problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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4. Let $D$, $E$, and $F$ be points on the sides $BC$, $AB$, and $AC$ of $\triangle ABC$, respectively, such that $AE = AF$, $BE = BD$, and $CF = CD$. Given that $AB \cdot AC = 2BD \cdot DC$, $AB = 12$, and $AC = 5$. Then the inradius $r$ of $\triangle ABC$ is $\qquad$ | 4.2.
As shown in Figure 6, let
$$
\begin{array}{l}
A E=A F=x, \\
B E=B D=y, \\
C D=C F=z .
\end{array}
$$
Then $A B=x+y$,
$$
\begin{array}{l}
A C=x+z, \\
B C=y+z .
\end{array}
$$
From $A B \cdot A C=2 B D \cdot D C$
$$
\begin{array}{l}
\Rightarrow(x+y)(x+z)=2 y z \\
\Rightarrow x^{2}+x y+x z=y z \\
\text { Also } A ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) In the set of numbers $1,2, \cdots, 2009$, what is the maximum number of numbers that can be selected such that the sum of any two selected numbers is divisible by 100? | Three, let the $n$ numbers that meet the conditions be,
$$
a_{1}, a_{2}, \cdots, a_{n} \text {, }
$$
Take any three of these numbers (let them be $a_{k} \backslash a_{m} \backslash a_{f}$). Then
$$
\begin{array}{l}
a_{k}+a_{m}=100 k_{1}, \\
a_{k}+a_{4}=100 k_{2}, \\
a_{m}+a_{4}=100 k_{3},
\end{array}
$$
where $k_{1} ... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive real numbers $a$, $b$, $c$ satisfy
$$
(1+a)(1+b)(1+c)=8 \text {. }
$$
Then the minimum value of $a b c+\frac{9}{a b c}$ is | -1.10 .
$$
\begin{array}{l}
\text { Given } 8=(1+a)(1+b)(1+c) \\
\geqslant 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}=8 \sqrt{a b c} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$
Equality holds if and only if $a=b=c=1$.
It is easy to verify that the function
$$
f(x)=x+\frac{9}{x}
$$
is decreasing on $(0,3)$.
Si... | 10 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a finite arithmetic sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1$, and a common difference of 2, the arithmetic mean of all its terms is 2011. If one term is removed, the arithmetic mean of the remaining terms is an integer. Then the number of ways to remove a term is $\qquad$. | 4.3.
According to the problem, we have
$$
\frac{1}{n}\left[n+\frac{n(n-1)}{2} \times 2\right]=2011 \text {. }
$$
Solving this, we get $n=2011$.
Thus, the sum of all terms in the sequence is $2011^{2}$.
Suppose that after removing the $k$-th term from the sequence, the arithmetic mean of the remaining terms is an inte... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 2, in plane $\alpha$, $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are on opposite sides of line $l$, with no common points with $l$, and are symmetric about line $l$. Now, if plane $\alpha$ is folded along line $l$ to form a right dihedral angle, then the six points $A, B, C, A_{1}, B_{1},... | 7.11.
Notice that after the fold, the three sets of four points
$$
\left(A, B, A_{1}, B_{1}\right) γ\left(B, C, B_{1}, C_{1}\right) γ\left(C, A, C_{1}, A_{1}\right)
$$
are all coplanar, therefore, these six points can determine
$$
C_{6}^{3}-3\left(C_{4}^{3}-1\right)=11 \text { (planes). }
$$ | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) A company printed a batch of T-shirts, each T-shirt can have three different colors: red, yellow, and blue, and four different patterns. Now, this batch of T-shirts is to be distributed to $n$ new employees, with each employee receiving exactly 4 T-shirts with different patterns. Try to find the minim... | Four, the minimum value of $n$ is 19.
When $n=18$, the answer scenario shown in Table 1 does not meet the requirements.
[Note] In Table 1, (1), (2), (3), (4) are patterns, $A_{1}, A_{2}, \cdots, A_{18}$ are members, and $A$, $B$, $C$ represent red, yellow, and blue colors, respectively.
The proof below shows that when ... | 19 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In $\triangle A B C$, it is known that $A B=A C=2$, and there are 100 different points $P_{1}, P_{2}, \cdots, P_{100}$ on side $B C$. Let $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C(i=1,2, \cdots, 100)$.
Find the value of $m_{1}+m_{2}+\cdots+m_{100}$. | Solve As shown in Figure 2, since $\triangle A B C$ is an isosceles triangle, applying the property we get
$$
\begin{aligned}
& A P_{i}^{2} \\
= & A B^{2}-B P_{i} \cdot P_{i} C .
\end{aligned}
$$
Therefore, $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C=A B^{2}=4$.
Thus, $m_{1}+m_{2}+\cdots+m_{100}=400$. | 400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given points $A, B, C, D$ lie on the same circle, and $BC = DC = 4, AC$ intersects $BD$ at point $E, AE = 6$. If the lengths of segments $BE$ and $DE$ are both integers, find the length of $BD$. | Apply the property to $\triangle B C D$ to get
$$
C E^{2}=C D^{2}-B E \cdot E D=16-6 \cdot E C \text {. }
$$
Solving, we get $E C=2$.
From $B E \cdot E D=A E \cdot E C=12$, and
$$
B D<B C+C D=8 \text {, }
$$
we solve to get $B D=4+3=7$. | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. A circle is filled with 12 positive integers, each taken from $\{1,2, \cdots, 9\}$ (each number can appear multiple times on the circle). Let $S$ represent the sum of all 12 numbers on the circle. If the sum of any three consecutive numbers on the circle is a multiple of 7, then the number of possible values for $S$... | 4.9.
For any three consecutive numbers $a_{k}, a_{k+1}, a_{k+2}$ on a circle, $a_{k}+a_{k+1}+a_{k+2}$ can be 7, 14, or 21.
For any four consecutive numbers on the circle, if they are $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$, since
$$
a_{k}+a_{k+1}+a_{k+2} \text { and } a_{k+1}+a_{k+2}+a_{k+3}
$$
are both multiples of 7, i... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \text { and } \\
a_{n+2}=\left|a_{n+1}\right|-a_{n}
\end{array}\left(n \in \mathbf{N}_{+}\right) .
$$
Let $\left\{a_{n}\right\}$'s sum of the first $n$ terms be $S_{n}$. Then $S_{100}=$ | - 1. 89.
From the given, $a_{k+9}=a_{k}$.
Then $S_{100}=a_{1}+11\left(a_{1}+a_{2}+\cdots+a_{9}\right)=89$. | 89 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $n<100$. Then the largest integer $n$ such that the expansion of $(a+b)^{n}$ has three consecutive terms with coefficients in arithmetic progression is $\qquad$ . . . | 3. 98.
Let the coefficients of three consecutive terms in the expansion of $(a+b)^{n}$ be $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1} (1 \leqslant k \leqslant n-1)$.
By the problem, we have $2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1}$.
Expanding and rearranging according to the... | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Among the positive integers less than 20, each time three numbers are taken without repetition, so that their sum is divisible by 3. Then the number of different ways to do this is $\qquad$ . | 4.327.
Divide these 19 numbers into three categories based on the remainder when divided by 3:
$$
\begin{array}{l}
A_{1}: 3,6,9,12,15,18 ; \\
A_{2}: 2,5,8,11,14,17 ; \\
A_{3}: 1,4,7,10,13,16,19 .
\end{array}
$$
Thus, the number of ways to satisfy the conditions of the problem are only four scenarios.
(1) Choose any t... | 327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Place 10 numbers on a given circle such that their total sum is 200, and the sum of any three consecutive numbers is not less than 58. Then the maximum value of the largest number among all sets of 10 numbers that satisfy the above requirements is $\qquad$ | 8. 26 .
Let the maximum number in all placements be $A$. Then
$$
A+3 \times 58 \leqslant 200 \Rightarrow A \leqslant 26 \text {. }
$$
In fact, $26,6,26,26,6,26,26,6,26,26$ satisfies. | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Let
$$
P=x^{4}+6 x^{3}+11 x^{2}+3 x+31 \text {. }
$$
Find the integer value(s) of $x$ that make $P$ a perfect square. | 10. Since $P=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, therefore, when $x=10$, $P=131^{2}$ is a perfect square.
Next, we only need to prove: there are no other integer $x$ that satisfy the requirement.
(1) $x>10$.
If $P0$, then $P>\left(x^{2}+3 x\right)^{2}$.
Therefore, $\left(x^{2}+3 x\right)^{2}\left(x^{2}+3 x+1\right)... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1}-2 a_{n}=2^{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the smallest positive integer $n$ for which $a_{n}>10$ holds is $\qquad$ . . | 3.3.
Given $a_{n+1}-2 a_{n}=2^{n+1}$, we know $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=1$.
Thus, the sequence $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with a common difference of 1. Also, $a_{1}=2$, so $\frac{a_{n}}{2^{n}}=n \Rightarrow a_{n}=n \times 2^{n}$. Therefore, $\left\{a_{n}\right\}$... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{0,1, \cdots, 6\}(i=0,1,2,3)$, and $a_{3} \neq 0$.
If positive integers $m γ n \in A$, and $m+n=2010(m>n)$, then the number of positive integers $m$ that satisfy the condition is $\q... | 6. 662 .
According to the problem, we know that $m$ and $n$ are four-digit numbers in base 7, and the largest four-digit number in base 7 is
$$
6 \times 7^{3}+6 \times 7^{2}+6 \times 7+6=2400,
$$
the smallest one is $1 \times 7^{3}=343$.
Since $m+n=2010(m>n)$, therefore,
$$
1006 \leqslant m \leqslant 1667 \text {. }
... | 662 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange the real solutions of the equation $x^{3}-3[x]=4$ in ascending order to get $x_{1}, x_{2}, \cdots, x_{k}$. Then the value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}$ is $\qquad$ ( $[x]$ denotes the greatest integer less than or equal to the real number $x$). | 8. 15 .
Notice that $x-1<[x] \leqslant x$.
Therefore, when $x \geqslant 3$,
$$
\begin{array}{l}
x^{3}-3[x] \geqslant x^{3}-3 x=x\left(x^{2}-3\right) \\
\geqslant 3 \times 6=18 ;
\end{array}
$$
When $x \leqslant-3$,
$$
\begin{array}{l}
x^{3}-3[x]<x^{3}-3(x-1)=x\left(x^{2}-3\right)+3 \\
\leqslant-3 \times 6+3=-15 .
\end... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If the positive integer $m$ makes it true that for any set of positive numbers $a_{1} γ a_{2} γ a_{3} γ a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positiv... | 9.3.
Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
It is verified that $m=1, m=2$ do not meet t... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. There are three numbers arranged in sequence: $3, 9, 8$. For any two adjacent numbers, the difference between the right number and the left number is written between these two numbers, resulting in a new sequence $3, 6, 9, -1, 8$, which is called the first operation; after the second similar operation, a new sequenc... | For convenience, let the sequence of $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, the newly added numbers are $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}$. Therefore, the sum of the newly added numbers is
$$
\begin{array}{l}
\left(a_{2}-a_{1}\right) + \left(a_{3}-a_{2}\right) + \cdots + ... | 520 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
19. There are 10 red, 10 black, and 10 white balls. Now, all of them are to be placed into two bags, A and B, with the requirement that each bag must contain balls of all three colors, and the product of the number of balls of each color in bags A and B must be equal. How many ways are there to do this? | 19. Let the number of red, black, and white balls in bag A be $x$, $y$, and $z$ respectively. Then $1 \leqslant x, y, z \leqslant 9$, and
$$
x y z=(10-x)(10-y)(10-z) \text {, }
$$
i.e., $x y z=500-50(x+y+z)+5(x y+y z+z x)$.
Thus, $5 \mid x y z$.
Therefore, one of $x, y, z$ must be 5.
Assume $x=5$, substituting into eq... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
20. Let the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1), \operatorname{Rt} \triangle A B C$ with $A(0,1)$ as the right-angle vertex, and sides $A B, B C$ intersecting the ellipse at points $B, C$. If the maximum area of $\triangle A B C$ is $\frac{27}{8}$, find the value of $a$. | 20. Let $l_{A B}: y=k x+1(k>0)$. Then $l_{A C}: y=-\frac{1}{k} x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ \frac{x^{2}}{a^{2}}+y^{2}=1,\end{array}\right.$,
$\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0$
$\Rightarrow x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$.
Thus, $|A B|=\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. If the arithmetic mean of two positive numbers is $2 \sqrt{3}$, and the geometric mean is $\sqrt{3}$, what is the difference between these two numbers? | 1. Let $x, y$ represent two numbers. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3
\end{array}\right. \\
\Leftrightarrow|x-y|=\sqrt{(x-y)^{2}} \\
=\sqrt{(x+y)^{2}-4 x y}=6,
\end{array}
$$
which means the difference between the two numbers is 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. It is known that fresh shiitake mushrooms contain $90 \% \sim 99 \%$ water, while dried shiitake mushrooms contain $30 \% \sim 45 \%$ water. Then, under the influence of drying, by what maximum factor can the weight of fresh shiitake mushrooms be reduced? | 2. Let the weights of fresh mushrooms, baked mushrooms be $m_{1} \mathrm{~g}, m_{2} \mathrm{~g}$, and the weight of dried mushrooms be $x \mathrm{~g}$ (unknown). Then the range of the proportion of dried mushrooms in fresh mushrooms and baked mushrooms is
$$
\begin{array}{l}
0.01=1-0.99 \\
\leqslant \frac{x}{m_{1}} \le... | 70 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $A$, circle $\Gamma$ is externally tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at points $B$ and $C$ respectively. Extend the chord $B A$ of circle $\Gamma_{1}$ to intersect circle $\Gamma_{2}$ at point $D$, extend the chord $C A$ of circle... | 5. First, prove that quadrilateral $B C G F$ is a rectangle:
In fact, let $K L, B M, C N$ be the common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, $\Gamma_{1}$ and $\Gamma$, $\Gamma_{2}$ and $\Gamma$ (Figure 2). According to the inscribed angle and the angle between a tangent and a chord, we have
$$
\angle A B... | 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Let $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ be a function, and for any positive integers $m, n$, we have
$$
f(f(m)+f(n))=m+n .
$$
Find the value of $f(2011)$. | Since $f(x)$ is a function from the set of positive integers to the set of positive integers, let
$$
f(1)=p\left(p \in \mathbf{N}_{+}\right) \text {. }
$$
If $p>1$, then $p \geqslant 2$.
$$
\text { Let } p=1+b\left(b \in \mathbf{N}_{+}\right), f(b)=c\left(c \in \mathbf{N}_{+}\right) \text {. }
$$
On one hand,
$$
\beg... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic function $y=x^{2}+b x+c$, the vertex of its graph is $D$, and it intersects the positive x-axis at points $A$ and $B$ from left to right, and the positive y-axis at point $C$. If $\triangle A B D$ and $\triangle O B C$ are both isosceles right triangles (where $O$ is the origin), then $b+2 c=$ | 2. 2 .
From the given information, we have
$$
\begin{array}{l}
C(0, c) γ A\left(\frac{-\dot{b}-\sqrt{b^{2}-4 c}}{2}, 0\right), \\
B\left(\frac{-b+\sqrt{b^{2}-4 c}}{2}, 0\right), D\left(-\frac{b}{2},-\frac{b^{2}-4 c}{4}\right) .
\end{array}
$$
Draw $D E \perp A B$ at point $E$. Then $2 D E=A B$,
which means $2 \times ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The value of the positive integer $n$ that makes $2^{n}+256$ a perfect square is $\qquad$ | 3. 11.
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$, if it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a natural number). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, then $k=1$, thus, $n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2, given that $A B$ is the diameter of $\odot O$, chord $C D$ intersects $A B$ at point $E$, a tangent line through $A$ intersects the extension of $C D$ at point $F$, and $D$ is the midpoint of $E F$. If $D E=\frac{3}{4} C E, A C=$ $8 \sqrt{5}$, then $A B=$ . $\qquad$ | 4.24.
Let $C E=4 x, A E=y$. Then
$$
D F=D E=3 x, E F=6 x \text {. }
$$
Connect $A D, B C$.
Since $A B$ is the diameter of $\odot O$ and $A F$ is the tangent of $\odot O$, we have
$$
\angle E A F=90^{\circ}, \angle A C D=\angle D A F \text {. }
$$
Since $D$ is the midpoint of the hypotenuse $E F$ of the right triangl... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}
$$
The value equals $\qquad$ [5]
(2009, Beijing Middle School Mathematics Competition (Grade 8)) | Solution 1
$$
p=\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \text {. }
$$
Then $p^{2}=2 x+2 \sqrt{x^{2}-(2 \sqrt{x-1})^{2}}$
$$
=2 x+2 \sqrt{(x-2)^{2}} \text {. }
$$
Since $1 \leqslant x \leqslant 2$, therefore,
$$
\begin{array}{l}
p^{2}=2 x+2 \sqrt{(x-2)^{2}} \\
=2 x+2(2-x)=4 .
\end{array}
$$
And $p \geqslant 0$, s... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $f(x)$ be a continuous even function, and when $x>0$, $f(x)$ is a strictly monotonic function. Then the sum of all $x$ that satisfy $f(x) = f\left(\frac{x+3}{x+4}\right)$ is ( ).
(A) -3
(B) -8
(C) 3
(D) 8 | 8. B.
When $f(x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $x=\frac{x+3}{x+4}$, we get $x^{2}+3 x-3=0$. At this point,
$$
x_{1}+x_{2}=-3 \text {. }
$$
Since $f(x)$ is a continuous even function, another scenario is $f(-x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $-x=\frac{x+3}{x+4}$, which gives
$$
x^{2}+5 x+3=0 \text {. }
... | -8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 6 Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ [6] $(2008$, National Junior High School Mathematics Competition) | Solve: Given $a=\frac{\sqrt{5}-1}{2} \Rightarrow 2 a+1=\sqrt{5}$.
Square both sides and simplify to get
$$
a^{2}+a=1 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a(a+1)(a-1)} \\
=\frac{1-a^{3}}{\left(a^{2}+a\right)(a-1)} \\
=-\... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. If the constant term in the expansion of $\left(a \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{6}$ is -160, then $\int\left(3 x^{2}-1\right) \mathrm{d} x=$. $\qquad$ | $$
\begin{array}{l}
T_{r+1}=\mathrm{C}_{6}^{r}(a \sqrt{x})^{6-r}\left(-\frac{1}{\sqrt{x}}\right)^{r} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{\frac{6-r}{2}-\frac{r}{2}} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{3-r} .
\end{array}
$$
Let $3-r=0$. Then $r=3$.
The constant term is
$$
\begin{array}{l}
-\mathrm{C}_{6}^{3}... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a=\sqrt{5}-1$. Then the value of $2 a^{3}+7 a^{2}-2 a$ -12 is $\qquad$ [7]
(2010) "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | Given $a=\sqrt{5}-1 \Rightarrow a+1=\sqrt{5}$. Squaring both sides, we get
$$
a^{2}+2 a=4 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=2 a\left(a^{2}+2 a\right)+3 a^{2}-2 a-12 \\
=3 a^{2}+6 a-12 \\
=3\left(a^{2}+2 a\right)-12=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $M=(5+\sqrt{24})^{2 n}\left(n \in \mathbf{N}_{+}\right), N$ be the fractional part of $M$. Then the value of $M(1-N)$ is $\qquad$ | 10.1.
Since $(5+\sqrt{24})^{2 n}+(5-\sqrt{24})^{2 n}$ is a positive integer, and $0<(5-\sqrt{24})^{2 n}<1$, therefore,
$$
\begin{array}{l}
N=1-(5-\sqrt{24})^{2 n} \\
M(1-N)=(5+\sqrt{24})^{2 n}(5-\sqrt{24})^{2 n}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. As shown in Figure $2, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$. | 16. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), \( C(0, c) \), and assume \( b > c \). The line \( l_{P B} \) is given by \( y - b = \frac{y_{0} - b}{x_{0}} x \), which can be rewritten as
\[
\left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0.
\]
The distance from the circle's center \((1, 0)\) to \( P B \) is ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x$ be a positive integer, and $x<50$. Then the number of $x$ such that $x^{3}+11$ is divisible by 12 is $\qquad$. | $$
\begin{array}{l}
x^{3}+11=(x-1)\left(x^{2}+x+1\right)+12 \\
=(x-1) x(x+1)+(x-1)+12 .
\end{array}
$$
Notice that $6 \mid(x-1) x(x+1)$, so $6 \mid(x-1)$;
also, $2 \mid \left(x^{2}+x+1\right)$, hence $12 \mid (x-1)$.
Since $x<50$, we have
$$
x-1=0,12,24,36,48 \text {. }
$$
Thus, $x=1,13,25,37,49$, a total of 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation
$$
x^{2}-2009 x+2011=0
$$
real numbers $m$ and $n$ satisfy
$$
\begin{array}{l}
2009 m x_{1}+2009 n x_{2}=2009, \\
2010 m x_{1}+2010 n x_{2}=2010 .
\end{array}
$$
Then $2011 m x_{1}+2011 n x_{2}=$ $\qquad$ | 2. -2009 .
From the given, we know
$$
\begin{array}{l}
=2010 \times 2009-2011 \times 2009=-2009 . \\
\end{array}
$$ | -2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 5, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and $G$ is the midpoint of $C D$. A line through $G$ intersects $A C$ and $B C$ at points $P$ and $Q$, respectively. Then the value of $\frac{C A}{C P}+\frac{C B}{C Q}$ is $\qquad$. | 4. 4 .
As shown in Figure 12, draw $A E / / P Q$ intersecting $C D$ at point $E$, and $B F / / P Q$ intersecting the extension of $C D$ at point $F$. Then $A E / / B F$.
Since $A D = B D$, therefore, $E D = F D$.
$$
\begin{array}{c}
\text { Also, } \frac{C A}{C P} = \frac{C E}{C G}, \\
\frac{C B}{C O} = \frac{C F}{C ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the set $M$ is a subset of $\{1,2, \cdots, 2011\}$, and the sum of any four elements in $M$ cannot be divisible by 3. Then $|M|_{\text {max }}=$ $\qquad$ | - 1. 672.
Consider the set $A=\{3,6,9, \cdots, 2010\}$,
$$
\begin{array}{l}
B=\{1,4,7, \cdots, 2011\}, \\
C=\{2,5,8, \cdots, 2009\} .
\end{array}
$$
If $M \cap A \neq \varnothing$, then $|M \cap B|<3,|M \cap C|<3,|M \cap A|<4$. Therefore, $|M|<10$.
Now assume $M \cap A=\varnothing$.
If $|M \cap B| \geqslant 2$, then ... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If $n \in \mathbf{N}, n \geqslant 2, a_{i} \in\{0,1, \cdots, 9\}$
$$
\begin{array}{l}
(i=1,2, \cdots, n), a_{1} a_{2} \neq 0 \text {, and } \\
\sqrt{a_{1} a_{2} \cdots a_{n}}-\sqrt{a_{2} a_{3} \cdots a_{n}}=a_{1}, \\
\end{array}
$$
then $n=$ $\qquad$, where $\overline{a_{1} a_{2} \cdots a_{n}}$ is the $n$-digit num... | 2. 2 .
Let $x=\overline{a_{2} a_{3} \cdots a_{n}} \in \mathbf{N}_{+}$. Then $\overline{a_{1} a_{2} \cdots a_{n}}=10^{n-1} a_{1}+x$.
By the given condition, $\sqrt{10^{n-1} a_{1}+x}-\sqrt{x}=a_{1}$. Therefore, $10^{n-1} a_{1}+x=x+a_{1}^{2}+2 a_{1} \sqrt{x}$.
Thus, $10^{n-1}-2 \sqrt{x}=a_{1}$.
If $n \geqslant 3$, then
$... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. For any positive integer $n$, let $a_{n}$ be the smallest positive integer such that $n \mid a_{n}$!. If $\frac{a_{n}}{n}=\frac{2}{5}$, then $n=$ $\qquad$ . | 4.25.
From $\frac{a_{n}}{n}=\frac{2}{5} \Rightarrow a_{n}=\frac{2 n}{5} \Rightarrow 51 n$.
Let $n=5 k\left(k \in \mathbf{N}_{+}\right)$. If $k>5$, then $5 k \mid k!$.
Thus, $a_{\mathrm{n}} \leqslant k<\frac{2 n}{5}$, a contradiction.
Clearly, when $k=2,3,4$, $a_{n} \neq \frac{2 n}{5}$.
Also, $a_{25}=10=\frac{2}{5} \ti... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. On a straight line, three points $A$, $B$, and $C$ are arranged in sequence, and $A B=6, A C=24, D$ is a point outside the line, and $D A$ $\perp A B$. When $\angle B D C$ takes the maximum value, $A D=$ $\qquad$ . | 5. 12 .
Let $\angle B D C=\theta\left(\theta<90^{\circ}\right), \triangle B C D$'s circumcircle $\odot O$ has a radius of $R$. Then $\sin \theta=\frac{B C}{2 R}$.
When $R$ decreases, $\theta$ increases.
Therefore, when $\odot O$ is tangent to $A D$ at point $D$, $\theta$ is maximized.
At this time, $A D^{2}=A B \cdot ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that on the side $C A$ of $\angle A C B$ there are 2011 points $A_{1}, A_{2}, \cdots, A_{2011}$, and on the side $C B$ there are 2011 points $B_{1}, B_{2}, \cdots, B_{2011}$, satisfying
$$
\begin{array}{l}
A_{1} A_{2}=A_{2} A_{3}=\cdots=A_{2010} A_{2011}, \\
B_{1} B_{2}=B_{2} B_{3}=\cdots=B_{2010} B_{2011} .
\... | 7.2010 .
$$
\begin{array}{l}
\text { Let } \angle A C B=\alpha, C A_{1}=a, C B_{1}=b, \\
A_{i} A_{i+1}=s, B_{i} B_{i+1}=t(i=1,2, \cdots, 2010), \\
S_{\text {trapezoid } A_{i} A_{i+1} B_{i+2} B_{i}}=S_{i}(i=1,2, \cdots, 2009) . \\
\text { Then } S_{i}=\frac{1}{2}\{[a+(i+1) s][b+(i+1) t]- \\
\quad[a+(i-1) s][b+(i-1) t]\}... | 2010 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given positive integers $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!+c! \text {. }
$$
then $c\left(a^{5}+b^{5}+c^{2}\right)+3=$ $\qquad$ | 8.2011 .
$$
\begin{array}{l}
\text { Given }(a!)(b!)=a!+b!+c! \\
\Rightarrow(a!-1)(b!-1)=c!+1 \text {. }
\end{array}
$$
Assume without loss of generality that $a \geqslant b$. Clearly, $c > a$. Then $(a!-1) \mid(c!+1)$.
Also, $c!+1=\frac{c!}{a!}(a!-1)+\frac{c!}{a!}+1$, so $(a!-1) \left\lvert\,\left(\frac{c!}{a!}+1\rig... | 2011 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. At a concert, there are 20 singers who will perform. For each singer, there is a set of other singers (possibly an empty set) such that he wishes to perform later than all the singers in this set. Question: Is there a way to have exactly 2,010 ways to order the singers so that all their wishes are satisfied? | 1. Such examples exist.
A ranking of singers that satisfies everyone's wishes is called "good".
If for a set of wishes of $k$ singers there exist $N$ good rankings, then $N$ is called "achievable by $k$ singers" (or simply "$k$-achievable").
Next, we prove that 2010 is "20-achievable".
First, we prove a lemma.
Lemma ... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. (50 points) Given that $p$ is a prime number, the fractional part of $\sqrt{p}$ is $x$, and the fractional part of $\frac{1}{x}$ is $\frac{\sqrt{p}-31}{75}$. Find all prime numbers $p$ that satisfy the condition. | 3. Let $p=k^{2}+r$, where $k, r$ are integers, and satisfy $0 \leqslant r \leqslant 2 k$.
Since $\frac{\sqrt{p}-31}{75}$ is the fractional part of $\frac{1}{x}$, we have $0 \leqslant \frac{\sqrt{p}-31}{75} < 1$. Let
$\frac{1}{x}=\frac{1}{\sqrt{p}-k}=N+\frac{\sqrt{p}-31}{75}(N \geqslant 1)$.
Then $\frac{\sqrt{p}+k}{r}=N... | 2011 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. (50 points) A scientist stored the design blueprint of a time machine on a computer, setting the file opening password as a permutation of $\{1,2, \cdots, 64\}$. They also designed a program that, when eight positive integers between 1 and 64 are input each time, the computer will indicate the order (from left to ri... | 8. Prepare $n^{2}(n=8)$ cards, the front side of which are numbered $1,2, \cdots, n^{2}$, and the back side corresponds to the position of the number in the password (counting from the left). Of course, the operator does not know the numbers on the back side in advance.
First, divide the $n^{2}$ cards into $n$ groups ... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the set $M=\{2,0,11\}$. If $A \varsubsetneqq M$, and $A$ contains at least one even number, then the number of sets $A$ that satisfy the condition is $\qquad$ . | 1. 5.
The sets $A$ that satisfy the conditions are $\{2\}, \{0\}, \{2,0\}, \{2,11\}, \{0,11\}$. There are 5 in total. | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζε¦δΈοΌ
5. In the array of numbers shown in Figure 1, the three nu... | 5.18.
From the problem, we have
$$
\begin{array}{l}
a_{11}+a_{13}=2 a_{12}, a_{21}+a_{23}=2 a_{22}, \\
a_{31}+a_{33}=2 a_{32}, a_{12}+a_{32}=2 a_{22} .
\end{array}
$$
Thus, the sum of all nine numbers is $9 a_{n}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{1}{p}, & x=\frac{q}{p} ; \\
0, & x \neq \frac{q}{p},
\end{array}\right.
$$
where $p$ and $q$ are coprime, and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$ and $f(x)>\frac{1}{5}$ is $\qquad$ . | 7.5.
Obviously, $x=\frac{q}{p}$ (otherwise, $f(x)=0$). At this point, from $f(x)=\frac{1}{p}>\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$. When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$.
Therefore, there are 5 values of $x$ that satisfy the condition. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $p$ and $q$ are both prime numbers, and $7p+q$, $2q+11$ are also prime numbers. Then $p^{q}+q^{p}=$ $\qquad$ . | 8. 17 .
Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be an odd number.
Therefore, one of $p$ or $q$ must be even (which can only be 2).
Clearly, $q \neq 2$ (otherwise, $2 q+11=15$, which is not a prime number). Thus, $p=2$.
At this point, $14+q$ and $2 q+11$ are both prime numbers.
If $q=3 k+1\left(k \... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 As shown in Figure 7, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3$, the distance from point $P$ to line $A N$ is $P D=12, A D=$
30, and a line is drawn through $P$ intersecting
$A N$ and $A M$ at points
$B$ and $C$ respectively. Find the minimum
value of the area of... | Solve As shown in Figure 7, it can be proven: when the line moves to $B_{0} P=P C_{0}$, the area of $\triangle A B C$ is minimized.
Draw $C_{0} Q / / A N$, intersecting $C B$ at point $Q$.
Thus, $\triangle P C_{0} Q \cong \triangle P B_{0} B$.
Also, $S_{\triangle P C_{0} C} \geqslant S_{\triangle P C_{0} Q}=S_{\triangl... | 624 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. From the set $\{1,2, \cdots, 10\}$, any two non-adjacent numbers are taken and multiplied. Then the sum of all such products is equal to
| 5. 990 .
Take any two numbers, multiply them, and then find their sum:
$$
\begin{array}{l}
S_{1}=\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \\
=\frac{1}{2} \sum_{k=1}^{10} k(55-k)=\frac{1}{2} \sum_{k=1}^{10}\left(55 k-k^{2}\right) \\
=\frac{1}{2}(55 \times 55-385)=1320,
\end{array}
$$
Among them, t... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given positive integers $a_{1}, a_{2}, \cdots, a_{18}$ satisfying
$$
\begin{array}{l}
a_{1}<a_{2}<\cdots<a_{18}, \\
a_{1}+a_{2}+\cdots+a_{18}=2011 .
\end{array}
$$
Then the maximum value of $a_{9}$ is | 7.193.
To maximize $a_{9}$, $a_{1}, a_{2}, \cdots, a_{8}$ should be as small as possible, and $a_{10}, a_{11}, \cdots, a_{18}$ should be as close to $a_{9}$ as possible. Therefore, we take $a_{1}, a_{2}, \cdots, a_{8}$ to be $1, 2, \cdots, 8$, respectively, with their sum being 36.
Let $a_{9}=n$. Then
$$
\begin{array}... | 193 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4\left(n \in \mathbf{N}_{+}\right)$, let $d_{n}=\left(a_{n}, a_{n+1}\right)$. Then the maximum value of $d_{n}$ is $\qquad$ | 8.433.
Given $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$.
Then $d_{n} \mid\left[-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right)\right]$,
and $\square$
$$
\begin{array}{l}
d_{n} \mid\left(3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}... | 433 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find all odd prime numbers $p$ such that
$$
p \mid \sum_{k=1}^{2011} k^{p-1} .
$$ | If $p>2011$, then for $k(1 \leqslant k \leqslant 2011)$ by Fermat's Little Theorem we have
$k^{p-1} \equiv 1(\bmod p)$.
Thus $\sum_{k=1}^{2011} k^{p-1} \equiv 2011 \not\equiv 0(\bmod p)$, a contradiction.
Therefore, $p \leqslant 2011$.
Let $2011=p q+r(0 \leqslant r < q)$, then if $r>q$, we have $q=r$, so
$$
2011=p q+r=... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The equation concerning $x, y$
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}
$$
has $\qquad$ groups of positive integer solutions $(x, y)$. | 3. 12 .
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}$, we get
$$
\begin{array}{l}
x y-2011 x-2011 y-2011=0 \\
\Rightarrow(x-2011)(y-2011) \\
\quad=2011 \times 2012 \\
=2^{2} \times 503 \times 2011 .
\end{array}
$$
Thus, the positive integer solutions of the original equation are
$$
(2+1)(1+1)(1+1)=12 \t... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. For $n$ consecutive natural numbers, if each number is written in its standard prime factorization form, and each prime factor appears an odd number of times, such $n$ consecutive natural numbers are called a βconsecutive $n$ strange groupβ (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \... | 5.7.
For example: $29, 30, 31, 32, 33, 34, 35$ is a consecutive 7 peculiar group.
Below is the proof: When $n \geqslant 8$, there does not exist a consecutive $n$ peculiar group.
Otherwise, one of them (denoted as $m$) is divisible by 8, so among these $n$ numbers, there must be $m-4$ or $m+4$, which is divisible by... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=(\sqrt{3}+\sqrt{2})^{2 n}\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
Let $b_{n}=a_{n}+\frac{1}{a_{n}}$.
(1) Find the recurrence relation between $b_{n+2} γ b_{n+1} γ b_{n}$;
(2) Find the unit digit of the integer part of $a_{2011}$. | -γ(1) From the given, we have
$$
\begin{array}{l}
b_{n}=(\sqrt{3}+\sqrt{2})^{2 n}+(\sqrt{3}-\sqrt{2})^{2 n} \\
\Rightarrow b_{n}=(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n} .
\end{array}
$$
Then $b_{n+2}=[(5+2 \sqrt{6})+(5-2 \sqrt{6})]$.
$$
\begin{array}{l}
{\left[(5+2 \sqrt{6})^{n+1}+(5-2 \sqrt{6})^{n+1}\right]-} \\
(5+2 \... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, let $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ be a partition of the set $A=\{1,2, \cdots, 29\}$, and the sum of any elements in $A_{i}(i=1,2$, $\cdots, n)$ does not equal 30. Find the minimum possible value of $n$.
[Note] If the non-empty subsets $A_{1}, A_{2}, \cdots, A_{n}$ $\left(n \in \mathbf{N}_{+}, n \geqslan... | The minimum value of $n$ is 3.
First, we decompose
$$
\begin{array}{l}
A_{1}=\{1,2, \cdots, 7\}, \\
A_{2}=\{10,11, \cdots, 15,21,22\}, \\
A_{3}=\{8,9,16,17, \cdots, 20,23,24, \cdots, 29\}
\end{array}
$$
which satisfies the conditions.
Next, we prove that $n=2$ does not satisfy the conditions.
Assume, for contradiction... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 As shown in Figure 6, in the regular nonagon ABCDEFGHI, it is known that $A E=1$. Then the length of $A B + A C$ is
The translation preserves the original text's line breaks and format. | Solving: Since the sum of the interior angles of a regular nonagon is
$$
(9-2) \times 180^{\circ}=1260^{\circ} \text {, }
$$
and each interior angle is $140^{\circ}$, therefore,
$$
\angle C A B=\left(180^{\circ}-140^{\circ}\right) \div 2=20^{\circ} \text {. }
$$
Connecting $A H$, and drawing $H M$ and $G N$ perpendic... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9.3. There is a convex 2011-gon on the blackboard. Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? | 9.3.4016.
Use induction to prove: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. We can sequentially draw $2n-6$ diagonals as follows: $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_... | 4016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let four distinct real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2011}-c^{2011}\right)\left(a^{2011}-d^{2011}\right)=2011, \\
\left(b^{2011}-c^{2011}\right)\left(b^{2011}-d^{2011}\right)=2011 . \\
\text { Then }(a b)^{2011}-(c d)^{2011}=(\quad) .
\end{array}
$$
Then $(a b)^{2011}-(c d)^{2011}=(\quad... | 4. B.
From $a \neq b$, we know $a^{2011} \neq b^{2011}$.
Thus, $a^{2011}$ and $b^{2011}$ are the two distinct real roots of the quadratic equation in $x$:
$$
\left(x-c^{2011}\right)\left(x-d^{2011}\right)=2011,
$$
which is
$$
x^{2}-\left(c^{2011} \div d^{2011}\right) x+(c d)^{2011}-2011=0.
$$
By Vieta's formulas, we g... | -2011 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Figure 2 is a part of the HZ district map. A river runs through the district, with the two banks being the broken lines $A-B-C$ and $D-O-E$, and there are two locations $M$ and $N$. Two bridges perpendicular to the riverbanks and roads are to be built to connect $M$ and $N$ to both banks of the river, making the tot... | 6. B.
Translate point $M(-16,12)$ 3 units to the left to get point $M^{\prime}(-13,12)$, and translate point $N(11,-1)$ 3 units upwards to get point $N^{\prime}(11,2)$. Connect $M^{\prime} N^{\prime}$, which intersects $A B$ and $B C$ at points $P$ and $Q$ respectively.
Then the minimum length of the road is
$$
M^{\pr... | 32 | Geometry | MCQ | Yes | Yes | cn_contest | false |
4. Given a prime number $p$ such that $p^{3}-6 p^{2}+9 p$ has exactly 30 positive divisors. Then the smallest value of $p$ is $\qquad$ . | 4. 23 .
Obviously, when $p=2$ or 3, it does not meet the requirements of the problem.
Therefore, $p>3$.
Also, $p^{3}-6 p^{2}+9 p=p(p-3)^{2}$, at this point,
$(p, p-3)=(p, 3)=1$.
Since $p$ has two factors, $(p-3)^{2}$ has 15 factors.
And $15=5 \times 3$, to make $p$ the smallest, $p-3$ is also even, so it can only be
... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y$ satisfy
$$
17\left(x^{2}+y^{2}\right)-30 x y-16=0 \text {. }
$$
then the maximum value of $f(x, y)=\sqrt{16 x^{2}+4 y^{2}-16 x y-12 x+6 y+9}$ is . $\qquad$ | $-1.7$.
From $17\left(x^{2}+y^{2}\right)-30 x y-16=0$, we get
$$
\begin{array}{l}
(x+y)^{2}+16(x-y)^{2}=16 . \\
\text { Let }\left\{\begin{array}{l}
x+y=4 \cos \theta, \\
x-y=\sin \theta
\end{array}(\theta \in \mathbf{R})\right. \\
\Rightarrow\left\{\begin{array}{l}
x=2 \cos \theta+\frac{1}{2} \sin \theta, \\
y=2 \cos ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A$ be the set of values of $m$ for which the roots of the equation in $x$
$$
2(m+1) x^{2}-\left(m^{2}+m+16\right) x+8 m=0
$$
are both integers. Then $|A|=$
$\qquad$ . | 2. 2 .
If $m=-1$, then $x=-\frac{1}{2}$, which does not meet the requirement. If $m \neq-1$, then we can get $x_{1}=\frac{m}{2}, x_{2}=\frac{8}{m+1}$. According to the problem, $m$ is an even number, and $(m+1) \mid 8$, so $m=0$ or -2.
Therefore, $|A|=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=3, a_{n+2}=3 a_{n+1}-2 a_{n}-1 \text {, }
$$
where, $n \in \mathbf{N}_{+}, S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$. Then the maximum value of $S_{n}$ is $\qquad$ . | 4. 8 .
From the given information, we have
$$
\begin{array}{l}
a_{n+2}-a_{n+1}-1=2\left(a_{n+1}-a_{n}-1\right) \\
=2^{n}\left(a_{2}-a_{1}-1\right)=-2^{n} \\
\Rightarrow a_{n}-a_{n-1}=1-2^{n-2} \\
\Rightarrow a_{n}=n+3-2^{n-1}\left(n \in \mathbf{N}_{+}\right)
\end{array}
$$
Therefore, $a_{3}=2$.
Also, when $n \geqslan... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. From the set $\{1,2, \cdots, 2011\}$, any two different numbers $a, b$ are selected such that $a+b=n$ (where $n$ is some positive integer) with a probability of $\frac{1}{2011}$. Then the minimum value of $ab$ is | 6.2010.
Let the number of ways such that $a+b=n$ be $k$. Then
$$
\frac{k}{\mathrm{C}_{2011}^{2}}=\frac{1}{2011} \Rightarrow k=1005 \text {. }
$$
Consider $a b$ to be as small as possible, and the number of ways such that $a+b=n$ is 1005.
Take $n=2011$. Then
$$
1+2010=2+2009=\cdots=1005+1006 \text {. }
$$
At this poi... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Each point in the plane is colored with one of $n$ colors, while satisfying:
(1) Each color has infinitely many points, and they are not all on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist four points of ... | Given $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arc $\overparen{A B}$ (including point $A$ but not $B$), the arc $\overparen{B C}$ (including point $B$ but not $C$), and the arc $\overparen{C A}$ (including point $C$ but not $A$) with colors 1, 2, and 3, respec... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
---
The above text translated into English, preserving the original text's line breaks and format, is as follows:
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, ... | Solve: It is easy to know that $x=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $d=(a, b), a=d a_{0}, b=d b_{0},\left(a_{0}, b_{0}\right)=1$.
Then $x=\frac{d^{2} a_{0}^{2} b_{0}^{2}}{a_{0}^{2}+b_{0}^{2}} \in \mathbf{N}_{+}$.
By $\left(a_{0}^{2}+b_{0}^{2}, a_{0}^{2} b_{0}^{2}\right)=1$, we know $\left(a_{0}^{2}+b_{0}^{2}\right)... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given 10 points on a circle, color six of them black and the remaining four white. They divide the circumference into arcs that do not contain each other. Rule: arcs with both ends black are labeled with the number 2; arcs with both ends white are labeled with the number $\frac{1}{2}$; arcs with ends of diffe... | Mark all the black points with $\sqrt{2}$, and all the white points with $\frac{1}{\sqrt{2}}$, then the number marked on each arc is exactly the product of the numbers at its two ends. Therefore, the product of the numbers marked on all these arcs is the square of the product of the numbers marked on all the points, i.... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Question 4 Let $P$ be a polynomial of degree $3n$, such that
$$
\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2, \\
P(1)=P(4)=\cdots=P(3 n-2)=1, \\
P(2)=P(5)=\cdots=P(3 n-1)=0 .
\end{array}
$$
$$
\text { If } P(3 n+1)=730 \text {, find } n \text {. }
$$ | Solving, we know
$$
\sum_{k=0}^{3 n+1}(-1)^{k} \dot{\mathrm{C}}_{3 n+1}^{k} P(3 n+1-k)=0,
$$
which means $729+2 \sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{C}_{3 n+1}^{3 j+1}+\sum_{j=0}^{n}(-1)^{3 j} \mathrm{C}_{3 n+1}^{3 j}=0$
Using the multi-section formula, we can find
$$
\begin{array}{l}
\sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $s_{1}, s_{2}, \cdots, s_{n}$, satisfying
$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \cdots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010} .
$$ | 1. Suppose the positive integer $n$ satisfies the condition, and let $s_{1}=39$.
Thus, $n \geqslant 39$.
Below, we provide an example to show that $n=39$ satisfies the condition.
Take 39 different positive integers:
$$
2,3, \cdots, 33,35,36, \cdots, 40,67,
$$
which satisfy the given equation.
In conclusion, the minimu... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Find the smallest positive integer $n$, such that there exist rational coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$ | 3. Since $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$, then $n \leqslant 5$.
Thus, it only needs to be proven that $x^{2}+7$ is not equal to the sum of squares of no more than four rational coefficient polynomials.
Assume there exist four rational coefficient polynomials $f_{1}, f_{2}, f_{3}, f_{4}$ (some of which may be 0... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a_{1}, a_{2}, \cdots$ be integers, for any positive integer $n$ we have
$$
a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] \text {. }
$$
If $2001 a_{199}$, find the smallest positive integer $n(n>1)$, such that $200 \mathrm{l} a_{n}$. | 3. From the given, we have
$$
a_{\mathrm{T99}}=198\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right] \text {. }
$$
From $2001 a_{199}$, we get
$$
100 \left\lvert\, 99\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right]\right. \text {. }
$$
Thus, $a_{2}$ is an even number.
Let $a_{2}=2 m$. Then
$$
1001[(m-1)... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $\triangle A B C$ with the three angles $\angle A, \angle B, \angle C$ opposite to the sides of lengths $a, b, c$, and $\angle A=2 \angle B$. Then $\frac{a^{2}}{b(b+c)}=$ $\qquad$ . | 1. 1.
As shown in Figure 4, it is easy to know,
$$
\begin{array}{l}
\triangle A B C \backsim \triangle D A C \\
\Rightarrow \frac{A B}{D A}=\frac{A C}{D C}=\frac{B C}{A C} .
\end{array}
$$
Let $B D=x, D C=y$. Then
$$
\begin{array}{l}
\frac{c}{x}=\frac{b}{y}=\frac{a}{b} \Rightarrow \frac{b+c}{x+y}=\frac{a}{b} \\
\Righ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $0<a<b<c<d<500$, and $a+d=b+c$. Also, $bc-ad=93$.
Then the number of ordered quadruples of integers ( $\dot{a}, b, c$,
$d)$ that satisfy the conditions is . $\qquad$ | 7.870.
Since $a+d=b+c$, we set
$$
(a, b, c, d)=(a, a+x, a+y, a+x+y) \text {, }
$$
where $x$ and $y$ are integers, and $0<x<y$.
Then $93=b c-a d$
$$
=(a+x)(a+y)-a(a+x+y)=x y \text {. }
$$
Therefore, $(x, y)=(1,93)$ or $(3,31)$.
First case
$$
(a, b, c, d)=(a, a+1, a+93, a+94) \text {, }
$$
where $a=1,2, \cdots, 405$;... | 870 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. The number of prime pairs $(a, b)$ that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is $\qquad$. | 9.2.
If $a=b$, then the equation is equivalent to $a^{2a}=(3a+1)^2$. But $a \mid a^{2a}, a \mid (3a+1)^2$, which is a contradiction, so $a \neq b$.
Assume without loss of generality that $a > b$. Since the difference between the two factors on the right side of the original equation is $a-b$, $a$ can only divide one o... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Each cell of a $2011 \times 2011$ grid is labeled with an integer from $1,2, \cdots, 2011^{2}$, such that each number is used exactly once. Now, the left and right boundaries, as well as the top and bottom boundaries of the grid, are considered the same, forming a torus (which can be viewed as the surface of a "doug... | 6. Let $N=2011$.
Consider a general $N \times N$ table.
When $N=2$, the conclusion is obvious, and the required $M=2$. An example is shown in Table 1.
Table 1
\begin{tabular}{|l|l|}
\hline 1 & 2 \\
\hline 3 & 4 \\
\hline
\end{tabular}
When $N \geqslant 3$, first prove:
$M \geqslant 2 N-1$.
Starting from a state where... | 4021 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.1. A table of numbers consisting of $n$ rows and 10 columns, with each element being an integer from $0 \sim 9$, satisfies the following condition: for any row $A$ and any two columns $B$ and $C$, there exists another row $D$ such that $D$ differs from $A$ only in the numbers in columns $B$ and $C$. Prove: $n \geqsl... | 10.1. Let $R_{0}$ be the first row. Arbitrarily select $2 m$ columns $C_{1}$, $C_{2}, \cdots, C_{2 m}$. By the given condition, there exists a row $R_{1}$, which differs from $R_{0}$ only in $C_{1}$ and $C_{2}$; further, there exists a row $R_{2}$, which differs from $R_{1}$ only in $C_{3}$ and $C_{4}$; $\cdots \cdots$... | 512 | Combinatorics | proof | Yes | Yes | cn_contest | false |
10.2. Nine real-coefficient quadratic polynomials
$$
x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \cdots, x^{2}+a_{9} x+b_{9}
$$
satisfy: The sequences $a_{1}, a_{2}, \cdots, a_{9}$ and $b_{1}, b_{2}, \cdots, b_{9}$ are both arithmetic sequences. It is known that the polynomial obtained by adding these nine polynomials h... | $$
\begin{array}{l}
\text { 10. 2. Let } P_{i}(x)=x^{2}+a_{i} x+b_{i}(i=1,2, \cdots, 9), \\
P(x)=P_{1}(x)+P_{2}(x)+\cdots+P_{9}(x) .
\end{array}
$$
Notice that
$$
P_{i}(x)+P_{10-i}(x)=2 P_{5}(x)=\frac{2}{9} P(x)
$$
has real roots, denoted as $x_{0}$.
Then $P_{i}\left(x_{0}\right)+P_{10-i}\left(x_{0}\right)=0$.
Thus, ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10.7. For any positive integers $a, b (a>b>1)$, define the sequence $x_{n}=\frac{a^{n}-1}{b^{n}-1}(n=1,2, \cdots)$. It is known that the defined sequence does not have $d$ consecutive terms consisting of prime numbers. Find the minimum value of $d$.
untranslated part:
ε°δΈι’ηζζ¬ηΏ»θ―ζθ±ζοΌθ―·δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ
Note: The la... | 10.7. The minimum value of $d$ is 3.
When $a=4, b=2$, $x_{1}=3, x_{2}=5$ are prime numbers.
Therefore, the minimum value of $d$ is greater than 2.
Next, we prove: there cannot be three consecutive terms all being prime numbers.
In fact, a stronger conclusion can be proven:
For $n \geqslant 2, x_{n}$ and $x_{n+1}$ cann... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example $8 A, B, C, D, E$ five people participated in an exam, which has seven questions, all of which are true or false questions. The scoring criteria are: for each question, 1 point is awarded for a correct answer, 1 point is deducted for a wrong answer, and no points are awarded or deducted for unanswered questions... | Let $x_{k}=\left\{\begin{array}{ll}1, & \text { if the conclusion of the } k \text {th question is correct; } \\ -1, & \text { if the conclusion of the } k \text {th question is incorrect, }\end{array}\right.$ where $k=1,2, \cdots, 7$.
At this point, if the conclusion is judged to be correct (i.e., marked with a β $\c... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. If the length, width, and height of a rectangular prism are all prime numbers, and the sum of the areas of two adjacent sides is 341, then the volume of this rectangular prism $V=$ $\qquad$ . | 3. 638.
Let the length, width, and height of the rectangular prism be $x, y, z$. From the problem, we have
$$
\begin{array}{l}
x(y+z)=341=11 \times 31 \\
\Rightarrow(x, y+z)=(11,31),(31,11) .
\end{array}
$$
Since $y+z$ is odd, one of $y, z$ must be 2 (let's assume $z=2$).
Also, $11-2=9$ is not a prime number, so
$$
(... | 638 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let
$$
\begin{array}{l}
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \\
\text { then } f(1)+f(2)+\cdots+f(50)=
\end{array}
$$ | $$
\begin{array}{l}
x^{2}-53 x+196=(x-4)(x-49) . \\
\text { Therefore, when } 4 \leqslant x \leqslant 49, f(x)=0 . \\
\text { Then } f(1)+f(2)+\cdots+f(50) \\
=f(1)+f(2)+f(3)+f(50)=660 \text {. }
\end{array}
$$ | 660 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $t=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$. Then the sum of all real solutions of the equation $(t-1)(t-2)(t-3)=0$ with respect to $x$ is $\qquad$ | 3.4.
Notice that
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}
$$
is a monotonically decreasing function.
When $x=0,1,3$, its values are $3, 2, 1$ respectively, so the three roots of the equation are $x=0,1,3$, and their sum is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Two boxes are filled with black and white balls. The total number of balls in both boxes is 25. Each time, a ball is randomly drawn from each box. The probability that both balls are black is $\frac{27}{50}$, and the probability that both balls are white is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$.... | 5.26.
Let the first box contain $x$ balls, of which $p$ are black, and the second box contain $25-x$ balls, of which $q$ are black. Then
$$
\begin{array}{l}
\frac{p}{x} \cdot \frac{q}{25-x}=\frac{27}{50}, \\
50 p q=27 x(25-x) .
\end{array}
$$
Thus, $x$ is a multiple of 5.
Substituting $x=5$ and $x=10$ into the two e... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The equation $x+y+z=2011$ satisfies $x<y<z$ to prove: the number of solutions $(x, y, z)$ is $\qquad$ groups. | 5. 336005 .
The equation $x+y+z=2011$ has $\mathrm{C}_{2010}^{2}$ sets of positive integer solutions. In each solution, $x, y, z$ cannot all be equal, and there are $3 \times 1005$ sets of solutions where two of $x, y, z$ are equal. Therefore, the number of solutions that meet the requirements is
$$
\frac{C_{2010}^{2}... | 336005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. For any real numbers $x, y, z$ not all zero, we have
$$
\begin{array}{l}
-6 x y + 18 z x + 36 y z . \\
\leqslant k\left(54 x^{2} + 41 y^{2} + 9 z^{2}\right) .
\end{array}
$$
Then the minimum value of the real number $k$ is | 6.1.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{c}
{\left[(x+y)^{2}+(z+x)^{2}+(y+z)^{2}\right]\left(4^{2}+8^{2}+10^{2}\right)} \\
\geqslant[4(x+y)+8(z+x)+10(y+z)]^{2} .
\end{array}
$$
Simplifying, we get $-11r$.
$$
54 x^{2}+41 y^{2}+9 z^{2} \geqslant-6 x y+18 z x+36 y z \text {. }
$$
When $x: y: z=1:... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Given real numbers $a, b, c, d$ satisfy $a^{4}+b^{4}=c^{4}+d^{4}=2011, a c+b d=0$. Find the value of $a b+c d$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given $a c+b d=0 \Rightarrow a c=-b d$.
Raising both sides to the fourth power, we get $a^{4} c^{4}=b^{4} d^{4}$.
Since $c^{4}=2011-d^{4}, b^{4}=2011-a^{4}$, we have,
$$
\begin{array}{l}
a^{4}\left(2011-d^{4}\right)=\left(2011-a^{4}\right) d^{4} \\
\Rightarrow 2011 a^{4}-a^{4} d^{4}=2011 d^{4}-a^{4} d^{4} .
\end{array}... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $C D$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle A B C$, $\odot O$ is its circumcircle, $\odot O_{1}$ is internally tangent to arc $\overparen{A C}$ and tangent to $A B$ and $C D$, with $E$ being the tangency point on side $A B$; $\odot O_{2}$ is internally tangent to arc $\o... | Prompt: From Example 8, we get $A F=A C, B E=B C$. Then prove that $C E$ bisects $\angle A C D$, and $C F$ bisects $\angle D C B$. Thus, $\frac{A E}{E D} \cdot \frac{D F}{F B} \cdot \frac{B C}{C A}=\frac{A C}{C D} \cdot \frac{C D}{C B} \cdot \frac{B C}{C A}=1$. | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
Example 4 As shown in Figure 2, let $M$ be the centroid of $\triangle A B C$, and a line through $M$ intersects sides $A B$ and $A C$ at points $P$ and $Q$, respectively, and
$$
\frac{A P}{P B}=m, \frac{A Q}{Q C}=n \text {. }
$$
Then $\frac{1}{m}+\frac{1}{n}=$ | Solution 1 As shown in Figure 2, draw lines through points $C$ and $B$ parallel to $PQ$ intersecting $AD$ or its extension at points $E$ and $F$. Then,
$$
\begin{array}{l}
\frac{1}{m}=\frac{P B}{A P}=\frac{F M}{A M}, \\
\frac{1}{n}=\frac{Q C}{A Q}=\frac{E M}{A M} .
\end{array}
$$
Since $D$ is the midpoint of side $BC$... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a b+b c+c a+2(a+b+c)=8045, \\
a b c-a-b-c=-2 .
\end{array}\right.
$$
then $a+b+c=$ $\qquad$ | $$
-1.2012 .
$$
Note that
$$
\begin{array}{l}
(a+1)(b+1)(c+1) \\
=a b c+a b+b c+c a+a+b+c+1 \\
=8045+(-2)+1=8044 .
\end{array}
$$
Since $a, b, c$ are positive integers, we have
$$
a+1 \geqslant 2, b+1 \geqslant 2, c+1 \geqslant 2 \text{. }
$$
Therefore, 8044 can only be factored as $2 \times 2 \times 2011$. Hence $a... | 2012 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a class, there are two types of students: one type always lies, and the other type never lies. Each student knows what type the other students are. During a gathering today, each student has to state what type the other students are, and all students together said "liar" 240 times. At a similar gathering yesterda... | 2. 22 .
Consider four possible scenarios:
(1) If student $A$ is a liar and student $B$ is not a liar, then $A$ will say $B$ is a liar;
(2) If student $A$ is a liar and student $B$ is also a liar, then $A$ will say $B$ is not a liar;
(3) If student $A$ is not a liar and student $B$ is also not a liar, then $A$ will say... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. The positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that $n+1$ is four times the sum of the other two divisors. Then $n=$ | 5.95.
Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p \neq q$).
In the first case, all positive divisors are $1, p, p^{2}, p^{3}$, then $1+p^{3}=4\left(p+p^{2}\right)$, but $p \nmid \left(1+p^{3}\right)$, which is a contradictio... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Given a positive integer $n$ that satisfies the following conditions:
(1) It is an eight-digit number, and all its digits are 0 or 1;
(2) Its first digit is 1;
(3) The sum of the digits in the even positions equals the sum of the digits in the odd positions.
How many such $n$ are there? | 10.35.
From the fact that the sum of the digits in the even positions equals the sum of the digits in the odd positions, we know that the number of 1s in the even positions equals the number of 1s in the odd positions.
Since the first digit is fixed, there are only three positions in the odd positions that can be fre... | 35 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. Given real numbers $a, b, c$ satisfy
$$
\frac{a(b-c)}{b(c-a)}=\frac{b(c-a)}{c(b-a)}=k>0,
$$
where $k$ is some constant. Then the greatest integer not greater than $k$ is . $\qquad$ | 12. 0 .
Let the substitution be $x=a b, y=b c, z=c a$. Then
$$
x-z=k(y-x), y-x=k(y-z) \text {. }
$$
Thus, $x-z=k^{2}(y-z)$. Therefore,
$$
\begin{array}{l}
(y-x)+(x-z)+(z-y)=0 \\
\Leftrightarrow(y-z)\left(k^{2}+k-1\right)=0(x \neq y \neq z) \\
\Rightarrow k^{2}+k-1=0 .
\end{array}
$$
Also, $k>0 \Rightarrow k=\frac{-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given that 15 rays share a common endpoint. Question: What is the maximum number of obtuse angles (considering the angle between any two rays to be the one not greater than $180^{\circ}$) that these 15 rays can form? | 15. First, it is explained that constructing 75 obtuse angles is achievable.
The position of the rays is represented by their inclination angles, with 15 rays placed near the positions of $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$. Within each group, the five rays are sufficiently close to each other (as shown in F... | 75 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. (40 points) Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$ | 1. The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0 \text {. }
$$
Let $x=-1$. Then $1-3+6-7+3=0$.
Therefore, $x+1$ is a factor.
$$
\begin{array}{l}
\text { Hence } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0 .
\end{array}
$$
... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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