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 |
|---|---|---|---|---|---|---|---|---|---|
7. Given odd prime numbers $x, y, z$ satisfying
$$
x \mid \left(y^{5}+1\right), y \mid \left(z^{5}+1\right), z \mid \left(x^{5}+1\right) \text {. }
$$
Find the minimum value of the product $x y z$. (Cheng Chuanping, problem contributor) | 7. Let's assume $x$ is the minimum of $x, y, z$.
(1) If $x=3$, then
$$
3^{5}+1=244=2^{2} \times 61 \Rightarrow z=61.
$$
Since $3 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{3}$.
Clearly, $5 \nmid (61^{5}+1)$.
After calculation, we find $11 \mid (61^{5}+1)$.
Thus, $y_{\text{min}}=11$.
Therefore, $(x y z)_{\min}=3 \time... | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,936 |
3. If $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{s}^{\alpha_{s}}$, then
$$
\varphi(n)=n\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \cdots\left(1-\frac{1}{p_{s}}\right) \text {. }
$$
Find the smallest positive integer $n$ such that $\varphi(n)=\frac{2^{5}}{47} n$.
(Weng Shiyou, problem contribu... | 3. From the problem, we have
$$
47\left(p_{1}-1\right)\left(p_{2}-1\right) \cdots\left(p_{s}-1\right)=2^{5} p_{1} p_{2} \cdots p_{s} \text {. }
$$
Thus, one of \( p_{1}, p_{2}, \cdots, p_{s} \) must be 47. Without loss of generality, let \( p_{1}=47 \). Then
$$
\begin{array}{l}
46\left(p_{2}-1\right)\left(p_{3}-1\righ... | 47 \times 23 \times 11 \times 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,937 |
2. As shown in Figure 2, a circle with radius 1 passes through vertex $A$ of $\triangle A B C$, and is tangent to side $B C$ at point $D$, intersecting sides $A B$ and $A C$ at points $E$ and $F$ respectively. If $E F$ bisects $\angle A F D$, and $\angle A D C=80^{\circ}$, does there exist a triangle that satisfies the... | 2. Existence.
It is easy to know that $\angle A E D=\angle A D C=80^{\circ}$,
$$
\begin{array}{c}
\angle A F D=100^{\circ}, \\
\angle B A D=\angle E F D=50^{\circ}, \\
\angle A B D=30^{\circ},
\end{array}
$$
and $A D=2 \sin 80^{\circ}$.
Let $\angle A C D=\alpha$. Then $20^{\circ}<\alpha<80^{\circ}$.
$$
\begin{array}{... | 1+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,938 |
4. Given $a_{1}, a_{2}, \cdots, a_{108}$ are 108 distinct positive integers, each not exceeding 2015. Prove that there exists a positive integer $k$ such that there are at least four different pairs $(i, j)$ satisfying
$$
a_{i}-a_{j}=k .
$$
(Proposed by the Problem Committee) | 4. Assume the contrary. Then for each positive integer $k$, there are at most three pairs $(i, j)$ such that $a_{i}-a_{j}=k$.
By symmetry, we may assume $a_{1} \sum_{i=1}^{107}\left(a_{i+1}-a_{i}\right)+\sum_{i=1}^{105}\left(a_{i+2}-a_{i}\right) \\
\geqslant 3(1+2+\cdots+53)+54=4347 .
\end{array}
$$
Contradiction.
Thu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,939 |
7. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let
$$
S_{n}=\left[1+\frac{1}{2}+\cdots+\frac{1}{n}\right](n=1,2, \cdots) .
$$
Prove: There exist infinitely many $n$ such that $\mathrm{C}_{n}^{S_{n}}$ is even.
(Proposed by the Problem Committee) | 7. Let $n=2^{k}(k \geqslant 3)$. Then
$$
\begin{array}{l}
1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{7}\right)+\cdots+ \\
\left(\frac{1}{2^{k-1}}+\frac{1}{2^{k-1}+1}+\cdots+\frac{1}{2^{k}-1}\right)+\frac{1}{2^{k}} \\
<1+\frac{5}{6}+1+\cdots+1+\frac{1}{2^{k}}<k . \\
\text { Henc... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,940 |
1. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1 \text {, }
$$
and $a_{2 k}=a_{2 k-1}+a_{k}, a_{2 k+1}=a_{2 k}(k=1,2, \cdots)$.
Prove: For any integer $n \geqslant 3$, we have $a_{2^{n}}<2^{\frac{n^{2}}{2}}$.
(Li Shenghong) | 1. By calculation, we get
$$
a_{4}=a_{3}+a_{2}=2 a_{2}=4 a_{1}=4 \text {. }
$$
For any integer $i \geqslant 2$, we have
$$
a_{2 i}-a_{2 i-2}=\left(a_{2 i-1}+a_{i}\right)-a_{2 i-1}=a_{i} \text {. }
$$
Since the sequence $\left\{a_{n}\right\}$ is non-decreasing, for all positive integers $m$, we have
$$
\begin{array}{l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,941 |
2. In $\triangle A B C$, $A B>A C, I$ is the incenter of $\triangle A B C$. Let $\Gamma$ denote the circle with diameter $A I$. Suppose circle $\Gamma$ intersects the circumcircle of $\triangle A B C$ at points $A$ and $D$, and $D$ lies on the arc $\overparen{A C}$ not containing point $B$. Draw a line through point $A... | 2. Connect $A D$.
From the conditions, we have
$$
\begin{array}{l}
\angle I D C=\angle I D E=\angle I A E \\
=\angle I A B+\angle B A E \\
=\frac{\angle B A C}{2}+\angle A B C .
\end{array}
$$
Notice that, $\angle I D A=90^{\circ}$.
$$
\begin{array}{l}
\text { Then } \angle I D C=\angle A D C-\angle A D I \\
=\left(1... | 48^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,942 |
3. Can the first 2015 positive integers $1, 2, \cdots$, 2015 be arranged in a certain order on a circle such that the sum of any two adjacent numbers is a multiple of 4, or a multiple of 7? Prove your conclusion.
(Tao Pingsheng) | 3. The following proof: It is possible to arrange $1,2, \cdots, 2015$ on a circle according to the requirements of the problem.
Define the sequences
$$
\begin{array}{l}
A=(1,3,5,7, \cdots, 2013,2015), \\
B=(8,4,2012,2008, \cdots, 16,12), \\
C=(2,2014,2010,2006, \cdots, 10,6) .
\end{array}
$$
In each of the above seque... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,943 |
6. In $\triangle A B C$, the lengths of the three sides $B C=a, C A=b$, $A B=c$ satisfy $c<b<a<2 c$. $P$ and $Q$ are two points on the sides of $\triangle A B C$, and the line $P Q$ divides $\triangle A B C$ into two parts with equal area. Find the minimum length of the segment $P Q$.
(Wang Changping) | 6. First consider the case where points $P$ and $Q$ are on $AB$ and $AC$ respectively.
Assume point $P$ is on $AB$ and point $Q$ is on $AC$.
Let $AP = x$, $AQ = y$.
By $S_{\triangle APQ} = \frac{1}{2} S_{\triangle ABC}$
$\Rightarrow xy \sin A = \frac{1}{2} bc \sin A \Rightarrow xy = \frac{1}{2} bc$.
By the cosine rule... | \sqrt{2ab} \sin \frac{C}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,944 |
7. In $\triangle A B C$, $A B>A C>B C$. The incircle of $\triangle A B C$ touches sides $A B$, $B C$, and $C A$ at points $D$, $E$, and $F$ respectively. The midpoints of segments $D E$, $E F$, and $F D$ are $L$, $M$, and $N$ respectively. Lines $N L$, $L M$, and $M N$ intersect rays $A B$, $B C$, and $A C$ at points $... | 7. As shown in Figure 3, let the line $DE$ intersect $AR$ at point $S$. By Menelaus' theorem, we have
$$
\frac{AD}{DB} \cdot \frac{BE}{EC} \cdot \frac{CS}{SA}=1 \text{. }
$$
Given that $AD=AF, DB=BE, EC=CF$, then
$$
\frac{CS}{CF}=\frac{SA}{AF} \text{. }
$$
Notice that $M, N$ are the midpoints of $EF, FD$ respectively... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,945 |
2. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1 \text {, }
$$
and $a_{2 k}=a_{2 k-1}+a_{k}, a_{2 k+1}=a_{2 k}(k=1,2, \cdots)$.
Prove: For any positive integer $n$, $a_{2^{n}}>2^{\frac{n 2}{4}}$.
(Li Shenghong) | 2. From the conditions, the sequence $\left\{a_{n}\right\}$ is non-decreasing.
Furthermore, for any integers $s, t(0 \leqslant t < s)$, we have
$$
a_{s+t} + a_{s-t} \geqslant 2 a_{s}.
$$
In particular, when $s=2^{m-2}(m \geqslant 2)$, we have
$$
a_{2^{m}}>2^{m-1} a_{2^{m-2}}.
$$
We will prove by mathematical inductio... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,946 |
4. Let $a_{1}, a_{2}, \cdots, a_{8}$ be eight given distinct positive integers, where the greatest common divisor of any three numbers is 1. Prove: there exists an integer $n \geqslant 8$, and $n$ distinct positive integers $m_{1}, m_{2}, \cdots, m_{n}$, such that the greatest common divisor of $m_{1}, m_{2}, \cdots, m... | 4. Let's assume $a_{1}<a_{2}<\cdots<a_{8}$.
Consider the product of any six numbers from $a_{1}, a_{2}, \cdots, a_{8}$, resulting in $\mathrm{C}_{8}^{6}=28$ positive integers.
Let $M$ denote the set of all distinct values among these 28 numbers.
Clearly, $|M| \geqslant 8$.
In fact, assuming $a_{1}<a_{2}<\cdots<a_{8}$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,947 |
5. As shown in Figure $2, E$ and $F$ are points on segments $A B$ and $A D$, respectively, and $B F$ intersects $D E$ at point $C$. If
$$
A E+E C=A F+F C,
$$
prove:
$$
A B+B C=A D+D C \text {. }
$$
(Supplied by Liu Bin) | 5. As shown in Figure 4, construct the angle bisectors of $\angle B E C$ and $\angle D F C$, intersecting at point $J$. On the extensions of $A E$ and $A F$, take points $C_{1}$ and $C_{2}$ respectively, such that $E C_{1}=E C$ and $F C_{2}=F C$.
From the given conditions, we have
$$
\begin{array}{l}
A C_{1}=A E+E C_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,948 |
6. Given an integer $n \geqslant 2$. For the set of integer points in the Cartesian coordinate system
$$
A=\{(a, b) \mid a, b \in\{1,2, \cdots, n\}\}
$$
each point is colored one of three colors: red, yellow, or blue, and it satisfies: for any $a, b \in\{1,2, \cdots, n-1\}$, if $(a, b)$ and $(a+1, b)$ are the same col... | 6. For $k=1,2, \cdots, n$, define the set of integer points
$$
A_{k}=\{(k, b) \mid b \in\{1,2, \cdots, n\}\} \text {. }
$$
For $A_{1}, A_{2}, \cdots, A_{n}$, color them sequentially (to complete the coloring of $A$). In these $n$ steps, the number of ways to color $A_{k}$ is denoted as $N_{k}(k=1,2, \cdots, n)$.
First... | 3^{n}\left(2^{n+1}-1\right)^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,949 |
8. Find all prime numbers $p$ such that there exists an integer-coefficient polynomial
$$
f(x)=x^{p-1}+a_{p-2} x^{p-2}+\cdots+a_{1} x+a_{0} \text {, }
$$
satisfying that $f(x)$ has $p-1$ consecutive positive integer roots, and $p^{2} \mid f(\mathrm{i}) f(-\mathrm{i})$, where $\mathrm{i}$ is the imaginary unit.
(Yang X... | 8. Let
$$
f(x)=(x-m)(x-m-1) \cdots(x-m-p+2) \text {, }
$$
where $m$ is a positive integer.
$$
\begin{array}{l}
\text { Then } f(\mathrm{i}) f(-\mathrm{i}) \\
=\left[\prod_{k=m}^{m+p-2}(\mathrm{i}-k)\right]\left[\prod_{k=m}^{m+p-2}(-\mathrm{i}-k)\right] \\
=\prod_{k=m}^{m+p-2}\left(k^{2}+1\right) .
\end{array}
$$
By $... | p \equiv 1(\bmod 4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,950 |
2. In $\square A B C D$, $A B<A C<B C$. From point $D$ draw tangents to the circumcircle $\Gamma$ of $\triangle A B C$, the points of tangency are $E$ and $F$. If segment $A D$ intersects $C E$, and $\angle A B F=\angle D C E$, find $\angle A B C$. | 2. Since point $D$ is outside the circle $\Gamma$, $\angle A B C$ is an acute angle.
Let $A^{\prime}$ be the second intersection point of line $D C$ with circle $\Gamma$.
From $B C>A C$, we know
$$
\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A \text {. }
$$
This indicates that point $A^{\prime}$ lies on... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,951 |
4. Let $S(k)$ denote the sum of the digits of the positive integer $k$. For positive integers $a, n$, if there exists a sequence of positive integers $a_{0}, a_{1}$, $\cdots, a_{n}$ satisfying
$$
a_{i+1}=a_{i}-S\left(a_{i}\right)(i=0,1, \cdots, n-1), a_{n}=a,
$$
then $a$ is called “$n$-good”. Does there exist, for eac... | 4. Yes.
Let $f(m)=m-S(m), f^{i}$ denote the $i$-th iteration of $f$. Then $f$ is non-decreasing, and $f(m+10)>f(m)$.
Take a positive integer $d$ such that
$$
\begin{array}{l}
10^{d}>20 d(n+1), k=10^{d} . \\
\text { Let } b_{0}=10^{k}-1, c_{0}=10^{k}-k, \\
b_{i}=f^{i}\left(b_{0}\right), c_{i}=f^{i}\left(c_{0}\right) .
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,953 |
8. A coin collector has 100 coins that look the same. The collector knows that 30 of them are genuine, and 70 are fake, and that all genuine coins weigh the same, while all fake coins weigh different and are heavier than the genuine ones. The collector has a balance scale that can be used to compare the weight of two g... | 8. First, it is clear that 70 weighings can certainly find at least one genuine coin.
In fact, each time, one coin is placed on each side of the balance. If they weigh the same, both are genuine; if they do not, the heavier one must be a counterfeit. Thus, each weighing either finds two genuine coins or one counterfei... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 728,954 |
1. The function $f(x)$ defined on $\mathbf{R}$ satisfies $f\left(\frac{2 a+b}{3}\right)=\frac{2 f(a)+f(b)}{3}(a, b \in \mathbf{R})$,
and $f(1)=1, f(4)=7$.
Then $f(2015)=$ | $$
-, 1.4029 .
$$
From the function $f(x)$ being concave or convex in reverse on $\mathbf{R}$, we know its graph is a straight line, $f(x)=a x+b$.
Given $f(1)=1$ and $f(4)=7$, we have
$$
\begin{array}{l}
a=2, b=-1 \Rightarrow f(x)=2 x-1 \\
\Rightarrow f(2015)=4029 .
\end{array}
$$ | 4029 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,955 |
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$, both satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
From the problem, we know that for any $A(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking
$$
(x, y)=(1,0),(0,1),(-1,-1) \text {, }
$$
we get that the fixed point $B(a, b)$ satisfies the necessary conditions
$$
\left\{\begin{array}{l}
a \leqslant 1, \\
b \leqslant 1, \\
a+b \geqslant-1
\end{array} \quad ... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,956 |
3. Define the function on $\mathbf{R}$
$$
f(x)=\left\{\begin{array}{ll}
\log _{2}(1-x), & x \leqslant 0 ; \\
f(x-1)-f(x-2), & x>0 .
\end{array}\right.
$$
Then $f(2014)=$ | 3. 1.
First, consider the method for finding the function values on the interval $(0,+\infty)$.
If $x>3$, then
$$
\begin{array}{l}
f(x)=f(x-1)-f(x-2) \\
=f(x-1)-(f(x-1)+f(x-3)) \\
=-f(x-3) .
\end{array}
$$
Thus, for $x>6$, we have
$$
f(x)=-f(x-3)=f(x-6) \text {. }
$$
Therefore, $f(x)$ is a periodic function with a p... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,957 |
1. At a conference, 500 delegates participate, and each delegate knows more than 400 people. Prove: it is always possible to find six delegates such that every two of them know each other. | Assume that at most five representatives know each other, let's denote the set they form as
$$
P=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\} .
$$
Then, for the other 495 representatives, each of them knows at most four people in $P$, which means that there is at least one person in $P$ that they do not know. Sinc... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,958 |
6. The set $X \backslash Y=\{a \mid a \in X, a \notin Y\}$ is called the difference set of set $X$ and set $Y$. Define the symmetric difference of sets $A$ and $B$ as
$$
A \Delta B=(A \backslash B) \cup(B \backslash A) \text {. }
$$
If two non-empty finite sets $S$ and $T$ satisfy $|S \Delta T|=1$, then the minimum va... | 6. 3 .
If $|S \Delta T|=1$, without loss of generality, assume
$$
\left\{\begin{array}{l}
|S \backslash T|=1, \\
|T \backslash S|=0,
\end{array}\right.
$$
i.e., $T$ is a proper subset of $S$, thus,
$$
\begin{array}{l}
|S|>|T| \geqslant 1 . \\
\text { Hence } k \geqslant|T|+1+|T| \geqslant 3 .
\end{array}
$$
$$
\text ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,959 |
8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there always exists a row where the cells intersecting with these two columns are both... | 8. 5 .
Construct an application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."
Let the $i(i=1,2, \cdots, m)$-th kind of book be purchased by $x_{i}$ p... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,960 |
11. (20 points) The hyperbola $C: x^{2}-y^{2}=1$ and its symmetric curve $C^{\prime}$ with respect to the line $l: y=k x-1$ have common points. Find the set of real values of $k$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation and should not be inclu... | 11. From $\left\{\begin{array}{l}y=k x-1, \\ x^{2}-y^{2}=1\end{array}\right.$, eliminating $y$ yields the quadratic equation in $x$:
$$
\left(1-k^{2}\right) x^{2}+2 k x-2=0 \text {. }
$$
When $\Delta=4 k^{2}+8\left(1-k^{2}\right) \geqslant 0$, i.e., $-\sqrt{2} \leqslant k \leqslant \sqrt{2}$, the line $l$ intersects t... | k \in(-\infty,-1) \cup\left(-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) \cup(1,+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,961 |
446 Let $a, b, c > 0, \lambda \geqslant 1$. Prove:
$$
\sum \frac{a^{2}}{a^{2}+\lambda a b+b^{2}} \geqslant \frac{3}{\lambda+2},
$$
where, “$\sum$” denotes the cyclic sum. | To prove:
$$
\begin{array}{l}
\sum \frac{1}{\frac{b^{2}}{a^{2}}+\frac{\lambda b}{a}+1} \geqslant \frac{3}{\lambda+2} . \\
\text { Let } \frac{b}{a}=\frac{y z}{x^{2}}, \frac{c}{b}=\frac{z x}{y^{2}}, \frac{a}{c}=\frac{x y}{z^{2}}(x, y, z>0) .
\end{array}
$$
Then equation (1) can be transformed into:
$$
\sum \frac{x^{4}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,962 |
Given positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ less than 1, satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, and $p>0, q \geqslant 1$, $-1<p-q<0$. Prove:
$$
\frac{1}{x_{1}^{p}-x_{1}^{q}}+\frac{1}{x_{2}^{p}-x_{2}^{q}}+\cdots+\frac{1}{x_{n}^{p}-x_{n}^{q}} \geqslant \frac{n^{q+1}}{n^{q-p}-1} \text {. }
$$ | $$
\begin{array}{l}
\sum_{i=1}^{n} 1 \\
=\sum_{i=1}^{n}\left[\frac{1^{q+1}}{x_{i}^{q-1}\left(x_{i}^{p-q+1}-x_{i}\right)}\right]^{\frac{1}{q+1}} \text {. } \\
x_{i}^{q+1}\left(x_{i}^{p-q+1}-x_{i}\right)^{\frac{1}{q+1}} \\
\leqslant\left[\sum_{i=1}^{n} \frac{1^{q+1}}{x_{i}^{q-1}\left(x_{i}^{p-q+1}-x_{i}\right)}\right]^{\... | \frac{n^{q+1}}{n^{q-p}-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,963 |
In $\triangle A B C$, if $1 \leqslant k \leqslant 2$, prove:
$$
\frac{1}{k-\cos A}+\frac{1}{k-\cos B}+\frac{1}{k-\cos C} \geqslant \frac{6}{2 k-1} \text {. }
$$ | Prove that using “$\sum$” and “川” to represent cyclic sums and cyclic products, respectively, it is known that equation (1) is equivalent to
$$
\begin{array}{l}
\left(2 k^{2}+2 k\right) \sum \cos A- \\
(4 k+1) \sum(\cos A \cdot \cos B)+ \\
6 \prod \cos A-3 k^{2} \geqslant 0 .
\end{array}
$$
In $\triangle ABC$, the fol... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,964 |
Example 1 Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ be real numbers. Prove: there exist $\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n} \in\{1,-1\}$, such that
$$
\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} \varepsilon_{i} a_{i}\right)^{2} \leqslant(n+1) \sum_{i=1}^{n} a_{i}^{2} \text {... | 【Analysis】By the Cauchy inequality, we have
$$
n \sum_{i=1}^{n} a_{i}^{2} \geqslant\left(\sum_{i=1}^{n}\left|a_{i}\right|\right)^{2} \geqslant\left|\sum_{i=1}^{n} a_{i}\right|^{2} .
$$
It is sufficient to prove by mathematical induction that there exist $\varepsilon_{1}, \varepsilon_{2}$, $\cdots, \varepsilon_{n} \in\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,965 |
Question As shown in Figure 1, in $\triangle A B C$, $A D \perp B C$ at point $D, D E \perp A C$ at point $E, M$ is the midpoint of $D E$, and $A M \perp B E$ at point $F$. Prove: $\triangle A B C$ is an isosceles triangle.
(Tenth Northern Mathematical Olympiad Invitational) | Proof 1 As shown in Figure 1, take the midpoint $N$ of $AD$. Connect $MN$.
Then $MN$ is the midline of $\operatorname{Rt} \triangle ADE$.
Thus, $MN \perp DE$.
From the given $AD \perp BC, AF \perp BE$, we get
$\angle DBE = \angle DAF$;
From the given $DE \perp AC, AF \perp BE$, we get
$\angle FAE = \angle FEM$.
Therefo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,966 |
In $\triangle A B C$, it is given that $\angle B C A=90^{\circ}$, and $D$ is the foot of the altitude from vertex $C$. Let $X$ be a point on the segment $C D$ internally, $K$ be a point on the segment $A X$ such that $B K=B C$, and $L$ be a point on the segment $B X$ such that $A L=A C$. Let $M$ be the intersection of ... | Proof 1 As shown in Figure 1, let $H$ be the orthocenter of $\triangle XAB$, $AX \perp HB$ at point $F$, and $BX \perp HA$ at point $E$. Connect $HK$, $HL$, $DK$, and $DL$.
Since $H$, $E$, $D$, and $B$ are concyclic, we have:
$A L^{2}=A C^{2}=A D \cdot A B$
$\Rightarrow \triangle A D L \backsim \triangle A L B$
$\Right... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,967 |
Question 2 In a convex quadrilateral $ABCD$, it is known that $\angle ABC = \angle CDA = 90^{\circ}$, $H$ is the foot of the perpendicular from $A$ to $BD$, and points $S, T$ are on sides $AB, AD$ respectively, such that $H$ is inside $\triangle SCT$, and
$$
\begin{array}{l}
\angle CHS - \angle CSB = 90^{\circ}, \\
\an... | Proof: As shown in Figure 3, extend $CB$ to point $E$ such that $BE = BC$; extend $CD$ to point $F$ such that $DF = DC$.
Connect $AE, SE, AF, TF, HE, HF$. First, prove that $CH \perp ST$.
Draw $SP_{1} \perp CH$ at point $P_{1}$, $TP_{2} \perp CH$ at point $P_{2}$.
Since $\angle SHP_{1} = 180^{\circ} - \angle CHS$
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,968 |
Given $I, O$ are the incenter and circumcenter of $\triangle ABC$, respectively. Circle $\Gamma_{A}$ passes through points $B, C$, and is tangent to the incircle of $\triangle ABC$. Similarly, define circles $\Gamma_{B}$ and $\Gamma_{C}$. Let circles $\Gamma_{B}$ and $\Gamma_{C}$ intersect at two distinct points $A, A'... | Prove that defined by point $A^{\prime}$, the line $A A^{\prime}$ is the radical axis of circles $\Gamma_{B}$ and $\Gamma_{C}$.
Consider the radical center of circles $\Gamma_{A}$, $\Gamma_{B}$, and $\Gamma_{C}$.
By Monge's theorem, the lines $A A^{\prime}$, $B B^{\prime}$, and $C C^{\prime}$ intersect at a point, let ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,970 |
Question 2 Given that circle $\Gamma$ passes through points $B$ and $C$, and is tangent to the incircle $\odot I$ of $\triangle ABC$ at point $K$, and $D$ is the point where $\odot I$ touches side $BC$. Prove: the line $KD$ passes through the excenter $I_{A}$ of $\triangle ABC$ opposite to $\angle A$. | Prove that as shown in Figure 2, let the line $I_{A} D$ intersect the circle $\odot I$ at point $K^{\prime}$, and prove that the circumcircle of $\triangle B C K^{\prime}$ is tangent to $\odot I$ at point $K^{\prime}$.
Let the midpoint of $I_{A} D$ be $H$, and the point where the excircle opposite $\angle A$ touches s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,971 |
Question 4 Given that circle $\Gamma$ passes through points $B$ and $C$, and is tangent to the incircle $\odot I$ of $\triangle ABC$ at point $K$. Circle $\Gamma$ intersects lines $AB$ and $AC$ at points $Y$ and $X$, respectively. The excenter of $\triangle AXY$ opposite $\angle A$ is $J$, and $D$ is the point where $\... | With $A$ as the inversion center and the power of $A$ to circle $\Gamma$ as the inversion power, perform an inversion transformation.
Then circle $\Gamma$, line $AB$, and $AC$ remain unchanged. Therefore, circle $\odot I$ is transformed into another circle $\Gamma^{\prime}$ that is tangent to $AB$, $AC$, and circle $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,972 |
Example 3 Let integer $n \geqslant 4$. Prove: any triangle can be divided into $n$ isosceles triangles. ${ }^{[1]}$
(Sixth Chern Shiing-shen Cup National High School Mathematics Olympiad) | 【Analysis】Obviously, an isosceles triangle can be divided into four smaller isosceles triangles by three midlines, similar to Example 2, it is only necessary to verify that the proposition holds for three starting points.
First, prove three lemmas.
Lemma 1 Any right triangle can be divided into two isosceles triangles.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,973 |
Example 3 As shown in Figure 2, given two circles $\odot O_{1}$ and $\odot O_{2}$ with unequal radii intersecting at points $M$ and $N$, and $\odot O_{1}$ and $\odot O_{2}$ are internally tangent to $\odot O$ at points $S$ and $T$ respectively. Prove: $O M \perp M N$ if and only if points $S$, $N$, and $T$ are collinea... | Prove that, as shown in Figure 3, draw the common tangents of the two circles through points \( S \) and \( T \), intersecting at point \( P \).
By the tangent segment theorem, we have
\[
P S = P T \Rightarrow P S^{2} = P T^{2}
\]
\(\Rightarrow\) Point \( P \) has equal power with respect to \(\odot O_{1}\) and \(\odot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,974 |
Question: Draw a line $l$ with a slope of $\frac{1}{3}$, intersecting the ellipse $C$: $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$ at points $A$ and $B$ (as shown in Figure 1), and the point $P(3 \sqrt{2}, \sqrt{2})$ is above and to the left of line $l$. Prove: The incenter of $\triangle P A B$ lies on a fixed line.
[1]
(2011... | To prove that the incenter of $\triangle P A B$ lies on a fixed line, what about its centroid?
Let $l_{A B}: y=\frac{1}{3} x+b$, point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
centroid $G\left(x_{G}, y_{G}\right)$.
By combining the equation of line $A B$ with the equation of ellipse $C$, we get
$2 x^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,975 |
1. Let real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=3, a^{2}+b^{2}+c^{2}=4 . \\
\text { Then } \frac{a^{2}+b^{2}}{2-c}+\frac{b^{2}+c^{2}}{2-a}+\frac{c^{2}+a^{2}}{2-b}=(\quad \text {. }
\end{array}
$$
(A) 0
(B) 3
(C) 6
(D) 9 | -,1. D.
From the given, we have
$$
\begin{array}{l}
\frac{a^{2}+b^{2}}{2-c}+\frac{b^{2}+c^{2}}{2-a}+\frac{c^{2}+a^{2}}{2-b} \\
=\frac{4-c^{2}}{2-c}+\frac{4-a^{2}}{2-a}+\frac{4-b^{2}}{2-b}=9 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,976 |
2. In rectangle $A B C D$, it is known that $A D=5, A B=$ $10, E, F$ are two points outside the rectangle, $B E=D F=4$, $A F=C E=3$. Then $E F=(\quad)$.
(A) $4 \sqrt{15}$
(B) 15
(C) $\sqrt{221}$
(D) $10 \sqrt{2}$ | 2. C.
As shown in Figure 3, it is easy to see that $\angle A F D=\angle B E C=90^{\circ}$.
Then $\triangle B E C \cong \triangle D F A \Rightarrow \angle D A F=\angle B C E$.
Extend $F A$ and $E B$ to intersect at point $G$.
Since $\angle G A B=90^{\circ}-\angle D A F=\angle A D F$, $\angle G B A=90^{\circ}-\angle C B... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,977 |
3. If the parabola $y=x^{2}+b x+c$ intersects the $x$-axis at only one point, and passes through points $A(m, n), B(m-8, n)$, then $n=(\quad)$.
(A) 8
(B) 12
(C) 16
(D) 24 | 3. C.
According to the problem, we have
$$
\begin{array}{l}
n=m^{2}+b m+c=(m-8)^{2}+b(m-8)+c \\
\Rightarrow b=8-2 m .
\end{array}
$$
Notice that, the parabola $y=x^{2}+b x+c$ intersects the $x$-axis at only one point.
Then $b^{2}-4 c=0 \Rightarrow c=\frac{1}{4} b^{2}=(4-m)^{2}$.
Thus, $n=m^{2}+b m+c$
$$
=m^{2}+(8-2 m... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,978 |
4. Given $O$ is the origin, point $A$ in the first quadrant lies on the graph of the inverse proportion function $y=\frac{1}{x}(x>0)$, and point $B$ in the second quadrant lies on the graph of the inverse proportion function $y=-\frac{4}{x}$ $(x<0)$, and $O A \perp O B$. Then the value of $\tan \angle A B O$ is $(\quad... | 4. A.
Draw $A C \perp x$-axis at point $C$, and $B D \perp x$-axis at point $D$.
From $O A \perp O B$, we get $\angle A O B=90^{\circ}$.
Thus, $\triangle A O C \backsim \triangle O B D$.
Therefore, $\tan \angle A B O=\frac{O A}{O B}=\sqrt{\frac{S_{\triangle A O C}}{S_{\triangle O B D}}}$
$=\sqrt{\frac{O C \cdot A C}{O... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,979 |
1. Let $a, b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is | From the given, we have $a b=-1, a+b=1$,
$$
\begin{array}{l}
a^{2}=a+1, b^{2}=b+1 . \\
\text { Therefore, } 3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2} \\
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11 .
\end{array}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,980 |
3. Given that the circumcenter of acute $\triangle A B C$ is $O, A O$ intersects $B C$ at point $D, E$ and $F$ are the circumcenters of $\triangle A B D$ and $\triangle A C D$ respectively. If $A B>A C, E F=B C$, then $\angle C-\angle B=$ $\qquad$ . | 3. $60^{\circ}$.
As shown in Figure 4, draw $E M \perp B C$
at point $M$, $F N \perp B C$ at point $N$, and $F P \perp E M$ at point $P$.
It is easy to see that $M$ and $N$ are the midpoints of sides $B D$ and $C D$, respectively.
$$
\begin{array}{l}
\text { Also, } E F=B C, \\
P F=M N \\
= \frac{1}{2} B C=\frac{1}... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,981 |
4. Fill the numbers $1,2, \cdots, 36$ in a $6 \times 6$ grid, with each cell containing one number, such that the numbers in each row are in increasing order from left to right. Then the minimum value of the sum of the six numbers in the third column is $\qquad$ | 4. 63.
Let the six numbers filled in the third column, arranged in ascending order, be $A, B, C, D, E, F$.
Since the row where $A$ is located needs to fill in two numbers smaller than $A$ before it, then $A \geqslant 3$; since the row where $B$ is located needs to fill in two numbers smaller than $B$, and $A$ and the... | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,982 |
II. (25 points) As shown in Figure 1, the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ intersect at point $E$, and $AC \perp BD, AB = AC$. A line $DF \perp BD$ is drawn through point $D$, intersecting the extension of $BA$ at point $F$. The angle bisector of $\angle BFD$ intersects $AD$ and $BD$ at points... | (1) Take a point $P$ on side $BE$ such that
$$
\angle BAP = \angle DAC.
$$
Then $\triangle BAP \cong \triangle CAD \Rightarrow AP = AD$.
Since $AE \perp PD$, we have
$$
\begin{array}{l}
\triangle ADE \cong \triangle APE \Rightarrow \angle PAE = \angle DAE \\
\Rightarrow \angle PAE = \angle BAP = \angle DAC \\
\Rightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,983 |
1. Let $a$ and $b$ be distinct real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ . | $$
-, 1.4
$$
From the given conditions and the symmetry of the quadratic function graph, we get
$$
\begin{array}{l}
\frac{a+b}{2}=-\frac{a}{2} \Rightarrow 2 a+b=0 \\
\Rightarrow f(2)=4+2 a+b=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,985 |
Example 5 Let $f: \mathbf{R}_{+} \rightarrow \mathbf{R}$ satisfy that for any $x, y \in \mathbf{R}_{+}$, we have
$$
\frac{f(\sqrt{x})+f(\sqrt{y})}{2}=f\left(\sqrt{\frac{x+y}{2}}\right) .
$$
Then for any $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, we have
$$
\frac{1}{n} \sum_{k=1}^{n} f\left(\sqrt{x_{k}}\right)=f... | 【Analysis】Use mathematical induction as in Example 4.
First, prove: For any positive integer $k$, the proposition holds for $n=2^{k}$.
When $n=2$, the proposition has already been established.
If the proposition holds for $n=2^{k}$, then when $n=2^{k+1}$,
$$
\begin{array}{l}
\frac{1}{n} \sum_{i=1}^{n} f\left(\sqrt{x_{i... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,986 |
2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then $\frac{1}{\sin \alpha}+\cos ^{4} \alpha=$ $\qquad$ | 2. 2 .
From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$.
$$
\begin{array}{l}
\text { Then } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha \\
=(1+\sin \alpha)+\left(1-\cos ^{2} \alpha\right) \\
=2+\sin \alpha-\cos ^{2} \alpha=2 .
\end{array}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,987 |
3. Given the complex number sequence $\left\{z_{n}\right\}$ satisfies
$$
z_{1}=1, z_{n+1}=\overline{z_{n}}+1+n \mathrm{i}(n=1,2, \cdots),
$$
where $\mathrm{i}$ is the imaginary unit, and $\overline{z_{n}}$ denotes the complex conjugate of $z_{n}$. Then the value of $z_{2015}$ is $\qquad$ . | 3. $2015+1007 \mathrm{i}$.
From the given, for all positive integers $n$, we have
$$
\begin{array}{l}
z_{n+2}=\overline{z_{n+1}}+1+(n+1) \mathrm{i} \\
==\overline{\overline{z_{n}}+1+n \mathrm{i}}+1+(n+1) \mathrm{i}=z_{n}+2+\mathrm{i} \\
\text { Therefore, } z_{2015}=z_{1}+1007(2+\mathrm{i})=2015+1007 \mathrm{i} .
\end... | 2015+1007 \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,988 |
In the plane rectangular coordinate system $x O y$, the set of points
$$
\begin{aligned}
K= & \{(x, y) \mid(|x|+|3 y|-6) . \\
& (|3 x|+|y|-6) \leqslant 0\}
\end{aligned}
$$
corresponds to a plane region whose area is $\qquad$ | 6. 24 .
Let $K_{1}=\{(x, y)|| x|+| 3 y |-6 \leqslant 0\}$.
First, consider the part of the point set $K_{1}$ in the first quadrant, at this time, $x+3 y \leqslant 6$. Therefore, these points correspond to $\triangle O C D$ and its interior in Figure 2.
By symmetry, the region corresponding to the point set $K_{1}$ is... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,989 |
8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant$ $9)$, if $a>b, b<c, c>d$, then $\overline{a b c d}$ is called a $P$ class number; if $a<c, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbe... | 8. 285.
Let the sets of all P-type numbers and Q-type numbers be denoted as $A$ and $B$, respectively. Let the set of all P-type numbers whose unit digit is zero be denoted as $A_{0}$, and the set of all P-type numbers whose unit digit is not zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \... | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,990 |
Example 6 Let $S_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$. Prove:
$$
\max \left\{\frac{1}{p}, \frac{1}{q}\right\} \leqslant S_{p}+S_{q}-S_{p q} \leqslant 1 .
$$ | 【Analysis】We need to prove: $\frac{1}{p} \leqslant S_{p}+S_{q}-S_{p q} \leqslant 1$.
First, apply mathematical induction on the main variable $p$ to prove:
$$
\frac{1}{p} \leqslant S_{p}+S_{q}-S_{p q} \text {. }
$$
When $p=1$, $1 \leqslant S_{1}+S_{q}-S_{q}=S_{1}$, which is obviously true.
Assume the inequality holds ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,991 |
Three. (50 points) As shown in Figure 1, $\triangle A B C$ is inscribed in $\odot O$, $P$ is a point on arc $\overparen{B C}$, and point $K$ lies on segment $A P$ such that $B K$ bisects $\angle A B C$. The circle $\Gamma$ passing through points $K, P, C$ intersects side $A C$ at point $D$. Connecting $B D$ intersects ... | Proof 1: Let $CF$ intersect the circle $\Gamma$ at point $L$ (different from $C$), and connect $PB, PC, BL, KL$.
At this point, points $C, D, L, K, E, P$ all lie on the circle $\Gamma$.
Combining the fact that points $A, B, P, C$ are concyclic, we have
\[
\begin{array}{l}
\angle FEB = \angle DEP = 180^{\circ} - \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,992 |
2. Given that $K$, $L$, $M$ are points on the sides $BC$, $CA$, $AB$ of $\triangle ABC$ respectively, and that $AK$, $BL$, $CM$ are concurrent. Prove: Among $\triangle ALM$, $\triangle BMK$, $\triangle CKL$, there exist two triangles such that the sum of their inradii is greater than or equal to the inradius of $\trian... | 2. Let $a=\frac{B K}{K C}, b=\frac{C L}{L A}, c=\frac{A M}{M B}$.
By Ceva's Theorem, we know $a b c=1$.
Without loss of generality, assume $a \geqslant 1$. Then at least one of $b, c$ is no greater than 1. Among the pairs $(a, b)$ and $(b, c)$, at least one satisfies the condition that the first component is no less t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,994 |
4. Given a fixed circle $\Gamma$, $A, B, C$ are three fixed points on circle $\Gamma$. A real number $\lambda \in (0,1)$ is determined, and $P$ is a moving point on circle $\Gamma$ different from $A, B, C$. $M$ is a point on segment $CP$ such that $CM = \lambda CP$. Let the second intersection point of the circumcircle... | 4. Let $\chi(a, b)$ denote the directed angle between lines $a$ and $b$. Suppose $D$ is a point on segment $AB$ such that $BD = \lambda BA$.
Below, we prove: either point $Q$ coincides with $D$, or $\chi(DQ, QB) = \chi(AB, BC)$.
This indicates that point $Q$ lies on a fixed circle passing through point $D$ and tangen... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,995 |
2. Let $n>1$ be a positive integer. Consider the numbers $\frac{1}{n}, \frac{2}{n}, \cdots, \frac{n-1}{n}$ in their simplest forms. Let $f(n)$ denote the sum of the numerators of these simplest fractions. If $f(n)$ and $f(2015 n)$ have different parities, find all possible values of $n$.
---
The translation preserves... | 2. $n$ is all positive integers greater than 1.
Let $n=2^{t} m(t \geqslant 0,(m, 2)=1)$.
Consider the simplest form of the fraction $\frac{k}{n}$, its numerator is even if and only if $2^{t+1} \mid k$.
From $1,2, \cdots, n-1$ there are exactly $\frac{m-1}{2}$ multiples of $2^{t+1}$, thus
$$
f(n) \equiv n-1-\frac{m-1}... | n \text{ is all positive integers greater than 1} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,996 |
Example 7 Prove: A positive reduced proper fraction can be written as the sum of the reciprocals of a finite number of distinct positive integers.
(28th IMO Preliminary Question) | 【Analysis】Let the positive irreducible proper fraction be $\frac{q}{p}((p, q)=1$, $0<\frac{q}{p}<1)$. We need to prove that it can be expressed as the sum of the reciprocals of positive integers, and the number of terms in the sum is no more than $p$.
We use mathematical induction on $p$.
1. **Base Case**: When $p=1$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,997 |
4. Given a positive integer $N \geqslant 3$. If a set of $N$ points, each either red or blue, on the coordinate plane has each point with a unique x-coordinate, then the set of points is called "acceptable". $P(x)$ is a polynomial, and $A$ is an acceptable set. If the upper side of the graph of $P(x)$ contains no red p... | 4. $k_{\min }=N-2$.
First, we point out that $k=N-2$ satisfies the condition.
Choose any $N-1$ points from the set $A$, and construct a polynomial of degree no more than $N-2$ such that its graph passes through these points. Clearly, this polynomial distinguishes $A$.
Next, we show that a polynomial of degree less th... | k_{\min }=N-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,998 |
8. Given positive integers $a, b$ satisfying $a<b<2a$. A rectangle on the coordinate plane with vertices at integer points and sides parallel to the coordinate axes is called a "standard rectangle". A standard rectangle with an area of 1 is called a "square". Some squares on the coordinate plane are colored red. It is ... | 8. $\alpha_{\max }=\frac{1}{2 a^{2}-2 a b+b^{2}}$.
(1) First, construct an example where $\alpha \leqslant \frac{1}{D}$, where
$$
D=a^{2}+(b-a)^{2}=2 a^{2}-2 a b+b^{2} \text {. }
$$
Arbitrarily color one grid red, then translate it by $n \boldsymbol{\beta}+m \boldsymbol{\gamma}(m, n \in \mathbf{Z}, \boldsymbol{\beta}=... | \frac{1}{2 a^{2}-2 a b+b^{2}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,999 |
$$
\begin{array}{l}
\text { 1. Let } A=\left\{x \mid x^{2}<\log _{a} x\right\} , \\
B=\left\{y \mid y=2^{x}, x \in(-\infty,-1)\right\} .
\end{array}
$$
If $A \subsetneq B$, then the range of values for $a$ is | $-1 .\left(0, \frac{1}{16}\right) \cup\left[\mathrm{e}^{\frac{1}{2_{0}}},+\infty\right)$.
When $0 < a < 1$, $A \neq \varnothing, x^{2} \geqslant \log _{a} \frac{1}{2} \Rightarrow a \in\left(0, \frac{1}{16}\right)$.
When $a>1$, $A \neq \varnothing, x^{2} \geqslant \log _{a} x$.
Assume the function $y=x^{2}$ and $y=\log ... | a \in\left(0, \frac{1}{16}\right) \cup\left[\mathrm{e}^{\frac{1}{2}},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,000 |
2. Given the line $l: y=x-a$ intersects the parabola $C$: $x^{2}=2 p y(p>0)$ at points $M$ and $N$, and the circle passing through $M$ and $N$ intersects the parabola $C$ at two other distinct points $E$ and $F$. Then the cosine of the inclination angle of the line $E F$ is $\qquad$. | 2. $-\frac{\sqrt{2}}{2}$.
Let points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right), E\left(x_{3}, y_{3}\right)$, $F\left(x_{4}, y_{4}\right)$, and the circle passing through points $M, N, E, F$ be
$$
x^{2}+y^{2}+d x+e y+f=0 \text {. }
$$
Then the coordinates of points $M, N, E, F$ satisfy the system of equa... | -\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,001 |
4. A permutation $a_{1}, a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n(n \geqslant 2)$ is called a "good permutation" if there exists a unique $i \in\{2,3, \cdots, n\}$ such that $a_{i}<a_{i-1}$. If the number of good permutations is $p_{n}$, then $\sum_{i=2}^{n} \frac{p_{i}}{2^{i}}=$ $\qquad$ . | 4. $n-3+\frac{n+3}{2^{n}}$.
From the problem, we know
$$
a_{1}a_{i}, a_{i}a_{i}$, thus, we only need to exclude the case where the selected set is exactly $\{1,2, \cdots, i-1\}$.
$$
\begin{array}{l}
\text { Then } p_{n}=\sum_{i=2}^{n}\left(\mathrm{C}_{n}^{i-1}-1\right)=2^{n}-n-1 \text {. } \\
\text { Therefore, } \sum... | n-3+\frac{n+3}{2^{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,003 |
5. In the tetrahedron $ABCD$, it is known that $AD \perp BC$,
```
\begin{array}{l}
AD=6, BC=2, \text { and } \\
AB+BD=AC+CD=t(t \in[8,+\infty)) .
\end{array}
``` | 5. $[2 \sqrt{6},+\infty)$.
Draw $B F \perp A D$ at point $F$, and connect $C F$.
Since $A D \perp B C$, we know $A D \perp$ plane $B F C$.
Therefore, $C F \perp A D$.
Let $A B=x, A C=y$.
From $A B^{2}-B D^{2}=A F^{2}-D F^{2}=A C^{2}-C D^{2}$
$\Rightarrow x=y \Rightarrow B F=C F$.
Then $V=V_{\text {pyramid } A-B F C}+V... | [2 \sqrt{6},+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,004 |
Example 8 Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=1, a_{2}=2$,
$a_{n+2}=\left\{\begin{array}{ll}5 a_{n+1}-3 a_{n}, & a_{n+1} a_{n} \text { is even; } \\ a_{n+1}-a_{n}, & a_{n+1} a_{n} \text { is odd. }\end{array}\right.$ Prove: $a_{n} \neq 2016$. | 【Analysis】(1) Conduct a mathematical experiment:
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=7, \\
a_{4}=29, a_{5}=22, a_{6}=23, \\
a_{7}=49, a_{8}=26, a_{9}=-17, \\
\cdots \cdots
\end{array}
$$
It can be observed that the sequence has a pattern: every three numbers form a group, with the parity being odd, even, odd.
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,005 |
6. Let $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ be an arithmetic sequence, and
$$
\sum_{i=1}^{n}\left|a_{i}+j\right|=2028(j=0,1,2,3) \text {. }
$$
Then the maximum value of the number of terms $n$ is | 6. 52.
Since the equation $|x|=|x+1|=|x+2|$ has no solution, we have $n \geqslant 2$ and the common difference is not 0.
Assume the general term of the sequence is $a-k d(1 \leqslant k \leqslant n, d>0)$.
Construct the function $f(x)=\sum_{k=1}^{n}|x-k d|$.
The given condition is equivalent to $f(x)=2028$ having at le... | 52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,006 |
7. Given $a, b, c \geqslant 0, t \geqslant 1$, satisfying
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{2}, \\
\sqrt{a+\frac{1}{2}(b-c)^{2}}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{6 t}}{2} .
\end{array}\right.
$$
Then $a^{2 t}+b^{2 t}+c^{2 t}=$ $\qquad$ . | 7. $\frac{1}{12}$.
Assume $b \geqslant c$, let $\sqrt{2 b}=x+y, \sqrt{2 c}=x-y$.
Then $b-c=2 x y, 2 a=1-2 x^{2}-2 y^{2} \geqslant 0$.
Thus, $0 \leqslant x \leqslant \frac{\sqrt{2}}{2}$.
$$
\begin{array}{l}
\text { Hence } \sqrt{2 a+(b-c)^{2}}+\sqrt{2 b}+\sqrt{2 c} \\
=\sqrt{\left(1-2 x^{2}\right)\left(1-2 y^{2}\right)... | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,007 |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Then the sum of the squares of all natural numbers $m$ that satisfy
$$
[(2 m+1)\{\sqrt{2 m+1}\}]=m
$$
is $\qquad$ . | 8.0.
Let $2 m+1=n$. Then $2[n\{\sqrt{n}\}]=n-1$. Note that,
$$
\begin{array}{l}
2 n\{\sqrt{n}\}-21$ when, by $n=2[n\{\sqrt{n}\}]+1$ being odd, we know $n \geqslant 3$.
Thus, $\left|4 n-(2[\sqrt{n}]+1)^{2}\right| \leqslant \frac{4}{\sqrt{n}}+\frac{1}{n}<3$.
Therefore, $4 n-(2[\sqrt{n}]+1)^{2}= \pm 1$.
This leads to
$$
... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,008 |
11. (20 points) Let $\alpha, \beta$ be any two interior angles of a triangle. If there exist $\alpha, \beta$ such that the inequality
$$
\cos \sqrt{\alpha}+\cos \sqrt{\beta}>a+\cos \sqrt{\alpha \beta}
$$
holds, find the range of the real number $a$.
保留了源文本的换行和格式。 | 11. From the problem, we know $\alpha, \beta>0$, and $0<\sqrt{\alpha}<1, \sqrt{\beta}>1$, then
$$
\begin{array}{l}
\sqrt{\alpha} \sqrt{\beta} \leqslant\left(\frac{\sqrt{\alpha}+\sqrt{\beta}}{2}\right)^{2} . \\
\text { Let } \frac{\sqrt{\alpha}+\sqrt{\beta}}{2}=t .
\end{array}
$$
From $(x+y)^{2} \leqslant 2\left(x^{2}+... | a \in (-\infty, 1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,009 |
(40 points) As shown in Figure 1, $H$ is the orthocenter of acute $\triangle ABC$, and points $M$ and $N$ are on sides $AB$ and $AC$ respectively, such that $\angle HMB = \angle HNC = 60^{\circ}$. Line $HM$ intersects the extension of $CA$ at point $P$, and line $HN$ intersects the extension of $BA$ at point $Q$. $O$ i... | (1) From $\triangle H P N \backsim \triangle H Q M$, we know that points $N, M, P, Q$ are concyclic.
Let $T$ be the orthocenter of $\triangle H M N$.
Since $\angle T H M = \angle O H N$, we have
$\angle P Q H - \angle O H N = \angle N M H - \angle T H M = 90^{\circ}$.
Thus, $H O \perp P Q$.
(2) As shown in Figure 2, co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,010 |
Three. (50 points) Let $a, b, c$ be rational numbers, and $a+b+c$ and $a^{2}+b^{2}+c^{2}$ be equal integers. Prove: There exist integers $u, v$ such that $a b c v^{3}=u^{2}$, where $(u, v)=1$.
---
The translation maintains the original text's formatting and line breaks. | Let $a+b+c=a^{2}+b^{2}+c^{2}=t$. Then $t \geqslant 0$. By the Cauchy-Schwarz inequality, we have
$$
3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2} \text {. }
$$
Thus, $0 \leqslant t \leqslant 3$, i.e., $t \in\{0,1,2,3\}$.
(1) If $t=0$, then $a=b=c=0$. In this case, take $u=0, v=1$, satisfying $a b c v^{3}=u^{2}$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,011 |
Four. (50 points) Let $A$, $B$, and $C$ be any three $n$-element subsets of the set $M=\{1,2, \cdots, 3n\}$, and
$$
A \cap B \cap C=\varnothing, A \cup B \cup C=M.
$$
Does there exist $a \in A, b \in B, c \in C$ such that the sum of two of these numbers equals the third? | To prove by contradiction: There exist $a \in A, b \in B$, $c \in C$, such that the sum of two of these numbers equals the third. Assume there is some partition
$$
M=A \cup B \cup C, A \cap B \cap C=\varnothing,
$$
such that the three $n$-element sets $A, B, C$ do not contain such three elements.
$$
\begin{array}{l}
\... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,012 |
Example 9 Polynomial
$$
f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} .
$$
If there exist $n+1$ consecutive integers $m, m+1, \cdots, m+n$, such that $f(m+k) \in \mathbf{Z}(k=0,1, \cdots, n)$, then $f(x)$ is an integer-valued polynomial (when $x \in \mathbf{Z}$, $f(x) \in \mathbf{Z}$). | 【Analysis】Apply mathematical induction on $n$.
When $n=0$, $f(x)=c \in \mathbf{Z}$, the proposition holds.
Assume the proposition holds when $n=k$.
Then when $n=k+1$, $f(x)$ takes integer values at $k+2$ consecutive integers $m, m+1, \cdots, m+k+1$.
$$
\text { Let } f^{*}(x)=f(x+1)-f(x) \text {. }
$$
Thus, $f^{*}(x)=n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,013 |
449 Proof: For any odd number of real numbers $a_{1}$, $a_{2}$, \cdots, $a_{k}$ (where $k$ is an odd constant, and $a_{k} \neq 0$), there exists a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any $x \in \mathbf{R}$, we have
$$
a_{1} f(x)+a_{2} f(f(x))+\cdots+a_{k} \underbrace{f(f(\cdots}_{k \uparrow} ... | Prove: Take $f(x)=p x(p \neq 0)$. Then
$$
\underbrace{f(f(\cdots}_{k \uparrow} f(x) \cdots))=p^{k} x \text {. }
$$
Thus, the conclusion of the problem is converted to
$$
a_{1} p x+a_{2} p^{2} x+\cdots+a_{k} p^{k} x=x
$$
always holds, which means
$$
a_{k} p^{k}+\cdots+a_{2} p^{2}+a_{1} p-1=0 .
$$
The above equation i... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,014 |
Given $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $M$, and the common tangent line through point $M$ is line $l$. On line $l$, take two points $A$ and $C$. Through points $A$ and $C$, draw tangents to $\odot O_{2}$ other than line $l$, and let these tangents intersect at point $B$; then through poin... | Prove the following by first presenting two well-known conclusions.
Lemma 1 A convex quadrilateral $ABCD$ is a tangential quadrilateral if and only if $AB + CD = AD + BC$.
Lemma 2 For three circles, the three exsimilicenters of each pair of circles are collinear.
Now, let's return to the original problem.
We will dis... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,015 |
451 Find all integer solutions to the indeterminate equation
$$
\left(x^{2}-4\right)\left(y^{2}-4\right)=t^{2}+12
$$ | If $x=0$, then
$t^{2}+4 y^{2}=4 \Rightarrow t=0, y= \pm 1$ or $t= \pm 2, y=0$.
If $y=0$, similarly,
$t=0, x= \pm 1$ or $t= \pm 2, x=0$.
If $t=0$, then $x= \pm 1, y=0$ or $x=0, y= \pm 1$.
Consider the non-zero integer solutions of the equation. It is easy to see that if the equation has non-zero integer solutions, then ... | (x, y, t)=(0, \pm 1,0),(0,0, \pm 2),( \pm 1,0,0) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,016 |
Let $m, n, p \in \mathbf{Z}_{+} (m < n < p)$. There are $p$ lamps arranged in a circle, and initially, the pointer points to one of the lamps. Each operation can either rotate the pointer $m$ positions clockwise and then change the on/off state of the lamp it reaches, or rotate the pointer $n$ positions clockwise and t... | For convenience, let's call the lamp initially pointed to by the pointer as lamp 0, and the lamps in the clockwise direction as lamp 1, lamp 2, ..., lamp \( p-1 \). The numbering of the lamps is modulo \( p \), so lamp \( p \) is the same as lamp 0, and so on. The congruence symbol used for the numbering of the lamps i... | (n-m, p) \leq 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,017 |
Example: $10 f(x)$ is a continuous function defined on the interval $[0,2015]$, and $f(0)=f(2015)$. Find the minimum number of real number pairs $(x, y)$ that satisfy the following conditions:
(1) $f(x)=f(y)$;
(2) $x-y \in \mathbf{Z}_{+}$.
$(2015$, Peking University Mathematics Summer Camp) | 【Analysis】Consider a continuous function $f(x)$ on the interval $[0, n]$ that satisfies $f(0)=f(n)$. Let the minimum number of real number pairs $(x, y)$ that satisfy (1) and (2) be $F(n)$.
Obviously, $F(1)=1, F(2)=2$.
We will prove: $F(n)=n\left(n \in \mathbf{Z}_{+}\right)$.
Assume that for $n \leqslant k-1$, equation... | 2015 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,018 |
Example 1 Prove: $\sum_{k=0}^{n} k^{2} \mathrm{C}_{n}^{k}=2^{n-2} n(n+1)$. | Prove that in fact,
$$
\begin{array}{l}
\sum_{k=0}^{n} k^{2} \mathrm{C}_{n}^{k}=\sum_{k=2}^{n} k(k-1) \mathrm{C}_{n}^{k}+\sum_{k=1}^{n} k \mathrm{C}_{n}^{k} \\
=n(n-1) \sum_{k=2}^{n} \mathrm{C}_{n-2}^{k-2}+n \sum_{k=1}^{n} \mathrm{C}_{n-1}^{k-1} \\
=n(n-1) \sum_{k=0}^{n-2} \mathrm{C}_{n-2}^{k}+n \sum_{k=0}^{n-1} \mathr... | 2^{n-2} n(n+1) | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,019 |
Example 2 Proof: $\sum_{k=1}^{n-r+1} k \mathrm{C}_{n-k}^{r-1}=\mathrm{C}_{n+1}^{r+1}$. | $$
\begin{array}{l}
\text { Left side }=\sum_{k=1}^{n-r+1} k\left(\mathrm{C}_{n-k+1}^{r}-\mathrm{C}_{n-k}^{r}\right) \\
=\sum_{k=1}^{n-r+1} \mathrm{C}_{n-k+1}^{r}=\mathrm{C}_{n+1}^{r+1} .
\end{array}
$$ | \mathrm{C}_{n+1}^{r+1} | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,020 |
Example 3 In $\triangle ABC$ with a fixed perimeter, it is known that $AB=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ intersecting the c... | (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis.
Let $|CA| + |CB| = 2a (a > 3)$ be a constant. Then the locus of point $C$ is an ellipse with foci at $A$ and $B$.
Thus, the focal distance $2c = |AB| = 6$.
Notice,
$$
\begin{ar... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,021 |
Example 4 As shown in Figure 1, in the Cartesian coordinate system $x O y$, the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its left and right foci at $F_{1}(-c, 0)$ and $F_{2}(c, 0)$, respectively. Let $A$ and $B$ be two points on the ellipse located above the $x$-axis, with $A F_{1} \parallel B F_{... | Prove that if we establish a polar coordinate system with $F_{1}$ as the pole and $F_{1} x$ as the polar axis, then the polar equation of the ellipse is
$$
\rho=\frac{e p}{1-e \cos \theta}\left(e=\frac{c}{a}, p=\frac{a^{2}}{c}-c=\frac{b^{2}}{c}\right) .
$$
Thus, $\rho=\frac{b^{2}}{a-c \cos \theta}$.
Let $\angle A F_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,022 |
Example 6: Through the right focus $F$ of the ellipse $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, draw two perpendicular chords $A B$ and $C D$. Let the midpoints of $A B$ and $C D$ be $M$ and $N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point, and find this fixed point;
(2) If the slopes of the c... | (1) Proof: Establish a polar coordinate system with $F$ as the pole and $F x$ as the polar axis. Then the equation of the ellipse is $\rho=\frac{e p}{1+e \cos \theta}$.
Let point $A\left(\rho_{1}, \alpha\right)\left(\rho_{1}=\frac{e p}{1+e \cos \alpha}\right)$,
$B\left(\rho_{2}, \alpha+\pi\right)\left(\rho_{2}=\frac{e ... | \frac{4}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,023 |
Second question Let $S=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}(n \geqslant 2)$, where $A_{1}, A_{2}, \cdots, A_{n}$ are $n$ distinct finite sets, satisfying that for any $A_{i}, A_{j} \in S$, there is $A_{i} \cup A_{j} \in$ S. If $k=\min _{1 \leqslant i \leqslant n}\left|A_{i}\right| \geqslant 2$ ( $|X|$ denotes the... | Proof 1 proves a stronger conclusion: For any set $A_{i}(1 \leqslant i \leqslant n)$, there exists $x \in A_{i}$, such that $x$ belongs to at least $\frac{n}{\left|A_{i}\right|}$ of the sets $A_{1}$, $A_{2}, \cdots, A_{n}$.
If the intersection of $A_{1}, A_{2}, \cdots, A_{n}$ with $A_{i}$ is non-empty, then by the pri... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,024 |
Question 1 In a cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $E$, $AD$ and $BC$ intersect at point $F$, and $G$, $H$ are the midpoints of sides $AB$, $CD$ respectively. Prove that $FE$ is tangent to the circumcircle of $\triangle EGH$. ${ }^{[1]}$
This is both a problem from the 50th IMO... | To prove as shown in Figure 1, extend $GH$ to intersect $EF$ at point $P$.
To prove the conclusion, it suffices to show that $PE$ is tangent to the circumcircle of $\triangle EGH$, which can be done by proving
$$
\angle PEH = \angle PGE \text{ or } \frac{GP}{HP} = \frac{EG^2}{EH^2}.
$$
For this purpose, connect $FG$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,025 |
Question 2 Let $M, N, P$ be points on the sides $AB, BC, CA$ of $\triangle ABC$, respectively, such that quadrilateral $CPMN$ is a parallelogram. $AN$ and $MP$ intersect at point $R$, $BP$ and $MN$ intersect at point $S$, $AN$ and $BP$ intersect at point $Q$. Let $T$ be the projection of point $Q$ onto side $AB$. Prove... | Prove as shown in Figure 2, connect $A S$ and $R S$.
From $M N / / A C \Rightarrow \frac{M S}{S N}=\frac{A P}{P C}=\frac{A P}{M N}=\frac{A R}{R N}$
$\Rightarrow S R / / A M \Rightarrow S R / / A B$.
At this point, $S_{\triangle T S R}=S_{\triangle A S R}$.
Therefore, $S_{\text {quadrilateral } T S Q R}=S_{\triangle T S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,026 |
Example 3 A sequence of positive integers satisfies: when written in binary, they have exactly 2013 digits, and the number of 0s is greater than the number of 1s. Let $n$ be the number of positive integers that satisfy this condition, and $s$ be the sum of these positive integers. Prove: when written in binary, $n+s$ h... | Notice that, in the binary representation of positive integers, the leading digit must be 1.
From the condition, we know that the number of 0s is at least 1007 and at most 2012.
By $\mathrm{C}_{a}^{b}=\mathrm{C}_{a}^{a-b}$, we get
$$
\begin{aligned}
n & =\sum_{i=1007}^{2012} \mathrm{C}_{2012}^{i}=\frac{1}{2} \sum_{i=0... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,027 |
Example 1 Let $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfy
$$
f\left(x^{3}\right)+f\left(y^{3}\right)=(x+y) f\left(x^{2}\right)+f\left(y^{2}\right)-f(x y) \text {. }
$$
Find the analytical expression of the function $f(x)$. | 【Analysis】(1) Observing that $x$ and $y$ are relatively scattered, one can remove a variable by setting one of them to 0, thereby observing the other variable. Based on this, finding the value of $f(0)$ becomes the starting point of this problem.
(2) Let $x=y=0$. Then
the left side of equation (1) $=2 f(0)=0=$ right si... | f(x)=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,028 |
Example 3 Let $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfy
$$
\begin{array}{l}
f(x f(x)+f(x) f(y)+y-1) \\
=f(x f(x)+x y)+y-1 .
\end{array}
$$
Find the analytical expression of the function $f(x)$. | 【Analysis】Let one of the variables be 0, which can simplify the equation (since when a variable multiplies a function, if the variable is 0, the entire expression becomes 0, effectively ignoring the function).
$$
\begin{array}{l}
\text { Let } x=0, f(0)=s. \text { Then } \\
f(s f(y)+y-1)=s+y-1 .
\end{array}
$$
In the ... | f(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,029 |
Example 5 Let $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfy for any real numbers $x$, $y$,
$$
f(x+f(x+y))+f(x y)=x+f(x+y)+y f(x) \text {. }
$$
Find the analytical expression of the function $f(x)$. | 【Analysis】Let $x=y=0$. Then $f(f(0))=0$.
Let $f(0)=s, x=0, y=s$. Then
$2 s=s^{2} \Rightarrow s_{1}=2, s_{2}=0$.
This gives us the points for case analysis.
(1) When $s=2$, let $x=0, x=-y$, we get
$$
\begin{array}{l}
f(f(y))=f(y)+2 y-2, \\
f(x+2)+f\left(-x^{2}\right)=x+2-x f(x) .
\end{array}
$$
In equation (2), let $x=... | f(x)=x \text{ or } f(x)=2-x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,030 |
2. Find all pairs of positive integers $(x, y)$ such that
$$
\sqrt[3]{7 x^{2}-13 x y+7 y^{2}}=|x-y|+1 .
$$ | 2. $(x, y)=(1,1)$ or
$$
\{x, y\}=\left\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\right\},
$$
where, integer $m \geqslant 2$.
Let $(x, y)$ be any pair of positive integers satisfying equation (1).
If $x=y$, then
$$
x^{\frac{2}{3}}=1 \Rightarrow x=1 \Rightarrow(x, y)=(1,1) \text {. }
$$
If $x \neq y$, by symmetry, we can a... | (x, y)=(1,1) \text{ or } \{x, y\}=\left\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\right\}, \text{ where } m \geqslant 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,031 |
1. As shown in Figure 1, in the acute triangle $\triangle ABC$, $AB > AC$, $O$ is the circumcenter, and $D$ is the midpoint of side $BC$. A circle is drawn with $AD$ as its diameter, intersecting sides $AB$ and $AC$ at points $E$ and $F$, respectively. A line $DM$ is drawn through point $D$ parallel to $AO$, intersecti... | 1. Connect $D E$ and $D F$, draw $O N \perp A B$, intersecting $A B$ at point $N$.
By the problem, $D E \perp A B, D F \perp A C$.
Thus, $O N / / D E$.
Since $D M / / A O$, we have
$\angle E D M=\angle A O N$.
Also, $O$ is the circumcenter of $\triangle A B C$, so
$\angle A O N=\angle A C B$.
Thus, $\angle E D M=\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,032 |
3. Color each cell of a $12 \times 12$ grid either black or white, such that any $3 \times 4$ or $4 \times 3$ subgrid contains at least one black cell. Find the minimum number of black cells.
（Liang Yingde provided the problem） | 3. The minimum number of black cells required is $n=12$.
First, prove that $n \geqslant 12$.
Since a $12 \times 12$ grid can be divided into $\frac{12 \times 12}{3 \times 4}=12$ non-overlapping $3 \times 4$ sub-grids (excluding the boundaries), according to the problem statement, there must be at least 12 black cells.... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,033 |
5. How many different right-angled triangles with integer side lengths have an area that is 999 times their perimeter (considering congruent triangles as the same)? (Provided by Lin Chang)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result d... | 5. Let the three sides of a right-angled triangle be \(a, b, c\) (with \(c\) being the hypotenuse).
From the Pythagorean triple formula, we have
\[
a = k \cdot 2uv, \quad b = k(u^2 - v^2), \quad c = k(u^2 + v^2),
\]
where the greatest common divisor of the three sides \(k\) is a positive integer, \(u\) and \(v\) are co... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,034 |
8. Given an integer $n \geqslant 2$. On the blackboard, there are $n$ sets, and the following operation is performed: select two sets $A$ and $B$ on the blackboard that do not contain each other, erase them, and then write $A \cap B$ and $A \cup B$. This is called one operation. Continue operating until any two sets on... | 8. First, prove that the number of operations cannot exceed
$$
\mathrm{C}_{n}^{2}=\frac{n(n-1)}{2} \text {. }
$$
When there are $n$ sets on the blackboard, consider the number of pairs of sets that are in a containment relationship. After each operation, this number increases by at least 1.
Assume transforming sets $A... | \frac{n(n-1)}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,035 |
1. Given a positive integer $n$, real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{k=1}^{n} x_{k}$ is an integer. Let
$$
d_{k}=\min _{m \in \mathbf{Z}}\left|x_{k}-m\right|(1 \leqslant k \leqslant n) .
$$
Find the maximum value of $\sum_{k=1}^{n} d_{k}$.
(Cold Gangsong, problem contributor) | 1. Let's assume $x_{1}, x_{2}, \cdots, x_{n} \in (0,1]$. Otherwise, a translation transformation of an integer can be applied to $x_{i} (i=1,2, \cdots, n)$, which does not affect the conclusion of the problem.
Let $\sum_{i=1}^{n} x_{i}=t$. Then, $0 \leqslant t \leqslant n (t \in \mathbf{Z}_{+})$.
Assume $x_{1}, x_{2}, ... | \left[\frac{n}{2}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,036 |
2. As shown in Figure 1, circles $\Gamma_{1}$ and $\Gamma_{2}$ are internally tangent at point $T, M$ and $N$ are two points on circle $\Gamma_{1}$ different from $T$, and two chords $A B$ and $C D$ of circle $\Gamma_{2}$ pass through points $M$ and $N$ respectively. If segments $A C$, $B D$, and $M N$ intersect at the... | 2. As shown in Figure 2, extend $T M$ and $T N$ to intersect the circle $\Gamma_{2}$ at points $E$ and $F$, respectively. Connect $E F$.
Thus, $M N / / E F$.
Therefore, $\frac{T M}{T N}=\frac{M E}{N F}$.
Combining with the intersecting chords theorem, we have
$$
\frac{T M^{2}}{T N^{2}}=\frac{T M}{T N} \cdot \frac{M E}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,037 |
3. Let integer $n \geqslant 2$, positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$
\left(\sum_{i=1}^{n} \frac{1}{1-x_{i}}\right)\left(\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\right) \leqslant \frac{n}{2} .
$$ | 3. Note that,
$$
\begin{array}{l}
\text { 2 } \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}=\sum_{i=1}^{n}\left(x_{i} \sum_{j \neq i} x_{j}\right) \\
=\sum_{i=1}^{n} x_{i}\left(1-x_{i}\right) .
\end{array}
$$
Therefore, the original inequality is equivalent to
$$
\left(\sum_{i=1}^{n} \frac{1}{1-x_{i}}\right)\left[\su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,038 |
5. Let the area of convex quadrilateral $ABCD$ be $S, AB=$ $a, BC=b, CD=c, DA=d$. Prove: For any permutation $(x, y, z, w)$ of $a, b, c, d$, we have
$$
S \leqslant \frac{1}{2}(x y+z w)
$$ | 5. Note that, for the side lengths $a, b, c, d$ of the convex quadrilateral $ABCD$, there are $4! = 24$ permutations.
In fact, depending on whether the side lengths $x, y$ are adjacent, we only need to consider the following two cases.
(1) If $x, y$ are adjacent sides of the convex quadrilateral $ABCD$, without loss o... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,039 |
6. For the sequence $a_{1}, a_{2}, \cdots, a_{m}$, define the sets
$$
\begin{array}{l}
A=\left\{a_{i} \mid 1 \leqslant i \leqslant m\right\}, \\
B=\left\{a_{i}+2 a_{j} \mid 1 \leqslant i, j \leqslant m, i \neq j\right\} .
\end{array}
$$
Let $n$ be a given integer greater than 2. For all strictly increasing arithmetic ... | 6. When $n=3$, the required minimum value is 5; when $n \geqslant 4$, the required minimum value is $2 n$.
First, we prove a lemma.
Lemma When $n \geqslant 4$, for an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}$ with common difference $d$, we have
$$
B=\left\{3 a_{1}+k d \mid 1 \leqslant k \leqslant 3 n-4, k \in ... | 5 \text{ when } n=3, 2n \text{ when } n \geqslant 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,040 |
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