle bisector $A L$ of $\angle B A C$ intersecting $B C$ at point $L$, and connect $L M$ and $L N$.
By $D E \parallel B C$, we have
$$
\beg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,231 |
Three. (50 points) Let $S=\{1,2, \cdots, 100\}$. Find the largest integer $k$, such that the set $S$ has $k$ distinct non-empty subsets with the property: for any two different subsets among these $k$ subsets, if their intersection is non-empty, then the smallest element in their intersection is different from the larg... | For a finite non-empty set of real numbers $A$, use $\min A$ and $\max A$ to denote the smallest and largest elements of set $A$, respectively.
Consider all subsets of set $S$ that contain 1 and have at least two elements.
Notice that, $\min \left(A_{i} \cap A_{j}\right)=1<\max A_{i}$.
Therefore, $k_{\max } \geqslant ... | 2^{99}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,232 |
1. As shown in Figure $1, \odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. Extend $O_{1} A$ to intersect $\odot O_{2}$ at point $C$, and extend $O_{2} A$ to intersect $\odot O_{1}$ at point $D$. Draw $B E / / O_{2} A$, intersecting $\odot O_{1}$ at another point $E$. If $D E / / O_{1} A$, prove: $D C \pe... | 1. Auxiliary lines as shown in Figure 3.
$$
\begin{array}{l}
\text { Given } \angle O_{1} D A=\angle O_{1} A D \\
=\angle O_{2} A C=\angle O_{2} C A,
\end{array}
$$
we know that $O_{1} 、 O_{2} 、 C 、 D$ are concyclic.
Thus, $\angle D O_{1} O_{2}+\angle D C O_{2}=180^{\circ}$.
Since $B E / / A D$, we have $A B=D E$.
The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,234 |
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Suppose $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}$, and $\left[x_{1}\right],\left[x_{2}\right], \cdots,\left[x_{n}\right]$ is a permutation of 1, 2, ..., n, where $n \geqslant 2$ is a given integer. Find the maximum and minimum values of $\su... | 2. Since $\left[x_{1}\right],\left[x_{2}\right], \cdots,\left[x_{n}\right]$ is a permutation of $1,2, \cdots, n$, for any $i=1,2, \cdots, n$, we have $1 \leqslant x_{i}x_{i+1}-x_{i}-1$, thus
$$
\begin{array}{l}
S>\sum_{i=1}^{n-1}\left(x_{i+1}-x_{i}-1\right)=x_{n}-x_{1}-(n-1) \\
>-(2 n-1) .
\end{array}
$$
Since $S$ is ... | n-1 \text{ and } 2-2n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,235 |
3. Among $n$ students, each student knows exactly $d$ boys and $d$ girls (knowing is mutual). Find all possible pairs of positive integers $(n, d)$.
(Wang Xinmao) | 3. With these $n$ students as vertices and the acquaintance relationships as edges, construct a simple graph $G=(V, E)$. Let there be $x$ girls and $y$ boys, with $x, y>d$.
Then the number of edges between boys and girls is
$$
x d=y d \Rightarrow x=y=\frac{n}{2} \geqslant d+1 \text {; }
$$
The number of edges between ... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,236 |
4. For an integer $m \geqslant 4$, define $T_{m}$ as the number of sequences $a_{1}, a_{2}, \cdots, a_{m}$ that satisfy the following conditions:
(1) For each $i=1,2, \cdots, m, a_{i} \in\{1,2,3,4\}$;
(2) $a_{1}=a_{m}=1, a_{2} \neq 1$;
(3) For each $i=3,4, \cdots, m, a_{i} \neq a_{i-1}, a_{i} \neq a_{i-2}$.
Prove: Ther... | 4. By enumeration, we know $T_{4}=T_{5}=T_{6}=6$.
For $n \geqslant 7$, we calculate the number of positive integer sequences $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ that satisfy the conditions in two cases.
【Case 1】If $a_{2}, a_{3}, \cdots, a_{n-1}$ are all not equal to 1, then sequentially, element $a_{2}$ has ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,237 |
Example 6 Let $f: \mathbf{R} \rightarrow \mathbf{R}$, and for any $x, y \in \mathbf{R}$ we have
$$
\begin{array}{l}
f(x+y)+f(x-y)+f(2 x) \\
=4 f(x) f\left(\frac{x+y}{2}\right) f\left(\frac{y-x}{2}\right)-1 .
\end{array}
$$
Given $f(1)=0$.
(1) Prove: $f(x)$ is an even periodic function;
(2) Find $f(2005)$ and $f(2006)$... | (1) Proof For equation (1), let $x=y=0$, we get
$f(0)=1$ or $f(0)=-\frac{1}{2}$.
Let $x=y$, then
$2 f(2 x)+f(0)=4 f(0) f^{2}(x)-1$.
If $f(0)=-\frac{1}{2}$, let $x=\frac{1}{2}$, then
$$
f(2 x)=-f^{2}(x)-\frac{1}{4}<0 \Rightarrow f(1)<0,
$$
which is a contradiction.
Therefore, it can only be $f(0)=1$.
In equation (1), l... | f(2005)=0, f(2006)=-1 | Algebra | proof | Yes | Yes | cn_contest | false | 728,238 |
5. Let $a$ be a positive integer that is not a perfect square, and let $r$ be a real root of the equation $x^{3}-2 a x+1=0$. Prove that $r+\sqrt{a}$ is an irrational number.
(Li Shenghong) | 5. Suppose $r+\sqrt{a}$ is a rational number.
Let $y=r+\sqrt{a}$. Then $r=y-\sqrt{a}$.
Substitute into the given equation:
$$
\begin{array}{l}
(y-\sqrt{a})^{3}-2 a(y-\sqrt{a})+1=0 \\
\quad \Rightarrow y^{3}+a y+1-\left(3 y^{2}-a\right) \sqrt{a}=0 .
\end{array}
$$
Since $a$ is not a perfect square, i.e., $\sqrt{a}$ is... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,239 |
6. As shown in Figure 2, in acute triangle $\triangle ABC$, $AB > AC$, and $D$, $E$ are the midpoints of sides $AB$, $AC$ respectively. The circumcircle of $\triangle ADE$ intersects the circumcircle of $\triangle BCE$ at point $P$ (other than point $E$), and the circumcircle of $\triangle ADE$ intersects the circumcir... | 6. As shown in Figure 4, connect $DE$, $PD$, $QE$, $PB$, $QC$, $PE$, and $QD$. Let $QD$ intersect $AC$ at point $R$.
By the inscribed angle theorem, we have
$$
\begin{array}{l}
\angle APD = \pi - \angle AED = \pi - \angle ACB, \\
\angle BPD = \angle BPE - \angle EPD \\
= (\pi - \angle ACB) - \angle BAC = \angle ABC, \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,240 |
7. For a finite non-empty set of real numbers $X$, denote the number of elements in $X$ as $|X|$, then
$$
f(X)=\frac{1}{|X|} \sum_{a \in X} a .
$$
The pair of sets $(A, B)$ satisfies
$$
A \cup B=\{1,2, \cdots, 100\}, A \cap B=\varnothing \text {, }
$$
and $1 \leqslant|A| \leqslant 98$.
For any $p \in B$, let
$$
A_{p}... | 7. Let $S=\left(f\left(A_{p}\right)-f(A)\right)\left(f\left(B_{p}\right)-f(B)\right)$.
From the definition of $f$ and the conditions given in the problem, we have
$$
\begin{array}{l}
(|A|+1) f\left(A_{p}\right)=|A| f(A)+p, \\
(|B|-1) f\left(B_{p}\right)=|B| f(B)-p . \\
\text { Then } S=\left(\frac{|A| f(A)+p}{|A|+1}-f... | \frac{625}{196} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,241 |
8. Let $n$ be a positive integer, and $S$ be the set of all numbers in $\{1,2, \cdots, n\}$ that are coprime with $n$. Denote
$$
\begin{array}{l}
S_{1}=S \cap\left(0, \frac{n}{3}\right], S_{2}=S \cap\left(\frac{n}{3}, \frac{2 n}{3}\right), \\
S_{3}=S \cap\left(\frac{2 n}{3}, n\right] .
\end{array}
$$
If the number of ... | 8. Let $|X|$ denote the number of elements in a finite set $X$.
For each positive integer $n$, define $A(n)$ as the set of all integers that are coprime with $n$, and for each integer $k$, define
$$
A_{k}(n)=A(n) \cap\left(\frac{k-1}{3} n, \frac{k}{3} n\right] \text {. }
$$
Then for any integer $x$, we have
$$
\begin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,242 |
1. Let $x, y$ be positive real numbers. Find the minimum value of
$$
x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}
$$
(Gu Bin, Jin Aiguo) | 1. Let $f(x, y)=x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.
If $x \geqslant 1, y \geqslant 1$, then
$$
f(x, y) \geqslant x+y \geqslant 2 \text {; }
$$
If $0<x<1, 0<y<1$, then
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{1-y}{x} = \left(x + \frac{1-x}{y}\right) + \left(y + \frac{1-y}{x}\right) \geqslant 2.
$$
For $0<x<1, y \... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,243 |
2. As shown in Figure 1, given that $A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two points on the arc $\overparen{A B}$, $P, Q$ are the circumcenters of $\triangle O A C$ and $\triangle O B D$ respectively. Prove: $C P \cdot C Q = D P \cdot D Q$. | 2. Connect $O P, O Q, A P, A D, B Q, B C$.
Let $\angle B A D=\alpha, \angle A B C=\beta$.
From the problem, we know
$$
\begin{array}{l}
\angle O A P=\angle A O P=\frac{1}{2} \angle A O C=\angle A B C=\beta, \\
\angle O B Q=\angle B O Q=\frac{1}{2} \angle B O D=\angle B A D=\alpha .
\end{array}
$$
Thus, $\angle P A D=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,244 |
3. Let the sequence of sets $A_{1}, A_{2}, \cdots$ satisfy that for any positive integer $j$, there are only finitely many positive integers $i$ such that $A_{i} \subseteq A_{j}$. Prove: There exists a sequence of positive integers $a_{1}, a_{2}, \cdots$ such that for any positive integers $i, j$, $a_{i} \mid a_{j}$ if... | 3. Let $p_{1}, p_{2}, \cdots$ be all the prime numbers arranged in ascending order.
For $i \in \mathbf{Z}_{+}$, let $S_{i}=\left\{j \in \mathbf{Z}_{+} \mid A_{j} \subseteq A_{i}\right\}$.
By the given condition, $S_{i}$ is a finite set, and $i \in S_{i}$.
Let $a_{i}=\prod_{j \in S_{i}} p_{j}$.
Next, we prove that the ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,245 |
4. Given a positive integer $n$, let $a_{1}, a_{2}, \cdots, a_{n}$ be a sequence of non-negative integers. If the arithmetic mean of some consecutive terms (which can be a single term) is not less than 1, then these terms are called a "dragon", where the first term is called the "dragon head" and the last term is calle... | 4. The minimum value of $\sum_{i=1}^{n} a_{i}$ is $\left[\frac{n}{2}\right]+1$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
First, provide a construction:
When $n=2 k-1$, let $a_{k}=k$, and all other terms be 0;
When $n=2 k$, let $a_{k}=k, a_{2 k}=1$, and all other terms be 0.
It is easy... | \left[\frac{n}{2}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,246 |
6. Given an integer $n \geqslant 2$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy:
(1) $\sum_{i=1}^{n} x_{i}=0$;
(2) $\left|x_{i}\right| \leqslant 1(i=1,2, \cdots, n)$.
Find the maximum value of $\min \left|x_{i}-x_{i+1}\right|$. | 6. Let $A=\min _{1 \leqslant i \leqslant n-1}\left|x_{i}-x_{i+1}\right|$.
(i) When $n$ is even, from condition (2) we know that for $1 \leqslant i \leqslant n-1$, we have
$$
\left|x_{i}-x_{i+1}\right| \leqslant\left|x_{i}\right|+\left|x_{i+1}\right| \leqslant 2 \text {. }
$$
Thus, $A \leqslant 2$.
Taking $x_{i}=(-1)^{... | \frac{2 n}{n+1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,248 |
Example 7 Let the function $f(x)$ be defined for all $x>0$ and satisfy:
(i) $f(x)$ is strictly increasing on the interval $(0,+\infty)$;
(ii) For all $x>0$, $f(x)>-\frac{1}{x}$;
(iii) For all $x>0$, $f(x) f\left(f(x)+\frac{1}{x}\right)=1$.
(1) Find $f(1)$.
(2) Find a function $f(x)$ that satisfies the above conditions. | (1) Let $f(1)=a$. Then by (iii), setting $x=1$, we get
$$
a f(a+1)=1 \Rightarrow f(a+1)=\frac{1}{a} \text {. }
$$
Now let $x=a+1$, we have
$$
\begin{array}{l}
f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1 \\
\Rightarrow f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a \\
\Rightarrow f\left(\frac{1}{a}+\frac{1}{a+1}\right)=f(1)... | f(x)=\frac{1-\sqrt{5}}{2 x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,249 |
7. In the plane, given that $O$ is the center of the equilateral $\triangle ABC$, and points $P, Q$ satisfy $\overrightarrow{O Q}=2 \overrightarrow{P O}$. Prove:
$$
P A+P B+P C \leqslant Q A+Q B+Q C \text {. }
$$ | 7. Let the midpoints of $BC$, $CA$, and $AB$ be $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Since $\triangle ABC$ and $\triangle A_{1}B_{1}C_{1}$ are homothetic with respect to point $O$ with a homothety ratio of $-\frac{1}{2}$, under this transformation, $P \rightarrow Q$.
Thus, $QA + QB + QC = 2(PA_{1} + PB_{1} + PC... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,250 |
8. Given a real number $q$ satisfying $1<q<2$. Let the binary representation of a positive integer $n$ be
$$
n=a_{0}+a_{1} 2+a_{2} 2^{2}+\cdots+a_{k} 2^{k},
$$
where $a_{i} \in\{0,1\}, i=0,1, \cdots, k$.
Define the sequence $\left\{x_{n}\right\}$ such that
$$
x_{n}=a_{0}+a_{1} q+a_{2} q^{2}+\cdots+a_{k} q^{k} \text {.... | 8. From $x_{2^{k}}=q^{k}$, we know that the sequence $\left\{x_{n}\right\}$ has no upper bound.
For any positive integer $n$, let $m$ be the smallest positive integer such that $x_{m}>x_{n}$.
Next, we prove: $x_{m} \leqslant x_{n}+1$.
Clearly, $m>1$.
We discuss two cases.
(1) If $m-1$ is even, let
$$
m-1=a_{1} 2^{1}+a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,251 |
In $\triangle A B C$, as shown in Figure 1, it is known that $\angle B, \angle C$ are acute angles, $A D \perp B C, D E \perp A C, M$ is the midpoint of $D E$. If $A M \perp B E$ at point $F$, prove: $\triangle A B C$ is an isosceles triangle. | Given $A D \perp B C, A M \perp B E$, we know that points $A, B, D, F$ are concyclic.
Connecting $D F$, we have $\angle A B D=\angle D F M$.
In the right triangle $\triangle A M E$, by the projection theorem, we know
$M E^{2}=M F \cdot A M$.
Since $D M=M E$, it follows that
$D M^{2}=M F \cdot A M$.
Therefore, $\triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,252 |
$$
\begin{array}{l}
a_{1}=1,\left(n^{2}+1\right) a_{n-1}^{2}=(n-1)^{2} a_{n}^{2}(n \geqslant 2) . \\
\text { Prove: } \frac{1}{a_{1}}+\frac{1}{2 a_{2}}+\cdots+\frac{1}{n a_{n}} \leqslant 1+\sqrt{1-\frac{n^{2}}{a_{n}^{2}}} .
\end{array}
$$ | Because $a_{1}=1$, so, $\frac{1}{a_{1}}=1$.
By Cauchy-Schwarz inequality we have
$$
\begin{array}{l}
\left(\frac{1}{2 a_{2}}+\frac{1}{3 a_{3}}+\cdots+\frac{1}{n a_{n}}\right)^{2} \\
\leqslant\left(\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}\right)\left(\frac{1}{a_{2}^{2}}+\frac{1}{a_{3}^{2}}+\cdots+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,253 |
Three, does there exist infinitely many distinct positive integers $x, y$ such that $\left(x+y^{2}\right) \mid\left(x^{2}+y\right)$? Prove your conclusion. | For any $w \in \mathbf{Z}_{+}$, let $x=w^{2}(w+1), y=w+1$. Then
\[
\begin{array}{l}
x+y^{2}=w^{3}+2 w^{2}+2 w+1, \\
x^{2}+y=w^{6}+2 w^{5}+w^{4}+w+1 \\
=\left(w^{3}+2 w^{2}+2 w+1\right)\left(w^{3}-w+1\right) .
\end{array}
\]
Thus, $\left(x+y^{2}\right) \mid\left(x^{2}+y\right)$.
Therefore, there exist infinitely many d... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,254 |
Four, in an election, there are 12 candidates, and each member of the electoral committee casts 6 votes. It is known that any two members' votes have at most 2 candidates in common. Find the maximum number of members in the committee.
(Proposed by the Problem Committee) | The maximum number of committee members is 4.
Let the number of committee members be $k$, and the candidates be represented by $1,2, \cdots, 12$. Each person's vote is a set $A_{i}(1 \leqslant i \leqslant k)$, and the number of votes each candidate receives is $m_{i}(1 \leqslant i \leqslant 12)$. Then,
$$
\sum_{i=1}^{1... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,255 |
Five, As shown in Figure 2, in $\square A B C D$, it is known that $I$ is the incenter of $\triangle B C D$, and $H$ is the orthocenter of $\triangle I B D$. Prove:
$$
\angle H A B=\angle H A D .
$$ | Five, as shown in Figure 3, connect $D H$ and $B H$, extend $D I$, and let it intersect lines $B H$ and $A B$ at points $F$ and $E$ respectively.
Since $I$ is the incenter of $\triangle B C D$ and $A E / / D C$, we have $\angle B D E = \angle C D E = \angle A E D$.
Also, since $H$ is the orthocenter of $\triangle B D I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,256 |
Six, let $x, y, z, w \in \mathbf{R}$, and $x+2 y+3 z+4 w=1$. Find the minimum value of $s=x^{2}+y^{2}+z^{2}+w^{2}+(x+y+z+w)^{2}$.
(Weng Shiyou) | Six, notice that,
$$
1=2(x+y+z+w)-x+0 y+z+2 w \text {. }
$$
Then $1^{2} \leqslant\left[2^{2}+(-1)^{2}+0^{2}+1^{2}+2^{2}\right]$.
$$
\left[(x+y+z+w)^{2}+x^{2}+y^{2}+z^{2}+w^{2}\right] \text {. }
$$
Thus, $s \geqslant \frac{1^{2}}{2^{2}+(-1)^{2}+0^{2}+1^{2}+2^{2}}=\frac{1}{10}$.
When $y=0$, and $\frac{x+y+z+w}{2}=-x=z=... | \frac{1}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,257 |
Seven, Prove: There exist infinitely many positive integers $n$, such that $3^{n}+2$ and $5^{n}+2$ are both composite.
(Jiao Heping) | Seven, notice that $3^{n}+2=3\left(3^{n-1}-1\right)+5$.
By Fermat's Little Theorem, $3^{4} \equiv 1(\bmod 5)$.
Thus, let $n-1=4 r\left(r \in \mathbf{Z}_{+}\right)$.
Then $51\left(3^{n}+2\right)$.
When $n=4 r+1$, $3^{n}+2$ is composite.
Also, $5^{n}+2=5\left(5^{n-1}-1\right)+7$, then by Fermat's Little Theorem, $5^{6} \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,258 |
Eight, A and B take turns to blow up the same balloon. The balloon will burst if and only if the volume of gas inside $V>2014$ (unit: milliliters). A starts by blowing in 1 milliliter of gas, and it is agreed that the volume of gas blown in each time is not less than the volume of gas blown in the previous time, nor mo... | Let the volume of the balloon be denoted as $A$.
First, prove the conclusion:
When $A \in\left[2^{k+1}-3,2^{k+1}-2\right)\left(k \in \mathbf{Z}_{+}\right)$, player A has a winning strategy;
When $A \in\left[2^{k+1}-2,2^{k+2}-3\right)\left(k \in \mathbf{Z}_{+}\right)$, player B has a winning strategy.
Use mathematical ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 728,259 |
Example 8 A function $f(x)$ defined on $\mathbf{R}$ satisfies
$$
\begin{array}{l}
f(x)+f(1-x)=1, \\
f\left(\frac{x}{3}\right)=\frac{1}{2} f(x)(x \geqslant 0),
\end{array}
$$ | $$
f\left(x_{1}\right) \leqslant f\left(x_{2}\right)\left(0 \leqslant x_{1}<x_{2} \leqslant 1\right) .
$$
Find the value of $f\left(\frac{1}{2011}\right)$.
【Analysis】This problem requires some observation and patient calculation.
Solution In equations (2) and (1), let $x=0$, we get
$$
f(0)=0, f(1)=1 \text {. }
$$
In ... | \frac{1}{128} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,260 |
1. Does there exist a positive real number $a$ such that the inequality
$$
|\cos x|+|\cos a x|>\sin x+\sin a x
$$
always holds? | 1. Does not exist.
When $01$, let $a x=t, b=\frac{1}{a}$. Then the original inequality becomes $|\cos b t|+|\cos t|>\sin b t+\sin t$.
This reduces to the previous case.
Therefore, there does not exist a positive real number $a$. | proof | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,261 |
2. Betya and Vasya are playing a game on an $n \times n$ chessboard. At the start, all squares are white except for one corner square which is black and contains a "rook". Betya starts, and the two players take turns moving the rook several squares horizontally or vertically. The squares the rook passes over and lands ... | 2. Betya has a winning strategy.
Betya's strategy: At each move, move the rook vertically to the farthest possible square.
Below is an explanation of how this strategy can lead to victory.
If at some point the rook is in the same connected region of white squares, then the white squares at that moment are called "reac... | Betya has a winning strategy | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,262 |
3. Rational numbers $a, b$ have decimal expansions that are repeating decimals with the smallest period of 30. It is known that the decimal expansions of $a-b$ and $a+kb$ have the smallest period of 15. Find the smallest possible value of the positive integer $k$.
| 3. $k_{\min }=6$.
Notice that, $a, b, a-b, a+k b$ can simultaneously become pure repeating decimals by multiplying them by a power of 10. Therefore, assume they are all pure repeating decimals.
Since the decimal part of a rational number is a pure repeating decimal with a period of $T$ if and only if it can be writte... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,263 |
5. If for any positive divisor $a$ of a positive integer $n$ we have $(a+1) \mid (n+1)$, then $n$ is called a "good number". Find all good numbers. | 5. Good numbers are 1 and all odd primes.
Let $n$ be a good number.
Since $1 \mid n$, and $(1+1) \mid (n+1)$, therefore, $n$ is an odd number.
Obviously, any odd prime is a good number.
If there is an odd composite number $n = ab$ that is a good number, without loss of generality, assume $3 \leqslant b \leqslant a$.
Th... | 1 \text{ and all odd primes} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,264 |
6. The sphere $\Gamma$ passing through the vertex $S$ of the tetrahedron $SABC$ intersects the edges $SA$, $SB$, and $SC$ at points $A_1$, $B_1$, and $C_1$, respectively. The intersection of the sphere $\Gamma$ and the circumsphere $\Gamma_1$ of the tetrahedron $SABC$ (a circle) lies on a plane parallel to the base $AB... | 6. It suffices to prove that
$S A_{2} \cdot S A=S B_{2} \cdot S B=S C_{2} \cdot S C$.
From $A A_{1}=S A_{2}, B B_{1}=S B_{2}, C C_{1}=S C_{2}$, we know
Equation (1) $\Leftrightarrow A A_{1} \cdot A S=B B_{1} \cdot B S=C C_{1} \cdot C S$.
Let $l$ be the line connecting the center of sphere $\Gamma$ and the center of the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,265 |
7. Initially, two polynomials are written on the blackboard:
$$
x^{3}-3 x^{2}+5 、 x^{2}-4 x \text {. }
$$
If there are polynomials $f(x)$ and $g(x)$ on the blackboard, then the following can be written on the blackboard:
$$
\begin{array}{l}
f(x) \pm g(x) 、 f(x) g(x) 、 \\
f(g(x)) 、 c f(x)(c \in \mathbf{R}) .
\end{array... | 7. No.
For polynomials $f(x)$ and $g(x)$, suppose there exists $x_{0}$ such that $f^{\prime}\left(x_{0}\right)=g^{\prime}\left(x_{0}\right)=0$.
Then $(f \pm g)^{\prime}\left(x_{0}\right)=0, c f^{\prime}\left(x_{0}\right)=0$, $(f g)^{\prime}\left(x_{0}\right)=f\left(x_{0}\right) g^{\prime}\left(x_{0}\right)+f^{\prime}\... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,266 |
8. Two people play a card game with a total of $n$ cards, where a "greater than" relationship is defined between any two cards (this relationship is not necessarily transitive). Initially, all cards are arbitrarily divided into two piles, one for each player. In each round, both players reveal the top card of their pil... | 8. Suppose we divide $n$ cards into two piles belonging to two people (the cards in each pile are in order from top to bottom, and one pile can be an empty set), there are $N$ ways to do this.
Next, construct a directed graph with $N$ vertices.
If a state $A$ can be transformed into state $B$ with one operation, there ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,267 |
1. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=(\quad)$.
(A) $\frac{3}{2}$
(B) $\frac{1}{2}$
(C) 1
(D) 2 | - 1. C.
From the given, we have
$$
\sqrt{x}-\sqrt{y}=\sqrt{2-\sqrt{3}},
$$
which means $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Squaring both sides, we get
$$
x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}.
$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,268 |
2. Use several $2 \times 3 \times 5$ small rectangular prisms to form a cube with an edge length of 90 by aligning them in the same direction. Then the number of small rectangular prisms that a diagonal of the cube passes through is ( ).
(A) 66
(B) 22
(C) 68
(D) 24 | 2. A.
Notice that, $[2,3,5]=30$, where $[x, y, z]$ represents the least common multiple of the positive integers $x, y, z$.
Therefore, we only need to find the number of small rectangular prisms a diagonal of a cube with edge length 30 passes through.
By the principle of inclusion-exclusion, we have
$$
\begin{array}{... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,269 |
3. Given $0 \leqslant a, b, c \leqslant 1$. Then
$$
\frac{a}{b c+1}+\frac{b}{c a+1}+\frac{c}{a b+1}
$$
The maximum value is ( ).
(A) 1
(B) 2
(C) 3
(D) $\frac{3}{2}$ | 3. B.
When $a+b+c=0$, we have $a=b=c=0$, then the original expression $=0$.
When $a+b+c>0$, by
$$
a+b+c \leqslant 1+b+c \leqslant 2(1+b c),
$$
we know $\frac{a}{b c+1} \leqslant \frac{2 a}{a+b+c}$.
Similarly, $\frac{b}{c a+1} \leqslant \frac{2 b}{a+b+c}, \frac{c}{a b+1} \leqslant \frac{2 c}{a+b+c}$.
Adding the three ... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 728,270 |
Example 9 Let $f(n)$ be a function defined on $\mathbf{N}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}$ we have
$$
f(m+n)-f(m)-f(n)=0 \text{ or } 1 \text{, }
$$
and $f(2)=0, f(3)>0, f(6000)=2000$.
Find $f(5961)$. | Solve: From $0=f(2) \geqslant 2 f(1) \Rightarrow f(1)=0$;
From $f(3)-f(2)-f(1)=0$ or 1
$$
\Rightarrow 0 \leqslant f(3) \leqslant 1 \text {. }
$$
But $f(3)>0$, hence $f(3)=1$.
By the problem statement, we know
$$
\begin{array}{l}
f(3 n+3)=f(3 n)+3+0 \text { or } 1 \\
\Rightarrow f(3(n+1)) \geqslant f(3 n)+1 .
\end{arra... | 1987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,271 |
4. The equation $3^{x}+4^{x}+5^{x}=6^{x}$ has $(\quad)$ solutions.
(A) 3
(B) 2
(C) 1
(D) 0 | 4.C.
It is easy to know that $x=3$ is a solution to the original equation.
When $x>3$, we have
$$
\begin{array}{l}
\left(\frac{3}{6}\right)^{x}>\left(\frac{3}{6}\right)^{3},\left(\frac{4}{6}\right)^{x}>\left(\frac{4}{6}\right)^{3},\left(\frac{5}{6}\right)^{x}>\left(\frac{5}{6}\right)^{3} . \\
\text { Then }\left(\frac... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,272 |
5. Let $x$ and $y$ be real numbers. Then the algebraic expression
$$
2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5
$$
has the minimum value of ( ).
(A) -8
(B) 8
(C) 0
(D) -10 | 5. D.
Notice that,
$$
\text { the original expression }=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10 \text {. }
$$
Therefore, when $x=2, y=-1$, the minimum value of the given algebraic expression is -10. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,273 |
6. In the Cartesian coordinate system, given four fixed points $A(-3,0), B(1,-1), C(0,3), D(-1,3)$ and a moving point $P$. Then the minimum value of $|P A|+|P B|+|P C|+|P D|$ is ( ).
(A) $4 \sqrt{2}$
(B) $3 \sqrt{2}+2 \sqrt{5}$
(C) 5
(D) $2 \sqrt{2}+3 \sqrt{5}$ | 6. B.
As shown in Figure 2, let $A C$ and $B D$ intersect at point $F$. Then
$$
\begin{array}{l}
|P A|+|P C| \geqslant|A C| \\
=|F A|+|F C|, \\
|P B|+|P D| \geqslant|B D| \\
=|F B|+|F D| .
\end{array}
$$
Therefore, when the moving point $P$ coincides with $F$,
$$
|P A|+|P B|+|P C|+|P D|
$$
reaches the minimum value ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,274 |
1. Let $72 \mid \overline{a 679 b}$. Then $a+b=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Notice that, $72=8 \times 9, (8,9)=1$.
From $8 \mid \overline{a 679 b} \Rightarrow 8 \mid \overline{79 b} \Rightarrow b=2$.
From $9 \mid \overline{a 679 b} \Rightarrow 9 \mid (a+6+7+9+b)$
$\Rightarrow 9 \mid (a+6+7+9+2) \Rightarrow a=3$.
Therefore, $a+b=5$. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,275 |
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