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3. In $\triangle A B C$, it is known that $D$ is a point on side $B C$, $\frac{B D}{D C}=\frac{1}{3}$, $E$ is the midpoint of side $A C$, $A D$ intersects $B E$ at point $O$, and $C O$ intersects $A B$ at point $F$. Find the ratio of the area of quadrilateral $B D O F$ to the area of $\triangle A B C$ | For $\triangle A B C$ and point $O$, applying Ceva's Theorem, we get
$$
\frac{A F}{F B}=3, \frac{A F}{A B}=\frac{3}{4} \text {. }
$$
For $\triangle A D C$ and the transversal $F O C$, we get
$$
\begin{array}{l}
\frac{A O}{O D}=4, \frac{A O}{A D}=\frac{4}{5} . \\
\text { Hence } \frac{S_{\triangle A F O}}{S_{\triangle ... | \frac{1}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,826 |
4. Let $P$ be a point inside $\triangle A B C$ such that $\angle B P A = \angle C P A$, and let $G$ be a point on segment $A P$. The line $B G$ intersects side $A C$ at point $E$, and the line $G C$ intersects side $A B$ at point $F$. Prove: $\angle B P F = \angle C P E$. | Hint: Draw $A M / / P B$, intersecting line $P F$ at point $M$, $A N / / P C$, intersecting line $P E$ at point $N$. Prove $\triangle P A M$ $\xlongequal{\mathscr{\triangle} \triangle P A N}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,827 |
Example 2 Let real numbers $0<x_{1}<x_{2}<\cdots<x_{n}, \lambda_{i} \geqslant 0$
$$
\begin{array}{l}
(i=1,2, \cdots, n), \text { and } \sum_{i=1}^{n} \lambda_{i}=1 \text {. Then } \\
\left(\sum_{i=1}^{n} \lambda_{i} x_{i}\right)\left(\sum_{i=1}^{n} \lambda_{i} x_{i}^{-1}\right) \leqslant A^{2} G^{-2},
\end{array}
$$
$(... | Note that, the inequality is homogeneous. Therefore, without loss of generality, let $G=1$, so we have $x_{n}=\frac{1}{x_{1}}$.
Then for any $x \in\left[x_{1}, x_{n}\right]=\left[x_{1}, \frac{1}{x_{1}}\right]$, there is $x+\frac{1}{x} \leqslant x_{1}+\frac{1}{x_{1}}$.
Hence $\sum_{i=1}^{n} \lambda_{i} x_{i}+\sum_{i=1}^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,828 |
Example 3 Given an integer $n \geqslant 2$, let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers. Prove:
$$
\left(\frac{\sum_{j=1}^{n} \sqrt[j]{a_{1} a_{2} \cdots a_{j}}}{\sum_{j=1}^{n} a_{j}}\right)^{\frac{1}{n}}+\frac{\sqrt[n]{a_{1} a_{2} \cdots a_{n}}}{\sum_{j=1}^{n} \sqrt[j]{a_{1} a_{2} \cdots a_{j}}} \leqsla... | Prove that if $\sqrt[j]{a_{1} a_{2} \cdots a_{j}}=x_{j}(j=1,2, \cdots, n)$, and let $x_{0}=1$. Then
$$
a_{j}=\frac{x_{j}^{j}}{x_{j-1}^{j-1}}(j=1,2, \cdots, n) .
$$
Since the left side of equation (6) is homogeneous with respect to $a_{1}, a_{2}, \cdots, a_{n}$, we can assume without loss of generality that
$$
\sum_{j=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,829 |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its positive divisors $1, 2, 17, 34, 59, 118, 1003, 2006$ is 3240. Find the smallest good number... | Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $\alpha_{i} \in \mathbf{Z}_{+}, p_{i} (i=1,2, \cdots, k)$ are prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$.
From $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=8=2^{3}$, we know $k=1,2,3$.
(1) If $k=1$, then
$\alpha_{1}=7, n=p^{7}$ ($p$ is a prime number), $\sum_{i=... | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,830 |
Example 4 Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in \mathbf{R}$.
$$
\text { Let } c_{i}=\left|a b_{i}+a_{i} b-a_{i} b_{i}\right|(i=1,2, \cdots, n) \text {, }
$$
where, $a=\frac{1}{n} \sum_{i=1}^{n} a_{i}, b=\frac{1}{n} \sum_{i=1}^{n} b_{i}$. | Prove: $\left(\sum_{i=1}^{n} c_{i}\right)^{2} \leqslant\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)$.
Proof:
$$
\begin{array}{l}
x_{i}=a_{i}-a, y_{i}=b_{i}-b(i=1,2, \cdots, n) . \\
\text { Then } \sum_{i=1}^{n} x_{i}=0, \sum_{i=1}^{n} y_{i}=0 \text {. } \\
\text { Hence } c_{i}=\left|a b_{... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,831 |
Question 3 Let positive real numbers $a, b, c$ satisfy $ab + bc + ca = 48$. Try to find the minimum value of $f = \left(a^{2} + 5\right)\left(b^{2} + 5\right)\left(c^{2} + 5\right)$. | Given $A=48, k=5$, hence
$$
f \geqslant 5(48-5)^{2}=9245 \text {. }
$$
When $a=5,\{b, c\}=\left\{\frac{43+\sqrt{97}}{12}, \frac{43-\sqrt{97}}{12}\right\}$,
the equality holds.
Therefore, $f_{\min }=9245$. | 9245 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,833 |
2. In $\triangle A B C$, it is known that $\angle A C B$ is an obtuse angle, $C F$ is the median on side $A B$, and $B E$ is the altitude on side $A C$. If $C F=B E$, then $\angle A C F=(\quad)$.
(A) $15^{\circ}$
(B) $30^{\circ}$
(C) $45^{\circ}$
(D) $60^{\circ}$ | 2. B.
As shown in Figure 3, draw $F D \perp A C$ at point $D$. Then $F D=\frac{1}{2} B E=\frac{1}{2} C F$.
Thus, $\angle A C F=30^{\circ}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,834 |
3. Given real numbers $a, b, c$ satisfy
$$
a^{2}+b^{2}-4 a \leqslant 1, \quad b^{2}+c^{2}-8 b \leqslant-3,
$$
and $c^{2}+a^{2}-12 c \leqslant-26$.
Then the value of $(a+b)^{c}$ is ( ).
(A) 1
(B) 8
(C) 9
(D) 27 | 3. D.
From the condition, we have
$$
\begin{array}{l}
\left(a^{2}+b^{2}-4 a\right)+\left(b^{2}+c^{2}-8 b\right)+\left(c^{2}+a^{2}-12 c\right) \leqslant-28 \\
\Rightarrow(a-1)^{2}+(b-2)^{2}+(c-3)^{2} \leqslant 0 \\
\Rightarrow a=1, b=2, c=3 \\
\Rightarrow(a+b)^{c}=3^{3}=27 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,835 |
6. Given $a_{1}, a_{2}, \cdots, a_{10}$ are ten different positive integers, satisfying
$$
\left|a_{i+1}-a_{i}\right|=2 \text { or } 3,
$$
where, $i=1,2, \cdots, 10$, with the convention $a_{11}=a_{1}$.
If $a_{1}, a_{2}, \cdots, a_{10}$ have the maximum number $M$ and the minimum number $m$, then the maximum value of ... | 6. B.
Assuming without loss of generality that $m=a_{1}=1$, and the maximum value $M=a_{j}$.
Thus, $a_{j}-a_{1}$
$$
\begin{array}{l}
=\left(a_{j}-a_{j-1}\right)+\left(a_{j-1}-a_{j-2}\right)+\cdots+\left(a_{2}-a_{1}\right) \\
\leqslant\left|a_{j}-a_{j-1}\right|+\left|a_{j-1}-a_{j-2}\right|+\cdots+\left|a_{2}-a_{1}\righ... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,836 |
Example 4 Let $n$ be a positive integer, $\varphi(n)$ denote the number of positive integers not exceeding $n$ and coprime to $n$, and $\tau(n)$ denote the number of all positive divisors of $n$. Find all positive integers $n$ that have exactly two distinct prime factors, such that
$$
\varphi(\tau(n))=\tau(\varphi(n)) ... | Let $n=p^{\alpha} q^{\beta}$, where $p, q$ are primes, $p<q$, and $\alpha, \beta \in \mathbf{Z}_{+}$.
$$
\begin{array}{l}
\text { Then } \varphi(n)=p^{\alpha} q^{\beta}\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right) \\
=p^{\alpha-1} q^{\beta-1}(p-1)(q-1), \\
\tau(n)=(\alpha+1)(\beta+1) .
\end{array}
$$
Thus, $\va... | n=2^{r-1} \times 3^{r-1}\left(r\right. \text{ is a prime, } \left.t \in \mathbf{Z}_{+}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,837 |
8. As shown in Figure 1, in the right $\triangle A B C$, it is known that $\angle A C B$ $=90^{\circ}, A C=21, B C=$ 28, and a square $A B D E$ is constructed outward with $A B$ as one side. The angle bisector of $\angle A C B$ intersects $D E$ at point $F$. Then the length of line segment $D F$ is $\qquad$ | 8. 15 .
As shown in Figure 6, let $CF$ intersect $AB$ and $AD$ at points $G$ and $O$ respectively.
Then $\angle BCO = 45^{\circ}$
$= \angle OAB$.
Thus, $O, A, C, B$
are concyclic.
Hence $\angle ABO$
$$
= \angle OCA = 45^{\circ}.
$$
Therefore, $OA = OB$.
Thus, $O$ is the center of the square $ABDE$.
By symmetry, $DF ... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,838 |
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2011}{d}}$ is $\qquad$ . | 9.7.
Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,839 |
10. Given that $a$ and $b$ are real numbers, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
20 x+a>0, \\
15 x-b \leqslant 0
\end{array}\right.
$$
has only the integer solutions $2, 3, 4$. Then the maximum value of $ab$ is . $\qquad$ | 10. -1200 .
From the conditions, we know that $-\frac{a}{20}<x \leqslant \frac{b}{15}$.
The integer solutions of the inequality system are only $2, 3, 4$, so $1 \leqslant-\frac{a}{20}<2,4 \leqslant \frac{b}{15}<5$, which means $-40<a \leqslant-20,60 \leqslant b<75$.
Therefore, $-3000<a b \leqslant-1200$.
Thus, when $a... | -1200 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,840 |
1. Given
$$
f(x)=\left\{\begin{array}{ll}
\left(\frac{1}{4}\right)^{x}, & x \in(-\infty, 1) ; \\
\log _{\frac{1}{2}} x, & x \in[1,+\infty) .
\end{array}\right.
$$
Then $f(f(-1))=()$.
(A) 2
(B) -2
(C) $\frac{1}{4}$
(D) $-\frac{1}{2}$ | $$
-1 . B \text {. }
$$
From the problem, we know that
$$
f(f(-1))=f(4)=\log _{\frac{1}{2}} 4=-2 \text {. }
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,841 |
2. “Real numbers $a, b, c, d$ form an arithmetic sequence in order” is a ( ) condition for “$a+d=b+c$”.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | 2. A.
Sufficiency is obvious.
Necessity: a counterexample is $3+1=2+2$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,842 |
3. If the equation $f(x)-2=0$ has a solution in the interval $(-\infty, 0)$, then the graph of $y=f(x)$ could be $(\quad)$.
| 3. D.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,843 |
4. Let the angle between vectors $a$ and $b$ be $60^{\circ}$, and $|a|=1$, $|\boldsymbol{a}-2 \boldsymbol{b}|=\sqrt{13}$. Then $\boldsymbol{b}=(\quad)$.
(A) $\sqrt{2}$
(B) $\frac{3}{2}$
(C) $2 \sqrt{2}$
(D) 2 | 4. D.
$$
\begin{array}{l}
\text { Given }|a-2 b|=\sqrt{13} \\
\Rightarrow(a-2 b)^{2}=13 \\
\Rightarrow|a|^{2}-4|a||b| \cos +4 b^{2}=13 \\
\Rightarrow|b|=2 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,844 |
12. Non-empty set
$$
A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}
a x-2 y+8 \geqslant 0, \\
x-y-1 \leqslant 0, \\
2 x+a y-2 \leqslant 0
\end{array}\right\} .\right.\right.
$$
When $(x, y) \in A$, the function $z=y-x$ has both a maximum value and a minimum value, then the range of the real number $a$ is $\qquad... | 12. $[2,+\infty)$.
When $a>2$, the region enclosed by the function graph and the Cartesian coordinate system is a triangle, which obviously satisfies the condition;
When $a=2$, the function $z=y-x$ attains its minimum and maximum values on the boundaries $x-y-1=0$ and $2x-2y+8=0$, respectively. | [2,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,847 |
14. (25 points) Given the ellipse $G: \frac{x^{2}}{4}+y^{2}=1$, line
$l$ intersects the ellipse $G$ at points $A$ and $B$, and $|A B|=2$. Determine the positional relationship between line $l$ and the circle $\odot O: x^{2}+y^{2}=1$, and provide a proof. | 14. When the slope of line $l$ does not exist, given $|A B|=2$, we can assume $A(0,1), B(0,-1)$, i.e., the equation of line $l$ is $x=0$, at this time, line $l$ intersects with $\odot O$.
When the slope of line $l$ exists, let its equation be
$$
y=k x+m \text {. }
$$
Combining with the ellipse equation, we get
$$
\lef... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,848 |
Example 6 Let $S(k)$ be the sum of all positive integers not exceeding $k$ and coprime to $k$. Prove: For any positive integers $m, n$ ($n$ is odd), there exist positive integers $x, y$, such that
$$
m \mid x, 2 S(x)=y^{n} \text{. [4] }
$$
(2012, Croatian National Team Selection Exam) | Prove that from $(a, k)=(k-a, k)$, we know that positive integers not exceeding $k$ and coprime with $k$ appear in pairs, and the sum of each pair is $k$. Therefore,
$$
S(k)=\frac{k \varphi(k)}{2} \Rightarrow 2 S(x)=x \varphi(x) .
$$
For any positive integer $m$, let
$$
m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,849 |
15. (25 points) Given that for any $x \geqslant 1$, we have $\ln x - a\left(1 - \frac{1}{x}\right) \geqslant 0$.
Find the range of real number $a$.
| 15. From the problem, we know that $x \ln x - a(x-1) \geqslant 0$.
Let $f(x) = x \ln x - a(x-1) \ (x \geqslant 1)$.
(1) When $a \leqslant 1$, since $x \geqslant 1$, we have
$$
f'(x) = \ln x + 1 - a \geqslant 0 + (1 - a) \geqslant 0.
$$
Thus, $f(x)$ is monotonically increasing on the interval $[1, +\infty)$, i.e., $f(... | a \leqslant 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,850 |
16. (25 points) Given $A \subseteq\{1,2, \cdots, 2014\}$, let real numbers $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} 、 x_{1} 、 x_{2} 、 x_{3}$ satisfy
(i) $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} \in\{-1,0,1\}$ and not all are 0;
(ii) $x_{1}, x_{2} 、 x_{3} \in A$;
(iii) If $x_{i}=x_{j}$, then $\lambda_{i} \lambda_{j} \ne... | 16. (1) Construct a good set $A$ with 503 elements.
Let $A=\{1,3,5, \cdots, 1005\}$.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are all non-zero, then
$$
\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3} \equiv x_{1}+x_{2}+x_{3} \equiv 1(\bmod 2) \text {. }
$$
Thus, $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_... | 503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,851 |
1. Given sets
$$
A=\{1,3,5,7,9\}, B=\{2,4,6,8,10\} \text {. }
$$
If set $C=\{x \mid x=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is $\qquad$ . | ,- 1.9 .
Since $a$ is odd and $b$ is even, all elements in set $C$ are odd.
Also, the minimum value of $a+b$ is 3, and the maximum value is 19, and all odd numbers between 3 and 19 can be obtained, thus, set $C$ contains 9 elements. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,852 |
2. Given the function
$$
f(x)=\left\{\begin{array}{ll}
0, & x<0, \\
1, & x \geqslant 0 .
\end{array}\right.
$$
Then $f(f(x))=$ $\qquad$ | 2. 1 .
Since for any $x \in \mathbf{R}$, we have $f(x) \geqslant 0$, therefore, $f(f(x))=1$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,853 |
3. Given
$$
\sin \alpha+\sqrt{3} \sin \beta=1, \cos \alpha+\sqrt{3} \cos \beta=\sqrt{3} \text {. }
$$
Then the value of $\cos (\alpha-\beta)$ is $\qquad$ . | 3. 0 .
Squaring and adding the two known equations, we get
$$
\begin{array}{l}
4+2 \sqrt{3}(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=4 \\
\Rightarrow \cos (\alpha-\beta)=0
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,854 |
6. Let $O$ be the circumcenter of $\triangle A B C$, and it satisfies
$$
\overrightarrow{O A}+\overrightarrow{O B}=\overrightarrow{O C}
$$
Then $\angle A C B=$ | 6. $\frac{2 \pi}{3}$.
From $\overrightarrow{O A}+\overrightarrow{O B}=\overrightarrow{O C}$, we know that quadrilateral $O A C B$ is a parallelogram.
Also, $|O A|=|O B|=|O C|$, so, $\square O A C B$ is a rhombus, and $\angle O A C=\angle O B C=\frac{\pi}{3}$.
Thus, $\angle A C B=\frac{2 \pi}{3}$. | \frac{2 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,855 |
7. In $\triangle A B C$, if $\tan \frac{A}{2}+\tan \frac{B}{2}=1$, then the minimum value of $\tan \frac{C}{2}$ is $\qquad$ . | 7. $\frac{3}{4}$.
From the given, we know
$$
\begin{array}{l}
\tan \frac{C}{2}=\frac{1}{\tan \frac{A+B}{2}}=\frac{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}} \\
=1-\tan \frac{A}{2} \cdot \tan \frac{B}{2} \\
\geqslant 1-\left(\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{2}\right)^{2} \\
=1... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,856 |
8. A person tosses a coin, with the probability of landing heads up and tails up being $\frac{1}{2}$ each. Construct the sequence $\left\{a_{n}\right\}$, such that $a_{n}=\left\{\begin{array}{ll}1, & \text { if the } n \text {th toss is heads; } \\ -1, & \text { if the } n \text {th toss is tails. }\end{array}\right.$ ... | 8. $\frac{13}{128}$.
From $S_{2} \neq 0$, we know that the first two tosses are either both heads or both tails.
When the first two tosses are both heads, by $S_{8}=2$, we know that in the next six tosses, there must be three heads and three tails, the probability of which is
$$
P_{1}=\left(\frac{1}{2}\right)^{2} \ma... | \frac{13}{128} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,857 |
9. If positive integers $m, n$ satisfy $\frac{(m+n)!}{n!}=5040$, then the value of $m!n$ is $\qquad$ . | 9. 144 .
$$
\begin{array}{l}
\text { Given } \frac{(m+n)!}{n!} \\
=(m+n)(m+n-1) \cdots(n+1), \\
5040=10 \times 9 \times 8 \times 7,
\end{array}
$$
we know $\left\{\begin{array}{l}m+n=10, \\ n+1=7\end{array} \Rightarrow\left\{\begin{array}{l}m=4, \\ n=6 .\end{array}\right.\right.$
Therefore, $m!n=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,858 |
1. If $n=2012 \times 2014$, then the number of positive divisors of $n$ is $\qquad$, and the sum of all positive divisors is $\qquad$. | Notice that, $n=2^{3} \times 19 \times 53 \times 503$. Therefore, the number of positive divisors of $n$ is
$$
(3+1)(1+1)(1+1)(1+1)=32 \text {. }
$$
The sum of all positive divisors is
$$
\begin{array}{l}
\left(1+2+2^{2}+2^{3}\right)(1+19)(1+53)(1+503) \\
=8164800 .
\end{array}
$$ | 32, 8164800 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,859 |
10. Let the monotonic increasing sequence $\left\{a_{n}\right\}$ consist of positive integers, and $a_{7}=120, a_{n+2}=a_{n}+a_{n+1}\left(n \in \mathbf{Z}_{+}\right)$. Then $a_{8}=$ . $\qquad$ | 10. 194 .
From $a_{n+2}=a_{n}+a_{n+1}$, we get $a_{7}=5 a_{1}+8 a_{2}=120, a_{8}=8 a_{1}+13 a_{2}$. Since $(5,8)=1$, and $a_{1}, a_{2}$ are both positive integers, it follows that $8\left|a_{1}, 5\right| a_{2}$.
Let $a_{1}=8 k, a_{2}=5 m\left(k, m \in \mathbf{Z}_{+}\right)$.
Then $k+m=3$.
Also, $a_{1}<a_{2}$, so, $k=1... | 194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,860 |
One, (15 points) Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, and $a_{n+1}=2 S_{n}+\frac{1}{2}\left(n \in \mathbf{Z}_{+}\right)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$.
(2) Insert $n$ real numbers between $a_{n}$ and $a_{n... | (1) Since $a_{n+1}=2 S_{n}+\frac{1}{2}(n \geqslant 1)$, we have
$$
a_{n}=2 S_{n-1}+\frac{1}{2}(n \geqslant 2) \text {. }
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
a_{n+1}-a_{n}=2\left(S_{n}-S_{n-1}\right)=2 a_{n} \\
\Rightarrow a_{n+1}=3 a_{n}(n \geqslant 2) .
\end{array}
$$
Since $\left\{a_{n}\rig... | \frac{15}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,861 |
II. (15 points) Let the internal angles of $\triangle ABC$ be $\angle A, \angle B, \angle C$ with their opposite sides being $a, b, c$ respectively, and they satisfy
$$
\sin A + \sin B = (\cos A + \cos B) \sin C.
$$
(1) Prove that $\triangle ABC$ is a right triangle;
(2) If $a + b + c = 1 + \sqrt{2}$, find the maximum ... | (1) From equation (1), and combining the sine and cosine laws, we get
$$
\begin{array}{l}
a+b=\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right) c \\
\Rightarrow(a+b)\left(a^{2}+b^{2}\right)=(a+b) c^{2} .
\end{array}
$$
Since $a+b>0$, we have $a^{2}+b^{2}=c^{2}$.
Therefore, $\triangle A B C$ ... | \frac{1}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 728,862 |
Three. (15 points) As shown in Figure 2, let $H$ be the orthocenter of acute $\triangle ABC$. Draw a perpendicular from $H$ to $BH$, intersecting $AB$ at point $D$. Draw a perpendicular from $H$ to $CH$, intersecting $AC$ at point $E$. Draw a perpendicular from $C$ to $BC$, intersecting line $DE$ at point $F$. Prove: $... | Extend $C F$ and $H E$ to intersect at point $G$, connect $A H$, and intersect $D E$ at point $M$.
Since $D H \perp B H, A C \perp B H$, therefore, $D H / / A C$.
Similarly, $E H / / A B$.
Thus, quadrilateral $A D H E$ is a parallelogram.
Therefore, $M$ is the midpoint of $A H$.
By $G C \perp B C, A H \perp B C$, we kn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,863 |
Five. (20 points) Given the functions
$$
\begin{array}{l}
f(x)=x \ln x, \\
g(x)=-x^{2}+a x-3(a \in \mathbf{R}) .
\end{array}
$$
(1) If for any $x \in(0,+\infty)$, the inequality $f(x) \geqslant \frac{1}{2} g(x)$ always holds, find the range of $a$;
(2) Prove that for any $x \in(0,+\infty)$, we have
$$
\ln x>\frac{1}{\m... | (1) When $x>0$,
$$
\begin{array}{l}
f(x) \geqslant \frac{1}{2} g(x) \\
\Leftrightarrow \ln x \geqslant \frac{1}{2}\left(-x^{2}+a x-3\right) \\
\Leftrightarrow a \leqslant 2 \ln x+x+\frac{3}{x} . \\
\text { Let } h(x)=2 \ln x+x+\frac{3}{x}(x>0) .
\end{array}
$$
Then $a \leqslant h(x)_{\text {min }}$.
By $h^{\prime}(x)=... | (-\infty, 4] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,864 |
2. The lengths of two sides of a triangle are 20 and 15. The following ( ) cannot be the perimeter of the triangle.
(A) 52
(B) 57
(C) 62
(D) 67
(E) 72 | 2. E.
Let the length of the third side of the triangle be $x$.
Then $20-15<x<20+15 \Rightarrow 5<x<35$
$$
\Rightarrow 40<20+15+x<70 \text {. }
$$
Therefore, the perimeter cannot be 72. | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,865 |
2. Let $n$ be the smallest positive integer satisfying the following conditions:
(1) $n$ is a multiple of 75;
(2) $n$ has exactly 75 positive divisors (including 1 and itself).
Find $\frac{n}{75}$.
(Eighth American Mathematical Invitational) | ```
Given: $n=75 k=3 \times 5^{2} k$.
\[
\begin{array}{l}
\text { By } 75=3 \times 5 \times 5 \\
=(2+1)(4+1)(4+1),
\end{array}
\]
we know the number of prime factors is at most three.
By discussing the number of prime factors, we can solve to get
\[
\left(\frac{n}{75}\right)_{\min }=432 \text {. }
\]
``` | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,867 |
11. Isabella drew two circles with radii of 2 and 3 on a piece of paper, and then drew the common tangents to these two circles, getting exactly $k$ tangents. Then $k$ has $(\quad)$ possible cases.
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6 | 11. D.
When two circles are concentric, $k=0$; when two circles are internally tangent, $k=1$; when two circles intersect, $k=2$; when two circles are externally tangent, $k=3$; when two circles are separated, $k=4$. Therefore, there are 5 possible values. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,868 |
12. The parabolas $y=a x^{2}-2$, $y=4-b x^{2}$ intersect the coordinate axes at exactly four points, and the area of the kite formed by these four points is 12. Then the value of $a+b$ is ( ).
(A) 1
(B) 1.5
(C) 2
(D) 2.5
(E) 3 | 12. B.
Let the parabola $y=a x^{2}-2$ intersect the $x$-axis at points $A\left(x_{1}, 0\right)$ and $C\left(x_{2}, 0\right)$, and intersect the $y$-axis at point $B(0,-2)$; the parabola $y=4-b x^{2}$ intersects the $x$-axis at points $A\left(x_{1}, 0\right)$ and $C\left(x_{2}, 0\right)$, and intersects the $y$-axis at... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,869 |
20. Two isosceles triangles $T$ and $T'$ are not congruent, but they have equal perimeters and equal areas. The side lengths of triangle $T$ are $5, 5, 8$, and the side lengths of triangle $T'$ are $a, a, b$. Then the number ( ) that is closest to $b$ is:
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8 | 20. A.
From the problem, we know that the perimeter of triangle $T$ is 18, and the area is 12, i.e., $\square$
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 a+b=18, \\
\frac{1}{2} b \sqrt{a^{2}-\left(\frac{b}{2}\right)^{2}}=12
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2 a+b=18, \\
b \sqrt{4 a^{2}-b^{2}}... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,870 |
3. Find the smallest positive integer $m$ such that $5 m$ is a fifth power of a positive integer, $6 m$ is a sixth power of a positive integer, and $7 m$ is a seventh power of a positive integer.
(2013, Irish Mathematical Olympiad) | Let
$$
m=2^{a} \times 3^{b} \times 5^{c} \times 7^{d} M(a, b, c, d \in \mathbf{N}),
$$
where $M$ is a positive integer that does not contain the prime factors $2, 3, 5, 7$.
By the problem's condition, to make $m$ as small as possible, we take $M=1$.
Since $5m$ is a fifth power of a positive integer, all the exponents o... | 2^{35} \times 3^{35} \times 5^{84} \times 7^{90} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,871 |
25. As shown in Figure 1, in the upper half-plane of the coordinate plane, there are several circles forming a set. These circles are all tangent to the $x$-axis. The outermost layer (the $L_{0}$ layer) consists of two externally tangent circles with radii $70^{2}$ and $73^{2}$. On $\bigcup_{j=0}^{k-1} L_{j}(k \geqslan... | 25. D.
Let the larger circle in layer $L_{0}$ be $X$, with radius $x$, and the smaller circle be $Y$, with radius $y$. The circle in layer $L_{1}$ is $Z$, with radius $z$.
Let the distance between the two points where circle $X$ and $Z$ touch the $x$-axis be $a$, and the distance between the two points where circle $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,872 |
2. Given $a_{1}, a_{2}, \cdots$ as an integer sequence, satisfying that the arithmetic mean of any consecutive terms is a perfect square, i.e., for each positive integer $n, k, \frac{a_{n}+a_{n+1}+\cdots+a_{n+k-1}}{k}$ is the square of an integer. Prove: the sequence is a constant sequence (all $a_{i}(i=1,2, \cdots)$ a... | 2. Prove the equivalent proposition of this problem:
If the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}$ satisfies
$(f(m)-f(n))(m-n)$
is always a perfect square, then $f$ must be of the form $A^{2} x+B$, where $A, B$ are integers.
First, because for each prime $p$ and positive integer $n$, $p(f(n+p)-f(n))$ is ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,873 |
4. Given that $n$ is a positive even number, and real numbers $c_{1}, c_{2}, \cdots, c_{n-1}$ satisfy
$$
\sum_{i=1}^{n-1}\left|c_{i}-1\right|<1 \text {. }
$$
Prove: the equation
$$
2 x^{n}+\sum_{i=1}^{n-1}(-1)^{i} c_{i} x^{i}+2=0
$$
has no real roots. | 4. Let $P(x)=2 x^{n}+\sum_{i=1}^{n-1}(-1)^{i} c_{i} x^{i}+2$.
Since for all integers $i(1 \leqslant i \leqslant n-1)$, $c_{i} \in(0,2)$, for all $x \leqslant 0$, we have
$$
P(x) \geqslant 2>0 \text {. }
$$
We will now prove that for all $x>0$, $P(x)>0$.
Let $c_{i}=1+\varepsilon_{i}(i=1,2, \cdots, n-1)$.
Then $\sum_{i... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,875 |
6. For a prime $p$, if a subset $S$ of the residues modulo $p$ satisfies:
(1) There does not exist a non-zero residue $\alpha$ modulo $p$ such that $S=$ $\left\{1, \alpha, \alpha^{2}, \cdots\right\}$ (all in the sense of modulo $p$);
(2) There do not exist $a, b, c \in S$ (which can be the same), such that
$$
a+b \equi... | 6. First, prove a stronger conclusion.
Generalize condition (2) " $0 \notin S+S-S$ " to " $0 \notin a_{1} S+a_{2} S+\cdots+a_{k} S$ ", where $a_{1}$, $a_{2}, \cdots, a_{k}$ are fixed integers, and the sum $a_{1}+a_{2}+\cdots+a_{k}$ is not 0 (in the original problem, $k=3$, and $\left(a_{1}, a_{2}, a_{3}\right) = (1,1,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,876 |
4. Find all positive integers $n$ such that $n(n+2)(n+4)$ has at most 15 positive divisors.
(2010, Slovenia National Team Selection Exam) | If $n$ is even, let $n=2k$, then
$$
n(n+2)(n+4)=2^{3} k(k+1)(k+2) \text{.}
$$
Since $k$, $k+1$, and $k+2$ must include at least one even number and one multiple of 3, it can be determined that only $n=4$ meets the condition.
If $n$ is odd, considering that $n$, $n+2$, and $n+4$ are pairwise coprime, and using $\tau(n... | n=1,2,3,4,5,7,9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,878 |
2. In $\triangle A B C$, it is known that the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and the altitude from $A$ to side $B C$ is $A D=\frac{1}{2} a$. Then the range of $\frac{(b+c)^{2}}{2 b c}$ is . $\qquad$ | 2. $[2,1+\sqrt{2}]$.
Notice,
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{1}{2} a A D=\frac{1}{4} a^{2} \\
\Rightarrow 2 b c \sin A=a^{2}=b^{2}+c^{2}-2 b c \cos A \\
\Rightarrow(b+c)^{2}=2 b c(1+\sin A+\cos A) \\
\Rightarrow \frac{(b+c)^{2}}{2 b c}=1+\sqrt{2} \sin \left(A+\frac{\pi}{4}\right) \... | [2,1+\sqrt{2}] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,879 |
3. Choose three different angles from $1^{\circ}, 2^{\circ}, \cdots, 179^{\circ}$ to form the three interior angles of a triangle. There are $\qquad$ different ways to do this. | 3.2611.
Notice that, the equation $x+y+z=180$ has $\mathrm{C}_{179}^{2}=$ 15931 sets of positive integer solutions, among which, the solution where $x=y=z$ is 1 set, and the solutions where exactly two of $x, y, z$ are equal are $88 \times 3=$ 264 sets.
Therefore, the number of selection methods is $\frac{15931-1-264}... | 2611 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,880 |
4. Given complex numbers $z_{1}$ and $z_{2}$ satisfy $\arg \frac{z_{1}}{z_{2}}=\frac{5 \pi}{6}$, and $\left|z_{1}\right|=\left|z_{1}+z_{2}\right|$. Then the minimum value of $\frac{\left|2 z_{1}+t z_{2}\right|}{\left|z_{2}\right|}$ is $\qquad$ | 4. $\frac{\sqrt{3}}{3}$.
From Fig. 5, we can easily see that
$$
\begin{array}{l}
\left|\frac{z_{1}}{z_{2}}\right|=\frac{\sqrt{3}}{3} . \\
\text { Then } \frac{z_{1}}{z_{2}} \\
=\frac{\sqrt{3}}{3}\left(\cos \frac{5 \pi}{6}+\mathrm{i} \sin \frac{5 \pi}{6}\right) . \\
\text { Therefore, } \frac{\left|2 z_{1}+t z_{2}\righ... | \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,881 |
5. The equation
$$
\left(\sin ^{4} x+\sin ^{2} x\right)^{3}-\left(\cos ^{4} x+\cos ^{2} x\right)^{3}=2 \cos 2 x
$$
has the solution set $\qquad$ . | 5. $\left\{\left.\frac{(2 k+1) \pi}{4} \right\rvert\, k \in \mathbf{Z}\right\}$.
The original equation is
$$
\begin{array}{l}
\left(\sin ^{4} x+\sin ^{2} x\right)^{3}-\left(\cos ^{4} x+\cos ^{2} x\right)^{3} \\
=2\left(\cos ^{2} x-\sin ^{2} x\right) \\
=\left(\sin ^{2} x+\cos ^{2} x+1\right)\left(\sin ^{2} x-\cos ^{2}... | \left\{\left.\frac{(2 k+1) \pi}{4} \right\rvert\, k \in \mathbf{Z}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,882 |
6. Given real numbers $x, y$ satisfy $x^{2}-x y-2 y^{2}=1$. Then the minimum value of $2 x^{2}+y^{2}$ is $\qquad$ .
| 6. $\frac{2+2 \sqrt{3}}{3}$.
Notice,
$$
\begin{array}{l}
x^{2}-x y-2 y^{2}=1 \Leftrightarrow(x+y)(x-2 y)=1 . \\
\text { Let } x+y=m, x-2 y=n .
\end{array}
$$
Then $x=\frac{2 m+n}{3}, y=\frac{m-n}{3}$.
Thus $2 x^{2}+y^{2}=\frac{1}{9}\left(9 m^{2}+3 n^{2}+6 m n\right)$
$$
=\frac{1}{9}\left(9 m^{2}+\frac{3}{m^{2}}+6\rig... | \frac{2+2 \sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,883 |
7. As shown in Figure 2, given points $A, B, C$ on the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
$A B$ passes through the origin $O, A C$ passes through the right focus $F$. If
$$
B F \perp A C \text {, and } \overrightarrow{A F}=2 \overrightarrow{F C} \text {, }
$$
then the eccentricity of the e... | 7. $\frac{\sqrt{5}}{3}$.
Let the left focus of the ellipse be $F^{\prime}$.
By symmetry, we know $A F^{\prime} \perp A C$, and $\left|A F^{\prime}\right|=|B F|$.
Let $\left|A F^{\prime}\right|=|B F|=m,|A F|=n$.
Then $|C F|=\frac{n}{2}$.
Thus, $\left\{\begin{array}{l}m+n=2 a, \\ m^{2}+n^{2}=4 c^{2}, \\ m^{2}+\left(n+\f... | \frac{\sqrt{5}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,884 |
8. Let $f_{0}(x)=|x|-2015$,
$$
f_{n}(x)=\left|f_{n-1}(x)\right|-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then the number of zeros of the function $y=f_{2015}(x)$ is
$\qquad$ | 8.4031 .
From the graph, it is easy to see that the function $y=f_{1}(x)$ has 4 zeros, the function $y=f_{2}(x)$ has 6 zeros, $\cdots \cdots$ and so on, the function $y=f_{2014}(x)$ has 4030 zeros. However, the intersection point of the function $y=f_{2014}(x)$ with the y-axis is $(0,1)$, therefore, the function $y=f_... | 4031 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,885 |
9. (16 points) Given the quadratic function $f(x)$ satisfies $|f(0)| \leqslant 2,|f(2)| \leqslant 2,|f(-2)| \leqslant 2$. When $x \in[-2,2]$, find the maximum value of $y=|f(x)|$.
| $$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
f(0)=c, \\
f(2)=4 a+2 b+c, \\
f(-2)=4 a-2 b+c
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a=\frac{f(2)+f(-2)-2 f(0)}{8}, \\
b=\frac{f(2)-f(-2)}{4}, \\
c=f(0)
\end{array},\right. \\
\Rightarrow f(x)=a x^{2}+b x+c \\
=\frac{x^{2}+2 x}{8} f(2)+\frac{x^... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,886 |
10. (20 points) As shown in Figure 3, the ellipses $C_{1}$ and $C_{2}$, with centers at the origin and foci on the $x$-axis and $y$-axis respectively, both pass through the point $M(0, -\sqrt{2})$, and the eccentricities of ellipses $C_{1}$ and $C_{2}$ are both $\frac{\sqrt{2}}{2}$. Two lines with slopes $k$ and $k'$ a... | 10. Notice that the standard equations of ellipses $C_{1}$ and $C_{2}$ are
$$
\frac{x^{2}}{4}+\frac{y^{2}}{2}=1, \frac{y^{2}}{2}+x^{2}=1 \text {; }
$$
The line $l_{M P}: y=k x-\sqrt{2}$.
From $\left\{\begin{array}{l}\frac{x^{2}}{4}+\frac{y^{2}}{2}=1, \\ y=k x-\sqrt{2}\end{array}\right.$
$\Rightarrow\left(2 k^{2}+1\rig... | (0, \sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,887 |
11. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{0}=1, \\
a_{n+1}-a_{n}=\sum_{k=0}^{n}[2(n-k)+1] a_{k}(n \in \mathbf{N}) .
\end{array}
$$
Prove: When $n \geqslant 1$, we have
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{7}{10} .
$$ | 11. Notice,
$$
\begin{array}{l}
a_{n+1}-a_{n}=\sum_{k=0}^{n}[2(n-k)+1] a_{k}, \\
a_{n+2}-a_{n+1}=\sum_{k=0}^{n+1}[2(n-k)+3] a_{k} \\
\Rightarrow\left(a_{n+2}-a_{n+1}\right)-\left(a_{n+1}-a_{n}\right) \\
\quad=2 \sum_{k=0}^{n} a_{k}+a_{n+1} \\
\Rightarrow a_{n+2}-3 a_{n+1}-a_{n}=2 \sum_{k=0}^{n-1} a_{k} \\
\Rightarrow a... | \frac{7}{10} | Algebra | proof | Yes | Yes | cn_contest | false | 728,888 |
Example 1 Given four non-coincident parallel planes. Prove: There exists a regular tetrahedron, whose four vertices lie on these four planes respectively.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, the instructio... | 【Analysis】Let the four known planes be $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$, respectively. Thus, the problem is transformed into finding points $P_{i}$ in the plane $\alpha_{i} (i=1, 2, 3, 4)$ such that the tetrahedron $P_{1} P_{2} P_{3} P_{4}$ is a regular tetrahedron. Let the distance between plane $\alph... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,889 |
Example 1 Let $x, y, z > 0$, and $x+y+z=1$. Find
$$
f(x, y, z)=\sum \frac{x(2 y-z)}{1+x+3 y}
$$
the maximum value. | Notice,
$$
\begin{array}{l}
f(x, y, z)=\sum \frac{x(2 y-z)}{1+x+3 y} \\
=\sum \frac{x(2 y-z)}{2+(2 y-z)} .
\end{array}
$$
Consider the function $f(x)=\frac{x}{x+2}$.
It is easy to see that, $f^{\prime}(x)=\frac{2}{(x+2)^{2}}>0$,
$$
f^{\prime \prime}(x)=\frac{-4}{(x+2)^{3}}<0 \text {. }
$$
Therefore, $f(x)$ is a conve... | \frac{1}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,891 |
Example 4 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ be all positive numbers, and $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}} .
$$
(Fourth China Mathematical Olympiad) | Prove that by Cauchy's inequality, $\sum_{i=1}^{n} \sqrt{x_{i}} \leqslant \sqrt{n}$.
Strengthen the inequality to be proved:
$$
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sqrt{n}}{\sqrt{n-1}}.
$$
Consider the function $f(x)=\frac{1}{\sqrt{1-x}}$.
It is easy to see that $f^{\prime}(x)=\frac{1}{2}(1-x)^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,892 |
1. For a finite set of points $S$ in the plane, if for any two distinct points $A, B$ in $S$, there exists a point $C$ in $S$ such that $A C = B C$, then the point set $S$ is called "balanced"; if for any three distinct points $A, B, C$ in $S$, there does not exist a point $P$ in $S$ such that $P A = P B = P C$, then t... | 1. (1) For odd $n \geqslant 3$, consider the set $S$ of $n$ vertices of a regular $n$-gon.
The following explains: The point set $S$ is balanced.
In fact, all points in the point set $S$ lie on a circle $\Gamma$. For any two different points $A$ and $B$ in $S$, the points $A$ and $B$ divide the circle $\Gamma$ into two... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,893 |
Example 4 Proof: For any pyramid with a base that is a regular $n$-sided polygon, there does not exist a section that is a regular $(n+1)$-sided polygon.
(22nd All-Russian Mathematical Olympiad) | 【Analysis】Assume the regular $(n+1)$-gon $B_{1} B_{2} \cdots B_{n+1}$ is a section of the $n$-pyramid $S-A_{1} A_{2} \cdots A_{n}$, with the base $A_{1} A_{2} \cdots A_{n}$ being a regular $n$-gon.
Next, we discuss the following three cases:
$$
n=5, n=2 k-1(k>3), n=2 k(k>2) \text {. }
$$
Since the $n$-pyramid has $n+1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,894 |
3. For a sequence of real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define its "value" as $\max _{1 \leqslant i \leqslant n}\left\{\left|x_{1}+x_{2}+\cdots+x_{i}\right|\right\}$. Given $n$ real numbers, David and George want to arrange these $n$ numbers into a sequence with low value. On one hand, diligent David examines ... | 3. $c=2$.
If the initial numbers given are $1, -1, 2, -2$, then David arranges these four numbers as $1, -2, 2, -1$, and George can get the sequence $1, -1, 2, -2$.
Thus, $D=1, G=2$.
Therefore, $c \geqslant 2$.
We now prove: $G \leqslant 2 D$.
Let the $n$ real numbers be $x_{1}, x_{2}, \cdots, x_{n}$, and assume David... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,895 |
4. Find all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$, such that for all integers $m, n$, we have
$$
\begin{array}{r}
f(f(m)+n)+f(m) \\
=f(n)+f(3 m)+2014 .
\end{array}
$$ | 4. Let the function $f$ satisfy equation (1).
Let $C=1007$, and define the function $g: \mathbf{Z} \rightarrow \mathbf{Z}$ such that for all $m \in \mathbf{Z}$, we have
$$
g(m)=f(3 m)-f(m)+2 C.
$$
In particular, $g(0)=2 C$.
Thus, for all $m, n \in \mathbf{Z}$, we have
$$
f(f(m)+n)=g(m)+f(n).
$$
By mathematical induc... | f(n)=2n+1007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,896 |
2. The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{0}=1, x_{1}=6, \\
x_{2}=x_{1}+\sin x_{1}=5.72 \cdots, \\
x_{3}=x_{2}+\cos x_{2}=6.56 \cdots .
\end{array}
$$
In general,
if $x_{n} \geqslant x_{n-1}$, then $x_{n+1}=x_{n}+\sin x_{n}$;
if $x_{n}<x_{n-1}$, then $x_{n+1}=x_{n}+\cos x_{n}$. For any $n... | 2. $c_{\min }=3 \pi$.
Firstly, note: if $x\frac{5 \pi}{2}$.
Thus, $\cos x_{k-1}2 \pi$, it is easy to prove by mathematical induction that $x_{n}>2 \pi,\left\{x_{n}\right\}$ is monotonically increasing.
Therefore, $x_{n+1}=x_{n}+\sin x_{n}(n \geqslant 3)$.
If $x_{n}<c<3 \pi\left(n \in \mathbf{Z}_{+}\right)$, then
$x_{n... | 3 \pi | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,898 |
5. As shown in Figure 1, given an acute triangle $\triangle ABC$ with its circumcircle $\odot O$, the orthocenter and incenter are $H$ and $I$, respectively. The midpoint of $AH$ is $M$, and it satisfies $AO \parallel MI$. Let the extension of $AH$ intersect $\odot O$ at point $D$, and the lines $AO$ and $OD$ intersect... | 5. (1) Clearly, $\angle O A D=\angle O D A$.
Since $A D \perp B C$, we have
$$
\begin{array}{l}
\angle O A D+\angle A P C=90^{\circ}, \\
\angle O D A+\angle D Q C=90^{\circ} .
\end{array}
$$
Also, $\angle D Q C=\angle P Q O$, thus
$$
\angle O D A+\angle P Q O=90^{\circ} \text {. }
$$
Therefore, $\angle P Q O=\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,899 |
Example 5 Proof: In space, there do not exist four points $A, B,$ $C, D$, such that $AB=CD=8, AC=BD=10, AD=$ $BC=13$. | 【Analysis】Construct $\triangle A B C$, such that
$$
A B=8, A C=10, B C=13 \text {. }
$$
Then construct $\triangle B C D$, such that $C D=8, B D=10$.
Let the midpoint of side $B C$ be $M$. It is clear that, regardless of whether $A, B, C, D$ are coplanar, we always have $A D \leqslant A M+M D$.
Since $A M$ and $M D$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,900 |
8. Given a graph $G$ with eight vertices.
(1) If any induced subgraph (a subgraph that retains the adjacency relationships of these five vertices in the original graph, hereinafter referred to as a subgraph) containing five vertices contains a complete subgraph of order three, prove: Graph $G$ contains a complete subgr... | 8. The answer to (2) is negative.
As shown in Figure 7,
a counterexample is given, where every
three consecutive vertices form a triangle
(a complete subgraph of order 3).
In fact, since any six vertices chosen from the octagon must include three that are connected, any subgraph containing six vertices must contain... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,901 |
2. In $\square A B C D$, $\angle B<90^{\circ}, A B<B C$. From point $D$ draw tangents to the circumcircle $\Gamma$ of $\triangle A B C$, the points of tangency are $E$ and $F$. It is known that $\angle E D A=\angle F D C$. Find $\angle A B C$ | 2. Let $A D$ intersect the circle $\Gamma$ at point $C^{\prime}$.
Since $\angle E D C^{\prime}=\angle F D C$, points $E$ and $F$ as well as $C^{\prime}$ and $C$ are symmetric with respect to the angle bisector of $\angle E D F$, so,
$$
C^{\prime} D=C D=B A=C C^{\prime} \text {. }
$$
Thus, $\triangle C^{\prime} C D$ i... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,902 |
1. Let $p, q \in \mathbf{R}_{+}$, and satisfy
$$
\begin{array}{l}
\log _{9} p=\log _{12} q=\log _{16}(p+q) . \\
\text { Then } \frac{q}{p}=\text {. }
\end{array}
$$ | $-1 . \frac{1+\sqrt{5}}{2}$.
By the change of base formula, we get
$$
\begin{array}{l}
\frac{\ln p}{2 \ln 3}=\frac{\ln q}{2 \ln 2+\ln 3}=\frac{\ln (p+q)}{4 \ln 2} . \\
\text { Then } \frac{\ln p-2 \ln q}{2 \ln 3-2(2 \ln 2+\ln 3)}=\frac{\ln (p+q)}{4 \ln 2} \\
\Rightarrow 2 \ln q-\ln p=\ln (p+q) \\
\Rightarrow q^{2}=p(p+... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,903 |
2. Given the function
$$
f(x)=\cos ^{2} x-a \sin x+2
$$
has a maximum value of 5. Then the value of the real number $a$ is $\qquad$ . | 2. $\pm 3$.
Notice,
$$
f(x)=-\left(\sin x+\frac{a}{2}\right)^{2}+3+\frac{a^{2}}{4} \text {. }
$$
When $-1 \leqslant \frac{a}{2} \leqslant 1$,
$$
f(x)_{\max }=3+\frac{a^{2}}{4}=5
$$
$\Rightarrow a= \pm 2 \sqrt{2} \notin[-2,2]$ (discard);
When $\frac{a}{2} \geqslant 1$,
$$
f(x)_{\max }=2+a=5 \Rightarrow a=3 ;
$$
When ... | \pm 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,904 |
3. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}},
$$
where, $\left[a_{n}\right] 、\left\{a_{n}\right\}$ represent the integer part and the fractional part of the positive number $a_{n}$, respectively. Then $a_{2015}=$ $\qquad$ | 3. $3021+\sqrt{3}$.
From the given, we have
$$
\begin{array}{l}
a_{1}=1+(\sqrt{3}-1), \\
a_{2}=1+\frac{1}{\sqrt{3}-1}=2+\frac{\sqrt{3}-1}{2} .
\end{array}
$$
It is easy to see that $a_{2 k+1}=3 k+1+(\sqrt{3}-1)$,
$$
a_{2 k+2}=3 k+2+\frac{\sqrt{3}-1}{2} \text { . }
$$
Therefore, $a_{2015}=3022+(\sqrt{3}-1)=3021+\sqrt... | 3021+\sqrt{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,905 |
7. Let the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1$ pass through the fixed point $P(1,2)$. Then the minimum value of $m+n$ is $\qquad$ . | 7. $\left(1+2^{\frac{2}{3}}\right)^{\frac{3}{2}}$.
From the given, we have $\frac{1}{m^{2}}+\frac{4}{n^{2}}=1$.
Let $m=\frac{1}{\cos \alpha}, n=\frac{2}{\sin \alpha}$.
Then $(m+n)^{2}=\left(\frac{1}{\cos \alpha}+\frac{2}{\sin \alpha}\right)^{2}$
$=\sec ^{2} \alpha+4 \operatorname{csc}^{2} \alpha+\frac{4}{\sin \alpha \... | \left(1+2^{\frac{2}{3}}\right)^{\frac{3}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,906 |
8. A math competition problem: the probabilities of A, B, and C solving it alone are $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$, respectively, where $a$, $b$, and $c$ are all single-digit numbers. Now A, B, and C are solving this problem independently at the same time. If the probability that exactly one of them s... | 8. $\frac{4}{15}$.
According to the problem, we have
$$
\begin{array}{l}
\frac{1}{a} \cdot \frac{b-1}{b} \cdot \frac{c-1}{c}+\frac{a-1}{a} \cdot \frac{1}{b} \cdot \frac{c-1}{c}+ \\
\frac{a-1}{a} \cdot \frac{b-1}{b} \cdot \frac{1}{c}=\frac{7}{15} \\
\Rightarrow 15[(a-1)(b-1)+(b-1)(c-1)+ \\
\quad(c-1)(a-1)]=7 a b c \\
\... | \frac{4}{15} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,907 |
Example 7 In a cubic box with an edge length of 1, do there exist three non-overlapping regular tetrahedra with edge lengths of 1? | 【Analysis】As shown in Figure 8, let $M$ and $N$ be the midpoints of the edges $D D_{1}$ and $B B_{1}$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$. Since the distance between the lines $A A_{1}$ and $M N$ is equal to $\frac{\sqrt{2}}{2}$, which is exactly the distance between the opposite edges of a regular tetrahedro... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,908 |
One, (40 points) Given positive integers $n, k (n>k)$, and $x_{1}, x_{2}, \cdots, x_{k}>0$. Find $x_{k+1}, x_{k+2}, \cdots, x_{n}>0$, such that $\sum_{1 \leqslant i, j \leqslant n} \frac{x_{i}}{x_{j}}$ is minimized. | Let $A=\sum_{i=1}^{k} x_{i}, B=\sum_{i=1}^{k} \frac{1}{x_{i}}$.
$$
\begin{array}{l}
\text { Then } \sum_{1 \leqslant i, j \leqslant n} \frac{x_{i}}{x_{j}}=\left(\sum_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{x_{i}}\right)-n \\
=\left(A+\sum_{i=k+1}^{n} x_{i}\right)\left(B+\sum_{i=k+1}^{n} \frac{1}{x_{i}}\righ... | x_{k+1}=x_{k+2}=\cdots=x_{n}=\sqrt{\frac{A}{B}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,909 |
441 Given that $a$ and $b$ are both positive numbers. Prove:
$$
\prod_{k=1}^{2015}\left(a^{k}+b^{k}\right) \geqslant \prod_{m=1}^{403}\left(a^{m+806}+b^{m+806}\right)^{5} .
$$ | Prove that since $a, b$ are both positive numbers, we have:
$$
\begin{array}{l}
a^{m} b^{m+4 \times 403} + a^{m+4 \times 403} b^{m} \geqslant 2 a^{m+806} b^{m+806}. \\
\text { Then } \left(a^{m} + b^{m}\right)\left(a^{m+4 \times 403} + b^{m+4 \times 403}\right) \\
= a^{2 m+4 \times 403} + a^{m} b^{m+4 \times 403} + a^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,911 |
Given positive integers $a$, $b$, $c$, $d$ satisfy
$$
a+2b+3c+4d \leqslant 10
$$ | Proof:
$$
\frac{1}{1+3 a}+\frac{1}{1+3 b^{2}}+\frac{1}{1+3 c^{3}}+\frac{1}{1+3 d^{4}} \geqslant 1 \text {. }
$$
The original inequality is equivalent to
$$
\frac{a}{1+3 a}+\frac{b^{2}}{1+3 b^{2}}+\frac{c^{3}}{1+3 c^{3}}+\frac{d^{4}}{1+3 d^{4}} \leqslant 1 \text {. }
$$
By the Cauchy-Schwarz inequality, we have
$$
(1+... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,912 |
Given $\odot O$ is the circumcircle of Rt $\triangle ABC$, $AB$ is the diameter of $\odot O$, and $\odot Q$ is tangent to sides $AC$ and $BC$. Prove: $\odot Q$ is tangent to $\odot O$ if and only if the radius of $\odot Q$ $=AC + BC - AB$. | Prove the necessity.
Given that $\odot Q$ is tangent to $\odot O$.
Let the circumradius and inradius of $\triangle ABC$ be $R$ and $r$, and the radius of $\odot Q$ be $x$.
It is easy to know that $r=\frac{1}{2}(AC+BC-AB)$.
As shown in Figure 4, connect $CQ$ and extend it to intersect $\odot O$ at point $S$, and connect... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,913 |
Example 9 There are seven tetrahedra on the table. For any three of these tetrahedra, there exists a horizontal plane that cuts out three triangles of equal area from them. Question: Is there a plane that can cut out triangles of equal area from all seven tetrahedra? | 【Analysis】For the convenience of discussion, we quantify the cross-sectional areas. Let the base areas and heights of each pyramid be denoted as $S_{i}$ and $H_{i} (i=1,2, \cdots, 7)$. Let $S_{i}(x)$ represent the area of the cross-section at the $i$-th pyramid when the horizontal plane is at a distance $x$ from the ta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,914 |
Example 1 Can the numbers
$$
1,1,2,2,3,3,4,4,5,5, \cdots, 1986,1986
$$
be rearranged in a row so that between each pair of $i$'s there are exactly $i$ numbers? Please prove your conclusion. | Assume that there exists a rearrangement satisfying the problem's conditions. Consider the sum of the positions of all numbers after rearrangement. Let the positions from left to right after rearrangement be denoted as the 1st, 2nd, ..., 3972nd positions.
On one hand, for any two $i$ ($i=1,2, \cdots, 1986$), the posit... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,915 |
Example 2 There are 15 football teams participating in a single round-robin tournament, meaning each pair of teams plays only one match. Prove: there is at least one match where the sum of the number of games played by the two opposing teams before this match is odd, and construct a scenario where there is only one suc... | Prove that for a certain match, consider the sum of the number of matches played by one of the teams before.
If it is odd, then this team's match is called an "O-type match". Thus, each of the 15 teams has played 7 O-type matches.
If for every match, the sum of the number of matches played by the opposing teams befor... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,916 |
Question 1 Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ be complex numbers. Is there necessarily a $k \in\{1,2, \cdots, n\}$, such that
$$
\sum_{i=1}^{n}\left|a_{i}-a_{k}\right| \leqslant \sum_{i=1}^{n}\left|b_{i}-a_{k}\right| \text { ? }
$$ | In fact, at this point, $k$ does not exist.
For example, take
$$
\begin{array}{c}
b_{1}=b_{2}=\cdots=b_{n}=0, \\
a_{j}=\mathrm{e}^{\frac{2 x j_{j}}{n}}(j=1,2, \cdots, n) . \\
\text { Then } \sum_{i=1}^{n}\left|a_{i}-a_{k}\right|>\sum_{i=1}^{n}\left|b_{i}-a_{k}\right| .
\end{array}
$$ | not found | Inequalities | proof | Yes | Yes | cn_contest | false | 728,917 |
Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ be complex numbers. Is there definitely a $k \in\{1,2, \cdots, n\}$, such that
$$
\sum_{i=1}^{n}\left|a_{i}-a_{k}\right|^{2} \leqslant \sum_{i=1}^{n}\left|b_{i}-a_{k}\right|^{2} ?
$$ | The conclusion is that $k$ does not necessarily exist.
For example, take
$$
\begin{array}{l}
a_{1}=n, a_{2}=a_{3}=\cdots=a_{n}=0, \\
b_{1}=b_{2}=\cdots=b_{n}=1 .
\end{array}
$$
When $k=1$, $(n-1) n^{2} \leqslant(n-1)^{2} n$;
When $k \neq 1$, $n^{2} \leqslant n$.
Neither holds. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,918 |
Question 4 Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ be complex numbers. Then there must exist $k \in\{1,2, \cdots, n\}$, such that
$$
\begin{aligned}
& \sum_{i=1}^{n}\left|a_{i}-a_{k}\right| \leqslant \sum_{i=1}^{n}\left|b_{i}-a_{k}\right| \\
\text { or } & \sum_{i=1}^{n}\left|b_{i}-b_{k}\right| \... | This is the famous Levy Theorem.
Levy Theorem: On the plane, given $n$ red points and $n$ blue points, the sum of the distances between points of different colors is not less than the sum of the distances between points of the same color.
The problem at the beginning of the article is a special case of the Levy Theore... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,919 |
Question 5 Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ be complex numbers. Prove: there must exist $k \in\{1,2, \cdots, n\}$, such that
$$
\begin{array}{l}
\sum_{i=1}^{n}\left|a_{i}-a_{k}\right|^{2} \leqslant \sum_{i=1}^{n}\left|b_{i}-a_{k}\right|^{2}, \\
\text { or } \quad \sum_{i=1}^{n}\left|b_{i}... | To prove for \( j, k \in \{1, 2, \cdots, n\} \), let
\[
\begin{aligned}
f_{j k}= & \left(\left|b_{j}-a_{k}\right|^{2}-\left|a_{j}-a_{k}\right|^{2}\right)+ \\
& \left(\left|a_{j}-b_{k}\right|^{2}-\left|b_{j}-b_{k}\right|^{2}\right) .
\end{aligned}
\]
To prove problem 5, it suffices to show that there exists an integer ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,920 |
Question 1 As shown in Figure 1, $M$ and $N$ are the midpoints of the arcs $\overparen{BC}$ and $\overparen{AC}$ of the circumcircle $\Gamma$ of acute triangle $\triangle ABC$ $(\angle A < \angle B)$. A line through point $C$ parallel to $MN$ intersects the circle $\Gamma$ at point $P$, and $I$ is the incenter. Extend ... | In the problem, $M$ and $N$ are common points. From $\overparen{N P}=\overparen{C M}$, it is easy to get $\overparen{A P}=\overparen{P B}$, which means $P$ is the midpoint of arc $\overparen{A C B}$, and thus the intersection point of the external angle bisector of $\angle A C B$ with the circle $\Gamma$. Therefore, th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,921 |
Question 4 As shown in Figure $4, \triangle A B C$ has an incircle $\odot I$ that is tangent to side $A B$ at point $X, \angle A C B$'s exterior angle bisector intersects the circumcircle $\Gamma$ of $\triangle A B C$ at point $P$, extending $P I$, it intersects the circumcircle $\Gamma$ at point $T$. Prove: $\angle C ... | Let the circumradius and inradius be $R$ and $r$, respectively, and the distance between the two centers be $d$.
Extend $C I$, intersecting the circle $\Gamma$ at point $Q$. As shown in Figure 5, connect $P Q$.
In fact, by $\angle X I T = \angle I P Q$, and according to Euler's formula $2 R r = R^{2} - d^{2} = P I \cdo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,922 |
Example 1 As shown in Figure 1, given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, $M$ is the midpoint of the arc $\overparen{A B}$ of $\odot O_{1}$ inside $\odot O_{2}$. The chord $M P$ of $\odot O_{1}$ intersects $\odot O_{2}$ at point $Q$. A tangent line $l_{1}$ to $\odot O_{1}$ is drawn thr... | 【Analysis】Let $X$ be the common tangent point on $\odot O_{2}$. It can be observed that $X$ is the Miquel point of the complete quadrilateral $K J C Q I P$. Therefore, $X$, $I$, $K$, and $J$ are concyclic. Let $X T$ be the tangent line of $\odot O_{2}$. We need to prove that $X T$ is the tangent line of the circumcircl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,923 |
Example 2 Mannheim's Theorem: If a fixed point $A$ is outside a fixed circle $\odot O$, and points $B$ and $C$ are on the tangents $AJ$ and $AK$ of $\odot O$ respectively, and $BC$ is tangent to $\odot O$ at point $D$, then the circumcircle of $\triangle ABC$ is tangent to a certain fixed circle. ${ }^{[1]}$ | 【Analysis】As shown in Figure 2, draw $HI \perp AO$ through point $O$, and draw a perpendicular from point $H$ to $AB$, intersecting $AO$ at point $E$. It is not difficult to see through a special case that this theorem is equivalent to proving that the circumcircle of $\triangle ABC$ is tangent to $\odot E$. Find the p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,924 |
Example 3 As shown in Figure 3, given that $O, G$ are the circumcenter and centroid of $\triangle ABC$ respectively, $D$ is the midpoint of $BC$, the circle with $BC$ as its diameter intersects the altitude $AH$ at point $E$, $EG$ intersects $OD$ at point $F$, $FK \parallel OB, FL \parallel OC, KM \perp BC, NL \perp BC... | 【Analysis】Obviously, $M N$ is bisected by $O D$. First, calculate to get $M D \perp D N$. Then, let $X$ be the common tangent point on $\odot I$, and prove that $X$ is the Miquel point of $M, D, N$ with respect to $\triangle A B C$, obtaining that $A, M, X, N$ are concyclic. Finally, prove that the tangent line $T X$ o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,925 |
Example 4 As shown in Figure 4, let $D$ be any point on the circumcircle $\odot O$ of $\triangle A B C$, and the tangent line to $\odot O$ through $D$ be $l$. Prove: the lines symmetric to the tangent line $l$ with respect to the three sides of $\triangle A B C$ intersect to form $\triangle H I J$ whose circumcircle is... | 【Analysis】Let the common tangent point be $X$. Through attempts, we find that the circumcircle of $\triangle X C I$ and the circumcircle of $\triangle H A X$ pass through the point $D$'s reflection over $A C$, denoted as $D^{\prime}$. The following steps can be carried out accordingly.
Proof Let the reflection of poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,926 |
Example 5 Feuerbach's Theorem: The nine-point circle of $\triangle A B C$ is tangent to the incircle. $[2]$ | 【Analysis】As shown in Figure 5, let $M, L, P$ be the midpoints, $D, E, F$ be the points of tangency, and the incircle be $\odot I$. The Feuerbach theorem states that the circumcircle of $\triangle M L P$ is tangent to $\odot I$.
Proof: Let $C R \perp A I$, intersecting $A B$ at point $V$.
Obviously, $R$ is the midpoi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,927 |
1. Given a rectangle $R$ containing $n$ points, no line segment connecting any two points is parallel to the sides of rectangle $R$. Rectangle $R$ is divided into smaller rectangles whose sides are parallel to those of $R$, such that none of the given points lie inside any of the smaller rectangles. Prove: Rectangle $R... | 1. Let rectangle $R$ be divided into $k$ smaller rectangles, and the set of all vertices of these smaller rectangles can be divided into three pairwise disjoint subsets $A$, $B$, and $C$:
$A$ contains the vertices of the original rectangle $R$, and each such vertex is a vertex of one of the smaller rectangles;
$B$ cont... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,928 |
2. There are $2^{m}$ cards, and the number written on each card is 1. Perform the following operation: each time, choose two different cards. If the numbers on these two cards are $a$ and $b$, then erase these two numbers and write the number $a+b$ on both cards. Prove: After $2^{m-1} m$ operations, the sum of the numb... | 2. Let the product of all numbers on the cards after the $k$-th operation be $P_{k}$. Suppose in the $(k+1)$-th operation, the numbers $a$ and $b$ are replaced by two $(a+b)$. Then in $P_{k+1}$, $ab$ is replaced by $(a+b)^{2}$, while other factors remain unchanged.
Since $(a+b)^{2} \geqslant 4ab$, we have $P_{k+1} \geq... | 4^{m} | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,929 |
4. Two $2 \times 1$ dominoes are glued along their longer sides to form a "tetromino," such that the midpoint of one domino's longer side is a vertex of the other domino. This results in two types of tetrominoes based on their orientation, referred to as S-tetromino and Z-tetromino (as shown in Figure 1).
If a lattice... | 4. Suppose the polygon $P$ is the union of some squares on an infinite chessboard, and each square of the chessboard is colored black or white as shown in Figure 2.
Regardless of how the S-tetrominoes and Z-tetrominoes cover the polygon $P$, each S-tetromino covers an even number of black squares, while each Z-tetromi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,930 |
6. There are infinitely many cards, each with a real number written on it. For each real number $x$, there is exactly one card with the number $x$ written on it. Two players each select a set of 100 cards, denoted as $A$ and $B$, such that the sets are disjoint. Formulate a rule to determine which of the two players wi... | 6. There are 100 ways.
To prove the more general case, where each person selects $n$ cards, there are $n$ ways that satisfy the conditions. Let $A>B$ or $B>A$ if and only if $a_{k}>b_{k}$.
Such rules satisfy all three conditions, and different $k$ correspond to different rules, hence there are at least $n$ different r... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,931 |
4. If the set $S=\{1,2,3, \cdots, 16\}$ is arbitrarily divided into $n$ subsets, then there must exist a subset in which there are elements $a, b, c$ (which can be the same), satisfying $a+b=c$. Find the maximum value of $n$.
【Note】If the subsets $A_{1}, A_{2}, \cdots, A_{n}$ of set $S$ satisfy the following condition... | 4. First, when $n=3$, assume there exists a partition of the set that does not satisfy the condition. Thus, there must be a subset with at least six elements, without loss of generality, let
$$
A=\left\{x_{1}, x_{2}, \cdots, x_{6}\right\}\left(x_{1}<x_{2}<\cdots<x_{6}\right) \text {. }
$$
Then $x_{6}-x_{1}, x_{6}-x_{2... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,933 |
5. As shown in Figure 1, given points $D$, $E$, and $F$ are on the sides $AB$, $BC$, and $CA$ of acute $\triangle ABC$, respectively. If
$$
\begin{array}{l}
\angle EDC = \angle CDF, \\
\angle FEA = \angle AED, \\
\angle DFB = \angle BFE,
\end{array}
$$
Prove: $CD$, $AE$, and $BF$ are the three altitudes of $\triangle ... | 5. Note that,
$$
\begin{array}{l}
\frac{B E}{E C}=\frac{S_{\triangle B D E}}{S_{\triangle C D E}}=\frac{B D \sin \angle B D E}{C D \sin \angle C D E}, \\
\frac{A F}{F C}=\frac{S_{\triangle A D F}}{S_{\triangle C D F}}=\frac{A D \sin \angle A D F}{C D \sin \angle C D F}. \\
\text { Therefore, } \frac{B E}{E C} \cdot \fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,934 |
6. A shape obtained by removing one unit square from a $2 \times 2$ grid is called an "L-shape". In an $8 \times 8$ grid, place $k$ L-shapes, each of which can be rotated, but each L-shape must cover exactly three unit squares of the grid, and any two L-shapes must not overlap. Additionally, no other L-shape can be pla... | 6. On one hand, using six straight lines to divide the large square grid into 16 smaller $2 \times 2$ grids, each $2 \times 2$ grid must have at least two cells covered by an L-shape. Therefore, the number of cells covered is no less than 32. Thus, the number of L-shapes is no less than $\left[\frac{32}{3}\right]+1$ $=... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,935 |
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