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
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class | __index_level_0__ int64 0 742k |
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10.4. In $\triangle A B C$, circles $\omega_{B}$ and $\omega_{C}$ are the excircles that are tangent to sides $A C$ and $A B$, respectively, and also tangent to the extensions of the other two sides. Circles $\omega_{B}^{\prime}$ and $\omega_{B}$ are symmetric with respect to the midpoint of side $A C$; circles $\omega... | 10.4. Proof 1: As shown in Figure 3.
Let the midpoints of $AC$ and $AB$ be $B_{0}$ and $C_{0}$, respectively, and let circle $\omega_{A}$ be the third excircle of $\triangle ABC$; the intersection points of circles $\omega_{B}^{\prime}$ and $\omega_{C}^{\prime}$ be $P$ and $Q$. Let $AB = c$, $BC = a$, and $CA = b$. De... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,829 |
10.5. Place 16 rooks on an $8 \times 8$ chessboard. How many pairs of rooks can attack each other (a pair of rooks can attack each other if they are in the same row or column and there are no other rooks between them)? | 10.5.16 Correct.
It should be noted that if there are $a$ rooks in a row, then there are $a-1$ pairs of rooks that can attack each other in that row. Therefore, the number of rooks that can attack each other horizontally will not be less than 8 pairs.
Similarly, the number of rooks that can attack each other vertical... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,830 |
10.7. Given positive integers $x, y$ satisfy $2 x^{2}-1=$ $y^{15}$. Prove: If $x>1$, then $x$ is divisible by 5. | 10.7. Note: In this solution, all letters represent integers.
Let $t=y^{5}$, from $2 x^{2}-1=y^{15}$, we get $t^{3}+1=(t+1)\left(t^{2}-t+1\right)=2 x^{2}$.
Since $t^{2}-t+1$ is always odd, therefore,
either $t+1=2 u^{2}, t^{2}-t+1=v^{2}$;
or $t+1=6 u^{2}, t^{2}-t+1=3 v^{2}$.
This is because, if $a=\left(t+1, t^{2}-t+1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,831 |
10.8. On an infinitely large sheet of white square paper, a finite number of squares are colored black, and each black square has an even number (0, 2, or 4) of white squares sharing a common edge with it. Prove: it is possible to color the remaining white squares red or green such that for each black square, the numbe... | 10.8. The argument for this problem proceeds as follows: Connect the centers of adjacent black squares with line segments, then the "union" of all these line segments divides the plane into several regions. Since the degree of each vertex is even, the plane can be colored with two colors (yellow and blue) such that any... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,832 |
Example 7 As shown in Figure 7, $\odot O$ is the incircle of Rt $\triangle A B C$, and $A^{\prime} 、 B^{\prime} 、 C^{\prime}$ are the points of tangency on the three sides. Prove: The Euler line of $\triangle A^{\prime} B^{\prime} C^{\prime}$ bisects the hypotenuse $A B$. | Explanation: As shown in Figure 7, connect $O A^{\prime}, O B^{\prime}, O C^{\prime}$, and the ray $C O$ intersects $A B$ and $A^{\prime} B^{\prime}$ at $N$ and $S$. It is easy to see that quadrilateral $A^{\prime} C B^{\prime} O$ is a square, so $C S = S O$, $S A^{\prime} = S B^{\prime}$, and the centroid $G$ of $\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,833 |
11.1. Let $a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}$ be distinct numbers such that the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{50}\right|
\end{array}
$$
has a finite number of r... | 11.1.49.
$$
\begin{array}{l}
\text { Let } f(x)=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right|- \\
\left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{50}\right| \text {. } \\
\end{array}
$$
Thus, the original equation is $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{100}$ be the elements of the... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,834 |
11.3. Let the three excircles of $\triangle ABC$ touch the sides $BC$, $CA$, $AB$ at points $A'$, $B'$, $C'$, respectively. The circumcircles of $\triangle A'B'C$, $\triangle AB'C'$, $\triangle A'BC'$ intersect the circumcircle of $\triangle ABC$ again at points $C_1$, $A_1$, $B_1$, respectively. Prove that $\triangle ... | 11.3. As shown in Figure 5, draw perpendiculars from the points of tangency $A'$ and $C'$ of the excircle on the sides of the triangle. These perpendiculars pass through the centers of the excircles $I_1$ and $I_3$ and intersect at some point $O$. Since $I_1$ and $I_3$ lie on the external angle bisector of $\angle B$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,835 |
11.4. Let positive integers $x, y, z (x>2, y>1)$ satisfy the equation $x^{y}+1=z^{2}$. Let $p$ denote the number of distinct prime factors of $x$, and let $q$ denote the number of distinct prime factors of $y$. Prove: $p \geqslant q+2$. | 11.4. From the problem, we know that $(z-1)(z+1)=x^{y}$.
When $x$ is odd, $(z-1, z+1)=1$;
When $x$ is even, $(z-1, z+1)=2$.
In the first case, we have $z-1=u^{y}, z+1=v^{y}$, where $u$ and $v$ are positive odd numbers. This gives us $v^{y}-u^{y}=2$.
On the other hand, since $v>u, y>1$, we have
$$
v^{y}-u^{y}=(v-u)\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,836 |
11.5 - Does there exist a bounded function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that $f(1)>0$, and for all $x, y \in \mathbf{R}$, we have
$$
f^{2}(x+y) \geqslant f^{2}(x)+2 f(x y)+f^{2}(y)
$$
holds? | 11.5. Does not exist.
For any $x_{1} \neq 0$, let $y_{1}=\frac{1}{x_{1}}$, then
$$
\begin{array}{l}
f^{2}\left(x_{1}+y_{1}\right) \geqslant f^{2}\left(x_{1}\right)+2 f(1)+f^{2}\left(y_{1}\right) \\
\geqslant f^{2}\left(x_{1}\right)+a,
\end{array}
$$
where $a=2 f(1)>0$.
Let $x_{n}=x_{n-1}+y_{n-1}, y_{n}=\frac{1}{x_{n}... | proof | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,837 |
11.6. Is it possible to place 12 rectangular parallelepipeds $P_{1}, P_{2}, \cdots, P_{12}$ in space, such that their edges are parallel to the $O x$, $O y$, and $O z$ axes, and $P_{2}$ intersects with all the parallelepipeds except $P_{1}$ and $P_{3}$ (i.e., has at least one common point); $P_{3}$ intersects with all ... | 11.6. Not possible.
Assuming this is possible.
It should be noted that in the case under consideration, two cuboids intersect if and only if their projections on the three coordinate axes all intersect.
Consider the following four pairs of cuboids
$$
\left(P_{1}, P_{2}\right),\left(P_{4}, P_{5}\right),\left(P_{7}, P_{... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,838 |
11.7. For a quadrilateral $A B C D$ that is tangential to a circle with center $O$, prove: the point $O$ is the intersection of the line segments joining the midpoints of the opposite sides of the quadrilateral $A B C D$ if and only if $O A \cdot O C=O B \cdot O D$ | 11. 7. As shown in Figure 7, let the midpoints of sides $AB$ and $CD$ be $X$ and $Y$ respectively, and the intersection of lines $AB$ and $CD$ be point $P$.
Let point $O$ be the intersection of the line segments connecting the midpoints of the opposite sides of quadrilateral $ABCD$. Since the quadrilateral formed by t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,839 |
11.8. Around a round table, there are 100 representatives from 25 countries, with 4 representatives from each country. Prove: It is possible to divide them into four groups, such that each group contains 1 representative from each country, and no two representatives in any group are adjacent to each other at the table. | 11.8. First, we prove a lemma.
Lemma: Suppose there are $n$ pairs of acquaintances, each pair coming from $n$ different countries, with 2 people from each country. Then, they can be divided into two groups such that each group contains one representative from each country, and no acquaintances are in the same group.
... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,840 |
1. If $(2 x-1)(3 x+5)=0$, then $3 x^{2}+\frac{7}{2} x-\frac{3}{2}=(\quad)$.
(A) 1
(B) $-\frac{5}{2}$
(C) $\frac{38}{3}$
(D) 1 or $\frac{38}{3}$ | $$
-1 . \mathrm{A} .
$$
The original equation is $6 x^{2}+7 x-5=0$. After dividing by 2 and rearranging, we get
$$
3 x^{2}+\frac{7}{2} x-\frac{3}{2}=1 \text {. }
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,841 |
2. From a point on a plane, draw $n$ rays, generating 42 angles, then $n=$. .
(A) 22
(B) 21
(C) 20
(D) 7 | 2. D.
Each pair of rays forms two angles, with a total of $n(n-1)$ ways to form angles, so $n(n-1)=42$. Rearranging gives
$$
n^{2}-n-42=0 \text {. }
$$
Solving the equation and taking the positive root, we get $n=7$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,842 |
3. Let $x+y=1$. If
$$
M=\frac{x^{100}}{x^{2}+y}+\frac{y^{2}}{x+y^{2}}, N=\frac{x^{100}}{x+y^{2}}+\frac{y^{2}}{x^{2}+y} \text {, }
$$
then the relationship between $M$ and $N$ is $($ ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Cannot be determined | 3. B.
Given $x+y=1$, we have
$$
\begin{array}{l}
x^{2}+y=x^{2}-x+(x+y)=1-x y, \\
x+y^{2}=(x+y)-y+y^{2}=1-x y .
\end{array}
$$
Thus, $M=N=\frac{x^{100}+y^{2}}{1-x y}$.
Note: When $x+y=1$, the equation $t^{2}-t+1=0$ has no solution, so $x y \neq 1$. Therefore, the denominator is not 0. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,843 |
Example 1 Find $\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}$.
(23rd Putnam Mathematical Competition)
Analysis: In this problem, there is a variable $k^{2}$ before $\mathrm{C}_{n}^{k}$, therefore, it is necessary to use identity (III) twice to transform to $n$, then the binomial theorem can be applied to solve it. | Solution: From the basic identities (III) and (V), we get
$$
\begin{array}{l}
\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}=n \sum_{k=1}^{n} k \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1) \mathrm{C}_{n-1}^{k-1}+n \sum_{k=1}^{n} \mathrm{C}_{n-1}^{k-1} \\
=n(n-1) \sum_{k=2}^{n} \mathrm{C}_{n-2}^{k-2}+n \times 2^{n-1} \\
=n... | n(n+1) 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,844 |
4. As shown in Figure 1, point $C$ is any point on segment $A B$, $\triangle D A C$ and $\triangle E C B$ are both equilateral triangles, connect $A E$ intersecting $C D$ at point $M$, connect $B D$ intersecting $C E$ at point $N$. When point $C$ moves on segment $A B$, among the following 4 equations:
$$
\begin{array}... | 4.D.
As shown in Figure 1, by rotating $\triangle C B D$ counterclockwise around point $C$ by $60^{\circ}$, we obtain $\triangle C E A$. Thus,
$$
\begin{array}{l}
A E=D B, \text { and } \angle C A E=\angle C D B . \\
\text { Also, } \angle A C M=\angle D C N=60^{\circ}, A C=D C, \text { hence } \\
\triangle A C M \con... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,845 |
5. There are 4 squares with side lengths of $a$, $a+3$, $a+5$, and $a+6$, and the sum of their areas is denoted as $M$; there are also 4 squares with side lengths of $a+1$, $a+2$, $a+4$, and $a+7$, and the sum of their areas is denoted as $N$. The relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(... | 5. B.
$$
\begin{array}{l}
M=a^{2}+(a+3)^{2}+(a+5)^{2}+(a+6)^{2} \\
=4 a^{2}+28 a+70, \\
N=(a+1)^{2}+(a+2)^{2}+(a+4)^{2}+(a+7)^{2} \\
=4 a^{2}+28 a+70 .
\end{array}
$$
Therefore, $M=N$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,846 |
6. There is a sequence of numbers arranged in a row, the 1st number is 1, the 2nd number is also 1, starting from the 3rd number, each number is the sum of the previous two numbers. Then the remainder when the 2025th number is divided by 4 is $(\quad)$.
(A) 0
(B) 1
(C) 2
(D) 3 | 6. C.
The remainders when this sequence of numbers is divided by 4 are $1,1,2,3,1,0,1,1,2$, $3,1,0, \cdots$, with a period of 6. Also, $2025=337 \times 6+3$, so the 2025th number is congruent to the 3rd number modulo 4, with a remainder of 2. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,847 |
1. $\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\cdots+\frac{2005^{2}+1}{2005^{2}-1}=$ | $$
=, 1.2005 \frac{1003502}{2011015} .
$$
First, consider the general term of the sum
$$
\begin{array}{l}
a_{n}=\frac{n^{2}+1}{n^{2}-1}=1+\frac{2}{(n-1)(n+1)} \\
=1+\frac{(n+1)-(n-1)}{(n-1)(n+1)}=1+\frac{1}{n-1}-\frac{1}{n+1}
\end{array}
$$
Taking $n=2,3, \cdots, k$, and summing up, we have
$$
\begin{array}{l}
\frac{... | 2005 \frac{1003502}{2011015} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,848 |
3. As shown in Figure 3, in isosceles right $\triangle ABC$, $D$ is a trisection point on the hypotenuse, $AD=2DB$, $E$ is the midpoint of $BC$, and $CD$ and $AE$ intersect at point $F$. Then, an equation satisfied by $FA$, $FC$, and $FE$ is | 3. $F C^{2}=F A \cdot F E$.
It is only necessary to prove: In the right triangle $\triangle A C E$, $C F$ is the altitude to the hypotenuse.
As shown in Figure 5, connect
$D E$, given $\angle D B E=$
$\angle D A C=45^{\circ}$ and $B E$
$=E C$, we know
$$
\frac{D B}{D A}=\frac{B E}{A C}=\frac{1}{2} \text {. }
$$
Thus,... | F C^{2}=F A \cdot F E | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,850 |
4.6 Four table tennis players are dressed in 4 different colors for a performance match, with 2 in red, 2 in yellow, 1 in blue, and 1 in black. Each performance selects 3 players to appear, and matches are only played between players with different colored outfits, with the specific rules being:
(1) If the "3-person gr... | 4.44.
Represent the 2 athletes in red as two parallel lines $l_{1}$, $l_{2}$; the 2 athletes in yellow as two other parallel lines $l_{3}$, $l_{4}$; and the athletes in blue and black as intersecting lines $l_{5}$, $l_{6}$, which intersect with $l_{1}$, $l_{2}$, $l_{3}$, $l_{4}$. This results in Figure 6, where there ... | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,851 |
One, (20 points) Given $a^{2}+a b+a c<0$. Prove: the equation $a x^{2}+b x+c=0$ must have two distinct real roots, one greater than 1 and one less than 1.
| $$
\begin{array}{l}
\text { Given the quadratic function } y=x^{2}+b x+a c. \\
\text { From } a^{2}+a b+a c=0. \\
\text { Therefore, the equation } a x^{2}+b x+c=0 \text { is a quadratic equation, and it has two distinct real roots } x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}. \\
\text { Also, } \left(1-x_{1}\right)... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,852 |
II. (20 points) As shown in Figure 4, three equal circles are externally tangent to each other at points $A$, $B$, and $C$. In the region enclosed by the arcs $\overparen{A B}$, $\overparen{B C}$, and $\overparen{C A}$, take any point $P$ and connect $P A$,
$P B$, $P C$. Try to determine:
Can a right-angled triangle be... | II. $P A, P B, P C$ can form an acute triangle. The proof is as follows:
From the fact that equal circles are externally tangent to each other, the three lines connecting the centers form an equilateral triangle, with $A, B, C$ being the midpoints of their respective sides. Therefore, by connecting $A B, B C, C A$, we... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,853 |
Three. (20 points) A tetrahedron has 6 edges and 4 triangular faces. Given that the 6 edge lengths are exactly the positive integers $n, n+$ $1, n+2, n+3, n+4, n+5$. If the perimeter of a certain triangular face is a multiple of 3, then this triangle is painted red; otherwise, it is painted yellow. Question: What is th... | Three, at most 3 yellow triangles.
Since 6 consecutive positive integers divided by 3 must have remainders of 2 zeros, 2 ones, and 2 twos, denoted as $a, a, b, b, c, c$. If there are 4 yellow triangles, take any yellow $\triangle ABC$, then it must have two sides with the same remainder when divided by 3, denoted as $a... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,854 |
Example 2 Let $P_{n}(k)$ be the number of permutations of the set $\{1,2, \cdots, n\}$ that keep $k$ points fixed. Prove that:
$$
\sum_{k=0}^{n} k P_{n}(k)=n!\text {. }
$$
(28th IMO) | Prove: The number of ways to keep $k$ points fixed is $\mathrm{C}_{n}^{k}$. Since the remaining $n-k$ points are not fixed, we have
$$
\begin{array}{l}
P_{n}(k)=\mathrm{C}_{n}^{k} P_{n-k}(0) . \\
\text { Therefore, } \sum_{k=0}^{n} k P_{n}(k)=\sum_{k=0}^{n} k \mathrm{C}_{n}^{k} P_{n-k}(0) \\
=n \sum_{k=1}^{n} \mathrm{C... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,855 |
1. If the point $A(3,5)$ is symmetric to a point on the $x$-axis with respect to the line $l: y=k x$, then $k$ is ( ).
(A) $\frac{-1 \pm \sqrt{5}}{2}$
(B) $\pm \sqrt{3}$
(C) $\frac{-1 \pm \sqrt{30}}{4}$
(D) $\frac{-3 \pm \sqrt{34}}{5}$ | -、1.D.
Let point $A(3,5)$ have its symmetric point with respect to the line $l: y=k x$ as $P(a, 0)$. Then the midpoint $M\left(\frac{3+a}{2}, \frac{5}{2}\right)$ of $A P$ must lie on the line $l$, and $A P \perp l$, i.e., $\frac{5}{2}=k \cdot \frac{3+a}{2}$, and $k \cdot \frac{5-0}{3-a}=-1$.
Eliminating $a$ and rearran... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,856 |
2. Let $a \in \mathbf{R}, A=\{x|x \in \mathbf{R}| x-a \mid, \leqslant 1\}$, $B=\left\{x|x \in \mathbf{R}| x-1 \mid, \leqslant a^{2}\right\}$. If $A$ is not a proper subset of $B$, then the range of values for $a$ is ( ).
(A) $-1 \leqslant a \leqslant 1$
(B) $a \leqslant-2$ or $a>1$
(C) $-2<a \leqslant 1$
(D) $-2 \leqsl... | 2. C.
Obviously, $A=\{x \mid x \in \mathbf{R}, a-1 \leqslant x \leqslant a+1\}$,
$$
B=\left\{x \mid x \in \mathbf{R}, 1-a^{2} \leqslant x \leqslant a^{2}+1\right\},
$$
both $A$ and $B$ are non-empty.
Assuming $A$ is a proper subset of $B$, then we have
$$
\left\{\begin{array} { l }
{ a ^ { 2 } + 1 > a + 1 , } \\
{ 1 ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,857 |
3. If $a$ is a complex number, and $\left(a^{2}-a\right)^{2}=1$, then $a^{3}-2 a^{2}$ can take ( ) different values.
(A)2
(B) 3
(C) 4
(D) 6 | 3. B.
From the given, we have $a^{4}-2 a^{3}+a^{2}=1$, so $a^{3}-2 a^{2}=\frac{1-a^{2}}{a}($ obviously $a \neq 0)$. When $a^{2}-a=1$, $1-a^{2}=-a$, then $a^{3}-2 a^{2}=-1$; when $a^{2}-a=-1$, $a^{2}=a-1, a=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}$, then $a^{3}-2 a^{2}=\frac{1-(a-1)}{a}=-1+\frac{2}{a}$.
Thus, $a^{3}-2 a^{2}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,858 |
4. Let $a$ be a real number such that the graph of the function $y=f(x)=a \sin 2x + \cos 2x + \sin x + \cos x$ is symmetric about the line $x=-\pi$. Let the set of all such $a$ be denoted by $S$. Then $S$ is ( ).
(A) empty set
(B) singleton set (contains only one element)
(C) finite set with more than one element
(D) i... | 4. A.
By symmetry, we have
$$
f(-x)=f(x-2 \pi) \text {. }
$$
Since $f(x)$ is a periodic function with a period of $2 \pi$, we have
$$
f(x)=f(x-2 \pi) \text {. }
$$
Therefore, $f(-x)=f(x)$.
Thus, $-a \sin 2 x+\cos 2 x-\sin x+\cos x$
$$
=a \sin 2 x+\cos 2 x+\sin x+\cos x \text {, }
$$
which simplifies to $\sin x(2 a ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,859 |
5. Let $a$ be a real number. If the inequality $|\cos 2 x| \geqslant a \sin x$ holds for all $x$ in the closed interval $\left[-\frac{\pi}{3}, \frac{\pi}{6}\right]$, then the range of values for $a$ is ( ).
(A) $\{0\}$
(B) $[-1,0]$
(C) $\left[0, \frac{\sqrt{2}}{2}\right]$
(D) $[0,1]$ | 5.D.
Let $t=\sin x$, given $x \in\left[-\frac{\pi}{3}, \frac{\pi}{6}\right]$, we know $t \in\left[-\frac{\sqrt{3}}{2}, \frac{1}{2}\right]$.
The original inequality is equivalent to
$$
\left|1-2 t^{2}\right| \geqslant a t\left(-\frac{\sqrt{3}}{2} \leqslant t \leqslant \frac{1}{2}\right),
$$
which is equivalent to
$$
\... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,860 |
$6 . \tan 10^{\circ}+\frac{1}{\sin 40^{\circ}}$ The value is ( ).
(A) $\sqrt{2}$
(B) 2
(C) $\sqrt{3}$
(D) $\frac{3 \sqrt{3}}{2}$ | 6.C.
$$
\begin{array}{l}
\tan 10^{\circ}+\frac{1}{\sin 40^{\circ}}=\cot 80^{\circ}+\frac{1}{\sin 40^{\circ}} \\
=\frac{\cos 80^{\circ}+2 \cos 40^{\circ}}{\sin 80^{\circ}}=\frac{2 \cos 60^{\circ} \cdot \cos 80^{\circ}+2 \cos 40^{\circ}}{\sin 80^{\circ}} \\
=\frac{\cos 20^{\circ}+\cos 40^{\circ}}{\sin 80^{\circ}}=\frac{2... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,861 |
1. The last four digits of the number $7^{355}$ are $\qquad$ | II. 1.1943.
This problem is to find the remainder of $7^{355}$ divided by $10^{4}$.
$$
\begin{array}{l}
7^{355}=7 \times 49^{1 \pi}=7 \times 49 \times(50-1)^{2 \times 88} \\
=7 \times 49 \times(2400+1)^{28} \\
\equiv 7 \times 49 \times\left(1+2400 \mathrm{C}_{88}^{1}\right) \\
\equiv 343 \times[1+2400 \times(100-12)] \... | 1943 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,862 |
2. Let $\angle X O Y=30^{\circ}$, point $A$ is inside this angle, draw $A B \perp O X, A C \perp O Y ; B, C$ are the feet of the perpendiculars. If $A C=3, O B=8$, then the length of $A O$ is $\qquad$ . | 2. $\frac{14 \sqrt{3}}{3}$.
As shown in Figure 1, with $O$ as the origin and $OX$ as the positive direction of the $x$-axis, a Cartesian coordinate system is established. Let the coordinates of point $A$ be $\left(x_{A}, y_{A}\right)$, where $x_{A}=OB=8$. Clearly, the equation of the line $OY$ is
$$
y=x \tan 30^{\circ... | \frac{14 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,863 |
3. Let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube, and point $M$ be the midpoint of edge $A A_{1}$. Then the plane angle $\theta$ (acute angle) of the dihedral angle $B_{1}-M C-A_{1}$ is $\qquad$ (express in radians or inverse trigonometric functions). | Figure 2
3. $\frac{\pi}{4}$.
As shown in Figure 2, let's assume $A B$ $=2$, then
$$
\begin{array}{l}
A M=M A_{1}=1, \\
M C^{2}=1^{2}+(2 \sqrt{2})^{2}=9, \\
B_{1} C^{2}=(2 \sqrt{2})^{2}=8, \\
M B_{1}^{2}=1^{2}+2^{2}=5 .
\end{array}
$$
Let $O_{1}$ be the midpoint of $A_{1} C_{1}$. Clearly, the projection of point $B_{1... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,864 |
4. Let $a, b$ be complex numbers, their conjugate complex numbers are $\bar{a}, \bar{b}$, and their moduli are $|a|, |b|$. If $a+2 \bar{b}=\mathrm{i}$, $\bar{a} \cdot \bar{b}=-5-\mathrm{i}$, then $|a|^{2}$ is $\qquad$ . | 4.8 or 13.
From $2 \bar{a} \cdot \bar{b}=-10-2 \mathrm{i}$ and $a+2 \bar{b}=\mathrm{i}$, eliminating $2 \bar{b}$, we get $\bar{a}(\mathrm{i}-a)=-10-2 \mathrm{i}$,
Taking the conjugate of both sides of the above equation, we get $a(-\mathrm{i}-\bar{a})=-10+2 \mathrm{i}$, which is
$$
\begin{array}{l}
\text { i } \bar{a}... | 8 \text{ or } 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,865 |
Example 3 Find $S=\sum_{k=0}^{n} \frac{(\dot{m}+k)!}{k!}$.
(1962, Beijing Mathematical Competition (Grade 12)) | \begin{aligned} \text { Sol } & : S=\sum_{k=0}^{n} \frac{(m+k)!}{k!} \\ = & m!\left[1+\frac{m+1}{1!}+\frac{(m+1)(m+2)}{2!}+\cdots\right. \\ & \left.+\frac{(m+1)(m+2) \cdots(m+n)}{n!}\right] \\ = & m!\left(\mathrm{C}_{m}^{0}+\mathrm{C}_{m+1}^{1}+\mathrm{C}_{m+2}^{2}+\cdots+\mathrm{C}_{m+n}^{n}\right) \\ = & m!\left(\mat... | \frac{(n+m+1)!}{n!(m+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,866 |
5. If $a=\frac{13}{6}, b=\log _{3} 10, c=\log _{4} 20$, then the size relationship between $a 、 b 、 c$ is $\qquad$ . | 5. $a>c>b$.
Notice that $a>c \Leftrightarrow \frac{13}{6}>\log _{4} 20 \Leftrightarrow 4^{\frac{13}{6}}>20 \Leftrightarrow 4^{\frac{1}{6}}>\frac{5}{4}$ $\Leftrightarrow 2^{\frac{1}{3}}>\frac{5}{4} \Leftrightarrow 2>\frac{5^{3}}{4^{3}}=\frac{125}{64} \Leftrightarrow 2 \times 64=128>125$.
Below we prove $b10 \\
\Leftrig... | a>c>b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,867 |
6. In a decimal positive integer, if it contains an even number (including zero) of the digit 8, it is called a "good number", otherwise it is called a "non-good number". Then, the number of all "good numbers" with length (number of digits) not exceeding $n$ ($n$ is a positive integer) is $\qquad$ | 6. $\frac{1}{2}\left(8^{n}+10^{n}\right)-1$.
Let the number of "good numbers" of length $n$ be $a_{n}$, then $a_{1}=8$.
For $n \geqslant 2$, regarding $a_{n}$, on one hand, appending the digit 8 to the end of a "non-good number" of length $n-1$ turns it into a "good number" of length $n$, and there are $9 \times 10^{n... | \frac{1}{2}\left(8^{n}+10^{n}\right)-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,868 |
Three. (20 points) Let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube with edge length 2, and point $M$ is the midpoint of edge $A A_{1}$. A sphere is constructed passing through points $M, B_{1}, C, D_{1}$. Try to find the radius $R$ of this sphere. | Three, as shown in Figure 3, connect $B_{1} C$, $C D_{1}$, and $D_{1} B_{1}$. Let the body diagonal $A C_{1}$ intersect the plane of the equilateral $\triangle B_{1} C D_{1}$ at point $O_{1}$. Clearly, point $O_{1}$ is the center (centroid) of the equilateral $\triangle B_{1} C D_{1}$ (by symmetry), and $A C_{1} \perp$... | \frac{\sqrt{11}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,869 |
Four. (20 points) Given $10^{\circ}<\theta<50^{\circ}, a=$ $\sin \left(\theta+15^{\circ}\right), b=\cos \left(2 \theta-15^{\circ}\right)$. Prove:
$$
a^{2} \leqslant \frac{\sqrt{2} b+\sqrt{2-2 b^{2}}-1}{2 \sqrt{2} b},
$$
and determine all values of $\theta$ for which equality holds. | First, $\cos \left(2 \theta+30^{\circ}\right)$
$$
=1-2 \sin ^{2}\left(\theta+15^{\circ}\right)=1-2 a^{2} \text {. }
$$
Therefore, $\sin \left(2 \theta+30^{\circ}\right)=\sqrt{1-\cos ^{2}\left(2 \theta+30^{\circ}\right)}$
$$
=\sqrt{1-\left(1-2 a^{2}\right)^{2}}=2 a \sqrt{1-a^{2}} \text {. }
$$
Also, $\sin \left(2 \the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,870 |
Five. (20 points) Let two different tangent lines $l_{1}$ and $l_{2}$ be drawn from the point $P(1, -1)$ to the parabola $y = x^{2}$. Another tangent line to the parabola is $l$, and the angle (acute) between $l$ and $l_{1}$ is equal to the angle (acute) between $l$ and $l_{2}$. Try to find the equation of $l$.
---
T... | Let the tangent point of the tangent line $l_{1}$ or $l_{2}$ with the parabola $y=x^{2}$ be $\left(x_{0}, x_{0}^{2}\right)$. It is easy to know that the slope of the tangent line is $k=2 x_{0}$, and the equation of the tangent line is $y-x_{0}^{2}=2 x_{0}\left(x-x_{0}\right)$, i.e.,
$$
y=2 x_{0} x-x_{0}^{2} \text {. }
... | y=\frac{-5 \pm \sqrt{41}}{4} \cdot x-\frac{33 \mp 5 \sqrt{41}}{32} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,871 |
Let $\mathbf{R}$ be the set of all real numbers. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x, y \in \mathbf{R}$, we have
$$
f(x f(y))=x y f(x+y) .
$$ | Given the problem, for any $x, y \in \mathbf{R}$, we have
$$
f(x f(y)) = x y f(x + y).
$$
Taking $x = 0$ in equation (1), we get $f(0) = 0$.
Taking $x = -y$ in equation (1), we get
$$
f(-y f(y)) = 0.
$$
Assume there exists an $x_{0} \in \mathbf{R}$, and $x_{0} \neq 0$, such that $f(x_{0}) \neq 0$.
Taking $y$ as $-y f... | f(x) = 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,872 |
II. (50 points) Given that the incircle $\odot I$ of $\triangle ABC$ touches side $BC$ at point $D$, and $P$ is any interior point on side $BC$. Let the incenter of $\triangle ABP$ and $\triangle ACP$ be $I_{1}$ and $I_{2}$, respectively.
(1) Prove that $\angle I_{1} D I_{2}=90^{\circ}$ (i.e., $I_{1}, P, D, I_{2}$ are ... | II. (1) As shown in Figure 4, connect $I_{1}P$, $I_{2}P$, and $I_{1}I_{2}$. Since $PI_{1}$ bisects $\angle APB$ and $PI_{2}$ bisects $\angle APC$, it is easy to see that $\angle I_{1}PI_{2}=90^{\circ}$. Therefore, we only need to prove that $I_{1}$, $P$, $D$, and $I_{2}$ are concyclic, which is equivalent to proving
$$... | \left(\frac{1}{2}, +\infty\right) | Geometry | proof | Yes | Yes | cn_contest | false | 715,873 |
Three, (50 points) Prove: A convex heptagon on the Cartesian plane with grid points (each vertex is a grid point — points with integer coordinates) contains at least four grid points in its interior.
untranslated part:
(Each vertex is a grid point — points with integer coordinates)
The phrase is already in English,... | Three, first, let's assume that the interiors of the sides of the grid convex heptagon $A B C D E F G$ do not contain any grid points (otherwise, if the interior of $F G$ contains a grid point $H$, then use the heptagon $A B C D E F H$ to replace the original heptagon, and since the number of grid points is finite, thi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,874 |
In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $G$. The incircle of $\triangle ABC$ touches $AB$ at point $E$, and the incircle of $\triangle ABD$ touches $AB$ at point $F$. If
$$
2 AE \cdot BE = AC \cdot BC, \quad 2 AF \cdot BF = AD \cdot BD,
$$
prove that: $AD \cdot BG = BC \cdot AG$. | Prove: As shown in Figure 1, in $\triangle ABC$, let the other two tangency points of the incircle be $M$ and $N$, and let
$$
\begin{array}{l}
x = AE, y = BE, \\
z = CM.
\end{array}
$$
Then $AB = x + y, BC = y + z, CA = z + x$.
From the given condition $2 AE \cdot BE = AC \cdot BC$, we have
$$
2xy = (z + x)(y + z);
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,875 |
In $\triangle A B C$, $A B=A C, M 、 N$ are the midpoints of sides $A B 、 A C$ respectively, and $P$ is an intersection point of line $M N$ with the circumcircle of $\triangle A B C$. Find the range of $w=\frac{P B}{P C}+\frac{P C}{P B}$. | Solution: As shown in Figure 2, let $Q$ be the other intersection point of line $MN$ with the circumcircle of $\triangle ABC$, and connect $CQ$.
Since $AB = AC$, and $M, N$ are the midpoints of sides $AB, AC$ respectively, it follows that quadrilateral $PBCQ$ is an isosceles trapezoid, hence
$$
QC = PB, \quad NQ = PM.... | 3 < w < 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,876 |
Example 2 The vertex of the parabola $y=a x^{2}+b x+c$ is $(4,-11)$, and the x-coordinates of its two intersections with the $x$-axis are one positive and one negative. Then the positive number(s) among $a 、 b 、 c$ is/are ( ).
(A) only $a$
(B) only $b$
(C) only $c$
(D) $a 、 b$ | Explanation: Given the vertex is $(4,-11)$, the parabola intersects the $x$-axis at two points. It is easy to see that the graph opens upwards, i.e., $a>0$. Let the $x$-coordinates of the two intersection points of the parabola with the $x$-axis be $x_{1}$ and $x_{2}$, i.e., $x_{1}$ and $x_{2}$ are the two roots of the... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,878 |
Example 13 Prove:
$$
\sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^{k} \mathrm{C}_{n+1}^{k} \mathrm{C}_{2 n-2 k}^{n}=n+1
$$ | Prove: Since
$$
\begin{array}{l}
\left(1-x^{2}\right)^{n+1}=\sum_{k=0}^{n+1}(-1)^{k} \mathrm{C}_{n+1}^{k} x^{2 k}, \\
(1-x)^{-n-1}=\sum_{h=0}^{\infty} \mathrm{C}_{n+h}^{n} x^{h},
\end{array}
$$
multiplying, consider the coefficient of $x^{n}$. Since $2 k+h=n$, then $h=n-2 k$.
Also, $0 \leqslant h \leqslant n$, then $0... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,879 |
1. Given the function $f(x)=[x[x[x]]]$, where $[x]$ denotes the greatest integer not exceeding $x$. If its domain is $[0,4]$, then its range is $\qquad$ . | Ni.1. $\{0,1,8,9,12,13,14,27,28,29,33,34,35, 36,40,41,42,43,64\}$.
(1) When $0 \leqslant x<1$, $[x]=0$, then $f(x)=0$.
(2) When $1 \leqslant x<2$, $[x]=1$, then $f(x)=1$.
(3) When $2 \leqslant x<3$, let $x=2+\alpha, \alpha \in[0,1)$. Then
$$
\begin{array}{l}
f(x)=[(2+\alpha)[(2+\alpha) \times 2]] \\
=[(2+\alpha)(4+[2 \... | \{0,1,8,9,12,13,14,27,28,29,33,34,35,36,40,41,42,43,64\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,880 |
3. If $x, y, z, t \in \mathbf{N}_{+}$, and $1 \leqslant x<y \leqslant z$ $<t \leqslant 100$, then the maximum and minimum values of $\frac{x}{y}+\frac{z}{t}$ are $\qquad$. | 3. $\frac{19601}{9900}, \frac{1}{5}$.
It is easy to know, $\frac{x}{y}+\frac{z}{t} \leqslant \frac{98}{99}+\frac{99}{100}=\frac{19601}{9900}$.
$$
\text { Also, } \frac{x}{y}+\frac{z}{t} \geqslant \frac{1}{y}+\frac{z}{100} \geqslant 2 \sqrt{\frac{z}{100 y}} \geqslant \frac{1}{5} \text {. }
$$
Equality in (1) holds if ... | \frac{19601}{9900}, \frac{1}{5} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,882 |
4. Given $P(2,1)$, draw a line $l$ through point $P$ that intersects the positive $x$-axis and $y$-axis at points $A$ and $B$ respectively. Then the equation of the line $l$ that minimizes the perimeter of $\triangle A O B$ (where $O$ is the origin) is $\qquad$ | $$
4.3 x+4 y-10=0 \text {. }
$$
As shown in Figure 2, let $\angle B A O = \theta, t = \tan \frac{\theta}{2}$. Then the perimeter of $\triangle A O B$ is
$$
\begin{aligned}
s= & (2+\cot \theta)+ \\
& (1+2 \tan \theta)+ \\
& \left(\frac{1}{\sin \theta}+\frac{2}{\cos \theta}\right) \\
= & 3+\frac{1+\cos \theta}{\sin \the... | 3 x+4 y-10=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,883 |
5. Given $a, b \in \mathbf{N}_{+}$, the equations $x^{2}-a x+b=0$ and $x^{2}-b x+a=0$ have roots that are positive integers. Then all possible values of $a, b$ are $\qquad$ . | 5. $\left\{\begin{array}{l}a=5, \\ b=6 ;\end{array} \quad\left\{\begin{array}{l}a=6, \\ b=5 ;\end{array} \quad\left\{\begin{array}{l}a=4 ; \\ b=4 .\end{array}\right.\right.\right.$
Let the roots of the equations $x^{2}-a x+b=0$ and $x^{2}-b x+a=0$ be $x_{1}, x_{2}$ and $x_{3}, x_{4}\left(x_{1}, x_{2}, x_{3}, x_{4} \in... | a=5, b=6; a=6, b=5; a=4, b=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,884 |
Three. (20 points) Given the hyperbola $\Gamma: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If for any $\triangle A B C$ with vertices all on the hyperbola $\Gamma$, the orthocenter of $\triangle A B C$ is also on the hyperbola $\Gamma$. Explore the condition that the hyperbola $\Gamma$ should satisfy. | Let $A(a \sec \alpha, b \tan \alpha), B(a \sec \beta, b \tan \beta)$, $C(a \sec \gamma, b \tan \gamma)$, and the centroid $H(a \sec \theta, b \tan \theta)$.
Then, $\boldsymbol{A B}=(a(\sec \beta-\sec \alpha), b(\tan \beta-\tan \alpha))$,
$$
\boldsymbol{C H}=(a(\sec \theta-\sec \gamma), b(\tan \theta-\tan \gamma)) \text... | a=b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,886 |
Four. (20 points) Given the function $f_{n}(x)=n^{2} x^{2}(1-$ $x)^{n}, x \in[0,1], n \in \mathbf{N}_{+}$. If the maximum value of $f_{n}(x)$ is denoted as $a_{n}$, try to find the minimum term of the sequence $\left\{a_{n}\right\}$. | It is known that $f_{n}(0)=f_{n}(1)=0$.
When $x \in(0,1)$,
$$
\begin{array}{l}
f_{n}(x)=n^{2} x^{2}(1-x)^{n} \\
=4 \cdot \frac{n x}{2} \cdot \frac{n x}{2} \cdot \underbrace{(1-x)(1-x) \cdots(1-x)}_{n \uparrow} \\
\leqslant 4\left[\frac{\frac{n x}{2}+\frac{n x}{2}+(1-x) n}{n+2}\right]^{n+2}=4\left(\frac{n}{n+2}\right)^{... | \frac{4}{27} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 715,887 |
Five. (20 points) Given that the three sides of $\triangle A B C$ are $a, b, c$, and they satisfy
$$
a b c=2(a-1)(b-1)(c-1) .
$$
(1) Does there exist a $\triangle A B C$ with all sides being integers? If so, find the side lengths; if not, explain the reason.
(2) If $a>1, b>1, c>1$, find the minimum perimeter of $\trian... | (1) Suppose $a \geqslant b \geqslant c$, obviously $c \geqslant 2$.
If $c \geqslant 5$, then $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5}$.
From $a b c=2(a-1)(b-1)(c-1)$, we get
$\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \geqslant\left(\... | 4,5,6 \text{ or } 3,7,8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,888 |
One, (50 points) In $\triangle A B C$, $D$ is a point on side $B C$, and the inradii of $\triangle A B D$ and $\triangle A C D$ are equal. Then $S_{\triangle A B C}=A D^{2}$ if and only if $\angle A=90^{\circ}$. | Given as in Figure 3,
\[
\begin{array}{l}
B C=a, C A=b, \\
A B=c, A D=x .
\end{array}
\]
From the fact that the inradii of $\triangle A B D$ and $\triangle A C D$ are equal, we have
\[
\begin{array}{l}
\frac{S_{\triangle A B D}}{S_{\triangle A C D}}=\frac{A B+B D+A D}{A C+C D+A D} \\
=\frac{c+x+B D}{b+x+C D}=\frac{B D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,889 |
Example 14 Find $\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k}}{k+1}$. | Solution: From $(1+x)^{11}=\sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k}$, we have
$$
\begin{array}{l}
\int_{0}^{1}(1+x)^{11} \mathrm{~d} x=\int_{0}^{1} \sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k} \mathrm{~d} x . \\
\text { and } \int_{0}^{1} \sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k} \mathrm{~d} x \\
=\left.\left(\sum_{k=0}^... | \frac{1365}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,890 |
II. (50 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a_{2}=1$, and $a_{n+1} a_{n-1}=a_{n}^{2}+n a_{n} a_{n-1}(n=2,3, \cdots)$.
(1) Find $a_{n}$;
(2) Find $\sum_{k=2}^{n} \frac{a_{k}}{(k-2)!}$. | (1) From $a_{n+1} a_{n-1}=a_{n}^{2}+n a_{n} a_{n-1}$, we can get
$$
\frac{a_{n+1}}{a_{n}}=\frac{a_{n}}{a_{n-1}}+n \text {. }
$$
Let $b_{n}=\frac{a_{n+1}}{a_{n}}$, thus, $b_{n}=b_{n-1}+n, b_{1}=\frac{a_{2}}{a_{1}}=1$.
$$
\begin{array}{l}
\text { Hence } b_{n}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)+b_{1} \\
=\sum_{i... | \frac{(n+1)!}{2^{n-1}}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,891 |
Three. (50 points) Given $a, b, c, d \in \mathbf{R}_{+}, k_{1}, k_{2}, k_{3}, k_{4} \in \mathbf{N}_{+}$, and
$$
\begin{array}{l}
b+c+d=k_{1} a, c+d+a=k_{2} b, \\
d+a+b=k_{3} c, a+b+c=k_{4} d .
\end{array}
$$
Try to find the set of values for $k_{1}+k_{2}+k_{3}+k_{4}$. | Three, note that
$$
\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4} \\
=\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d} .
\end{array}
$$
Since the above expression is symmetric in $a, b, c, d$, we can assume without loss of generality that $a \geqslant b \geqslant c \geqslant d$. Then $k_{1}=\frac{b+c+d}{a} \... | \{12,13,14,16,18,20,25,26,27,28,33,50\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,892 |
Given $p$ is a positive integer and a constant. Find the maximum and minimum positive integers $n$, for which there exists a unique positive integer $k$ satisfying
$$
\frac{p+1}{2 p+1}<\frac{n}{n+k}<\frac{p}{2 p-1} .
$$ | Solution: Since $p$, $n$, and $k$ are all positive integers, transforming the original inequality yields
$$
\frac{n(p-1)}{p}<k<\frac{n p}{p+1} \text{. }
$$
For $n$, there is a unique positive integer $k$ that satisfies equation (1), so,
$$
\frac{n p}{p+1}-\frac{n(p-1)}{p} \leqslant 2 \text{. }
$$
Solving this, we get... | 2 p(p+1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,893 |
In $\triangle A B C$, $\angle B A C=100^{\circ}, \angle A B C=$ $50^{\circ}$, point $D$ is on $B C$, $\angle B A D=50^{\circ}$, point $E$ is on $A C$, $\angle A B E=20^{\circ}$. Find the degree measure of $\angle A D E$. | Solution: As shown in Figure 1, construct a regular $\triangle A D F$ inside $\triangle A B D$, and draw the angle bisector of $\angle A F D$ to intersect $B E$ at $G$. Connect $G A$, $G D$, and $B F$. It is easy to see that
$$
\begin{array}{l}
D F=D A=D B, \\
\angle B D F=20^{\circ} .
\end{array}
$$
Therefore, $\angl... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,894 |
159 Let the continuous function $f: \mathbf{R} \rightarrow \mathbf{R} \backslash\{0\}$, and for any $x, y \in \mathbf{R}$ satisfy
$$
\begin{array}{l}
f^{2}(x) f(2 y)+f^{2}(y) f(2 x) \\
=2 f(x) f(y) f(x+y) .
\end{array}
$$
Find $f(x)$. | Let $y=0$, we get
$f^{2}(x) f(0)+f^{2}(0) f(2 x)=2 f^{2}(x) f(0)$.
Thus, $f(2 x)=\frac{f^{2}(x)}{f(0)}, f(2 y)=\frac{f^{2}(y)}{f(0)}$.
Substituting equation (2) into equation (1) yields
$$
f(x) f(y)(f(x) f(y)-f(0) f(x+y))=0.
$$
From equation (3) and $f(x) \neq 0, f(y) \neq 0$, we have
$$
f(x) f(y)-f(0) f(x+y)=0.
$$
L... | f(x)=b c^{x}(b \neq 0, c>0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,895 |
Given $x, y$ are positive numbers satisfying $x+y=1$, and $n$ is an integer no less than 2. Prove:
$$
\frac{x^{n}}{x+y^{3}}+\frac{y^{n}}{x^{3}+y} \geqslant \frac{2^{4-n}}{5} .
$$ | Proof: Since the function $z=t^{m}$ (where $m$ is a constant greater than 0) is an increasing function on $(0,+\infty)$, we have
$$
\begin{array}{l}
(x-y)\left(x^{n+3}-y^{n+3}\right) \geqslant 0, \\
(x-y)\left(x^{n-1}-y^{n-1}\right) \geqslant 0 .
\end{array}
$$
Therefore,
$$
2\left(\frac{x^{n+1}}{x+y^{3}}+\frac{y^{n+1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,896 |
Example 16 Try to express $\sum_{k=0}^{n} \frac{(-1)^{k} \mathrm{C}_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$ in the form $\frac{P(n)}{Q(n)}$. Here $P(n) 、 Q(n)$ are two polynomials with integer coefficients. | Let the original expression be $S(n)$, then
$$
\begin{array}{l}
S(n)=\sum_{k=0}^{n} \frac{(-1)^{k} \mathrm{C}_{n}^{k}}{k^{3}+9 k^{2}+26 k+24} \\
=\sum_{k=0}^{n} \frac{(-1)^{k} n!}{k!(n-k)!(k+2)(k+3)(k+4)} \\
=\sum_{k=0}^{n} \frac{(-1)^{k}(n+4)!}{(k+4)!(n-k)!} \text {. } \\
{\left[\frac{k+1}{(n+1)(n+2)(n+3)(n+4)}\right]... | \frac{1}{2(n+3)(n+4)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,897 |
Proposition 1 Let the incenter of $\triangle A B C$ be $I, A I$ intersect the circle of $\triangle B I C$ at another point $A^{\prime}, B I$ intersect the circle of $\triangle C I A$ at another point $B^{\prime}, C I$ intersect the circle of $\triangle A I B$ at another point $C^{\prime}$. Then we have
$$
I A^{\prime} ... | Proof: Let the circumradius of $\triangle ABC$ be $R$. In $\triangle ABI$, by the Law of Sines, we have
$$
\begin{array}{l}
\frac{IA}{\sin \angle ABI} \\
=\frac{AB}{\sin \angle AIB} \\
=\frac{AB}{\sin \left(\pi-\frac{A}{2}-\frac{B}{2}\right)} \\
=\frac{c}{\cos \frac{C}{2}}=4 R \sin \frac{C}{2}.
\end{array}
$$
Therefor... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,898 |
Proposition 2 Let the orthocenter of acute triangle $\triangle ABC$ be $H, AH$ intersects the circumcircle of $\triangle BHC$ at another point $A^{\prime}, BH$ intersects the circumcircle of $\triangle CHA$ at another point $B^{\prime}, CH$ intersects the circumcircle of $\triangle AHB$ at another point $C^{\prime}$. T... | Proof: Let the circumradius of $\triangle ABC$ be $R$. As shown in Figure 2, let $CH$ intersect $AB$ at $F$. In the right triangle $\triangle AHF$,
$$
\begin{array}{c}
HA = \frac{AF}{\cos \angle BAH} \\
= \frac{b \cos A}{\cos \left(\frac{\pi}{2} - B\right)} \\
= 2R \cos A.
\end{array}
$$
Similarly, we have
$HB = 2R \c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,899 |
Proposition 3 Let the centroid of $\triangle A B C$ be $G, A G$ intersects the circle of $\triangle B G C$ at another point $A^{\prime}, B G$ intersects the circle of $\triangle C G A$ at another point $B^{\prime}, C G$ intersects the circle of $\triangle A G B$ at another point $C^{\prime}$. Then we have
$$
G A^{\prim... | Proof: As shown in Figure 3, let $A G, B G, C G$ intersect $B C, C A, A B$ at points $D, E, F$ respectively. Denote $B C=a, C A=b, A B=c$, $A D=m_{a}, B E=m_{b}, C F=m_{c}$.
By the intersecting chords theorem, we have
$$
G D \cdot D A^{\prime}=B D \cdot D C.
$$
Thus, $D A^{\prime}=\frac{B D \cdot D C}{G D}$
$$
=\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,900 |
Example 1 Given $a, b, c > 0$, and
$$
\left\{\begin{array}{l}
a^{2}+a b+\frac{b^{2}}{3}=25, \\
\frac{b^{2}}{3}+c^{2}=9, \\
a^{2}+a b a^{2}+a c+c^{2}=16 .
\end{array}\right.
$$
Try to find the value of $a b+2 b c+3 a c$.
Analysis: It is not easy to solve this problem directly. Observing the numbers on the right side of... | Solution: Establish a rectangular coordinate system as shown in Figure 1. Then
$$
\begin{array}{l}
A\left(-\frac{\sqrt{3}}{3} b, c\right), \\
B\left(\frac{\sqrt{3}}{2} a, \frac{a+2 c}{2}\right) .
\end{array}
$$
Given $O A=3$,
$$
\begin{array}{l}
O B=\sqrt{\frac{3}{4} a^{2}+\left(\frac{a^{2}+4 a c+4 c^{2}}{4}\right)} \... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,901 |
Example 2 Given $n(n \geqslant 6)$ points on a circle, connect each pair of points with a line segment, such that no three segments intersect at the same point inside the circle. Find the number of triangles determined by these segments whose vertices are inside the circle.
Translate the above text into English, pleas... | Solution: Let the set of all triangles that meet the conditions be $A$, and $\triangle A_{1} A_{2} A_{3} \in A$. Extend the three sides of $\triangle A_{1} A_{2} A_{3}$ to intersect the circle at points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ (as shown in Figure 2). Let the set of all ways to choose 6 points from th... | \mathrm{C}_{n}^{6} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,902 |
Example 3 Any 6 people in the world must include at least 3 people who are either all mutual strangers or all mutual acquaintances. | Proof: Let the set of people among the remaining 5 who know $A$ be $F$, and the set of people who do not know $A$ be $S$. According to the pigeonhole principle, at least one of the sets $F$ or $S$ has 3 people, let's assume it is $F$.
If the 3 people $B$, $C$, $D$ in $F$ do not know each other, then the proposition is... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,903 |
Example 3 Given two quadratic functions $y_{1}$ and $y_{2}$, when $x$ $=\alpha(\alpha>0)$, $y_{1}$ reaches its maximum value of 5, and $y_{2}=25$; also, the minimum value of $y_{2}$ is $-2, y_{1}+y_{2}=x^{2}+16 x+$ 13. Find the value of $\alpha$ and the analytical expressions of the quadratic functions $y_{1}$ and $y_{... | Explanation: Since when $x=\alpha(\alpha>0)$, $y_{1}$ reaches its maximum value of 5, the vertex coordinates of $y_{1}$ are $(\alpha, 5)$. Therefore, the quadratic function $y_{1}$ can be set as $y_{1}=m(x-\alpha)^{2}+5$ $(m<0)$. Thus,
$$
y_{2}=x^{2}+16 x+13-m(x-\alpha)^{2}-5 \text {. }
$$
When $x=\alpha$, $y_{2}=25$,... | y_{1}=-2 x^{2}+4 x+3, y_{2}=3 x^{2}+12 x+10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,905 |
Example 1 As shown in Figure 1, in $\triangle A B C$, $O$ is the circumcenter, the three altitudes $A D, B E, C F$ intersect at point $H$, line $E D$ and $A B$ intersect at point $M$, line $F D$ and $A C$ intersect at point $N$. Prove:
(1) $O B \perp D F, O C \perp D E$;
(2) $O H \perp M N$. | Proof: Since the center $K$ of the nine-point circle lies at the midpoint of the Euler line, i.e., the midpoint of $O H$, it follows that $O K$ is the line connecting the centers of the circumcircle and the nine-point circle of $\triangle A B C$. By the radical axis theorem, to prove (2), it is sufficient to show that ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,906 |
Example 2 In a convex quadrilateral $ABCD$, the two diagonals intersect at point $O$, the centroids of $\triangle AOB$ and $\triangle COD$ are $M_{1}$ and $M_{2}$, and the orthocenters of $\triangle BOC$ and $\triangle AOD$ are $H_{1}$ and $H_{2}$. Prove: $M_{1} M_{2} \perp H_{1} H_{2}$. | Proof: As shown in Figure 2, construct the two altitudes $A A_{1}$ and $D D_{1}$ of $\triangle A O D$, and the two altitudes $B B_{1}$ and $C C_{1}$ of $\triangle B O C$.
Since $\angle A A_{1} B = 90^{\circ} = \angle B B_{1} A$, points $A, B, B_{1}, A_{1}$ are concyclic with the center being the midpoint $E$ of $A B$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,907 |
Example 3 In a convex pentagon $A B C D E$, $A B=B C$, $\angle B C D=\angle E A B=90^{\circ}$, $P$ is a point inside the pentagon such that $A P \perp B E$, $C P \perp B D$. Prove that $B P \perp D E$. | Proof 1: As shown in Figure 3, draw $PH \perp DE$ at point H. Since
$$
\begin{aligned}
\angle PFD & = \angle PGE \\
& = 90^{\circ} = \angle PHD \\
& = \angle PHE,
\end{aligned}
$$
therefore, points $F, D, H, P$ and $P, H, E, G$ are concyclic, denoted as circles $\odot M_{1}$ and $\odot M_{2}$, respectively.
Also, $BF ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,908 |
Example 4 Take a point $C$ inside $\angle A O B$, draw $C D \perp O A$ at point $D$, draw $C E \perp O B$ at point $E$, then draw $D N \perp O B$ at point $N$, and draw $E M \perp O A$ at point $M$. Prove: $O C \perp M N$. | Proof: As shown in Figure 4, draw $C H \perp M N$ at point $H$. Since
$$
\begin{array}{l}
\angle C D M=\angle C E N \\
=90^{\circ},
\end{array}
$$
therefore, $C, D, M, H$ and $C, H, N, E$ are respectively concyclic, denoted as $\odot O_{1}$ and $\odot O_{2}$. From
$$
\angle D M E=90^{\circ}=\angle D N E,
$$
we know t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,909 |
Example 5 Let $O$ be the circumcenter of acute $\triangle A B C$, $T$ be the circumcenter of $\triangle B O C$, and $M$ be the midpoint of side $B C$. Take points $D, E$ on sides $A B, A C$ respectively, such that $\angle A D M = \angle A E M = \angle B A C$. Prove: $A T \perp D E$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 5 ... | Proof: As shown in Figure 5, since $O$ is the circumcenter of $\triangle ABC$ and $T$ is the circumcenter of $\triangle BOC$, it is known that points $O$, $M$, and $T$ are collinear, and $OT \perp BC$.
Extending $DM$ and $AC$ to intersect at point $G$, and extending $EM$ and $AB$ to intersect at point $F$, connecting ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,910 |
Example 1 If $x, y, z \in (0,1)$, then
$$
x(1-y)+y(1-z)+z(1-x)<1 .
$$ | Proof: Since $x, y, z \in (0,1)$, therefore,
$$
1-x, 1-y, 1-z \in (0,1).
$$
Without loss of generality, we call two numbers of the form $a$ and $1-a$ a pair of complementary numbers. The arrangement order of complementary numbers is $XYZ$, where $X=x$ or $X=1-x$, $Y=y$ or $Y=1-y$, $Z=z$ or $Z=1-z$. Construct the follo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,911 |
Example 3 If $x_{i} \in(0,1), \overline{x_{i}}=1-x_{i}(i=1,2$, $\cdots, n)$, let
$$
\begin{array}{l}
f(0)=\varphi(\underbrace{0 \cdots 0}_{n \uparrow})=\prod_{i=1}^{n} \overline{x_{i}}=\overline{x_{n}} \cdots \overline{x_{2}} \overline{x_{1}}, \\
f(1)=\varphi(\underbrace{0 \cdots 01}_{n-1 \uparrow})=\overline{x_{n}} \c... | Proof: It is easy to prove that $\sum_{k=0}^{2^{n}-1} f(k)=1$. Since $f(k)>0$ $\left(k=0,1, \cdots, 2^{n}-1\right), \sum_{k=p}^{q} f(k)$ is the sum of some terms in $\sum_{k=0}^{2^{n}-1} f(k)$, hence $\sum_{k=p}^{q} f(k) \leqslant 1$. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,912 |
Question 1 Let $a, b, c$ be positive real numbers, and $a^{2}+b^{2}+c^{2}+a b c=4$. Prove:
$$
3 a b c \leqslant a b+b c+a c \leqslant a b c+2 \text {. }
$$ | Proof: From $4=a^{2}+b^{2}+c^{2}+a b c$
$$
\geqslant a b c+3(a b c)^{\frac{2}{3}},
$$
it follows that $a b c \leqslant 1$, hence
$$
a b+a c+b c \geqslant 3(a b c)^{\frac{2}{3}} \geqslant 3 a b c.
$$
From the given condition, among $a, b, c$, there must be exactly two numbers that are either both not greater than 1 or... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,914 |
Question 2 Proof: For any real numbers $a>1$, $b>1$, we have
$$
\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8 \text {. }
$$ | $$
\begin{array}{l}
\frac{a^{2}}{b-1}+4(b-1) \geqslant 4 a, \\
\frac{b^{2}}{a-1}+4(a-1) \geqslant 4 b . \\
\text { (1) + (2) gives } \frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8 .
\end{array}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,915 |
Example 4 The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$. | Explanation: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, with $x_{1}<x_{2}$, and let $C(0, c)$, where $c \neq 0$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1}<0<x_{2}$, and it must be that $\angle A C B=90^{\circ}$. The graph can only be as shown in Figure 2 or Figure 3. In either ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,916 |
Question 3 Let non-negative numbers $a, b, c, d$ satisfy $ab + bc + cd + ad = 1$. Prove:
$$
\begin{array}{l}
\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{b+d+a}+ \\
\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}^{[1]} .
\end{array}
$$ | Prove: Let $a+b+c+d=s$.
Let $f(x)=\frac{x^{3}}{s-x}, x \in[0, s)$.
By the AM-GM inequality, we have
$$
\frac{x^{3}}{s-x}+\frac{s-x}{18}+\frac{1}{12} \geqslant \frac{x}{2} \text {. }
$$
Since $a, b, c, d \in[0, s)$, we substitute $x$ with $a, b, c, d$ in equation (1) and add the resulting 4 inequalities, then we have
$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,917 |
Question 4 Let $x, y, z \in \mathbf{R}_{+}$, then
$$
\frac{x^{3}}{y z}+\frac{y^{3}}{z x}+\frac{z^{3}}{x y} \geqslant x+y+z \text{. }
$$ | Prove: Since $\frac{x^{3}}{y z}+y+z \geqslant 3 x$,
$$
\frac{y^{3}}{z x}+z+x \geqslant 3 y, \frac{z^{3}}{x y}+x+y \geqslant 3 z \text {, }
$$
adding the three inequalities yields
$$
\frac{x^{3}}{y z}+\frac{y^{3}}{z x}+\frac{z^{3}}{x y} \geqslant x+y+z .
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,918 |
Question 5 Let $a, b, c \in \mathbf{R}_{+}$, and $\sqrt{a}, \sqrt{b}, \sqrt{c}$ form the three sides of a triangle, $f(x)=x^{2}-(a+b+c) x+ab+bc+ac$. Prove: $f(x)=0$ has no real roots. | Proof: As shown in Figure 1, let $O$ be the incenter of $\triangle ABC$. Then
$$
\begin{array}{l}
OE \perp AB, \\
OF \perp BC, \\
OG \perp AC .
\end{array}
$$
$$
\begin{array}{l}
\text{Let } AB = \sqrt{c} \\
= x + y, \\
BC = \sqrt{a} = y + z, \\
AC = \sqrt{b} = x + z \left(x, y, z \in \mathbf{R}_{+}\right) .
\end{array... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,919 |
Question 6 Let $x, y, z \geqslant 0$. Prove:
$$
\begin{array}{l}
x(x-z)^{2}+y(y-z)^{2} \\
\geqslant(x-z)(y-z)(x+y-z)^{2} .
\end{array}
$$ | Prove: Notice
$$
\begin{array}{l}
x(x-z)^{2}+y(y-z)^{2}- \\
(x-z)(y-z)(x+y-z) \\
= x(x-z)(x-y)+y(y-z)(y-x)+ \\
\quad z(z-y)(z-x),
\end{array}
$$
According to the symmetry of the above equation, without loss of generality, assume $x \geqslant y \geqslant z \geqslant 0$. Then
$$
\begin{array}{l}
x(x-z)(x-y)+y(y-x)(y-z)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,920 |
Question 7 Let $a, b, c$ be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8^{[3]} .
\end{array}
$$ | Prove: Let $s=a+b+c$,
$$
f(t)=\frac{(t+s)^{2}}{2 t^{2}+(s-t)^{2}}, t \in[0, s) \text {. }
$$
Since $f(t)=\frac{1}{3}+\frac{2}{3} \times \frac{4 s t+s^{2}}{3\left(t-\frac{s}{3}\right)^{2}+\frac{2}{3} s^{2}}$
$$
\leqslant \frac{1}{3}+\frac{4 s t+s^{2}}{s^{2}}=4\left(\frac{1}{3}+\frac{t}{s}\right),
$$
thus $f(a)+f(b)+f(... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false | 715,921 |
1. On the three sides of the equilateral $\triangle A B C$, select 6 points in the following manner: On side $B C$, select points $A_{1}$ and $A_{2}$; on side $C A$, select points $B_{1}$ and $B_{2}$; on side $A B$, select points $C_{1}$ and $C_{2}$, such that the sides of the convex hexagon $A_{1} A_{2} B_{1} B_{2} C_... | 1. As shown in Figure 1, take a point $P$ inside the equilateral $\triangle ABC$ such that $\triangle A_{1} A_{2} P$ is an equilateral triangle. Then, from $A_{1} P / / C_{2} C_{1}$ and $A_{1} P = C_{2} C_{1}$, we know that quadrilateral $A_{1} P C_{1} C_{2}$ is a rhombus.
Similarly, quadrilateral $A_{2} B_{1} B_{2} P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,922 |
2. Let $a_{1}, a_{2}, \cdots$ be a sequence of integers, where there are infinitely many positive integers and infinitely many negative integers. If for every positive integer $n$, the integers $a_{1}, a_{2}, \cdots, a_{n}$ yield $n$ distinct remainders when divided by $n$, prove that each integer appears exactly once ... | 2. From the problem, for any positive integer $n, a_{1}, a_{2}, \cdots, a_{n}$ form a complete residue system modulo $n$.
If $i\max \{i, j\}$, then $a_{1}, a_{2}, \cdots, a_{n}$ includes every integer between $a_{i}$ and $a_{j}$, so $x$ appears in $a_{1}, a_{2}, \cdots, a_{n}$.
In summary, each integer appears exactly... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,923 |
3. Positive real numbers $x, y, z$ satisfy $xyz \geqslant 1$. Prove:
$$
\begin{array}{l}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \\
\geqslant 0 . \quad \text { (Provided by Korea) }
\end{array}
$$
(Provided by Korea) | 3. The original inequality can be transformed into
$$
\frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{2}+z^{2}}+\frac{x^{2}+y^{2}+z^{2}}{y^{5}+z^{2}+x^{2}}+\frac{x^{2}+y^{2}+z^{2}}{z^{5}+x^{2}+y^{2}} \leqslant 3 \text {. }
$$
By the Cauchy-Schwarz inequality and the condition \(xyz \geqslant 1\), we have
$$
\begin{array}{l}
\left(x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,924 |
4. The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1(n=1,2,3, \cdots) \text {. }
$$
Find all positive integers that are coprime with every term of this sequence.
(Poland provided) | 4. The only positive integer that satisfies the condition is 1.
The following is the proof: For any prime number $p$, it must be a divisor of some term in the sequence $\left\{a_{n}\right\}$.
For $p=2$ and $p=3$, they are divisors of $a_{2}=48$.
For every prime number $p$ greater than 3, since
$$
(2, p)=1,(3, p)=1,(6,... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,925 |
5. Given a convex quadrilateral $A B C D, B C=A D$, and $B C$ is not parallel to $A D$. Let points $E$ and $F$ be on the interiors of sides $B C$ and $A D$, respectively, such that $B E=D F$. Line $A C$ and $B D$ intersect at point $P$, line $B D$ and $E F$ intersect at point $Q$, and line $E F$ and $A C$ intersect at ... | 5. As shown in Figure 2, let the perpendicular bisectors of line segments $AC$ and $BD$ intersect at point $O$.
Below is the proof: When points $E$ and $F$ vary, the circumcircle of $\triangle PQR$ passes through point $O$.
Since $OA = OC$, $OB = OD$, and $AD = BC$, we have:
$\triangle ODA \cong \triangle OBC$, meani... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,926 |
Example 5 Let the quadratic function $y=x^{2}+2 a x+\frac{a^{2}}{2}(a<0)$ have its vertex at $A$, and intersect the $x$-axis at points $B$ and $C$. When $\triangle A B C$ is an equilateral triangle, the value of $a$ is $\qquad$ | Explanation: The graph of the quadratic function
$$
y=x^{2}+2 a x+\frac{a^{2}}{2}(a<0)
$$
is shown in Figure 4. Let \( A\left(-a,-\frac{a^{2}}{2}\right) \),
$$
B\left(x_{1}, 0\right), C\left(x_{2}, 0\right) \text {. }
$$
Let the intersection of the axis of symmetry with the \( x \)-axis be \( D \). Then
$$
\begin{arra... | a=-\sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,927 |
6. A math competition consists of 6 problems, where any two problems were solved correctly by more than $\frac{2}{5}$ of the participants. However, no participant solved all 6 problems. Prove: At least two participants solved exactly 5 problems.
(Romania provided) | 6. Let there be $n$ participants, and let $p_{ij}$ denote the number of participants who answered both the $i$-th and $j$-th questions correctly $(1 \leqslant i < j \leqslant 6)$. It is given that for any $i$, the number of participants who answered the $i$-th question correctly is more than $\frac{2}{5} n$.
Thus, $p_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,928 |
2. Given an infinite sequence of real numbers $a_{0}, a_{1}, a_{2}, \cdots$ satisfying the condition $a_{n}=\left|a_{n+1}-a_{n+2}\right|, n \geqslant 0$, where $a_{0} 、 a_{1}$ are two distinct positive numbers. Is this sequence bounded? | 2. This sequence is unbounded.
Obviously, each term of the sequence $\left\{a_{n}\right\}$ is non-negative.
If there exists $a_{n}=a_{n+1}=c$, then $a_{n-1}=0, a_{n-2}=a_{n-3}=$ $c, \cdots$, eventually leading to $a_{0}$ and $a_{1}$ being either both equal to $c$, or one equal to $c$ and the other equal to 0. This is ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,929 |
3. Does there exist a function $f: \mathbf{Q} \rightarrow\{-1,1\}$, such that if $x$ and $y$ are two different rational numbers, and satisfy $x y = 1$ or $x+y \in\{0,1\}$, then $f(x) f(y)=-1$? Prove your conclusion. | 3. Existence.
Let $x=\frac{a}{b}$ be a positive real number, where $a$ and $b$ are coprime positive integers. Consider the sequence of successive remainders obtained by applying the Euclidean algorithm to the pair $(a, b)$. Let $(u \bmod v)$ denote the smallest non-negative remainder of $u$ modulo $v$. This sequence c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,930 |
5. Let $a, b, c > 0$, and $ab + bc + ca = 1$. Prove the inequality:
$$
\sqrt[3]{\frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a} \leqslant \frac{1}{abc}.
$$ | 5. By Jensen's inequality, we have
$$
\left(\frac{u+v+w}{3}\right)^{3} \leqslant \frac{u^{3}+v^{3}+w^{3}}{3},
$$
where $u, v, w$ are all positive real numbers.
$$
\text { Let } u=\sqrt[3]{\frac{1}{a}+6 b}, v=\sqrt[3]{\frac{1}{b}+6 c}, w=\sqrt[3]{\frac{1}{c}+6 a} \text {, }
$$
then we have
$$
\begin{array}{l}
\sqrt[3]... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,931 |
7. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, $n>1$, and let $g_{n}$ denote their geometric mean. Let $A_{1}, A_{2}, \cdots, A_{n}$ denote the arithmetic means given by
$$
A_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}, k=1,2, \cdots, n,
$$
and let $G_{n}$ denote the geometric mean of $A_{1}, A_{2}, \cdots,... | 7. For convenience, let $A_{0}=0$, then for $k=1,2, \cdots, n$, we have
$$
\frac{a_{k}}{A_{k}}=\frac{k A_{k}-(k-1) A_{k-1}}{A_{k}}=k-(k-1) \frac{A_{k-1}}{A_{k}} .
$$
Let $x_{1}=1, x_{k}=\frac{A_{k-1}}{A_{k}}, k=2,3, \cdots, n$, then we have
$$
\begin{array}{c}
\sqrt[n]{\frac{G_{n}}{A_{n}}}=\sqrt[n]{\frac{A_{1} A_{2} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,933 |
2. Let circle $\Gamma$ and line $l$ not intersect, $AB$ be a diameter of circle $\Gamma$, and perpendicular to line $l$, point $B$ is closer to line $l$ than point $A$. Take any point $C (C \neq A, B)$ on circle $\Gamma$, line $AC$ intersects line $l$ at point $D$, line $DE$ is tangent to circle $\Gamma$ at point $E$, ... | 2. As shown in Figure 1, let $C F$ intersect the circle $\Gamma$ at point $H$. Since the diameter $A B \perp l$, the problem is equivalent to proving $G H \parallel l$.
Let $A B \perp l$, with the foot of the perpendicular being point $X$, then
$$
\begin{array}{l}
\angle A X F=\angle A E F \\
=90^{\circ} .
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,934 |
3. Let point $O$ be the circumcenter of acute $\triangle ABC$, $\angle B < \angle C$, and line $AO$ intersects side $BC$ at point $D$. The circumcenters of $\triangle ABD$ and $\triangle ACD$ are points $E$ and $F$, respectively. Extend $BA$ and $CA$, and take points $G$ and $H$ on the extensions such that $AG = AC$ an... | 3. As shown in Figure 2, let $\angle ABC = \beta, \angle ACB = \gamma$, then $\beta = 90^{\circ}$.
Therefore, $\triangle ABD$ is an obtuse triangle, $\triangle ACD$ is an acute triangle, point $E$ is outside $\triangle ABD$, point $F$ is inside $\triangle ACD$, and points $O$ and $E$ are on opposite sides of $AB$.
Sin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,935 |
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