problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
23. The function $y=\sin \left[2\left(x-\frac{\pi}{3}\right)+\varphi\right]$ is an even function, and $0<\varphi<\pi$. Then $\varphi=$ $\qquad$ , and its interval of monotonic decrease is $\qquad$ . | 23. $\frac{\pi}{6},\left[k \pi-\frac{\pi}{2}, k \pi\right](k \in \mathbf{Z})$.
Since $y=\sin \left[2\left(x-\frac{\pi}{3}\right)+\varphi\right]=\sin \left(2 x-\frac{2 \pi}{3}+\varphi\right)$ is an even function, we have
$$
-\frac{2 \pi}{3}+\varphi=k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) \text {. }
$$
Also, because $0<\... | \frac{\pi}{6},\left[k \pi-\frac{\pi}{2}, k \pi\right](k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,230 |
24. The general term formula of the sequence $1,2,3,1,2,3, \cdots$ is
$a_{n}=$ $\qquad$ , and the sum of the first $n$ terms $S_{n}=$ $\qquad$ . (Express each using a single formula) | $24.2+\frac{2}{\sqrt{3}} \sin \frac{2(n-2) \pi}{3}, 2 n-\frac{4}{3} \sin ^{2} \frac{n \pi}{3}$ Observing, we know that this sequence is a periodic sequence with a period of 3. | 2+\frac{2}{\sqrt{3}} \sin \frac{2(n-2) \pi}{3}, 2 n-\frac{4}{3} \sin ^{2} \frac{n \pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,231 |
25. Given the first 10 terms of the sequence $\left\{a_{n}\right\}$ are
$$
\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \text {. }
$$
Then, the 2006th term of this sequence is $\qquad$, and the sum of the first 2006 terms is $\qquad$. | 25. $\frac{53}{64}, 998 \frac{55}{64}$.
For the sequence, group it as follows:
$$
\left(\frac{1}{2}\right),\left(\frac{1}{3}, \frac{2}{3}\right),\left(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}\right),\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right), \cdots \text {. }
$$
Each group has 1 term, 2 terms, 3 te... | \frac{53}{64}, 998 \frac{55}{64} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,232 |
1. Given $A=-\frac{2001 \times 2002 \times 2003}{2004 \times 2005 \times 2006}$,
$$
\begin{array}{l}
B=-\frac{2001 \times 2003 \times 2005}{2002 \times 2004 \times 2006}, \\
C=-\frac{2001 \times 2003 \times 2006}{2002 \times 2004 \times 2005} .
\end{array}
$$
Then the size relationship of $A$, $B$, and $C$ is $(\quad)... | 1. A.
Obviously $AB$. Similarly, $A>C, B>C$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,233 |
2. Given $\sqrt{x^{2}+2008}+\sqrt{x^{2}-2006}=$ 2007. Then
$$
3013 \sqrt{x^{2}-2006}-3005 \sqrt{x^{2}+2008}
$$
is ( ).
(A) 2008
(B) 2009
(C) 2010
(D) 2011 | 2.C.
Let $a=\sqrt{x^{2}+2008}, b=\sqrt{x^{2}-2006}$, then $a^{2}-b^{2}=4014$,
which means $(a+b)(a-b)=4014$.
Given $a+b=2007$. Therefore, $a-b=2$.
Solving, we get $a=1004.5, b=1002.5$.
Thus, the original expression $=3013 b-3005 a=8 b-3005(a-b)$
$$
=8 \times 1002.5-3005 \times 2=2010 \text {. }
$$ | 2010 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,234 |
3. As shown in Figure 1, points $D$ and $E$ are on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at point $O$. The areas of $\triangle OBE$, $\triangle OBC$, and $\triangle OCD$ are $15$, $30$, and $24$, respectively. Then $AE: BE=(\quad)$.
(A) $5: 2$
(B) $2: 1$
(C) $5: 4$
(D) $9: 5$ | 3. B.
As shown in Figure 3, draw $D F / / E C$ intersecting $A B$ at point $F$.
Since $\triangle O B E$ and $\triangle O B C$ have the same height, we have
$$
\frac{O E}{O C}=\frac{S_{\triangle O B E}}{S_{\triangle O H C}}=\frac{1}{2}.
$$
Similarly, $\frac{O B}{O D}=\frac{S_{\triangle O B C}}{S_{\triangle D D}}=\fra... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,235 |
4. Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-2006=0$. Then the algebraic expression
$$
\alpha+\frac{2006}{1+\frac{2006}{1+\frac{2006}{\beta}}}=(\quad) .
$$
(A) -1
(B) 0
(C) 1
(D) 2006 | 4. B.
From the relationship between roots and coefficients, and the definition of the roots of the equation, we get $\alpha+\beta=1, \beta^{2}-\beta-2006=0$.
Thus, $\beta+2006=\beta^{2}$.
Dividing both sides by $\beta$ gives $1+\frac{2006}{\beta}=\beta$.
Therefore, the original expression $=\alpha+\frac{2006}{1+\frac{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,236 |
5. Given two points $A(3,1), B(-3,-5)$ in the Cartesian coordinate system $x O y$, find a point $P$ on the $x$-axis such that the value of $|P A-P B|$ is maximized. Then the maximum value of $|P A-P B|$ is ( ).
(A) $4 \sqrt{2}$
(B) $5 \sqrt{2}$
(C) $2 \sqrt{13}$
(D) $5 \sqrt{3}$ | 5.C.
As shown in Figure 4, construct the point $A_{1}(3,-1)$, which is the reflection of point $A$ over the $x$-axis.
The line $A_{1} B$ intersects the $x$-axis at $P$. Then point $P$ maximizes the value of $\mid P A - P B \mid$.
Assume that $P_{1}$ is any point on the $x$-axis different from point $P$, and connect ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,237 |
6. Point $P$ in the Cartesian coordinate system has coordinates $(a, b)$, where $a$ and $b$ satisfy
$$
\left\{\begin{array}{l}
(a+b)^{2 a}(b+a)^{2 b}=(a+b)^{14}, \\
(a-b)^{2 a+7}(a-b)^{6-2 b}=(a-b)^{11} .
\end{array}\right.
$$
Then the number of possible positions of point $P$ in the Cartesian coordinate system is ( )... | 6.C.
From equation (1), we get
$$
(a+b)^{2 a+2 b}=(a+b)^{14} \text {. }
$$
Thus, $2 a+2 b=14$ or $a+b=0, \pm 1$.
When $a+b=0$, at least one of $a$ and $b$ is 0 or a negative number, making $(a+b)^{2 a}$ or $(b+a)^{2 b}$ undefined.
When $a+b= \pm 1$, $2 a+2 b$ is always even, so equation (1) always holds.
Therefore, $... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,238 |
4. Given real numbers $a$ and $b$ satisfy $a^{2}+a b+b^{2}=1$, and $t=a b$ $-a^{2}-b^{2}$. Try to find the range of $t$.
(2001, National Junior High School Mathematics Competition) | (Tip: Treat $t$ as a parameter, construct a quadratic equation with $a, b$ as its roots. Use the discriminant to find the range of $t$: Answer: $-3 \leqslant t$ $\left.\leqslant-\frac{1}{3}.\right)$ | -3 \leqslant t \leqslant -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,239 |
1. Given $k$ is an irrational number, $b d f>0$,
$$
\frac{a-c}{b-d}=\frac{c-e}{d-f}=\frac{e-k a}{f-k b}=\frac{\sqrt{2}}{2} \text {. }
$$
Then the value of $\frac{a^{2}+c^{2}+e^{2}+a c f+b c e+a d e}{b^{2}+d^{2}+f^{2}+3 b d f}$ is | $=1 . \frac{1}{2}$.
Since $b d f>0$, it follows that $b \neq 0, d \neq 0, f \neq 0$.
Also, $k$ is an irrational number, so $k \neq 1, 1-k \neq 0$. By the property of proportions, we have
$$
\frac{(a-c)+(c-e)+(e-k a)}{(b-d)+(d-f)+(f-k b)}=\frac{\sqrt{2}}{2},
$$
which simplifies to $\frac{a-k a}{b-k b}=\frac{a(1-k)}{b(1... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,240 |
2. Given that $D$ is a point on side $B C$ of $\triangle A B C$,
$$
\begin{array}{l}
\angle B=10^{\circ}, \angle C=20^{\circ}, \angle A D C=85^{\circ}, A B=10, \\
C D=m \text {. Then } B D=
\end{array}
$$ | 2. $\frac{10 m}{10-m}$.
As shown in Figure 5, on $AB$, take $AE = AC$, and connect $DE$.
From the given information, we have
$$
\begin{array}{l}
\angle BAD = \angle ADC - \angle B = 85^{\circ} - 10^{\circ} = 75^{\circ}, \\
\angle CAD = 180^{\circ} - \angle ADC - \angle C = 75^{\circ}.
\end{array}
$$
Thus, $\angle BAD... | \frac{10 m}{10 - m} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,241 |
3. Given that $x, y$ are integers, and
$$
y=\frac{4012}{\sqrt{x+2005}-\sqrt{x-2007}} \text {. }
$$
Then the maximum value of $y$ is $\qquad$ . | 3.2006 .
Rationalizing the denominator of the given equation, we get
$$
y=\sqrt{x+2005}+\sqrt{x-2007} \text {. }
$$
Let $m=\sqrt{x+2005}, n=\sqrt{x-2007}$.
Then $y=m+n, m^{2}=x+2005, n^{2}=x-2007$ $(m>n \geqslant 0)$, obviously $m, n$ are integers.
Eliminating $x$ yields $m^{2}-n^{2}=4012$, that is,
$$
(m+n)(m-n)=2^{... | 2006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,242 |
4. In the inscribed $\triangle A B C$ of $\odot O$, $A B=10, A C$ $=2$, point $P$ is the midpoint of the major arc opposite chord $B C$, $P Q \perp A B$ at $Q$. Then the length of $A Q$ is $\qquad$ | 4.4 or 6.
If point $A$ is on the major arc corresponding to chord $BC$, as shown in Figure 6, extend $BA$ to point $D$ such that $AD = AC$. Connect $PA$, $PB$, $PC$, and $PD$. Then
$$
AD = 2.
$$
Since $\overparen{PB} = \overparen{PC}$, we have
$$
PB = PC, \angle PBC = \angle PCB.
$$
Also, quadrilateral $APBC$ is a c... | 4.4 \text{ or } 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,243 |
One, (20 points) According to the "Personal Income Tax Law of the People's Republic of China," citizens do not need to pay tax on their monthly wages and salaries up to 800 yuan, and the portion exceeding 800 yuan is the taxable income for the whole month. In October 2005, the Standing Committee of the National People'... | Let's assume that Mr. A's total monthly salary and wage income in January 2006 is $x$ yuan, the tax due is $y_{1}$ yuan, the tax due in the past is $y_{2}$ yuan, and the underpaid tax is $y$ yuan, then $y=y_{2}-y_{1}$.
When $0 \leqslant x \leqslant 1600$, $y_{1}=0$.
When $1600<x \leqslant 2100$, $y_{1}=5 \% \cdot(x-160... | 1590 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,244 |
II. (25 points) As shown in Figure 2, $BC$ is the diameter of semicircle $O$, and $D$ is the midpoint of $\overparen{AC}$. The diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $E$.
(1) Prove that $AC \cdot BC = 2 BD \cdot CD$;
(2) If $AE = 3$ and $CD = 2\sqrt{5}$, find the lengths of chord $AB$ and di... | (1) As shown in Figure 8, connect $O D$, intersecting $A C$ at point $F$.
Since $D$ is the midpoint of $\overparen{A C}$, we have
$$
\begin{array}{l}
\overparen{A D}=\overparen{C D}, \\
\angle 1=\angle 2=\angle 3, \\
O D \perp A C, \\
A F=F C=\frac{1}{2} A C .
\end{array}
$$
Also, $B C$ is the diameter of the semicirc... | AB=6, BC=10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,245 |
Three, (25 points) Find the smallest positive integer $k$, such that from the 4006 numbers $1,2, \cdots, 4006$, any $k$ different numbers chosen will always include 4 numbers whose sum is 8013. | Three, from $1,2, \cdots, 4006$ take out the last 2005 numbers: $2002,2003, \cdots, 4006$,
then the sum of any four different numbers is not less than
$$
2002+2003+2004+2005=8014>8013 \text {. }
$$
Thus, $k \geqslant 2006$.
Let $x_{1}, x_{2}, \cdots, x_{2006}$ be any 2006 numbers from $1,2, \cdots, 4006$.
First, divi... | 2006 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,246 |
2. For a point $(x, y)$ on the line $y=2 x+1$, after translating it by the vector $\boldsymbol{a}\left(x_{0}, y_{0}\right)$, we get a point $\left(x_{1}, y_{1}\right)$ on the line $y=2 x-1$, with $x+x_{0}=x_{1}, y+y_{0}=y_{1}$. Then the value of $2 x_{0} - y_{0}$ ( ).
(A) equals 1
(B) equals 2
(C) equals 3
(D) cannot b... | 2.B.
Substitute $x+x_{0}=x_{1}, y+y_{0}=y_{1}$ into the line $y=2 x-1$ to get $y+y_{0}=2\left(x+x_{0}\right)-1$, which is
$$
y=2 x+\left(2 x_{0}-y_{0}-1\right) \text {. }
$$
Equation (1) should be the line $y=2 x+1$. Comparing the two equations, we get
$$
2 x_{0}-y_{0}-1=1 \text {. }
$$
Thus, $2 x_{0}-y_{0}=2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,248 |
5. Let the equation $a x^{2}+(a+2) x+9 a=0$ with respect to $x$ have two distinct real roots $x_{1} 、 x_{2}$, and $x_{1}<\frac{2}{5}$. Then the range of $a$ is:
(A) $a>-\frac{2}{5}$
(B) $-\frac{2}{5}<a<-\frac{2}{7}$
(C) $a<-\frac{2}{7}$
(D) $-\frac{2}{11}<a<0$
(2002, National Junior High School Mathematics Competition) | Answer: D.)
The text above has been translated into English, preserving the original text's line breaks and format. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,250 |
4. Given the parabola $y=a x^{2}+b x+c(a \neq 0)$ intersects the $x$-axis at points $P, Q$ which are on opposite sides of the $y$-axis, and the circle with diameter $P Q$ intersects the $y$-axis at $M_{1}(0,4)$ and $M_{2}(0,-4)$. If the vertex of the parabola is $\left(-\frac{b}{2 a},-\frac{1}{4 a}\right)$, then the cu... | 4. B.
As shown in Figure 3, let the x-coordinates of the points where the parabola intersects the x-axis be $x_{1}$ and $x_{2}$. Then, $\frac{c}{a}=x_{1} x_{2}<0$. In the right triangle $\triangle P M_{i} Q (i=1,2)$, $M_{i} O$ is the altitude to the hypotenuse, so we have
$$
|O P| \cdot|O Q|=\left|O M_{i}\right|^{2},
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,251 |
5. Given the sets
$$
\begin{array}{l}
M_{1}=\{x \mid(x+2)(x-1)>0\}, \\
M_{2}=\{x \mid(x-2)(x+1)>0\} .
\end{array}
$$
Then the set equal to $M_{1} \cup M_{2}$ is ( ).
(A) $\left\{x \mid\left(x^{2}-4\right)\left(x^{2}-1\right)>0\right\}$
(B) $\left\{x \mid\left(x^{2}+4\right)\left(x^{2}-1\right)>0\right\}$
(C) $\left\{x... | 5.B.
From $M_{1}$ we have $x1$.
From $M_{2}$ we have $x2$.
Since the union of $x>1$ and $x2$ are respectively
$$
\begin{array}{l}
|x|>1, \\
|x|>2,
\end{array}
$$
thus, the union of equations (1) and (2) is $|x|>1$, which is $x^{2}-1>0$.
Therefore, $\left(x^{2}+4\right)\left(x^{2}-1\right)>0$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,252 |
6. For $a>3$, let
$$
\begin{array}{l}
x=3 \sqrt{a-1}+\sqrt{a-3}, \\
y=\sqrt{a}+3 \sqrt{a-2} .
\end{array}
$$
Then the relationship between $x$ and $y$ is ().
(A) $x>y$
(B) $x<y$
(C) $x=y$
(D) cannot be determined | 6. B.
From $\sqrt{a}>\sqrt{a-2}, \sqrt{a-1}>\sqrt{a-3}$, adding them we get
$$
\begin{array}{l}
\sqrt{a}+\sqrt{a-1}>\sqrt{a-2}+\sqrt{a-3} \\
\Rightarrow \frac{1}{\sqrt{a}-\sqrt{a-1}}>\frac{1}{\sqrt{a-2}-\sqrt{a-3}}>0 \\
\Rightarrow \sqrt{a}-\sqrt{a-1}<\sqrt{a-2}-\sqrt{a-3} \\
\Rightarrow \sqrt{a}+\sqrt{a-3}<\sqrt{a-1}... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,253 |
1. For a regular triangular prism with base edge length and side edge length both 1, a section is made through one side of the base and the midpoint of the line connecting the centers of the top and bottom bases. What is the area of the section? $\qquad$ . | $=.1 . \frac{4 \sqrt{3}}{9}$.
As shown in Figure 4, let $P$ be the midpoint of the line segment $O_{1} O$ connecting the centers of the upper and lower bases. Draw a plane through $A B$ and $P$ that intersects the extension of $C C_{1}$ at point $E$. Connect $A E$ to intersect $A_{1} C_{1}$ at point $F$, and connect $B... | \frac{4 \sqrt{3}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,254 |
2. Given the complex numbers $z_{1}, z_{2}$ satisfy
$$
\left|z_{1}-z_{2}\right|=\sqrt{3}, z_{1}^{2}+z_{1}+q=0, z_{2}^{2}+z_{2}+q=0 \text {. }
$$
then the real number $q=$ $\qquad$ | 2.1.
From $\left|z_{1}-z_{2}\right|=\sqrt{3}$, we know $z_{1} \neq z_{2}$.
By the definition of the roots of the equation, $z_{1}$ and $z_{2}$ are the two imaginary roots of the quadratic equation $x^{2}+x+q=0$, then we have
$$
\Delta=1-4 q<0 \text{. }
$$
Solving the equation, we get $z_{1,2}=\frac{-1 \pm \sqrt{4 q-1... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,255 |
3: Given the function $f(x)=\sqrt{1-x}(x \leqslant 1)$. Then the coordinates of the intersection point of the function $f(x)$ and its inverse function $f^{-1}(x)$ are $\qquad$ | 3. $(1,0),(0,1),\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right)$.
The inverse function of $f(x)=\sqrt{1-x}(x \leqslant 1)$ is $f^{-1}(x)=1-x^{2}(x \geqslant 0)$.
Solve the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2} .
\end{array}\right.
$$
From equation (1), we get
$$
y^{4}=1-2 ... | (1,0),(0,1),\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,256 |
4. A line segment of length 18 is randomly divided into three segments. The probability that these three segments can form a triangle is $\qquad$ . | 4.0.25.
Let the lengths of two sides of a triangle be $x, y$, then
$$
\left\{\begin{array}{l}
9<x+y<18, \\
0<x<9, \\
0<y<9 .
\end{array}\right.
$$
This forms the medial triangle of Rt $\triangle A O B$ (as shown in Figure 5), the shaded area is $\frac{1}{4}$ of the entire $\triangle A O B$. Therefore, the probability... | 0.25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,257 |
5. Given a spatial quadrilateral $ABCD$ with diagonals $AC$ $=10 \, \text{cm}, BD=6 \, \text{cm}, M, N$ are the midpoints of $AB, CD$ respectively. If the angle formed by the skew lines $AC, BD$ is $60^{\circ}$, then the length of $MN$ is $\qquad$ | 5. $\sqrt{19} \mathrm{~cm}$ or $7 \mathrm{~cm}$.
As shown in Figure 6, take the midpoint $E$ of $B C$, and connect $M E$ and $N E$. By the Midline Theorem of a triangle, we have
$$
\begin{array}{l}
M E=\frac{1}{2} A C=5, \\
N E \perp \frac{1}{2} B D=3 .
\end{array}
$$
By the definition of the angle formed by skew lin... | \sqrt{19} \mathrm{~cm} \text{ or } 7 \mathrm{~cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,258 |
6. Given that $A$ is a subset of $S=\{1,2,3,4,5,6\}$ with at least 2 elements, and $a, b$ are two distinct elements in $A$. When $A$ ranges over $S$ and $a, b$ range over $A$, the total sum of the product $ab$ is $M=$ $\qquad$ | 6.2800 .
For any $\{a, b\} \subset S$ (assuming $a<b$), the number of $A$ that satisfies $\{a, b\} \subseteq A \subseteq S$ is $2^{6-2}=2^{4}$. Therefore, the contribution of $\{a, b\}$ to the total sum $M$ is $a b \times 2^{4}$.
When $a$ takes all values of $1,2,3,4,5$ and $b$ takes all values of $2,3,4,5,6$, we have... | 2800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,259 |
Three. (20 points) Through the point $M(3,2)$ inside the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, draw a line $A B$ intersecting the ellipse at points $A$ and $B$, and draw a line $C D$ intersecting the ellipse at points $C$ and $D$. Draw tangents to the ellipse at $A$ and $B$ intersecting at point $P$, and draw t... | Three, as shown in Figure 7, the tangent line equations through points $A, B, C, D$ are respectively
$$
\begin{aligned}
l_{P A} & : \frac{x_{A} x}{25}+\frac{y_{A} y}{9} \\
& =1, \\
l_{P B} & : \frac{x_{B} x}{25}+\frac{y_{B} y}{9} \\
& =1, \\
l_{Q C} & : \frac{x_{C} x}{25}+\frac{y_{C} y}{9} \\
& =1, \\
& l_{Q n}: \frac{... | \frac{3 x}{25}+\frac{2 y}{9}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,260 |
Example 10 Let $x, y, z > 0$. Try to prove:
$$
\sum \frac{(x+y-z)^{2}}{(x+y)^{2}+z^{2}} \geqslant \frac{3}{5} .
$$ | Explanation 1: $\sum \frac{(x+y-z)^{2}}{(x+y)^{2}+z^{2}} \geqslant \frac{3}{5}$
$$
\Leftrightarrow f(x, y, z)=\sum \frac{(x+y) z}{(x+y)^{2}+z^{2}} \leqslant \frac{6}{5} \text {. }
$$
Assume without loss of generality that $x \geqslant y \geqslant z>0, x+y+z=1$. First, we prove:
$$
\begin{array}{l}
f(x, y, z) \leqslant... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,261 |
$$
\begin{array}{l}
M=\left\{x \mid x^{2}-8 x+k<0\right\}, \\
N=\left\{x \mid x^{2}-6 x+5<0\right\}, \\
P=\left\{x \mid x^{2}-10 x+16<0\right\},
\end{array}
$$
satisfy "if $a \in M$, then $a \in N \cup P$". Find the range of the real number $k$. | $$
\begin{array}{l}
M: 4-\sqrt{16-k}k \geqslant 7, \\
16>k \geqslant 0 .
\end{array}\right.\right.
\end{array}
$$
Solving the inequalities corresponding to sets $M$, $N$, and $P$ yields $7 \leqslant k<16$.
Therefore, the range of real number $k$ is $\{x \mid 7 \leqslant k<16\}$. | 7 \leqslant k<16 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,262 |
Five. (20 points) Arrange all positive divisors of 8128 that are less than itself in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$. Prove that:
$$
\begin{array}{l}
\frac{a_{2}}{2\left(a_{1}^{2}+a_{2}^{2}\right)}+\frac{a_{3}}{3\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)}+\cdots+ \\
\bar{n} \overline{\left(a_{1}^{2}+a_... | Five, from $8128=1 \times 8128=2 \times 4064=4 \times 2032$
$$
=8 \times 1016=16 \times 508=32 \times 254=64 \times 127 \text {, }
$$
we know that $\left\{a_{n}\right\}$ has a total of 13 values, which are
$$
1,2,4,8,16,32,64,127,254,508,1016,2032,4064 .
$$
At the same time, we have
$$
\begin{array}{l}
\sum_{k=1}^{13... | \frac{8127}{8128} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,263 |
One, (50 points) In $\triangle A B C$, $\angle A=45^{\circ}, \angle B=60^{\circ}, O$ is the circumcenter of the triangle. A line through point $A$ parallel to $O B$ intersects the extension of $C B$ at point $D$. Find the value of $\frac{B C}{D B} \sin D$. | As shown in Figure 8, connect radii $O A, O C$, and let $B O$ intersect $A C$ at point $E$.
Since $O$ is the circumcenter of $\triangle A B C$,
we have
$$
\begin{array}{l}
\angle B O C=2 \angle A=90^{\circ}, \\
\angle A O C=2 \angle B=120^{\circ} . \\
\text { Therefore, } \angle O B C=\angle O C B \\
=45^{\circ}, \\
\... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,264 |
$$
f(n)=\left\{\begin{array}{ll}
1, & n=1 \text { when; } \\
2, & 1<n \leqslant 3 \text { when; } \\
3, & 3<n \leqslant 6 \text { when; } \\
\cdots \cdots . & \\
m, & \frac{m(m-1)}{2}<n \leqslant \frac{m(m+1)}{2} \text { when; } \\
\cdots \cdots . &
\end{array}\right.
$$
If $S_{n}=\sum_{k=1}^{n} f(k)=2001$, find the v... | By $1^{2}+2^{2}+\cdots+17^{2}=1785$,
$$
1^{2}+2^{2}+\cdots+18^{2}=2109,
$$
and $1785<2001<2109$,
we know that the last term of the sum $S_{n}=f(1)+f(2)+\cdots+f(n)$ appears in the 18th row, thus, $f(n)=18$.
Also, $1+2+\cdots+17=153$, so $S_{n}$ can be divided into two parts: the first part is the sum of the numbers i... | 165 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,265 |
Three. (50 points) Given $x \in(0, \pi)$. Prove: $\sum_{k=1}^{48} \frac{|\sin (x+k)|}{x+k}>1$.
| Three, as shown in Figure 9, using three lines with inclination angles of $\pm \frac{\pi}{6}$ and $\frac{\pi}{2}$ to divide the unit circle into six equal parts, then the arc length of each part is $\frac{\pi}{3}$,
$$
12$, so, for 7 points $B_{i}(i=1,2, \cdots, 7)$ on the circle with an interval of 1:
$$
x+1, x+2, \cdo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,266 |
Given a cyclic hexagon $A B C D E F$ with unequal sides, and $\frac{A B}{B C} \cdot \frac{C D}{D E} \cdot \frac{E F}{F A}=1$. Prove that $A D, B E, C F$ are concurrent. | Proof: As shown in Figure 1, connect $A D$ and $B E$ intersecting at point $G$, connect $C G$ and extend it to intersect the circle at point $F'$, then connect $A F'$ and $E F'$.
In $\triangle A B G$ and $\triangle E D G$,
since $\angle B A G = \angle D E G$,
$\angle A G B = \angle E G D$,
then $\triangle A B G \backs... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,267 |
Find all integer solutions that satisfy the following conditions:
(1) $a, b, c$ are positive integers;
$$
\text{(2) } a+b+c=[a, b, c] \text{.}
$$ | Solution: First, we discuss the case where $(a, b, c)=1$.
Assume $a \leqslant b \leqslant c$. Since $a+b+c=[a, b, c]$, then $c \mid (a+b)$. Therefore, we have $c=a+b$ or $2c=a+b$.
When $c=a+b$, since $2a+2b=[a, b, c]$, then $b \mid (2a+2b)$, which implies $b=a$ or $b=2a$.
- When $b=a$, $c=2a$. Given $(a, b, c)=1$, we... | a=n, b=2n, c=3n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,268 |
Let $a, b, c, d$ be positive real numbers, and satisfy $a^{2}+b^{2}+c^{2}+d^{2}=4$. Then
$$
\begin{array}{l}
a+b+c+d \geqslant \frac{2}{3}(a b+b c+c d+d a+a c+b d) \\
\geqslant a b c+b c d+c d a+d a b \geqslant 4 a b c d .
\end{array}
$$ | Proof: According to the problem, we have
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{2}(a+b+c+d) \sqrt{a^{2}+b^{2}+c^{2}+d^{2}} \\
=\frac{1}{2} \sqrt{(a+b+c+d)^{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)} . \\
\text { Also, }(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(a b+ \\
b c+c d+d a+c a+b d) .
\end{array}
$$
It is easy to ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,269 |
In $\triangle A B C$, $A B>A C$, and $A F 、 A F^{\prime}$ are the angle bisectors of $\angle B A C$ and its exterior angle, respectively. A semicircle is constructed with $F F^{\prime}$ as its diameter, and point $P$ lies on the semicircle and inside $\triangle A B C$. Prove that:
$$
\angle A P B-\angle A C B=\angle A ... | Proof: As shown in Figure 2, connect $P F^{\prime}$ and extend $B P$ to point $Q$.
Since $A F$ bisects $\angle B A C$ and $A F^{\prime}$ bisects the exterior angle of $\angle B A C$, the circle with $F F^{\prime}$ as its diameter is the Apollonius circle of $\triangle A B C$.
Using the properties of the Apollonius circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,270 |
Example 15 Let $S=\frac{1}{\sqrt[3]{4}}+\cdots+\frac{1}{\sqrt[3]{1000}}$. Prove that $146<S<147$. | Explain: Using the method of partial fraction summation, first prove
$$
\begin{array}{l}
\frac{3}{2}\left(\sqrt[3]{(k+1)^{2}}-\sqrt[3]{k^{2}}\right)<\frac{1}{\sqrt[3]{k}} \\
<\frac{3}{2}\left(\sqrt[3]{k^{2}}-\sqrt[3]{(k-1)^{2}}\right) .
\end{array}
$$
The terms of the partial fraction are precisely the antiderivative ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,274 |
Example 3 Try to determine all rational numbers $r$, such that the equation $r x^{2}+(r+2) x+r-1=0$ has roots and only integer roots.
(2002, National Junior High School Mathematics League) | Explanation: If $r=0$, then the original equation is equivalent to $2 x=1$. Thus, $x=\frac{1}{2}$. This contradicts the problem statement, so $r \neq 0$.
Let $x_{1} 、 x_{2}$ be the integer roots of the quadratic equation in $x$
$$
r x^{2}+(r+2) x+r-1=0
$$
By the relationship between roots and coefficients, we have
$$
... | r=-\frac{1}{3} \text{ or } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,276 |
Example 1 Place the numbers $1,2, \cdots, 8$ in the 8 squares around the perimeter of a $3 \times 3$ chessboard (as shown in Figure 1), such that the sum of the absolute values of the differences between adjacent numbers (numbers in squares that share a common edge) is maximized. Find this maximum value. | To make the description convenient, we define the difference between two adjacent numbers as the larger number minus the smaller number, thus avoiding the concept of absolute value.
To maximize the sum of the 8 differences, we want each difference to be as large as possible. For example, the largest number 8 should be... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,277 |
Proposition 1 Fill the $2n$ numbers from $1 \sim 2n$ into the $2n$ cells of the circular ring in Figure 3, so that the sum of the differences (the larger number minus the smaller number) of the numbers in adjacent cells (cells sharing a common edge) is maximized. This maximum value is $2n^2$ | Prove: By alternately arranging the larger $n$ numbers $n+1, n+2, \cdots$, $2n$ and the smaller $n$ numbers $1, 2, \cdots, n$ in the $2n$ cells of the circular arrangement in Figure 3, the total sum of the $2n$ differences can be maximized (for the same reason as in the solution of Example 1), and its maximum value is
... | 2n^2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,278 |
Example 2 Fill the numbers $1,2, \cdots, 9$ into the squares of a $3 \times 3$ chessboard (as shown in Figure 1), so that the sum of the absolute values of the differences between numbers in adjacent (sharing a common edge) squares is maximized. Find this maximum value. | Solution: To avoid the concept of absolute value, we now define each difference as the larger number minus the smaller number.
To maximize the sum of the 12 differences, we hope each difference is as large as possible, which means the minuend should be as large as possible and the subtrahend as small as possible.
Not... | 58 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,279 |
Proposition 2 Fill the $2n+1$ numbers from $1 \sim 2n+1$ into the $2n+1$ cells of the figure 5 (a ring formed by a regular $n$-sided polygon, divided into $2n$ cells, with a regular $n$-sided polygon cell in the middle), so that the sum of the differences (the larger number minus the smaller number) of the numbers in a... | Proof: Following the solution idea of Example 2, the center of Figure 5 is a regular $n$-sided polygon, which has $n$ adjacent cells. Therefore, the largest number $2 n+1$ should be filled in this regular $n$-sided polygon. It is adjacent to the $n$ surrounding cells, which should be filled with $1,2, \cdots, n$. The r... | \frac{1}{2} n(7 n+1) | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,280 |
Example 3 Fill the numbers $1 \sim 8$ into the 8 squares surrounding the four edges of a $3 \times 3$ chessboard (as shown in Figure 1), so that the sum of the differences (the larger number minus the smaller number) of the adjacent (sharing a common edge) two numbers in these 8 squares is minimized. Find this minimum ... | To make the sum of the 8 differences as small as possible, we hope that each difference is as small as possible. For example, the largest number 8 should have 7 and 6 on its sides, making the differences 1 and 2, which are the smallest (as shown in Figure 6). Therefore, 5 should be placed next to 6 to minimize the diff... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,281 |
Proposition 3 Fill the $2n$ numbers from $1 \sim 2n$ into the $2n$ cells of Figure 3, such that the sum of the differences (the larger number minus the smaller number) of adjacent cells (cells sharing a common edge) is minimized. This minimum value is $4n-2$. | Prove: Following the thought process of Example 3, arrange the $2n$ numbers from 1 to $2n$ in ascending order in the $2n$ cells of Figure 3. This way, we get the minimum value
$$
\begin{array}{l}
{[2n-(2n-1)]+[(2n-1)-(2n-2)]+\cdots+} \\
(2-1)+(2n-1) \\
=[2n+2n+(2n-1)+(2n-2)+\cdots+2]- \\
\quad[(2n-1)+(2n-2)+\cdots+2+1+... | 4n-2 | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,282 |
Example 4 Fill the 9 cells in Figure 8 with the numbers $1 \sim 9$ respectively, so that the sum of the differences (the larger number minus the smaller number) of the adjacent (having a common edge) two numbers in these 9 cells is maximized. Find this maximum value. | To maximize the sum of the 9 differences, we hope that the 9 minuends are as large as possible and the 9 subtrahends are as small as possible. For example, the largest number 9 should be adjacent to the smallest number 1 and the second smallest number 2, the second largest number 8 should be adjacent to 1, and then to ... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,283 |
Proposition 4 Fill the $2 n+1$ numbers from $1 \sim 2 n+1$ into the $2 n+1$ cells of Figure 3, so that the total sum of the differences (the larger number minus the smaller number) of adjacent cells (with a common edge) is maximized. This maximum value is $2 n(n+1)$. | Prove: As in Example 4, interleave $2 n+1, 2 n, \cdots, n+2$ with $1, 2, \cdots, n$, and insert the number $n+1$ between $n$ and $n+2$. This ensures that the numbers $2 n+1, 2 n, \cdots, n+2$ are the $n$ larger numbers, each serving as a minuend twice, while the $n$ smaller numbers $1, 2, \cdots, n$ each serve as a sub... | 2 n(n+1) | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,284 |
Given $x, y, z$ are positive real numbers. Prove:
$$
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4} .
$$ | Proof: Let $m=2 x+y+z>0$,
$$
n=x+2 y+z>0, p=x+y+2 z>0 \text {. }
$$
Then $x=\frac{3 m-n-p}{4}$,
$$
\begin{array}{l}
y=\frac{3 n-m-p}{4}, \\
z=\frac{3 p-m-n}{4} .
\end{array}
$$
Let $M=\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z}$.
$$
\begin{array}{l}
\text { Then } M=\frac{3 m-n-p}{4 m}+\frac{3 n-m-p}{4 n}+\... | \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,285 |
Example 4 Let $a$ be a real number greater than zero. It is known that there exists a unique real number $k$, such that the quadratic equation in $x$
$$
x^{2}+\left(k^{2}+a k\right) x+1999+k^{2}+a k=0
$$
has two roots that are both prime numbers. Find the value of $a$.
(1999, National Junior High School Mathematics Co... | Let the two prime roots of the equation be $p, q$. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
p+q=-k^{2}-a k, \\
p q=1999+k^{2}+a k . \\
\text { Adding (1) and (2) gives } p+q+p q=1999, \text { i.e., } \\
(p+1)(q+1)=2^{4} \times 5^{3} .
\end{array}
$$
From equation (3), it is clear... | 2 \sqrt{502} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,287 |
Example 2 Let $a, b, c, d$ all be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+ \\
\frac{c}{d+2 a+3 b}+\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3} .
\end{array}
$$
(34th IMO Shortlist) | Proof: Let $m=b+2c+3d>0$,
$$
\begin{array}{l}
n=c+2d+3a>0, \\
p=d+2a+3b>0, \\
q=a+2b+3c>0 .
\end{array}
$$
Then $a=\frac{7n+p+q-5m}{24}$,
$$
\begin{array}{l}
b=\frac{7p+q+m-5n}{24}, \\
c=\frac{7q+m+n-5p}{24}, \\
d=\frac{7m+n+p-5q}{24} .
\end{array}
$$
Hence $\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+$
$$
\begin{array}{l}
\... | \frac{2}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,288 |
Example 3 Let $x, y, z$ be positive real numbers. Prove that:
$$
\frac{z^{2}-x^{2}}{x+y}+\frac{x^{2}-y^{2}}{y+z}+\frac{y^{2}-z^{2}}{z+x} \geqslant 0 .
$$ | Proof: Let \( m = x + y > 0, n = y + z > 0, p = z + x > 0 \). Then
\[
\begin{array}{l}
x = \frac{m + p - n}{2}, y = \frac{m + n - p}{2}, \\
z = \frac{n + p - m}{2} \text{. } \\
\text{ Hence } \frac{z^2 - x^2}{x + y} + \frac{x^2 - y^2}{y + z} + \frac{y^2 - z^2}{z + x} \\
= \frac{(n - m) p}{m} + \frac{(p - n) m}{n} + \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,289 |
In a singles round-robin table tennis tournament (i.e., each pair of players competes in one match, with no ties), $n(n \geqslant 2)$ players participate. After the tournament, outstanding players are determined. For players $A$ and $B$, if $A$ beats $B$ or $A$ indirectly beats $B$ (i.e., there exists a player $C$ such... | Proof: After the competition, let all participants rest in the waiting room. Then, arbitrarily designate a participant $A_{1}$, and have $A_{1}$ along with all his defeated opponents (i.e., those defeated by $A_{1}$) leave the waiting room. After this, if there are no more participants in the waiting room, then $A_{1}$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,290 |
Property 2 In a regular hexagon $A B C D E F$, if the line containing diagonal $A D$ intersects diagonals $C E$ and $B F$ at points $G$ and $H$, then
$$
\begin{array}{l}
\frac{A B}{B C} \cdot \frac{C G}{G E} \cdot \frac{E F}{F A}=1 ; \\
\frac{C F}{F D} \cdot \frac{D B}{B E} \cdot \frac{E G}{G C}=1 ; \\
\frac{D G}{G A} ... | Proof: Only the formula is proved, the proof of other formulas is similar.
For $\triangle A C G$
and the transversal $B D E$, applying Menelaus' theorem, we have
$$
\frac{A B}{B C} \cdot \frac{C E}{E G} \cdot \frac{G D}{D A}=1 .
$$
For $\triangle A G E$ and the transversal
$C D F$, applying Menelaus' theorem, we have... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,292 |
Property 3 In a complete hexagon $A B C D E F$, the circumcircles of triangles $\triangle A B E, \triangle B C D, \triangle A C F, \triangle D E F$ concur at a point (this point is called the Miquel point). | Proof: As shown in Figure 4, let the circumcircles of $\triangle BCD$ and $\triangle DEF$ intersect at point $D$ and another point $M$.
Let the projections of point $M$ onto lines $CB$, $CD$, and $BD$ be $P$, $Q$, and $R$, respectively. By the Simson line theorem, points $P$, $Q$, and $R$ are collinear.
Similarly, th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,293 |
Property 6 In a complete quadrilateral, if a diagonal line intersects with the other two diagonal lines, then this line is harmonically divided by the other two diagonal lines.
As shown in Figure 7, in the complete quadrilateral $A B C D E F$, if the diagonal $A D$ intersects with the diagonals $B F$ and $C E$ at poin... | Prove: Only prove that when $B F / X C E$, we have $\frac{A M}{A N}=\frac{M D}{N D}$. The other two formulas can be similarly proved.
$$
\begin{array}{c}
\text { Let } \angle C A N=\alpha, \angle N A E=\beta, A B=b, A C= \\
c, A M=m, A D=d, A N=n, A F=f, A E=e .
\end{array}
$$
Taking $A$ as the viewpoint, apply the an... | \frac{A M}{A N}=\frac{M D}{N D} | Geometry | proof | Yes | Yes | cn_contest | false | 717,296 |
Property 7 In a complete quadrilateral $A B C D E F$, the midpoints $M, N, P$ of the three diagonals $A D, B F, C E$ are collinear. | Proof: As shown in Figure 8, take the midpoints $Q, R, S$ of $CD$, $BD$, and $BC$ respectively.
Thus, in $\triangle ACD$, points $M$, $R$, and $Q$ are collinear;
in $\triangle BCF$, points $S$, $R$, and $N$ are collinear;
in $\triangle BCE$, points $S$, $Q$, and $P$ are collinear.
By the properties of parallel lines, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,297 |
Example 5 Given that $p$ and $q$ are both prime numbers, and that the quadratic equation in $x$
$$
x^{2}-(8 p-10 q) x+5 p q=0
$$
has at least one integer root. Find all pairs of prime numbers $(p, q)$.
(2005, National Junior High School Mathematics Competition) | Explanation: According to the sum of the roots of the equation being \(8 p-10 q\) and the product of the roots being \(5 p q\), if the equation has one positive integer root, then the other root must also be a positive integer. Let's assume the two positive integer roots of the equation are \(x_{1}\) and \(x_{2}\left(x... | (7,3),(11,3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,298 |
Property 10 In: In a complete quadrilateral $A B C D E F$, $G$ is a point on the line containing diagonal $A D$, connect $B G$, $C G$, $E G$, $F G$. If $\angle A G C=\angle A G E$, then $\angle A G B=\angle A G F$ | Proof: As shown in Figure 11, point $G$ is on the extension of $DA$. Draw a line $a \perp AD$ through point $G$, and draw lines $BM \perp a$ at point $M$, intersecting $CD$ at point $M_{1}$ and $CG$ at point $M_{2}$. Draw line $FN \perp a$ at point $N$, intersecting $DE$ at point $N_{1}$ and $GE$ at point $N_{2}$.
Then... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,301 |
2. Given that $m$ is a real number, and $\sin \alpha, \cos \alpha$ satisfy the quadratic equation $3 x^{2}-m x+1=0$ in $x$. Then $\sin ^{4} \alpha+$ $\cos ^{4} \alpha$ is ( ).
(A) $\frac{2}{9}$
(B) $\frac{1}{3}$
(C) $\frac{7}{9}$
(D) 1 | 2.C.
From the relationship between roots and coefficients, we know that $\sin \alpha \cdot \cos \alpha=\frac{1}{3}$. Therefore, we have
$$
\begin{array}{l}
\sin ^{4} \alpha+\cos ^{4} \alpha \\
=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2(\sin \alpha \cdot \cos \alpha)^{2}=\frac{7}{9} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,303 |
3. For the equation in $x$, $\left|\frac{x^{2}}{x-1}\right|=a$, there are only two distinct real roots. Then the range of the constant $a$ is ( ).
(A) $a>0$
(B) $a \geqslant 4$
(C) $2<a<4$
(D) $0<a<4$ | 3.D.
When $a>0$, the original equation becomes $\frac{x^{2}}{x-1}= \pm a$. Rearranging gives
$$
x^{2}-a x+a=0,
$$
or $x^{2}+a x-a=0$.
Since the discriminant of equation (2) is greater than 0, equation (2) has two distinct real roots.
Since the original equation has only two distinct real roots, the discriminant of e... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,304 |
4. Let $b>0, a^{2}-2 a b+c^{2}=0, b c>a^{2}$. Then the size relationship of the real numbers $a, b, c$ is ( ).
(A) $b>c>a$
(B) $c>a>b$
(C) $a>b>c$
(D) $b>a>c$ | 4. A.
From $bc > a^2$ and $b > 0$, we know $c > 0$.
From $2ab = a^2 + c^2$ and $b > 0$, we know $a > 0$.
From $a^2 - 2ab + c^2 = 0$, we know $b^2 - c^2 = (a - b)^2 \geqslant 0$.
Thus, $b \geqslant c$.
If $b = c$, from $a^2 - 2ab + c^2 = 0$, we know $a = b$.
Thus, $a = b = c$. This contradicts $bc > a^2$, so $b > c$.
F... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,305 |
5. $a$ and $b$ are rational numbers, and satisfy the equation
$$
a+b \sqrt{3}=\sqrt{6} \times \sqrt{1+\sqrt{4+2 \sqrt{3}}} \text {. }
$$
Then the value of $a+b$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 5. B.
Since $\sqrt{6} \times \sqrt{1+\sqrt{4+2 \sqrt{3}}}=\sqrt{6} \times \sqrt{1+(1+\sqrt{3})}$ $=\sqrt{12+6 \sqrt{3}}=3+\sqrt{3}$,
thus, $a+b \sqrt{3}=3+\sqrt{3}$, which means $(a-3)+(b-1) \sqrt{3}=0$.
Since $a, b$ are rational numbers, then $a=3, b=1$, hence $a+b=4$. | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,306 |
6. Arrange the integers that satisfy the condition "contain at least one digit 0 and are multiples of 4" in ascending order:
$$
20,40,60,80,100,104, \cdots \text {. }
$$
Then the 158th number in this sequence is ( ).
(A) 2000
(B) 2004
(C) 2008
(D) 2012 | 6.C.
Among positive integers, the characteristic of being a multiple of 4 is that the last digit is a multiple of 4, which includes 7 forms containing the number 30: 00, 04, 08, 20, 40, 60, 80, and 18 scenarios not containing the number 30.
Obviously, there are only 4 two-digit numbers that satisfy the condition; the... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,307 |
1. The sum of the x-coordinates of the x-axis intercepts of the graph of the function $y=x^{2}-2006|x|+2008$ is $\qquad$ | $=.1 .0$.
The original problem is transformed into solving the equation
$$
x^{2}-2006|x|+2008=0
$$
We are to find the sum of all real roots.
If a number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1). Therefore, the sum of all real roots of the equation is 0, i.e., ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,308 |
Example 6 Let $a, b, c$ be distinct real numbers, and satisfy the relation
$$
b^{2}+c^{2}=2 a^{2}+16 a+14,
$$
and $bc=a^{2}-4a-5$.
Find the range of values for $a$.
(2006, Hunan Province Junior High School Mathematics Competition) | From the given equations, we have
$$
\begin{array}{l}
(b+c)^{2} \\
=2 a^{2}+16 a+14+2\left(a^{2}-4 a-5\right) \\
=4(a+1)^{2} .
\end{array}
$$
Thus, \( b+c = \pm 2(a+1) \).
Therefore, we can construct a quadratic equation with roots \( b \) and \( c \):
$$
x^{2} \pm 2(a+1) x + a^{2} - 4 a - 5 = 0 .
$$
Since equation (... | a > -1 \text{ and } a \neq \frac{1 \pm \sqrt{21}}{4}, a \neq -\frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,309 |
3. The real number $x$ that makes $\sqrt{x^{2}}+4+\sqrt{(8-x)^{2}}+16$ take the minimum value is estimated to be
| 3. $\frac{8}{3}$.
Approximately 5
As shown in Figure 5, in the Cartesian coordinate system $x(0)$, let $A(0,-2), B(8, 4)$. $P(x, 0)$, then
$$
\begin{array}{l}
|P A|=\sqrt{x^{2}+4}, \\
|P B|=\sqrt{(8-x)^{2}+16} .
\end{array}
$$
Thus, $|P A|+|P B| \geqslant|A B|$
$$
=\sqrt{8^{2}+6^{2}}=10 \text {. }
$$
Equality holds i... | \frac{8}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,311 |
The coordinates of the top points are $0(0,0), A(100,0)$, $B(100,100), C(0,100)$. If a lattice point $P$ is inside the square $O A B C$,
then the lattice point $P$ is called a "good point". The number of good points inside the square $O A B C$ is . $\qquad$ | 4. 197.
As shown in Figure 6, through point $P$, draw
$P I, P E, P F, P G$ perpendicular to sides
$O A, A B, B C, O C$ at $D, E$,
$F, G$. It is easy to see that
$$
\begin{array}{l}
P F+P D=100, \\
P E+P G=100 \text {. } \\
\text { If| } S_{\text {эхи }} \cdot S_{\text {эми: }} \\
=S_{\triangle P U} \cdot S_{\triangle ... | 197 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,312 |
One, (20 points) Given the equation in terms of $x$
$$
x^{2}+2(a+2 b+3) x+\left(a^{2}+4 b^{2}+99\right)=0
$$
has no distinct real roots. How many ordered pairs of positive integers $(a, b)$ satisfy this condition? | Given that $x^{2}+2(a+2 b+3) x+\left(a^{2}+4 b^{2}+99\right)$ $=0$ has no two distinct real roots, therefore,
$$
\Delta=[2(a+2 b+3)]^{2}-4\left(a^{2}+4 b^{2}+99\right) \leqslant 0 \text {. }
$$
Simplifying to $2 a b+3 a+6 b \leqslant 45$, then
$$
(a+3)(2 b+3) \leqslant 54 \text {. }
$$
Since $a, b$ are positive integ... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,313 |
II. (25 points) As shown in Figure $1 . D$ is the midpoint of the base $BC$ of isosceles $\triangle ABC$, and $E, F$ are points on $AC$ and its extension, respectively. Given that $\angle EDF=90^{\circ}, ED=DF=1, AD=5$. Find the length of segment $BC$. | II. As shown in Figure 7, draw $EG \perp AD$ at point $G$. Draw $FH \perp AD$ at point $H$. Then $\angle EDG = \angle DFH$. Therefore,
$\mathrm{Rt} \triangle EDG \cong \mathrm{Rt} \triangle DFH$.
Let $EG = x, DG = y$. Then
$$
DH = x, FH = y \text{. }
$$
II. $x^{2} + y^{2} = 1$.
Also, $\mathrm{Rt} \triangle AEG \sim \ma... | \frac{10}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,314 |
Three. (25 points) As shown in Figure 2, in parallelogram $ABCD$, the angle bisector of $\angle A$ intersects the extensions of $BC$ and $DC$ at points $E$ and $F$, respectively. Points $O$ and $O_1$ are the circumcenters of $\triangle CEF$ and $\triangle ABE$, respectively. Prove:
(1) $O$, $E$, and $O_1$ are collinear... | (1) As shown in Figure 8, connect $O E, O F, O A, O E$.
Since quadrilateral $A B C D$ is a
parallelogram, therefore,
$$
\angle A B E = \angle E C F.
$$
Also, since points $O$ and $O_1$ are the circumcenters of $\triangle C E F$ and $\triangle A B E$ respectively, we have
$$
\begin{array}{l}
O E = O F, O_1 A = O_1 E, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,315 |
Three. (25 points) Let $p$ be a positive integer, and $p \geqslant 2$. In the Cartesian coordinate system, the line segment connecting point $A(0, p)$ and point $B(p, 0)$ passes through $p-1$ lattice points $C_{1}(1, p-1), \cdots$, $C_{i}(i, p-i), \cdots, C_{p-1}(p-1,1)$. Prove:
(1) If $p$ is a prime number, then the l... | (1) Let $P(a, b)$ represent the lattice points within $\triangle O A B$, where $a, b$ are positive integers.
Assume the conclusion is not true. Then point $P$ is located on some line segment $O C_{i}$ (as shown in Figure 9). Draw $P E \perp O B$ at point $E$, and draw $C_{i} F \perp O B$ at point $F$. We know $\frac{b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,316 |
1. With $A B$ as a side, construct an equilateral triangle $\triangle A B F$ outside the regular pentagon $A B C D E$. Then $\angle C F E$ equals ( ).
(A) $36^{\circ}$
(B) $48^{\circ}$
(C) $72^{\circ}$
(D) $108^{\circ}$ | 1. B.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,317 |
2. If $b_{1}, b_{2}$ satisfy the inequality $\left(x-a_{1}\right)\left(x-a_{2}\right)<0$ with respect to $x$, and $b_{1}<b_{2}, a_{1}<a_{2}$, then which of the following conclusions is correct? ( ).
(A) $a_{1}<b_{1}<b_{2}<a_{2}$
(B) $b_{1}<a_{1}<b_{2}<a_{2}$
(C) $a_{1}<b_{1}<a_{2}<b_{2}$
(D) $b_{1}<a_{1}<a_{2}<b_{2}$ | 2.A.
$$
a_{1}<b_{1}<a_{2}, a_{1}<b_{2}<a_{2} .
$$ | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,318 |
3. If $a+b+c=0, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-4$, then the value of $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$ is ( ).
(A) 3
(B) 8
(C) 16
(D) 20 | 3.C. From $a+b+c=0$, we know $\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=0$. Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}=16$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,319 |
Example 7 Given that $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=a^{2}, \\
x+y+z=b,
\end{array}\right.
$$
where $a, b$ are real constants, and $|b| \leqslant \sqrt{3}|a|$. Try to find the range of values for $x$. | Explanation: According to the problem, the original system of equations is equivalent to
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=a^{2}-x^{2}, \\
y+z=b-x .
\end{array} \text { (considering } x\right. \text { as a parameter) }
$$
Since $y z=\frac{(y+z)^{2}-\left(y^{2}+z^{2}\right)}{2}$
$$
=\frac{1}{2}\left[(b-x)^{2}-\lef... | \left[\frac{b-\sqrt{2\left(3 a^{2}-b^{2}\right)}}{3}, \frac{b+\sqrt{2\left(3 a^{2}-b^{2}\right)}}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,320 |
4. As shown in Figure 1, in trapezoid $A B C D$, $A D / / B C, A D$ $=2, A C=4, B C=6, B D=8$. Then the area of trapezoid $A B C D$ is ( ).
(A) $4 \sqrt{15}$
(B) 16
(C) $8 \sqrt{15}$
(D) 32 | 4. A.
As shown in Figure 6, draw $DE // AC$ intersecting the extension of $BC$ at point $E$. Then the area of $\triangle AC$ is equal to the area of $\triangle BDE$ while $BD=BE=8, DE=4$. Therefore, $S_{\triangle ADEE}=4\sqrt{15}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,321 |
5. If $(x+1)(y+1)=x^{2} y+x y^{2}=6$, then, $x^{2}+y^{2}$ equals $(\quad)$.
(A) 6
(B) 5
(C) 4
(D) 3 | 5.B.
From $\left\{\begin{array}{l}x y+(x+y)=5, \\ x y(x+y)=6,\end{array}\right.$ we get $\left\{\begin{array}{l}x+y=3, \\ x y=2\end{array}\right.$ or $\left\{\begin{array}{l}x+y=2, \\ x y=3\end{array}\right.$ (discard). Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=9-4=5$. | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,322 |
6. A line divides a parallelogram into two congruent parts, then there are () such lines.
(A) 1
(B) 2
(C) 3
(D) Infinite
untranslated part: "元数" should be "Infinite" in this context. | 6.D.
Any line passing through the center of symmetry satisfies the condition. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,323 |
7. If $a^{2}-14 a+1=0$, then the tens digit of $a^{4}+\frac{1}{a^{4}}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 7.B.
$$
a+\frac{1}{a}=14, a^{2}+\frac{1}{a^{2}}=14^{2}-2=194 \text{, }
$$
- $a^{4}+\frac{1}{a^{4}}=194^{2}-2$.
Then $194^{2}-2=(200-6)^{2}-2 \equiv 34(\bmod 100)$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,324 |
8. If the equation in terms of $x$
$$
|x+1|+|x-1|=a
$$
has real roots, then the range of values for the real number $a$ is (. ).
(A) $a \geqslant 0$
(B) $a>0$
(C) $a \geqslant 1$
(D) $a \geqslant 2$ | 8. D.
$$
\begin{array}{l}
|x+1|+|x-1| \\
=\left\{\begin{array}{cc}
2 x & x \geqslant 1 ; \\
2, & -1<x<1, \\
-2 x & x \leqslant-1 .
\end{array}\right.
\end{array}
$$
Therefore, $a \geqslant 2$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,325 |
10. Given that $a, b, 2$ are the side lengths of a triangle, and $a, b$ are the roots of the equation
$$
\left(3 x^{2}-4 x-1\right)\left(3 x^{2}-4 x-5\right)=12
$$
the perimeter of the triangle can only be ( ).
(A) $\frac{10}{3}$ or $\frac{8}{3}$
(B) $\frac{14}{3}$ or $\frac{10}{3}$
(C) $\frac{16}{3}$ or $\frac{14}{3}... | 10.D.
From $\left(3 x^{2}-4 x\right)^{2}-6\left(3 x^{2}-4 x\right)-7=0$, we get $\left(3 x^{2}-4 x-7\right)\left(3 x^{2}-4 x+1\right)=0$.
Solving, we find $x_{1}=-1, x_{2}=\frac{7}{3}, x_{3}=1, x_{4}=\frac{1}{3}$. If $a \neq b$, from $a+b>2,|a-b|2$, we know $a=b=\frac{7}{3}$.
Thus, the perimeter of the triangle is $\f... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,327 |
12. If the integer part of $\frac{1}{\sqrt{17-12 \sqrt{2}}}$ is $a$, and the fractional part is $b$, then, $a^{2}-$ $a b+b^{2}$ is $\qquad$ . | $12.47-18 \sqrt{2}$.
It is easy to know that $a=5, b=2(\sqrt{2}-1)$.
Therefore, $a^{2}-a b+b^{2}=(a+b)^{2}-3 a b$ $=17+12 \sqrt{2}-30(\sqrt{2}-1)=47-18 \sqrt{2}$. | 47-18 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,329 |
13. If $n^{2}+100$ can be divided by $n+10$, then, the maximum positive integer value of $n$ that satisfies the condition is $\qquad$ | 13. 190.
Since $n^{2}+100$ and $n^{2}-100$ are both divisible by $n+10$, therefore, 200 is divisible by $n+10$.
Hence, the largest positive integer that satisfies the condition is 190. | 190 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,330 |
Example 8 When $m$ satisfies what conditions, the equation
$$
(m+3) x^{2}-4 m x+2 m-1=0
$$
has roots of opposite signs and the absolute value of the negative root is greater than the positive root? | Explanation: For equation (1), the conditions for the roots to have opposite signs and the absolute value of the negative root to be greater than the positive root are
$$
\left\{\begin{array}{l}
m+3 \neq 0, \\
\Delta=16 m^{2}-4(m+3)(2 m-1)>0, \\
x_{1}+x_{2}=\frac{4 m}{m+3}<0, \\
x_{1} x_{2}=\frac{2 m-1}{m+3}<0 .
\end{a... | -3<m<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,331 |
14. As shown in Figure 4, in: parallelogram
$ABCD$, points $E$ and
$F$ are on: $AD$ and $CD$ respectively.
$AF$ and $CE$ intersect at point $O$.
The following statements are given:
(1) If $S_{\triangle ABF}=S_{\triangle HEC}$, then $AF=CE$;
(2) If $AF=CE$, then $S_{\triangle ARF}=S_{\triangle BEC}$;
(3) If $AF=C... | 14.(2)(3)(4) . Let the distances from point $B$ to $A F$ and $C E$ be $h_{1}$ and $h_{2}$, respectively. From the above identity, we know that $A F \cdot h_{1}=C E \cdot h_{2}$. Therefore, $A F=C E \Leftrightarrow h_{1}=h_{2} \Leftrightarrow \angle A O B=\angle B O C$. The correct conclusions are (2), (3), (4). | (2)(3)(4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,332 |
15. The school offers four extracurricular interest classes in Chinese, Math, Foreign Language, and Natural Science for students to voluntarily sign up for. The number of students who want to participate in the Chinese, Math, Foreign Language, and Natural Science interest classes are 18, 20, 21, and 19, respectively. I... | Three, 15. The number of people not attending the Chinese interest class is 7, the number of people not attending the Math interest class is 5, the number of people not attending the Foreign Language interest class is 4, and the number of people not attending the Natural Science interest class is 6. Therefore, the maxi... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,333 |
16. As shown in Figure 5, in $\triangle ABC$, $AB = AC$, $AD \perp BC$, with the foot of the perpendicular being $D$. $E$ and $G$ are the midpoints of $AD$ and $AC$, respectively. $DF \perp BE$, with the foot of the perpendicular being $F$. Prove:
$$
FG = DG.
$$ | 16. As shown in Figure 7, connect $A F$ and $C F$.
Since $A D \perp B C$,
$D F \perp B E$,
then $\angle F D B = \angle F E D$,
$\frac{D F}{E F} = \frac{B D}{E D}$.
Thus, $\angle F D C = \angle F E A$,
$\frac{D F}{E F} = \frac{D C}{E A}$.
Therefore, $\triangle D F C \sim \triangle E F A$.
Hence, $\angle D F C = \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,334 |
2. When $\theta$ takes all real numbers, the area of the figure enclosed by the line
$$
x \cos \theta+y \sin \theta=4+\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)
$$
is ( ).
(A) $\pi$
(B) $4 \pi$
(C) $9 \pi$
(D) $16 \pi$ | 2.1).
The equation of the line becomes $(x-1) \cos \theta+(y-1) \sin \theta=4$. Therefore, the distance from point $A(1,1)$ to the line is
$$
d=\frac{4}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=4 \text {. }
$$
Thus, when $\theta \in \mathbf{R}$, the line
$$
x \cos \theta+y \sin \theta=4+\sqrt{2} \sin \left(\theta+\f... | 16 \pi | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,336 |
3. In the sequence $\left\{a_{n}\right\}$, adjacent terms $a_{n} 、 a_{n+1}$ are the roots of the equation $x^{2}+3 n x+b_{n}=0$. It is known that $a_{11}=-17$. Then the value of $b_{s 1}$ is ( ).
(A) 5800
(B) 5840
(C) 5860
(D) 6000 | 3. B.
Given that $a_{n}+a_{n+1}=-3 n$. Then
$$
\begin{array}{l}
a_{n+2}-a_{n}=\left(a_{n+2}+a_{n+1}\right)-\left(a_{n+1}+a_{n}\right) \\
=-3(n+1)-(-3 n)=-3 .
\end{array}
$$
Therefore, $a_{1}, a_{3}, \cdots, a_{2 n+1}$ and $a_{2}, a_{4}, \cdots, a_{2 n}$ are both arithmetic sequences with a common difference of -3. He... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,337 |
4. Given that $a, b, c, d$ are four different rational numbers, and $(a+c)(a+d)=1,(b+c)(b+d)=1$. Then the value of $(a+c)(b+c)$ is $(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -1 | 4. D.
From the problem, we have
$$
\begin{array}{l}
a^{2}+(c+d) a+c d-1=0 . \\
b^{2}+(c+d) b+c d-1=0 .
\end{array}
$$
Then \(a, b\) are the roots of the equation \(x^{2}+(c+d) x+c d-1=0\). Therefore, we have
$$
a+b=-(c+d), \quad a b=c d-1 .
$$
Thus, \((a+c)(b+c)=a b+(a+b) c+c^{2}\).
$$
\text { Hence }(a+c)(b+c)=-1 \... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,338 |
5. Let the function $f(x)=x^{2}+6 x+8$. If $f(b x+c)=4 x^{2}+16 x+15$, then $c-2 b=(\quad)$.
(A) 3
(B) 7
(C) -3
(D) -7 | 5. C.
Take $x=-2$, we have $f(c-2b)=16-16 \times 2+15=-1$. And when $x^{2}+6x+8=-1$, we have $x=-3$. Therefore,
$$
c-2b=-3 \text{. }
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,339 |
6. Given that $p$, $p+14$, and $p+q$ are all prime numbers, and $p$ has a unique value corresponding to it. Then $q$ can only be ( ).
(A) 40
(B) 44
(C) 74
(D) 86 | 6. A.
$q$ can only be 40. When $p=3k$, $p$ can only equal 3, which meets the requirement; when $p=3k+1$, $p+14$ is not a prime number; when $p=3k+2$, $p+40$ is not a prime number. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,340 |
1. The line $l$ passes through the focus of the parabola $y^{2}=a(x+1)(a>0)$ and is perpendicular to the $x$-axis. If the segment cut off by $l$ on the parabola is 4, then $a=$ $\qquad$ . | $=.1 .4$.
Since the parabolas $y^{2}=a(x+1)$ and $y^{2}=a x$ have the same length of the focal chord with respect to their directrix, the general equation $y^{2}=a(x+1)$ can be replaced by the standard equation $y^{2}=a x$ for solving, and the value of $a$ remains unchanged. Using the formula for the length of the latu... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,341 |
Example 9 Given the equation
$$
x^{2}-11 x+30+m=0
$$
has two real roots both greater than 5. Find the range of values for $m$.
| Let $y=x-5$, then the original equation is equivalently transformed into
$$
y^{2}-y+m=0 \text {. }
$$
Thus, the two real roots of the original equation being both greater than 5 is equivalent to equation (1) having two positive real roots.
Therefore, according to the theorem, we have
$$
\left\{\begin{array}{l}
\Delta=... | 0<m \leqslant \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,342 |
2. Let $a>b>0$. Then the minimum value of $a^{4}+\frac{32}{b(a-b)}$ is . $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.48.
$$
\begin{array}{l}
a^{4}+\frac{32}{b(a-b)} \geqslant a^{4}+\frac{32}{\frac{(b+a-b)^{2}}{4}}=a^{4}+\frac{128}{a^{2}} \\
=a^{4}+\frac{64}{a^{2}}+\frac{64}{a^{2}} \geqslant 3 \sqrt[3]{2^{12}}=48\left(a=2 \text { when equality holds } b^{3}\right) .
\end{array}
$$ | null | Algebra | proof | Yes | Yes | cn_contest | false | 717,343 |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=3, a_{n+1}=u_{n}^{2}-(n+1) a_{n}+1 \text {. }
$$
Then the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is $\qquad$ | 3. $\frac{n(n+5)}{2}$.
By induction, we can obtain $a_{n}=n+2$. Then we have
$$
a_{n+1}=(n+2)^{2}-(n+1)(n+2)+1=n+3 \text {. }
$$
And $a_{1}=3=1+2$ satisfies the condition. Therefore, $a_{n}=n+2$.
Thus, the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is
$$
S_{n}=3+4+\cdots+(n+2)=\frac{n(n+5)}{2} ... | \frac{n(n+5)}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,344 |
4. The number of right-angled triangles with integer side lengths and area (numerically) equal to the perimeter is $\qquad$ | 4.2.
Let the legs of a right-angled triangle be $a, b$, and $c = \sqrt{a^{2}+b^{2}} (a \leqslant b)$, then we have
$$
\frac{1}{2} a b = a + b + \sqrt{a^{2} + b^{2}}.
$$
Thus, $\frac{1}{2} a b - a - b = \sqrt{a^{2} + b^{2}}$.
Squaring both sides and simplifying, we get
$$
a b - 4 a - 4 b + 8 = 0.
$$
Then, $(a-4)(b-4)... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,345 |
5. A chord $AB$ is drawn through a focus $F$ of the ellipse $\frac{x^{2}}{6^{2}}+\frac{y^{2}}{2^{2}}=1$. If $|A F|=m,|B F|=n$, then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ | 5.3.
As shown in Figure 2, draw $A A_{1}$ perpendicular to the directrix at point $A_{1} . A E \perp x$-axis at point $E$.
$$
\begin{array}{l}
\text { Since } \frac{c}{a}=e=\frac{|F A|}{\left|A A_{1}\right|} \\
=\frac{|F A|}{|D F|+|F E|} \\
=\frac{m}{m \cos \alpha+\frac{b^{2}}{c}} .
\end{array}
$$
Therefore, $\frac{1... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,346 |
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