problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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First question As shown in Figure 1, in the acute triangle $\triangle ABC$, $AB > AC$, and $M, N$ are two different points on side $BC$ such that $\angle BAM = \angle CAN$. Let the circumcenters of $\triangle ABC$ and $\triangle AMN$ be $O_{1}$ and $O_{2}$, respectively. Prove: $O_{1}, O_{2}, A$ are collinear. | Proof 1 As shown in Figure 2, let the other intersection of $AB$ with $\odot O_{2}$ be $P$, and the other intersection of $AC$ with $\odot O_{2}$ be $Q$. Connect $PQ, QM$.
From $\angle BAM = \angle CAN$
$\Rightarrow \angle BAN = \angle CAM = \angle QAM$.
Also, $\angle AQM = \angle ANM = \angle ANB$, so $\triangle BAN \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,788 |
The second question: Prove that the set
$$
A=\left\{2,2^{2}, \cdots, 2^{n}, \cdots\right\}
$$
satisfies:
(1) For each $a \in A$ and $b \in \mathbf{N}_{+}$, if $b<2 a-1$, then $b(b+1)$ is definitely not a multiple of $2 a$;
(2) For each $a \in \bar{A}\left(\bar{A}\right.$ denotes the complement of $A$ in $\mathbf{N}_{+... | Proof 1 (1) Since $a \in A$, we can assume $a=2^{k}\left(k \in \mathbf{N}_{+}\right)$.
Assume $2 a \mid b(b+1)$.
Since $b$ and $b+1$ are one odd and one even, and $2 a$ is a power of 2, therefore,
$2 a \mid b$ or $2 a \mid(b+1)$.
Hence $2 a \leqslant b$ or $2 a \leqslant b+1$, both contradicting $b<2a-1$.
(2) Let $a=2^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,789 |
Question 4 Let $S_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}(n$ is a positive integer). Prove: For any real numbers $a, b$ satisfying $0 \leqslant a<b \leqslant 1$, the sequence $\left\{S_{n}-\left[S_{n}\right]\right\}$ has infinitely many terms in $(a, b)([x]$ denotes the greatest integer not exceeding the real number $x$). | Proof 1 Without loss of generality, let $a>0, b<N$, for all
$$
\frac{1}{k_{n}}<b-a, \quad n>N,
$$
the sequence
$$
\left\{x_{t} \mid t \in\left[k_{n}, k_{n+1}-1\right] \cap \mathbf{N}_{+}\right\}
$$
is monotonically increasing, and satisfies
$$
x_{t+1}-x_{t}=\frac{1}{t+1}<b-a, \quad t \in\left[k_{n}, k_{n+1}-1\right] \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,790 |
1. If $-3<x<-1$, then simplifying $|2-| 1+x||$ yields ().
(A) $1-x$
(B) $-3+x$
(C) $3-x$
(D) $3+x$ | $\begin{array}{l}\text { i.1. D. } \\ \text { Original expression }=|2+(1+x)|=|3+x|=3+x \text {. }\end{array}$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,791 |
2. If the vertex of the parabola $y=x^{2}-4 x+m$ is on the $x$-axis, then the value of $m$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 2. D.
From the condition, the discriminant of $x^{2}-4 x+m=0$ is $\Delta=4^{2}-4 m=0 \Rightarrow m=4$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,792 |
3. The side length of rhombus $A B C D$ is 1, and its area is $\frac{7}{9}$. Then the value of $A C+B D$ is ( ).
(A) $\frac{4}{3}$
(B) $\frac{16}{9}$
(C) $\frac{8}{3}$
(D) $\frac{32}{9}$ | 3. C.
Let $A C=2 a, B D=2 b$. Then
$$
\begin{array}{l}
a^{2}+b^{2}=1,2 a b=\frac{7}{9} \Rightarrow(a+b)^{2}=\frac{16}{9} \\
\Rightarrow a+b=\frac{4}{3} \Rightarrow A C+B D=\frac{8}{3} .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,793 |
4. In a convex quadrilateral $ABCD$, it is known that $AB=2AD$, $BC=1$, $\angle ABC=\angle BCD=60^{\circ}$, $\angle ADC=$ $90^{\circ}$. Then the length of $AB$ is ( ).
(A) $\frac{3-\sqrt{3}}{4}$
(B) $\frac{3-\sqrt{3}}{2}$
(C) $2-\sqrt{3}$
(D) $2 \sqrt{3}-3$ | 4. B.
As shown in Figure 2, extend $B A$ and $C D$ to intersect at $E$. Then $\triangle E B C$ is an equilateral triangle.
Let $A B=2 x$. Then
$$
\begin{array}{l}
A D=x, \\
A E=1-2 x .
\end{array}
$$
In the right triangle $\triangle E A D$,
$$
\begin{aligned}
\frac{x}{1-2 x}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\
\Rig... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,794 |
5. For an activity group, if five 13-year-old members leave, or if five 17-year-old members join (the two scenarios do not occur simultaneously), the average age of the members increases by 1 year. Then the original number of members in this activity group is ( ).
(A) 10
(B) 12
(C) 14
(D) 16 | 5. A.
Let the original number of members in the activity group be $x$, and the total sum of their ages be $y$.
From the problem, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ \frac { y - 65 } { x - 5 } = \frac { y } { x } + 1 , } \\
{ \frac { y + 85 } { x + 5 } = \frac { y } { x } + 1 }
\end{array} \Righta... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,795 |
6. For a positive integer, if it reads the same forward and backward, it is called a "palindrome" (such as $1, 11, 121$ are palindromes, while 110 is not a palindrome). If all palindromes are arranged in ascending order:
$1,2, \cdots, 9,11,22, \cdots$.
Then the 2012th palindrome is ( ).
(A) 1011101
(B) 1013101
(C) 1021... | 6. B.
Since there are 9 one-digit palindromic numbers, 9 two-digit palindromic numbers, 90 three-digit palindromic numbers, 90 four-digit palindromic numbers, 900 five-digit palindromic numbers, and 900 six-digit palindromic numbers, there are a total of 1998 palindromic numbers from $1 \sim 999999$.
Thus, $1000001, ... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,796 |
1. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-2 x-m=0$, and $2 x_{1}+x_{2}=0$. Then the value of $m$ is $\qquad$ . | By the relationship between roots and coefficients, we know $x_{1}+x_{2}=2$.
Also, $2 x_{1}+x_{2}=0$, then
$$
\begin{array}{l}
x_{1}+2=0 \\
\Rightarrow x_{1}=-2 \\
\Rightarrow(-2)^{2}-2 \times(-2)-m=0 \\
\Rightarrow m=8 .
\end{array}
$$ | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,797 |
Example 3 Given that $P$ is a point inside an acute triangle $\triangle ABC$ and minimizes $P A+P B+P C$. Determine the position of point $P$ and prove your conclusion. ${ }^{[1]}$
(2009, National Junior High School Mathematics Competition, Tianjin Preliminary) | Solve As shown in Figure 5, construct equilateral triangles $\triangle A C B^{\prime}$ and $\triangle B C A^{\prime}$ outwardly on sides $A C$ and $B C$, respectively, and connect $B B^{\prime}$ and $A A^{\prime}$, intersecting at point $P$. Then point $P$ is the desired point.
The proof is as follows.
It is easy to pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,798 |
2. In $\triangle A B C$, it is known that $\angle A C B=45^{\circ}, D$ is any point on side $A B$ other than points $A$ and $B$, and the circumcenters of $\triangle A B C$, $\triangle A D C$, and $\triangle B D C$ are $O$, $O_{1}$, and $O_{2}$, respectively. Then the degree measure of $\angle O_{1} O O_{2}$ is $\qquad$... | 2. $135^{\circ}$.
As shown in Figure 3, let the lines $O O_{1}$ and $O O_{2}$ intersect $A C$ and $B C$ at points $E$ and $F$, respectively. Then
$$
\begin{array}{l}
\mathrm{OO}_{1} \perp A C, \\
\mathrm{OO}_{2} \perp B C .
\end{array}
$$
Thus, points $C$, $E$, $O$, and $F$ are concyclic.
$$
\begin{array}{l}
\text { ... | 135^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,799 |
3. Let $a, b$ be positive real numbers, $m$ be a positive integer, and satisfy
$$
\left\{\begin{array}{l}
a+b \leqslant 14, \\
a b \geqslant 48+m .
\end{array}\right.
$$
Then the value of $m$ is | 3.1.
Notice,
$$
\begin{array}{l}
14 \geqslant a+b \geqslant 2 \sqrt{a b} \geqslant 2 \sqrt{48+m} \\
\geqslant 2 \sqrt{48+1}=14 .
\end{array}
$$
Therefore, all equalities hold. Hence \( m=1 \). | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,800 |
4. In a ball game competition, there are eight teams participating, and each pair of teams has to play a match. A team gets 2 points for a win, 1 point for a draw, and 0 points for a loss. If a team wants to ensure it enters the top four (i.e., its points must exceed those of at least four other teams), then the minimu... | 4. 11.
Since there are eight teams, there will be $\frac{8 \times 7}{2}=28$ matches, totaling $28 \times 2=56$ points.
If the top five teams draw with each other and each win against the bottom three teams, and the bottom three teams draw with each other, then there will be five teams each with 10 points, and the oth... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,801 |
1. (20 points) Given the parabola $y=x^{2}$ and the line $y=(k+2) x-(2 k-1)$.
(1) Prove: regardless of what real number $k$ is, the parabola and the line always have two different intersection points;
(2) Let the two different intersection points of the parabola and the line be $A\left(x_{1}, y_{1}\right), B\left(x_{2}... | $$
\begin{array}{l}
\text { Three, 1. (1) From } \\
\left\{\begin{array}{l}
y=x^{2}, \\
y=(k+2) x-(2 k-1),
\end{array}\right.
\end{array}
$$
eliminating $y$ yields
$$
\begin{array}{l}
x^{2}-(k+2) x+(2 k-1)=0 . \\
\text { Since } \Delta=(k+2)^{2}-4(2 k-1) \\
=k^{2}-4 k+8=(k-2)^{2}+4>0,
\end{array}
$$
therefore, the pa... | k=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,802 |
2. (25 points) As shown in Figure 1, given that circle $\odot A$ and circle $\odot B$ intersect at points $C$ and $D$, extend $AC$ to intersect circle $\odot B$ at point $E$, and extend $BC$ to intersect circle $\odot A$ at point $F$. Prove: $C$ is the incenter of $\triangle DEF$. | 2. As shown in Figure 4, connect $A D$, $A F$, $B D$, and $B E$.
Then $\angle A F C = \angle A C F = 180^{\circ} - \angle A C B = 180^{\circ} - \angle A D B$
$$
\Rightarrow \angle A F B + \angle A D B = 180^{\circ}
$$
$\Rightarrow A$, $D$, $B$, and $F$ are concyclic.
Similarly, $A$, $D$, $B$, and $E$ are concyclic.
Th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,803 |
3. (25 points) Write the 90 positive integers $10, 11, \cdots, 99$ on the blackboard, and erase $n$ of them so that the product of all the remaining numbers on the blackboard has a units digit of 1. Find the minimum value of $n$.
| 3. If the unit digit of the product of all remaining numbers on the blackboard is 1, then all even numbers between $10 \sim 99$ must be erased, and numbers with a unit digit of 5 must also be erased.
Thus, the unit digit of the remaining numbers must be one of $1, 3, 7, 9$.
Notice that the unit digit of $11 \times 13... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,804 |
1. Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$, and draw perpendiculars from $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then $\overrightarrow{P A} \cdot \overrightarrow{P B}=$ $\qquad$ | - 1. -1 .
Solution 1 Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$.
Then $l_{P A}: y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right)$, which is $y=-x+2 x_{0}+\frac{2}{x_{0}}$.
Solving the above equation with $y=x$ yields point $A\left(x_{0}+\frac{1}{x_{0}}, x_{0}+\frac{1}{x_{0}}\right)$.
Also, point $B\lef... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,805 |
2. Let $\triangle A B C$ have interior angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and satisfy
$$
\begin{array}{l}
a \cos B-b \cos A=\frac{3}{5} c . \\
\text { Then } \frac{\tan A}{\tan B}=
\end{array}
$$ | 2.4.
Solution 1 From the given and the cosine rule, we have
$$
\begin{array}{l}
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{3}{5} c^{2} . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\fra... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,806 |
3. Let $x, y, z \in [0,1]$. Then
$$
M=\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}
$$
The maximum value of $M$ is $\qquad$ . | 3. $\sqrt{2}+1$.
Let's assume $0 \leqslant x \leqslant y \leqslant z \leqslant 1$. Then
$$
\begin{array}{l}
M=\sqrt{y-x}+\sqrt{z-y}+\sqrt{z-x} . \\
\text { By } \sqrt{y-x}+\sqrt{z-y} \\
\leqslant \sqrt{2[(y-x)+(z-y)]}=\sqrt{2(z-x)} \\
\Rightarrow M \leqslant(\sqrt{2}+1) \sqrt{z-x} \leqslant \sqrt{2}+1 .
\end{array}
$$... | \sqrt{2}+1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,807 |
4. For the parabola $y^{2}=2 p x(p>0)$, the focus is $F$, and the directrix is $l$. Points $A$ and $B$ are two moving points on the parabola, and they satisfy $\angle A F B=\frac{\pi}{3}$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the projection of $M$ on $l$. Then the maximum value of $\frac{|M N|}{|A B... | 4. 1 .
Solution 1 Let $\angle A B F=\theta\left(0<\theta<\frac{2 \pi}{3}\right)$. Then by the Law of Sines, we have
$$
\frac{|A F|}{\sin \theta}=\frac{|B F|}{\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}} .
$$
Thus, $\frac{|A F|+|B F|}{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}=... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,808 |
Example 4 Given that $\triangle XYZ$ is an isosceles right triangle with legs of length 1 $\left(\angle Z=90^{\circ}\right)$, its three vertices are on the three sides of isosceles right $\triangle ABC\left(\angle C=90^{\circ}\right)$. Find the maximum possible length of the legs of $\triangle ABC$.
$(2002$, Shanghai J... | (1) As shown in Figure 7, if vertex $Z$ is on the hypotenuse $A B$, take the midpoint $M$ of $X Y$, and connect $C M$, $Z M$, $C Z$, and draw the altitude $C N$ from $C$ to $A B$. Then
$$
\begin{array}{l}
C N \leqslant C Z \leqslant C M+M Z \\
=\frac{1}{2} X Y+\frac{1}{2} X Y=X Y=\sqrt{2} .
\end{array}
$$
Thus, $C A=\... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,809 |
5. Let two regular tetrahedra $P-ABC$ and $Q-ABC$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-ABC$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-ABC$ is . $\qqua... | 5.4.
As shown in Figure 6, connect $P Q$. Then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$.
Connect $C H$ and extend it to intersect $A B$ at point $M$. Then $M$ is the midpoint of side ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,810 |
7. The sum of all positive integers $n$ that satisfy $\frac{1}{4}<\sin \frac{\pi}{n}<\frac{1}{3}$ is . $\qquad$ | 7. 33 .
By the convexity of the sine function, we know that when $x \in\left(0, \frac{\pi}{6}\right)$, $\frac{3}{\pi} x < \sin x < x$. For example, $\frac{3}{\pi} \times \frac{\pi}{12}=\frac{1}{4}$, $\sin \frac{\pi}{10} < \frac{3}{\pi} \times \frac{\pi}{9}=\frac{1}{3}$.
Therefore, the positive integer values of $n$ t... | 33 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,811 |
8. A certain intelligence station has four different passwords $A, B, C, D$, and uses one of them each week. Each week, one of the three passwords not used in the previous week is chosen with equal probability. If the first week uses password $A$, then the probability that the seventh week also uses password $A$ is $\q... | 8. $\frac{61}{243}$.
Let $P_{k}$ represent the probability of using the $A$ type codebook in the $k$-th week. Then the probability of not using the $A$ type codebook in the $k$-th week is $1-P_{k}$.
$$
\begin{array}{l}
\text { Hence } P_{k+1}=\frac{1}{3}\left(1-P_{k}\right)\left(k \in \mathbf{N}_{+}\right) \\
\Rightar... | \frac{61}{243} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,812 |
9. (16 points) Given the function
$$
f(x)=a \sin x-\frac{1}{2} \cos 2 x+a-\frac{3}{a}+\frac{1}{2} \text {, }
$$
where $a \in \mathbf{R}$, and $a \neq 0$.
(1) If for any $x \in \mathbf{R}$, $f(x) \leqslant 0$, find the range of values for $a$.
(2) If $a \geqslant 2$, and there exists $x \in \mathbf{R}$ such that $f(x) ... | 9. (1) $f(x)=\sin ^{2} x+a \sin x+a-\frac{3}{a}$. Let $t=\sin x(-1 \leqslant t \leqslant 1)$. Then $g(t)=t^{2}+a t+a-\frac{3}{a}$.
By the given conditions,
$$
\left\{\begin{array}{l}
g(-1)=1-\frac{3}{a} \leqslant 0, \\
g(1)=1+2 a-\frac{3}{a} \leqslant 0 .
\end{array}\right.
$$
Solving for $a$, we get the range of $a$ ... | [2,3] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,813 |
10. (20 points) Given a sequence $\left\{a_{n}\right\}$ whose terms are all non-zero real numbers, and for any positive integer $n$ we have
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} \text {. }
$$
(1) When $n=3$, find all sequences of three terms $a_{1}, a_{2}, a_{3}$ that satisfy... | 10. (1) When $n=1$, $a_{1}^{2}=a_{1}^{3}$.
By $a_{1} \neq 0$, we get $a_{1}=1$.
When $n=2$, $\left(1+a_{2}\right)^{2}=1+a_{2}^{3}$.
By $a_{2} \neq 0$, we get $a_{2}=2$ or -1.
When $n=3$, $\left(1+a_{2}+a_{3}\right)^{2}=1+a_{2}^{3}+a_{3}^{3}$.
If $a_{2}=2$, we get $a_{3}=3$ or -2;
If $a_{2}=-1$, we get $a_{3}=1$.
In su... | a_{n}=\left\{\begin{array}{ll}
n, & 1 \leqslant n \leqslant 2012 ; \\
(-1)^{n} 2012, & n \geqslant 2013 .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,814 |
11. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x O y$, the side length of rhombus $A B C D$ is 4, and $|O B| = |O D| = 6$.
(1) Prove that $|O A||O C|$ is a constant;
(2) When point $A$ moves on the semicircle $M: (x-2)^{2} + y^{2} = 4$ $(2 \leqslant x \leqslant 4)$, find the trajectory of poi... | 11. (1) From $|A B|=|A D|=|C B|=|C D|$, $|O B|=|O D|$, we know that points $O$, $A$, and $C$ are collinear.
As shown in Figure 7, connect $B D$. Then $B D$ is the perpendicular bisector of segment $A C$. Let the foot of the perpendicular be $K$.
Thus, $|O A||O C|$
$$
\begin{array}{l}
=(|O K|-|A K|)(|O K|+|A K|) \\
=|O... | (5,5) \text{ and } (5,-5) | Geometry | proof | Yes | Yes | cn_contest | false | 725,815 |
一、(40 points) As shown in Figure 2, in the acute triangle $\triangle ABC$, $AB > AC$, and $M, N$ are two different points on side $BC$ such that $\angle BAM = \angle CAN$. Let the circumcenters of $\triangle ABC$ and $\triangle AMN$ be $O_{1}$ and $O_{2}$, respectively. Prove that $O_{1}, O_{2}, A$ are collinear. | As shown in Figure 8, connect $A O_{1}$ and $A O_{2}$. Draw a perpendicular line $A P$ from point $A$ to $A O_{1}$, intersecting the extension of $B C$ at point $P$. Then $A P$ is a tangent to $\odot O_{1}$.
Thus, $\angle B=\angle P A C$.
Since $\angle B A M=\angle C A N$, we have,
$$
\begin{array}{l}
\angle A M P=\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,816 |
Sure, here is the translated text:
```
(40 points) Try to prove: the set
$$
A=\left\{2,2^{2}, \cdots, 2^{n}, \cdots\right\}
$$
satisfies
(1) for each $a \in A$ and $b \in \mathbf{N}_{+}$, if $b<2 a-1$, then $b(b+1)$ is definitely not a multiple of $2 a$;
(2) for each $a \in \bar{A}\left(\bar{A}\right.$ denotes the co... | (1) For any $a \in A$, let $a=2^{k}\left(k \in \mathbf{N}_{+}\right)$. Then $2 a=2^{k+1}$.
If $b$ is any positive integer less than $2 a-1$, then $b+1 \leqslant 2 a-1$.
Since $b$ and $b+1$ include one odd number, which does not contain the prime factor 2, and one even number, which contains the power of 2 at most $k$ t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,817 |
Three. (50 points) Let $P_{0}, P_{1}, \cdots, P_{n}$ be $n+1$ points on a plane, and the minimum distance between any two of them is $d$ $(d>0)$. Prove:
$$
\left|P_{0} P_{1}\right|\left|P_{0} P_{2}\right| \cdots\left|P_{0} P_{n}\right|>\left(\frac{d}{3}\right)^{n} \sqrt{(n+1)!} .
$$ | Three, Proof 1 Without loss of generality, assume
$$
\left|P_{0} P_{1}\right| \leqslant\left|P_{0} P_{2}\right| \leqslant \cdots \leqslant\left|P_{0} P_{n}\right| .
$$
First, we prove: For any positive integer $k$, we have
$$
\left|P_{0} P_{k}\right|>\frac{d}{3} \sqrt{k+1} \text {. }
$$
Obviously, $\left|P_{0} P_{k}\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,818 |
Example 5 As shown in Figure 9, in the right triangle $\triangle ABC$, $\angle C = 90^{\circ}, CD \perp AB$, the angle bisector of $\angle B$ intersects $CD$ and $CA$ at points $E$ and $F$, respectively. $G$ is the midpoint of $EF$, and $CG$ is connected. Let the perimeters of $\triangle CFG$, $\triangle BED$, and $\tr... | Solve: From $\angle C E F=\frac{1}{2} \angle B+\angle B C D$
$$
=\frac{1}{2} \angle B+\angle A=\angle C F E \text {, }
$$
we get $C E=C F$.
Also, since $G$ is the midpoint of $E F$, then $C G \perp E F$.
It is also easy to see that
$$
\begin{array}{l}
\text { Rt } \triangle C F G \backsim \mathrm{Rt} \triangle C E G \... | \frac{9}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,819 |
Four. (50 points) Let $S_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$ (where $n$ is a positive integer). Prove: For any real numbers $a, b$ satisfying $0 \leqslant a < b \leqslant 1$, the sequence $\left\{S_{n}-\left[S_{n}\right]\right\}$ has infinitely many terms in $(a, b)$ (where $[x]$ denotes the greatest integer less tha... | Proof 1 (1) For any $n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
S_{2^{n}}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^{n}} \\
=1+\frac{1}{2}+\left(\frac{1}{2^{1}+1}+\frac{1}{2^{2}}\right)+\cdots+\left(\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^{n}}\right) \\
>1+\frac{1}{2}+\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,820 |
2. Prove: Among the vertices of a regular $2 n-1(n \geqslant 3)$-sided polygon, any $n$ points chosen must include three points that form an isosceles triangle.
(Zou Jin) | 2. First consider the case where $n>4$.
Assume that it is possible to select $n$ vertices from a regular $(2n-1)$-gon $A_{1} A_{2} \cdots A_{2 n-1}$ such that no three vertices form an isosceles triangle. Color the $n$ selected vertices red and the remaining $n-1$ vertices blue.
Without loss of generality, let $A_{1}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,822 |
3. Let $E$ be a given $n$-element set, and $A_{1}, A_{2}, \cdots, A_{k}$ be $k$ distinct non-empty subsets of $E$, satisfying: for any $1 \leqslant i<j \leqslant k$, either $A_{i} \cap A_{j}=\varnothing$ or one of $A_{i}$ and $A_{j}$ is a subset of the other. Find the maximum value of $k$.
(Cold Gangsong, problem contr... | 3. The maximum value of $k$ is $2 n-1$.
Let $E=\{1,2, \cdots, n\}$.
For example, when $1 \leqslant i \leqslant n$, let $A_{i}=\{i\}$;
when $n+1 \leqslant i \leqslant 2 n-1$, let $A_{i}=\{x \in E \mid 1 \leqslant x \leqslant i-n+1\}$.
The $2 n-1$ sets $A_{i}$ thus selected satisfy the conditions.
We prove by mathematic... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,823 |
4. Given $P$ as any point inside acute $\triangle ABC$, points $E$ and $F$ are the projections of $P$ onto sides $AC$ and $AB$, respectively. The extensions of $BP$ and $CP$ intersect the circumcircle of $\triangle ABC$ at points $B_1$ and $C_1$, respectively. Let the circumradius and inradius of $\triangle ABC$ be $R$... | 4. As shown in Figure 1, draw $P D \perp B C$ at point $D$, connect $A P$ and extend it to intersect the circumcircle of $\triangle A B C$ at point $A_{1}$, connect $D E, D F, A_{1} B_{1}, A_{1} C_{1}$.
Figure 1
From the fact that points $P, D, B, F$ are concyclic, we know
$\angle P D F = \angle P B F$.
From the fact t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,824 |
5. In an acute triangle $\triangle ABC$, $H$ is the orthocenter, $O$ is the circumcenter (points $A$, $H$, and $O$ are not collinear), point $D$ is the projection of $A$ on side $BC$, and the perpendicular bisector of segment $AO$ intersects line $BC$ at point $E$. Prove: The midpoint of segment $OH$ lies on the circum... | 5. As shown in Figure 3, let the midpoints of $AO$ and $HO$ be $F$ and $N$ respectively. Extend $HD$ to intersect the circumcircle of $\triangle ABC$ at point $H'$. Connect $FN$, $DN$, $BH$, $BH'$, and $OH'$.
Figure 3
Since $H$ is the orthocenter, $\angle CBH' = \angle CAH' = \angle CBH$.
Therefore, $D$ is the midpoin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,825 |
7. For an $n \times n$ grid, we call cells that share a common edge "adjacent." Initially, each cell contains +1. An operation on the grid involves selecting one cell, not changing the number in this cell, but changing the sign of the numbers in all adjacent cells. Find all positive integers $n \geqslant 2$, such that ... | 7. The desired result is all even numbers.
First, denote the cell in the $i$-th row and $j$-th column as $A_{ij}$, where $i, j \in \{1, 2, \cdots, n\}$.
First, we prove: when $n$ is even, all the numbers in the cells can be changed to -1.
When $n = 2k \left(k \in \mathbf{N}_{+}\right)$, as shown in Figure 5, color t... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,827 |
8. Find all prime numbers $p$ such that there exist infinitely many positive integers $n$ satisfying
$$
p \mid \left[n^{n+1}+(n+1)^{n}\right] .
$$
(Proposed by Yonggao Chen) | 8. Let $p$ be an odd prime.
For any positive integer $n$, $n^{n+1}+(n+1)^{n}$ is odd, hence, $p \neq 2$.
Consider the case $p \geqslant 3$, prove that there are infinitely many $n$ satisfying the condition.
Method 1 To make the remainder of $(n+1)^{n}$ when divided by $p$ determined, take $n=p k-2$ and $n$ is odd, t... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,828 |
1. Given that $\alpha$ is an acute angle of a right triangle. Then $\sqrt{\left(1-\frac{\tan \alpha}{\sin \alpha}\right)^{2}}=(\quad)$.
(A) $1-\frac{\tan \alpha}{\sin \alpha}$
(B) $\frac{\tan \alpha}{\sin \alpha}-1$
(C) $1+\frac{\tan \alpha}{\sin \alpha}$
(D) Cannot be determined | $-1 . B$
From the definition of trigonometric functions, we get $\tan \alpha>\sin \alpha$.
Then $\sqrt{\left(1-\frac{\tan \alpha}{\sin \alpha}\right)^{2}}=\frac{\tan \alpha}{\sin \alpha}-1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,829 |
Example 6 In the interior of rectangle $ABCD$ (excluding the boundary), there is a point $P$, which is at a distance of 1 from vertex $A$ and sides $BC$, $CD$. Find the range of the area of rectangle $ABCD$.
$(2008$, New Knowledge Cup Shanghai Junior High School Mathematics Competition) | Solve as shown in Figure 10, introduce variables $x, y$.
Then $x^{2}+y^{2}=P A^{2}=1$, and $x+y>P A=1$.
Thus, $S_{\text {rectangle } A B C D}=(1+x)(1+y)$
$$
=1+x+y+x y>2 \text {, }
$$
and when $x$ (or $y) \rightarrow 0$, $S \rightarrow 2$.
Also, $(x+y)^{2} \leqslant 2\left(x^{2}+y^{2}\right)=2$, then $x+y \leqslant \s... | 2<S_{\text {rectangle } A B C D} \leqslant \frac{3}{2}+\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,830 |
2. If the graph of the quadratic function $y=(x-3)^{2}-2$ intersects the $x$-axis at two points $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$, then $\frac{\sqrt{x_{1} x_{2}}+x_{1}}{x_{2}+\sqrt{x_{1} x_{2}}}+\frac{\sqrt{x_{1} x_{2}}-x_{2}}{x_{1}-\sqrt{x_{1} x_{2}}}=(\quad)$.
(A) $\frac{12 \sqrt{13}}{13}$
(B) $\frac{10 \... | 2. C.
According to the problem, we have $y=x^{2}-6 x+7$.
By Vieta's formulas, we get
$$
\begin{array}{l}
x_{1}+x_{2}=6, x_{1} x_{2}=7 . \\
\text { Therefore, the original expression }=\frac{\sqrt{x_{1}}}{\sqrt{x_{2}}}+\frac{\sqrt{x_{2}}}{\sqrt{x_{1}}}=\frac{x_{1}+x_{2}}{\sqrt{x_{1} x_{2}}}=\frac{6 \sqrt{7}}{7} .
\end{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,831 |
3. If the two real roots of the equation $x^{2}-m x+n+1=0$ are both positive integers, then $m^{2}+n^{2}$ is ( ).
(A) Composite number
(B) Prime number
(C) Perfect square
(D) Even number | 3. A.
Let the two positive integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=m, \\
x_{1} x_{2}=n+1 .
\end{array}
$$
Therefore, $m^{2}+n^{2}=\left(x_{1}+x_{2}\right)^{2}+\left(x_{1} x_{2}-1\right)^{2}$
$$
=\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right) \text {. }
$$
Since $x... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,832 |
4. In an acute $\triangle A B C$, $a$, $b$, and $c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $\angle A=60^{\circ}$. Then $\frac{c}{a+b}+\frac{b}{a+c}=(\quad)$.
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) None of the above answers | Figure 5
4. A.
As shown in Figure 5, construct $C D \perp$
$A B$ at point $D$. Then
$$
\begin{array}{c}
A D=\frac{1}{2} b \\
C D=\frac{\sqrt{3}}{2} b \\
B D=c-\frac{1}{2} b \\
\Rightarrow \frac{a^{2}}{}=\left(\frac{\sqrt{3}}{2} b\right)^{2}+\left(c-\frac{1}{2} b\right)^{2}=b^{2}+c^{2}-b c \\
\Rightarrow \frac{c}{a+b}+... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,833 |
5. In $\triangle A B C$, it is known that $A B=13, B C=12$, $C A=5$, $D$ is the midpoint of side $A B$, $D E \perp A B$ and intersects the angle bisector of $\angle A C B$ at point $E$. Then the length of $D E$ is ( ).
(A) $\frac{60}{13}$
(B) $\frac{11}{2}$
(C) 6
(D) $\frac{13}{2}$ | 5. D.
As shown in Figure 6, connect $C D$. It is easy to know that $\angle A C B=90^{\circ}$.
Since $C E$ bisects $\angle A C B$, we get $\angle E C B=45^{\circ}$.
Thus, $\alpha=45^{\circ}+\angle B$.
And since $C D=B D$, we have
$$
\begin{array}{l}
\angle E C D \\
=45^{\circ}-\angle B .
\end{array}
$$
Since $D E \pe... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,834 |
6. Given the function $f(x)=\sqrt{x^{2}+2}(x>0)$. Then the integer part of $N=f(1002)+f(1003)+\cdots+f(2005)$ is ( ).
(A) 1506500
(B) 1509514
(C) 4010
(D) 3013 | 6. B.
Notice,
$$
\begin{array}{l}
f(x)-x=\sqrt{x^{2}+2}-x>0, \\
f(x)-x=\sqrt{x^{2}+2}-x=\frac{2}{\sqrt{x^{2}+2}+x} \\
1002+1003+\cdots+2005 \text {, }
$$
and $\square$
$$
\begin{aligned}
N- & (1002+1003+\cdots+2005) \\
= & \left(\sqrt{1002^{2}+2}-1002\right)+ \\
& \left(\sqrt{1003^{2}+2}-1003\right)+\cdots+ \\
& \lef... | 1509514 | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,835 |
1. Given 10 pairwise distinct positive integers $a_{1}$, $a_{2}, \cdots, a_{10}$ that satisfy the conditions
$$
\begin{array}{l}
a_{2}=a_{1}+a_{5}, a_{3}=a_{2}+a_{6}, \\
a_{4}=a_{3}+a_{7}, a_{6}=a_{5}+a_{8}, \\
a_{7}=a_{6}+a_{9}, a_{9}=a_{8}+a_{10} .
\end{array}
$$
then the minimum possible value of $a_{4}$ is | Ni, 1.20.
It is easy to get
$$
\begin{array}{l}
a_{4}=a_{3}+a_{7} \\
=a_{1}+a_{5}+a_{6}+a_{6}+a_{8}+a_{10} \\
=\left(a_{1}+a_{10}\right)+3\left(a_{5}+a_{8}\right) .
\end{array}
$$
To make $a_{4}$ the smallest, then $a_{5}$ and $a_{8}$ should be as small as possible. And they are all different, so let's take $a_{5}=1, ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,836 |
2. As shown in Figure 1, in rectangle $A B C D$, it is known that $A B=1$, point $M$ is on diagonal $A C$, $\frac{A M}{A C}=\frac{1}{4}$, line $l$ passes through point $M$ and is perpendicular to $A C$, intersecting side $A D$ at point $E$ and side $A B$ at point $G$. If line $l$ divides the rectangle into two parts wi... | 2. $\frac{\sqrt{14}}{7}$.
It is easy to get $\triangle A E G \backsim \triangle D C A \Rightarrow \frac{A G}{A D}=\frac{A E}{D C}$.
Since $D C=A B=1$, we have
$$
\begin{array}{l}
\frac{S_{\triangle A E G}}{S_{\text {rectangle } A B C D}}=\frac{\frac{1}{2} A E \cdot A G}{A D \cdot D C}=\frac{1}{7} \\
\Rightarrow A E^{... | \frac{\sqrt{14}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,837 |
3. As shown in Figure 2, given the line $y=\frac{1}{2} x$ intersects the parabola $y=-\frac{1}{4} x^{2}+6$ at points $A$ and $B$, point $P$ moves on the parabola above line $A B$. When the area of $\triangle P A B$ is maximized, the coordinates of point $P$ are $\qquad$ | 3. $\left(-1, \frac{23}{4}\right)$.
As shown in Figure 7, construct $P M \perp x$-axis intersecting line $A B$ at point $M$.
Let $P\left(a,-\frac{1}{4} a^{2}+6\right)$.
Then $M\left(a, \frac{1}{2} a\right)$.
Thus, $P M=-\frac{1}{4} a^{2}-\frac{1}{2} a+6$.
By solving $y=\frac{1}{2} x$ and $y=-\frac{1}{4} x^{2}+6$ simul... | \left(-1, \frac{23}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,838 |
4. As shown in Figure 3, given a circle with radius 1 and center at $M(0,1)$, point $B(0,2)$, and point $A$ is on the negative half of the $x$-axis. $D$ is the midpoint of $O A$, and $A B$ intersects $\odot M$ at point $C$. If quadrilateral $B C D M$ is a parallelogram, then $\sin \angle A B D=$ $\qquad$ | 4. $\frac{\sqrt{10}}{10}$.
As shown in Figure 8, connect $O C$.
Then $\angle O C A=90^{\circ}$.
Therefore, quadrilateral $B C D M$ is a parallelogram
$$
\begin{array}{l}
\Rightarrow C D / / M B \Rightarrow C D \perp O A \\
\Rightarrow O C=O A \Rightarrow \angle C A O=45^{\circ} \\
\Rightarrow O A=O B \Rightarrow A D=1... | \frac{\sqrt{10}}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,839 |
Example 1 There are 2011 points in space and no three points are collinear. Now, connect each pair of points with a line of one color, such that for any point, any two lines originating from that point are of different colors. How many different colors of lines are needed at least? Prove your conclusion. If the 2011 po... | To generalize, replace 2011 with $n(n \geqslant 2)$, and denote the minimum number of colors for the line segments as $f(n)$. It is easy to see that
$$
\begin{array}{l}
f(2)=1, f(3)=3, f(4)=3, \\
f(5)=5, \cdots \cdots
\end{array}
$$
Below, we prove that in the general case,
$$
f(2 n+1)=2 n+1, f(2 n)=2 n-1 \text {. }
$... | 2011 | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,841 |
II. (25 points) As shown in Figure 4, given the parabola
$$
y=a x^{2}+b x+8\left(a>\frac{1}{2}\right)
$$
passes through point $D(5,3)$, intersects the $x$-axis at points $B\left(x_{1}, 0\right)$ and $C\left(x_{2}, 0\right)$, and $S_{\triangle B C D}=3$. A line $l \perp C D$ is drawn through point $D$ and intersects th... | (1) From $S_{\triangle B C D}=3$, we get
$$
\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-32 a}}{|a|}=2 \text {. }
$$
Also, from $25 a+5 b+8=3$, we get $a=1, b=-6$.
Thus, the equation of the parabola is $y=x^{2}-6 x+8$.
It is easy to find that $C(4,0)$.
Let $P(t, 0)$.
As shown in Figure 9, draw $D E \perp x$-axis at poin... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,842 |
Three. (25 points) Let the pairs of positive integers $(m, n)$, where both are no more than 1000, satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \text {. }
$$
Find the number of all such pairs $(m, n)$. | Three, 1706.
$$
\begin{array}{l}
\text { Given } \frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \\
\Rightarrow \sqrt{2} n-1<m<\sqrt{2}(n+1) .
\end{array}
$$
For each $n$, the number of integers in the above range is
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1,
\end{array}
$$
where $[... | 1706 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,843 |
1. The function $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)+\arcsin x$. Then the solution set of $f(x)+f\left(2-x^{2}\right) \leqslant 0$ is $\qquad$ | $-、 1 .\{-1\}$.
It is known that the function $f(x)$ is a monotonically increasing odd function defined on $[-1,1]$.
$$
\begin{array}{l}
\text { Given } f(x)+f\left(2-x^{2}\right) \leqslant 0 \\
\Rightarrow f(x) \leqslant f\left(x^{2}-2\right) \\
\Rightarrow-1 \leqslant x \leqslant x^{2}-2 \leqslant 1 \\
\Rightarro... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,844 |
2. $\frac{\sqrt{2} \cos 55^{\circ}-\sin 20^{\circ}}{\sqrt{2} \cos 5^{\circ}+\sin 20^{\circ}}=$ | $\begin{array}{l}2.2-\sqrt{3} . \\ \frac{\sqrt{2} \cos 55^{\circ}-\sin 20^{\circ}}{\sqrt{2} \cos 5^{\circ}+\sin 20^{\circ}} \\ =\frac{\cos 10^{\circ}-\sin 10^{\circ}-\sin 20^{\circ}}{\cos 40^{\circ}+\sin 40^{\circ}+\sin 20^{\circ}} \\ =\frac{\sin 80^{\circ}-\sin 20^{\circ}-\sin 10^{\circ}}{\cos 40^{\circ}+2 \sin 30^{\c... | 2-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,845 |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}^{2}-2\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
and $a_{1}=a, a_{2012}=b(a 、 b>2)$.
Then $a_{1} a_{2} \cdots a_{2011}=$
(express in terms of $a, b$). | 3. $\sqrt{\frac{b^{2}-4}{a^{2}-4}}$.
From $a_{n+1}^{2}-4=a_{n}^{2}\left(a_{n}^{2}-4\right)=\cdots$
$$
=a_{n}^{2} a_{n-1}^{2} \cdots a_{1}^{2}\left(a_{1}^{2}-4\right) \text {, }
$$
then $b^{2}-4=a_{2011}^{2} a_{2010}^{2} \cdots a_{1}^{2}\left(a^{2}-4\right)$.
Therefore, $a_{1} a_{2} \cdots a_{2011}=\sqrt{\frac{b^{2}-4... | \sqrt{\frac{b^{2}-4}{a^{2}-4}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,846 |
4. Given a right triangle $\triangle A B C$ with the two legs $A C=4$, $B C=3$, and $D$ is a moving point on the hypotenuse $A B$. Now, fold the right triangle along $C D$ to form a right dihedral angle $A-C D-B$. Then, when $A B$ reaches its minimum value, the size of the dihedral angle $B-A C-D$ is | 4. $\arctan \sqrt{2}$.
As shown in Figure 1, construct $A E \perp C D$ at point $E$, and $B F \perp C D$ at point $F$.
Let $\angle A C D=\theta$.
From the folded figure, we have
$$
\begin{array}{l}
|A B|^{2} \\
=|A E|^{2}+|E F|^{2}+|B F|^{2} \\
=(4 \sin \theta)^{2}+(3 \cos \theta)^{2}+ \\
|4 \cos \theta-3 \sin \theta|... | \arctan \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,847 |
5. Given the equation of the ellipse $\Gamma$ as $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ with a slope of $k_{1}\left(k_{1} \neq 0\right)$ intersects the ellipse at points $A$ and $B$. Let $R(1,0)$, and extend $A R$ and $B R$ to intersect the ellipse at points $C$ and $D$, re... | 5. $\frac{4}{7}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$, $D\left(x_{4}, y_{4}\right)$.
Then the line $l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substituting into the ellipse equation and eliminating $x$ yields
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=... | \frac{4}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,848 |
For $n \in \mathbf{N}_{+}$, define
$$
S(n)=\left[\frac{n}{10^{[18 n]}}\right]+10\left(n-10^{[i \mid n]}\left[\frac{n}{10^{\left[1 / B^{n}\right]}}\right]\right),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then, among $1,2, \cdots, 2012$, the number of positive integers $n$ that sati... | 6. 108.
Let $t=10^{[\lg n]}$. Then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right) \text {. }
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the first digit of $n$.
We will discuss the cases below.
(1) If $n$ is a one-dig... | 108 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,849 |
7. Given a regular $n$-sided polygon ($n$ being an odd number) inscribed in a unit circle, let $S$ be the set of distances between any two distinct vertices of the regular $n$-sided polygon. Then the value of $P=\prod_{i \in S} i$ is $\qquad$ . | 7. $\sqrt{n}$.
Let $n=2 m+1\left(m \in \mathbf{N}_{+}\right)$, and denote the polygon as $P_{0} P_{1} \cdots P_{n-1}$. Place the regular polygon in the complex plane with the center at the origin, and the vertices correspond to the complex numbers
$$
P_{k}=\omega^{k}\left(\omega=\mathrm{e}^{\frac{2 \pi i}{n}}\right) .... | \sqrt{n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,850 |
8. Let $p$ be a given odd prime, for $i=1$, $2, \cdots, p-1$, denote $r_{i}$ as the remainder of $i^{p}$ modulo $p^{2}$. Then
$$
r_{1}+r_{2}+\cdots+r_{p-1}=
$$ | 8. $\frac{p^{3}-p^{2}}{2}$.
It is easy to see that
$$
\begin{array}{l}
r_{i}+r_{p-i} \equiv i^{p}+(p-i)^{p} \\
\equiv \mathrm{C}_{p}^{p-1} p \cdot i^{p-1} \equiv 0\left(\bmod p^{2}\right) .
\end{array}
$$
Since $r_{i}, r_{p-i} \in\left(0, p^{2}\right)$, therefore, $r_{i}+r_{p-i}=p^{2}$.
$$
\text { Hence } \sum_{i=1}^... | \frac{p^{3}-p^{2}}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,851 |
9. (16 points) Let $n$ be a given positive integer, and let
$$
\begin{array}{l}
a_{k}=2 \cos \frac{\pi}{2^{n-k}}(k=0,1, \cdots, n-1) . \\
\text { Prove: } \prod_{k=0}^{n-1}\left(1-a_{k}\right)=\frac{(-1)^{n-1}}{1+a_{0}} .
\end{array}
$$ | $$
\begin{array}{l}
1-a_{k}=1-2 \cos \frac{\pi}{2^{n-k}}=\frac{1-4 \cos ^{2} \frac{\pi}{2^{n-k}}}{1+2 \cos \frac{\pi}{2^{n-k}}} \\
=-\frac{1+2 \cos \frac{\pi}{2^{n-k-1}}}{1+2 \cos \frac{\pi}{2^{n-k}}}=-\frac{1+a_{k+1}}{1+a_{k}} . \\
\text { Therefore, } \prod_{k=0}^{n-1}\left(1-a_{k}\right)=\prod_{k=0}^{n-1}\left(-\fra... | \frac{(-1)^{n-1}}{1+a_{0}} | Algebra | proof | Yes | Yes | cn_contest | false | 725,853 |
10. (20 points) The function $f(x)$ defined on $[0,1]$ satisfies: $f(0)=f(1)$, and for any $x, y \in [0,1]$ $(x \neq y)$, we have $|f(x)-f(y)|<|x-y|$. Find the smallest real number $m$, such that for any $x, y \in [0,1]$, we have
$$
|f(x)-f(y)|<m .
$$ | 10. (1) Prove: For all $x, y \in [0,1]$, we have
$$
|f(x)-f(y)|\frac{1}{2}$, without loss of generality, assume $0 \leqslant x < y \leqslant 1$. Then
$$
\begin{array}{l}
|f(x)-f(y)| \\
=|f(x)-f(0)+f(1)-f(y)| \\
\leqslant|f(x)-f(0)|+|f(1)-f(y)| \\
<x+1-y<\frac{1}{2} .
\end{array}
$$
(2) For the function
$$
f(x)=\left\{\... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,854 |
11. (20 points) Given the hyperbola $C: \frac{x^{2}}{4}-y^{2}=1$, a line $l$ passing through the point $M(1,-1)$ intersects the right branch of the hyperbola $C$ at points $A$ and $B$, and intersects the $x$-axis at point $N$. Let
$$
\overrightarrow{M A}=\lambda_{1} \overrightarrow{A N}, \overrightarrow{M B}=\lambda_{2... | 11. Let the line $l: y+1=k(x-1)$. Then
$$
N\left(\frac{1}{k}+1,0\right) \text {. }
$$
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Substituting the equation of line $l$ into the equation of the hyperbola $C$ yields
$$
\begin{array}{l}
\left(1-4 k^{2}\right) x^{2}+8 k(k+1) x-4(k+1)^{2}-4 \\
=0 .
\end{a... | \left(-\frac{74}{35}, 2\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,855 |
One. (40 points) Given $\triangle A B C$ with its circumcircle $\odot O$ passing through points $B$ and $C$, the tangents at points $B$ and $C$ intersect at point $T$. On the rays $B T$ and $C T$, take points $M$ and $N$ such that $B M = B C = C N$. The lines through points $M$ and $N$ intersect the extensions of $A C$... | As shown in Figure 2, let $AD$ be the angle bisector of $\angle BAC$, intersecting $BC$ at point $D$. Connect $DQ$ and $DP$.
From $BM = CN = BC$, we get $BC \parallel EF$.
Thus, $\angle BMF = \angle TBC = \angle BAC$, $\angle BFM = \angle ABC$.
Therefore, $\triangle MFB \sim \triangle ABC$.
$$
\begin{array}{l}
\text{T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,856 |
For the given positive integer $k$, let
$$
S_{n}^{(j)}=\sum_{i=0}^{+\infty} \mathrm{C}_{n}^{j+i k} \quad (j=0,1, \cdots, k-1) \text {. }
$$
Convention: When $m \geqslant n$, $\mathrm{C}_{n}^{m}=0$.
Prove: $\left(\sum_{j=0}^{k-1} S_{n}^{(j)} \cos \frac{2 j \pi}{k}\right)^{2}+\left(\sum_{j=0}^{k-1} S_{n}^{(j)} \sin \fra... | Let $D_{j}=\{j+m k \mid m \in \mathbf{N}, j+m k \leqslant n\}$.
Then $S_{n}^{(j)}=\sum_{p \in D_{j}} \mathrm{C}_{n}^{p}, \bigcup_{j=0}^{k-1} D_{j}=\{0,1, \cdots, n\}$.
Let $a=\sum_{j=0}^{k-1} S_{n}^{(j)} \cos \frac{2 j \pi}{k}, b=\sum_{j=0}^{k-1} S_{n}^{(j)} \sin \frac{2 j \pi}{k}$.
Then $a+\mathrm{i} b$
Thus, $|a+\ma... | a^{2}+b^{2}=\left(2 \cos \frac{\pi}{k}\right)^{2 n} | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,857 |
Three. (50 points) Define the function $f: \mathbf{N}_{+} \rightarrow\{0,1\}$, satisfying
$$
\begin{array}{l}
f(1)=f(2)=1, \\
f(n)=1-f(n-1) f\left(\left[\frac{n}{3}\right]\right),
\end{array}
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
For (1) $n=120$; (2) $n=\frac{3^{2012}-1}{2}$; (... | It is known that $f(3)=0$, and for all $k \in \mathbf{N}_{+}$, $f(3 k)=1-f(3 k-1) f(k)$.
Then $f(3 k) f(k)=f(k)+f(3 k)-1$,
$f(3 k+1)=1-f(3 k) f(k)$
$=2-f(k)-f(3 k)$,
$f(3 k+2)=1-f(3 k+1) f(k)$
$=1-f(k)+f(3 k) f(k)=f(3 k)$.
If $f(k)=0$, then
$$
f(3 k)=f(3 k+1)=f(3 k+2)=1 \text {. }
$$
Since $f(3)=0$, we have
$$
f(3 k+3... | 0, 1, 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,858 |
Four. (50 points) There are 2012 scholars attending a mathematics conference, some of whom know each other, and the following conditions are satisfied:
(1) Each person knows at least 671 of the others;
(2) For any two people $A$ and $B$, if $A$ and $B$ do not know each other, then they can always be indirectly connecte... | Consider 2012 scholars as 2012 points \(a_{1}, a_{2}, \cdots, a_{2012}, a_{i}\) where \(a_{i}\) and \(a_{j}\) are adjacent if and only if \(a_{i}\) and \(a_{j}\) know each other. This forms a graph \(G\).
Take the longest path \(P_{0}\), without loss of generality, assume
\[
P_{0}=\left\{a_{1}, a_{2}, \cdots, a_{k}\rig... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,859 |
Given two distinct positive real numbers $x, y$ that satisfy $x^{4}-y^{4}=x^{5}-y^{5}$. Prove:
$$
1<x+y<\frac{8}{5} .
$$ | $$
\begin{array}{l}
(x+y)\left(x^{4}-y^{4}\right)=x^{5}-y^{5}+x y\left(x^{3}-y^{3}\right) \\
\Rightarrow x+y=\frac{x^{5}-y^{5}}{x^{4}-y^{4}}+x y \cdot \frac{x^{3}-y^{3}}{x^{4}-y^{4}} \\
=1+x y \cdot \frac{x^{2}+x y+y^{2}}{(x+y)\left(x^{2}+y^{2}\right)} \\
=1+\frac{x y}{x+y}\left(1+\frac{x y}{x^{2}+y^{2}}\right)>1 . \\
... | 1<x+y<\frac{8}{5} | Algebra | proof | Yes | Yes | cn_contest | false | 725,860 |
For the expression $\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}$, when written as a decimal, find the digit before the decimal point. | Let $a_{n}=\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{n}-\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
Then $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
The last digits are
$$
\begin{array}{l}
1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7 \\
7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9 \\
8,7,5,2,7,9,6,5,1,6,7,3,0... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,861 |
Given two circles intersecting at points $A$ and $B$. Please construct a line segment $P Q$ passing through point $A$ and intersecting the two circles at points $P$ and $Q$, such that $P Q$ is the longest.
Construction: Let $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. Connect $B O_{1}$ and let it i... | Proof As shown in Figure 1, let the radii of $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ be $r_{1}$ and $r_{2}$, respectively, and $\left|O_{1} O_{2}\right|=a$. Establish a complex plane with $O_{1} O_{2}$ as the real axis.
For any line segment $P Q$ passing through point $A$, we have
$$
\angle B O_{1} P=2 \angle... | 2a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,862 |
The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+c\left(n \in \mathbf{N}_{+}\right),
\end{array}
$$
where $c$ is a constant greater than 0.
(1) If $c=1$, find the general term of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence has an upper bound,... | (1) It is easy to know that
$$
a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+1 \geqslant 1 \text {. }
$$
Therefore, $a_{n} \geqslant 1\left(n \in \mathbf{N}_{+}\right)$.
Then $\left(a_{n+1}-1\right)^{2}=\left(a_{n}-1\right)^{2}+2$.
Let $b_{n}=\left(a_{n}-1\right)^{2}$. Then $\left\{b_{n}\right\}$ is an arithmetic sequence with t... | (0,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,864 |
Example 4 The family of sets $\Omega$ consists of 11 five-element sets $A_{1}, A_{2}$, $\cdots, A_{11}$, where the intersection of any two sets is not empty. Let $A=\bigcup_{i=1}^{11} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets in $\Omega$ that contain the element $x_{i}... | It is known that $\sum_{i=1}^{n} k_{i}=55$.
Notice that, the $k_{i}$ sets containing $x_{i}$ form
$$
\mathrm{C}_{k_{i}}^{2}=\frac{k_{i}\left(k_{i}-1\right)}{2}
$$
pairs of sets. Since the intersection of any two sets is not empty, the sum $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all pairs of sets, with some re... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,865 |
Example 1 When $n$ is any real number and $k$ is a certain specific integer, the equation
$$
n(n+1)(n+2)(n+3)+1=\left(n^{2}+k n+1\right)^{2}
$$
holds. Then $k=$ $\qquad$ . [1]
(2010, Taiyuan Junior High School Mathematics Competition) | 【Analysis】Since the left side of the given equation is a polynomial and the right side is in the form of a product, we only need to factorize the left side. The method of factorization is to use the overall idea and the complete square formula to handle it.
Solution Note that,
$$
\begin{array}{l}
n(n+1)(n+2)(n+3)+1 \\
... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,866 |
Example 2 Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {. }
$$
Then there exists an integer $k$, such that the following equations hold for:
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$;
(2... | 【Analysis】The condition equation in this problem is a fractional equation, which is relatively complex. The key to solving this problem is to simplify the relationship between the letters $a, b, c$. First, eliminate the denominator and rearrange the condition equation into the form $f(a, b, c)=0$, then factorize $f(a, ... | 2 | Algebra | proof | Yes | Yes | cn_contest | false | 725,867 |
For example, $5 n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ have a sum of 2009. If these $n$ numbers can be divided into 41 groups with equal sums and also into 49 groups with equal sums, find the minimum value of $n$.
| Let the 41 groups be $A_{1}, A_{2}, \cdots, A_{41}$, where the sum of the numbers in each group is 49, and we call such groups "A-type groups"; and the 49 groups be $B_{1}, B_{2}, \cdots, B_{49}$, where the sum of the numbers in each group is 41, and we call such groups "B-type groups".
Clearly, each term $x_{k}$ belo... | 89 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,869 |
Example 6 In the Mathematical Olympiad training team, there are 30 members, each of whom has the same number of friends in the team. It is known that in a test, everyone's scores are different. If a member scores higher than the majority of their friends, they are called a "pro". Question: What is the maximum number of... | Let each team member have $k$ friends, and this exam has produced $m$ experts, the best-performing member of the team, is the best in their $k$ "friend pairs," and is naturally an expert. Each of the other experts is at least the best in $\left[\frac{k}{2}\right]+1 \geqslant \frac{k+1}{2}$ (where [x] denotes the greate... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,870 |
1. Prove the following problems:
(1) For each positive integer $n(n \geqslant 3)$, there exist $n$ integer points in the plane, not all on a straight line, such that the distance between any two points is a positive integer;
(2) There exist infinitely many integer points in the plane such that no three points are colli... | (1) For any integer $n$, choose $n$ distinct prime numbers $p_{1}, p_{2}, \cdots, p_{n}$, and let
$$
\begin{array}{l}
m=p_{1} p_{2} \cdots p_{n}, \\
a_{i}=\left(p_{1} p_{2} \cdots p_{i}\right)^{2}-\left(p_{i+1} p_{i+2} \cdots p_{n}\right)^{2}
\end{array}
$$
where, $i=1,2, \cdots, n-1$.
Take point $M(0, m)$ on the $y$-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,871 |
3. Given 133 positive integers $a_{1}, a_{2}, \cdots, a_{133}$. If there are at least 799 pairs of numbers that are coprime, prove: there must exist four numbers $a, b, c, d$, such that
$$
(a, b)=(b, c)=(c, d)=(d, a)=1 .
$$ | Prompt: Use the graphical construction method.
Use points $A_{1}, A_{2}, \cdots, A_{133}$ to represent these 133 positive integers. If $\left(a_{i}, a_{j}\right)=1$, then let the corresponding points $A_{i}$ and $A_{j}$ be adjacent. Thus, we obtain a simple graph $G$ of order 133, and let the degree of point $A_{i}$ be... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,872 |
4. Let the set $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$, satisfying $(a+b) \mid a b$. | First, by enumeration, we obtain 23 pairs $(a, b)$, each of which satisfies $(a+b) \mid a b$.
Construct a 50-order graph $G$ (with the set $S$ of numbers $1,2, \cdots, 50$ as vertices, and if two numbers $a, b$ belong to the above pairs, then let $a, b$ be adjacent). Thus, the graph $G$ has exactly 23 edges (isolated ... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,873 |
Example 1 Find all positive integer solutions of the system of equations
$$
\left\{\begin{array}{l}
a^{3}-b^{3}-c^{3}=3 a b c, \\
a^{2}=2(b+c)
\end{array}\right.
$$ | From equation (1) we get
$$
(a-b-c)\left[a^{2}+(b-c)^{2}+a b+b c+c a\right]=0 \text {. }
$$
Since the second factor on the left side cannot be zero, then
$$
a=b+c=\frac{1}{2} a^{2} .
$$
Thus, the unique positive integer solution to the system of equations is
$$
(a, b, c)=(2,1,1) .
$$ | (a, b, c)=(2,1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,874 |
Example 2 Let $x_{1}, x_{2}, x_{3}$ be the three roots of the equation $x^{3}+x+2=0$. Then the determinant
$$
D=\left|\begin{array}{lll}
x_{1} & x_{2} & x_{3} \\
x_{2} & x_{3} & x_{1} \\
x_{3} & x_{1} & x_{2}
\end{array}\right|=(\quad \text {. }
$$
(A) -4
(B) -1
(C) 0
(D) 2 | From Vieta's formulas, we have $x_{1}+x_{2}+x_{3}=0$.
By the definition of the determinant, we know
$$
D=3 x_{1} x_{2} x_{3}-\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}\right)=0 .
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,875 |
Example 4 Proof: The cube roots of two different prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic sequence.
(2nd USA Mathematical Olympiad) | Proof by contradiction.
Assume $p, q, r$ are distinct primes, $\sqrt[3]{p}, \sqrt[3]{q}, \sqrt[3]{r}$ are terms in an arithmetic sequence with the first term $a$ and common difference $d$. That is, there exist $l, m, n \in \mathbf{N}$ such that
$$
\begin{array}{l}
\sqrt[3]{p}=a+l d, \sqrt[3]{q}=a+m d, \sqrt[3]{r}=a+n d... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,877 |
Example 3 Given
$$
a^{2}(b+c)=b^{2}(a+c)=2010 \text {, and } a \neq b \text {. }
$$
Then $c^{2}(a+b)=$ $\qquad$ [2]
$(2010$, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】The given condition equation has the same structure as the algebraic expression to be evaluated. According to the known condition equation, it is impossible to determine the values of $a$, $b$, and $c$. We can only conjecture that there is an intrinsic relationship between $a$, $b$, and $c$. By constructing a... | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,878 |
Example 5 Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation
$$
x^{3}-6 x^{2}+a x+a=0
$$
Find all real numbers $a$ such that
$$
\left(x_{1}-1\right)^{3}+\left(x_{2}-2\right)^{3}+\left(x_{3}-3\right)^{3}=0
$$
holds, and for each such $a$, find the corresponding $x_{1}, x_{2}, x_{3}$. | Solve: By Vieta's formulas, we know $x_{1}+x_{2}+x_{3}=6$.
$$
\begin{array}{l}
\text { Hence }\left(x_{1}-1\right)+\left(x_{2}-2\right)+\left(x_{3}-3\right)=0 \\
\quad \Rightarrow 3\left(x_{1}-1\right)\left(x_{2}-2\right)\left(x_{3}-3\right)=0 \\
\quad \Rightarrow\left\{\begin{array} { l }
{ x _ { 1 } = 1 , } \\
{ a =... | \frac{5}{2}, \frac{16}{3}, \frac{27}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,879 |
Example 6 If the function $f(x)$ satisfies
$$
\begin{array}{l}
f(x+y)=f(x)+f(y)+xy(x+y), \\
f^{\prime}(0)=1,
\end{array}
$$
find the analytical expression of the function $f(x)$.
(2000, Shanghai Jiao Tong University Admissions Exam for Recommended Students) | Notice,
$$
\begin{array}{l}
x y(x+y)=(-x)(-y)(x+y), \\
-x-y+(x+y)=0 \\
\text { Hence }(-x)^{3}+(-y)^{3}+(x+y)^{3}=3 x y(x+y) . \\
\text { By } f(x+y)=f(x)+f(y)+x y(x+y) \\
\Rightarrow f(x+y) \\
\quad=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] \\
\Rightarrow f(x+y)-\frac{1}{3}(x+y)^{3} \\
\quad=f(x)-\frac{1... | f(x)=\frac{1}{3} x^{3}+x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,880 |
Example 7 Let $a, b, c \geqslant 1$. Prove:
$$
\begin{array}{l}
a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}+3 a b c \\
\geqslant a^{3}+b^{3}+c^{3}+3 a^{2} b^{2} c^{2}
\end{array}
$$ | Prove that in fact,
$$
\begin{array}{l}
a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-3 a^{2} b^{2} c^{2} \\
=\frac{1}{2}(a b+b c+c a)\left[(a-c)^{2} b^{2}+(b-a)^{2} c^{2}+(c-b)^{2} a^{2}\right] \\
\geqslant \frac{1}{2}(a+b+c)\left[(a-c)^{2}+(b-a)^{2}+(c-b)^{2}\right] \\
=a^{3}+b^{3}+c^{3}-3 a b c .
\end{array}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,881 |
2. Let $x_{1}, x_{2}, x_{3} \in \mathbf{R}_{+}$. Prove:
$$
\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\frac{x_{1}}{x_{3}} \leqslant\left(\frac{x_{1}}{x_{2}}\right)^{3}+\left(\frac{x_{2}}{x_{3}}\right)^{3}+\left(\frac{x_{3}}{x_{1}}\right)^{3} .
$$ | Notice,
$$
\begin{array}{l}
\frac{x_{2}}{x_{1}}=\frac{x_{2}}{x_{3}} \cdot \frac{x_{3}}{x_{1}} \cdot 1 \leqslant \frac{1}{3}\left(\frac{x_{2}}{x_{3}}\right)^{3}+\frac{1}{3}\left(\frac{x_{3}}{x_{1}}\right)^{3}+\frac{1}{3}, \\
\frac{x_{3}}{x_{2}}=\frac{x_{1}}{x_{2}} \cdot \frac{x_{3}}{x_{1}} \cdot 1 \leqslant \frac{1}{3}\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,882 |
4. Find all positive integer triples $(x, y, z)$, such that:
$$
x^{3}+y^{3}+z^{3}-3 x y z=2012 \text{. }
$$ | Prompt: From the above equation, we have
$$
\begin{array}{l}
(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \\
=4024 .
\end{array}
$$
Also, $4024=2^{3} \times 503$, and
$$
\begin{array}{l}
(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \equiv 0(\bmod 2), \\
\text { then }\left\{\begin{array}{l}
x+y+z=k, \\
(x-y)^{2}+(y-z)^{2}+(z-x)... | (169,167,167) \text{ and its permutations, } (671,671,670) \text{ and its permutations} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,883 |
Question 1 Find a triplet of integers $(l, m, n)$ $(1<l<m<n)$, such that $\sum_{k=1}^{l} k 、 \sum_{k=l+1}^{m} k 、 \sum_{k=m+1}^{n} k$ form a geometric sequence. ${ }^{[1]}$
(The 9th China Southeast Mathematical Olympiad) It has been found through exploration that this problem can be generalized to:
Question 2 Does ther... | The conclusion is affirmative.
In fact, for any positive integer $a$, let
$$
t_{n}=\frac{1}{2}\left[(2 a+1)^{n}-1\right] \text {. }
$$
At this point, $\left\{t_{n}\right\}$ is clearly a strictly increasing sequence of integers.
$$
\begin{aligned}
\text { Also, } & M_{n}=\sum_{k=t_{n-1}+1}^{t_{n}} k \\
= & \frac{1}{2}\... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,884 |
2. Acute $\triangle A B C$ is inscribed in $\odot O$ with radius $R$, $H$ is the orthocenter of $\triangle A B C$, and the extension of $A O$ intersects $B C$ at point $M$. If $O H \perp A O, B C=10, O A=6$, then $O M=$ $\qquad$ - | 2. $\frac{11}{3}$.
The auxiliary line is shown in Figure 3.
Then $\triangle A O H \backsim \triangle O N M$, and $A H=C F=2 O N$.
Thus $\frac{O M}{O N}=\frac{A H}{A O}=\frac{C F}{A O}=\frac{2 O N}{A O}$
$\Rightarrow O M=\frac{2 O N^{2}}{A O}=\frac{2\left(O B^{2}-B N^{2}\right)}{A O}=\frac{11}{3}$. | \frac{11}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,886 |
3. The graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $A$ and $B$, with the vertex at $C$. If $\triangle A C B$ is a right triangle, then the value of the discriminant is $\qquad$. | 3. 4 .
As shown in Figure 4.
From the problem, we know $\Delta=b^{2}-4 a c>0$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$.
Then $\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,887 |
4. As shown in Figure 1, the radius of the semicircle $\odot 0$ is $1, A C \perp A B$ at point $A, B D \perp A B$ at point $B$, and $A C=2, B D=3, P$. is any point on the semicircle. Then the maximum area of the closed figure $A B D P C$ is $\qquad$ | 4. $\frac{5+\sqrt{5}}{2}$.
As shown in Figure 5, construct $C E \perp B D$ at point $E$. Then
$\triangle C E D \cong \triangle C A O$.
Thus, $\angle O C D=90^{\circ}$.
When the area $S$ of pentagon $A B D P C$ is maximized, to minimize the area of $\triangle C P D$, the height $P H$ of $\triangle C P D$ must be minimi... | \frac{5+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,888 |
Example 4 Let real numbers $x, y, z$ simultaneously satisfy
$$
\left\{\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}\right.
$$
Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-2)^{2}$. ${ }^{[3]}$
(1st Youth Mathematical Week (Zonghu Cup) Mathematical Competition) | Solve: From the given, we have
$$
\left\{\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
(x-2)(y-2)(z-2)
$$
$$
\begin{aligned}
= & -6(x-2)(y-2)(z-2)(x+1)^{2}(y+1)^{... | 4017 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,889 |
5. Two natural numbers $x$ and $y$ sum to 111, such that the equation
$$
\sqrt{x} \cos \frac{\pi y}{2 x}+\sqrt{y} \sin \frac{\pi x}{2 y}=0
$$
holds. Then a pair of natural numbers $(x, y)$ that satisfies the condition is
$\qquad$ | 5. $(37,74)$.
To make the equation hold, we need to consider the values of trigonometric functions at special angles.
Observe $\cos \frac{\pi y}{2 x}$.
When $y=2 x$, $\cos \frac{\pi y}{2 x}=\cos \pi=-1$. Let's experiment with the original equation by setting $y=2 x$.
The left side $=\sqrt{x} \cos \frac{2 x \pi}{2 x}+\... | (37,74) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,890 |
II. (15 points) As shown in Figure 2, in $\triangle ABC$, it is known that $\angle ACB=90^{\circ}, AC=3, BC=4$. With point $B$ as the center, $\triangle ABC$ is rotated clockwise so that point $A$ lands on point $A_{1}$ on the extension of $CB$, and at this time, point $C$ lands on point $C_{1}$. Connect $AA_{1}$ and $... | It is known that, $\angle A B A_{1}=\angle C B C_{1}$.
Therefore, $\triangle A B A_{1} \backsim \triangle C B C_{1}$, and
$\angle O A B=\angle O C B$.
Thus, points $B, C, A, O$ are concyclic.
Connecting $B O$. Then $\angle A O B=90^{\circ}, O$ is the midpoint of $A B$, and points $B, O, C_{1}, A_{1}$ are also concyclic... | \frac{105}{52} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,891 |
Three, (15 points) (1) If integers $a, b, c$ satisfy the equation $a^{2}+b^{2}=2 c^{2}-2$, prove: 144 $| a b c$.
(2) Try to write a set of positive integer solutions for the indeterminate equation $a^{2}+b^{2}=2 c^{2}-2$, and verify 144 $| a b c$ for this solution. | Three, (1) Since $144=3^{2} \times 4^{2}$, we only need to prove that 9|abc and 16|abc.
First, prove: 9|abc.
Note that the square of an integer not divisible by 3 leaves a remainder of 1 when divided by 3, and the square of an integer divisible by 3 is still divisible by 3; thus, an integer that leaves a remainder of 2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,892 |
Four, (15 points) In triangles with side lengths all being positive integers,
which one has more: triangles with a perimeter of 2009 or triangles with a perimeter of 2012? Explain your reasoning. | There are two types of triangles in equal numbers.
Let the positive integers $k \geqslant n \geqslant m$ be the side lengths of a triangle with a perimeter of 2009. Then the positive integers $k+1, n+1, m+1$ are the side lengths of a triangle with a perimeter of 2012. Such triangles exist because
$$
\begin{array}{l}
m+... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,893 |
Five. (15 points) In an acute triangle $\triangle ABC$, $O$ is the circumcenter, and $I$ is the incenter. The lines $AI$, $BI$, and $CI$ intersect the circumcircle of $\triangle ABC$ at points $A_1$, $B_1$, and $C_1$ respectively. Prove:
$$
\frac{S_{\triangle ABC}}{S_{\triangle A_1 B_1 C_1}}=\frac{2 r}{R},
$$
where $R... | Five, since $I$ is the incenter, it is easy to know that $A_{1}$ is the midpoint of arc $\overparen{B C}$, $B_{1}$ is the midpoint of arc $\overparen{A C}$, and $C_{1}$ is the midpoint of arc $\overparen{A B}$. As shown in Figure 6, connect $A B_{1}$, $B_{1} C$, $C A_{1}$, $A_{1} B$, $B C_{1}$, and $C_{1} A$.
Then $A_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,894 |
1. $a_{1}, a_{2}, a_{3}, \cdots$ is an arithmetic sequence, where $a_{1}>0, s_{n}$ represents the sum of the first $n$ terms. If $S_{3}=S_{11}$, in $S_{1}, S_{2}, S_{3}, \cdots$ the largest number is $S_{k}$, then $k=$ $\qquad$ . | -1.7 .
Let the common difference be $d$. Then
$$
\begin{array}{l}
a_{n}=a_{1}+(n-1) d . \\
\text { By } S_{3}=S_{11} \Rightarrow d=-\frac{2}{13} a_{1}<0 . \\
\text { Hence } a_{n}=a_{1}+(n-1)\left(-\frac{2}{13} a_{1}\right) \\
=\frac{a_{1}}{13}(15-2 n),
\end{array}
$$
and the largest positive integer $n$ for which $a_... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,895 |
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