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Example 5. (1MO-17-5) Prove: On the unit circle, 1975 points can be placed such that the straight-line distance between any two points is a rational number.
| Proof Our method is to use Pythagorean numbers to construct the points mentioned (i.e., to find them directly).
$$
\text { Let } \theta_{n}=\operatorname{arctg} \frac{n^{2}-1}{2 n}, \quad(1 \leqslant 5 \leqslant 1975)
$$
Then $\sin \theta_{n}=\frac{n^{2}-1}{n^{2}+1}, \cos \theta_{n}=\frac{2 n}{n^{2}+1}$
are rational n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,652 |
Example 6. (1MO-27 Preliminary Question) Let $A, B, C$ be three points on the edge of a circular pool, with $B$ due west of $C$, and $A B C$ forming an equilateral triangle with side lengths of 86 meters. A swimmer starts from $A$ and swims directly to $B$. After swimming $x$ meters, he reaches point $\boldsymbol{E}$, ... | Given the figure, since $\triangle AEF$ is an equilateral triangle, we have
$$
AF = AE = x.
$$
By symmetry, $FG = DE = y$. Also, since
$$
AE \cdot EB = DE \cdot EG,
$$
we have
$$
\begin{aligned}
& x(86 - x) \\
= & y(x + y).
\end{aligned}
$$
Notice that if $x$ is odd, then $x(86 - x)$ is also odd, while $y(x + y)$ is... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,653 |
Example 7. (1MO-24-3) Let $a, b, c$ be three positive integers, any two of which have no common divisor greater than 1. Prove:
$$
2 a b c-b c-a c-a b
$$
is the largest integer that cannot be expressed as $x b c+y c a+z a b$ (where $x$, $y$, $z$ are non-negative integers) | The problem requires us to do two things:
First, prove that \(2abc - bc - ac - ab\) cannot be expressed in the given form; then prove that any integer greater than \(2abc - bc - ac - ab\) can be expressed in the given form.
Let's start with the first part.
We use proof by contradiction. If there exist non-negative inte... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,654 |
Prove: Only using coins of 2 cents and 5 cents, any amount greater than 3 cents can be exchanged.
保留源文本的换行和格式，翻译结果如下：
Prove: Only using coins of 2 cents and 5 cents, any amount greater than 3 cents can be exchanged. | This problem can be proved by induction. However, applying the aforementioned theorem will be more straightforward.
In the theorem, take $a=2, b=5$, then $(a, b)=1$. According to the theorem, any positive integer greater than $2 \times 5-2-5=3$ can be expressed as
$$
2 x+5 y,
$$
where $x, y$ are non-negative integers.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,655 |
Example 9. Prove that any integer greater than 36 can be expressed as the sum of two numbers with square factors.
保留源文本的换行和格式，所以翻译结果如下：
Example 9. Prove that any integer greater than 36 can be expressed as the sum of two numbers with square factors. | To prove that a positive integer with a square factor means it can be divided by a square number greater than 1. Of course, perfect squares are numbers with square factors, but the converse is not true. A number without a square factor means it cannot be divided by a square number greater than 1.
For example, 4, 12, 2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,656 |
Example 3. If $x>0, y>0$, and $21 \lg(x-2 y)=\lg x+\lg y$, then $x: y$ is ( )
(A) 4. (B) 1 .
(C) 1 or 4 .
(D) $\frac{1}{4}$. | Many students often transform the given condition
$21 g(x-2 y)=\lg x+\lg y$ into
$\lg (x-2 y)^{2}=\lg x y$. However, this transformation brings changes. Before the transformation, the conditions required $x-2 y>0, x>0, y>0$, while after the transformation, they became $x-2 y \neq 0, x y>0$, which weakens the conditions... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 703,657 |
Example 10. Prove: Integers greater than 11 can be expressed as the sum of two composite numbers. | Proof: In the theorem, take $a=2, b=3$. Clearly, $(a, b)=1$. By the theorem, any integer greater than $2 \times 3-2-3$ can be expressed as
$$
2 x+3 y,
$$
where $x, y$ are non-negative integers.
The trouble lies in: first, $x$ and $y$ may be 0; second, even if $x, y$ are not 0, they can be 1, thus $2 x, 3 y$ are not nec... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,658 |
Example 8. $m$ is an integer, and the equation
$$
9 x^{2}+(2-12 m) x+4 m^{2}+2 m-12=0
$$
has integer solutions. | Substitute (13), $x_{1}$ is not an integer.
ii) Let $k_{1}-1=3 k_{2}$,
then $k_{1}=3 k_{2}+1$.
(16)
Substitute (16) into (12), (13), (14), we get
$$
\begin{array}{l}
m=-144 k_{2}^{2}-90 k_{2}-11, \\
x_{1}=-96 k_{2}^{2}-52 k_{2}-5, \\
x_{2}=-96 k_{2}^{2}-68 k_{2}-10 .
\end{array}
$$
$$
\text { iii) Let } k_{1}+1=3 k_{2... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,659 |
Example 1. For what integer values of $m$ does the equation $2 x^{2}+(1-2 m) x+m^{2}-2 m=0$ have integer solutions? | Solve $\because \Delta=-4 m^{2}+12 m+1$,
$$
\begin{array}{l}
\therefore x=-\frac{2 m-1 \pm \sqrt{\Delta}}{4} . \\
\because x, m \text { are integers, } \\
\therefore \Delta=-4 m^{2}+12 m+1 \geqslant 0 .
\end{array}
$$
Solving, we get $\frac{3-\sqrt{10}}{2} \leqslant m \leqslant \frac{3+\sqrt{10}}{2}$.
Thus, $m=0,1,2,3... | m=0,1,2,3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,663 |
Example 2. For what integer values of $m$, does the equation $m x^{2}+(m-2) x+m-3=0$ ($m \neq 0$) have integer solutions? And find these solutions. | Solving by imitating Example 1, we get: when $m=1$, the equation has integer solutions $x=2, -1$;
when $m=3$, the equation has an integer solution $x=0$.
Summary $1^{\circ}$ When $\Delta=A m^{2}+B m+C$ with $A<0$, let $\Delta \geqslant 0$, solving yields $\alpha \leqslant m \leqslant \beta$, determine the integers ... | m=1, x=2, -1; m=3, x=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,664 |
Example 3. For what integer values of $m$ does the equation $x^{2}-m x+m+1=0$ have integer solutions.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve $\Delta=m^{2}-4(m+1)=m^{2}-4 m-4$.
If we follow the method of "Summary $1^{\circ}$, and let $\Delta \geqslant 0$, we get $m \leqslant 2-2 \sqrt{2}$ or $m \geqslant 2+2 \sqrt{2}$. To determine the values of $m$ that make $\Delta$ a perfect square in the set $(-\infty, 2-2 \sqrt{2}) \cup(2+2 \sqrt{2},+\infty)$ is c... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,665 |
Example 4. For what integer values of $m$ does the equation $x^{2}-(m-1) x-2 m^{2}-5 m-3=0$ have integer solutions?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solving by imitating Example 3, we get
$$
\left\{\begin{array}{l}
m=-1, \\
k=2 .
\end{array}\right.
$$
Substituting into (2), we get $x=0,-2$.
Therefore, when $m=-1$, the equation has integer solutions.
Summary $2^{\circ}$ When $\Delta=A m^{2}+B m+C$ and
$A>0$, and $A$ is a perfect square, the discriminant can be comp... | m=-1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,666 |
Example 5. For what value of $m$ does the equation $\left(m^{2}-1\right) x^{2}$ $-6(3 m-1) x+72=0$ have positive integer solutions? | Solve $\because \Delta=[6(m-3)]^{2}$,
$$
\begin{aligned}
\therefore \quad x & =\frac{6(3 m-1) \pm 6(m-3)}{2\left(m^{2}-1\right)} \\
& =\frac{2[(3 m-1) \pm(m-3)]}{m^{2}-1} .
\end{aligned}
$$
Thus, $x_{1}=\frac{12}{m+1}, \quad x_{2}=\frac{6}{m-1}$.
$\because x_{1}>0, x_{2}>0, m$ is an integer,
$\therefore$ For $x_{1}$, ... | m=2,3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,667 |
Example 6. For what integer values of $m$ does the equation $4 x^{2}-4(m+1) x-3 m^{2}+10 m-3=0$ have two integer solutions? Two positive integer solutions? | Solve $\because \Delta=16(2 m-2)^{2}$,
$$
\begin{array}{l}
\therefore x_{1}=\frac{3 m-1}{2}=m+\frac{m-1}{2}, \\
x_{2}=\frac{-m+3}{2}=1-\frac{m-1}{2}, \\
\text { let } \quad m-1=2 k, \quad(k \text { is an integer })
\end{array}
$$
Let $m-1=2 k$, ($k$ is an integer)
Then $m=2 k+1$.
(3) Substituting into (1), (2) gives ... | m=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,669 |
Example 7. $m$ is an integer, find the integer solutions of the equation $4 x^{2}+(2-4 m) x$ $+m^{2}+4 m-1=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve $\Delta=(2-4 m)^{2}-4 \times 4\left(m^{2}+4 m-1\right)$ $=20(-4 m+1)$.
Let $-4 m+1=5 u^{2}$ ( $u$ is a positive integer),
then $x=\frac{2 m-1+5 u}{4}$,
or $x=\frac{2 m-1-5 u}{4}$.
$$
m=\frac{1-5 u^{2}}{4}=-u^{2}-\frac{(u-1)(u+1)}{4} .
$$
From (3), we know that $u-1$ and $u+1$ must have one that is a multiple of... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,670 |
Example 1. For a regular $\triangle ABC$, the vertex $B$ has coordinates $(m, 0)$ $\left(m > a \text{ is a constant}\right)$, and point $A$ moves along the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$. Assuming $A$, $B$, and $C$ are arranged counterclockwise, find the equation of the trajectory of point $C$. | $$
\begin{array}{l}
\text { Let } A, B, C \text { be the points corresponding to the complex numbers } x_{0}+y_{0} i \text {, } \\
\rightarrow \\
m, x+y i \text {, then } B A_{1}\left(x_{0}-m\right)+y_{0} i \text {, } \\
B C: \quad(x-m)+y i \text {. } \\
\end{array}
$$
Since $\overrightarrow{B C}$ rotated counterclock... | \frac{(x+m-\sqrt{3} y)^{2}}{4 a^{2}} +\frac{[y+\sqrt{3}(x-m)]^{2}}{4 b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,671 |
Example $2 . \square O A B C$ has its vertex $O$ fixed at the origin, $A$ moves on the line segment $x=1, y=t(-1 \leqslant t \leqslant 1)$, and $C$ moves on the unit circle $\odot O$. Find the trajectory of $B$. | Let $A, B, C$ points correspond to the complex numbers $1+t i$
$$
\begin{array}{l}
(-1 \leqslant t \leqslant 1), x+y i, \cos \theta+i \sin \theta(0 \\
\leqslant\theta \leqslant 2 \pi) \\
\vec{O} \quad \overrightarrow{O C}, \text { for } \square O A B C, \overrightarrow{O B}=\overrightarrow{O A}+\overrightarrow{A B} \\
... | (x-1)^{2}+(y-t)^{2}=1 \cdot(-1 \leqslant t \leqslant 1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,672 |
Example 3. For the equation $t^{2} + z t + z i = 0$ with $z$ and $i$ as coefficients, it always has a real root $\alpha$. Find the locus equation of point $z$.
保持源文本的换行和格式如下：
Example 3. For the equation $t^{2} + z t + z i$ $=0$ with $z$ and $i$ as coefficients, it always has a real root $\alpha$. Find the locus equat... | Let $z=x+y i$, we need to find the relationship between $x$ and $y$. Substituting $\lambda a^{2}+\alpha z+z i=0$, and applying the condition for equality of complex numbers, we get $\boldsymbol{a}^{2}+\boldsymbol{\alpha} x-y=0, a y+x=0$. If $y=0$, then $x=0$; assuming $y \neq 0$, then $a=-\frac{x}{y}$. Substituting int... | x^{2}(1-y)=y^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,673 |
For example, the vertices $B$ and $A$ of a regular $\triangle A B P$ with side length $a$ slide on the $x$-axis and $y$-axis respectively $(A, B, P$ are arranged clockwise). Find the trajectory of vertex $P$ in the first quadrant. | Let $P, A, B$ correspond to the complex numbers $x+yi, 2mi$, $2n$, respectively. Then the midpoint $C$ of $AB$ corresponds to the complex number $n+mi$, and $\overrightarrow{CP}=\overrightarrow{CB} \cdot \sqrt{3} \vec{i}$, i.e.,
$$
\begin{aligned}
& (x+yi)-(n+mi) \\
= & (2n-n-mi) \sqrt{3i} .
\end{aligned}
$$
Rearrangi... | x^{2}+y^{2}-\sqrt{3} xy=\frac{a^{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,674 |
Example 1. In $\triangle A B C$, prove that
$$
\begin{array}{l}
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \\
\cdot \cos B \cos C=1
\end{array}
$$ | Consider the system of equations
$$
\left\{\begin{array}{l}
-x + y \cos C + z \cos B = 0, \\
x \cos C - y + z \cos A = 0, \\
x \cos B + y \cos A - z = 0.
\end{array}\right.
$$
Since it has a non-zero solution \((\sin A, \sin B, \sin C)\), the determinant of the coefficient matrix \(D = 0\). Expanding this determinant ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,675 |
Example 2. If $a x_{1}+b y_{1}=1, b x_{1}+c y_{1}=1$,
$c x_{1}+a y_{1}=1$, prove that
$$
a b+b c+c a=a^{2}+b^{2}+c^{2}
$$ | Consider the system of equations
$$
\left\{\begin{array}{l}
a x+b y-z=0, \\
c x+a y-z=0, \\
b x+c y-z=0 .
\end{array}\right.
$$
It has a non-zero solution $x=x_{1}, y=y_{1}, z=1$. Its coefficient determinant $D=0$, expanding which yields the equation to be proved. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,676 |
Example 2. Simplify $\sin ^{3} 10^{\circ}+\sin ^{3} 30^{\circ}+\sin 50^{\circ}$ $+\sin ^{3} 70^{\circ}$. | Apply the formula
$$
\begin{array}{l}
\sum_{i=1}^{n} \sin ^{3} a_{0} \\
=\frac{3 \sin \left(a_{1}+\frac{n-1}{2} d\right) \sin \frac{n}{2} d}{4 \sin \frac{d}{2}} \\
-\frac{\operatorname{sins}\left(a_{1}+\frac{n-1}{2} d\right) \sin \frac{3 n}{2} d}{4 \sin \frac{3 d}{2}}
\end{array}
$$
It is known that the original expre... | \frac{3}{8}\left(\frac{1}{\sin 10^{\circ}}-2\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,678 |
Example 5. Given $A=6 \lg p+\lg q$, where $p, q$ are prime numbers, and satisfy $q-p=29$.
Prove: $3<A<4$.
The text is translated while preserving the original line breaks and format. | Prove that if $p, q$ are prime numbers and $q-p=29$, then: $p$ and $q$ must be one odd and one even (otherwise $p-q$ would be even). The only number that is both prime and even is 2, so $p=2$. Thus, it follows that $q=31$.
$$
\begin{array}{l}
A=6 \lg p+\lg q \\
=6 \lg 2+\lg 31 \\
=\lg 2^{6} \cdot 31 \\
=\lg 1984 . \\
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,679 |
Example 1. On a circle with diameter $AB$, there are two points $M, N$. Let the intersection point of $AM$ and $BN$ be $P$. Prove:
$$
AP \cdot AM
$$
$$
+ BP \cdot BN = AB^2
$$ | How was this proposition derived? When posing the problem,
in addition to using the conditions in this example, the conditions that $P Q \perp A B$ and line segments $A N, B M$ were also used, thus leading to:
(1) The sum of two equations
$$
\begin{array}{c}
\triangle A P Q \backsim \triangle A B M \Rightarrow \frac{A ... | A P \cdot A M + B P \cdot B N = A B^2 | Geometry | proof | Yes | Yes | cn_contest | false | 703,680 |
Given: In the circle $\odot O$, the diagonals of the inscribed quadrilateral $ABCD$ are
$$
\begin{array}{l}
\text { } AC \text { and } BD. \\
\quad \text { Prove: } AC \cdot BD \\
=AD \cdot BC \\
+AB \cdot CD .
\end{array}
$$ | Analysis: Let $B D={ }_{B} K + D K$.
$$
\text{Then } \begin{aligned}
A C \cdot B D & =A C(E K+D K) \\
& =A C \cdot B K+A C \cdot D K .
\end{aligned}
$$
Assume $A \mathcal{C} \cdot B D=A D \cdot B C+A B \cdot C D$ holds, then
$$
\begin{array}{l}
A C \cdot B K+A C \cdot D K \\
=A D \cdot B C+A B \cdot C D \text{. } \\
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,682 |
Example 4. As shown in the figure, $C$ is a point on the diameter $A B$, $C E$ is perpendicular to the tangent $D E$, and $D$ is the point of tangency. Prove: $A B \cdot C E$ $=A C \cdot B C+C D^{2}$.
Analysis: To make the equation useful, we need to substitute terms containing the common factor $C D$. Extend $D C$ to... | $$
\begin{array}{l}
\text{Since } \triangle D F G \sim \triangle C E D, \therefore \frac{D G}{C D}=\frac{D F}{C E} \text{.} \\
\text{Since } A B=D G, \therefore \frac{A B}{C D}=\frac{D F}{C E} \text{.} \\
\text{Since } A B \cdot C E=C D \cdot D F \text{,} \\
D F=C F+C D \text{,} \\
\therefore A B \cdot C E=C D(C F+C D)... | A B \cdot C E=A C \cdot B C+C D^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 703,683 |
1. The Distance Formula from a Point to a Line
Prove that the distance formula from point $P_{0}\left(x_{0}, y_{0}\right)$ to the line $l: A x+B y$ $+C=0$ is
$$
d=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} .
$$ | Proof: Let the inclination angle of line $l$ be $\theta_{1}$, and the parametric equation of the line $l^{\prime}$ passing through point $P_{0}^{\prime}\left(x_{0}, y_{0}\right)$ and perpendicular to $l$ is
$$
l^{\prime}:\left\{\begin{array}{l}
x=x_{0}+t \cos \theta_{2}, \\
y=y_{0}+t \sin \theta_{2}.
\end{array}\right.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,685 |
2. Coordinate Formula for a Point Dividing a Line Segment in a Given Ratio
Let $P_{1}\left(x_{1}, y_{1}\right)$ and $P_{2}\left(x_{2}, y_{2}\right)$ be two known points, and $P(x, y)$ be a point dividing the line segment $P_{1} P_{2}$, and $\frac{\overline{P_{1} P}}{\overline{P P_{2}}}=\lambda(\lambda \neq-1)$ is a co... | Proof: Let the parametric equation of the line $P_{1} P$ be $\left\{\begin{array}{l}x=x_{1}+t \cos \theta, \\ y=y_{1}+t \sin \theta .\end{array}\right.$ Then $t_{1}=\frac{x-x_{1}}{\cos \theta}=\frac{y-y_{1}}{\sin \theta}$. The parametric equation of the line $P_{1} P_{2}$ is
$$
\left\{\begin{array}{l}
x_{2}=x_{1}+t_{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,686 |
3. The area formula of a triangle
Let the vertices of $\triangle A B C$ be $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2},{ }_{2} y,\right) C\left(x_{3}, y_{3}\right)$, prove that
$$
S_{\triangle A B C}=\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x & y & 1
\end{array}\right|
$$ | As shown in Figure 6, let $\triangle ABC$ have $\angle BAC = \alpha$, the inclination angle of side $AB$ be $\beta$, and the inclination angle of side $AC$ be $\gamma$. Therefore, we have:
The parametric equation of line $AB$ is
$$
\left\{\begin{array}{l}
x_{2}=x_{1}+t_{1} \cos \beta, \\
y_{2}=y_{1}+t_{1} \sin \beta .... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,687 |
$2^{\circ}$ Let the line $\left(L^{\prime}\right)$ passing through the fixed point $P_{0}\left(x_{0}, y_{0}\right)$ intersect the conic section $f(x, y)=A x^{2}+B x y+C y^{2}$ $+D x+E y+F=0$ at points $A$ and $B$. Find the length of the chord $AB$. | Solve As shown in Figure 8, according to the relationship between $t^{\prime}$ and $t$ (•),
it is known that $\left|P_{0} A\right|=\sqrt{a^{2}+b^{2}}\left|t_{1}^{\prime}\right|$,
$$
|P \circ B|=\sqrt{a^{2}+b^{2}}\left|t_{2}^{\prime}\right| \text {, }
$$
Therefore, $|A B|=\left|P_{0} B-P_{0} A\right|=\left|\sqrt{a^{2}+... | |AB| = \sqrt{a^{2}+b^{2}} \sqrt{\left(t_{1}^{\prime}+t_{2}^{\prime}\right)^{2}-4 t_{1}^{\prime} t_{2}^{\prime}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,689 |
Example 2. The parametric equation of line $L^{\prime}$ is $\left\{\begin{array}{l}x=1+t^{\prime}, \\ y=5+\sqrt{3} t^{\prime}\end{array}\right.$, it intersects the circle $x^{2}+y^{2}=16$ at points $A, B$. Find the length of the chord $AB$. | Solution 1 Using the formula $\left(L_{A B}\right) \cdot$ From Example 1, we know that the line
$$
x=1+\frac{t}{2},
$$
$L'$ in standard form is
Substituting it into
$$
x^{2}+y^{2}=16 \text { gives } t^{2}+(1+5 \sqrt{3}) t+10=0 \text {. }
$$
Therefore, $|A B|=\sqrt{[-(1+5 \sqrt{3})]^{2}-4} \times 10$
$$
=\sqrt{36+10 \s... | \sqrt{36+10 \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,691 |
1. Find the locus equation of the midpoint of a moving chord passing through a fixed point
Proposition 1 Prove that the locus equation of the midpoint of the chord intercepted by a moving line passing through a fixed point $P(h, k)$ on a conic section $f(x, y)$ $=A x^{2}+B x y+C y^{2}+D x+E y+F=0$ is still a conic sec... | Proof: Let the parametric equations of a line passing through point $P_{0}\left(x_{0}, y_{0}\right)$ be
\[
\left(\begin{array}{l}
x=x_{0}+t \cos \theta, \\
y=y_{0}+t \sin \theta
\end{array}\right.
\]
($t$ is a parameter, $\theta$ is the angle of the line)
Substituting (1) into $f(x, y)=0$ and simplifying, we get
\[
\b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,692 |
Proposition 2 Find the equation of the locus of the midpoints of parallel chords with slope $k$ of the conic section $f(x, y)=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | From the proof of Proposition 1, equation (3) shows that when the line varies, the midpoint \( P_{\mathrm{E}}\left(x_{0}^{\prime}\right., y_{0} \) of the chord it intercepts also varies. Therefore, equation (3) becomes the relationship that the coordinates of the midpoints of the chords should satisfy:
$$
\begin{array}... | (2 A+B k) x+(B+2 C k) y+D+E k=0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,693 |
Example 4. The locus of the midpoints of parallel chords of the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}(a>0$, b>0) is. | Given the equation (8), with $A=b^{2}, C=a^{2}$, $B=D=E=0, F=-a^{2} b^{2}$, the required trajectory equation is $2 b^{2} x+2 a^{2} k y=0$.
Therefore, when $k \neq 0$, it is $y=-\frac{b^{2}}{a^{2} k} x$; when $k=0$, it is $x=0$. They are all straight line equations passing through the center of the ellipse (i.e., the o... | y=-\frac{b^{2}}{a^{2} k} x \text{ or } x=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,694 |
Example. Prove: If all interior angles of a hexagon are $120^{\circ}$, and the sides are in G.P., then there is one and only one such hexagon, which is a regular hexagon.
| To prove: $\because \angle 2+\angle \angle=180^{\circ}-120^{\circ}=60^{\circ}$,
$$
\angle 1+\angle 4=2 \times 120^{\circ}-60^{\circ}=180^{\circ} \text {, }
$$
$\therefore$ opposite sides are parallel. Similarly, all pairs of opposite sides are parallel.
According to (*), we get:
$$
\begin{array}{l}
\quad(x q)^{2}=x^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,696 |
Example 1. Let $a \neq 0$ be a real number, and $\lim _{n \rightarrow \infty}\left(\sqrt{\frac{6}{n}}+a^{2}+a\right) n=-a$, find the value of $a$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
=\frac{6}{\lim_{n \rightarrow \infty} \sqrt{\frac{6}{n}+a^{2}}-a}=\frac{6}{\sqrt{a^{2}}-a}, \\
\therefore \frac{6}{\sqrt{a^{2}}-a}=-a, \text { solve for } a=0 \text { (discard): } \\
a=-\sqrt{3} \text { is the solution. }
\end{array}
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,697 |
Example 2. Let $\lim _{x \rightarrow 1} \frac{a \sqrt{x}+b}{x-1}=1$, find the values of $a$ and $b$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Since the limit exists, as $x \rightarrow 1$, the denominator $x-1$ $\rightarrow 0$, and at the same time, the numerator $a \sqrt{x}+b \rightarrow 0$, i.e., $\lim _{x \rightarrow 1}(a \sqrt{x}+b)=a+b=0, \quad \therefore b=$ $-a$, substituting into the original expression we get $\lim _{x \rightarrow 1} \frac{a \sqrt{x-... | null | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 703,698 |
Proposition: The inverse function of $y=f(x)$ is $y=f^{-1}(x)$. Let the solution sets of the equations $f(x)=f^{-1}(x)$, $x=f(x)$, and $x=f^{-1}(x)$ be $M_{1}$, $M_{2}$, and $M_{3}$, respectively. Then
$$
I_{2} \subseteq M_{1}, M_{3}=M_{1} \text {, and } M_{2}=M_{3} \text {. }
$$ | Proof: Let $x_{1} \in M_{2}$, i.e., $x_{1}=f\left(x_{1}\right)$. Apply the $f^{-1}$ function to both sides,
$$
f^{-1}\left(x_{1}\right)=f^{-1}\left[f\left(x_{1}\right)\right]=x_{1},
$$
i.e., $x_{1} \in M_{3}$, hence $M_{2} \subseteq M_{3}$.
Similarly, we can prove $M_{3} \subseteq M_{2}$, so $M_{2}=M_{3}$.
For any $x_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,699 |
For example, the function $(f x)=\left(1+\frac{x}{2}\right)^{2}-2$, $x \in[-2,+\infty)$, find the solution set of the equation $f(x)=f^{-1}(x)$. | We can first draw the graph, noting that $f(x)$ is a parabola, and on $[-2,+\infty)$, it is intuitively clear that $y:=f(x)$ and $y=x$ intersect at only two points. Applying the proposition we have proven, to find the solution set of $f(x)=f^{-1}(x)$, we can first find the solution set of its subset $x=f(x)$.
$$
\text... | \{-2,2\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,700 |
Let $a_{1}, a_{2}, \cdots, a_{n}$ be given real numbers, not all zero, and $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers. If the inequality
$$
\begin{array}{l}
\quad r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+ \\
r_{n}\left(x_{n}-a_{n}\right) \\
\leqslant \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{... | Let $x_{1}=x_{2}=\cdots=x_{n}=0$, from (1) we have
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \\
\quad \geqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}
\end{array}
$$
Take $x_{1}=2 x_{1}, x_{2}=x_{2}, \cdots, x_{n}=2 a_{n}$, from (1) we also get
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdo... | r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} = \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 703,706 |
Let $C_{1}, C_{2}$ be concentric circles, with the radius of $C_{2}$ being twice that of $C_{1}$. Quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is inscribed in $C_{1}$. Extend $A_{1} A_{1}$ to intersect circle $C_{2}$ at $B_{1}$, extend $A_{1} A_{2}$ to intersect circle $C_{2}$ at $B_{2}$, extend $A_{2} A_{3}$ to intersect c... | Let the common center be $O$, and connect $O A_{1}, O B_{1}$, and $O B_{2}$. Applying the generalized Ptolemy's theorem in quadrilateral $O A_{1} B_{1} B_{2}$, we have
$$
\begin{aligned}
O B_{1} & \cdot A_{1} B_{2} \leqslant O A_{1} \cdot B_{1} B_{2} + O B_{2} \cdot A_{1} B_{1} \\
& \because O B_{1} = O B_{2} = 2 O A_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,707 |
$$
a_{1}, a_{2}, \cdots, a_{n}
$$
Among these numbers, if the arithmetic mean of a segment of numbers $a_{k}, \cdots, a_{k+i-1}$ is greater than 1988, then we call this segment of numbers a “dragon”, and $a_{k}$ is called the “dragon head” of this dragon. (If a single term $a_{m}>1988$, then this single term is also a... | Proof: Let
$$
a_{k}, a_{k+1}, \cdots a_{k+1}
$$
be the shortest dragon with $a_{k}$ as the head. We will prove that any term $a_{m}(k1988 \text {, }
$$
i.e., $a_{m}, \cdots, a_{A+1}$ is a dragon, so $\boldsymbol{a}_{m}$ can also be the head.
In sequence (1): pick out the shortest dragon with the first term that can ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,708 |
(i) Let three positive real numbers $a, b, c$ satisfy
$$
\left(a^{2}+b^{2}+c^{2}\right)^{2}>2\left(a^{4}+b^{4}+c^{4}\right) \text {. }
$$
Prove that $a, b, c$ are the lengths of the sides of a triangle.
(ii) Let $n$ positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the inequality
$$
\begin{array}{c}
\left(a_... | Prove H (1) has
$$
\begin{array}{l}
\left(a_{1}^{2}+\cdots+a_{n-1}^{2}\right)^{2}+2\left(a_{1}^{2}+\cdots+a_{n-1}^{2}\right) a_{n}^{2} \\
+a_{n}^{4}>(n-1)\left(a_{1}^{4}+\cdots+a_{n-1}^{4}\right)+(n-1) a_{n}^{4}, \\
(n-2) a_{n}^{4}-2\left(a_{1}^{2}+\cdots+a_{n-1}^{2}\right) a_{n}^{2}+ \\
(n-1)\left(a_{1}^{4}+\cdots+a_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,709 |
Five, given three tetrahedra \(A_{i} B_{i} C_{i} D_{i} (i=1, 2,3)\), through points \(B_{i}, C_{i}, D_{i}\) construct planes \(\alpha_{i}, \beta_{i}, \gamma_{i} (i=1,2,3)\) perpendicular to edges \(A_{i} B_{i}, A_{i} C_{i}\), \(A_{1} D_{i}\) respectively \((i=1,2,3)\). If the planes \(\alpha_{i}, \beta_{i}, \gamma_{i} ... | Connect $E A_{i}, E B_{i}, E C_{i}, E D_{i}(i$, $=1,2,3)$, then $E B_{i} \perp A B_{i}, E C_{i} \perp A C_{i}$, $E D_{i} \perp A D_{i}(i=1,2,3)$.
$$
\therefore A_{i}, B_{i}, C_{i} \text {, }
$$
$D_{i}, E$ are five points on a sphere
$H A_{1} E$ is the diameter of this sphere $(i=1,2,3)$.
$\therefore E$ is the common po... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,710 |
10. Find the number of ways to divide the set $\{1,2, \cdots, n\}$ into three subsets $A_{1}, A_{2}, A_{3}$. The subsets can be empty, and must satisfy the following conditions:
(1) After arranging the elements of each subset in ascending order, the parity of adjacent elements is different;
(2) If the sets $A_{1}, A_{2... | 10. Let $A_{1}$ be a set containing the element 1, and let $A_{2}, A$ be two sets such that the smallest element of $A_{2}$ is less than the smallest element of $A_{3}$. Now, numbers are sequentially allocated to subsets that always satisfy certain conditions to construct a partition that meets condition (2). The numbe... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,713 |
11. Given an equilateral $\triangle A B C$, with its vertices arranged counterclockwise. Find the locus of the centroid of the equilateral $\triangle A^{\prime} B^{\prime} C^{\prime}$, where the vertices are also arranged counterclockwise, and points $A, B^{\prime}$, $C^{\prime}, A^{\prime}, B, C^{\prime}$, and $A^{\pr... | 11. Angles are measured in the clockwise direction. Let $a, b, c$ be variable lines determined by the following conditions:
1. $A \in a, B \in b, C \in c$;
2. $\angle(a, b)=\angle(b, c)=\angle(c, a)$
$$
=\frac{\pi}{3} \text {; }
$$
3. $a, b$ and $c$ do not intersect at a single point.
The relations $A^{\prime} \in b \... | S=C \setminus \{P\} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,714 |
1. There are three consecutive positive integers, the sum of whose reciprocals is $\frac{47}{60}$. Find these three numbers. | Let the three numbers be $x, x+1, x+2$, then we have
$$
\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}=\frac{47}{60} .
$$
Eliminating the denominators and rearranging, we get
$$
47 x^{3}-180 x^{2}-219 x-26=0 \text {. }
$$
This equation is generally not solved in high school, and even if it can be solved, the calculation is ... | 3,4,5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,717 |
Example 14. Find the number of integer solutions $(x, y, z)$ that satisfy $0<x<y<z$, and $\sqrt{1984}=\sqrt{ } x+\sqrt{ } y+\sqrt{ } z$. | $$
\text { Stele } \begin{aligned}
& \sqrt{1984}=8 \sqrt{31} \\
& =\sqrt{31}+2 \sqrt{31}+5 \sqrt{31} \\
& =\sqrt{31}+\sqrt{2} \times 31+\sqrt{5} \times 31 \\
& =\sqrt{31}+3 \sqrt{31}+4 \sqrt{31} \\
& =\sqrt{31}+\sqrt{3} \times 31+\sqrt{2} \times 1 .
\end{aligned}
$$
There are two kinds of correct statements in total.
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,722 |
oi1. Find the number of positive integer solutions for the equation
$$
x+y+z=20
$$ | Solving for $z$: $=1$, substituting into (1), we get $x+y=19$. For this boundary, when $x$ takes the values $1, 2, \cdots, 18$ $=2$, there are 17 sets of positive integer solutions. Continuing this way, when $z=18$, there is only 1 set of positive integer solutions. Therefore, the total number of positive integer solut... | 171 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,724 |
Example 4. Find the indefinite solution of the equation
$$
x+y+z=20
$$ | Find the number of positive integer solutions for (1). For this, we have 20 a's and 19 spaces (denoted by $s$) arranged in one line as follows:
asasasasasasasasasasasasasasasasas asisa.
Now, choose any 2 out of the 19 s's, and remove the rest of the $s$'s, for example:
caaaasaaaaaasaaauaaaaa. This results in three seg... | 231 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,727 |
For example. Solve the equation: $\sqrt{x+3}+\sqrt{2 x-1}=4$. | Solve for $(\sqrt{x+3}+\sqrt{2 x-1})^{2}$
$$
\begin{array}{c}
=\left(\sqrt{2} \cdot \sqrt{x+3}+\frac{1}{\sqrt{2}}\right. \\
\cdot \sqrt{2 x-1})^{2}+(1-2) \cdot\left(3+\frac{1}{2}\right),
\end{array}
$$
i.e., $16=\left(\sqrt{2} \cdot \sqrt{x+3}+\frac{1}{\sqrt{2}} \sqrt{2 x-1}\right)^{2}$
$$
-\frac{7}{2} \text {, }
$$
... | x=52-8 \sqrt{39} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,729 |
Example 6. Prove that the number of non-negative integer solutions to the equation
$$
2 x+3 y=n
$$
is equal to $\left(\frac{n}{6}\right)$ or $\left(\frac{n}{6}\right)+1$. | Prove that when $n=1,2,3,4,5,6$, the conclusion can be directly verified as correct.
Now assume that when $n=k$, the equation $2 x+3 y=k$ has a number of non-negative integer solutions equal to $\left(\frac{k}{6}\right)$ or $\left(\frac{k}{6}\right)+1$. Then, when $n=k+6$, the equation becomes
$$
2 x+3 y=k+6
$$
For t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,730 |
List 7. Let $x+y+z+w=20$. (8) Try to find (i) the number of positive integer solutions to equation (8); the number of groups;
(iii) the number of positive integer solutions to (8) satisfying $x \geq 7, y=8$;
(iv) the number of positive integer solutions to (8) satisfying $x \leqslant 6$;
(v) the number of positive inte... | (i) From the last statement of Example 4, the number of positive integer solutions of equation (8) is $C_{2}^{4-1}=C_{1}^{3}=969$.
(ii) Let $x^{\prime}=x-6$, then equation (8) becomes $x^{\prime}+y+z+w=14$.
Now, we find the number of positive integer solutions of (9) under the condition that $x \geqslant 7$ for equati... | 483 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,731 |
15. The boss hands the letters to be printed to the secretary, one at a time, and places each on top of the pile. The secretary, whenever free, takes the top letter to type. One day, there are 9 letters to be typed, and the boss hands them over in the order of the first, second, $\cdots$, to the ninth. During lunch, th... | 15. ( 704$)$
We consider two cases.
The first case: The ninth letter is finally typed by the secretary before noon.
This may be the order of the subset of the set $T=\{1,2$, ..., 6, 7, 9\} of possible subsets. In fact, the secretary can type a letter as soon as it arrives if its number is not in the subset, and not t... | 704 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,734 |
Example 1. $A A^{\prime}$ is the median of the inscribed $\triangle A B C$ in $\odot O$, and $P B, P C$ are tangents to $\odot O$. Prove: $\angle B A A^{\prime} = \angle C A P$. (Figure 1)
Here, we need to prove that $\angle B A A^{\prime} = \angle C A P$. Since these angles do not belong to the same triangle or quadr... | Prove: Connect $O B$, let the radius of the circumcircle of $\triangle A B C$ be $R$, then
$$
\begin{array}{l}
A A^{\prime 2}=\frac{1}{4}\left(2 \cdot 4 R^{2} \sin ^{2} B\right. \\
\left.\quad+2 \cdot 4 R^{2} \sin ^{2} C-4 R^{2} \sin ^{2} A\right) \\
=R^{2}\left(2 \sin ^{2} B+2 \sin ^{2} C-\sin ^{2} A\right)
\end{array... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,735 |
Example 3. (Shanghai Xuhui District 1982 Math Imperial Case Question) $O A, O B$ are two perpendicular radii of $\odot O$, $M$ is the midpoint of $A B$, $M C \parallel O A$ intersects $\overparen{B C}$ at $C$. Prove that $\overparen{A C} = \frac{1}{3} \overparen{A B}$. (Figure 3)
Arcs are often converted to angles for... | Prove that since $M$ is the midpoint of $AB$, $OM \perp AB$, so $MC \parallel OA$.
$\angle 1=\angle 2=45^{\circ}, \angle OMC=90^{\circ}+45^{\circ}$
$=135^{\circ}, OA=r. OM=\frac{\sqrt{2}}{2} r$.
In $\triangle OMC$, by the Law of Sines,
$\frac{OC}{\sin \angle UMC}=\frac{OM}{\sin \angle MCD}$,
i.e., $\sin \angle MCD=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,737 |
Example 4. (Jiehu City 1984 Hua Math Competition Question) Let the three sides of $\triangle ABC$ be $a, b, c$, and $\frac{1}{c-a}-\frac{1}{c+b}=1$. What kind of triangle is $\triangle ABC$? | From the given, we can derive that $b(c+b)-a(c-a)$ $=(c-a)(c+b)$,
which means $b^{2}+a^{2}-c^{2}=-a b$, and $\frac{b^{2}+a^{2}-c^{2}}{2 a b}=-\frac{1}{2}$. Therefore, $\cos C=-\frac{1}{2}$. Hence, this triangle is an obtuse triangle. | \text{obtuse triangle} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,738 |
Example 5. (Hungarian Mathematical Olympiad Problem) Given the area of $\triangle ABC$ is $S$, and $\angle C=\alpha$. What kind of triangle will make the side $c$ opposite to $\angle C$ the shortest? (Figure 4) | According to the cosine theorem,
$$
\begin{array}{l}
c^{2}=a^{2}+b^{2}-2 a b \cos C \\
=(a-b)^{2}+2 a b \\
\cdot(1-\cos a) .
\end{array}
$$
The area $S=\frac{1}{2} a b \sin \alpha$. Therefore,
$$
c^{2}=(a-b)^{2}+4 S \frac{1-\cos \alpha}{\sin \alpha} \text {. }
$$
The second term on the right is a constant. If $c$ is... | c^{2}=4 S \frac{1-\cos \alpha}{\sin \alpha} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,739 |
Example 4. Solve the equation:
$$
\sqrt{x^{2}-5 x+1}+\sqrt{x^{2}-6 x+6}=2 \text{. }
$$ | $$
\begin{array}{l}
\text { Solve } \sqrt{x^{2}-5 x+1}+\sqrt{x^{2}-6 x+6}=2 , \\
\text { and }\left(x^{2}-5 x+1\right)=-\left(x^{2}-6 x+6\right) \\
\quad=x-5,
\end{array}
$$
(1) We get $\sqrt{x^{2}-5 x+1}-\sqrt{x^{2}-6 x+6}=\frac{x-5}{2}$,
Then $2 \sqrt{x^{2}-5 x+1}-2 \sqrt{x^{2}-6 x+6}$
$$
\begin{array}{l}
=x-5 . \\
... | x=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,740 |
Column 6. With the side $BC$ of the equilateral $\triangle ABC$ as the diameter, draw a circle. Take a point $E$ on $BC$ such that $BE=\frac{1}{3} BC$, and connect $AE$ and extend it to intersect the circle at $D$. Determine the shape of $\triangle ABC$. (Figure 5) | Given $B E=\frac{2 R}{3}$, then $E O=\frac{R}{3}$.
In $\triangle A B E$,
$$
\begin{array}{l}
\frac{A B}{\sin \angle B E A}=\frac{B E}{\sin \angle 1} \\
\Rightarrow \frac{2 R}{\sin \angle B E A}=\frac{\frac{2 R}{3}}{\sin \angle 1}
\end{array}
$$
$$
\Rightarrow \frac{\sin \angle 1}{\sin \angle B E A}=\frac{1}{3} \text {.... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,741 |
Example 7. (26th IMO problem) In $\triangle ABC$, a circle with center $O$ passes through vertices $A$ and $C$, and intersects segments $AB$ and $BC$ at points $K$ and $N$ respectively, where $K$ and $N$ are distinct. The circumcircles of $\triangle ABC$ and $\triangle BKN$ intersect at $B$ and another point $M$. What ... | Prove that extending $K N, A C$ to intersect at $P$, and connecting $B P$.
$K M$. Let $B P$
intersect the circumcircles of
$\triangle A B C$ and
$\triangle B K N$ at
$M_{1}$ and $M_{2}$, respectively.
The radius of $\odot O$ is
$R$. According to the power of a point theorem, we prove that $M_{1}$,
$M_{2}$ coincide, i.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,742 |
Find all positive integer solutions to the equation $w! = x! + y! + z!$ . (1983 Canadian Mathematical Olympiad problem) | Let's assume $x \leqslant y \leqslant z$, obviously $w \geqslant z+1$.
Then we have
$$
(z+1)!\leqslant w!=x!+y!+z!\leqslant 3 z!,
$$
which implies $z+1 \leqslant 3$, or $z \leqslant 2$. At this point, $x, y$ are both no greater than 2.
We can consider the following four cases:
$$
\begin{array}{l}
z=1, x=1, y=1, \\
z=2... | x=2, y=2, z=2, w=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,743 |
Example 2. Given the system of equations
$$
\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=0, \\
a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=0, \\
a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=0 .
\end{array}\right.
$$
The coefficients satisfy the following conditions:
(1) $a_{11}, a_{22}, a_{33}$ are positive;
(2... | Prove that $x_{1}=x_{2}=x_{3}=0$ is obviously a solution to the system of equations.
Below, we prove that the system of equations has no non-zero solutions.
Assume that at least one of $x_{1}, x_{2}, x_{3}$ is not equal to 0. We will discuss this in two cases.
(i) One of the $x_{\mathrm{i}}$ is positive.
Without loss o... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,744 |
Example 5. In $\triangle ABC$, $a, b, c$ are the lengths of its three sides. Prove that:
$$
\begin{array}{l}
a^{2} b(a-b)+b^{2} c(b-c) \\
+c^{2} a(c-a) \geqslant 0 .
\end{array}
$$
and specify when equality holds. (24th IMO problem) | $\begin{array}{c}\text { Proof } \text { Without loss of generality, assume } a \geqslant b \geqslant c>0, \\ \text { Let } x=\frac{a+b-c}{2}, y=\frac{a+c-b}{2}, z=a+\frac{a+b-c}{2}, \\ \text { Then } \quad a=y+z, b=x+z, c=x+y, \\ \text { If } x=0, y>0, c>0 . \\ a^{2} b(c-b)+b^{2} c(b-c)+c^{2} a(c-a) \\ =(y+z)^{2}(x+z)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,747 |
Example 1. Given the sequence $\left\{a_{n}\right\}$ with the recurrence relation $\left\{\begin{array}{l}a_{1}=\text { b } \\ a_{n+1}=c a_{\mathrm{n}}+\cdot d .\end{array} \quad\right.$ where $c \neq 1$. Prove: The explicit formula for this sequence is:
$$
a_{n}=\frac{b c^{n}+(d-b) c^{n-1}-d}{c-1} .(n \in N)
$$ | Prove that given the initial conditions $\left\{\begin{array}{l}a_{1}=b, \\ a_{n+1}=c a_{n}+d\end{array}\right.$, we have:
$$
\begin{array}{l}
a_{2}=c a_{1}+d=b c+d, \\
a_{3}=c a_{2}+d=b c^{2}+c d+d, \\
a_{4}=c a_{3}+d=b c^{3}+c^{2} d+c d+d, \\
\cdots \cdots
\end{array}
$$
$$
\begin{aligned}
a_{n}= & c a_{n-1}+d=b c^{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,749 |
Example 5. Solve the equation: $2 x+1+x \sqrt{x^{2}+2}$ $+(x+1) \sqrt{x^{2}+2 x+3}=0$. | Thus, we have
\[
\left\{\begin{array}{l}
2 x+1+u x+(x+1) v=0, \\
v^{2}-u^{2}=2 x+1 .
\end{array}\right.
\]
Solving this system of equations for \( u \) and \( v \) is simpler than directly solving the original equation, but it is still a quadratic equation, making it somewhat complicated. Therefore, consider further ex... | x=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,751 |
Example. If $n \in N$, prove:
$$
a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{n}=\frac{a^{n+1}-b^{n+1}}{a-b} .
$$ | $$
\begin{array}{l}
a^{n}+a^{n-1} \cdot b^{2}+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{2}=\frac{a^{n+1}-b^{n+1}}{a-b} . \\
\text { For } \\
(a-b)\left[a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a \cdot b^{n-1}\right. \\
\left.+b^{n}\right]=a^{n+1}-b^{n+1}
\end{array}
$$
$$
\begin{array}{l}
a^{n+1}+a^{n} b+a^{x}-1 b^{2}+\cdots+a^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,752 |
Example 8. Prove,
\[
\begin{array}{l}
\sqrt{\frac{(111 \cdots 1)}{2 n}-\frac{(222 \cdots 2)}{n}} \\
=\frac{333 \cdots 3}{n \text { digits }} . \\
\end{array}
\] | $\begin{array}{l}\text { Prove } \because \frac{111 \cdots 1}{2 n}=\frac{10^{22}-1}{9} \text {. } \\ \frac{222 \cdots 2}{n \text { digits }}=2 \cdot \frac{\left(10^{2}-1\right)}{9} \text {, } \\ \text { then } \frac{(111 \cdots 1)}{2 n 10}-\frac{(222 \cdots 2)}{n 14 i} \\ =\frac{10^{2 n}-1-2 \times 10^{2}+2}{9}=\left(\... | \frac{333 \cdots 3}{n \text { digits }} | Number Theory | proof | Yes | Yes | cn_contest | false | 703,756 |
Example 1. Find the general term formula of the sequence $1,2,3,1,2,3, \cdots$ | We use a periodic function to represent the sequence as follows:
$$
a_{\mathrm{s}}=2+\frac{2}{\sqrt{3}} \sin \frac{2\left(\frac{2 i}{3}-2\right)}{3}
$$
or $a_{n}=2+\frac{1}{\sqrt{3}} \operatorname{tg} \frac{n-2}{3} \pi$.
Using the inverse trigonometric function, we get
$$
a_{n}=2+\frac{3}{\pi} \operatorname{arctg}\lef... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,757 |
Example. Find the general term formula of the sequence $-3,-2,-1,0,1,2$, $3,-3,-2,-1,0,1,2,3, \cdots$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve Theorem 2, where $s=7, p_{\mathrm{i}}=1$
$$
\begin{array}{l}
(i=1,2, \cdots, 7), l_{1}=-3, l_{2}=-2, l_{3}= \\
-1, l_{4}=0, l_{5}=1, l_{6}=2, l_{7}=3, T=7 \text {. } \\
\text { Hence } a_{\mathrm{n}}=\sum_{\mathrm{i}=1}^{7} l_{\mathrm{i}}\left\{\left(\frac{n+7-(i-1)-1}{7}\right)-\left(\frac{n+7-i-1}{7}\right)\rig... | a_{\mathrm{n}}=\frac{7}{\pi} \operatorname{arctg}\left(\operatorname{tg} \frac{n+3}{7} \pi\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,759 |
Example 1. The unit digit sequence formed by 2: 2, $4,8,6,2,4,8,6, \cdots$, try to find the general term formula of this sequence. | The essence of the problem is to find the general term formula of the sequence $\left\{a_{\mathrm{n}}\right\}$, where $D_{n+4}=a_{n}$, and $a_{1}=2, a_{2}=4, a_{3}=8, a_{4}=6$.
Since the characteristic equation of this sequence is $\lambda^{4}-1=0$, the characteristic roots are $\lambda_{1}=1, \lambda_{2}=-1, \lambda_... | a_{\mathrm{n}}=c_{1}+(-1)^{n} c_{2}+A \cos \frac{n \pi}{2}+B \sin \frac{n \pi}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,760 |
Example 2 The last two digits of $36^{\circ}$ are $36,95,56,16$, $76, \cdots$, thus its tens digit is a periodic sequence with a period of 5: $3,9,5,1,7,3,9,5$, $1,7, \cdots$, try to find its general term formula. | The characteristic equation of the factor sequence is $\lambda^{\circ}-1=0$, whose characteristic roots are $\lambda_{1}=1, \lambda_{2,3}=\cos \frac{2 \pi}{5} \pm i \sin \frac{2 \pi}{5}$. $\lambda_{4,8}=\cos \frac{4 \pi}{5} \pm i \sin \frac{4 \pi}{5}$.
Therefore, we can assume its general term formula as:
$$
\begin{arr... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,761 |
Example 6. Solve the equation:
$$
(x+3)^{4}+(x+1)^{4}=82
$$ | Let $y=x+2$,
the original equation becomes $(y+1)^{4}+(y-1)^{4}=82$.
After simplification, we get $y^{4}+6 y^{2}-40=0$.
Solving for $y$ yields $y= \pm 2$.
From $x+2= \pm 2$, we obtain the solutions of the original equation:
$$
x_{1}=0, x_{2}=4 .
$$ | x_{1}=0, x_{2}=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,762 |
Example 2. Calculate using synthetic division
$$
\left(3 x^{4}-5 x^{3}-6 x^{2}-x+4\right) \div\left(2 x^{2}-3 x+1\right) \text {. }
$$ | \[
\begin{array}{l}
\because f(x)=\left(2 x^{2}-3 x+1\right) q(x)+r(x) \\
=\left(x^{2}-\frac{3}{2} x+\frac{1}{2}\right) \cdot 2 q(x)+r(x) \text{.} \\
3-5-6-1+4 \left\lvert\, \frac{3}{2}-\frac{1}{2}\right. \\
-\frac{3}{2}+\frac{1}{4} \quad+\frac{33}{8} \\
\frac{\frac{9}{2}-\frac{3}{4}-\frac{99}{8}}{3-\frac{1}{2}-\frac{3... | q(x) = \frac{3}{2} x^{2} - \frac{1}{4} x - \frac{33}{8}, \quad r(x) = -\frac{105}{8} x + \frac{65}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,764 |
Example. The maximum and minimum values of the function $y=\cos ^{2} x+2 p \sin x+q$ are 10 and 7, respectively. Find the values of $p$ and $q$. | $$
\begin{array}{l}
\text { Solve } y=-(\sin x-p)^{2}+p^{2}+q+1 . \\
\text { i) If } -p>1, y_{\text { max }}=-(-1-p)^{2} \\
+p^{2}+q+1=-2 p+q, \quad y_{\text { min }}=-(1-p)^{2} \\
+p^{2}+q+1=2 p+q .
\end{array}
$$
Solve $-2 p+q=10, 2 p+q=7$, we get $p=-\frac{3}{4}$ which contradicts $p<-1$.
ii) If $-p<-1$, similarly,... | p=-1+\sqrt{3}, q=5+2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,766 |
Solve the equation:
$$
36 x^{4}-84 x^{3}+37 x^{2}+14 x-3=0 \text {. }
$$ | $$
\begin{array}{l}
\text { Sol } \because \Delta=b^{3}+43(2 n d-6 c) \\
=(-84)^{3}+4 \times 36(2 \times 36 \\
\times: 4-(-84) \times 37 \text { J } \\
=0 \text {, } \\
\end{array}
$$
$\therefore$ The equation can be transformed into
$$
\left(x^{2}+\frac{b}{2 a} x\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}} \cdot\left(x^{2}... | x_{1}=-\frac{1}{3}, x_{2}=\frac{3}{2}, x_{3}=1, x_{4}=\frac{1}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,773 |
Go to the room, take a card at random, and ask which number is most likely to be the unit digit of the number drawn by Jack and Jill.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 12
I
A
12. The list below, the numbers in the table represent the units digit of the sum of two numbers.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline 2 & 3 & 4 & 3 & 6 & 7 & 8 & 0 & 3 \\
3 & 4 & 0 & 3 & 7 & 3 & 9 & 1 & 1 \\
4 & 5 & 6 & 7 & 3 & 9 & 0 & 1 & 2 \\
5 & 6 & 1 & 3 & 3 & 0 & 1 & 2 & 3... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,779 |
Example 8. Solve the equation:
$$
6 x^{4}-13 x^{3}+12 x^{2}-13 x+6=0 .
$$ | Since $x=0$ is not a root of the original equation, we divide both sides of the equation by $x^{2}$, getting
$$
\begin{array}{l}
6 x^{2}-13 x+12-\frac{13}{x}+\frac{6}{x^{2}}=0, \\
6\left(x^{2}+\frac{1}{x^{2}}\right)-13\left(x+\frac{1}{x}\right)+12=0 .
\end{array}
$$
Let $y=x+\frac{1}{x}$, then $x^{2}+\frac{1}{x^{2}}=y... | x_{1}=\frac{2}{3}, x_{2}=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,784 |
17. If $|x|+x+y=10, x+|y|-y=12$, then $x+y=$
(A) -2 .
(B) $2 \cdot$ (C) $\frac{18}{5}$
(D) $\frac{22}{3}$
(E) 22 . | 17
17. Given the equations $|x|+x+y=10$,(1) $x+|y|-y=12$. (2)
If $x \leqslant 0$, then from (1) we get $y=10$, and from (2) we get $x=12$, which is a contradiction. Therefore, $x>0$. In this case, equation (1) becomes
$$
2 x+y=10 .
$$
If $y \geqslant 0$, from (2) we get $x=12$, and from (3) we get $y=-14$, which is a... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 703,785 |
18. The most critical stage of a professional bowling competition. The 5th place player competes with the 3rd place winner, then the winner competes with the 1st place player, the winner gets the 2nd prize, and the loser gets the 3rd prize. How many different prize-winning sequences are there?
(A) 10. (B) 16. (C) 24. (... | 18
j3
18. There are 2 possible outcomes in each match, with a total of four matches. For each match result, the order of awarding is different, thus there are $2^{4}=16$ different awarding sequences. Therefore, the correct choice is (B). | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 703,786 |
!9. Simplify $\frac{b x\left(a^{2} x^{2}+2 a^{2} y^{2}+b^{2} y^{2}\right)}{b x+a y}$
$$
+\frac{a y\left(a^{2} x^{2}+2 b^{2} x^{2}+b^{2} y^{2}\right)}{b x+a y}
$$
(A) $a^{2} x^{2}+b^{2} y^{2}$.(B) $(a x+b y)^{2}$.
(C) $(a x+b y)(b x+a y)$. (D) $2\left(a^{2} x^{2}+b^{2} y^{2}\right)$.
(E) $(b x+a y)^{2}$. | 19
B
19. The hint from the options suggests that the numerator can be divisible by $b x + a y$. For this, the original expression in the numerator can be transformed as follows:
$$
\begin{aligned}
& b x\left(a^{2} x^{2}+b^{2} y^{2}\right)+a y\left(a^{2} x^{2}+b^{2} y^{2}\right) \\
+ & b x\left(2 a^{2} y^{2}\right)+a y\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 703,787 |
22. How many integers $x$ make the triangle with sides $10, 24, x$ have all acute angles?
(A) 4. (B) 5. (C)6. (D) 7. (E) More than 7. | 22
22. In any $\triangle ABC$, if $a^{2}<b^{2}+c^{2}$, then the angle opposite to side $a$ is acute. By the Pythagorean theorem, we have
$$
\begin{array}{l}
x^{2}<10^{2}+24^{2}=26^{2}, 24^{2}<10^{2}+x^{2} \\
\Rightarrow 476<x^{2} . \\
\therefore \quad 22 \leqslant x<26 .
\end{array}
$$
The integer values that satisfy ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,790 |
23. The six edges of the tetrahedron $ABCD$ are 7, 13, 18, 27, 36, and 41 units long, respectively. If $AB$ is 41 units long, then the length of $CD$ is
(A) 7. (B) 13. (C) 18. (D) 27. (E) 36. | 23
B
23. In tetrahedron $ABCD$, we look at the edge of length 7 and the two triangular faces that share this edge. In these two triangles, the difference between the other two sides must be less than 7 (otherwise, they cannot form a triangle).
Among the given numbers, only the pairs $(13,18)$ and $(36,41)$ satisfy the... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,791 |
24. An isosceles trapezoid is circumscribed around a circle, with the longer base being 16, and one of the base angles being $\arcsin 0.8$. The area of the trapezoid is
(A) 72. (B) 75. (C) 80. (D) 90.
(E) Not uniquely determined. | 24
24. The non-parallel sides of an isosceles trapezoid are tangents to a circle, and the sum of the two bases equals the sum of the two non-parallel sides. As shown in the figure, let $x, y$, then
$$
\left\{\begin{array}{l}
2 y+1.2 x=2 x \\
y+1.2 x=16
\end{array}\right.
$$
Solving it, we get $x=10, y=4$.
$\therefore ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,792 |
26. Let $p, q$ satisfy $\log _{4}(p)=\log _{12}(q)$ $=\log _{18}(p+q)$. Then $\frac{q}{p}$
(A) $\frac{4}{3}$
(B) $\frac{1}{2}(1+\sqrt{3})$.
(C) $\frac{8}{5}$.
(D) $\frac{1}{2}(1+\sqrt{5})$.
(E) $\frac{16}{y}$ | 26
D
26. Let the given equation be $t$, then we have $p=9^{t}$, $q=12^{t}$, and $16^{t}=p+q=9^{t}+12^{t}$. Dividing the last equation by $9^{t}$, we get $\frac{16^{\mathrm{t}}}{9^{\mathrm{t}}}=\left(\frac{4^{\mathrm{t}}}{3^{\mathrm{t}}}\right)^{2}=\left(\frac{12^{\mathrm{t}}}{9^{\mathrm{t}}}\right)^{2}=\left(\frac{q}{p... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 703,794 |
27. As shown in the figure,
$$
A B \perp B C, B C \perp C D \text {, }
$$
$B C$ is tangent to the circle $O$ with $A D$ as the diameter. In which of the following cases is the area of $A B C D$ an integer?
(A) $A B=3, C D=1$.
(B) $A B=5, C D=2$.
(C) $A B=7, C D=3$.
(D) $A B=8, C D=4$.
(E) $A B=11, C D=5$. | 27
D
27. $ABCD$ is a trapezoid, then its area is $\frac{BC(AB+CD)}{2}$. In the figure, $BCDN$ is a rectangle, $BN = CD, \angle OEB=90^{\circ}$,
$\therefore OE \parallel CD, M$ is the midpoint of $BC$.
Also, $\because BE^2 = BN \cdot BA$,
$\therefore BC = 2 \sqrt{CD \cdot BA}$.
And $S = (AB + CD) \sqrt{AB \cdot CD}$.
Wh... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,796 |
30. Let $f(x)=4 x-x^{2}$. Given $x_{0}$, consider how many real numbers $x_{0}$, make the sequence $x_{0}, x_{1}, x_{2}, \cdots H$ take infinitely many different values.
(A)0.(B)1 or 2.(C)3, 1, or 6.
(D)More than 6 but finite. (E)Infinitely many. | 30
E
30. Notice that when $x_{0}=0$, from $f(0)=4 \cdot 0-0^{2}=0$ we get the constant sequence $0,0, \cdots, 0$.
Solving $4 x-x^{2}=0$ yields $x=0, x=4$.
When $x_{0}=4$, from $f(4)=4 \times 4-4^{2}=0$ we get the sequence $4,0,0, \cdots$.
Solving $4 x-x^{2}=4$ yields $x=2$.
When $x_{0}=2$, from $f(2)=4 \times 2-2^{2}=4... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 703,799 |
1. A commercial lock with 10 buttons, which can be opened by pressing the correct five numbers, regardless of the order. The figure below is an example using $\{1,2,3,6,9\}$ as its combination. If these locks are reprogrammed to allow combinations of one to nine digits, how many additional combinations (i.e., not using... | 1. ( 770$)$
There are $2^{10}$ ways to choose from ten buttons. From this, we need to subtract the case where all 10 digits are selected, the case where no digits are selected, and the case where exactly five digits are selected. Therefore, the total number is
$$
2^{10}-1-1-C_{10}^{5}=1024-2-252=770 .
$$ | 770 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,800 |
3 . If $\mathrm{iog}_{2}\left(\log _{8} x\right)=\log _{8}\left(\log _{2} x\right)$, find $\left(\log _{2} x\right)^{2}$. | $$
\begin{array}{l}
3 .(27) \\
\because \log _{8} x=\frac{1}{\log _{x} 3}=\frac{1}{3 \log _{x} 2}=\frac{1}{3} \log _{2} x, \\
\log _{8}\left(\log _{2} x\right)=\frac{1}{3} \log _{2}\left(\log _{2} x\right),
\end{array}
$$
Let $y=\log _{2} x$, then the equation in the problem is equivalent to
$$
\log _{2} \frac{y}{3}=\... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,802 |
6. As shown in the figure, in a 5 $\times$ 5 square, there are 21 empty cells. It is known that positive integers can be filled in the empty cells so that each row and each column forms an arithmetic sequence. What number must be filled in the cell marked with a star?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
6. As shown in... | $6 .(142)$
Let $a$ and $b \%$ be the numbers that fill in the blanks. From $b-2a$, and the common difference of the first row being $2b-a-74$, we get
$$
2a+4(b-2a)=186
$$
and $a+2(2b-a-74)=103$.
Solving these equations, we can find the numbers that fill in the blanks in the figure. Therefore, the number that should be... | 142 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,805 |
Solve the system of equations:
$$
\sqrt{\bar{x}}+\sqrt{-\frac{\bar{y}}{x}}=\frac{5}{2},
$$
$$
x+y=10 \text {. }
$$ | Solve (1), $\frac{x+y}{\sqrt{x y}}=\frac{5}{2}$.
(2) Substituting (3), we get $\sqrt{x y}=4$,
$$
\therefore x y=16 \text {. }
$$
From (2) and (4), we obtain the system of symmetric equations
$$
\begin{array}{l}
x+y=10 \\
x y=16
\end{array}
$$ | x+y=10 \\ x y=16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,806 |
7. In $\triangle \mathrm{ABC}$, $\operatorname{tg} \angle \mathrm{CAB}=\frac{22}{7}$, a line is drawn from $\mathrm{A}$ to $\mathrm{BC}$ dividing $BC$ into segments of lengths 3 and 17. What is the area of $\triangle \mathrm{ABC}$? | 7. (110)
Let $x$ be the length of the altitude, intersecting $B C$ at $D$. Then, from $\angle B A C=\angle B A D+\angle D A C$, we get $\operatorname{arctg} \frac{22}{7}=\operatorname{arctg} \frac{3}{x}+\operatorname{arctg} \frac{17}{x}$.
Taking the tangent of both sides of (1) and using the $\operatorname{tg}(\alpha+... | 110 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,807 |
8. The function $f$ defined on ordered pairs of positive integers satisfies the following three properties:
$f(x, x)=x, f(x, y)=f(y, x)$ and $(x+$ $y) f(x, y)=y f(x, x+y)$. Try to compute $f(14$, 52 ). | 8. (364)
By property three, we get $f(x, x+y)=\frac{x+y}{y} f(x, y)$. We will repeatedly use this property, applying property two when the first number of the ordered pair is greater than the second, and finally applying property one, to get
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac... | 364 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,808 |
9. Find the smallest positive integer $n$, such that the last three digits of its cube are 888. | 9. (192)
If a positive integer ends with 8, then this number itself must end with 2. That is, it can be written in the form $10k + 2$. Thus,
$$
n^{3} = (10k + 2)^{3} = 1000k^{3} + 600k^{2} + 120k + 8.
$$
The term $120k$ determines the tens digit of $n^{3}$, which is 8. Therefore, $12k$ must end with 8, meaning $k$ is... | 192 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,809 |
10. The surface of a convex polyhedron is composed of 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex, one square, one octagon, and one hexagon meet. How many line segments connecting the vertices of the polyhedron lie inside the polyhedron, and not on the faces or edges of the polyhedron? | 10. $(840)$
Let $a, b, c, d$ be the number of vertices, edges, face diagonals, and body diagonals, respectively.
Since at each vertex there is one of each type of regular polygon, and the four vertices of each square must be four different vertices of the polyhedron, we have
$c=4 \times$ the number of squares $=4 \ti... | 840 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,810 |
11. Let $w_{1}, w_{4}, \cdots, m_{n}$ be complex numbers. If a straight line $l$ passes through points (complex numbers) $z_{1}: z_{2}, \cdots, z_{n}$, such that $\sum_{k=1}^{n}\left(z_{k}-w_{k}\right)=0$, then $l$ is called the “average line” of $w_{1}, w_{i}$, $\cdots, ~ w w_{\mathrm{n}}$.
For $w_{1}=32+170 i, w_{2}... | 11. (163)
Let $y=m x+b$ be the "average line" of the complex numbers $w_{\mathrm{K}}=u_{\mathrm{K}}+i v_{\mathrm{K}}$, where $u_{\mathrm{K}}, v_{\mathrm{K}} \in R, k=1,2, \cdots, n$. Assume that on the line $y=m x+b$, we select complex numbers $z_{\mathrm{K}}=x_{\mathrm{K}}+i y_{\mathrm{K}}\left(x_{\mathrm{K}}, y_{\ma... | 163 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,811 |
12. Let $P$ be an interior point of $\triangle A B C$, and extend $A P, B P, C P$ to intersect the opposite sides. In the figure, $a, b, c$, $d$ are the lengths of the respective line segments. Given that $a+b+c=43$, $d=3$, find $a b c=$ ? | 12. $(441)$
$$
\begin{array}{l}
\because \frac{S_{\triangle B P C}}{S_{\triangle B A C}}=\frac{d}{d+a}, \frac{S_{\triangle C P A}}{S_{\triangle C B A}}=\frac{d}{d+b}, \\
\frac{S_{\triangle A P B}}{S_{\triangle A C B}}=\frac{d}{d+c},
\end{array}
$$
and $S_{\triangle B P C}+S_{\triangle C P A}+S_{\triangle A P B}=S_{\t... | 441 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,812 |
13. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}+b x^{10}+1$, try to find the value of $a$.
| 13. ( 987 )
The roots of $x^{2}-x-1=0$ are $p=\frac{1}{2}(1+\sqrt{5})$, $q=\frac{1}{2}(1-\sqrt{5})$. They must be the roots of $\boldsymbol{a} x^{17}$
$+b x^{18}+1=0$.
$$
\therefore a p^{17}+b p^{16}=-1, a q^{17}+b q^{18}=-1 \text {. }
$$
Multiplying the first equation by $q^{16}$, and the second equation by $p^{16}$,... | 987 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,813 |
14. $C$ is the graph of $x y=1$, and the symmetric image of $C$ about the line $y=2 x$ is $C^{\prime}$. It is known that $C^{\prime}$ can be expressed as
$$
12 x^{2}+b x y+c y^{2}+d=0
$$
Use this form to find the value of $b c$ | 14. $(84)$ points), connect $P^{\prime} P^{\prime}$ (as shown in the figure).
$\because F P^{\prime \prime} \perp$ line $y=2 x$,
$\therefore$ the slope of $P P^{\prime}$ is $-\frac{1}{2}$,
i.e., $-\frac{y-v}{x-u}=-\frac{1}{2}$.
Also, $\because$ the midpoint of $P P^{\prime}$ is on $y=2 x$,
$\therefore \quad \frac{y+v}... | 84 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,814 |
1. Consider two concentric circles on the same plane with radii $R$ and $r (R > r)$. Let $P$ be a fixed point on the circumference of the smaller circle, and $B$ be a moving point on the circumference of the larger circle. The line $B P$ intersects the larger circle at another point $C$. The line $l$ passing through po... | 1. Let the centers of the two circles be $O$, and draw $OM$ perpendicular to $BC$, intersecting $BC$ at $M$. Let $AP=2t$, so $OM=t$. Thus,
$$
\begin{array}{l}
BC = 2BM = 2\sqrt{R^2 - t^2}, \\
BP^2 + CP^2 \\
= (BM - PM)^2 + (BM + PM)^2 \\
= 2(BM^2 + PM^2),
\end{array}
$$
which means $BP^2 + CP^2 = 2(R^2 - t^2) + 2(r^2 ... | 6R^2 + 2r^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,815 |
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