problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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3. If real numbers $m, n, p, q$ satisfy the conditions
$$
\begin{array}{l}
m+n+p+q=22, \\
m p=n q=100,
\end{array}
$$
then the value of $\sqrt{(m+n)(n+p)(p+q)(q+m)}$ is
$\qquad$ | 3. 220 .
From the given, we have
$$
\begin{aligned}
& (m+n)(n+p)(p+q)(q+m) \\
= & {[(m+n)(p+q)][(n+p)(q+m)] } \\
= & (200+m q+n p)(200+m n+p q) \\
= & 200^{2}+100\left(m^{2}+n^{2}+p^{2}+q^{2}+2 m n+\right. \\
& 2 m q+2 n p+2 p q) \\
= & 200^{2}+100\left[(m+n+p+q)^{2}-400\right] \\
= & {[10(m+n+p+q)]^{2} . }
\end{align... | 220 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) If the two sides of a right triangle, $x, y$, are both prime numbers, and make the algebraic expressions $\frac{2 x-1}{y}$ and $\frac{2 y+3}{x}$ both positive integers, find the inradius $r$ of this right triangle. | (1) If $x>y$, then
$$
1 \leqslant \frac{2 y+3}{x}<\frac{2 x+3}{x}<4 \text {. }
$$
It is easy to see that $\frac{2 y+3}{x}=1$ or 2.
$$
\begin{array}{l}
\text { (i) From } \frac{2 y+3}{x}=1 \\
\Rightarrow x=2 y+3 \\
\Rightarrow \frac{2 x-1}{y}=\frac{2(2 y+3)-1}{y}=4+\frac{5}{y} \\
\Rightarrow y=5, x=13 ;
\end{array}
$$
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If the function
$$
f(x)=a x+\sin x
$$
has perpendicular tangents on its graph, then the real number $a$ is
$\qquad$ . | 4.0.
Notice that, $f^{\prime}(x)=a+\cos x$.
If the function $f(x)$ has two perpendicular tangents, then there exist $x_{1}, x_{2} \in \mathbf{R}$, such that
$$
\begin{array}{l}
f^{\prime}\left(x_{1}\right) f^{\prime}\left(x_{2}\right)=-1 \\
\Leftrightarrow\left(a+\cos x_{1}\right)\left(a+\cos x_{2}\right)=-1 \\
\Leftr... | 0 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 The equation $2 x^{2}+5 x y+2 y^{2}=2007$ has $\qquad$ different integer solutions.
(2007, National Junior High School Mathematics League Sichuan Preliminary Competition) | The original equation can be transformed into
$$
(2 x+y)(x+2 y)=2007 \text {. }
$$
Since $x$ and $y$ are integers, without loss of generality, assume $x \leqslant y$, so,
$$
2 x+y \leqslant x+2 y \text {. }
$$
Notice that, $31[2 x+y)+(x+2 y)]$.
Thus, from equation (1), we get the system of equations
$$
\left\{\begin{... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Select $k$ numbers from 1 to 2012, such that among the selected $k$ numbers, there are definitely three numbers that can form the lengths of the sides of a triangle (the lengths of the three sides of the triangle must be distinct). What is the minimum value of $k$ that satisfies the condition? | Three, the problem is equivalent to:
Select $k-1$ numbers from $1,2, \cdots, 2012$, such that no three of these numbers can form the sides of a triangle with unequal sides. What is the maximum value of $k$ that satisfies this condition?
Now consider the arrays that meet the above conditions.
When $k=4$, the smallest th... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A five-digit number $\overline{a b c d e}$ satisfies
$$
ac>d, dd, b>e
$$
(for example, 37201, 45412), if its digits change with the position in a manner similar to the monotonicity of a sine function over one period, then this five-digit number is said to conform to the "sine rule". Therefore, there are $\qquad$ fiv... | 6.2892 .
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers among $a, b, c, d, e$. It is easy to see that $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$. In this case, there are $10-k$ ways to choose $(b, d)$, and $a, c, e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ g... | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
320 Suppose there are $2 k+1$ consecutive natural numbers,
where the sum of the squares of the $k$ larger natural numbers is equal to the sum of the squares of the remaining $k+1$ smaller natural numbers (such as $5^{2}=4^{2}+3^{2}$ $\left.(k=1), 14^{2}+13^{2}=12^{2}+11^{2}+10^{2}(k=2)\right)$. For convenience, such an... | Let
\[
\begin{array}{l}
(a+k)^{2}+(a+k-1)^{2}+\cdots+(a+1)^{2} \\
=a^{2}+(a-1)^{2}+\cdots+(a-k)^{2}
\end{array}
\]
be the $k$-th transformation method,
\[
w_{k}=(a+k, a-k)
\]
indicates that the largest number is $a+k$ and the smallest number is $a-k$.
Clearly, $a^{2}=4 \times(1+2+\cdots+k) a$, or $a=4 \times(1+2+\cdo... | 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Suppose the lengths of the two legs of a right triangle are $a$ and $b$, and the length of the hypotenuse is $c$. If $a$, $b$, and $c$ are all positive integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
(2010, National Junior High School Mathematics Competi... | Solve: By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Also, $c=\frac{1}{3} a b-(a+b)$, so
$$
c^{2}=\left[\frac{1}{3} a b-(a+b)\right]^{2} \text {. }
$$
Rearranging gives $a b-6(a+b)+18=0$.
Thus, $(a-6)(b-6)=18$.
Since $a$ and $b$ are both positive integers, and without loss of generality, let $a<b$, then,
$$... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 It is known that for any $x$,
$$
a \cos x + b \cos 2x \geqslant -1
$$
always holds. Find the minimum value of $a + b$.
(2009, Peking University Independent Admission Examination) | When $x=0$, $a+b \geqslant-1$. Taking $a=-\frac{4}{5}, b=-\frac{1}{5}$, then
$$
\begin{array}{l}
a \cos x+b \cos 2 x \\
=-\frac{2}{5}(\cos x+1)^{2}+\frac{3}{5} \\
\geqslant-\frac{2}{5}(1+1)^{2}+\frac{3}{5} \\
=-1 .
\end{array}
$$
Therefore, the minimum value of $a+b$ is -1. | -1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
B. If $a, b, c$ are positive numbers, and satisfy
$$
\begin{array}{c}
a+b+c=9, \\
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{10}{9}, \\
\text { then } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=
\end{array}
$$ | B. 7 .
From the given information, we have
$$
\begin{array}{l}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
=\frac{9-b-c}{b+c}+\frac{9-c-a}{c+a}+\frac{9-a-b}{a+b} \\
=\frac{9}{b+c}+\frac{9}{c+a}+\frac{9}{a+b}-3 \\
=9 \times \frac{10}{9}-3=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. A. As shown in Figure 4, the side length of square $ABCD$ is $2 \sqrt{15}, E, F$ are the midpoints of sides $AB, BC$ respectively, $AF$ intersects $DE, DB$ at points $M, N$. Then the area of $\triangle DMN$ is $\qquad$. | 7. A. 8 .
Connect $D F$. Let the side length of the square $A B C D$ be $2 a$. From the problem, we easily know
$$
\begin{array}{l}
\triangle B F N \backsim \triangle D A N \\
\Rightarrow \frac{A D}{B F}=\frac{A N}{N F}=\frac{D N}{B N}=\frac{2}{1} \\
\Rightarrow A N=2 N F \Rightarrow A N=\frac{2}{3} A F .
\end{array}
... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
B. Let $n$ be an integer, and $1 \leqslant n \leqslant 2012$. If $\left(n^{2}-n+3\right)\left(n^{2}+n+3\right)$ is divisible by 5, then the number of all $n$ is $\qquad$. | B. 1610.
Notice,
$$
\begin{array}{l}
\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \\
=n^{4}+5 n^{2}+9 \\
=(n-1)(n+1)\left(n^{2}+1\right)+5 n^{2}+10 .
\end{array}
$$
When $n$ is divided by 5, the remainder is 1 or 4, then $n-1$ or $n+1$ is divisible by 5, so
$$
51\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }... | 1610 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. A. 2 eighth-grade students and $m$ ninth-grade students participate in a single round-robin chess tournament, where each participant plays against every other participant exactly once. The scoring rule is: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both players get 1 ... | 9. A. 8 .
Let the number of draws be $a$, and the number of wins (losses) be $b$.
From the problem, we know
$$
2 a+3 b=130 \text {. }
$$
This gives $0 \leqslant b \leqslant 43$.
Also, $a+b=\frac{(m+1)(m+2)}{2}$
$$
\Rightarrow 2 a+2 b=(m+1)(m+2) \text {. }
$$
Thus, $0 \leqslant b=130-(m+1)(m+2) \leqslant 43$.
Therefo... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
B. Given that $n$ is even, and $1 \leqslant n \leqslant 100$. If there is a unique pair of positive integers $(a, b)$ such that $a^{2}=b^{2}+n$ holds, then the number of such $n$ is | B. 12.
From the given, we have $(a-b)(a+b)=n$, and $n$ is even, so $a-b$ and $a+b$ are both even.
Therefore, $n$ is a multiple of 4.
Let $n=4m$. Then $1 \leqslant m \leqslant 25$.
(1) If $m=1$, we get $b=0$, which contradicts that $b$ is a positive integer.
(2) If $m$ has at least two different prime factors, then the... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. A. Given integers $a, b$ satisfy $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$. | 13. A. Let $a-b=m$ (where $m$ is a prime number), and $ab=n^2$ (where $n$ is a non-negative integer).
When $b \neq 0$,
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive ... | 2017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
B. In a convex $n$-sided polygon, what is the maximum number of interior angles that can be $150^{\circ}$? Explain your reasoning. | B. Suppose in a convex $n$-sided polygon, there are $k$ interior angles equal to $150^{\circ}$. Then there are $n-k$ interior angles not equal to $150^{\circ}$.
(1) If $k=n$, then
$$
n \times 150^{\circ}=(n-2) \times 180^{\circ},
$$
we get $n=12$. Thus, in a regular dodecagon, all 12 interior angles are $150^{\circ}$.... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, a_{2}=4, a_{3}=9, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}(n=4,5, \cdots) .
\end{array}
$$
Then $a_{2011}=$ | $-、 1.8041$.
From the problem, we have
$$
a_{2}-a_{1}=3, a_{3}-a_{2}=5 \text {, }
$$
and $a_{n}-a_{n-1}=a_{n-2}-a_{n-3}(n \geqslant 4)$.
Thus, $a_{2 n}-a_{2 n-1}=3, a_{2 n+1}-a_{2 n}=5\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $a_{2 n+1}-a_{2 n-1}=8$.
Hence, $a_{2011}=\sum_{k=1}^{1005}\left(a_{2 k+1}-a_{2 k-1}\rig... | 8041 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given a function $f(n)$ defined on the set of positive integers satisfies the conditions:
(1) $f(m+n)=f(m)+f(n)+m n\left(m, n \in \mathbf{N}_{+}\right)$;
(2) $f(3)=6$.
Then $f(2011)=$ . $\qquad$ | 3.2023066.
In condition (1), let $n=1$ to get
$$
f(m+1)=f(m)+f(1)+m \text {. }
$$
Let $m=n=1$ to get
$$
f(2)=2 f(1)+1 \text {. }
$$
Let $m=2, n=1$, and use condition (2) to get
$$
6=f(3)=f(2)+f(1)+2 \text {. }
$$
From equations (3) and (2), we get
$$
f(1)=1, f(2)=3 \text {. }
$$
Substitute into equation (1) to get... | 2023066 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
has $\qquad$ solutions. | 4. 4.
The solutions to the equation $||x|-1|=1$ are $x=0$ or $\pm 2$;
The solutions to the equation $|||x|-1|-2|=2$ are $x= \pm 1$ or $\pm 5$;
The solutions to the equation $||||x|-1|-2|-3|=3$ are $x= \pm 3$ or $\pm 9$;
In general, the solutions to the equation
$$
|1 \cdots||| x|-1|-2|\cdots|-n \mid=n(n \geqslant 2)
$... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. 10 students stand in a row, and a red, yellow, or blue hat is to be given to each student. It is required that each color of hat must be present, and the hats of adjacent students must be of different colors. Then the number of ways to distribute the hats that meet the requirements is $\qquad$ kinds. | 8. 1530.
Generalize to the general case.
Let the number of ways to arrange $n$ students according to the given conditions be $a_{n}$. Then
$$
\begin{array}{l}
a_{3}=6, a_{4}=18, a_{n+1}=2 a_{n}+6(n \geqslant 3) . \\
\text { Hence } a_{n+1}+6=2\left(a_{n}+6\right) \\
\Rightarrow a_{n}=\left(a_{3}+6\right) \times 2^{n-3... | 1530 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the parabola
$$
y=x^{2}+(k+1) x+1
$$
intersects the $x$-axis at two points $A$ and $B$, not both on the left side of the origin. The vertex of the parabola is $C$. To make $\triangle A B C$ an equilateral triangle, the value of $k$ is $\qquad$ | $=1 .-5$.
From the problem, we know that points $A$ and $B$ are to the right of the origin, and
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{(k+1)^{2}-4} \text {. } \\
\text { Then } \frac{\sqrt{3}}{2} \sqrt{(k+1)^{2}-4}=\left|1-\left(\frac{k+1}{2}\right)^{2}\... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $A M$ is the median of $\triangle A B C$ on side $B C$, $P$ is the centroid of $\triangle A B C$, and a line $E F$ through point $P$ intersects sides $A B$ and $A C$ at points $E$ and $F$ respectively. Then $\frac{B E}{A E}+\frac{C F}{A F}=$ $\qquad$ | 3. 1 .
Draw $B G$ and $C K$ parallel to $A M$ through points $B$ and $C$ respectively, intersecting line $E F$ at points $G$ and $K$. Then
$$
\frac{B E}{A E}=\frac{B G}{A P}, \frac{C F}{A F}=\frac{C K}{A P} \text {. }
$$
Adding the two equations gives
$$
\frac{B E}{A E}+\frac{C F}{A F}=\frac{B G+C K}{A P} .
$$
In tr... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In $\triangle A B C$, $A C=B C, \angle A C B=$ $90^{\circ}, D, E$ are two points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$
(2006, Beijing Middle School Mathematics Competition (Grade 8)) | Solve as shown in Figure 1, construct square $C A H B$, and extend $C D$, $C E$ to intersect $A H$, $B H$ at points $G$, $F$ respectively.
Let $D E=x$.
Since $A C / / B F$
$$
\Rightarrow \frac{3+x}{4}=\frac{A C}{B F} \text {, }
$$
$B C / / A G$
$$
\begin{array}{c}
\Rightarrow \frac{4+x}{3}=\frac{B C}{A G} . \\
\text { ... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 In $\triangle A B C$, it is known that $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=2, C D=3$, then $S_{\triangle A B C}$ $=$ $\qquad$
(2007, Shandong Province Junior High School Mathematics Competition) | Solve as shown in Figure 5, with $AB$ as the axis of symmetry, construct the symmetric figure of $\triangle ADB$ as $\triangle AGB$, and with $AC$ as the axis of symmetry, construct the symmetric figure of $\triangle ADC$ as $\triangle AFC$, and extend $GB$ and $FC$ to intersect at point $E$. Then it is easy to know th... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given four distinct positive real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2012}-c^{2012}\right)\left(a^{2012}-d^{2012}\right)=2012, \\
\left(b^{2012}-c^{2012}\right)\left(b^{2012}-d^{2012}\right)=2012 . \\
\text { Then }(a b)^{2012}-(c d)^{2012}=(\quad) .
\end{array}
$$
(A) -2012
(B) -2011
(C) 2012... | - 1. A.
From the problem, we know that \(a^{2012}\) and \(b^{2012}\) are the two distinct real roots of the quadratic equation in \(x\):
\[
\left(x-c^{2012}\right)\left(x-d^{2012}\right)=2012,
\]
which is \(x^{2}-\left(c^{2012}+d^{2012}\right) x+(c d)^{2012}-2012=0\).
Thus, \(a^{2012} b^{2012}=(c d)^{2012}-2012\).
The... | -2012 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. There are $n$ people registered to participate in four sports competitions: A, B, C, and D. It is stipulated that each person must participate in at least one competition and at most two competitions, but competitions B and C cannot be registered for simultaneously. If in all different registration methods, there mu... | 5. B.
Use the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{J}}\right)$ to represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is recorded as 1 (for example, if participating in event 甲, ... | 172 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
4. Given that the circumradius of $\triangle A B C$ is $1, \angle A$, $\angle B$, and $\angle C$'s angle bisectors intersect the circumcircle of $\triangle A B C$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Then
$$
\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin ... | 4. 2 .
Connect $B A_{1}$. By the Law of Sines, we have
$$
\begin{array}{l}
A A_{1} \cos \frac{A}{2}=2 \sin \left(B+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Similarly, $B B_{1} \cos \frac{B}{2}=\sin C+\sin A$,
$$
C C_{1} \cos \frac{C}{2}=\sin A+\sin B \text {. }
$$
Therefore, the o... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)=a \sin [(x+1) \pi]+b \sqrt[3]{x-1}+2$, where $a$ and $b$ are real constants. If $f(\lg 5)=5$, then $f(\lg 20)=$ | 5. -1 .
Obviously,
$$
f(x)=a \sin [(x-1) \pi]+b \sqrt[3]{x-1}+2 .
$$
Let $t=x-1$. Then
$$
f(t+1)=a \sin \pi t+b \sqrt[3]{t}+2=g(t)+2,
$$
where, $g(t)=a \sin \pi t+b \sqrt[3]{t}$ is an odd function.
According to the problem,
$$
\begin{aligned}
5 & =f(\lg 5)=f(1-\lg 2) \\
& =g(-\lg 2)+2=-g(\lg 2)+2 .
\end{aligned}
$$
... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In the Cartesian coordinate system, $O$ is the origin, points $A(3, a) 、 B(3, b)$ make $\angle A O B=45^{\circ}$, where $a 、 b$ are integers, and $a>b$. Then the number of pairs $(a, b)$ that satisfy the condition is. | 6.6.
Let $\angle A O X=\alpha, \angle B O X=\beta$. Then $\tan \alpha=\frac{a}{3}, \tan \beta=\frac{b}{3}$.
Given $a>b$, we have
$$
\begin{array}{l}
1=\tan 45^{\circ}=\tan (\alpha-\beta) \\
=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{3(a-b)}{9+a b} .
\end{array}
$$
Rearranging gives $(a+3)(b... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (16 points) For an integer $k$, define the set
$$
S_{k}=\{n \mid 50 k \leqslant n<50(k+1), n \in \mathbf{Z}\} \text {. }
$$
How many of the 600 sets $S_{0}, S_{1}, \cdots, S_{599}$ do not contain any perfect squares? | 11. Notice,
$$
\begin{array}{l}
(x+1)^{2}-x^{2}=2 x+1 \leqslant 50(x \in \mathbf{N}) \\
\Leftrightarrow x \leqslant 24(x \in \mathbf{N}) .
\end{array}
$$
While $(24+1)^{2}=625 \in S_{12}$, thus, the square numbers in $S_{0}, S_{1}$, $\cdots, S_{12}$ do not exceed $25^{2}$, and each set consists of 50 consecutive non-n... | 439 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$. Then the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is | 6.3.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(a_{n}+1\right)}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \\
\Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\sum_{n=0}^{2011}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. $[x]$ represents the greatest integer not exceeding the real number $x$. The area of the figure formed by points satisfying $[x]^{2}+[y]^{2}=50$ on the plane is $\qquad$ . | 8. 12 .
First, consider the case in the first quadrant.
When $x>0, y>0$, from $[x]^{2}+[y]^{2}=50$, we get
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l }
{ 7 \leqslant x < 8 , } \\
{ 1 \leqslant y < 2 }
\end{array} \left\{\begin{array} { l }
{ 5 \leqslant x < 6 , } \\
{ 5 \leqslant y < 6 ; }
\end{array} \... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=$ $\qquad$ . | 3. 1 .
From the given, $\sqrt{2-\sqrt{3}}=\sqrt{x}-\sqrt{y}$.
Then $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{(\sqrt{3}-1)^{2}}}{\sqrt{2}}$ $=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Thus, $x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}$.
S... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { II. (25 points) (1) Given the parabola } \\
y=x^{2}-2 m x+4 m-8
\end{array}
$$
with vertex $A$, construct an inscribed equilateral $\triangle A M N$ (points $M, N$ are on the parabola). Find the area of $\triangle A M N$;
(2) If the parabola $y=x^{2}-2 m x+4 m-8$ intersects the $x$-axis at ... | (1) From the symmetry of the parabola and the equilateral triangle, we know that $M N \perp y$-axis.
As shown in Figure 8, let the axis of symmetry of the parabola intersect $M N$ at point $B$. Then
$$
A B=\sqrt{3} B M .
$$
Let $M(a, b)$ $(m<a)$. Then
$$
\begin{aligned}
& B M=a-m . \\
& \text { Also, } A B=y_{B}-y_{A... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a, b, c \in \mathbf{R}_{+}$, and $abc=4$. Then the minimum value of the algebraic expression $a^{a+b} b^{3b} c^{c+b}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 64.
It is easy to know that $a^{a+b} b^{3 b} c^{c+b}=a^{a} b^{2 b} c^{c} \cdot 4^{b}=a^{a}(2 b)^{2 b} c^{c}$.
And $a b c=4$ can be transformed into $a \cdot 2 b \cdot c=8$, considering $2 b$ as a whole.
Since the function $f(x)=\ln x$ is increasing on $(0,+\infty)$, for any $a, b \in(0,+\infty)$, we always have $(... | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given vectors
$$
a=(x-y+1, x-y), b=\left(x-y+1,10^{x}\right) \text {. }
$$
Then the number of all integer pairs $(x, y)$ that satisfy $a \cdot b=2012$ is $\qquad$ | 5.0.
From the problem, we have
$$
\begin{array}{l}
(x-y+1)^{2}+10^{x}(x-y)=2012 \\
\Rightarrow(x-y)\left(x-y+2+10^{x}\right)=2011 .
\end{array}
$$
Obviously, $x \neq 0$, otherwise, $y(y-3)=2011$, this equation has no integer solutions.
If $x>0$, then $10^{x}$ is a positive integer.
Thus, $x-y+2+10^{x}>x-y$.
We get th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) For a given positive integer $M$, define $f_{1}(M)$ as the square of the sum of the digits of $M$. When $n>1$ and $n \in \mathbf{N}$, $f_{n}\left(f_{n-1}(M)\right)$ represents the $r_{n}$-th power of the sum of the digits of $f_{n-1}(M)$, where, when $n$ is odd, $r_{n}=2$; when $n$ is even, $r_{n}=3$. Fi... | Let the positive integer $M=a_{1} a_{2} \cdots a_{m}$, where $a_{i} \in \mathbf{N}, a_{1}>0 (i=1,2, \cdots, m)$. Then
$$
f_{1}(M)=\left(a_{1}+a_{2}+\cdots+a_{m}\right)^{2} \equiv M^{2}(\bmod 9) .
$$
When $M=3^{2010}$, $f_{1}(M) \equiv M^{2} \equiv 0(\bmod 9)$.
By mathematical induction, it is easy to see that for $n \... | 729 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $a$ is a prime number less than 2012, and 2012 $-a$ is coprime with $a$. Then the number of $a$ that satisfies the condition is ( ).
(A) 1003
(B) 1004
(C) 1005
(D) 1006 | 5. B.
If $a$ is even, then $2012-a$ is also even. In this case, $a$ and $2012-a$ have a common divisor of 2, so they are not coprime. Therefore, $a$ cannot be even and must be odd.
Since $2012=503 \times 4$, and 503 is a prime number, we have
$$
a \neq 503 \times 1 \text { and } 503 \times 3 \text {. }
$$
There are 1... | 1004 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
$\qquad$ 1. A six-digit number $\overline{a b c d e f}$, when multiplied by 4, becomes $\overline{f a b c d e}$. The number of six-digit numbers that satisfy this condition is $\qquad$. | $=1.6$
Let $\overline{a b c d e}=x$. Then
$$
\begin{array}{l}
4(10 x+f)=100000 f+x \\
\Rightarrow x=2564 f .
\end{array}
$$
Since $f$ is a single digit and $x$ is a five-digit number, it is easy to see that,
$$
f=4,5,6,7,8,9 \text {. }
$$
Therefore, the six-digit numbers $\overline{a b c d e f}$ that satisfy the cond... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If $a-b=2, \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4$, then $a^{5}-b^{5}=$ | 3. 82 .
$$
\begin{aligned}
\text { Given } & \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4 \\
\Rightarrow & a(1-a)^{2}-b(1+b)^{2}=4 a b \\
\Rightarrow & a-2 a^{2}+a^{3}-b-2 b^{2}-b^{3}=4 a b \\
\Rightarrow & (a-b)-2\left(a^{2}+b^{2}\right)+\left(a^{3}-b^{3}\right)=4 a b \\
\Rightarrow & (a-b)-2\left[(a-b)^{2}+2 a b\right]+... | 82 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Find the integer part of $\left(\frac{1+\sqrt{5}}{2}\right)^{19}$. | Let $a=\frac{1+\sqrt{5}}{2}, b=\frac{1-\sqrt{5}}{2}$. Then $a+b=1, a^{2}=a+1, b^{2}=b+1$.
Thus $a^{2}+b^{2}=(a+1)+(b+1)$ $=(a+b)+2=3$, $a^{3}+b^{3}=\left(a^{2}+a\right)+\left(b^{2}+b\right)$ $=\left(a^{2}+b^{2}\right)+(a+b)=1+3=4$, $a^{4}+b^{4}=\left(a^{3}+a^{2}\right)+\left(b^{3}+b^{2}\right)$ $=\left(a^{3}+b^{3}\righ... | 9349 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Given that $D$ is a point inside $\triangle A B C$, $E$ is the midpoint of side $A C$, $A B=6, B C=10, \angle B A D=$ $\angle B C D, \angle E D C=\angle A B D$. Find the length of $D E$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | II. As shown in Figure 4, extend \( CD \) to point \( F \) such that \( DF = CD \), and connect \( AF \) and \( BF \).
Then \( AF \parallel DE \), and \( DE = \frac{1}{2} AF \).
Thus, \( \angle AFD = \angle EDC = \angle ABD \)
\(\Rightarrow A, F, B, D\) are concyclic
\(\Rightarrow \angle BFD = \angle BAD = \angle BCD \... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Place 27 balls numbered $1 \sim 27$ into three bowls, Jia, Yi, and Bing, such that the average values of the ball numbers in bowls Jia, Yi, and Bing are $15$, $3$, and $18$, respectively, and each bowl must contain no fewer than 4 balls. Then the maximum value of the smallest ball number in bowl Jia is $\qquad$ | 4. 10 .
Let there be $a$, $b$, and $c$ balls in bowls 甲, 乙, and 丙, respectively. Then,
$$
\begin{aligned}
a+b+c & =27, \\
15 a+3 b+18 c & =\frac{27 \times 28}{2} .
\end{aligned}
$$
From equation (2), we get
$$
5 a+b+6 c=126 \text {. }
$$
If $b \geqslant 6$, then the average value of the ball numbers in bowl 乙 is not... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for all non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Then the remainder when $a_{2012}$ is divided by 2012 is ـ. $\qquad$ | 5.1.
In equation (1), let $n=0$, we get
$$
a_{2 m}=4 a_{m}+2 m-2-a_{0} \text {. }
$$
In equation (2), let $m=0,1$, we get
$$
a_{0}=1, a_{2}=3 \text {. }
$$
In equation (1), let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right) \\
=\frac{1}{2}\left(4 a_{m}+2 m-3+3\right)=2 a_... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given a positive integer $k$ that satisfies for any positive integer $n$, the smallest prime factor of $n^{2}+n-k$ is no less than 11. Then, $k_{\text {min }}$ $=$ . $\qquad$ | 6.43 .
Notice,
$$
\begin{array}{l}
n^{2}+n \equiv 0(\bmod 2), \\
n^{2}+n \equiv 0,2(\bmod 3),
\end{array}
$$
$$
\begin{array}{l}
n^{2}+n \equiv 0,2,1(\bmod 5), \\
n^{2}+n \equiv 0,2,6,5(\bmod 7),
\end{array}
$$
and the smallest prime factor of $n^{2}+n-k$ is not less than 11, then
$$
\begin{aligned}
k & \equiv 1(\bmo... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given positive integers $a, b$ satisfy
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \text {. }
$$
Then $|10(a-5)(b-15)|+2=$ | 7.2012.
Notice,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{... | 2012 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a cyclic quadrilateral $A_{1} A_{2} A_{3} A_{4}$ with an inscribed circle $\odot I$ that is tangent to the sides $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, and $A_{4} A_{1}$ at points $B_{1}$, $B_{2}$, $B_{3}$, and $B_{4}$ respectively, then
$$
\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{... | 8. 8 .
As shown in Figure 5, let the radius of $\odot I$ be $r$,
$$
\begin{array}{l}
A_{i} B_{i}=A_{i} B_{i-1} \\
=a_{i},
\end{array}
$$
where $i=1,2,3,4, B_{0}=B_{4}$.
Then $A_{i} A_{i+1}=a_{i}+a_{i+1}$,
$$
B_{i} B_{i+1}=\frac{2 A_{i+1} B_{i} \cdot B_{i} I}{\sqrt{A_{i+1} B_{i}^{2}+B_{i} I^{2}}}=2\left(a_{i+1}^{-2}+r... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) How many nine-digit numbers $\overline{a_{1} a_{2} \cdots a_{9}}$ satisfy $a_{1} \neq 0$, all digits are distinct, and
$$
a_{1}+a_{3}+a_{5}+a_{9}=a_{2}+a_{4}+a_{6}+a_{8} .
$$ | 10. Let the missing digit be $a$, and
$$
\sum_{i=\pi} a_{i}=\sum_{i=m} a_{i}=t .
$$
Then $2 t=\sum_{i=1}^{9} a_{i}=45-a$.
Thus, $a \in\{1,3,5,7,9\}$.
If $a=1$, then $t=22$, we have
$$
\begin{array}{l}
22=2+3+8+9=2+4+7+9 \\
=2+5+6+9=2+5+7+8 \\
=3+4+6+9=3+4+7+8 \\
=3+5+6+8=4+5+6+7 \\
=0+5+8+9=0+6+7+9 ;
\end{array}
$$
I... | 120384 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Write down all positive integers from 1 to 10000 from left to right, then remove those numbers that are divisible by 5 or 7, and concatenate the remaining numbers to form a new number. Try to find:
(1) The number of digits in the new number;
(2) The remainder when the new number is divided by 11. | Three, (1) Obviously, the remainder of the remaining numbers when divided by 35 is
$$
\begin{array}{l}
1,2,3,4,6,8,9,11,12,13,16,17,18, \\
19,22,23,24,26,27,29,31,32,33,34,
\end{array}
$$
denoted as $a_{i}(i=1,2, \cdots, 24)$ in sequence.
Among the numbers from $1 \sim 9$, 7 numbers remain, denoted as $b_{i}$ $(i=1,2,... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Find all real roots of the equation
$$
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right)
$$ | Solve:
$$
\begin{array}{l}
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right) \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}=x^{2}+2 x+4 \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}-3=x^{2}+2 x+1 \\
\Rightarrow \frac{-2\left(x^{3}+1\right)^{2}}{x^{6}+x^{3}+1}=(x+1)^{2} \\
\Rightarrow(x+1)^{2}\le... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let $n$ be a positive integer, $[x]$ be the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$.
(1) Find all positive integers $n$ that satisfy
$$
\sum_{k=1}^{2013}\left[\frac{k n}{2013}\right]=2013+n
$$
(2) Find all positive integers $n$ that maximize $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right... | (1) Let $m=2013, d=(m, n), m=d m_{1}, n=d n_{1}$.
For $1 \leqslant k \leqslant m-1$, we have
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=\left[\frac{k n_{1}}{m_{1}}\right]+\left[\frac{(m-k) n_{1}}{m_{1}}\right] \text{. }
$$
When $m_{1} \mid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\rig... | 1006 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If $\frac{1}{a}+\frac{1}{b}=\frac{5}{a+b}$, then $\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=$ | 2. 7 .
$$
\begin{array}{l}
\text { Given } \frac{1}{a}+\frac{1}{b}=\frac{5}{a+b} \Rightarrow \frac{b}{a}+\frac{a}{b}=3 \\
\Rightarrow \frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=\left(\frac{b}{a}+\frac{a}{b}\right)^{2}-2=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If $x-1$ is a factor of $x^{3}+a x^{2}+1$, then the value of $a$ is $\qquad$ . | 3. -2 .
Let $x^{3}+a x^{2}+1=(x-1)\left(x^{2}-m x-1\right)$, that is
$$
\begin{array}{l}
x^{3}+a x^{2}+1 \\
=x^{3}-(m+1) x^{2}+(m-1) x+1 .
\end{array}
$$
By comparing, we get $m-1=0, a=-(m+1)$.
Thus, $m=1, a=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (25 points) Let the two intersection points of the functions $y=2x$ and $y=\frac{4}{x}$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}>x_{2}\right)$, and point $C(\sqrt{2},-2 \sqrt{2})$. Find the area of $\triangle ABC$. | From $\left\{\begin{array}{l}y=2 x, \\ y=\frac{4}{x},\end{array}\right.$ eliminating $y$ we get
$$
2 x=\frac{4}{x} \Rightarrow x^{2}=2 \Rightarrow x= \pm \sqrt{2} \text {. }
$$
Therefore, $A(\sqrt{2}, 2 \sqrt{2}), B(-\sqrt{2},-2 \sqrt{2})$.
Since point $C(\sqrt{2},-2 \sqrt{2})$, $\triangle A B C$ is a right triangle, ... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1} \leqslant \frac{a_{n+2}+a_{n}}{2}, a_{1}=1, a_{403}=2011 \text {. }
$$
Then the maximum value of $a_{5}$ is $\qquad$ | - 1.21.
Obviously, the sequence of points $\left(n, a_{n}\right)$ is arranged in a convex function. When the sequence of points is distributed on the line determined by the points $(1,1)$ and $(403,2011)$, $a_{5}$ takes the maximum value 21. | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $a_{1}=1, a_{2}=3$,
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} \text {, }
$$
when $m \geqslant n$, $a_{m}$ is divisible by 9. Then the minimum value of $n$ is $\qquad$ . | 5.5.
Notice,
$$
\begin{array}{l}
a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right) \\
=\cdots=(n+2)(n+1) n \cdots \cdots \cdot 3\left(a_{2}-a_{1}\right) \\
=(n+2)!.
\end{array}
$$
Thus, $a_{n}=a_{1}+\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)=\sum_{i=1}^{n} i!$.
Given $a_{1}=1, a_{2}=3$, we have
$$
a_{3}=9, a_{4}=33, a_... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Six college graduates apply to three employers. If each employer hires at least one of them, the number of different hiring scenarios is $\qquad$ .
| 7. 2100 .
The number of ways for three people to be hired is $\mathrm{A}_{6}^{3}=120$; the number of ways for four people to be hired is $\mathrm{C}_{6}^{4} \mathrm{C}_{4}^{2} \mathrm{~A}_{3}^{3}=15 \times 6 \times 6=540$ (ways); the number of ways for five people to be hired is $C_{6}^{5}\left(C_{5}^{3} A_{3}^{3}+\fr... | 2100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. (12 points) Given
$$
\begin{array}{l}
f(x, y) \\
=x^{3}+y^{3}+x^{2} y+x y^{2}-3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{array}
$$
and $x, y \geqslant \frac{1}{2}$. Find the minimum value of $f(x, y)$. | 12. When $x \neq y$, multiplying both sides of the function by $x-y$ gives
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-y^{4}\right)-3\left(x^{3}-y^{3}\right)+3\left(x^{2}-y^{2}\right) . \\
\text { Let } g(x)=x^{4}-3 x^{3}+3 x^{2} .
\end{array}
$$
Then $f(x, y)=\frac{g(x)-g(y)}{x-y}$ is the slope of the line segm... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than
1. Then
$$
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=
$$
$\qquad$ | 15.2011.
When the common ratio is 1, $\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011$.
When the common ratio is $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} ... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. Figure 3 is a defective $3 \times 3$ magic square, in which the sum of the three numbers in each row, each column, and each diagonal is equal. Then the value of $x$ is . $\qquad$ | $$
\begin{array}{l}
\text { Magic Square } \\
4017+2012 \\
=x-2003+x \\
\Rightarrow x=4016 \text {. } \\
\end{array}
$$ | 4016 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
1 \times 2 - 3 \times 4 + 5 \times 6 - 7 \times 8 + \cdots + \\
2009 \times 2010 - 2011 \times 2012 \\
= \quad .
\end{array}
$$ | $$
\text { II, 1. }-2025078 \text {. }
$$
Notice that,
$$
\begin{array}{l}
(n+2)(n+3)-n(n+1)=4 n+6 \\
=n+(n+1)+(n+2)+(n+3) .
\end{array}
$$
Therefore, the original expression is
$$
=-(1+2+\cdots+2012)=-2025078 .
$$ | -2025078 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The ten positive integers from $1 \sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\cdots, S_{10}=a_{1}+a_{2}+\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \cdots, S_{10}$, the maximum number of primes that can occur is . | 2. 7 .
Adding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \backslash S_{n}$ and $... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (10 points) Prove: For any three distinct numbers $a, b, c$, we have
$$
\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}
$$
is a constant. | $$
\begin{array}{l}
S=\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}, \\
p=a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a . \\
\text { Then } (a+b-c)^{2}=p+4 a b, \\
(b+c-a)^{2}=p+4 b c, \\
(c+a-b)^{2}=p+4 c a . \\
\text { Hence } S=\frac{p+4 a b}{(a-c)(b-c)}+\frac{p+4 b c}{(b-a)(c-a)... | 4 | Algebra | proof | Yes | Yes | cn_contest | false |
Four, (15 points) It is known that a positive integer $n$ can be expressed as the sum of 2011 identical natural numbers, and also as the sum of 2012 identical natural numbers. Determine the minimum value of $n$.
| Let $a_{1}, a_{2}, \cdots, a_{2011}$ be 2011 natural numbers with the same digit sum, and $b_{1}, b_{2}, \cdots, b_{2012}$ be 2012 natural numbers with the same digit sum, and they satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=n \\
=b_{1}+b_{2}+\cdots+b_{2012} .
\end{array}
$$
Since each of the numbers $a_{1... | 10055 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. The constant term in the expansion of $\left(x^{2}+x-\frac{1}{x}\right)^{6}$ is $\qquad$ (answer with a specific number). | 8. -5 .
From the conditions, the constant term is
$$
C_{6}^{2}(-1)^{4}+C_{6}^{3}(-1)^{3}=-5 .
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$. | 9. 0 .
From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$, we know
$$
S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4} \text {. }
$$
Then $a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. The number of positive integers not exceeding 2012 and having exactly three positive divisors is $\qquad$ . | 10. 14.
Let $1 \leqslant a \leqslant 2012$ and $a$ has only three positive divisors. Then $a$ is the square of a prime number, i.e., $a=p^{2} \leqslant 2012$.
Thus, $2 \leqslant p \leqslant 43$.
$$
\begin{array}{c}
\text { Hence } p=2,3,5,7,11,13,17,19,23, \\
29,31,37,41,43 .
\end{array}
$$ | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy
$$
a f(x)+b f(x-c)=1
$$
for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$. | Three, 13. (1) From the given conditions,
$$
f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1 \text {. }
$$
From $0 \leqslant x \leqslant \frac{\pi}{2} \Rightarrow \frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $\triangle A B C$ is an isosceles right triangle, $\angle A$ $=90^{\circ}$, and $\overrightarrow{A B}=a+b, \overrightarrow{A C}=a-b$.
If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then $S_{\triangle A B C}$ $=$ . $\qquad$ | 4. 1.
From the problem, we know that $A B \perp A C,|A B|=|A C|$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
(a+b) \cdot(a-b)=0, \\
|a+b|=|a-b|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
|a|=|b|, \\
a \cdot b=0 .
\end{array}\right.
\end{array}
$$
Since $|a|=1$, we have $|b|=1$.
According ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$ | 5. 1.
As shown in Figure 3, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$. Then
$$
\begin{array}{c}
O B=\frac{\sqrt{3}}{3} a, \\
M B=\frac{\sqrt{2}}{2} a . \\
\text { Therefore, } M O=\sqrt{M B^{2}-O B^{2}} \\
=\frac{\sqrt{6}}{6} a=\frac{1}{2} A O=A M . \\
\text { Hence, } \frac{A M}{M... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. For a certain game activity,
the rewards are divided into first, second, and third prizes (all participants in the game activity will receive a prize), and the corresponding winning probabilities form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prize money forms an arithm... | 7.500 .
From the problem, we know the probabilities of winning the first, second, and third prizes are
$$
P_{1}=a, P_{2}=2 a, P_{3}=4 a .
$$
From $P_{1}+P_{2}+P_{3}=1$, we get $a=\frac{1}{7}$.
Thus, $P_{1}=\frac{1}{7}, P_{2}=\frac{2}{7}, P_{3}=\frac{4}{7}$.
The prizes for winning the first, second, and third prizes a... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $p$ and $q$ be two different prime numbers. Then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p q$ is $\qquad$ | 8. 1 .
Since $p$ and $q$ are different prime numbers, by Fermat's Little Theorem, we have
$$
p^{q-1} \equiv 1(\bmod q) \text {. }
$$
Also, $q^{p-1} \equiv 0(\bmod q)$, thus
$$
p^{q-1}+q^{p-1} \equiv(\bmod q) \text {. }
$$
Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Therefore, $p^{q-1}+q^{p-1} \equiv 1(\bmod p q)... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Transporting utility poles from a construction site by the roadside along a straight road in the same direction to plant them 500 m away on the roadside, plant one at the 500 m mark, and then plant one every 50 m along the roadside. Knowing that the transport vehicle can carry a maximum of 3 poles at a time, to com... | 10. 14000 .
Assuming the completion of transporting and planting 21 utility poles, 3 poles each time, let the round trip distance for the $k(k=1,2, \cdots, 7)$-th time be $a_{k}$. Then
$$
a_{1}=2 \times 600=1200,
$$
and $a_{k+1}=a_{k}+2 \times 150(k=1,2, \cdots, 6)$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic... | 14000 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Four. (30 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=1+\frac{1}{a_{n... | (1) From
$$
\begin{array}{l}
a_{n}=2 a_{n} a_{n+1}+3 a_{n+1} \\
\Rightarrow a_{n+1}=\frac{a_{n}}{2 a_{n}+3} \Rightarrow a_{n}>0 . \\
\text { Hence } \frac{1}{a_{n+1}}-\frac{3}{a_{n}}=2 \Rightarrow \frac{1}{a_{n+1}}+1=3\left(\frac{1}{a_{n}}+1\right) .
\end{array}
$$
Therefore, $\left\{\frac{1}{a_{n}}+1\right\}$ is a ge... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that among the vertices of a regular 2012-gon, there exist $k$ vertices such that the convex $k$-gon formed by these $k$ vertices has no two parallel sides. Find the maximum value of $k$.
| 2. The maximum value of $k$ is 1509.
Let $A_{1}, A_{2}, \cdots, A_{2012}$ be the set of vertices of a regular polygon.
Consider the set of four vertices
$$
\begin{array}{l}
\left(A_{1}, A_{2}, A_{1000}, A_{1008}\right),\left(A_{3}, A_{4}, A_{100}, A_{1010}\right), \\
\cdots,\left(A_{1005}, A_{1000}, A_{2011}, A_{2012}... | 1509 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $n \geqslant 2$,
$$
A_{n}=\left\{x \in \mathbf{R} \left\lvert\, x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right]\right.\right\} .
$$ | Prove: $A=\underset{n \geqslant 2}{\cup} A_{n}$ is a finite set, and find the maximum and minimum elements of $A$.
(2010, Romanian Mathematical Olympiad (Final)) [Analysis] Start with simple cases in mathematical experiments.
Elements in $A_{2}$ satisfy $x=\left[\frac{x}{2}\right]$, and $x \in \mathbf{Z}$. Hence, $\fra... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is ( ).
(A) 209520
(B)209 519
(C)209 518
(D) 209517 | 4. B.
Let $3 \sqrt{2}+2 \sqrt{3}=a, 3 \sqrt{2}-2 \sqrt{3}=b$.
Then $a+b=6 \sqrt{2}, ab=6$
$$
\begin{array}{l}
\Rightarrow a^{2}+b^{2}=(a+b)^{2}-2ab=60 \\
\Rightarrow a^{6}+b^{6}=\left(a^{2}+b^{2}\right)^{3}-3(ab)^{2}\left(a^{2}+b^{2}\right) \\
\quad=209520 .
\end{array}
$$
Since $0<b<1$, we have
$$
209519<a^{6}<20952... | 209519 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. As shown in Figure 4, in the "dart-shaped" quadrilateral $ABCD$, $AB=4\sqrt{3}$, $BC=8$, $\angle A=\angle B=\angle C=30^{\circ}$. Then the distance from point $D$ to $AB$ is $\qquad$ | 3. 1.
As shown in Figure 9, extend $A D$ to intersect $B C$ at point $E$. Then $A E=B E$.
Draw $E F \perp A B$ at point $F$.
It is easy to see that $A F=B F=2 \sqrt{3}$,
$$
E F=2, A E=4 \text {. }
$$
Thus, $C E=4$.
Also, $\angle A D C=90^{\circ}$
$\Rightarrow D E=2$
$\Rightarrow D$ is the midpoint of $A E$
$\Rightarr... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 5, an equilateral $\triangle ABC$ with side length 26 is inscribed in a circle, and chord $DE \parallel BC$. The extension of $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respectively. If the lengths of segments $AF$ and $DF$ are both positive integers, then the chord $DE... | 4. 16 .
Let $A F=x, D F=y$. Then
$$
B F=x-26, D E=x-2 y, E F=x-y \text {. }
$$
By the secant theorem, we have
$$
\begin{array}{l}
A F \cdot B F=D F \cdot E F \\
\Rightarrow x(x-26)=y(x-y) \\
\Rightarrow x^{2}-(26+y) x+y^{2}=0 .
\end{array}
$$
Since $A F$ and $D F$ are both positive integers, we have
$$
\Delta=(26+y)... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Write out a sequence of consecutive positive integers starting from 1, then erase one of the numbers, so that the average of the remaining numbers is $43 \frac{14}{17}$. What is the number that was erased? | Three, suppose we have written $n$ consecutive positive integers 1, 2, ..., $n$. If the number $k$ is erased, then
$$
\begin{array}{l}
\frac{(1+2+\cdots+n)-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2}+\frac{n-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2} \leqslant 43 \frac{14}{17} \leqslant \frac{n}{2}+1 \... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $f(x)$ is a function defined on $\mathbf{R}$. If $f(0)=0$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant x^{2}, \\
f(x+16)-f(x) \geqslant 4 x^{2}+48 x+224,
\end{array}
$$
then $f(64)=$ $\qquad$ | 5. 19840.
Notice that,
$$
\begin{array}{l}
f(x+4)-f(x) \\
=(f(x+16)-f(x))-(f(x+16)- \\
f(x+12))-(f(x+12)-f(x+8))- \\
(f(x+8)-f(x+4)) \\
\geqslant\left(4 x^{2}+48 x+224\right)-(x+12)^{2}- \\
(x+8)^{2}-(x+4)^{2} \\
= x^{2} .
\end{array}
$$
Since $f(x+4)-f(x) \leqslant x^{2}$, we have,
$$
\begin{array}{l}
f(x+4)-f(x)=x... | 19840 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, $f(0)=2$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(5+2 x)=f(-5-4 x), \\
f(3 x-2)=f(5-6 x) .
\end{array}
$$
Find the value of $f(2012)$. | In equation (1), let $x=\frac{1}{2} y-\frac{3}{2}(y \in \mathbf{R})$, we get $f(2+y)=f(1-2 y)$.
In equation (2), let $x=\frac{1}{3} y+\frac{2}{3}(y \in \mathbf{R})$, we get $f(y)=f(1-2 y)$.
Thus, $f(2+y)=f(y)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2012)=f(2 \times 1006+0)=f(0)=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $x, y$ be real numbers. Then the minimum value of the algebraic expression
$$
2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5
$$
is $\qquad$ [1]
(2005, National Junior High School Mathematics League Wuhan CASIO Cup Selection Competition) | $$
\begin{array}{l}
\text { Original expression }= x^{2}+4 x y+4 y^{2}+x^{2}-4 x+4+ \\
y^{2}+2 y+1-10 \\
=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10 .
\end{array}
$$
Therefore, when $x=2, y=-1$, the minimum value of the required algebraic expression is -10. | -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
(2011, National High School Mathematics League Sichuan Province Preliminary Contest) | First, use the discriminant method to find $g(m)$.
Notice that the original function is $y-x=\sqrt{100-m x}$. Squaring both sides and rearranging, we get
$$
x^{2}+(m-2 y) x+y^{2}-100=0 \text {. }
$$
By $\Delta \geqslant 0$, we have $y \leqslant \frac{m}{4}+\frac{100}{m}$.
Thus, $g(m)=\frac{m}{4}+\frac{100}{m}$
$$
\geq... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4: From the numbers $1, 2, \cdots, 2012$, select a set of numbers such that the sum of any two numbers cannot be divisible by their difference. How many such numbers can be selected at most?
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities) | Solve: Divide $1,2, \cdots, 2012$ into
$$
\begin{array}{l}
(1,2,3),(4,5,6), \cdots, \\
(2008,2009,2010),(2011,2012)
\end{array}
$$
these 671 groups.
If at least 672 numbers are taken, then by the pigeonhole principle, there must be two numbers in the same group, let's say $a$ and $b$ ($a>b$).
Thus, $a-b=1$ or 2.
When ... | 671 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given that $a$ and $b$ are real numbers, and
$$
a^{2}+a b+b^{2}=3 \text {. }
$$
If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. ${ }^{[3]}$
$(2008$, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | Let $a^{2}-a b+b^{2}=k$.
From $\left\{\begin{array}{l}a^{2}+a b+b^{2}=3 \\ a^{2}-a b+b^{2}=k,\end{array}\right.$, we get $a b=\frac{3-k}{2}$.
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$=3+\frac{3-k}{2}=\frac{9-k}{2}$.
Since $(a+b)^{2} \geqslant 0$, then $\frac{9-k}{2} \geqslant 0$, which means $k \leqslant 9$.... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-x y+y^{2}$ is $\qquad$ [4]
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Solve: Completing the square for $x^{2}-x y+y^{2}$ yields
$$
\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}
$$
When $x$ and $y$ are such that one i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{6}$ are six different positive integers, taking values from $1,2, \cdots, 6$. Let
$$
\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
$$
\begin{aligned}
S \geqslant & \left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
& \left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the function $S=|x-2|+|x-4|$.
(1) Find the minimum value of $S$;
(2) If for any real numbers $x, y$ the inequality
$$
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
$$
holds. It is easy to find that the maximum value of $-y^{2}+2 y$ is 1. Therefore, $0 \leqslant m \leqslant 2$, and the maximum value of $m$ is 2. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) A fly and $k$ spiders are placed at some intersections of a $2012 \times 2012$ grid. An operation consists of the following steps: first, the fly moves to an adjacent intersection or stays in place, then each spider moves to an adjacent intersection or stays in place (multiple spiders can occupy the s... | The minimum value of $k$ is 2.
(1) First, prove that a single spider cannot catch the fly.
Establish a Cartesian coordinate system, then the range of the spider and fly's movement is
$$
\{(x, y) \mid 0 \leqslant x, y \leqslant 2012, x, y \in \mathbf{N}\}.
$$
For each point in the above set, there are at least two poi... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Sets $S_{1}, S_{2}, \cdots, S_{n}$ are pairwise distinct and satisfy the following conditions:
(1) $\left|S_{i} \cup S_{j}\right| \leqslant 2004(1 \leqslant i, j \leqslant n, i, j \in \mathbf{N}_{+})$;
(2) $S_{i} \cup S_{j} \cup S_{k}=\{1,2, \cdots, 2008\}(1 \leqslant i < j < k \leqslant n, i, j, k \in \mathb... | Let the complement of $S_{i}$ in $Y=\{1,2, \cdots, 2008\}$ be $A_{i}(1 \leqslant i \leqslant n)$. Then
$\left|A_{i} \cap A_{j}\right| \geqslant 4, A_{i} \cap A_{j} \cap A_{k}=\varnothing$.
Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$.
Consider the bipartite graph $X+Y$, where a vertex $A_{i} \in X$ is adjacent to... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 There are 10 sets of test papers, each set containing 4 questions, and at most one question is the same between any two sets. Among these test papers, what is the minimum number of different questions?
(2005, Taiwan Mathematical Olympiad) | Consider each test paper as a vertex, these vertices form a set $X$; consider each question as a vertex, these vertices form a set $Y$.
If a test paper $x \in X$ contains a question $y \in Y$, connect a line between the corresponding vertices $x$ and $y$. This results in a bipartite graph $X+Y$.
Let $Y$ have $n$ vert... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given any 2012 points in the plane. Prove: the distances between each pair of them take at least 32 different values. | Prove that with these 2012 points as vertices, connecting edges between points of the same distance, we obtain a graph $G$.
Since there is at most one point that is equidistant from three given points, the graph $G$ does not contain $K_{2,3}$.
By Theorem 2, the number of edges in graph $G$ is
$$
m \leqslant \frac{2012... | 32 | Combinatorics | proof | Yes | Yes | cn_contest | false |
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 |
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 |
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
$$
\begin{array}{l}
\alpha \beta=-36 \\
\Rightarrow(|\alpha|,|\beta|) \\
=(1,36)(2,18),(3,12),(4,9),(6,6) \\
\Rightarrow m= \pm 35, \pm 16, \pm 9, \pm 5,0 \\
\Rightarrow M=\{0\} \cup\{-5,5\} \cup\{-9,... | 31 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 6 Let $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can always be found two elements $a, b$ such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$. ${ }^{[2]}$
(2011, China Western Mathematical Olympiad) | Solve for when
$$
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 |
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 |
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 |
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 |
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