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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Proposition 8 For any positive integer $k$, if positive integers $m, n$ satisfy $mn=k^2+k+1$, then $m, n$ are both $\mathrm{T}$-type numbers. | For Proposition 8, without loss of generality, let $m \geqslant n$. We proceed by induction on $k$.
When $k=1$, $mn=3$, we have $m=3, n=1$, both of which are $\mathrm{T}$-type numbers.
When $k=2$, $mn=7$, we have $m=7, n=1$, both of which are also $\mathrm{T}$-type numbers.
Assume for $k \leq n$.
Therefore, $m \geqsl... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,578 |
Proposition 9 Every prime of the form $3 N+1$, is a $\mathrm{T}$-type number. | For Proposition 9, the proof starts with $p=3N+1$ being a prime number, which implies that $N$ is even.
Let $N=2M$, so $p=6M+1$.
Therefore, $\left(\frac{-3}{p}\right)_{\mathrm{L}}=1$.
Hence, for every positive integer $y$, the congruence
$$
x^{2} \equiv -3 y^{2} \pmod{p}
$$
has a solution $x$, i.e., $p \mid (x^{2} + 3... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,579 |
As shown in Figure 2, in $\triangle ABC$, $AB > AC$, $\odot O$ intersects side $BC$ and the extensions of $AC$ and $AB$ at points $D$, $E$, and $F$ respectively. $M$ is the midpoint of side $BC$, $AH \perp BC$ at point $H$, and $AO$ intersects lines $DE$ and $DF$ at points $K$ and $L$ respectively. Prove that quadrilat... | Proof As shown in Figure 2, connect $B O, B K, D O$. Then $O D \perp B C, O B$ bisects $\angle D B F$.
Connect $C L$ and extend it to intersect side $A B$ at point $N$.
Since $C D, C E$ are both tangents to $\odot O$, we have
$$
\begin{array}{l}
C D=C E \Rightarrow \angle C D K=\angle C E D=\frac{1}{2} \angle A C B . \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,581 |
Example 5 Given that $a$ and $b$ are real numbers, and
$$
a^{2}+a b+b^{2}=3 \text {. }
$$
If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. ${ }^{[3]}$
$(2008$, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | Let $a^{2}-a b+b^{2}=k$.
From $\left\{\begin{array}{l}a^{2}+a b+b^{2}=3 \\ a^{2}-a b+b^{2}=k,\end{array}\right.$, we get $a b=\frac{3-k}{2}$.
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$=3+\frac{3-k}{2}=\frac{9-k}{2}$.
Since $(a+b)^{2} \geqslant 0$, then $\frac{9-k}{2} \geqslant 0$, which means $k \leqslant 9$.... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,582 |
Given $x, y$ are real numbers, and
$$
\left(x+\sqrt{x^{2}+1}+1\right)\left(y+\sqrt{y^{2}+1}+1\right)=2 \text {. }
$$
Prove: (1) $x<0, y<0$;
(2) $x y=1$. | Proof (1) Let $\sqrt{x^{2}+1}+x=a$.
Then $\sqrt{x^{2}+1}-x=\frac{1}{a}$.
(1) - (2) gives $x=\frac{1}{2}\left(a-\frac{1}{a}\right)$.
Let $\sqrt{y^{2}+1}+y=b$. Then $y=\frac{1}{2}\left(b-\frac{1}{b}\right)$.
Notice,
$$
(1+a)(1+b)=2 \text {. }
$$
Also, $\sqrt{x^{2}+1}+x>|x|+x \geqslant 0$, so $a>0$.
Similarly, $b>0$.
Fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,583 |
327 Let real numbers $x_{i} \geqslant 0(i=1,2, \cdots, m)$, $n \geqslant 2, \sum_{i=1}^{m} x_{i}=S$. Prove: $\sum_{i=1}^{m} \sqrt[n]{\frac{x_{i}}{S-x_{i}}} \geqslant 2$, equality holds if and only if two of the $x_{i}$ are equal and non-zero, and the rest are 0. | Prove the lemma first using mathematical induction.
Lemma When $x, y \geqslant 0, n \geqslant 2\left(n \in \mathbf{N}_{+}\right)$,
$$
\left(x^{n}+y^{n}\right)^{2} \leqslant\left(x^{2}+y^{2}\right)^{n} \text {. }
$$
Proof When $n=2$, it is obviously true.
Assume that when $n=k(k \geqslant 2)$, the conclusion holds, i.e... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,584 |
In the acute triangle $\triangle ABC$, $AB \neq AC$, $\cos A = 2 \cos B \cdot \cos C$, $O$ is the circumcenter, and point $P$ lies on the ray $AO$. It is known that $\angle APB = \angle APC$. Prove:
$$
S_{\triangle PBC} = S_{\triangle ABC}.
$$ | Proof As shown in Figure 3, construct the circumcircle $\odot O$ of $\triangle A B C$, and let $A P$ intersect $B C$ at point $D$ and $\odot O$ at point $E$.
If $P B=P C$, then it is easy to see that
$$
\triangle A B P \cong \triangle A C P, A B=A C,
$$
which contradicts the given condition $A B \neq A C$. Therefore, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,585 |
Example 6 If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-x y+y^{2}$ is $\qquad$ [4]
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Solve: Completing the square for $x^{2}-x y+y^{2}$ yields
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also, }|x \pm y| \leqslant|x|+|y| \leqslant 1, \text { then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ are such that one i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,586 |
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{6}$ are six different positive integers, taking values from $1,2, \cdots, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{alig... | Since equation (1) is a cyclic expression about $x_{1}, x_{2}, \cdots, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{aligned}
S \geqslant & \left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
& \left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,587 |
1. Given the function $S=|x-2|+|x-4|$.
(1) Find the minimum value of $S$;
(2) If for any real numbers $x, y$ the inequality
$$
S \geqslant m\left(-y^{2}+2 y\right)
$$
holds, find the maximum value of the real number $m$. | (1) Using the absolute value inequality, the answer is 2.
(2) From the problem, we know that for any real number $y$,
$$
m\left(-y^{2}+2 y\right) \leqslant 2
$$
holds. It is easy to find that the maximum value of $-y^{2}+2 y$ is 1. Therefore, $0 \leqslant m \leqslant 2$, and the maximum value of $m$ is 2. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,588 |
2. The minimum value of the algebraic expression $\sqrt{x^{2}+1}+\sqrt{(12-x)^{2}+9}$ is ( ).
(A) 12
(B) 13
(C) 14
(D) 11 | Hint: Use the combination of numbers and shapes. Answer: B. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,589 |
3. Given $A(3,-1), B(-1,4), C(1,-6)$, find a point $P$ inside $\triangle A B C$ such that $P A^{2}+P B^{2}+P C^{2}$ is minimized. Then the coordinates of point $P$ are $\qquad$. | Let $P(x, y)$. Then
$$
\begin{array}{l}
P A^{2}+P B^{2}+P C^{2} \\
=3 x^{2}+3 y^{2}-6 x+6 y+64 \\
=3(x-1)^{2}+3(y+1)^{2}+58 .
\end{array}
$$
Therefore, $P(1,-1)$. | (1,-1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,590 |
Given the quadratic function
$$
y=a x^{2}+b x+c
$$
the graph intersects the $x$-axis at one point with coordinates $(8,0)$, and the vertex has coordinates $(6,-12)$. Find the analytical expression of the quadratic function. | Solution 1 From the problem, we have
$$
\left\{\begin{array} { l }
{ 8 ^ { 2 } a + 8 b + c = 0 } \\
{ - \frac { b } { 2 a } = 6 , } \\
{ \frac { 4 a c - b ^ { 2 } } { 4 a } = - 1 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=3, \\
b=-36, \\
c=96 .
\end{array}\right.\right.
$$
Therefore, the analytical express... | y=3 x^{2}-36 x+96 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,591 |
Example 1 Given the quadratic equation in $x$
$$
x^{2}+c x+a=0
$$
has two integer roots which are exactly 1 more than the roots of the equation
$$
x^{2}+a x+b=0
$$
Find the value of $a+b+c$.
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | 【Analysis】The problem gives the relationship between the roots of two quadratic equations. Let's assume the factored form of these two equations to find the coefficients of each term.
Solution: Let $\alpha, \beta (\alpha, \beta \in \mathbf{Z}, \alpha < \beta)$ be the two integer roots of equation (1). Then $\alpha-1, ... | -3 \text{ or } 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,592 |
Example 1 In a permutation $a_{1}, a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$, if there exists some $i$ such that $a_{i}=i$, then $i$ is called a "fixed point" of the permutation. If the number of permutations with exactly $k$ fixed points is $p_{n}(k)$, prove:
$$
\sum_{k=0}^{n} k p_{n}(k)=n! \text{. }
$$
(1987, IMO Sho... | Consider the bipartite graph $X+Y$, where,
$$
X=\{1,2, \cdots, n\} \text {. }
$$
and $Y=\left\{\sigma_{1}, \sigma_{2}, \cdots, \sigma_{n!}\right\}$ consists of all $n!$ permutations. An edge $\left\{i, \sigma_{j}\right\}$ exists in the graph if and only if $i \in X$ is a fixed point of the permutation $\sigma_{j} \in ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,593 |
Four. (50 points) A fly and $k$ spiders are placed at some intersections of a $2012 \times 2012$ grid. An operation consists of the following steps: first, the fly moves to an adjacent intersection or stays in place, then each spider moves to an adjacent intersection or stays in place (multiple spiders can occupy the s... | The minimum value of $k$ is 2.
(1) First, prove that a single spider cannot catch the fly.
Establish a Cartesian coordinate system, then the range of the spider and fly's movement is
$$
\{(x, y) \mid 0 \leqslant x, y \leqslant 2012, x, y \in \mathbf{N}\}.
$$
For each point in the above set, there are at least two poi... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 725,594 |
As shown in Figure 2, in $\triangle ABC$, $\angle ACB = 90^{\circ}$. Let the areas of the circumcircle $\odot O$ and the incircle $\odot I$ be $S_{\text{outer}}$ and $S_{\text{inner}}$, respectively. $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ are tangent to the sides $BC$, $CA$, and $AB$ of $\triangle ABC$ at thei... | Prove as shown in Figure 2, connect $O F, O H, O G$.
Let $A B=c, B C=a, C A=b$.
Since $\angle A C B=90^{\circ}$, we have
$a^{2}+b^{2}=c^{2}, O F=O G=O H=\frac{c}{2}$.
It is easy to see that quadrilateral $O D C E$ is a rectangle, and
$O D=E C=\frac{1}{2} A C=\frac{b}{2}$,
$O E=C D=\frac{1}{2} B C=\frac{a}{2}$.
Then $D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,595 |
Let
$$
A=\underbrace{88 \cdots 8}_{3 n \uparrow}, B=\underbrace{22 \cdots 2}_{2 n \uparrow}, C=\underbrace{33 \cdots}_{n \uparrow},
$$
where $n$ is a positive integer. Prove: $9 A+54 B+16 C+27$ is a perfect cube. | Proof from the conditions:
$$
\begin{array}{l}
A=\frac{8}{9}\left(10^{3 n}-1\right), \\
B=\frac{2}{9}\left(10^{2 n}-1\right), \\
C=\frac{3}{9}\left(10^{n}-1\right) .
\end{array}
$$
Therefore, $9 A+54 B+18 C+27$
$$
\begin{array}{l}
=8\left(10^{3 n}-1\right)+12\left(10^{2 n}-1\right)+6\left(10^{n}-1\right)+27 \\
=\left(... | \left(2 \times 10^{n}+1\right)^{3} | Number Theory | proof | Yes | Yes | cn_contest | false | 725,596 |
In Rt $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the median on the hypotenuse $A B$. Let the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ be $r$, $r_{1}$, and $r_{2}$, respectively, and $r_{1}+r_{2} \geqslant k r$. Find the maximum value of $k$.
In Rt $\triangle A B C$, $\angle C=90^{... | Solve As shown in Figure 3, let $A B=c, A C=b, B C=a$. Then $r=\frac{a+b-c}{2}$.
Since $C D$ is a median, we have
$$
C D=A D=B D=\frac{c}{2} \text {. }
$$
Thus, the perimeter of $\triangle B C D$ is
$$
\begin{array}{l}
a+\frac{c}{2}+\frac{c}{2}=a+c . \\
\text { Hence } r_{1}=\frac{S_{\triangle B C D}}{\frac{1}{2}(a+c)... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,597 |
For $n \in \mathbf{N}$, and $n \geqslant 2$, let $n$ distinct sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfy $\left|A_{i}\right|=n(i=1,2$, $\cdots, n)$, and for any $k(2 \leqslant k \leqslant n-1)$ sets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k}}$ we have
$$
\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}}\right| \ge... | Given that there is at least one element in $A_{2}$ that is not in $A_{1}$, we have
$$
\begin{array}{l}
S=\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right| \\
\geqslant\left|A_{1} \cup A_{2}\right| \geqslant n+1 .
\end{array}
$$
Also, for the set $T=\{1,2, \cdots, n+1\}$, its $n$ subsets $A_{1}, A_{2}, \cdots, A_{n... | S_{\text{min}} = n+1, S_{\max} = 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,598 |
Example 2 For each $r$-element subset of $\{1,2, \cdots, n\}$, take the smallest number. Prove: The average of all these smallest numbers is $\frac{n+1}{r+1}$.
$(1981$, IMO ) | Proof Consider the bipartite graph $X+Y$, where $X$ consists of all $(r+1)$-element subsets of $\{0,1, \cdots, n\}$, and $Y$ consists of all $r$-element subsets of $\{1,2, \cdots, n\}$. If an $r$-element subset $y \in Y$ is obtained by removing the smallest element from an $(r+1)$-element subset $x \in X$, then $\{x, y... | \frac{n+1}{r+1} | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,599 |
Example 3 In a travel group of $n$ people, among any three people, there are two who do not know each other. If they arbitrarily board two buses, then there must be one bus on which two people know each other. Prove: there exists a traveler who knows at most $\frac{2}{5} n$ people. $(2004$, Bulgarian Mathematical Olymp... | Prove that considering each of these $n$ people as a vertex, and if two people know each other, connect the corresponding vertices with an edge, we obtain a graph $G$.
Since in every group of three people, there are two who do not know each other, the graph $G$ does not contain a cycle of length 3.
Furthermore, if th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,600 |
Example 4 Let $S$ be a set of 2004 points in the plane, no three of which are collinear. Each pair of points in $S$ determines a line, and the set of all these lines is denoted by $L$. Prove that the points in $S$ can be colored with at most two colors such that for any two points $p, q$ in $S$, the number of lines in ... | Proof of a lemma first.
Lemma If $k \geqslant 3$, and any three points among the points $A_{1}, A_{2}, \cdots, A_{k}, B$ in the plane are not collinear, then among the lines $B A_{1}, B A_{2}, \cdots, B A_{k}$, there are exactly an even number of lines that cross
$$
C=\left\{A_{1} A_{2}, A_{2} A_{3}, \cdots, A_{k} A_{1... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,601 |
Example 5 Sets $S_{1}, S_{2}, \cdots, S_{n}$ are pairwise distinct and satisfy the following conditions:
(1) $\left|S_{i} \cup S_{j}\right| \leqslant 2004(1 \leqslant i, j \leqslant n, i, j \in \mathbf{N}_{+})$;
(2) $S_{i} \cup S_{j} \cup S_{k}=\{1,2, \cdots, 2008\}(1 \leqslant i < j < k \leqslant n, i, j, k \in \mathb... | Let the complement of $S_{i}$ in $Y=\{1,2, \cdots, 2008\}$ be $A_{i}(1 \leqslant i \leqslant n)$. Then
$\left|A_{i} \cap A_{j}\right| \geqslant 4, A_{i} \cap A_{j} \cap A_{k}=\varnothing$.
Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$.
Consider the bipartite graph $X+Y$, where a vertex $A_{i} \in X$ is adjacent to... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,602 |
Example 6 There are 10 sets of test papers, each set containing 4 questions, and at most one question is the same between any two sets. Among these test papers, what is the minimum number of different questions?
(2005, Taiwan Mathematical Olympiad) | Consider each test paper as a vertex, these vertices form a set $X$; consider each question as a vertex, these vertices form a set $Y$.
If a test paper $x \in X$ contains a question $y \in Y$, connect a line between the corresponding vertices $x$ and $y$. This results in a bipartite graph $X+Y$.
Let $Y$ have $n$ vert... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,603 |
Example 7 Given any 2012 points in the plane. Prove: the distances between each pair of them take at least 32 different values. | Prove that with these 2012 points as vertices, connecting edges between points of the same distance, we obtain a graph $G$.
Since there is at most one point that is equidistant from three given points, the graph $G$ does not contain $K_{2,3}$.
By Theorem 2, the number of edges in graph $G$ is
$$
m \leqslant \frac{2012... | 32 | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,604 |
Example 8 Let $X$ be a set with 56 elements. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any seven of these subsets has at least $n$ elements, then there must exist three of these 15 subsets whose intersection is non-empty. ${ }^{[2]}$
(2006, China Mathematical Olympiad) | First, we prove by contradiction that \( n = 41 \) satisfies the requirement.
Assume there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 41 elements, but the intersection of any three subsets is empty.
At this point, let \( Y = \{A_{1}, A_{2}, \cdots, A_{15}... | 41 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,605 |
Example 9 Fill each cell of an $m \times n$ grid with 0 or 1, such that any two rows have at least $\frac{n}{2}$ cells in which the numbers are different. Prove: $m \leqslant 2 n$.
(2003, Iran National Training Team Test) | Let $X=\{1,2, \cdots, m\}$,
$Y=\{1,2, \cdots, n\}$.
Consider the complete bipartite graph $X+Y$.
When 1 is filled at the $x$-th row and $y$-th column, color the edge $xy$ red; when 0 is filled, color it blue.
According to the problem, for any two vertices $x_{1}$ and $x_{2}$ in $X$, there are at least $\frac{n}{2}$ ve... | m \leqslant 2 n | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,606 |
Example 2 Given that when the value of $x$ is $2$, $m_{1}$, and $m_{2}$, the polynomial $a x^{2}+b x+c$ has values of $0$, $p_{1}$, and $p_{2}$ respectively. If $a>b>c$, and $p_{1} p_{2}-c p_{1}+a p_{2}-a c=0$, can it be guaranteed that: when the value of $x$ is $m_{1}+5$ and $m_{2}+5$, at least one of the values of th... | 【Analysis】When $x=2$, the value of the polynomial $a x^{2}+b x+c$ is 0, indicating that 2 is a zero of the polynomial. By Vieta's formulas, it is not difficult to find that the other zero is $\frac{c}{2 a}$. Therefore, the polynomial can be set in the form of its zeros.
Solution From the given information, we have
$$
a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,608 |
3. Given $n+1$ points $X_{0}, X_{1}, \cdots, X_{n}$ $(n \geqslant 2)$ in the plane. Prove: In $\triangle X_{0} X_{i} X_{j}(1 \leqslant i<j \leqslant n)$, the number of triangles with area 1 is less than or equal to $\frac{9}{8} n \sqrt{n}$. | Construct a graph $G$ with vertices $X_{1}, X_{2}, \cdots, X_{n}$, where $X_{i} X_{j}$ is an edge of $G$ if and only if the area of $\triangle X_{0} X_{i} X_{j}$ is 1. It is easy to verify that graph $G$ does not contain $K_{2,5}$.
Therefore, using Theorem 2, the number of edges in graph $G$ does not exceed
$$
\frac{n}... | \frac{9}{8} n \sqrt{n} | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,609 |
Example 1 Given $A=\left\{x \left\lvert\, \frac{2 x+1}{x-3} \geqslant 1\right.\right\}$, $B=\left\{y \mid y=b \arctan t,-1 \leqslant t \leqslant \frac{\sqrt{3}}{3}, b \leqslant 0\right\}$,
and $A \cap B=\varnothing$.
Find the range of real number $b$.
(2010, Nanjing University Characteristic Test) | Notice that,
$$
\begin{array}{l}
\frac{2 x+1}{x-3} \geqslant 1 \Leftrightarrow \frac{x+4}{x-3} \geqslant 0 \\
\Leftrightarrow x>3 \text { or } x \leqslant-4 .
\end{array}
$$
Therefore, $A=(-\infty,-4] \cup(3,+\infty)$.
From $-1 \leqslant t \leqslant \frac{\sqrt{3}}{3} \Leftrightarrow-\frac{\pi}{4} \leqslant \arctan t ... | b=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,610 |
Example 2 Given the sets
$$
\begin{array}{l}
M=\{(x, y) \mid x(x-1) \leqslant y(1-y)\}, \\
N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant k\right\} .
\end{array}
$$
If $M \subset N$, then the minimum value of $k$ is $\qquad$ .
(2007, Shanghai Jiao Tong University Independent Admission Examination) | Notice that,
$$
M=\left\{(x, y) \left\lvert\,\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}\right.\right\}
$$
represents a disk with center $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{2}}{2}$.
By $M \subset N \Rightarrow \sqrt{k} \geqslant \sqrt{2} \times \frac{... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,611 |
Example 3 Given the set of integers
$M=\left\{m \mid x^{2}+m x-36=0\right.$ has integer solutions $\}$, set $A$ satisfies the conditions:
(1) $\varnothing \subset A \subseteq M$;
(2) If $a \in A$, then $-a \in A$.
The number of all such sets $A$ is ( ).
(A) 15
(B) 16
(C) 31
(D) $32^{[1]}$
(2010, National High School Ma... | Let $\alpha, \beta$ be the integer roots of the equation
$$
x^{2}+m x-36=0
$$
Assume $|\alpha| \geqslant|\beta|$. Then
$$
\begin{array}{l}
\alpha \beta=-36 \\
\Rightarrow(|\alpha|,|\beta|) \\
=(1,36)(2,18),(3,12),(4,9),(6,6) \\
\Rightarrow m= \pm 35, \pm 16, \pm 9, \pm 5,0 \\
\Rightarrow M=\{0\} \cup\{-5,5\} \cup\{-9,... | 31 | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,612 |
Example 4 Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x<0$, $f(x)$ is monotonically increasing, $f(-1)=0$. Let
$$
\begin{array}{l}
\varphi(x)=\sin ^{2} x+m \cos x-2 m, \\
M=\left\{m \mid \text { for any } x \in\left[0, \frac{\pi}{2}\right], \varphi(x)<0\right\}, \\
N=\left\{m \mid \text { for any }... | Notice that,
$$
\begin{array}{l}
\varphi(x)=-\cos ^{2} x+m \cos x+(1-2 m) \\
=-t^{2}+m t+(1-2 m)=g(t),
\end{array}
$$
where $t=\cos x \in[0,1]$.
Then for any $x \in\left[0, \frac{\pi}{2}\right]$, we have
$\varphi(x)\max _{t \in[0,1]} \frac{1-t^{2}}{2-t}$.
And $\frac{1-t^{2}}{2-t}=4-\left(2-t+\frac{3}{2-t}\right) \leqs... | (4-2 \sqrt{2},+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,613 |
Example 5 Let $X$ be a set containing $n(n>2)$ elements, and $A, B$ be two disjoint subsets of $X$, containing $m, k (m, k \geqslant 1, m+k \leqslant n)$ elements respectively. Then the number of subsets of $X$ that do not contain $A$ or $B$ is ( ).
(A) $2^{n-m}+2^{n-k}-2^{n-m-k}$
(B) $2^{n-m-k}$
(C) $2^{n}-2^{n-m}-2^{... | Let the set of subsets of $X$ containing $A$ be $P(A)$, and the set of subsets of $X$ containing $B$ be $P(B)$. Then
$$
\begin{array}{l}
|P(A) \cup P(B)|=2^{n}-|P(A) \cup P(B)| \\
=2^{n}-(|P(A)|+|P(B)|-|P(A) \cap P(B)|) \\
=2^{n}-2^{n-m}-2^{n-k}+2^{n-k-m} .
\end{array}
$$ | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,614 |
Example 6 Let $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can always be found two elements $a, b$ such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$. ${ }^{[2]}$
(2011, China Western Mathematical Olympiad) | Solve for when
$$
M=\left\{2^{k}, 3 \times 2^{l} \mid k=0,1, \cdots, 10 ; l=0,1, \cdots, 9\right\}
$$
the condition is satisfied, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011$, a contradiction.
In summary, $|M|_{\max }=21$. | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,615 |
Example 7 Define a closed set: If $a, b \in S$, then $a+b \in S, a-b \in S$.
(1) Give an example of an infinite closed set that is a proper subset of $\mathbf{R}$;
(2) Prove: For any two closed sets $S_{1}, S_{2} \subset \mathbf{R}$, there exists $c \in \mathbf{R}$, but $c \notin S_{1} \cup S_{2}$.
(2003, Fudan Univers... | (1) The set of integers $\mathbf{Z}$.
(2) Proof by contradiction.
Assume there exist two closed sets $S_{1}, S_{2} \subset \mathbf{R}$, such that
$$
S_{1} \cup S_{2}=\mathbf{R} \text {. }
$$
Clearly, $S_{1} \nsubseteq S_{2}, S_{2} \nsubseteq S_{1}$.
Otherwise, suppose $S_{1} \subseteq S_{2}$. Then
$$
S_{1} \cup S_{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,616 |
Example 8 Question: (1) Can the set $\{1,2, \cdots, 96\}$ be represented as the union of its 32 three-element subsets, with the sum of the elements in each three-element subset being equal?
(2) Can the set $\{1,2, \cdots, 99\}$ be represented as the union of its 33 two-element subsets, with the sum of the elements in e... | (1) No.
Because $32 \times(1+2+\cdots+96)=48 \times 97$.
(2) Yes.
The sum of the elements of each binary subset is
$$
\frac{1+2+\cdots+99}{33}=150 \text {. }
$$
Divide $1,2, \cdots, 66$ into 33 groups, with the sum of each pair forming an arithmetic sequence with a common difference of 1:
$$
\begin{array}{l}
1+50,3+4... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,617 |
1. Given non-empty sets
$$
\begin{array}{l}
X=\{x \mid a+1 \leqslant x \leqslant 3 a-5\}, \\
Y=\{x \mid 1 \leqslant x \leqslant 16\} .
\end{array}
$$
Then the set of all values of $a$ for which $X \subseteq(X \cap Y)$ holds is $\qquad$
(2006, Fudan University Independent Admission Examination) | Answer: $a \in[0,7]$. | a \in [0,7] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,618 |
Example 3 The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$
(2003, National Junior High School Mathematics League) | 【Analysis】This is a comprehensive problem of algebra and geometry. The double perpendicularity in the problem is the key to solving it.
Solution Let $x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)$ be the two zeros of the parabola. Then
$$
\begin{array}{l}
y=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) \\
\Rightarro... | a c=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,619 |
2. Set $A$ satisfies: if $a \in A$, then $\frac{1}{1-a} \in A$. If $2 \in A$, then the set $A$ with the minimum number of elements that satisfies the condition is
(2009, Shanghai Jiao Tong University Independent Recruitment Exam) | Answer: $\left\{2,-1, \frac{1}{2}\right\}$. | \left\{2,-1, \frac{1}{2}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,620 |
3. Let the set
$$
A=\left\{y \left\lvert\, y=\sin \frac{k \pi}{4}(k=0, \pm 1, \pm 2, \cdots)\right.\right\} \text {. }
$$
Then the number of proper subsets of set $A$ is $\qquad$ (2009, Tongji University Independent Admission Examination) | Hint: Calculate $A=\left\{0, \pm 1, \pm \frac{\sqrt{2}}{2}\right\}$. The number of its proper subsets is $2^{5}-1=31$. | 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,621 |
4. Given the set $M=\left\{(x, y) \left\lvert\, y \geqslant \frac{1}{4} x^{2}\right.\right\}$,
$$
\begin{array}{l}
N=\left\{(x, y) \left\lvert\, y \leqslant-\frac{1}{4} x^{2}+x+7\right.\right\}, \\
D_{r}\left(x_{0}, y_{0}\right)=\left\{(x, y)\left|\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2} \leqslant r^{2}\right|... | Establish a Cartesian coordinate system, then $M \cap N$ represents a figure symmetric about the point $(1,4)$. Furthermore, $D_{r}\left(x_{0}, y_{0}\right)$ is the largest inscribed circle of the figure $M \cap N$. Therefore,
$$
r_{\max }=\sqrt{\frac{25-5 \sqrt{5}}{2}} .
$$ | r_{\max }=\sqrt{\frac{25-5 \sqrt{5}}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,622 |
Question As shown in Figure 1, $A B$ and $C D$ are two chords of different lengths in $\odot O$, and $A B$ intersects $C D$ at point $E$. $\odot I$ is internally tangent to $\odot O$ at point $F$, and is tangent to chords $A B$ and $C D$ at points $G$ and $H$, respectively. A line $l$ through point $O$ intersects $A B$... | $$
\begin{array}{l}
\text { Given } E P=E Q, M K / / A B \\
\Rightarrow \angle E P Q=\frac{1}{2} \angle H E G \\
=\angle I E G=\angle O M K \\
\Rightarrow I E / / P Q, \triangle O M K \sim \triangle I E G \\
\Rightarrow \frac{O K}{I G}=\frac{O M}{I E}=\frac{O F}{I F}=\frac{O F}{I G} \\
\Rightarrow O K=O F \\
\Rightarr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,623 |
The indeterminate equation given in [1] is
$$
x^{3}+y^{3}+z^{3}=x+y+z=3
$$
has only 4 sets of integer solutions:
$$
(1,1,1),(-5,4,4),(4,-5,4),(4,4,-5)
$$
The proof is provided. This paper further generalizes equation (1) to the form:
$$
\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{3}\left(x^{3}+y^{3}+z^{3}\right)
$$
wh... | Assume $x, y, z$ are any set of solutions to equation (2). Let
$$
x+y+z=3 w \text{. }
$$
Then equation (2) becomes
$$
x^{3}+y^{3}+z^{3}=3 w^{3} \text{. }
$$
Clearly, $w$ is an integer. Without loss of generality, let
$$
x=w-p, y=w-q, z=w-r \text{. }
$$
Substituting into equation (3) gives $p+q+r=0$.
Substituting int... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,624 |
2. Let 1000 students form a circle. Prove: there exists a positive integer $k(100 \leqslant k \leqslant 300)$, such that in this circle there are adjacent $2 k$ students, satisfying that the number of girls in the first $k$ students is the same as in the last $k$ students. | 2. Number the 1000 students sequentially as 1, 2, $\cdots, 1000$, and denote the $i$-th student as $a_{i}$. If the $i$-th student is a girl, let $a_{i}=1$; otherwise, let $a_{i}=0$, where
$a_{i+1000}=a_{i-1000}=a_{i}\left(i \in \mathbf{N}_{+}\right)$.
Let $S_{k}(i)=a_{i}+a_{i+1}+\cdots+a_{i+k-1}$.
We need to prove: the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,625 |
4. Find the largest positive integer $k$ such that the set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$, so that for all integers $n(n \geqslant 15)$ and all $i \in\{1,2, \cdots, k\}$, there exist two distinct elements in $A_{i}$ whose sum is $n$. | 4. The largest positive integer $k$ is 3.
When $k=3$, let
$A_{1}=\{1,2,3\} \cup\{3 m \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{3}=\{7,8,9\} \cup\{3 m-21 m \in \mathbf{Z}$, and $m \geqslant 4\}$.
Then the sum of two different elements ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,626 |
5. Let $m$ be a positive integer. Consider an $m \times m$ chessboard. At the center of some unit squares, there is an ant each. At time 0, each ant moves in a direction parallel to one of the edges of the chessboard at a speed of 1. When two ants meet while moving in opposite directions, they both turn $90^{\circ}$ cl... | 5. The latest moment for the last ant to fall off the chessboard is $\frac{3 m}{2}-1$.
When $m=1$, the conclusion is obviously true.
Thus, assume $m>1$.
A collision occurs when exactly two ants meet moving in opposite directions.
Initially, if one ant is placed in the unit square at the southwest corner facing east, a... | \frac{3 m}{2}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,627 |
6. Let $n$ be a positive integer,
$$
W=\cdots x_{-1} x_{0} x_{1} x_{2} \cdots
$$
is an infinite periodic sequence of letters $a$ and $b$, and the smallest positive period of $W$ is $N>2^{n}$. In $W$, if there exist indices $k \leqslant l$ such that $U=x_{k} x_{k+1} \cdots x_{l}$, then the finite non-empty sequence is ... | 6. Consider only non-empty letter sequences.
For any letter sequence $R$ of length $m$, the number of $i \in \{1,2, \cdots, N\}$ such that $R$ is the same as the subsequence $x_{i+1} x_{i+2} \cdots x_{i+m}$ of $W$ is called the "large number" of $R$, denoted by $\mu(R)$. Then $R$ appears in $W$ if and only if $\mu(R)>... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,628 |
7. On a square table composed of unit squares of size $2011 \times 2011$, a finite number of napkins are placed, each covering a $52 \times 52$ square. In each unit square, write the number of napkins covering it, and let the maximum number of unit squares with the same number be $k$. For all possible configurations of... | 7. The maximum value of $k$ is
$2011^{2}-\left[\left(52^{2}-35^{2}\right) \times 39-17^{2}\right]$
$=4044121-57392=3986729$.
Let $m=39$. Then $2011=52 m-17$.
Below is an example where there are 3986729 unit squares with the same number written in them.
Let the column numbers from left to right, and the row numbers from... | 3986729 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,629 |
Example 4 Let the edge lengths of a rectangular parallelepiped be $x, y, z (x < y < z)$. It is known that the sum of the lengths of all the edges is $p$, the sum of the areas of all the faces is $S$, and the length of the space diagonal is $d$. Then the edge lengths are given by the formulas: $x=\frac{1}{3}\left(\frac{... | 【Analysis】Observing the proof, it is not difficult to find that
$$
\frac{1}{3}\left(\frac{1}{4} p-\sqrt{d^{2}-\frac{1}{2}} S\right), \frac{1}{3}\left(\frac{1}{4} p+\sqrt{d^{2}-\frac{1}{2} S}\right)
$$
are the two zeros of a quadratic function. Therefore, the solution to the inequality should start from constructing th... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,630 |
1. (50 points) Given that $\odot O$ is the circumcircle of acute $\triangle ABC$, $\odot O_{1}$ is internally tangent to $\odot O$ at point $A$, and is tangent to side $BC$ at point $D$. Let the incenter of $\triangle ABC$ be $I$, and the circumcircle of $\triangle IBC$ be $\odot O_{2}$, which intersects $\odot O_{1}$ ... | 1. From the well-known properties, we know that $O_{2}$ is the midpoint of the arc $\overparen{B C}$ of $\odot O$, and points $A$, $I$, and $O_{2}$ are collinear.
As shown in Figure 1, let the second intersection point of $A D$ with $\odot O$ be $O_{2}^{\prime}$.
Since point $A$ is the center of homothety of $\odot O_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,631 |
2. (50 points) Given $a, b, c > 0$. Prove:
$$
\begin{array}{l}
\left(a^{3}+\frac{1}{b^{3}}-1\right)\left(b^{3}+\frac{1}{c^{3}}-1\right)\left(c^{3}+\frac{1}{a^{3}}-1\right) \\
\leqslant\left(a b c+\frac{1}{a b c}-1\right)^{3} .
\end{array}
$$ | 2. Since the sum of any two parentheses on the left side of the original inequality is positive, at most one of the three parentheses is non-positive.
Therefore, we may assume that all three parentheses are positive.
Let $k=abc, a^{3}=\frac{k x}{y}, b^{3}=\frac{k y}{z}, c^{3}=\frac{k z}{x}$.
Then the left side of the o... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,632 |
3. (50 points) Find the smallest prime $p$ that satisfies $(p, N)=1$, where $N$ is the number of all $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ that meet the following conditions:
(1) $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ is a permutation of $0,1, \cdots, 2012$;
(2) For any positive divisor $m$ of 2013 and a... | 3. From (2), we know that for any $i, j (0 \leqslant i < j \leqslant 2012)$ and $m=3^{\alpha} \times 11^{\beta} \times 61^{\gamma} (\alpha, \beta, \gamma \in \{0,1\})$, we have
$$
a_{i} \equiv a_{j}(\bmod m) \Leftrightarrow i \equiv j(\bmod m).
$$
Let
$$
\begin{array}{l}
A_{i}=\{x \in \mathrm{N} \mid 0 \leqslant x \le... | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,633 |
4. (50 points) Given a subset $X$ of $\{1,2, \cdots, n\}$ that satisfies: for any $a, b \in X$, if $\frac{a+b}{2} \in \mathbf{Z}$, then $\frac{a+b}{2} \in X$. Such a subset $X$ is called a "good subset". Let $A(n)$ be the number of good subsets of $\{1,2, \cdots, n\}$ (for example, $A(3)=7$, among the eight subsets of ... | 4. On the one hand, the good subsets of $\{1,2, \cdots, n+2\}$ can be divided into two categories: those that contain 1 and those that do not.
Since the good subsets of $\{1,2, \cdots, n+2\}$ that do not contain 1, when each element is reduced by 1, become the good subsets of $\{1,2, \cdots, n+1\}$, and the good subse... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,634 |
5. (50 points) Given two circles $\odot O$ and $\odot O^{\prime}$ with different radii and externally separated, the internal common tangent $l$ of $\odot O$ and $\odot O^{\prime}$ intersects the two external common tangents $l_{1}$ and $l_{2}$ at points $B$ and $C$, respectively. The circle $\odot O_{1}$, which is ext... | 5. Let the radii of $\odot O$ and $\odot O^{\prime}$ be $r$ and $r^{\prime}$, respectively, and assume $r < r^{\prime}$. The selection of $l$ is as shown in Figure 2.
Let the lines $l_{1}$ and $l_{2}$ intersect at point $A$, and the radii of $\odot O_{1}$ and $\odot O_{2}$ be $R_{1}$ and $R_{2}$, respectively.
Then $A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,635 |
6. (50 points) Given $a, b, c > 1$, and $a + b + c = 9$.
Prove:
$$
\sqrt{a b + b c + c a} \leqslant \sqrt{a} + \sqrt{b} + \sqrt{c} \text{. }
$$ | 6. Let $a=\frac{9 x^{2}}{x^{2}+y^{2}+z^{2}}, b=\frac{9 y^{2}}{x^{2}+y^{2}+z^{2}}$,
$$
c=\frac{9 z^{2}}{x^{2}+y^{2}+z^{2}}, x+y+z=1 \text {. }
$$
Then the original inequality becomes
$$
x^{2}+y^{2}+z^{2} \geqslant 9\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right) \text {. }
$$
Since $a \geqslant 1$, we have
$$
\begin{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,636 |
7. (50 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$
Find the units digit of $x_{2012}$. | 7. Clearly, $x_{2}=7$, and for any positive integer $n, x_{n}$ is a positive integer.
By the property of the floor function, we have
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}>x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \\
>4 x_{n}+\sqrt{11} x_{n}-1 .
\end{array}
$$
Multiplying both sides of the above inequality by ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,637 |
8. (50 points) For any permutation $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of $1,2, \cdots, n\left(n \in \mathbf{N}_{+}, n>2012\right)$, define:
$$
L=\sum_{i=1}^{n}\left|x_{i}-\sqrt{3} x_{i+1}\right| \text {, }
$$
where, $x_{n+1}=x_{1}$. Try to find $L_{\max } 、 L_{\min }$ and the number of all permutations $\left(... | 8. Clearly, $L$ is the sum of $n$ numbers with positive signs and $n$ numbers with negative signs from $i$ and $\sqrt{3} i$ (where $i=1,2, \cdots, n$).
To maximize $L$, the larger $n$ numbers among the $2n$ numbers should take positive signs, and the smaller $n$ numbers should take negative signs.
Let $\sqrt{3} i$ (w... | L_{\max} = \sqrt{3}\left[\frac{n(n+1)}{2} - a(a+1)\right] + \left[\frac{n(n+1)}{2} - b(b+1)\right], \quad L_{\min} = (\sqrt{3} - 1) \frac{n(n+1)}{2} + 4 - 2 \sqrt{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,638 |
1. Let $n$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers. Prove:
$$
\begin{array}{l}
\frac{1}{1+a_{1}}+\frac{a_{1}}{\left(1+a_{1}\right)\left(1+a_{2}\right)}+\cdots+ \\
\frac{a_{1} a_{2} \cdots a_{n-1}}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)} \leqslan... | 1. Notice that, $\frac{1}{1+a}=1-\frac{a}{1+a}$.
Therefore, for any $i(i=1,2, \cdots, n)$ we have
$$
\begin{array}{l}
\frac{a_{1}}{1+a_{1}} \cdots \cdot \frac{a_{i-1}}{1+a_{i-1}} \cdot \frac{1}{1+a_{i}} \\
=\frac{a_{1}}{1+a_{1}} \cdots \cdot \frac{a_{i-1}}{1+a_{i-1}}\left(1-\frac{a_{i}}{1+a_{i}}\right) \\
=\frac{a_{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,639 |
2. As shown in Figure 1, circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $T$, points $A$ and $E$ are on circle $\Gamma_{1}$, lines $A B$ and $D E$ are tangent to circle $\Gamma_{2}$ at points $B$ and $D$ respectively, and line $A E$ intersects $B D$ at point $P$. Prove:
(1) $\frac{A B}{A T}=\frac{... | 2. (1) As shown in Figure 3, extend $A T$ and $E T$ to intersect the circle $\Gamma_{2}$ at points $H$ and $G$, respectively, and connect $G H$.
It is easy to see that $A E \parallel G H$.
Therefore, $\triangle A T E \sim \triangle H T G$.
Thus, $\frac{A T}{T H} = \frac{E T}{T G}$.
Hence, $\frac{A H}{T H} = \frac{E G}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,640 |
Example 5 Given the real-coefficient equation
$$
x^{3}+a x^{2}+b x+c=0
$$
has three real roots. If $-2 \leqslant a+b+c \leqslant 0$, prove: this cubic equation has at least one root in the interval $[0,2] .{ }^{[4]}$
(34th Russian Mathematical Olympiad) | Let $p(x)=x^{3}+a x^{2}+b x+c, x_{1}, x_{2}, x_{3}$ be the three real roots of $p(x)$. Then
$$
\begin{array}{l}
p(x)=(x-x_{1})(x-x_{2})(x-x_{3}). \\
\text { Also }-2 \leqslant a+b+c \leqslant 0 \\
\Rightarrow-1 \leqslant p(1)=1+a+b+c \leqslant 1 \\
\Rightarrow|p(1)|=\left|(1-x_{1})(1-x_{2})(1-x_{3})\right| \leqslant 1 ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,641 |
3. Find all integer pairs $(a, b)$ such that there exists an integer $d > 1$ satisfying that for any positive integer $n$, $a^{n} + b^{n} + 1$ is a multiple of $d$.
(Chen Yonggao \quad supplied the problem) | 3. When $a, b$ are one odd and one even, $a^{n}+b^{n}+1$ is always a multiple of 2, satisfying the condition.
When $a, b$ are both odd or both even, by $d \mid\left(a^{n}+b^{n}+1\right)$, we know $d$ is an odd number.
$$
\begin{array}{l}
\text { Also, } a^{2}+b^{2}+1=(a+b)^{2}-2 a b+1, \\
d \mid (a+b+1), d \mid \left(... | (2 k, 2 l+1), (2 k+1, 2 l), (3 k+1, 3 l+1) (k, l \in \mathbf{Z}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,642 |
5. As shown in Figure $2, \triangle A B C$ has an incircle $\odot I$ that touches sides $A B$ and $A C$ at points $D$ and $E$ respectively, and $O$ is the circumcenter of $\triangle B C I$. Prove: $\angle O D B=\angle O E C$. | 5. Proof 1 The auxiliary lines are shown in Figure 4.
From $O$ being the circumcenter of $\triangle BCI$, we know
$\angle BOI = 2 \angle BCI = \angle BCA$.
Similarly, $\angle COI = \angle CBA$.
Thus, $\angle BOC = \angle BOI + \angle COI$
$= \angle BCA + \angle CBA = \pi - \angle BAC$.
Therefore, $A, B, O, C$ are conc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,644 |
6. A country has $n(n \geqslant 3)$ cities and two airlines. There is exactly one two-way flight between every pair of cities, and this two-way flight is operated exclusively by one of the airlines. A female mathematician wants to start from a city, pass through at least two other cities (each city is visited only once... | 6. Solution 1 Consider each city as a vertex, each flight route as an edge, and each airline as a color. Then, the country's flight network can be seen as a complete graph with $n$ vertices whose edges are colored with two colors.
From the condition, we know that any cycle contains edges of both colors, meaning that t... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,645 |
7. Let $a_{1} \leqslant a_{2} \leqslant \cdots$ be an infinite sequence of positive integers, and there exist positive integers $k, r$ such that $\frac{r}{a_{r}}=k+1$. Prove: There exists a positive integer $s$ such that $\frac{s}{a_{s}}=k$. | 7. Proof 1 Let $a_{r}=x$. Then $r=(k+1) x$.
Consider the following expression
$$
\frac{k}{a_{k}}, \frac{2 k}{a_{2 k}}, \cdots, \frac{x k}{a_{x k}} \text {. }
$$
Then there must be a term in expression (1) that equals $k$. Otherwise, assume none of the above terms equals $k$. Then
$$
a_{k} \neq 1, a_{2 k} \neq 2, \cdo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,646 |
8. How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? | 8 . Note Quality.
First, consider whether the combination number $\mathrm{C}_{2012}^{k}$ is a multiple of $p=503$.
If $p \nmid k$, then
$\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1}$
$\Rightarrow p\left|k \mathrm{C}_{4 p}^{k} \Rightarrow p\right| \math... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,647 |
1. At the beginning, there are 111 pieces of clay of equal weight on the table. Perform the following operations on the clay: First, divide a part or all of the clay into several groups, with the same number of pieces in each group, then knead the clay in each group into one piece. It is known that after $m$ operations... | 1. 2 .
Obviously, one operation can result in at most two different weights of clay blocks.
Below, we show that two operations can achieve the goal.
Assume without loss of generality that each block of clay initially weighs 1.
In the first operation, select 74 blocks and divide them into 37 groups, with two blocks in ... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,648 |
3. Cover the entire plane with black and white squares like an international chessboard. Now color all the white squares with red or blue such that originally adjacent white squares (sharing a vertex) are different colors. For any line segment $m$ in the plane, let $\delta(m)$ denote the difference between the sum of t... | 3. First, prove a lemma.
Lemma: The coordinate plane covered by vertical strip regions of equal width is colored with two colors, black and white, such that each strip region is colored with one color and adjacent strip regions are of different colors. Then, for any line segment parallel to $l$, the difference between... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,649 |
4. $S A_{1} A_{2} \cdots A_{n}$ is an $n$-sided pyramid with the convex polygon $A_{1} A_{2} \cdots A_{n}$ as its base. For each $i (i=1,2, \cdots, n)$, $X_{i}$ is a point on the plane of the base such that
$$
\triangle X_{i} A_{i} A_{i+1} \cong \triangle S A_{i} A_{i+1},
$$
and $X_{i}$ is on the same side of the line... | 4. Consider any point $P$ on the base.
First, construct a small sphere inside the pyramid that is tangent to the base at point $P$. Then, while keeping the sphere tangent to the base at point $P$, increase the radius of the sphere until it first touches the side face $S A_{1} A_{2}$ at point $Q$. Then,
$Q A_{1}=P A_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,650 |
5. Let $P(x)$ be a polynomial with real coefficients, and let real numbers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ satisfy $a_{1} a_{2} a_{3} \neq 0$, and for any $x \in \mathbf{R}$, we have
$$
P\left(a_{1} x+b_{1}\right)+P\left(a_{2} x+b_{2}\right)=P\left(a_{3} x+b_{3}\right).
$$
Prove: $P(x)$ has at least one rea... | 5. If $a_{1} \neq a_{3}$, take $x_{0}$ such that
$$
a_{1} x_{0}+b_{1}=a_{3} x_{0}+b_{3} \text {. }
$$
Then $P\left(a_{2} x_{0}+b_{2}\right)=0$, i.e., $P(x)$ has a real root.
Similarly, if $a_{2} \neq a_{3}$, the same result can be obtained.
$$
\text { Let } a_{1}=a_{2}=a_{3}=a \neq 0 \text {. }
$$
Suppose $P(x)$ is a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,651 |
Example 6 Let $f(x)$ represent a quartic polynomial in $x$. If $f(1)=f(2)=f(3)=0, f(4)=6$, $f(5)=72$, then the last digit of $f(2010)$ is $\qquad$. ${ }^{3}$
(2010, International Cities Mathematics Invitational for Youth) | 【Analysis】It is given in the problem that $1,2,3$ are three zeros of the function $f(x)$, so we consider starting from the zero point form of the function.
Solution From the problem, we know
$$
f(1)=f(2)=f(3)=0 \text {. }
$$
Let the one-variable quartic polynomial function be
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,652 |
6. Given $A_{1} 、 B_{1} 、 C_{1}$ are points on the sides $B C$ 、 $C A 、 A B$ of $\triangle A B C$, respectively, satisfying
$$
A B_{1}-A C_{1}=C A_{1}-C B_{1}=B C_{1}-B A_{1} \text {, }
$$
$O_{A} 、 O_{B} 、 O_{C}$ are the circumcenters of $\triangle A B_{1} C_{1} 、 \triangle A_{1} B C_{1} 、 \triangle A_{1} B_{1} C$, res... | 6. Let $I$ denote the incenter of $\triangle ABC$, and $A_{0}, B_{0}, C_{0}$ be the points where the incircle touches sides $BC, CA, AB$ respectively.
Let point $A_{1}$ lie on the segment $A_{0}B$ (other cases are discussed similarly).
Notice,
$$
\begin{array}{l}
C A_{0} + A C_{0} = C B_{0} + A B_{0} = C A, \\
C A_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,653 |
7. Given that $A$ is the set of vertices of a regular $(2n+1)$-gon. Two players, A and B, take turns starting with A, each removing one point from $A$. If after a player's move, any three remaining points in $A$ form the vertices of an obtuse triangle, that player wins. Question: Who has a winning strategy, A or B? | 7. Player B has a winning strategy.
B's strategy is: after each of his moves, if there are still $2k+1 \geqslant 5$ vertices left, then ensure that any semicircle on the circumcircle of $A$ has at least $k$ vertices.
At the beginning of the game, this condition is satisfied.
Assume it is Player A's turn, and the verti... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,654 |
8. Let $S_{n}=1!+2!+\cdots+n!$. Prove: there exists a positive integer $n$, such that $S_{n}$ has a prime factor greater than $10^{2012}$. | 8. For any prime $p$, any positive integer $n$, let $v_{p}(n)$ denote the exponent of the prime $p$ in the prime factorization of $n$.
Notice that, if $v_{p}(n) \neq v_{p}(k)$, then
$$
v_{p}(n \pm k)=\min \left\{v_{p}(n), v_{p}(k)\right\} \text {. }
$$
From this, we get the following lemma.
Lemma If there exists a pos... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,655 |
1. Solve the inequality
$$
(\sqrt{3}-\sqrt{2})^{\left(\log _{2} 3\right)^{4-x^{2}}} \leqslant(\sqrt{3}+\sqrt{2})^{-\left(\log _{3} 2\right)^{2 x-1}} \text {. }
$$ | 1. Notice,
$$
\begin{array}{l}
(\sqrt{3}-\sqrt{2})^{\left(\log _{2} 3\right)^{4-x^{2}}} \leqslant(\sqrt{3}+\sqrt{2})^{-\left(\log _{3} 2\right) 2 x-1} \\
\Leftrightarrow a^{\left(\log _{2} 3\right)^{4-x^{2}}} \leqslant a^{\left(\log _{3} 2\right) 2 x-1},
\end{array}
$$
where, $a=\sqrt{3}-\sqrt{2}=(\sqrt{3}+\sqrt{2})^{... | -1 \leqslant x \leqslant 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,656 |
2. On the base $AC$ of isosceles $\triangle ABC$, take a point $E$, and on the two legs $AB, BC$, take points $D, F$ respectively, such that $DE \parallel BC, EF \parallel AB$. If $BF: EF=2: 3$, what fraction of the area of $\triangle ABC$ is the area of $\triangle DEF$? | 2. As shown in Figure 1, quadrilateral $B D E F$ is a parallelogram, and
$$
\begin{array}{l}
\triangle A D E \\
\sim \triangle A B C, \\
\triangle E F C \\
\sim \triangle A B C,
\end{array}
$$
the similarity ratios are respectively
$$
\begin{array}{l}
\frac{D E}{B C}=\frac{2 x}{2 x+3 x}=\frac{2}{5}, \frac{F C}{B C}=\f... | \frac{6}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,657 |
3. Two investors jointly invest in a business. Then one of them adds 1 million rubles, which increases his share in the business by 0.04. When he adds another 1 million rubles, his share increases by 0.02. Question: How much more money must he add to increase his share by another 0.04? | 3. Suppose two investors have a total investment of $y$ million rubles. Among them, the one who increases the investment initially has $x$ million rubles, and the total increase is $z$ million rubles.
Thus, the amount to be increased is $z-2$.
From the given conditions, we have
$$
\begin{array}{l}
\left\{\begin{array}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,658 |
4. Solve the inequality
$$
\frac{1}{\sqrt{-x-2}}-\frac{1}{\sqrt{x+4}} \leqslant 1+\frac{1}{\sqrt{(x+4)(-x-2)}} .
$$ | 4. Note that,
$$
\begin{array}{l}
\frac{1}{\sqrt{-x-2}}-\frac{1}{\sqrt{x+4}} \leqslant 1+\frac{1}{\sqrt{(x+4)(-x-2)}} \\
\Leftrightarrow\left\{\begin{array}{l}
\sqrt{x+4}-\sqrt{-x-2} \leqslant 1+\sqrt{(x+4)(-x-2)}, \\
-x-2>0, \\
x+4>0 .
\end{array}\right.
\end{array}
$$
Consider the following two cases.
$$
\begin{arra... | -3 \leqslant x \leqslant -3 + 2\sqrt{\sqrt{5} - 2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,659 |
5. It is known that the numbers $54$ and $128$ are two terms of a geometric sequence. Find all natural numbers that can be encountered in this sequence. | 5. Let the natural number encountered in the sequence be $n$. Then, by the conditions of the problem, there exist $b, q \neq 0$ and integers $m, k$, such that
$$
\begin{array}{l}
\left\{\begin{array}{l}
54=2^{1} \times 3^{3}=b, \\
128=2^{7}=b q^{m}, \\
n=b q^{k}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
... | 2^{1} \times 3^{3}, 2^{3} \times 3^{2}, 2^{5} \times 3^{1}, 2^{7} \times 3^{0} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,660 |
6. In the spatial coordinate system, a curve's projections on the planes $x O z$ and $y O z$ satisfy the equations
$$
5 x + \cos z = 0, \quad z = \arctan \sqrt{y - 3} \text{. }
$$
Find a function $y = f(x)$ such that its graph consists of and only of the points that are the projections of the same curve under these co... | 6. For the point $(x, y, z)$ to lie on the given curve, it is necessary and sufficient that
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 5 x + \cos z = 0 , } \\
{ z = \arctan \sqrt { y - 3 } }
\end{array} \left\{\begin{array}{l}
\cos z=-5 x, \\
\sqrt{y-3}=\tan z, \\
-\frac{\pi}{2}<z<\frac{\pi}{2}
\end{array}\right... | y=\frac{1}{25 x^{2}}+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,661 |
7. Given $\left\{\begin{array}{l}25^{x}-13 \times 5^{x}+a<0, \\ 12 \sin ^{4} \pi x-\cos ^{4} \pi x=11\end{array}\right.$ for each value of $a$ there is at least one solution. Find all values of $a$.
For each value of $a$ there is at least one solution. Find all values of $a$. | 7. Notice,
$$
\begin{array}{l}
12 \sin ^{4} \pi x-\cos 4 \pi x=11 \\
\Leftrightarrow 3(1-c)^{2}-\left(2 c^{2}-1\right)-11=0(c=\cos 2 \pi x) \\
\Leftrightarrow c^{2}-6 c-7=0 \\
\Leftrightarrow(c-7)(c+1)=0 \\
\Leftrightarrow \cos 2 \pi x=-1 \\
\Leftrightarrow x=0.5+n(n \in \mathbf{Z}) .
\end{array}
$$
To ensure that for... | a<13 \sqrt{5}-5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,662 |
Example 7 Let $f(x)$ be a polynomial with integer coefficients, $f(0)=$
11. There exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that $f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010$.
Then the maximum value of $n$ is $\qquad$ (6)
$(2010$, Xin Zhi Cup Shanghai High School Mathem... | Let $g(x)=f(x)-2010$.
$$
\begin{array}{l}
\text { By } f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010, \text { we have } \\
g\left(x_{1}\right)=g\left(x_{2}\right)=\cdots=g\left(x_{n}\right)=0,
\end{array}
$$
That is, $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ distinct integer roots of the integer-... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,663 |
8. On the edge $AS$ of the tetrahedron $S-ABC$, mark points $M, N$ such that $AM=MN=NS$. If the areas of $\triangle ABC$, $\triangle MBC$, and $\triangle SBC$ are $1$, $2$, and $\sqrt{37}$, respectively, find the area of $\triangle NBC$. | 8. Let the areas of $\triangle A B C$, $\triangle M B C$, $\triangle N B C$, $\triangle S B C$ be $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, and let $h_{1}$, $h_{2}$, $h_{3}$, $h_{4}$ be the heights from these triangles to the common base $B C$, as shown in Figure 2.
Points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $S^{\prim... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,664 |
9. Write the quadratic trinomial $x^{2}+9 x+47$ on the blackboard. Tanya arbitrarily increases or decreases the coefficient of $x$ by 1, and then Vanya increases or decreases the constant term by a fixed number $m$, and then the operations are repeated. If the polynomial written on the blackboard has an integer root, V... | 9. (1) Yes; (2) No.
(1) Vanya's operations are independent of Tanya's, and in a finite number of steps (either his own or Tanya's), he can ensure that the value of the quadratic trinomial at the point \( x=1 \) is 0.
Initially, this value is \( 1+9+47=57 \). Next, Tanya can change this value by 1 each step, and Vanya ... | (1) Yes; (2) No | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,665 |
10. Given trapezoid $A B C D$ with bases $A D=3, B C=1$, the diagonals intersect at point $O$, two circles intersect base $B C$ at points $K, L$, these two circles are tangent at point $O$, and are tangent to line $A D$ at points $A, D$ respectively. Find $A K^{2}+D L^{2}$. | 10. As shown in Figure 4, draw the common tangent of the two circles through point $O$, intersecting the lower base $AD$ at point $P$. By the properties of tangents, we have
$$
A P=O P=D P \text {. }
$$
This indicates that $\triangle A O D$ is a right triangle and is similar to $\triangle C D B$, with the similarity r... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,666 |
1. Choose several numbers from $1,2, \cdots, 7$, such that the sum of the even numbers equals the sum of the odd numbers. Then the number of ways to choose the numbers that meet the condition is ( ) kinds.
(A) 6
(B) 7
(C) 8
(D) 9 | -1. B.
Notice that, $2+4+6=12$, so the sum of the numbers taken out is no more than 24.
$$
\begin{array}{l}
\text { Also } 12=2+4+6=5+7, \\
10=4+6=3+7, \\
8=2+6=1+7=3+5, \\
6=6=2+4=1+5, \\
4=4=1+3,
\end{array}
$$
Therefore, there are 7 ways to choose the numbers. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,667 |
2. Given that the sum of two different positive integers is 2012. Then the maximum value of their least common multiple $M$ is ( ).
(A) $1006^{2}$
(B) $1005 \times 1006$
(C) 1006
(D) $1005 \times 1007$ | 2. D.
Let two positive integers be $x, y(x>y)$, and
$$
(x, y)=d, x=d x_{0}, y=d y_{0}.
$$
Then $\left(x_{0}, y_{0}\right)=1$.
From the problem, we know $d\left(x_{0}+y_{0}\right)=2012$.
Then $M=d x_{0} y_{0}=\frac{2012 x_{0} y_{0}}{x_{0}+y_{0}}$.
From $\left(x_{0}+y_{0}\right) \times x_{0} y_{0}$, we know $\left(x_{0... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,668 |
3. As shown in Figure 1, the side length of square $ABCD$ is 4 units. A moving point $P$ starts from point $A$ and moves counterclockwise along the boundary of the square, advancing 5 units and then retreating 3 units. It is known that point $P$ moves forward or backward 1 unit per second. Let $x_{n}$ represent the dis... | 3. B.
Notice that, $2012=8 \times 251+4$.
Therefore, point $P$ moves counterclockwise from $A$ for 2012 seconds, covering a total of $2 \times 251+4=506$ units.
And $506=4 \times 4 \times 31+4 \times 2+2$, so point $P$ moves counterclockwise 31 full circles and reaches the midpoint of side $C D$.
Thus, $A P=2 \sqrt{5... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,669 |
4. Given positive integers $m, n$ can be written as
$$
a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}
$$
where $a_{i} (i=0,1,2,3)$ are positive integers from 1 to 7, and
$$
m+n=2012(m>n) .
$$
Then the number of pairs $(m, n)$ that satisfy the condition is $(\quad)$.
(A) 606
(B) 608
(C) 610
(D) 612 | 4. A.
There are $7^{4}=2401$ positive integers of the given form, the largest of which is
$$
7 \times 7^{3}+7 \times 7^{2}+7 \times 7+7=2800,
$$
and the smallest is
$$
1 \times 7^{3}+1 \times 7^{2}+1 \times 7+1=400 .
$$
Since $m+n=2012(m>n)$, we have
$$
1007 \leqslant m \leqslant 2012-400=1612 \text {. }
$$
Therefo... | 606 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,670 |
5. Divide the natural numbers from 1 to 30 into two groups, such that the product of all numbers in the first group $A$ is divisible by the product of all numbers in the second group $B$. Then the minimum value of $\frac{A}{B}$ is ( ).
(A) 1077205
(B) 1077207
(C) 1077209
(D) 1077211 | 5. A.
$$
\begin{array}{l}
\text { Given } A B=30 \times 29 \times \cdots \times 1 . \\
=2^{26} \times 3^{14} \times 5^{7} \times 7^{4} \times 11^{2} \times 13^{2} \times \\
17 \times 19 \times 23 \times 29 .
\end{array}
$$
Let $C=2^{13} \times 3^{7} \times 5^{4} \times 7^{2} \times 11 \times 13 \times 17 \times 19 \ti... | 1077205 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 725,671 |
6. As shown in Figure 2, the linear function $x+2 y-6=0$ intersects with the quadratic function $y=x^{2}$ at points $A$ and $B$. If $C$ is a point on the graph of $y=x^{2}$ such that the area of $\triangle A B C$ is $\frac{35}{8}$, then the coordinates of point $C$ cannot be ( ).
(A) $(-1,1)$
(B) $\left(\frac{1}{2}, \f... | 6. D.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(t, t^{2}\right)$.
From $\left\{\begin{array}{l}x+2 y-6=0, \\ y=x^{2},\end{array}\right.$ we solve to get
$(x, y)=(-2,4),\left(\frac{3}{2}, \frac{9}{4}\right)$.
Thus, $|A B|=\sqrt{\left(-2-\frac{3}{2}\right)^{2}+\left(4-\frac{9}{4}\right)^{2}}=\f... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,672 |
1. As shown in Figure 3, in the rectangular paper piece $A B C D$, $A B=4$, $A D=9$, point $E$ is on side $A B$, and $B E=1$. The paper is folded so that point $D$ coincides with point $E$, resulting in the crease $GH$. Then the area of the overlapping part $E F G H$ of the paper is | $=1 . \frac{295}{24}$.
Let $D H=x$. Then $E H=x, A H=9-x$.
In the right triangle $\triangle A E H$, we have
$$
3^{2}+(9-x)^{2}=x^{2} \text {. }
$$
Solving for $x$ gives $x=5$, so $D H=5$. Thus, $A H=4$.
Then, $\triangle A E H \sim \triangle B F E \sim \triangle C^{\prime} F G$
$$
\begin{array}{l}
\Rightarrow A E:... | \frac{295}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,673 |
Example 8 Given that $P(x)$ is a polynomial with integer coefficients, satisfying $P(17)=10, P(24)=17$. If the equation $P(n)=n+3$ has two distinct integer solutions $n_{1}, n_{2}$, find the value of $n_{1} n_{2}$. ${ }^{[7]}$
(2005, American Invitational Mathematics Examination) | 【Analysis】From the conditions of the problem, we cannot determine the zeros of the integer-coefficient polynomial $P(x)$. Let's construct another polynomial function $T(x)$ such that 17 and 24 are zeros of $T(x)$, and solve the problem using the zero-product property.
Solution Let $S(x)=P(x)-x-3$. Then
$S(17)=-10, S(24... | 418 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,674 |
2. Let positive integers $k_{1} \geqslant k_{2} \geqslant \cdots \geqslant k_{n}\left(n \in \mathbf{N}_{+}\right)$, and $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=2012$.
Then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is $\qquad$ | 2. 49 .
Notice that,
$$
2012=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{4}+2^{3}+2^{2} \text {. }
$$
Also, $2^{k+1}=2^{k}+2^{k}$, and $k+1 \leqslant 2 k(k \geqslant 1$ when $)$, then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is
$$
10+9+8+7+6+4+3+2=49 \text {. }
$$ | 49 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,675 |
3. In $\triangle A B C$, it is known that $A B=12, A C=8, B C$ $=13, \angle A$'s angle bisector intersects the medians $B E$ and $C F$ at points $M$ and $N$, respectively. Let the centroid of $\triangle A B C$ be $G$. Then
$$
\frac{S_{\triangle G M N}}{S_{\triangle A B C}}=
$$
$\qquad$ . | 3. $\frac{1}{168}$.
As shown in Figure 5, connect $E N$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{B M}{M E}=\frac{A B}{A E}=\frac{12}{4} \Rightarrow \frac{B G+G M}{G E-G M}=\frac{3}{1} \\
\Rightarrow \frac{B G+G M}{B G-2 G M}=\frac{3}{2} \Rightarrow \frac{3 G M}{3 B G}=\frac{3-2}{2 \times 3+2} ... | \frac{1}{168} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,676 |
4. Group all positive integers that are coprime with 2012 in ascending order, with the $n$-th group containing $2n-1$ numbers:
$$
\{1\},\{3,5,7\},\{9,11,13,15,17\}, \cdots \text {. }
$$
Then 2013 is in the $\qquad$ group. | 4.32.
Notice that, $2012=2^{2} \times 503$, where $2$ and $503$ are both prime numbers.
Among the positive integers not greater than 2012, there are 1006 multiples of 2, 4 multiples of 503, and 2 multiples of $2 \times 503$. Therefore, the numbers that are not coprime with 2012 are
$$
1006+4-2=1008 \text { (numbers).... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,677 |
One, (20 points) Find all integer pairs $(x, y) (x > y > 2012)$ that satisfy
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2012},
$$
and $x-y$ is maximized. | $$
\begin{array}{l}
x y-2012 x-2012 y=2012 \\
\Rightarrow(x-2012)(y-2012)=2012 \times 2013 .
\end{array}
$$
From $x-y$ being maximum, we know
$$
(x-2012)-(y-2012)
$$
is maximum. Therefore,
$$
\begin{array}{l}
\left\{\begin{array}{l}
x-2012=2012 \times 2013 \\
y-2012=1
\end{array}\right. \\
\Rightarrow(x, y)=(4052168,... | (4052168, 2013) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,678 |
II. (25 points) Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.
Arrange all positive integers that satisfy the follow... | II. Notice that, among three-digit numbers, the perfect squares are $100, 121, 144, 169, 196, 225, 256, 289$, $324, 361, 400, 441, 484, 529, 576, 625$, $676, 729, 784, 841, 900, 961$.
For each number in $M$, consider its leftmost three digits.
(1) The six-digit number with the leftmost three digits 100 is 100169, the f... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,679 |
Three. (25 points) As shown in Figure 4, let $D$ be the midpoint of the arc $\overparen{B C}$ of the circumcircle of $\triangle A B C$, and let point $X$ be on the arc $\overparen{B D}$. $E$ is the midpoint of the arc $\overparen{A X}$. A line $R T$ parallel to $D E$ is drawn through the incenter $I$ of $\triangle A B ... | Three, as shown in Figure 6, let $D R$ intersect the circumcircle of $\triangle A B C$ at point $S^{\prime}$, and $A X$ intersect $S^{\prime} E$ at point $T^{\prime}$. The other auxiliary lines are as shown in the figure.
Since $D$ is the midpoint of arc $\overparen{B C}$
$$
\begin{array}{l}
\Rightarrow \angle C S^{\pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,680 |
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