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Example 4 Prove: $\sin x>x-\frac{x^{3}}{6}\left(x \in\left(0, \frac{\pi}{2}\right)\right)$.
$(2010$, Nankai University Mathematics Talent Class Admission Exam) | Prove that for
$$
f(x)=\sin x-x+\frac{x^{3}}{6}\left(x \in\left(0, \frac{\pi}{2}\right)\right) \text {, }
$$
then $f(0)=0$,
$$
\begin{array}{l}
f^{\prime}(x)=\cos x-1+\frac{1}{2} x^{2} \\
f^{\prime \prime}(x)=-\sin x+x>0\left(x \in\left(0, \frac{\pi}{2}\right)\right) .
\end{array}
$$
Thus, $f^{\prime}(x)$ is monotoni... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 725,127 |
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 | 725,128 |
Example 6 Find the modulus of $2+2 e^{0.4 \pi i}+e^{1.2 \pi i}$.
(2009, Tsinghua University Independent Recruitment Examination) | Notice,
$$
\begin{array}{l}
12+2 \mathrm{e}^{0.4 \pi \mathrm{i}}+\mathrm{e}^{1.2 \pi \mathrm{i}} \\
=\sqrt{\left(2+2 \cos \frac{2 \pi}{5}+\cos \frac{6 \pi}{5}\right)^{2}+\left(2 \sin \frac{2 \pi}{5}+\sin \frac{6 \pi}{5}\right)^{2}} \\
=\sqrt{9+8 \cos \frac{2 \pi}{5}+4 \cos \frac{6 \pi}{5}+4 \cos \frac{4 \pi}{5}} \\
=\s... | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,129 |
Example 7 There are three points on the unit circle
$$
A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right) \text {. }
$$
If $x_{1}+x_{2}+x_{3}=y_{1}+y_{2}+y_{3}=0$, prove:
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=\frac{3}{2} \text {. }
$$
(2011, Peking University Ad... | Let \( x_{1}=\cos \alpha, y_{1}=\sin \alpha \),
\[
\begin{aligned}
x_{2}=\cos \beta, y_{2}=\sin \beta, \\
x_{3}=\cos \gamma, y_{3}=\sin \gamma,
\end{aligned}
\]
where \( 0 \leqslant \alpha<\beta<\gamma<2 \pi \).
From the problem, we have
\[
\begin{array}{l}
-\cos \gamma=\cos \alpha+\cos \beta, \\
-\sin \gamma=\sin \alp... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,130 |
Example 8 Does there exist a real number $x$, such that $\tan x + \sqrt{3}$ and $\cot x + \sqrt{3}$ are both rational numbers?
(2009, Peking University Independent Admission Examination) | If $\tan x + \sqrt{3}$ and $\cot x + \sqrt{3}$ are rational numbers, then there exist integers $p, q, s, t$ with $(p, q) = 1$ and $(s, t) = 1$ such that
$$
\tan x + \sqrt{3} = \frac{p}{q}, \quad \cot x + \sqrt{3} = \frac{s}{t},
$$
which implies
$$
\tan x = \frac{p}{q} - \sqrt{3}, \quad \cot x = \frac{s}{t} - \sqrt{3}.
... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,131 |
Find all ordered triples $(x, y, z)$ such that $x, y, z \in Q_{+}$, and $x+\frac{1}{y}$, $y+\frac{1}{z}$, $z+\frac{1}{x}$ are all integers.
(2010, International Invitational Competition for Young Mathematicians in Cities) | Let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}\left(a, b, c \in \mathbf{N}_{+}\right)$.
(1) If $a=b=c$, then
$$
(x, y, z)=(1,1,1).
$$
(2) If $a, b, c$ are exactly two equal, without loss of generality, let $a=b$. Then
$$
\begin{array}{l}
x+\frac{1}{y}=1+\frac{c}{a}, y+\frac{1}{z}=\frac{2 a}{c}, \\
z+\frac{1}{x}=\frac... | (1,1,1),\left(1, \frac{1}{2}, 2\right),\left(\frac{3}{2}, 2, \frac{1}{3}\right),\left(3, \frac{1}{2}, \frac{2}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,132 |
Example 4 The equation about $x, y$
$$
x^{2}+x y+2 y^{2}=29
$$
has ( ) groups of integer solutions $(x, y)$.
(A) 2
(B) 3
(C) 4
(D) infinitely many
(2009, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | The original equation can be regarded as a quadratic equation in $x$, which can be transformed into
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$
Since the equation has integer roots, $\Delta \geqslant 0$, and it must be a perfect square.
From $\Delta=y^{2}-4\left(2 y^{2}-29\right)=-7 y^{2}+116 \geqslant 0$, s... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,133 |
1. A. The positions of real numbers $a$, $b$, and $c$ on the number line are shown in Figure 1.
Figure 1
Then the algebraic expression
$$
\sqrt{a^{2}}-|a+b|+\sqrt{(c-a)^{2}}+|b+c|
$$
can be simplified to ( ).
(A) $2 c-a$
(B) $2 a-2 b$
(C) $-a$
(D) $a$ | - 1. A. C.
From the positions of real numbers $a$, $b$, $c$ on the number line, we know
$$
\begin{array}{l}
bc . \\
\text { Then } \sqrt{a^{2}}-|a+b|+\sqrt{(c-a)^{2}}+|b+c| \\
=-a+(a+b)+(c-a)-(b+c) \\
=-a .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,134 |
B. If $a=-2+\sqrt{2}$, then the value of $1+\frac{1}{2+\frac{1}{3+a}}$ is ( ).
(A) $-\sqrt{2}$
(B) $\sqrt{2}$
(C) 2
(D) $2 \sqrt{2}$ | B. B.
Notice,
$$
\begin{array}{l}
1+\frac{1}{2+\frac{1}{3+a}}=1+\frac{1}{2+\frac{1}{1+\sqrt{2}}} \\
=1+\frac{1}{2+\sqrt{2}-1}=1+\frac{1}{\sqrt{2}+1}=\sqrt{2} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,135 |
2. A. If the graph of the direct proportion function $y=a x(a \neq 0)$ and the graph of the inverse proportion function $y=\frac{b}{x}(b \neq 0)$ intersect at two points, and one of the intersection points has coordinates $(-3,-2)$, then the coordinates of the other intersection point are ( ).
(A) $(2,3)$
(B) $(3,-2)$
... | 2. A. D.
From $-2=a(-3),(-3)(-2)=b$, we get $a=\frac{2}{3}, b=6$.
Solving the system of equations $\left\{\begin{array}{l}y=\frac{2}{3} x, \\ y=\frac{6}{x},\end{array}\right.$ we get
$$
\left\{\begin{array} { l }
{ x = - 3 , } \\
{ y = - 2 ; }
\end{array} \quad \left\{\begin{array}{l}
x=3 \\
y=2
\end{array}\right.\r... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,136 |
B. In the Cartesian coordinate system $x O y$, the number of integer point coordinates $(x, y)$ that satisfy the inequality $x^{2}+y^{2} \leqslant 2 x+2 y$ is ( ).
(A) 10
(B) 9
(C) 7
(D) 5 | B. B.
From $x^{2}+y^{2} \leqslant 2 x+2 y$, we get
$$
0 \leqslant(x-1)^{2}+(y-1)^{2} \leqslant 2 \text {. }
$$
Since $x$ and $y$ are both integers, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ ( x - 1 ) ^ { 2 } = 0 , } \\
{ ( y - 1 ) ^ { 2 } = 0 ; }
\end{array} \left\{\begin{array}{l}
(x-1)^{2}=0, \\
(y-... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,137 |
3. A. If $a, b$ are given real numbers, and $1<a<b$, then the absolute value of the difference between the average and the median of the four numbers $1, a+1, 2a+b, a+b+1$ is ( ).
(A) 1
(B) $\frac{2a-1}{4}$
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$ | 3. A. D.
From the given, we know
$$
1<a+1<a+b+1<2 a+b .
$$
Therefore, the average of the four data points is
$$
\begin{array}{l}
\frac{1+(a+1)+(a+b+1)+(2 a+b)}{4} \\
=\frac{3+4 a+2 b}{4},
\end{array}
$$
The median is
$$
\begin{array}{l}
\frac{(a+1)+(a+b+1)}{2}=\frac{4+4 a+2 b}{4} . \\
\text { Then } \frac{4+4 a+2 b}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,138 |
B. As shown in Figure 2, in quadrilateral $A B C D$, $A C$ and $B D$ are diagonals, $\triangle A B C$ is an equilateral triangle, $\angle A D C=$ $30^{\circ}, A D=3, B D=5$. Then the length of $C D$ is ( ).
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{5}$
(D) 4.5 | B. B.
As shown in Figure 10, construct an equilateral $\triangle CDE$ with $CD$ as a side, and connect $AE$.
$$
\begin{aligned}
& \text{Given } AC=BC, \\
& CD=CE, \\
& \angle BCD \\
= & \angle BCA + \angle ACD \\
= & \angle DCE + \angle ACD = \angle ACE,
\end{aligned}
$$
we know that $\triangle BCD \cong \triangle AC... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,139 |
4. A. Xiao Qian and Xiao Ling each have several RMB notes with integer yuan denominations. Xiao Qian says to Xiao Ling: "If you give me 2 yuan, my money will be $n$ times yours"; Xiao Ling says to Xiao Qian: "If you give me $n$ yuan, my money will be 2 times yours", where $n$ is a positive integer. Then the number of p... | 4. A. D.
Let the amount of money Xiaolian has be $x$ yuan, and the amount of money Xiaoling has be $y$ yuan, where $x$ and $y$ are non-negative integers. From the problem, we have
$$
\left\{\begin{array}{l}
x+2=n(y-2), \\
y+n=2(x-n) .
\end{array}\right.
$$
Eliminating $x$, we get
$$
\begin{array}{l}
(2 y-7) n=y+4 \\
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,140 |
B. If the equation about $x$
$x^{2}-p x-q=0$ ( $p, q$ are positive integers)
has a positive root less than 3, then the number of such equations is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | B. C.
From the relationship between the roots and coefficients of a quadratic equation, we know that the product of the two roots is $-q0$.
Therefore, $3^{2}-3 p-q>0 \Rightarrow 3 p+q<9$.
Since $p, q$ are both positive integers, we have $p=1,1 \leqslant q \leqslant 5$; or $p=2,1 \leqslant q \leqslant 2$.
Thus, there a... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,141 |
5. A. A fair cube die has the numbers $1, 2, 3, 4, 5, 6$ on its six faces. When the die is rolled twice, let the remainders when the sum of the numbers on the top faces is divided by 4 be $0, 1, 2, 3$ with probabilities $P_{0}, P_{1}, P_{2}, P_{3}$, respectively. Then the largest among $P_{0}, P_{1}, P_{2}, P_{3}$ is $... | 5. A. D.
Rolling a die twice, the ordered pairs formed by the numbers on the two faces up total 36, and the remainders when their sums are divided by 4 are $0, 1, 2, 3$, with 9, 8, 9, and 10 ordered pairs respectively. Thus,
$$
P_{0}=\frac{9}{36}, P_{1}=\frac{8}{36}, P_{2}=\frac{9}{36}, P_{3}=\frac{10}{36} .
$$
There... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,142 |
B. On the blackboard, there are
$$
1, \frac{1}{2}, \cdots, \frac{1}{100}
$$
a total of 100 numbers. Each operation involves selecting two numbers $a$ and $b$ from the numbers on the blackboard, then deleting $a$ and $b$, and writing the number $a+b+a b$ on the blackboard. After 99 operations, the number left on the bl... | B. C.
Since $a+b+a b+1=(a+1)(b+1)$, the product of each number on the blackboard plus 1 remains unchanged before and after each operation.
Let the number left on the blackboard after 99 operations be $x$. Then
$$
\begin{array}{l}
x+1=(1+1)\left(\frac{1}{2}+1\right) \cdots \cdots\left(\frac{1}{100}+1\right) \\
=101 \\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,143 |
Example 5 When $x \leqslant y \leqslant z$, find the positive integer solutions of the equation
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{7}{8}
$$
(2007, Taiyuan Junior High School Mathematics Competition) | Given $x>0$, and $x \leqslant y \leqslant z$, we know $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}>0$.
Therefore, $\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x}$.
Thus, $\frac{1}{x}<\frac{7}{8} \leqslant \frac{3}{x}$.
Solving this, we get $\frac{8}{7}<x \leqslant \frac{24}{7}$.
There... | (2,3,24) \text{ and } (2,4,8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,144 |
B. If $a, b, c$ are positive numbers, and satisfy
$$
\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}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,146 |
7. A. As shown in Figure 4, the side length of square $ABCD$ is $2 \sqrt{15}, E, F$ are the midpoints of sides $AB, BC$ respectively, $AF$ intersects $DE, DB$ at points $M, N$. Then the area of $\triangle DMN$ is $\qquad$. | 7. A. 8 .
Connect $D F$. Let the side length of the square $A B C D$ be $2 a$. From the problem, we easily know
$$
\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 | 725,147 |
B. As shown in Figure $5, \odot O$ has a radius of $20, A$ is a point on $\odot O$. A rectangle $O B A C$ is constructed with $O A$ as the diagonal, and $O C=$ 12. Extend $B C$ to intersect $\odot O$ at points $D$ and $E$, then $C E-B D$ $=$ . $\qquad$ | B. $\frac{28}{5}$.
Let the midpoint of $D E$ be $M$, and connect $O M$. Then $O M \perp D E$.
Since $O B=\sqrt{20^{2}-12^{2}}=16$, we have,
$$
\begin{array}{l}
O M=\frac{O B \cdot O C}{B C}=\frac{16 \times 12}{20}=\frac{48}{5}, \\
C M=\sqrt{O C^{2}-O M^{2}}=\frac{36}{5}, B M=\frac{64}{5} .
\end{array}
$$
Therefore, ... | \frac{28}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,148 |
8. A. If the equation with respect to $x$
$$
x^{2}+k x+\frac{3}{4} k^{2}-3 k+\frac{9}{2}=0
$$
has two real roots $x_{1}$ and $x_{2}$, then $\frac{x_{1}^{2011}}{x_{2}^{2012}}=$ | 8. A. $-\frac{2}{3}$.
From the problem, we have
$$
\begin{array}{l}
\Delta=k^{2}-4\left(\frac{3}{4} k^{2}-3 k+\frac{9}{2}\right) \geqslant 0 \\
\Rightarrow(k-3)^{2} \leqslant 0 . \\
\text { Also, } (k-3)^{2} \geqslant 0, \text { so } \\
(k-3)^{2}=0 \Rightarrow k=3 .
\end{array}
$$
At this point, the equation is $x^{2... | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,149 |
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
$$
51\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }... | 1610 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,150 |
9. A. 2 eighth-grade students and $m$ ninth-grade students participate in a single round-robin chess tournament, where each participant plays against every other participant exactly once. The scoring rule is: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both players get 1 ... | 9. A. 8 .
Let the number of draws be $a$, and the number of wins (losses) be $b$.
From the problem, we know
$$
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 {. }
$$
Thus, $0 \leqslant b=130-(m+1)(m+2) \leqslant 43$.
Therefo... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,151 |
B. If positive numbers $x, y, z$ can be the lengths of the three sides of a triangle, then $(x, y, z)$ is called a "triangle number". If $(a, b, c)$ and $\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right)$ are both triangle numbers, and $a \leqslant b \leqslant c$, then the range of $\frac{a}{c}$ is $\qquad$. | B. $\frac{3-\sqrt{5}}{2}c \\
\frac{1}{c}+\frac{1}{b}>\frac{1}{a}
\end{array}\right. \\
\Rightarrow \frac{1}{c}+\frac{1}{c-a}>\frac{1}{c}+\frac{1}{b}>\frac{1}{a} \\
\Rightarrow \frac{1}{c}+\frac{1}{c-a}>\frac{1}{a} \\
\Rightarrow\left(\frac{a}{c}\right)^{2}-3\left(\frac{a}{c}\right)+1<0 \\
\Rightarrow \frac{3-\sqrt{5}}{... | \frac{3-\sqrt{5}}{2}<\frac{a}{c} \leqslant 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,152 |
10. A. As shown in Figure 6, quadrilateral $ABCD$ is inscribed in $\odot O$, $AB$ is the diameter, $AD=DC$, $BA$ and $CD$ are extended to intersect at point $E$, $BF \perp EC$ is drawn and intersects the extension of $EC$ at point $F$. If $AE=AO, BC=6$, then the length of $CF$ is $\qquad$ | 10. A. $\frac{3 \sqrt{2}}{2}$.
As shown in Figure 11, connect $A C$, $B D$, and $O D$.
Since $A B$ is the diameter of $\odot O$, we have
$$
\angle B C A = \angle B D A = 90^{\circ}.
$$
From the given conditions, $\angle B C F = \angle B A D$.
Thus, Rt $\triangle B C F \backsim$ Rt $\triangle B A D$
$$
\Rightarrow \fr... | \frac{3 \sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,153 |
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 | 725,154 |
Example 6 On the equation about $m$ and $n$
$$
5 m^{2}-6 m n+7 n^{2}=2011
$$
Does there exist an integer solution? If it exists, please write down one solution; if not, please explain the reason.
(2011, Beijing Middle School Mathematics Competition (Grade 8)) | No solution exists.
(1) If $m$ and $n$ have the same parity, then the left side of the given equation is even, which cannot equal 2011.
(2) If $m$ and $n$ have different parities, then $m+n$ and $m-n$ are both odd, and the original equation can be rewritten as
$$
4(m-n)^{2}+(m+n)^{2}+2 n^{2}=2011.
$$
We will discuss t... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,155 |
B. As shown in Figure 7, in the Cartesian coordinate system $x O y$,
$$
\begin{array}{l}
A O=8, A B=A C, \\
\sin \angle A B C=\frac{4}{5}
\end{array}
$$
$C D$ intersects the $y$-axis at $E$,
and $S_{\triangle C O E}=S_{\triangle A D E}$.
It is known that the image passing through points $B$, $C$, and $E$
is a parabola.... | B. From $\sin \angle A B C=\frac{A O}{A B}=\frac{4}{5}, A O=8$, we get $A B=10$.
By the Pythagorean theorem,
$B O=\sqrt{A B^{2}-A O^{2}}=6$.
Thus, $\triangle A B O \cong \triangle A C O$
$\Rightarrow C O=B O=6$.
Therefore, $A(0,-8), B(6,0), C(-6,0)$.
Let point $D(m, n)$.
From $S_{\triangle C O E}=S_{\triangle A D E}$
$... | y=\frac{2}{27} x^{2}-\frac{8}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,157 |
12. A. As shown in Figure $8, \odot O$ has a diameter $A B, \odot O_{1}$ passes through point $O$, and is internally tangent to $\odot O$ at point $B, C$ is a point on $\odot O$, $O C$ intersects $\odot O_{1}$ at point $D$, and $O D > C D$, point $E$ is on $O D$ and
$$
D C = D E, B E
$$
is extended to intersect $\odo... | 12. A. As shown in Figure 12, connect $B D$.
Since $O B$ is the diameter of $\odot O_{1}$, we have
$$
\angle O D B=90^{\circ} \text {. }
$$
Since $D C=D E$, $\triangle C B E$ is an isosceles triangle.
Let $B C$ intersect $\odot O_{1}$ at point $M$, and connect $O M$. Then
$$
\angle O M B=90^{\circ} \text {. }
$$
Sinc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,158 |
B. As shown in Figure 9, in the cyclic quadrilateral $ABCD$ inscribed in $\odot O$, $AC$ and $BD$ are its diagonals, and the midpoint $I$ of $AC$ is the incenter of $\triangle ABD$. Prove:
(1.) $OI$ is the tangent to the circumcircle of $\triangle IBD$;
$$
\begin{array}{l}
\text { (2) } AB + AD \\
= 2BD \text {. }
\end... | B. (1) As shown in Figure 13.
By the properties of the incenter of a triangle and the angles subtended by the same arc, we have
$$
\angle C I D=\frac{\angle B A D}{2}+\frac{\angle B D A}{2}=\angle C D I .
$$
Therefore, $C I=C D$.
Similarly, $C I=C B$.
Hence, $C$ is the circumcenter of $\triangle I B D$.
Connect $O A$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,159 |
13. A. Given integers $a, b$ satisfy $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$. | 13. A. Let $a-b=m$ (where $m$ is a prime number), and $ab=n^2$ (where $n$ is a non-negative integer).
When $b \neq 0$,
$$
\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}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive ... | 2017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,160 |
B. In a convex $n$-sided polygon, what is the maximum number of interior angles that can be $150^{\circ}$? Explain your reasoning. | B. Suppose in a convex $n$-sided polygon, there are $k$ interior angles equal to $150^{\circ}$. Then there are $n-k$ interior angles not equal to $150^{\circ}$.
(1) If $k=n$, then
$$
n \times 150^{\circ}=(n-2) \times 180^{\circ},
$$
we get $n=12$. Thus, in a regular dodecagon, all 12 interior angles are $150^{\circ}$.... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,161 |
14. A. Find all positive integers $n$, such that there exist positive integers $x_{1}, x_{2}, \cdots, x_{2012}$, satisfying
$$
x_{1}<x_{2}<\cdots<x_{2012} \text {, }
$$
and $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}=n$. | 14. A. Since $x_{1}, x_{2}, \cdots, x_{2012}$ are all positive integers, and $x_{1}<x_{2}<\cdots<x_{2012}$, we have
$$
x_{1} \geqslant 1, x_{2} \geqslant 2, \cdots, x_{2012} \geqslant 2012 .
$$
Then $n=\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}$
$$
\begin{array}{l}
\leqslant \frac{1}{1}+\frac{2}{2}+\... | 1,2, \cdots, 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,162 |
B. Divide $2,3, \cdots, n(n \geqslant 2)$ into two groups arbitrarily. If it is always possible to find numbers $a, b, c$ (which can be the same) in one of the groups such that $a^{b}=c$, find the minimum value of $n$. | B. When $n=2^{16}-1$, divide $2,3, \cdots, n$ into the following two arrays:
$$
\begin{array}{l}
\left\{2,3,2^{8}, 2^{8}+1, \cdots, 2^{16}-1\right\}, \\
\left\{4,5, \cdots, 2^{8}-1\right\} .
\end{array}
$$
In the first array, by
$$
3^{3}2^{16}-1 \text {, }
$$
we know that there do not exist numbers $a, b, c$ such tha... | 2^{16} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,163 |
Example 7 On the equation about $x, y$
$$
x^{2}+y^{2}=208(x-y)
$$
all positive integer solutions are $\qquad$
(2008, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | Since 208 is a multiple of 4, the square of an even number is divisible by 4 with a remainder of 0, and the square of an odd number is divisible by 4 with a remainder of 1, therefore, \(x\) and \(y\) are both even.
Let \(x = 2a, y = 2b\). Then \(a^2 + b^2 = 104(a - b)\).
Similarly, \(a\) and \(b\) are both even.
Let \(... | s = 20, t = 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,166 |
5. If for any $x \in \mathbf{R}$, the inequality $|x| \geqslant 4 a x$ always holds, then the range of real number $a$ is ( ).
(A) $a<-\frac{1}{4}$
(B) $|a| \leqslant \frac{1}{4}$
(C) $|a|<\frac{1}{4}$
(D) $a \geqslant \frac{1}{4}$ | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 725,169 |
6. Given three different planes $\alpha, \beta, \gamma$ and two non-coincident lines $m, n$, there are the following 4 propositions:
(1) $m / / \alpha, \alpha \cap \beta=n$, then $m / / n$;
(2) $m \perp \alpha, m / / n, n \subset \beta$, then $\alpha \perp \beta$;
(3) $\alpha \perp \beta, \gamma \perp \beta$, then $\al... | 6. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,170 |
8. Function
$$
f(x)=\left\{\begin{array}{ll}
\sin \pi x^{2}, & -1<x<0 ; \\
\mathrm{e}^{x-1}, & x \geqslant 0
\end{array}\right.
$$
satisfies $f(1)+f(a)=2$. Then the possible values of $a$ are ( ).
(A) 1 or $-\frac{\sqrt{2}}{2}$
(B) $-\frac{\sqrt{2}}{2}$
(C) 1
(D) 1 or $\frac{\sqrt{2}}{2}$ | 8. A | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,172 |
Example 8 Let $a$ be a prime number, $b$ be a positive integer, and $9(2 a+b)^{2}=509(4 a+511 b)$. Find the values of $a$ and $b$.
(2008, National Junior High School Mathematics Competition) | The original equation can be transformed into
$$
\left(\frac{6 a+3 b}{509}\right)^{2}=\frac{4 a+511 b}{509} \text {. }
$$
Let $m=\frac{6 a+3 b}{509}, n=\frac{4 a+511 b}{509}$.
Then $3 n-511 m+6 a=0$.
Also, $n=m^{2}$, so
$$
3 m^{2}-511 m+6 a=0 \text {. }
$$
From the original equation, we know that $(2 a+b)^{2}$ is div... | a=251, b=7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,176 |
17. (10 points) Given the quadratic function
$$
f(x)=x^{2}-t x+t(t \in \mathbf{R})
$$
satisfies:
(1) The solution set of the inequality $f(x) \leqslant 0$ has exactly one element; $\square$
(2) If $0<x_{2}<x_{1}$, the inequality $f\left(x_{2}\right)<$ $f\left(x_{1}\right)$ always holds. Let the sequence $\left\{a_{n}\... | Three, 17. (1) From condition (1), we have
$\Delta=t^{2}-4 t=0 \Rightarrow t=0$ or 4.
From condition (2), we know that $f(x)$ is an increasing function on $(0,+\infty)$. Therefore,
$$
t=0, f(x)=x^{2}, S_{n}=f(n)=n^{2}.
$$
Thus, $a_{n}=S_{n}-S_{n-1}=2 n-1(n \geqslant 2)$.
Also, $a_{1}=1$ fits, so $a_{n}=2 n-1$.
(2) Not... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,181 |
18. (12 points) In $\triangle A B C$, $a$, $b$, and $c$ are the sides opposite to the interior angles $\angle A$, $\angle B$, and $\angle C$, respectively, and satisfy $\sin A+\sqrt{3} \cos A=2$.
(1) Find the size of $\angle A$;
(2) Now, three conditions are given:
(1) $a=2$,(2) $\angle B=45^{\circ}$,(3) $c=\sqrt{3} b$... | 18. (1) According to the problem, we have
$$
\sin \left(A+\frac{\pi}{3}\right)=1
$$
Since $0<\angle A<\pi$, therefore, $\angle A=\frac{\pi}{6}$.
(2) Scheme 1 selects conditions (1) and (2).
From $\frac{a}{\sin A}=\frac{b}{\sin B} \Rightarrow b=2 \sqrt{2}$.
Also, $\sin C=\sin (A+B)$
$=\sin A \cdot \cos B+\cos A \cdot ... | \sqrt{3}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,182 |
19. (12 points) A head teacher conducted a survey on the learning enthusiasm and attitude towards class work of 50 students in the class. The statistical data is shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline & \begin{tabular}{l}
Actively Participate in \\
Class Work
\end{tabular} & \begin{tabular}{l}
N... | 19. (1) $\frac{11}{25} ; \frac{2}{5}$.
$$
\begin{array}{l}
\text { (2) Since } k^{2}=\frac{n(a d-b c)^{2}}{(a+b)(c+d)(a+c)(b+d)} \\
=\frac{50(17 \times 20-5 \times 8)^{2}}{25 \times 25 \times 22 \times 28}=11.688, \\
P\left(k^{2} \geqslant k_{0}\right)=0.001,
\end{array}
$$
Therefore, there is a $99.9\%$ confidence th... | 11.688 | Other | math-word-problem | Yes | Yes | cn_contest | false | 725,183 |
20. (12 points) In the geometric body shown in Figure 1, $E A \perp$ plane $A B C, D B \perp$ plane $A B C, A C \perp B C$, and $B C$ $=B D=\frac{3}{2} A E=a, A C=$ $\sqrt{2} a, A M=2 M B$.
(1) Prove: $C M \perp E M$;
(2) Find the angle between line $C D$ and plane $M C E$.
| 20. (1) It is easy to know, $A B=\sqrt{3} a$.
Draw $C M^{\prime} \perp A B$ intersecting $A B$ at point $M^{\prime}$. Then $B C^{2}=B M^{\prime} \cdot B A \Rightarrow B M^{\prime}=\frac{\sqrt{3}}{3} a$.
Thus, $A M^{\prime}=2 M^{\prime} B$, which means point $M$ coincides with $M^{\prime}$. Hence, $C M \perp A B$.
Sinc... | \arcsin \frac{\sqrt{6}}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 725,184 |
21. (12 points) Given two moving points $A$ and $B$, and a fixed point $M\left(x_{0}, y_{0}\right)$, all on the parabola $y^{2}=2 p x (p>0)$ ($A$ and $B$ do not coincide with $M$). Let $F$ be the focus of the parabola, and $Q$ be a point on the axis of symmetry. It is given that $\left(\overrightarrow{Q A}+\frac{1}{2} ... | 21. (1) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), Q(a, 0)$. Then, from $|\overrightarrow{F A}|, |\overrightarrow{F M}|, |\overrightarrow{F B}|$ forming an arithmetic sequence, we get $x_{0}=\frac{x_{1}+x_{2}}{2}$.
And $y_{1}^{2}=2 p x_{1}$,
$y_{2}^{2}=2 p x_{2}$.
(1) - (2) gives $\frac{y_{1}-y_{2}}{x... | (0,10] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,185 |
22. (12 points) Function
$$
f(x)=a \ln x+1(a>0) \text {. }
$$
(1) Prove: $f(x)-1 \geqslant a\left(1-\frac{1}{x}\right)$;
(2) On the interval $(1, \mathrm{e})$, $f(x)>x$ always holds, find the range of the real number $a$;
(3) When $a=\frac{1}{2}$, prove:
$$
\sum_{k=2}^{n+1} f(k)>2(n+1-\sqrt{n+1}) .
$$ | 22. (1) Let $g(x)=\ln x-1+\frac{1}{x}$. Then
$$
g^{\prime}(x)=\frac{1}{x}-\frac{1}{x^{2}} \text {. }
$$
By $g^{\prime}(x)=0$, we get $x=1$.
Thus, $g(x)$ is decreasing on $(0,1)$ and increasing on $(1,+\infty)$.
(Continued from page 32)
Therefore, the minimum value of $g(x)$ is $g(1)=0$.
Hence, $f(x)-1 \geqslant a\left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,186 |
Example 1 Let $p$ be a prime. If there exists a positive integer $n$, such that $p \|\left(2^{n}-1\right)$, prove:
$$
p \|\left(2^{p-1}-1\right) .
$$ | Notice,
$$
(p-1, p)=1, p \|\left(2^{n}-1\right) \text {. }
$$
By the lemma, we have
$$
\begin{array}{l}
p \|\left[2^{n(p-1)}-1\right] \\
\Rightarrow p \|\left[\left(2^{p-1}\right)^{n}-1\right] .
\end{array}
$$
Therefore, the number of factors of $p$ in $2^{p-1}-1$ is at most 1.
And by Fermat's Little Theorem, we know... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,187 |
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 | 725,188 |
2. The inequality
$$
\sin ^{2} x+a \cos x+a^{2} \geqslant 1+\cos x
$$
holds for all $x \in \mathbf{R}$. Then the range of the real number $a$ is
$\qquad$ . | 2. $a \geqslant 1$ or $a \leqslant-2$.
From the problem, we have
$$
\cos ^{2} x+(1-a) \cos x-a^{2} \leqslant 0
$$
for any $x \in \mathbf{R}$.
$$
\begin{array}{l}
\text { Let } f(t)=t^{2}+(1-a) t-a^{2}(-1 \leqslant t=\cos x \leqslant 1) . \\
\text { Then }\left\{\begin{array}{l}
f(1) \leqslant 0, \\
f(-1) \leqslant 0
... | a \geqslant 1 \text{ or } a \leqslant -2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,189 |
3. Given a function $f(n)$ defined on the set of positive integers satisfies the conditions:
(1) $f(m+n)=f(m)+f(n)+m n\left(m, n \in \mathbf{N}_{+}\right)$;
(2) $f(3)=6$.
Then $f(2011)=$ . $\qquad$ | 3.2023066.
In condition (1), let $n=1$ to get
$$
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 {. }
$$
Substitute into equation (1) to get... | 2023066 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,190 |
4. Equation
$$
|| \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 | 725,191 |
5. In a sphere with a radius of $10 \mathrm{~cm}$, there is a cube with edges of integer length $(\mathrm{cm})$. Then the maximum edge length of the cube is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $5.11 \mathrm{~cm}$.
Let the edge length of the cube be $a$. The diagonal of the cube $\sqrt{3} a$ is no greater than the diameter of the sphere 20, i.e.,
$$
\begin{array}{l}
a \leqslant \frac{20 \sqrt{3}}{3}\left(a \in \mathbf{N}_{+}\right) \Rightarrow a \leqslant 11 \\
\Rightarrow a_{\max }=11 .
\end{array}
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,192 |
6. The inner wall of a glass is formed by rotating the parabola
$$
y=x^{2}(-2 \leqslant x \leqslant 2)
$$
around the $y$-axis. Then the maximum radius of a sphere that can touch the bottom of the cup is $\qquad$ | 6. $\frac{1}{2}$.
Let the radius of the sphere be $r$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}+(y-r)^{2}=r^{2}, \\
y=x^{2}
\end{array}\right. \\
\Rightarrow x^{2}\left(1-2 r+x^{2}\right)=0 .
\end{array}
$$
According to the problem, the equation
$$
x^{2}+1-2 r=0
$$
has no non-zero real solutions.
Thus ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 725,193 |
7. Calculate:
$$
\begin{array}{l}
\frac{1}{\sin 45^{\circ} \cdot \sin 46^{\circ}}+\frac{1}{\sin 46^{\circ} \cdot \sin 47^{\circ}}+ \\
\cdots+\frac{1}{\sin 89^{\circ} \cdot \sin 90^{\circ}} \\
=
\end{array}
$$ | 7. $\frac{1}{\sin 1^{\circ}}$.
Notice,
$$
\begin{array}{l}
\frac{1}{\sin n^{\circ} \cdot \sin (n+1)^{\circ}} \\
=\frac{1}{\sin 1^{\circ}} \cdot \frac{\sin \left[(n+1)^{\circ}-n^{\circ}\right]}{\sin n^{\circ} \cdot \sin (n+1)^{\circ}} \\
=\frac{1}{\sin 1^{\circ}} \cdot \frac{\sin (n+1)^{\circ} \cdot \cos n^{\circ}-\cos... | \frac{1}{\sin 1^{\circ}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,194 |
8. 10 students stand in a row, and a red, yellow, or blue hat is to be given to each student. It is required that each color of hat must be present, and the hats of adjacent students must be of different colors. Then the number of ways to distribute the hats that meet the requirements is $\qquad$ kinds. | 8. 1530.
Generalize to the general case.
Let the number of ways to arrange $n$ students according to the given conditions be $a_{n}$. Then
$$
\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 | 725,195 |
1. (16 points) If $n$ is a positive integer greater than 2, find the minimum value of
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}
$$ | When $n=3$,
$$
\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60} \text {. }
$$
Assume that when $n=k(k \geqslant 3)$,
$$
\sum_{i=k+1}^{2 k} \frac{1}{i} \geqslant \frac{37}{60} \text {. }
$$
Then when $n=k+1$,
$$
\begin{array}{l}
\sum_{i=n+1}^{2 n} \frac{1}{i}=\sum_{i=k+2}^{2 k+2} \frac{1}{i} \\
=\sum_{i=k+1}^{2 k} \f... | \frac{37}{60} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,196 |
2. (20 points) Two points are randomly and independently chosen on a line segment, and the segment is then divided into three parts at these points. Question: What is the probability that the three new segments can form a triangle?
| Suppose the initial line segment is the interval $[0,1]$, and the two randomly chosen points are $x, y (0 < x < y < 1)$. The three new segments are $x, y - x, 1 - y$. For these three segments to form a triangle, they must satisfy the triangle inequality:
\begin{array}{l}
x + (y - x) > 1 - y , \\
x + ( 1 - y ) > y - x ,... | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,197 |
Example 2 Find all positive integers $n$ such that $\frac{2^{n}+1}{n^{2}}$ is an integer.
(31st IMO) | When $n=1,3$, it is obviously true.
When $n \neq 1,3$, let the prime $p$ be the smallest prime factor of $n$.
$$
\begin{array}{l}
\text { By } \frac{2^{n}+1}{n^{2}} \text { being an integer } \\
\Rightarrow p \mid\left(2^{n}+1\right) \\
\Rightarrow p \mid\left(2^{2 n}-1\right) .
\end{array}
$$
But by Fermat's Little T... | n=1,3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,198 |
3. (20 points) Given the sequence $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ satisfies $a_{0}=0, a_{1}=1, a_{2}=0$, when $n \geqslant 3$, we have
$$
a_{n}=\frac{2}{n-1} \sum_{i=0}^{n-2} a_{i} \text {. }
$$
Prove: For all integers $n \geqslant 3$, $a_{n}>\frac{n}{10}$. | 3. Proof 1 From the given, we have
$$
(n-1) a_{n}=2 \sum_{i=0}^{n-2} a_{i} \text {. }
$$
Substituting $n+1$ for $n$ in the above equation, we get
$$
n a_{n+1}=2 \sum_{i=0}^{n-1} a_{i} .
$$
Subtracting the two equations, we obtain
$$
n a_{n+1}-(n-1) a_{n}=2 a_{n-1} \text {. }
$$
The above equation holds for all integ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,199 |
1. The number of all integer solutions to the equation $\left(x^{2}+x-1\right)^{x+3}=1$ is ( ).
(A) 5
(B) 4
(C) 3
(D) 2 | 1. B.
The condition for the original equation to have integer solutions is and only is the following three:
(1) $x+3=0$, and $x^{2}+x-1 \neq 0$, in this case, $x=-3$ is an integer solution of the equation;
(2) $x^{2}+x-1=1$, solving gives $x=-2$ or 1, so the original equation has two integer solutions;
(3) $x^{2}+x-1=... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,200 |
2. In $\triangle A B C$, $a$, $b$, $c$ are the lengths of the sides opposite to $\angle A$, $\angle B$, $\angle C$ respectively. If $\angle B=60^{\circ}$, then the value of $\frac{c}{a+b}+\frac{a}{c+b}$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$ | 2. C.
Draw $A D \perp C D$ at point $D$.
In Rt $\triangle B D A$, since $\angle B=60^{\circ}$, we have $D B=\frac{c}{2}, A D=\frac{\sqrt{3}}{2} c$.
In Rt $\triangle A D C$, we have
$$
\begin{array}{l}
D C^{2}=A C^{2}-A D^{2} \\
\Rightarrow\left(a-\frac{c}{2}\right)^{2}=b^{2}-\frac{3}{4} c^{2} \\
\Rightarrow a^{2}+c^{2... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,201 |
3. As shown in Figure 1, in trapezoid $A B C D$, $A D / / B C, A D$ $=3, B C=9, A B=6, C D$ $=4$. If $E F / / B C$, and the perimeters of trapezoids $A E F D$ and $E B C F$ are equal, then the length of $E F$ is
(A) $\frac{45}{7}$
(B) $\frac{33}{5}$.
(C) $\frac{39}{5}$
(D) $\frac{15}{2}$ | 3. C.
From the given information, we have
$$
A D+A E+E F+F D=E F+E B+B C+C F \text {. }
$$
Then $A D+A E+F D=E B+B C+C F$
$$
=\frac{1}{2}(A D+A B+B C+C D)=11 \text {. }
$$
Since $E F \parallel B C$, we have
$E F \parallel A D, \frac{A E}{E B}=\frac{D F}{F C}$.
Let $\frac{A E}{E B}=\frac{D F}{F C}=k$. Then
$A E=\frac... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,202 |
4. Given $I$ is the incenter of acute $\triangle A B C$, and $A_{1} 、 B_{1}$ 、 $C_{1}$ are the reflections of point $I$ over $B C 、 C A 、 A B$ respectively. If point $B$ lies on the circumcircle of $\triangle A_{1} B_{1} C_{1}$, then $\angle A B C$ equals ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^... | 4. C.
Since $I A_{1}=I B_{1}=I C_{1}=2 r$ (where $r$ is the inradius of $\triangle A B C$), therefore, $I$ is the circumcenter of $\triangle A_{1} B_{1} C_{1}$.
Let the intersection of $I A_{1}$ and $B C$ be $D$. Then $I B=I A_{1}=2 I D$.
Thus, $\angle I B D=30^{\circ}$.
Similarly, $\angle I B A=30^{\circ}$.
Therefore... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,203 |
5. When $n=1,2, \cdots, 2012$, the quadratic function
$$
y=\left(n^{2}+n\right) x^{2}-(2 n+1) x+1
$$
intersects the $x$-axis, and the sum of the lengths of the segments intercepted on the $x$-axis is ( ).
(A) $\frac{2010}{2011}$
(B) $\frac{2011}{2012}$
(C) $\frac{2012}{2013}$
(D) $\frac{2013}{2014}$ | 5. C.
Solve $\left(n^{2}+n\right) x^{2}-(2 n+1) x+1=0$, we get $x_{1}=\frac{1}{n+1}, x_{2}=\frac{1}{n}$.
Then $d_{n}=\left|x_{1}-x_{2}\right|=\frac{1}{n}-\frac{1}{n+1}$.
Therefore, $d_{1}+d_{2}+\cdots+d_{2012}$
$$
\begin{array}{l}
=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{2... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,204 |
6. Calculate
$$
\begin{array}{l}
\frac{\left(3^{4}+4\right)\left(7^{4}+4\right)\left(11^{4}+4\right) \cdots\left(39^{4}+4\right)}{\left(5^{4}+4\right)\left(9^{4}+4\right)\left(13^{4}+4\right) \cdots\left(41^{4}+4\right)} \\
=(\quad) .
\end{array}
$$
(A) $\frac{1}{353}$
(B) $\frac{1}{354}$
(C) $\frac{1}{355}$
(D) $\frac... | 6. A.
Notice that,
$$
\begin{array}{l}
x^{4}+4=\left(x^{2}+2\right)^{2}-(2 x)^{2} \\
=\left(x^{2}+2 x+2\right)\left(x^{2}-2 x+2\right) \\
=\left[(x+1)^{2}+1\right]\left[(x-1)^{2}+1\right] .
\end{array}
$$
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{\left(2^{2}+1\right)\left(4^{2}+1\right) \cdots\... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,205 |
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
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{(k+1)^{2}-4} \text {. } \\
\text { Then } \frac{\sqrt{3}}{2} \sqrt{(k+1)^{2}-4}=\left|1-\left(\frac{k+1}{2}\right)^{2}\... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,206 |
3. Given that $A M$ is the median of $\triangle A B C$ on side $B C$, $P$ is the centroid of $\triangle A B C$, and a line $E F$ through point $P$ intersects sides $A B$ and $A C$ at points $E$ and $F$ respectively. Then $\frac{B E}{A E}+\frac{C F}{A F}=$ $\qquad$ | 3. 1 .
Draw $B G$ and $C K$ parallel to $A M$ through points $B$ and $C$ respectively, intersecting line $E F$ at points $G$ and $K$. Then
$$
\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} .
$$
In tr... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,208 |
1 Inscribed Triangle of a Triangle
Example 1 (1) The pedal triangle of an acute triangle is the "light path triangle";
(2) The "light path triangle" of an acute triangle is the pedal triangle;
(3) Given an acute triangle, find the inscribed triangle with the smallest perimeter.
Proof (1) Express the proposition in geom... | As shown in Figure 1, given an acute triangle $\triangle ABC$ with three altitudes $AD$, $BE$, and $CF$ intersecting at point $H$. Prove:
$\angle BDF$
$=\angle CDE$,
$\angle CED=\angle AEF$, $\angle AFE=\angle BFD$.
【Analysis】Using the perpendicular conditions of the triangle, leveraging the concept of four points bein... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,209 |
Example 2 Given that $\triangle XYZ$ is an isosceles right triangle with legs of length 1 $\left(\angle Z=90^{\circ}\right)$, and its three vertices are on the three sides of the isosceles right triangle $\triangle ABC\left(\angle C=90^{\circ}\right)$. Find the maximum possible length of the legs of $\triangle ABC$.
(2... | (1) As shown in Figure 3, if vertex $Z$ is on the hypotenuse $AB$, take the midpoint $M$ of $XY$, and connect $CM$, $ZM$, $CZ$, and draw the altitude $CN$ from $C$ to $AB$.
Then $C N \leqslant C Z \leqslant C M + M Z$
$$
=\frac{1}{2} X Y + \frac{1}{2} X Y = X Y = \sqrt{2} \text{. }
$$
Thus, $C A = \sqrt{2} C N \leqsla... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,210 |
Conclusion 4 As shown in Figure 4, let $D, E, F$ be the points of tangency of the incircle of $\triangle ABC$ with sides $BC, CA, AB$, respectively, and let line $FE$ intersect $BC$ at point $T$. Then $\frac{BD}{DC}=\frac{BT}{TC}$. | Prove that for $\triangle ABC$ and line $FET$, applying Menelaus' theorem yields
$$
\frac{AF}{FB} \cdot \frac{BT}{TC} \cdot \frac{CE}{EA}=1.
$$
Notice that, $AF = AE, BF = BD, CE = CD$.
Thus, $\frac{BD}{DC} = \frac{BT}{TC}$.
The above equation indicates that points $D$ and $T$ divide the side $BC$ internally and exter... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,211 |
II. (40 points) Given positive real numbers $x, y, z$ satisfying $x+y+z=1$. Try to find the minimum value of the real number $k$ such that the inequality
$$
\frac{x^{2} y^{2}}{1-z}+\frac{y^{2} z^{2}}{1-x}+\frac{z^{2} x^{2}}{1-y} \leqslant k-3 x y z
$$
always holds. | Let $x=y=z=\frac{1}{3}$. Then $k \geqslant \frac{1}{6}$.
Below is the proof:
$$
\frac{x^{2} y^{2}}{1-z}+\frac{y^{2} z^{2}}{1-x}+\frac{z^{2} x^{2}}{1-y} \leqslant \frac{1}{6}-3 x y z \text {. }
$$
Given $x>0, y>0, z>0$ and $x+y+z=1$, we know that inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{x y}{z(x+y)}+\f... | \frac{1}{6} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,212 |
Three. (50 points) Given that $p$ is a prime number greater than 3, the positive sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}^{2}+\left[1-\left(\frac{n+1}{n}\right)^{2 p+1}\right] a_{n+1} a_{n} \\
=\left(1+\frac{1}{n}\right)^{2 p+1} a_{n}^{2}(n=1,2, \cdots) .
\end{array}
$$
Prove: ... | $$
\left(a_{n+1}+a_{n}\right)\left[a_{n+1}-\left(\frac{n+1}{n}\right)^{2 p+1} a_{n}\right]=0 \text {. }
$$
Since $a_{n+1}+a_{n}>0$, we have,
$$
\begin{array}{l}
a_{n+1}=\left(\frac{n+1}{n}\right)^{2 p+1} a_{n} \\
\Rightarrow \frac{a_{n+1}}{a_{n}}=\left(\frac{n+1}{n}\right)^{2 p+1} \\
\Rightarrow a_{n}=n^{2 p+1} .
\end... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,213 |
Example 1 In $\triangle ABC$, $AB > AC$, the incircle $\odot I$ touches sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively, $M$ is the midpoint of side $BC$, $AH \perp BC$ at point $H$, the angle bisector $AI$ of $\angle BAC$ intersects lines $DE$, $DF$ at points $K$, $L$ respectively. Prove: $M$, $L$, $H$, $K... | Prove that, as shown in Figure 6, connect $B K$, connect $C L$ and extend it to intersect $A B$ at point $N$.
Then, by Conclusion 1, we know $B K \perp A K, C L \perp A L$, and $L$ is the midpoint of $C N$.
Since $M$ is the midpoint of side $B C$, we know $M L / / A B$.
Since $\angle B K A=\angle B H A=90^{\circ}$, we... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,215 |
Example 2 As shown in Figure 7, given $\triangle ABC$, $X$ is a moving point on line $BC$, and point $C$ is between points $B$ and $X$. The incircles of $\triangle ABX$ and $\triangle ACX$ intersect at two distinct points $P$ and $Q$. Prove that $PQ$ passes through a fixed point independent of $X$.
(45th IMO Shortlist ... | Proof: Let the incircles of $\triangle A B X$ and $\triangle A C X$ touch $B X$ at points $D$ and $F$, and touch $A X$ at points $E$ and $G$, respectively. Then $D E \parallel F G$, and $D E, F G$ are perpendicular to the angle bisector of $\angle A X B$.
Let line $P Q$ intersect $B X$ and $A X$ at points $M$ and $N$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,216 |
Conclusion 5 Let the incircle of non-isosceles $\triangle A B C(A B \neq A C)$ touch sides $B C, C A, A B$ at points $D, E, F$, respectively, and let the altitude $A P$ from $A$ to $B C$ intersect $F E$ at point $H$. Then $H$ is the orthocenter of $\triangle A B C$ if and only if $D H \perp F E$. | Proof As shown in Figure 5, without loss of generality, let $AB > AC$.
Sufficiency.
When $DH \perp FE$, let the line $FE$ intersect $BC$ at point $T$.
By Conclusion 4, we have
$$
\frac{BD}{DC}=\frac{BT}{TC} \Rightarrow \frac{BD}{BT}=\frac{DC}{CT}.
$$
Draw a line $F' E' \parallel FE$ through point $D$ intersecting $BH$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,217 |
Example 3 In $\triangle A B C$, $A B>A C$, the incircle touches side $B C$ at point $E$, and line $A E$ intersects the incircle at point $D$ (different from point $E$). Take a point $F$ on line segment $A E$, different from $E$, such that $C E=C F$. Connect $C F$ and extend it to intersect $B D$ at point $G$. Prove: $C... | Proof As shown in Figure 8, let the incircle touch $AB$ at point $P$, and touch $AC$ at point $Q$. Draw a tangent line to the incircle through point $D$.
Then, by the note to Conclusion 4, we know that the tangent line through point $D$ must pass through the intersection point $T$ of line $PQ$ and $BC$.
By $CF = CE$
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,218 |
Example 4 As shown in Figure $9, \odot O$ is the circumcircle of $\triangle A B C$, $A M$ and $A T$ are the median and angle bisector respectively, the tangents to $\odot O$ at points $B$ and $C$ intersect at point $P$, connect $A P$ and $B C$ intersecting $\odot O$ at points $D$ and $E$ respectively. Prove: $T$ is the... | (2006, National High School Mathematics League Fujian Province Preliminary
Competition)
Proof: Let the tangent line through point $A$ intersect the extensions of $PB$ and $PC$ at points $R$ and $Q$.
For quadrilateral $ABEC$, by conclusion 2, we have
$$
AB \cdot EC = BE \cdot AC.
$$
Applying Ptolemy's theorem, we get
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,219 |
Example 5 In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, the incircle $\odot O$ of $\triangle ABC$ touches the sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Connect $AD$ to intersect $\odot O$ at point $P$, and connect $BP$ and $CP$. If $\angle BPC=90^{\circ}$, prove:
$$
AE + AP ... | Proof: Auxiliary lines and points are marked as shown in Figure 10. $PC, PB$ intersect $\odot O$ at points $G, H$ respectively, and $GH$ intersects $PD$ at point $R$.
By Conclusion 3(2), we can assume that $AD, EH, FG$ are concurrent at point $K$.
Since $\angle GPH = 90^{\circ}$, the center $O$ lies on $GH$.
Because $... | AE + AP = PD | Geometry | proof | Yes | Yes | cn_contest | false | 725,220 |
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 $A C / / B F$
$$
\Rightarrow \frac{3+x}{4}=\frac{A C}{B F} \text {, }
$$
$B C / / A G$
$$
\begin{array}{c}
\Rightarrow \frac{4+x}{3}=\frac{B C}{A G} . \\
\text { ... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,221 |
Example 2 In an isosceles right triangle $\triangle ABC$, $AC=BC=1$, $M$ is the midpoint of side $BC$, $CE \perp AM$ at point $E$, and intersects $AB$ at point $F$. Then $S_{\triangle MBF}=$ $\qquad$
(2006, National Junior High School Mathematics League) | Solve As shown in Figure 2, construct square $C A D B$, and extend $C F$ to intersect $B D$ at point $G$. It is easy to see that,
Rt $\triangle A C M$
$\cong \mathrm{Rt} \triangle C B G$.
Since $M$ is the midpoint of side $B C$,
$$
\begin{array}{c}
B G=B M=C M . \\
\text { By } \triangle B G F \backsim \triangle A C F ... | \frac{1}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,222 |
Example 3 As shown in Figure 3, let $\triangle A B C$ be a right triangle, with point $D$ on the hypotenuse $B C$, and $B D=4 D C$. It is known that a circle passes through point $C$ and intersects $A C$ at point $F$, and is tangent to $A B$ at the midpoint $G$ of $A B$. Prove: $A D \perp B F$.
(1999, National
Junior ... | Solve as shown in Figure 3, construct a square $ACQP$ with $AC$ as the side length, and extend $AD$ to intersect $CQ$ at point $H$. Draw a perpendicular from $P$ to $AD$ intersecting $AC$ at point $E$. Then,
Rt $\triangle APE \cong$ Rt $\triangle CAH \Rightarrow CH = AE$.
Since $CH \parallel AB \Rightarrow \frac{CH}{AB... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,223 |
Example 3 As shown in Figure $5, E$ is a point on side $BC$ of equilateral $\triangle ABC$. An equilateral $\triangle AEF$ is constructed with $AE$ as a side, and $CF$ is connected. On the extension of $CF$, take a point $D$ such that $\angle DAF = \angle EFC$. Determine the shape of quadrilateral $ABCD$ and prove your... | Prove that quadrilateral $ABCD$ is a rhombus.
From $\angle BAE=60^{\circ}-\angle EAC$,
$\angle CAF=60^{\circ}-\angle EAC$,
we get $\angle BAE=\angle CAF$.
Since $AB=AC, AE=AF$, therefore,
$\triangle BAE \cong \triangle CAF \Rightarrow \angle BEA=\angle CFA$.
Notice that,
$\angle BEA=\angle ECA+\angle EAC=\angle EAC+60^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,224 |
Example 4 In the right triangle $\triangle ABC$, $\angle C=90^{\circ}, AC=$ 3, a square $ABEF$ is constructed outward from side $AB$, with the center of the square being $O$, and $OC=4\sqrt{2}$. Then the length of $BC$ is $(\quad)$.
(A) $3\sqrt{2}$
(B) 5
(C) $2\sqrt{5}$
(D) $\frac{9}{2}$ | Solve as shown in Figure 4, draw a perpendicular from point $E$ to the extension of line $CB$ and a perpendicular from point $F$ to the extension of line $CA$, intersecting at point $G$. It is easy to prove
$$
\begin{array}{l}
\cong \text { Rt } \triangle E F G \\
\cong \text { Rt } \triangle F A H \text {. } \\
\text ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,225 |
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 | 725,226 |
Example 6 As shown in Figure 6, in $\triangle A B C$, $\angle A B C=$ $45^{\circ}$, point $D$ is on side $B C$, $\angle A D C=60^{\circ}$, and $B D=$ $\frac{1}{2} C D$. $\triangle A C D$ is reflected over line $A D$ to get $\triangle A C^{\prime} D$, and $B C^{\prime}$ is connected.
(1) Prove: $B C^{\prime} \perp B C$;... | Solution (2) As shown in Figure 6, draw perpendiculars from point $A$ to $BC$, $C'D$, and $BC'$, with the feet of the perpendiculars being $E$, $F$, and $G$ respectively.
From $\angle ABC=45^{\circ}$ and the proven conclusion $BC' \perp BC$, we know that quadrilateral $AGBE$ is a square.
From $\angle ABC=45^{\circ}$,... | 75^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 725,227 |
Example 7 As shown in Figure 7, in quadrilateral $A B C D$, $A B=$ $B C, \angle A B C=\angle C D A=90^{\circ}, B E \perp A D$ at point $E, S_{\text {quadrilateral } A B C D}=8$. Then the length of $B E$ is ( ).
$\begin{array}{ll}\text { (A) } 2 & \text { (B) } 3\end{array}$
(C) $\sqrt{3}$
(D) $2 \sqrt{2}$
(2003, Wuhan ... | Solve As shown in Figure 7, draw a perpendicular line from point $B$ to $CD$, intersecting the extension of $DC$ at point $F$. Then
$$
\begin{array}{l}
\angle C B F=\angle A B E, \\
\angle A E B=\angle C F B=90^{\circ} .
\end{array}
$$
Since $A B=B C$, we have
$$
\triangle A B E \cong \triangle C B F \text {. }
$$
Th... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,228 |
Example 8 As shown in Figure 8, in quadrilateral $ABCD$, $AB=BC=CD$, $\angle ABC=90^{\circ}$, $\angle BCD=150^{\circ}$. Find the degree measure of $\angle BAD$.
(2003, Beijing
Municipal Junior High School Mathematics Competition
(Preliminary)) | As shown in Figure 8, draw a perpendicular line from point $A$ to $AB$, and then draw a perpendicular line from point $C$ to $BC$. The two perpendicular lines intersect at point $E$, and connect $DE$.
It is easy to see that quadrilateral $ABCE$ is a square, and $\triangle CDE$ is an equilateral triangle.
Thus, $AE = E... | 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,229 |
Example 9 Given that the longest diagonal of a regular octagon equals $a$, and the shortest diagonal equals $b$. Then the area of the regular octagon is ( ).
(A) $a^{2}+b^{2}$
(B) $a^{2}-b^{2}$
(C) $a+b$
(D) $a b$
(2009, Beijing Middle School Mathematics Competition) | Solve as shown in Figure 9, connect $A E$, $B D$, $D F$, $F H$, and $H B$. It is easy to see that quadrilateral $B D F H$ is a square, and its side length is exactly the shortest diagonal of the regular octagon $ABCDEFGH$:
$$
B H=B D=D F=F H=b,
$$
and $\triangle A B H \cong \triangle C D B \cong \triangle E F D \cong ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,230 |
Example 1 Given that the circumcircle of $\triangle ABC$ is $\Omega$, the incircle is $\omega$, and the radius of the circumcircle is $R$. Circle $\omega_{A}$ is internally tangent to $\Omega$ at point $A$ and externally tangent to circle $\omega$. Circle $\Omega_{A}$ is internally tangent to $\Omega$ at point $A$ and ... | Proof As shown in Figure 1, let the centers of circles $\omega$ and $\Omega$ be $I$ and $Q$, and the radii of circles $\omega$, $\omega_{A}$, and $\Omega_{A}$ be $r$, $r_{A}$, and $r_{a}$, respectively.
Since circles $\Omega$, $\Omega_{A}$, and $\omega_{A}$ are tangent at point $A$, points $A$, $P_{A}$, $Q_{A}$, and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,232 |
Example 3 As shown in Figure 4, let $D$ be the midpoint of arc $\overparen{B C}$ on the circumcircle $\Gamma$ of acute $\triangle A B C$, point $X$ lies on arc $\overparen{B D}$, $E$ is the midpoint of arc $\overparen{A B X}$, $S$ is a point on arc $\overparen{A C}$, line $S D$ intersects $B C$ at point $R$, and $S E$ ... | (2011, China Mathematical Olympiad)
The paper [3] proves the conclusion by multiple cyclic quadrilaterals and the method of coincidence, which is highly skillful but the proof is tortuous and hard to think of.
Proof: Connect $A D$ and $R T$ intersecting at point $I$. Connect $S A, S B, S C, C I, C D$.
Let $\angle C A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,234 |
Example 4 As shown in Figure 6, in square $A B C D$, $E$ and $F$ are points on sides $B C$ and $C D$ respectively, and $E F=B E+D F$. $A E$ and $A F$ intersect the diagonal $B D$ at points $M$ and $N$ respectively. Prove:
(1) $\angle E A F=45^{\circ}$;
(2) $M N^{2}=B M^{2}+D N^{2}$
【Analysis】This is a
very typical pro... | Proof (1) As shown in Figure 6, rotate $\triangle ABC$ $90^{\circ}$ around point $A$ to get $\triangle ADE_{1}$.
Thus, $AE=AE_{1}, \angle EAE_{1}=90^{\circ}$.
Also, $EF=BE+DF=E_{1}D+DF=E_{1}F$, so $\triangle AEF \cong \triangle AE_{1}F$.
Therefore, $\angle EAF=\angle E_{1}AF=\frac{1}{2} \angle EAE_{1}=45^{\circ}$.
(2) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,235 |
Original problem: Let the circumcircle of acute triangle $\triangle ABC$ be circle $\Gamma$, and let $l$ be a tangent line of circle $\Gamma$. Denote the symmetric lines of tangent line $l$ with respect to lines $BC$, $CA$, and $AB$ as $l_{a}$, $l_{b}$, and $l_{c}$, respectively. Prove: The circumcircle of the triangle... | Proof as shown in Figure 1.
Take the orthocenter $H$ of $\triangle ABC$, and the rays $AH, BH, CH$ intersect the circle $\Gamma$ at $A_1, B_1, C_1$ respectively. Let $l$ be the tangent to the circle $\Gamma$ at point $T$, and the reflections of $T$ about $BC, CA, AB$ are $A_2, B_2, C_2$ respectively. Let the tangent $l... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,236 |
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 | 725,237 |
2. A bag contains 4 identical balls, marked with numbers $1, 2, 3, 4$. After shaking the bag, a ball is randomly drawn, the number is noted, and the ball is put back. The bag is then shaken again, and a ball is randomly drawn. The probability that the number on the ball drawn the second time is not less than the number... | 2. D.
According to the problem, list all possible outcomes as shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline First & 1 & 2 & 3 & 4 \\
\hline 1 & $(1,1)$ & $(2,1)$ & $(3,1)$ & $(4,1)$ \\
\hline 2 & $(1,2)$ & $(2,2)$ & $(3,2)$ & $(4,2)$ \\
\hline 3 & $(1,3)$ & $(2,3)$ & $(3,3)$ & $(4,3)$ \\
\hline 4 & $(1... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,238 |
3. As shown in Figure 1, given a rectangular paper piece $A B C D, A B=3$, $A D=9$, fold it so that point $D$ coincides with point $B$, resulting in the crease $E F$. Then the length of $E F$ is ( ).
(A) $\sqrt{3}$
(B) $2 \sqrt{3}$
(C) $\sqrt{10}$
(D) $\frac{3 \sqrt{10}}{2}$ | 3. C.
Let $DE = x$. Then $BE = x, AE = 9 - x$.
In the right triangle $\triangle ABE$, we have
$$
3^{2} + (9 - x)^{2} = x^{2}.
$$
Solving for $x$ gives $x = 5$, so $DE = 5$.
Connecting $BD$ and intersecting $EF$ at point $G$, then $BD$ perpendicularly bisects $EF$.
In the right triangle $\triangle DEG$, we have
$$
DG ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,239 |
4. In a regular nonagon $A B C D E F G H I$, if the diagonal $A E=2$, then $A B+A C=(\quad)$.
(A) $\sqrt{3}$
(B) 2
(C) $\frac{3}{2}$
(D) $\frac{5}{2}$ | 4. B.
As shown in Figure 5, connect $A D$, and extend $A C$ to point $M$ such that $C M=A B$.
In the regular nonagon $ABCDEFGHI$, we have
$$
\begin{array}{l}
\angle A B C \\
=\frac{(9-2) \times 180^{\circ}}{9} \\
=140^{\circ},
\end{array}
$$
thus $\angle A C B=20^{\circ}$.
Therefore, $\angle A C D=140^{\circ}-20^{\c... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,240 |
5. There are $n$ people registered to participate in four sports competitions: A, B, C, and D. It is stipulated that each person must participate in at least one competition and at most two competitions, but competitions B and C cannot be registered for simultaneously. If in all different registration methods, there mu... | 5. B.
Use the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{J}}\right)$ to represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is recorded as 1 (for example, if participating in event 甲, ... | 172 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,241 |
6. If $\sqrt{x}-\frac{1}{\sqrt{x}}=-2$, then $x^{2}-\frac{1}{x^{2}}=$ | 6. $-24 \sqrt{2}$.
From the given, we know $0<x<1$, and
$$
\begin{array}{l}
\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}=x-2+\frac{1}{x}=4 \\
\Rightarrow x+\frac{1}{x}=6 . \\
\text { Also, }\left(x-\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)^{2}-4=32, \text { so } \\
x-\frac{1}{x}= \pm 4 \sqrt{2} \text { (positiv... | -24 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,242 |
7. If the four lines
$$
x=1, y=-1, y=3, y=k x-3
$$
enclose a convex quadrilateral with an area of 12, then the value of $k$ is $\qquad$. | 7.1 or -2.
From $\left\{\begin{array}{l}y=k x-3, \\ y=-1,\end{array}\right.$ we get the intersection point $A\left(\frac{2}{k},-1\right)$; from $\left\{\begin{array}{l}y=k x-3, \\ y=3,\end{array}\right.$ we get the intersection point $B\left(\frac{6}{k}, 3\right)$. According to the problem, points $A$ and $B$ are on t... | 1 \text{ or } -2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,243 |
8. As shown in Figure 2, a circle $\odot O$ with radius $r$ rolls without slipping along the broken line $A B C D E$. If $A B=B C=C D=D E=2 \pi r, \angle A B C=\angle C D E=150^{\circ}$, $\angle B C D=120^{\circ}$, then $\odot O$ rotates through _____ turns from point $A$ to point $E$. | 8. $4 \frac{1}{3}$.
From $A B=B C=C D=D E=2 \pi r$, we get that when $\odot O$ rolls along the line segments $A B, B C, C D, D E$, it rotates 1 week each;
From $\angle A B C=\angle C D E=150^{\circ}$, we get that when $\odot O$ passes through points $B$ and $D$, it rotates an additional $\frac{1}{12}$ week each;
Fro... | 4 \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,244 |
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