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 |
|---|---|---|---|---|---|---|---|---|---|
5. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with edge length 1, $P$ is a moving point on the line segment $A B_{1}$, and $Q$ is a moving point on the base $A B C D$. Then the minimum value of $P C_{1}+P Q$ is ( ).
(A) $1+\frac{\sqrt{2}}{2}$
(B) $\sqrt{3}$
(C) 2
(D) $\frac{1}{2}+\frac{\sqrt{5}}{2}$ | 5. (A).
As shown in Figure 2, $P$ is any point on $A B_{1}$, and $Q$ is a moving point on the base $A B C D$. It is easy to see that when point $Q$ is the projection of $P$ on $A B$, $P Q$ is minimized (the leg of a right triangle is less than the hypotenuse). Let
$$
A Q=x(0<P C_{1}>B_{1} C_{1}=1).
$$
Therefore, $y \g... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,943 |
6. Given in the sequence $\left.a_{n}\right\}$, $a_{1}=1, S_{n}$ is the sum of the first $n$ terms, and satisfies $S_{n}=n^{2} a_{n}(n=1,2, \cdots)$. Then the expression for $a_{n}$ is
(A) $\frac{1}{n+2}(n \geqslant 2)$
(B) $\frac{1}{n(n-1)}(n \geqslant 3)$
(C) $\frac{1}{2(n+1)}(n \geqslant 4)$
(D) $\frac{2}{n(n+1)}$ | 6. (D).
$$
\begin{array}{l}
\because a_{1}=S_{1}=1, S_{n}=n^{2} a_{n}=n^{2}\left(S_{n}-S_{n-1}\right), \\
\left(n^{2}-1\right) S_{n}=n^{2} S_{n-1}, \\
\therefore \frac{n+1}{n} S_{n}=\frac{n}{n-1} S_{n-1}(n \geqslant 2) .
\end{array}
$$
Let $x_{n}=\frac{n+1}{n} S_{n}$, then $x_{n-1}=\frac{n}{n-1} S_{n-1}$.
Thus $x_{n}=... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,944 |
1. In $\triangle A B C$, $A D \perp B C$ at $D$, and $\frac{A D}{B C}=\frac{1}{3}$. Then the maximum value of $\frac{A C}{A B}+\frac{A B}{A C}$ is . $\qquad$ | $=\sqrt{1} \cdot \sqrt{13}$.
Let $A B=c, B C=a, C A=b, A D=h$.
By the cosine rule, we get
$$
b^{2}+c^{2}=a^{2}+2 b c \cos A \text {. }
$$
By the principle of equal area, we get $a h=b c \sin A$.
$$
\begin{array}{l}
\because \frac{h}{a}=\frac{1}{3} \Rightarrow a^{2}=3 b c \sin A, \\
\therefore b^{2}+c^{2}=3 b c \sin A+... | \sqrt{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,945 |
2. Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ . | 2.3.-
From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ t... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,946 |
3. Set $A=\left\{x \mid \sqrt{a^{2}+4 x-x^{2}}>a(x+1)\right\}, B=$ $\left\{\left.x|| x-\frac{1}{2} \right\rvert\,<\frac{1}{2}\right\}$. When $A \subseteq B$, the range of values for $a$ is | $3.1 \leqslant a \leqslant \sqrt{2}$.
$$
\because\left|x-\frac{1}{2}\right|<\frac{1}{2}, \therefore 0<x<1 \text {. }
$$
Let $y=a(x+1), y=\sqrt{a^{2}+4 x-x^{2}}$, then
$$
(x-2)^{2}+y^{2}=a^{2}+4(y \geqslant 0) \text {. }
$$
As shown in Figure 3, it is easy to find the intersection points of the two curves are $E(0, a)... | 1 \leqslant a \leqslant \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,947 |
4. Given that line segment $A D / /$ plane $\alpha$, and the distance to plane $\alpha$ is 8, point $B$ is a moving point on plane $\alpha$, and satisfies $A B=10$. If $A D=21$, then the minimum distance between point $D$ and point $B$ is $\qquad$ . | 4.17.
As shown in Figure 4, let the projections of points $A$ and $D$ on plane $\alpha$ be $O$ and $C$, respectively, then we have $A O = C D = 8$.
$$
\begin{array}{l}
\because A B = 10, \\
\therefore O B = 6 \text{ (constant). }
\end{array}
$$
Therefore, the trajectory of point $B$ in $\alpha$ is a circle with $O$ a... | 17 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,948 |
5. Given that the polynomial $x^{2}-x-1$ divides the polynomial $a x^{5}+b x^{4}$ +1 . Then the real numbers $a=$ $\qquad$ ,$b=$ $\qquad$ . | $5 . a=3, b=-5$.
Given that $x^{2}-x-1$ can divide the polynomial $a x^{5}+b x^{4}+1$, there exists a polynomial $f(x)$ such that
$$
a x^{5}+b x^{4}+1=\left(x^{2}-x-1\right) f(x) \text {. }
$$
Let the two roots of the equation $x^{2}-x-1=0$ be $x_{1}$ and $x_{2}$. Then we have
$$
\begin{array}{l}
\left\{\begin{array}{... | a=3, b=-5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,949 |
Example 3 A sector-shaped plate, with a central angle of $120^{\circ}$ and a radius of 1, is to be trimmed to form a rectangle of maximum area (the axis of symmetry of the sector is perpendicular to one side of the rectangle). What is the area of the part that will be cut off?
(1958, Wuhan City High School Mathematics ... | Solution: As shown in Figure 3, let $OAB$ be a sector,
with $\angle AOB = 120^\circ$,
$OA = 1$, and $CDEF$ be the inscribed rectangle.
Let $CD = x$, $OD = y$, and draw a line through $O$ parallel to $ED$ intersecting the extension of $CD$ at $K$. Then
$$
DK = \frac{y}{2}, \quad KO = \frac{\sqrt{3}}{2} y,
$$
In $\trian... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,950 |
6. Let $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{2002}]$, where $[\sqrt{n}]$ denotes the greatest integer not exceeding $\sqrt{n}$. Then the value of $[\sqrt{S}]$ is | 6.242.
Let $k^{2} \leqslant n<(k+1)^{2}$, then $k \leqslant \sqrt{n}<k+1$, so
$$
\begin{array}{l}
{[\sqrt{n}]=k .} \\
\because(k+1)^{2}-k^{2}=2 k+1,
\end{array}
$$
$\therefore$ From $k^{2}$ to $(k+1)^{2}$ there are $2 k+1$ numbers, the integer parts of whose square roots are all $k$,
$$
\begin{array}{l}
\therefore\lef... | 242 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,951 |
Three. (20 points) Given that the three internal angle bisectors of $\triangle A B C$ are $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$. If the vectors $\overrightarrow{A A^{\prime}} 、 \overrightarrow{B B^{\prime}} 、 \overrightarrow{C C^{\prime}}$ satisfy the relation $\overrightarrow{A A^{\prime}}+$ $\overrightarrow{B ... | As shown in Figure 5, let
$$
|BC| = a, |CA| = b, |AB| = c.
$$
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\overrightarrow{CB'} = \frac{a}{a+c} \overrightarrow{CA}, \\
\overrightarrow{BB'} = \overrightarrow{BC} + \overrightarrow{CB'} \\
= \overrightarrow{BC} + \frac{a}{a+c} \overrightarrow{CA}.
\end{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,952 |
Four. (20 points) Given the sequence $\left\{a_{n}\right\}$, $S_{n}$ represents the sum of its first $n$ terms. If it satisfies the relation $S_{n}+a_{n}=n^{2}+3 n-1$, find the general formula for the sequence $\left\{a_{n}\right\}$, i.e., the expression for $a_{n}$. | Given that $S_{1}+a_{1}=3$, therefore $a_{1}=\frac{3}{2}$.
When $n \geqslant 2$, we have
$$
\begin{array}{l}
S_{n-1}+a_{n-1}=(n-1)^{2}+3(n-1)-1, \\
S_{n}+a_{n}=n^{2}+3 n-1 .
\end{array}
$$
Subtracting the two equations, we get $a_{n}=\frac{1}{2} a_{n-1}+n+1$.
Let $a_{n}+x n+y=\frac{1}{2}\left[a_{n-1}+x(n-1)+y\right]$.... | a_{n}=2 n-\frac{1}{2^{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,953 |
Five. (20 points) Given an ellipse with semi-major axis $a$ and semi-minor axis $b$, one endpoint of the minor axis is $O$. $P$ and $Q$ are any two points on the ellipse different from point $O$, and $O P \perp O Q$. If the projection of point $O$ on line segment $P Q$ is $M$, find the locus of point $M$.
---
Transla... | Five, as shown in Figure 6, establish a rectangular coordinate system, then the equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{(y-b)^{2}}{b^{2}}=1
$$
$(a>b>0)$.
That is, $b^{2} x^{2}+a^{2} y^{2}-2 a^{2} b y=0$.
Let $P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right)$, and the equation of $P Q$ is $y=k x+m... | x^{2}+y^{2}-\frac{2 a^{2} b y}{a^{2}+b^{2}}=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,954 |
1. (50 points) As shown in Figure 1, in the right triangle \( \triangle ABC \), \( AC > BC \), \( \angle C = 90^\circ \), \( O \) is the midpoint of the hypotenuse \( AB \), \( CH \) is the altitude from \( C \) to the hypotenuse \( AB \), and \( CH \) is extended to \( D \) such that \( CH = DH \). \( F \) is any poin... | As shown in Figure 7, connect $B D$. Since $B C=B D$, we have $\angle C B D=2 \angle C B A$.
$\because O C$ is the median of the hypotenuse $A B$ in the right $\triangle A B C$,
$$
\therefore A O=B O=C O \text {, }
$$
Thus, $\angle A O C=2 \angle C B A$,
then $\angle A O C$
$$
=\angle C B D \text {, }
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,955 |
Three. (50 points) There are 2002 points distributed on a circle. Now, they are arbitrarily colored white or black. If starting from a certain point and moving in any direction around the circle to any point, the total number of white points (including the point itself) is always greater than the number of black points... | From the problem, we know that a good point must be white. The following discussion is for the general case: there are $3n+1$ points on the circumference, which are colored black and white. Only when the number of black points $\leqslant n$, can we ensure that there must be a good point.
We prove this by mathematical i... | 667 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,957 |
As shown in Figure 1, quadrilateral $A O B C$ is a rectangle, with $A(0,2)$ and $B(4,0)$. Points $P$ and $Q$ start moving from point $O$ simultaneously. Point $P$ moves along the path $O A C B$, and point $Q$ moves along the path $O B C A$. $P Q$ intersects $O C$ at $R$. Point $P$ moves at a speed of 3 units per second... | Solution: If $\triangle A R O \sim \triangle R C B$, then
$\angle A O C=\angle O C B, A O=B C=2$.
Thus, $\frac{A O}{R C}=\frac{O R}{B C}$, which means $\frac{2}{R C}=\frac{O R}{2}$.
Hence, $R C \cdot O R=4$.
Also, $O C=\sqrt{2^{2}+4^{2}}=2 \sqrt{5}$,
which means $R C+O R=2 \sqrt{5}$.
From equations (1), (2), and $O R_{... | t=\frac{6(6-\sqrt{5})}{31} \text{ or } t=\frac{6(6+\sqrt{5})}{31} \text{ or } t=\frac{6}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,958 |
As shown in Figure 2, given a square $ABCD$, extend $BC$ and $DC$ to $M$ and $N$ respectively, such that $S_{\triangle QMN} = S_{\text{square } ABCD}$. Determine the degree measure of $\angle MAN$. | Solution: As shown in Figure 3, since $AD = AB$, we can rotate $\triangle ABM$ $90^{\circ}$ around point $A$ to the position of $\triangle ADE$. Connect $EM$. Then points $C$, $D$, and $E$ are collinear, $\triangle AME$ is an isosceles right triangle, and $\angle AME = 45^{\circ}$.
Let the side length of the square $A... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,959 |
For any $n, k \in \mathbf{N}$, let $S=1^{n}+2^{n}+3^{n}+\ldots$ $+k^{n}$. Find the remainder when $S$ is divided by 3. | Solution: $1^{\circ}$ When $n$ is odd, let's assume $n=2 l-1, l \in \mathbf{N}$. For $m \in \mathbf{N}$.
If $3 \nmid m$, then $m^{2} \equiv 1(\bmod 3) \Rightarrow m^{2 l} \equiv 1(\bmod 3) \Rightarrow$ $m^{2 l-1} \equiv m^{2(l-1)+1} \equiv m(\bmod 3)$;
If $3 \mid m$, then $m^{2 l-1} \equiv 0 \equiv m(\bmod 3)$.
Thus, ... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,960 |
Example 4 Given that $\triangle ABC$ is an equilateral triangle with side length 1, $D$ is a point on side $BC$ such that $BD=p$, and $r_{1}, r_{2}$ are the radii of the inscribed circles of $\triangle ABD$ and $\triangle ADC$ respectively. Express $r_{1} r_{2}$ in terms of $p$, and find the maximum value of $r_{1} r_{... | Solution: As shown in Figure 4, in $\triangle A B D$, by the cosine rule we have
$$
\begin{array}{l}
A D^{2}=p^{2}-p+1, \\
A D=\sqrt{p^{2}-p+1} .
\end{array}
$$
The area of a triangle is equal to the product of its semiperimeter and the inradius, so we have
$$
\begin{array}{l}
\frac{1+p+\sqrt{p^{2}-p+1}}{2} \cdot r_{1... | \frac{2-\sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,961 |
Let the natural numbers $x_{n}, y_{n}$ satisfy
$$
x_{n}+\sqrt{2} y_{n}=\sqrt{2}(3+2 \sqrt{2})^{2^{n}}(n \in \mathbf{N}) \text {. }
$$
Prove: $y_{n}-1$ is a perfect square $(n \in \mathbf{N})$. | Prove: Squaring both sides of equation (1), it is easy to prove that for any $n \in \mathbf{N}$, $x_{n}$ is even, and
$$
x_{n+1}=2 x_{n} y_{n}, y_{n+1}=\frac{1}{2} x_{n}^{2}+y_{n}^{2} .
$$
Use mathematical induction to prove $y_{n}^{2}=\frac{1}{2} x_{n}^{2}+1(n \in \mathbf{N})$. When $n=1$, from equation (1), we have ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,962 |
Example 5 Let $P$ be a point inside an acute $\triangle ABC$, and the feet of the perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$ are $D$, $E$, and $F$ respectively. Find (and prove) the point $P$ that minimizes $\overline{P D}^{2}+\overline{P E}^{2}+\overline{P F}^{2}$.
$(1990$, Zhejiang Province High School... | Solution: As shown in Figure 5, let $S$ be the area of $\triangle ABC$, with side lengths $a, b, c$, and $PD$
$$
\begin{aligned}
=x, PE & =y, PF=z, \\
w & =\overline{PD^{2}}+\overline{PE^{2}}+\overline{PF^{2}} \\
& =x^{2}+y^{2}+z^{2} .
\end{aligned}
$$
We have
$$
\begin{aligned}
S^{2}= & \left(S_{\triangle PBC}+S_{\tr... | w_{\text {min }}=\frac{4 S^{2}}{a^{2}+b^{2}+c^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,963 |
Example 6 Given that $\triangle ABC$ is an equilateral triangle with side length 1, and $O$ is its center. Try to answer: Among the line segments passing through point $O$ with both ends lying on the sides of $\triangle ABC$, which ones are the longest? Which ones are the shortest? What are their lengths? Prove your co... | Solution: As shown in Figure 6, taking $A$ as the origin and the line $A O$ as the $y$-axis, we establish a rectangular coordinate system. Then the equation of $A B$ is $y=\sqrt{3} x$, and the equation of $A C$ is $y=-\sqrt{3} x$. The equation of the line passing through point $O$ is
$$
y=k x+\frac{1}{\sqrt{3}} \text {... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,964 |
Example 3 Find the values of real numbers $x$ and $y$ such that
$$
(y-1)^{2}+(x+y-3)^{2}+(2 x+y-6)^{2}
$$
reaches its minimum value.
(2001, National Junior High School Mathematics League) | Solution: Original expression $=5 x^{2}+6 x y+3 y^{2}-30 x-20 y+46$
$$
\begin{array}{l}
=5\left(x+\frac{3}{5} y-3\right)^{2}+\frac{6}{5}\left(y-\frac{5}{6}\right)^{2}+\frac{1}{6} . \\
\text { When }\left\{\begin{array}{l}
x+\frac{3}{5} y-3=0, \\
y-\frac{5}{6}=0
\end{array}\right. \text {, the above expression reaches i... | x=\frac{5}{2}, y=\frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,965 |
Example $7 \triangle A B C$ is a known acute triangle, and the regular $\triangle D E F$ is circumscribed around $\triangle A B C$. Find the maximum value of the area of $\triangle D E F$.
(1983, Wuhu City, Anhui Province High School Mathematics Competition) | Solution: As shown in Figure 7, let $\angle A$ be the largest angle of $\triangle ABC$. From $S_{\text{SEEF}} = \frac{\sqrt{3}}{4} EF^2$, we know that $S_{\text{CDEF}}$ is maximized when $EF$ is maximized. Since $EF$ is uniquely determined by $\angle BAF$, we can choose $\angle BAF = \theta$ as the independent variable... | \frac{\sqrt{3}}{3} \left[b^2 + c^2 + 2bc \cos (120^{\circ} - \angle A)\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,966 |
Proposition: Any prime divisor $p$ of the sum of two coprime squares cannot be $p \equiv -1 (\bmod 4)$.
That is, any prime of the form $4k-1$ cannot be a divisor of the sum of two coprime squares. | Proof: Let $a, b$ be two coprime integers, and $a^{2}+b^{2}$ has a prime divisor $p$. Prove that $p$ cannot be a number of the form $4k-1$.
(1) Since $(a, b)=1$, then $a$ and $b$ cannot both be divisible by $p$.
(2) When one of $a, b$ (let it be $a$) is divisible by $p$, by $p \mid a, p \nmid b$, then $p \nmid b^{2}, p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,967 |
Example 1 Proof: The equation $x^{2}+5=y^{3}$ has no integer solutions. | Proof: Assume the equation has integer solutions $(x, y)$.
If $x$ is odd, then $x^{2}+5 \equiv 2(\bmod 4)$.
Thus, $y^{3} \equiv 2(\bmod 4)$.
Since $y \equiv 0(\bmod 2)$,
Therefore, $y^{3} \equiv 0(\bmod 4)$.
(1) and (2) are contradictory. Therefore, $x$ cannot be odd.
If $x$ is even, then we can set $x=2 n$.
By $y^{3}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,968 |
Example 2 Proof: The equation $x^{2}-7=y^{3}$ has no integer solutions.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: Suppose the equation has integer solutions $(x, y)$.
If $y$ is even, then $x^{2}=y^{3}+7 \equiv 3(\bmod 4)$. However, a square number cannot be of the form $4k+3$.
Therefore, $y$ cannot be even.
If $y$ is odd, then $x$ is even. We have
$$
\begin{array}{l}
x^{2}+1=y^{3}+8=(y+2)\left[(y-1)^{2}+3\right], \\
(y-1)^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,969 |
Example 3 Let $p$ be a prime of the form $4k+3$, and $x_{0}$, $y_{0}$, $z_{0}$, $t_{0}$ be any integer solution to the equation
$$
x^{2 p}+y^{2 p}+z^{2 p}=t^{2 p}
$$
Prove that at least one of $x_{0}$, $y_{0}$, $z_{0}$, $t_{0}$ is divisible by $p$. | Proof: Let $x_{0}, y_{0}, z_{0}, t_{0}$ be a set of integer solutions to equation (1), and assume that none of $x_{0}, y_{0}, z_{0}, t_{0}$ can be divided by $p$. Without loss of generality, let $\left(x_{0}, y_{0}, z_{0}, t_{0}\right)=1$. Then $x_{0}, y_{0}, z_{0}, t_{0}$ cannot all be even.
If $t_{0}$ is even, then ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,970 |
Example 1 Find the largest constant $c$, such that for real numbers $x, y$ satisfying $x>0, y>0, x^{2}+y^{2}=1$, it always holds that $x^{6}+y^{6} \geqslant c x y$.
| When $x=y=\frac{\sqrt{2}}{2}$, the inequality also holds, then,
$$
c \leqslant \frac{x^{6}+y^{6}}{x y}=\frac{1}{2} \text {. }
$$
Next, we prove that under the conditions $x>0, y>0, x^{2}+y^{2}=1$, the inequality
$$
x^{6}+y^{6} \geqslant \frac{1}{2} x y
$$
always holds.
Equation (1) is equivalent to
$$
\left(x^{2}+y^{... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,971 |
Example 2 The abscissa of a point moving in the positive direction of the $O x$ axis is $x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}}$, where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$. | Solution: Using the arithmetic-geometric mean inequality, we get
$$
\begin{aligned}
x(t)= & (\underbrace{t+1)^{2}+\cdots+(t+1)^{2}}_{5 \uparrow} \\
& +\frac{a}{2(t+1)^{5}}+\frac{a}{2(t+1)^{5}} \\
\geqslant & 7 \sqrt[7]{\frac{a^{2}}{4}} .
\end{aligned}
$$
Equality holds if and only if $(t+1)^{2}=\frac{a}{2(t+1)^{5}}$.
... | 2 \sqrt{\left(\frac{24}{7}\right)^{7}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,972 |
Example 3 Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
holds. | Solution: First, estimate the upper bound of $k$.
When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
First, we prove a... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,973 |
Example 4 Let $\lambda>0$, find the largest constant $c=c(\lambda)$, such that for all non-negative real numbers $x, y$, we have
$$
x^{2}+y^{2}+\lambda x y \geqslant c(x+y)^{2} .
$$ | Solution: We consider two cases for $\lambda$:
(i) When $\lambda \geqslant 2$, we have
$$
x^{2}+y^{2}+\lambda x y \geqslant x^{2}+y^{2}+2 x y=(x+y)^{2} \text {. }
$$
The equality holds when $x y=0$.
(ii) When $0<\lambda<2$, we have
$$
\begin{array}{l}
x^{2}+y^{2}+\lambda x y=(x+y)^{2}-(2-\lambda) x y \\
\geqslant(x+y)... | c(\lambda)=\left\{\begin{array}{ll}
1, & \lambda \geqslant 2, \\
\frac{2+\lambda}{4}, & 0<\lambda<2 .
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,974 |
Example 5 Let $a, b, c$ be the lengths of the three sides of a right-angled triangle, and $a \leqslant b \leqslant c$. Find the maximum constant $k$, such that
$$
\begin{array}{l}
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
\geqslant \text { kabc }
\end{array}
$$
holds for all right-angled triangles, and determine when equalit... | When $a=b$, i.e., $\triangle ABC$ is an isosceles right triangle, the original inequality becomes
$$
\begin{array}{l}
a^{2}(a+\sqrt{2} a)+a^{2}(\sqrt{2} a+a)+2 a^{2}(a+a) \\
\geqslant \sqrt{2} k a^{3},
\end{array}
$$
which simplifies to $k \leqslant 2+3 \sqrt{2}$. We conjecture that the maximum value of $k$ is $2+3 \s... | 2+3\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,975 |
Example 4 Let $a$, $b$, $c$ all be real numbers no less than 3. Then the minimum value of $\sqrt{a-2}+\sqrt{b+1}+|1-\sqrt{c-1}|$ is $\qquad$
(2001, Hope Cup Junior High School Mathematics Competition Second Trial) | Given: $\because a \geqslant 3, b \geqslant 3, c \geqslant 3$,
$$
\therefore a-2 \geqslant 1, b+1 \geqslant 4, c-1 \geqslant 2 \text {. }
$$
Then the original expression $=\sqrt{a-2}+\sqrt{b+1}+\sqrt{c-1}-1$.
Also, $\because$ the smaller the radicand, the smaller its arithmetic root,
$\therefore$ the minimum value of ... | 2+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,976 |
Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$ | Solution: First, we aim to eliminate the parameter $a$, and then it will be easier to find the minimum value of $k$. Using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n} .
\... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,977 |
Given the equation about $x$: $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7) \left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1} = \frac{3}{11}$, ... | Solution: (1) Let $\frac{x}{x-1}=t$, then $t \neq 1$, the original equation becomes $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes $-9 t+1=0$ or $-5 t+1=0$.
That is, $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Then $x=-\frac{1}{8}$ or $x=-\frac{1}{4}... | a \geqslant-\frac{53}{28} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,979 |
Question: A student participates in military training and engages in target shooting, which must be done 10 times. In the 6th, 7th, 8th, and 9th shots, he scored 9.0 points, 8.4 points, 8.1 points, and 9.3 points, respectively. The average score of his first 9 shots is higher than the average score of his first 5 shots... | Solution: From the given, the average score of the first 5 shots is less than
$$
\frac{9.0+8.4+8.1+9.3}{4}=8.7 \text{. }
$$
The total score of the first 9 shots is at most
$$
8.7 \times 9-0.1=78.2 \text{. }
$$
Therefore, the score of the tenth shot must be at least
$$
8.8 \times 10+0.1-78.2=9.9 \text{ (points). }
$$
... | 9.9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,980 |
Example 1 If $|x|<1, n$ is an integer greater than 1, $a=$ $(1-x)^{n}+(1+x)^{n}$, prove: $a<2^{n}$. | $$
\begin{array}{l}
a=\sum_{k=0}^{n} \mathrm{C}_{n}^{k}(-x)^{k}+\sum_{k=0}^{n} \mathrm{C}_{n}^{k} x^{k} \\
=\sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left[(-x)^{k}+x^{k}\right] \\
=2\left(\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{2} x^{2}+\mathrm{C}_{n}^{4} x^{4}+\cdots+\mathrm{C}_{n}^{2}\left[\frac{n}{2}\right] x^{2\left[\frac{n}{2}... | a < 2^{n} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,981 |
Example 2 For any natural number $n$ greater than 4, there is always $2^{n}$ $>n^{2}$. $\quad$. | Proof: $\because n>4$,
$$
\begin{aligned}
\therefore & 2^{n}=(1+1)^{n} \\
& =\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\cdots+\mathrm{C}_{n}^{n-2}+\mathrm{C}_{n}^{n-1}+\mathrm{C}_{n}^{n} \\
& =1+n+\frac{n(n-1)}{2}+\cdots+\frac{n(n-1)}{2}+n+1 \\
& \geqslant 2+2 n+2 \cdot \frac{n(n-1)}{2} \\
& =n^{2}+n+2>n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,982 |
Example 3 Proof: $\left(1+\frac{1}{n}\right)^{n} \leqslant 3-\frac{1}{2^{n-1}}(n \in \mathbf{N})$. | $\begin{array}{l}\text { Prove: }\left(1+\frac{1}{n}\right)^{n} \\ =\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{1} \frac{1}{n}+\mathrm{C}_{n}^{2} \frac{1}{n^{2}}+\cdots+\mathrm{C}_{n}^{n} \frac{1}{n^{n}} \\ =1+1+\frac{1}{2!} \cdot \frac{n(n-1)}{n^{2}}+\cdots \\ +\frac{1}{n!} \cdot \frac{n(n-1) \cdot \cdots \cdot 2 \cdot 1}{n^{n... | 3-\frac{1}{2^{n-1}} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,983 |
Example 4 Prove: For any positive number $n$, the inequality $(2 n+1)^{n} \geqslant(2 n)^{n}+(2 n-1)^{n}$ holds.
(21st All-Soviet Union Mathematical Competition) | Proof: By the binomial theorem, we have
$$
\begin{array}{l}
(2 n+1)^{n}-(2 n-1)^{n} \\
=2\left[\mathrm{C}_{n}^{1}(2 n)^{n-1}+\mathrm{C}_{n}^{3}(2 n)^{n-3}+\cdots\right] \\
\geqslant 2 \mathrm{C}_{n}^{1}(2 n)^{n-1}=(2 n)^{n} . \\
\text { Therefore, }(2 n+1)^{n} \geqslant(2 n)^{n}+(2 n-1)^{n} .
\end{array}
$$
The above ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,984 |
Example 5 Prove: $\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\cdots+\mathrm{C}_{n}^{n}>n \cdot 2^{\frac{n-1}{2}}$ $(n \geqslant 2)$. | Analysis: The left algebraic expression is similar to the expansion of the binomial theorem, which can be associated with the binomial theorem, and the problem can be solved easily.
Proof: $\because \mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\cdots+\mathrm{C}_{n}^{n}$
$$
\begin{aligned}
& =2^{n}-1=\frac{2^{n}-1}{2-1} \\
& =... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,985 |
Example 6 Let $n \geqslant 2, n \in \mathbf{N}, a+b>0, a \neq b$. Prove: $2^{n-1}\left(a^{n}+b^{n}\right)>(a+b)^{n}$. | Proof: $2^{n}\left(a^{n}+b^{n}\right)-2(a+b)^{n}$
$$
\begin{aligned}
= & \left(a^{n}+b^{n}\right)\left(\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\cdots+\mathrm{C}_{n}^{n-1}\right) \\
& -2\left(\mathrm{C}_{n}^{1} a^{n-1} b+\mathrm{C}_{n}^{2} a^{n-2} b^{2}+\cdots+\right. \\
& \left.\mathrm{C}_{n}^{n-1} a b^{n-1}\right). \\
\... | 2^{n-1}\left(a^{n}+b^{n}\right)>(a+b)^{n} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,986 |
Example 5 If $xy=1$, then the minimum value of the algebraic expression $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is $\qquad$ .
(1996, Huanggang City, Hubei Province, Junior High School Mathematics Competition) | $$
\text { Sol: } \begin{aligned}
\because & \frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\left(\frac{1}{x^{2}}\right)^{2}+\left(\frac{1}{2 y^{2}}\right)^{2} \\
& \geqslant 2 \cdot \frac{1}{x^{2}} \cdot \frac{1}{2 y^{2}}=1,
\end{aligned}
$$
$\therefore \frac{1}{x^{4}}+\frac{1}{4 y^{4}}$'s minimum value is 1.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,987 |
Example 7 Given a positive arithmetic sequence $\left\{a_{n}\right\}$, with common difference $d$ not equal to 0. Prove that when $n \geqslant 2$, $a_{m}^{n}+a_{3 m}^{n}>2 a_{2 m}^{n}(m \in \mathbf{N})$. | Proof: Constructing the binomial.
$\because\left\{a_{n}\right\}$ is an arithmetic sequence,
$\therefore a_{m} 、 a_{2 m} 、 a_{3 m}$ also form an arithmetic sequence.
Thus, $a_{m}+a_{3 m}=2 a_{2 m}$.
Let $a_{m}=a_{2 m}-d, a_{3 m}=a_{2 m}+d$. Then
$$
\begin{array}{l}
a_{m}^{n}+a_{3 m}^{n}=\left(a_{2 m}-d\right)^{n}+\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,988 |
Example 9 Given $a \geqslant 0, b \geqslant 0, n \in \mathbf{N}$. Prove:
$$
\frac{a^{n}+b^{n}}{2} \geqslant\left(\frac{a+b}{2}\right)^{n} \text {. }
$$ | Analysis: This problem is commonly handled using mathematical induction in many reference books. If a polynomial is constructed, the proof can be more concise.
Proof: When $a=0$ or $b=0$ or $n=1$, the equality holds.
Construct a binomial, and make the transformation
$$
\begin{aligned}
a & =\frac{a+b}{2}+\frac{a-b}{2},... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,990 |
Proposition 1 If the extensions of the two pairs of opposite sides of quadrilateral $ABCD$ intersect at $E$ and $F$, then the orthocenters of $\triangle BCE$, $\triangle DCF$, $\triangle ABF$, and $\triangle ADE$ are collinear. | Proof: By Miquel's theorem, let the common point of the circumcircles of $\triangle B C E$, $\triangle D C F$, $\triangle A B F$, and $\triangle A D E$ be $P$, then the Simson line of point $P$ with respect to these four triangles is the same line $l$. By Steiner's theorem (if the orthocenter of $\triangle A B C$ is $H... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,991 |
Proposition 2 If the extensions of the two pairs of opposite sides of quadrilateral $ABCD$ intersect at $E$ and $F$, then the circumcenters of $\triangle BCE$, $\triangle DCF$, $\triangle ABF$, and $\triangle ADE$ are concyclic. | Proof: As shown in Figure 1, by Miquel's theorem, let the common point of the circumcircles of $\triangle B C E$, $\triangle D C F$, $\triangle A B F$, and $\triangle A D E$ be $P$, and the circumcenters of these four triangles be $\mathrm{O}_{1}$, $\mathrm{O}_{2}$, $\mathrm{O}_{3}$, and $O_{4}$ respectively.
$\because... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,992 |
There is a person who rests for two days after working for eight days. Once, he rested on Saturday and Sunday. How many weeks at least will it take for him to rest on Sunday again?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
There is a person who rests for two days after working for eight days. Once, he rested on Saturd... | This article generalizes the problem and obtains the following results:
Generalization 1: A person works for $k$ consecutive days and then rests for $t$ days, where the $i$-th rest day and the $(i+1)$-th rest day in the $t$ rest days are separated by $j$ days, with $k, t \in \mathbf{Z}^{+}, j \in \mathbf{Z}^{+} \cup\{0... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,993 |
Promotion 2: A person works for $k$ consecutive days and then rests for $t$ days, where the $i$-th rest day and the $(i+1)$-th rest day are separated by $i$ days, $k, t \in \mathbf{Z}^{+}(1 \leqslant i \leqslant t-1)$. How many weeks at least will it take before they can rest again on the rest days $A_{1}, A_{2}, \cdot... | Solution: Let $P=7 x-[t+(t-1)+\cdots+(i+1)]$, $Q=k+t+\frac{t(t-1)}{2}$. For the rest day $A_{i}$, the smallest positive integer that makes $P \equiv 0(\bmod Q)$, the smallest positive integer that makes $P \equiv -t(\bmod Q)$, the smallest positive integer that makes $P \equiv -t-(t-1)(\bmod Q)$, $\cdots \cdots$, and t... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,994 |
1. Given $a=\sqrt{2}-1, b=2 \sqrt{2}-\sqrt{6}, c=\sqrt{6}-2$. Then, the size relationship of $a, b, c$ is ( ).
(A) $a<b<c$
(B) $b<a<c$
(C) $c<b<a$
(D) $c<a<b$ | - 1.(B).
Since $3>2 \sqrt{2}$, then $a-b=\sqrt{6}-1-\sqrt{2} \geqslant \sqrt{3+2 \sqrt{2}}-1$ $-\sqrt{2}=0$. Similarly, $c-a>0$. Therefore, $b<a<c$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,995 |
2. If $m^{2}=n+2, n^{2}=m+2(m \neq n)$, then the value of $m^{3}-$ $2 m n+n^{3}$ is ( ).
(A) 1
(B) 0
(C) -1
(D) -2 | 2.(D).
$$
\begin{array}{l}
\because m^{2}-n^{2}=n-m \text { and } n-m \neq 0, \\
\therefore m+n=-1 . \\
\text { Also, } \because m^{3}=m^{2} \cdot m=(n+2) m=m n+2 m, \\
n^{3}=n^{2} \cdot n=(m+2) n=m n+2 n, \\
\therefore m^{3}-2 m n+n^{3}=2(m+n)=-2 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,996 |
3. Given the quadratic function $y$ $=a x^{2}+b x+c$ whose graph is shown in Figure 1, and let $M=$ $|a+b+c|-\mid a-b+$ $c|+| 2 a+b|-| 2 a-$ $b 1$. Then ( ).
(A) $M>0$
(B) $M=0$
(C) $M<0$
(D) Cannot determine whether $M$ is positive, negative, or 0 | 3. (C).
According to the problem, we get $a>0, 0.2a-b>0$. From the graph, it is easy to see that $M=-2(a-b+c)<0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,997 |
Example 6 Given three non-negative numbers $a$, $b$, $c$ satisfying $3a + 2b + c = 5$ and $2a + b - 3c = 1$. If $m = 3a + b - 7c$, then the minimum value of $m$ is $\qquad$, and the maximum value of $m$ is $\qquad$
(2000, 14th Junior High School Mathematics Competition in Jiangsu Province) | Given the conditions, we have
$$
a=7 c-3, b=7-11 c \text { . }
$$
Substituting into \( m=3 a+b-7 c \), we get \( m=3 c-2 \).
Since \( a, b, c \) are non-negative,
$$
\therefore\left\{\begin{array}{l}
a=7 c-3 \geqslant 0, \\
b=7-11 c \geqslant 0, \\
c \geqslant 0 .
\end{array} \text { Solving, we get } \frac{3}{7} \leq... | -\frac{5}{7}, -\frac{1}{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,998 |
4. In right $\triangle ABC$, the area is 120, and $\angle BAC=90^{\circ}$. $AD$ is the median of the hypotenuse. A line $DE \perp AB$ is drawn from point $D$ to point $E$. Connect $CE$ to intersect $AD$ at point $F$. Then the area of $\triangle AFE$ is ( ).
(A) 18
(B) 20
(C) 22
(D) 24 | 4. (B).
As shown in Figure 6, draw $F G \perp A B$ at $G$, then $F G / / D E / / A C$. Therefore, we have
$$
\begin{array}{l}
\frac{F G}{\frac{1}{2} A C}=\frac{F G}{D E}=\frac{A G}{A E}, \\
\frac{F G}{A C}=\frac{E G}{A E} .
\end{array}
$$
(1) + (2) gives $\frac{3 F G}{A C}=\frac{A E}{A E}=1$.
Thus, $F G=\frac{1}{3} A... | 20 | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,999 |
$\odot O_{2}$ is externally tangent to $\odot O_{1}$ at point $A$, and an external common tangent of the two circles is tangent to $\odot O$ at point $B$. If $AB$ is parallel to another external common tangent of the two circles, then the ratio of the radii of $\odot O_{1}$ and $\odot O_{2}$ is ( ).
(A) $2: 5$
(B) $1: ... | 5.(C).
As shown in Figure 7, let another external common tangent line touch $\odot O_{1}$ at point $C$ and $\odot O_{2}$ at point $D$. Draw a perpendicular from $O_{1}$ to $O_{2} D$, with the foot of the perpendicular being $E$. The radii of the two circles are $r_{1}$ and $r_{2}$, respectively.
By symmetry, we have
$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,000 |
6. If for a natural number $n$ not less than 8, when $3n+1$ is a perfect square, $n+1$ can always be expressed as the sum of $k$ perfect squares, then the minimum value of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 6. (C).
From $3 n+1=a^{2}$, we know $3 \nmid a$. Therefore, $a=3 t \pm 1$. Thus, $3 n+1=9 t^{2} \pm 6 t+1 \Rightarrow n=3 t^{2} \pm 2 t$. Hence, $n+1=t^{2}+t^{2}+(t \pm 1)^{2}$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,001 |
3. Person A and Person B go to a discount store to buy goods. It is known that both bought the same number of items, and the unit price of each item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is 172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
Person A a... | 3.12.
Suppose each person bought $n$ items, among which $x$ items cost 8 yuan each, and $y$ items cost 9 yuan each. Then we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x + y = 2 n , } \\
{ 8 x + 9 y = 172 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=18 n-172, \\
y=172-16 n .
\end{array}\right.\right.... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,004 |
4. Let $N=23 x+92 y$ be a perfect square, and $N$ does not exceed 2,392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs.
| 4.27.
$\because N=23 x+92 y=23(x+4 y)$, and 23 is a prime number, $N$ is a perfect square not exceeding 2392,
$\therefore x+4 y=23 m^{2}$ ( $m$ is a positive integer) and
$N=23^{2} \cdot m^{2} \leqslant 2392$,
thus $m^{2} \leqslant \frac{2392}{23^{2}}=\frac{104}{23}<5$.
Solving, we get $m^{2}=1$ or 4.
When $m^{2}=1$, f... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,005 |
One, (20 points) Given that $a$, $b$, and $c$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a+b=8, \\
a b-c^{2}+8 \sqrt{2} c=48 .
\end{array}\right.
$$
Try to find the roots of the equation $b x^{2}+c x-a=0$. | (A Paper) One, obviously $a, b$ are the roots of the equation $x^{2}-8 x+c^{2}-8 \sqrt{2} c$ $+48=0$.
From the discriminant $\Delta \geqslant 0$, we get $c=4 \sqrt{2}$.
Thus, $\left\{\begin{array}{l}a b=16, \\ a+b=8 .\end{array}\right.$
Solving it, we get $a=b=4$.
Therefore, the original quadratic equation becomes $x^{... | x_{1}=\frac{-\sqrt{2}+\sqrt{6}}{2}, x_{2}=-\frac{\sqrt{2}+\sqrt{6}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,006 |
II. (25 points) As shown in Figure 4, in isosceles $\triangle ABC$, $P$ is any point on the base $BC$. Through $P$, two lines parallel to the legs are drawn, intersecting $AB$ and $AC$ at points $Q$ and $R$, respectively. Also, $P'$ is the reflection of $P$ about the line $RQ$. Prove: $\triangle P'QB \sim \triangle P'R... | $$
\begin{array}{l}
P^{\prime} B, P^{\prime} C, P^{\prime} Q, P^{\prime} R, \\
P^{\prime} P: \\
\because P Q / / A C, \\
\therefore Q P=Q B . \\
\quad \because P, P^{\prime} \text { are symmetric about } Q R
\end{array}
$$
are symmetric,
$$
\therefore Q P=Q P^{\prime} .
$$
Thus, $P^{\prime} Q=Q P=Q B$.
Therefore, $Q$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,007 |
Three. (25 points) Determine all rational numbers $r$ such that the equation $r x^{2}+(r+2) x+3 r-2=0$ has roots and only integer roots.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Three. (25 points) Determine all r... | If $r=0$, then the equation becomes $2 x-2=0$, solving for $x$ gives $x=1$. If $r \neq 0$, let the two roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$, then $x_{1}+x_{2}=-\frac{r+2}{r}, x_{1} x_{2}=\frac{3 r-2}{r}$.
We have $\quad x_{1} x_{2}-\left(x_{1}+x_{2}\right)=\frac{3 r-2}{r}+\frac{r+2... | 0, -\frac{2}{9}, \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,008 |
Example 7 Let $x, y$ be real numbers, and $x^{2} + xy + y^{2} = 3$. Find the maximum and minimum values of $x^{2} - xy + y^{2}$.
(1994, Huanggang City, Hubei Province Junior High School Mathematics Competition) | Solution: Let $x^{2}-x y+y^{2}=m$.
Also, $x^{2}+x y+y^{2}=3$,
Solving, we get $x+y= \pm \sqrt{\frac{9-m}{2}}, x y=\frac{3-m}{2}$.
Then $x, y$ are the two real roots of the equation $t^{2} \pm \sqrt{\frac{9-m}{2}} t+\frac{3-m}{2}=0$. Therefore,
$$
\Delta=\left( \pm \sqrt{\frac{9-m}{2}}\right)^{2}-4 \cdot \frac{3-m}{2} \... | 1 \leqslant m \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,009 |
II. (25 points) As shown in Figure 4, in isosceles $\triangle ABC$, $P$ is any point on the base $BC$. Through $P$, two lines parallel to the legs are drawn, intersecting $AB$ and $AC$ at points $Q$ and $R$, respectively. Also, $P'$ is the reflection of $P$ about the line $RQ$. Prove: $P'$ lies on the circumcircle of $... | From (A) Volume II, Question 2, we can prove that $\triangle P^{\prime} Q B \backsim \triangle P^{\prime} R C$.
We have $\angle P^{\prime} B A=\angle P^{\prime} C A$.
Therefore, $P^{\prime} 、 B 、 C 、 A$ are concyclic.
That is, $P^{\prime}$ lies on the circumcircle of $\triangle A B C$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,010 |
Three. (25 points) Determine all rational numbers $r$ such that the equation $r x^{2}+(r+2) x+r-1=0$ has roots and only integer roots.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Three. (25 points) Determine all rat... | If $r=0$, then the equation becomes $2 x-1=0$, solving for $x=\frac{1}{2}$ which is not an integer.
If $r \neq 0$. Let the two roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$, then $x_{1}+x_{2}=-\frac{r+2}{r}, x_{1} x_{2}=\frac{r-1}{r}$.
Thus, $2 x_{1} x_{2}-\left(x_{1}+x_{2}\right)=2\left(\fr... | -\frac{1}{3} \text{ and } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,011 |
II. (25 points) As shown in Figure 5, given that $P$ is a point on the hypotenuse $BC$ of $\mathrm{Rt} \triangle ABC$, $Q$ is the midpoint of $PC$, a perpendicular line is drawn from $P$ to $BC$, intersecting $AB$ at $R$, $H$ is the midpoint of $AR$, and a ray $HN \perp AB$ is drawn from $H$ towards the side where $C$ ... | Second, first prove $C Q>A H$.
$$
\begin{array}{l}
\because B C>A B, B R>B P, \\
\therefore A H=\frac{1}{2}(A B-B R)<\frac{1}{2}(B C-B P)=C Q .
\end{array}
$$
Therefore, the circle with $A$ as the center and $C Q$ as the radius must intersect a point $G$ on $H N$, i.e., $A G = C Q$.
As shown in Figure 10, connect $B G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,012 |
3. Given $a^{2}+b^{2}=6 a b$ and $a>b>0$. Then $\frac{a+b}{a-b}=$ | $3 \cdot \sqrt{2}$ | 3 \cdot \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,015 |
5. In an equilateral $\triangle A B C$ with side length 1 cm, $P_{0}$ is a point on side $B C$. Draw $P_{0} P_{1} \perp C A$ at point $P_{1}$, draw $P_{1} P_{2} \perp A B$ at point $P_{2}$, and draw $P_{2} P_{3} \perp B C$ at point $P_{3}$. If point $P_{3}$ coincides with point $P_{0}$, then the area of $\triangle P_{1... | 5. $\frac{\sqrt{3}}{12}$ | \frac{\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,017 |
II. (Full marks 15 points) Prove the identity:
$$
a^{4}+b^{4}+(a+b)^{4}=2\left(a^{2}+a b+b^{2}\right)^{2} \text {. }
$$ | \begin{aligned}= & 、 a^{4}+b^{4}+(a+b)^{4}-2\left(a^{2}+a b+b^{2}\right)^{2} \\ = & \left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}+\left(a^{2}+2 a b+b^{2}\right)^{2} \\ & -2\left(a^{2}+a b+b^{2}\right)^{2} \\ = & {\left[\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}+a b+b^{2}\right)^{2}\right] } \\ & +\left[\left(a^{2}+2 a b+b^{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,018 |
Three, (15 points) On the front of six pieces of paper, write down the integers 1, $2, 3, 4, 5, 6$, shuffle them, then flip the pieces of paper over, and on their backs, randomly write down the integers $1 \sim 6$ again. Then calculate the absolute value of the difference between the numbers written on the front and ba... | Three, suppose the numbers written on the front of six cards are
-1
$$
a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} ;
$$
The numbers written on the back correspond to $b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}$.
$\left(a_{i}, b_{i}\right)$ take values from $\left.1,2,3,4,5,6 . i=1,2,3,4,5,6.\right)$
Then the absolute value... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,019 |
Example 8 Given that $x, y, z$ are real numbers, and $x+y+z=$ $5, xy+yz+zx=3$. Try to find the maximum and minimum values of $z$. (10th Canadian High School Mathematics Competition) | Solution: $\because x+y+z=5$,
$$
\therefore y=5-x-z \text {. }
$$
Substitute equation (1) into $x y+y z+z x=3$, we get
$$
x(5-x-z)+(5-x-z) z+z x=3 \text {. }
$$
Rearranging gives $x^{2}+(z-5) x+\left(z^{2}-5 z+3\right)=0$.
$\because x$ is a real number,
$$
\therefore \Delta=(z-5)^{2}-4\left(z^{2}-5 z+3\right) \geqsla... | \frac{13}{3} \text{ and } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,020 |
Four. (15 points) As shown in Figure 2, in isosceles $\triangle ABC$, extend side $AB$ to point $D$, and extend side $CA$ to point $E$. Connect $DE$, such that $AD = BC = CE = DE$. Prove that $\angle BAC = 100^{\circ}$. | It is known that $\triangle A D E$ is an isosceles triangle, with its base angle $\angle E A D$ necessarily being an acute angle. Therefore, in the isosceles $\triangle A B C$, $\angle B A C$ is an obtuse angle and must be the vertex angle, so $A B$ and $A C$ are the legs, and $A B = A C$.
As shown in Figure 3, draw a... | 100^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 713,021 |
Five, (15 points) 1 and 0 alternate to form the following sequence of numbers:
$$
101,10101,1010101,101010101, \cdots
$$
Please answer, how many prime numbers are there in this sequence? And please prove your conclusion. | Obviously, 101 is a prime number.
Below is the proof that $N=\underbrace{101010 \cdots 01}_{k \uparrow 1}(k \geqslant 3)$ are all composite numbers (with $k-1$ zeros in between).
$$
\begin{aligned}
11 N= & 11 \times \underbrace{10101 \cdots 01}_{k \uparrow 1} \\
& =\underbrace{1111 \cdots 11}_{2 k \uparrow 1}=\underbra... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,022 |
3. Among the following four propositions:
(1) If $|a|=|b|$, then $a|a|=b|b|$;
(2) If $a^{2}-5 a+5=0$, then $\sqrt{(1-a)^{2}}=a-1$;
(3) If the solution set of the inequality $(m+3) x>1$ is $x>0$, then the equation has one positive root and one negative root, and the absolute value of the negative root is larger. The num... | 3. (C). Proposition 1 is not true; Propositions $2,3,4$ are true. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,030 |
Example 9 Under the conditions $x+2 y \leqslant 3, x \geqslant 0, y \geqslant 0$, the maximum value that $2 x+y$ can reach is $\qquad$
(2000, Hope Cup Junior High School Mathematics Competition Second Trial) | Solution: As shown in Figure 1, draw the line $x + 2y = 3$. The set of points satisfying the inequalities $x \geqslant 0, y \geqslant 0, x + 2y \leqslant 3$ is the region $\triangle ABO$ (including the boundaries) enclosed by the line and the $x$ and $y$ axes. To find the maximum value of $s = 2x + y$, we transform $s ... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,031 |
4. As shown in Figure 1, in quadrilateral $A B C D$, $\angle B A D=90^{\circ}, A B=B C=$ $2 \sqrt{3}, A C=6, A D=3$. Then the length of $C D$ is ( ).
(A) 4
(B) $4 \sqrt{2}$
(C) $3 \sqrt{2}$
(D) $3 \sqrt{3}$ | 4. (D).
As shown in Figure 9, draw $B E \perp A C$, with $E$ as the foot of the perpendicular, and connect $D E$. It can be obtained that $\angle B A E=30^{\circ}$, thus $\triangle A D E$ is an equilateral triangle, and $\angle A D C=90^{\circ}$.
Therefore, $C D=\sqrt{A C^{2}-A D^{2}}$ $=3 \sqrt{3}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,032 |
5. Given that the length of each side of a triangle
is a prime factor of 2001. Then, the number of different such triangles is ( ).
(A) 6
(B) 7
(C) 8
(D) 9 | 5. (B).
Hint: $2001=3 \times 23 \times 29, 3, 23$, and 29 are all prime numbers. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,033 |
6.12 A total of 12 pieces of chocolate of exactly the same size, each piece can be divided into at most two smaller pieces (which may not be equal). If these 12 pieces of chocolate can be evenly distributed among $n$ students, then $n$ can be ( ).
(A) 26
(B) 23
(C) 17
(D) 15 | 6. (D).
12 chocolate bars are evenly distributed among 15 students, each should receive $\frac{4}{5}$ of a bar. Therefore, each chocolate bar can be divided into two smaller pieces in a $4:1$ ratio, with 12 students each getting 1 larger piece, and the other 3 students each getting 4 smaller pieces. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,034 |
8. As shown in Figure 2, $D$, $E$, and $F$ are points on the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively, and $DE \parallel BA$, $DF \parallel CA$.
(1) To make quadrilateral $AFDE$ a rhombus, the additional condition needed is: $\qquad$
(2) To make quadrilateral $AFDE$ a rectangle, the additional condit... | 8. (1) $A D$ bisects $\angle B A C$, or $A D \perp E F$, or $\cdots \cdots$;
(2) $\angle B A C=90^{\circ}$, or $\cdots \cdots$. | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,036 |
9. The solution to the equation $\frac{x+1}{x+2}+\frac{x+8}{x+9}=\frac{x+2}{x+3}+\frac{x+7}{x+8}$ is | 9. $-\frac{11}{2}$.
Hint: The original equation can be transformed into
$$
(x+2)(x+3)=(x+8)(x+9) .
$$ | -\frac{11}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,037 |
11. As shown in Figure 3, the line $y=-2 x+6$ intersects the $x$-axis and $y$-axis at points $P$ and $Q$, respectively. If $\triangle P O Q$ is folded along $P Q$, point $O$ lands at point $R$. Then the coordinates of point $R$ are $\qquad$. | 11. $\left(\frac{24}{5}, \frac{12}{5}\right)$. | \left(\frac{24}{5}, \frac{12}{5}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,038 |
12. As shown in Figure 4, in the octagon $A B C D E F G H$, the areas of 4 squares are $25$, $144$, $48$, and $121$ square units, and $P R=13$ (units). Then the area of the octagon $=$ $\qquad$ square units. | Obviously, $S_{\triangle M B C}+S_{\triangle C D Q}$
As shown in Figure 10, draw $Q I \perp P S$ intersecting $P S$ at point $I$, and $B J \perp A P$ intersecting the extension of $A P$ at point $J$. It is easy to prove that Rt $\triangle P Q I \cong \mathrm{Rt} \triangle P B J$, thus $Q J = B J$. Therefore, $S_{\trian... | 428+66\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,039 |
13. As shown in Figure 5, let the sum of the two sides $AC$ and $BC$ of $\triangle ABC$ be $a$, $M$ is the midpoint of $AB$, $MC=MA=5$. Then the range of values for $a$ is $\qquad$. | $$
13.10A B=10 \text {. }
$$
From $x+y=a$ and $x^{2}+y^{2}=10^{2}$, we can get $xy = \frac{a^{2}-10^{2}}{2}$. Therefore, $x$ and $y$ are the two real roots of the quadratic equation $z^{2}-a z+\frac{a^{2}-10^{2}}{2} = 0$. Hence, $\Delta=a^{2}-4 \times \frac{a^{2}-10^{2}}{2} \geqslant 0$, which means
$$
a \leqslant 10 ... | a \leqslant 10 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,040 |
For example, if $a$ and $b$ are positive numbers, and the parabolas $y=x^{2}+ax+2b$ and $y=x^{2}+2bx+a$ both intersect the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
(2000, National Junior High School Mathematics League) | Solution: From the problem, we have
$$
\Delta_{1}=a^{2}-8 b \geqslant 0, \Delta_{2}=4 b^{2}-4 a \geqslant 0 \text {. }
$$
Thus, $a^{2} \geqslant 8 b$ and $b^{2} \geqslant a$.
Since $a$ and $b$ are both positive numbers,
$$
\therefore a^{4} \geqslant 64 b^{2} \geqslant 64 a \text {, i.e., } a \geqslant 4 \text {. }
$$
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,042 |
15. A store sells $A$ type exercise books, each retailing for 0.30 yuan, a dozen (12 books) for 3.00 yuan, and for purchases of more than 10 dozen, each dozen can be paid for at 2.70 yuan.
(1) Class 3(1) has a total of 57 students, each needing 1 $A$ type exercise book. If the class buys collectively, what is the minim... | Three, 15. (1) Can buy 5 dozen or 4 dozen plus 9 books, the former requires a payment of $3.00 \times 5 = 15.00$ (yuan), the latter only requires a payment of $3.00 \times 4 + 0.3 \times 9 = 14.70$ (yuan). Therefore, when the class buys together, the minimum payment required is 14.70 yuan.
(2) $227 = 12 \times 18 + 11$... | 14.70 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,044 |
16. Let $x_{1}, x_{2}$ be the real roots of the equation $2 x^{2}-4 m x+2 m^{2}+3 m-2=0$. For what value of $m$ is $x_{1}^{2}+x_{2}^{2}$ minimized? And find this minimum value.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 16. From the problem, we know the equation has real roots, $\Delta \geqslant 0$.
Given $-24 m+16 \geqslant 0$, then $m \leqslant \frac{2}{3}$.
From the relationship between roots and coefficients, we get
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}=2\left(\frac{3}{4}-m\right)^{2}+\frac{7}{8} . \\
\because m \leqslant \frac... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,045 |
2. As shown in Figure 1, a 10-meter-long ladder is leaning against a wall, with the top of the ladder 8 meters vertically above the ground. If the top of the ladder slides down 1 meter, then the distance the bottom of the ladder slides ( ).
(A) equals 1 meter
(B) is greater than 1 meter
(C) is less than 1 meter
(D) can... | 2. (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 | 713,049 |
3. Let $a, b$ be positive real numbers and $\frac{1}{a}+\frac{1}{b}-\frac{1}{a-b}=0$. Then, the value of $\frac{b}{a}$ is ( ).
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{1-\sqrt{5}}{2}$ (C) $\frac{-1+\sqrt{5}}{2}$
(D) $\frac{-1-\sqrt{5}}{2}$ | 3. (C).
From the given, we have $\frac{a+b}{a b}=\frac{1}{a-b}$, which means $a^{2}-b^{2}=a b$. Therefore, $\frac{a}{b}-\frac{b}{a}=1$. Solving this, we get $\frac{b}{a}=\frac{-1+\sqrt{5}}{2}$ (the negative value is discarded). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,050 |
4. If $x_{1}, x_{2}$ are two distinct real roots of the equation $x^{2}+2 x-k=0$, then $x_{1}^{2}+x_{2}^{2}-2$ is ( ).
(A) Positive
(B) Zero
(C) Negative
(D) Not greater than zero
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. (A)
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | A | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 713,051 |
Example 1 There is an even number of people sitting around a round table discussing a problem. After a break, they sit around the table in a different order to continue the discussion. Prove: At least two people will have the same number of people sitting between them before and after the break. | Analysis: It is obvious that the number of different seating arrangements for an even number of people sitting around a round table after a break is a huge number. Even with a small number of people, such as 8, the number of different seating arrangements for these 8 people around a round table is $7 \times 6 \times 5 ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,053 |
Example 2 Arrange the numbers $1,2, \cdots, 10$ randomly around a circle, prove: there must be three consecutive numbers, the sum of which is not less than 17. | Analysis: The randomness of the conditions in this problem is significant, as the numbers $1,2, \cdots, 10$ are not arranged in their original order but in any shuffled order on a circle. We cannot, nor is it necessary, to consider each case individually. Since the goal of the proof is the sum of three adjacent numbers... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,054 |
1. Given that $a, b, c$ are real numbers, and $x, y$ are any real numbers, set
$$
\begin{array}{l}
A=(a-b) x+(b-c) y+(c-a), \\
B=(b-c) x+(c-a) y+(a-b), \\
C=(c-a) x+(a-b) y+(b-c) .
\end{array}
$$
Prove: $A, B, C$ cannot all be positive, nor can they all be negative. | 1. Considering the whole of $A+B+C$, it is easy to see that $A+B+C=0$. Therefore, $A$, $B$, and $C$ cannot all be positive. They also cannot all be negative. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,055 |
2. For the arithmetic sequence $\left.\mid a_{n}\right\}$, the first term $a_{1}=8$, and there exists a unique $k$ such that the point $\left(k, a_{k}\right)$ lies on the circle $x^{2}+y^{2}=10^{2}$. Then the number of such arithmetic sequences is $\qquad$. | 2.17.
Obviously, the point $(1,8)$ is inside the circle, and the points on the circle $(1, \pm \sqrt{9 a})$
do not satisfy the general term formula of the arithmetic sequence
$$
a_{k}=a_{1}+(k-1) d .
$$
However, the points $\left(k, \pm \sqrt{100-k^{2}}\right)(k=2,3, \cdots, 10)$ on the circle
$$
k^{2}+a_{k}^{2}=100... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,058 |
3. In the tetrahedron $P-ABC$, $PA=PB=a$, $PC=AB=BC=CA=b$. And $a<b$. Then the range of $\frac{a}{b}$ is
保留了源文本的换行和格式。 | 3. $(\sqrt{2-\sqrt{3}}, 1)$.
Regardless of what positive number $b$ takes, $\triangle A B C$ always exists, and its height is $\frac{\sqrt{3}}{2} b$. For $\triangle P A B$ to exist, it should have $bP C$.
$$
\text{Rearranging and substituting, we get } \sqrt{a^{2}-\frac{b^{2}}{4}}>b-\frac{\sqrt{3}}{2} b,
$$
Squaring b... | (\sqrt{2-\sqrt{3}}, 1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,059 |
4. The moving point $A$ corresponds to the complex number $z=4(\cos \theta+\mathrm{i} \sin \theta)$, the fixed point $B$ corresponds to the complex number 2, and point $C$ is the midpoint of line segment $A B$. A perpendicular line to $A B$ through point $C$ intersects $O A$ at $D$. Then the equation of the locus of po... | 4. $\frac{(x-1)^{2}}{4}+\frac{y^{2}}{3}=1$.
From $C D$ being the perpendicular bisector of $A B$, we get $D A=D B$. Therefore, $D O+D B=D O+D A=O A=4$.
It is evident that the sum of the distances from point $D$ to the fixed points $U$ and $B$ is a constant 4, so its trajectory is an ellipse with foci at $O$ and $B$, ... | \frac{(x-1)^{2}}{4}+\frac{y^{2}}{3}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,060 |
6. There are 100 equally divided points on a circle. The number of obtuse triangles formed by these points as vertices is $\qquad$ . | $6.50 \times 49 \times 48$ (or 117600).
Let $A_{t}$ be the vertex of the obtuse angle. $\angle A_{\text{s}}$
intercepts an arc of $x$ equal parts, and the other two angles intercept arcs of $y, z$ equal parts, respectively, such that
$$
\left\{\begin{array}{l}
x+y+z=100, \\
x \geqslant 51, y \geqslant 1, z \geqslant 1 ... | 117600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,062 |
Three. (20 points) Given a parabola $y^{2}=2 p x(p>0)$ with a chord $A B$ of length $l$. Find the shortest distance from the midpoint $M$ of $A B$ to the $y$-axis, and determine the coordinates of point $M$ at this time.
Translate the above text into English, preserving the original text's line breaks and format, and ... | Three, from the parabolic equation $y^{2}=2 p x$, we can set the coordinates of $A$ and $B$ as $A\left(\frac{a^{2}}{2 p}, a\right), B\left(\frac{b^{2}}{2 p}, b\right)$. Then the coordinates of the midpoint $M$ are
$$
\left\{\begin{array}{l}
x=\frac{a^{2}+b^{2}}{4 p} . \\
y=\frac{a+b}{2} .
\end{array}\right.
$$
We have... | x_{\text {min }}=\left\{\begin{array}{ll}\frac{l^{2}}{8 p}, & 0<l<2 p \text {, } \\ \frac{l-p}{2}, & l \geqslant 2 p \text {, }\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,063 |
Four. (20 points) In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the center of square $A B C D$ is point $M$, and the center of square $A_{1} B_{1} C_{1} D_{1}$ is point $N$. Connect $A N, B_{1} M$.
(1) Prove that $A N$ and $B_{1} M$ are skew lines;
(2) Find the angle between $A N$ and $B_{1} M$. | (1) As shown in Figure 5, connect $B D$ and $B_{1} D_{1}$ to form the plane $B B_{1} D_{1} D$. Since point $N$ is in the plane $B B_{1} D_{1} D$ but not on the line $B_{1} M$; and the line $A A_{1} \parallel$ plane $B B_{1} D_{1} D$, point $A$ is not in the plane $B B_{1} D_{1} D$, hence $A N$ and $B_{1} M$ are skew li... | \arccos \frac{2}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 713,064 |
Five. (20 points) For positive real numbers $a, b, c$, prove:
$$
\frac{\sqrt{a^{2}+8 b c}}{a}+\frac{\sqrt{b^{2}+8 a c}}{b}+\frac{\sqrt{c^{2}+8 a b}}{c} \geqslant 9 .
$$ | Proof 1: Let $x=\frac{\sqrt{a^{2}+8 b c}}{a}>1$,
$$
y=\frac{\sqrt{b^{2}+8 a c}}{b}>1, z=\frac{\sqrt{c^{2}+8 a b}}{c}>1 \text {. }
$$
Then $\left(x^{2}-1\right)\left(y^{2}-1\right)\left(z^{2}-1\right)$
$$
\begin{aligned}
= & 8 \frac{b c}{a^{2}} \cdot 8 \frac{a c}{b^{2}} \cdot 8 \frac{a b}{c^{2}}=8^{3} . \\
8^{3}= & {[(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,065 |
2. A paper punch can be placed at any point on a plane. When it works, it can punch out points that are at an irrational distance from it. What is the minimum number of paper punches needed to punch out all points on the plane? | 2. Obviously, two paper punch machines are not enough. To prove that three paper punch machines are sufficient, it is enough to prove that among the distances from any point on the plane to these three points, at least one is an irrational number. Therefore, it is sufficient to construct a sum involving three distances... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,066 |
Given a sequence of positive integers
$$
\left\{\begin{array}{l}
a_{1}=1, a_{2}=2 . \\
a_{n+2}=a_{n+1}+a_{n}(n \geqslant 1) .
\end{array}\right.
$$
(1) Prove that the infinite set of integer points formed by consecutive terms of the sequence
$$
\left(a_{1}, a_{2}\right),\left(a_{3}, a_{4}\right), \cdots,\left(a_{2 k-1}... | (1) Use mathematical induction.
It is obvious that $\left\{\begin{array}{l}x=a_{1}=1 \\ y=a_{2}=2\end{array}\right.$.
is on the curve. Now assume the integer point $\left(a_{2 k-1}, a_{2 k}\right)$ is on the curve. Then we have
$$
a_{2 k-1}^{2}+a_{2 k-1} a_{2 k}-a_{2 k}^{2}+1=0 .
$$
By completing the square and transf... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,067 |
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