problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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A Coincidental Regular Hexagon in a Triangle
Wan Xiren
(Township Government of Baimiao, Yuanjiang City, Hunan Province, 413105)
Theorem In any $\triangle ABC$, construct regular $\triangle BCA_{1}$, regular $\triangle CAB_{1}$, and regular $\triangle ABC_{1}$ outwardly on its three sides. Let their centroids be $O_{1}... | Proof: As shown in the figure, with the midpoint of $BC$ as the origin and the line $BC$ as the $x$-axis, establish a rectangular coordinate system. The coordinates of each point can be set as $A(b, c), B(-a, 0)$, and $C(a, 0)$. Here, $a > 0$, $c > 0$, and $b$ is a real number. It is easy to find that
$$
\begin{array}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,959 |
Example 5 Solve the equation
$$
5 x^{2}+10 y^{2}-12 x y-6 x-4 y+13=0 \text {. }
$$
(1994, Tianjin City Junior High School Mathematics Competition) | Solution: Rearrange the original equation as an equation in terms of $x$, we get
$$
5 x^{2}+(-12 y-6) x+\left(10 y^{2}-4 y+13\right)=0.
$$
For the original equation to have real solutions, it is necessary that
$$
\begin{aligned}
\Delta & =(-12 y-6)^{2}-4 \times 5\left(10 y^{2}-4 y+13\right) \\
& =-56(y-2)^{2} \geqslan... | x=3, y=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,960 |
A Result of a Closed Broken Line
Escape Brave
(Sichuan Shuichuan Chongqing Printing Factory No. 6, 632160)
Theorem: A closed broken line with $n$ sides can divide the plane into at most $P(n)$ parts, then $(n \geqslant 3)$
$$
P(n)=\left\{\begin{array}{l}
\frac{n(n-3)}{2}+2,(n \text { is odd }) \\
\frac{n(n-4)}{2}+3 .(n... | Prove that if a closed polyline does not intersect itself, (1) is naturally correct. If the number of intersection points $F>0$, viewing the self-intersection points as points in a graph, the number of points is $F$, and each point has a degree of 4, it is known that
$$
4 F=2 I \text{. }
$$
$I$ is the number of edges (... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,961 |
For example, how many parts can a closed polyline with 6 vertices divide the plane into?
Please translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Solution: According to the theorem
$$
\begin{aligned}
& P(6)=\frac{6(6-4)}{2} \\
+3= & 9 .
\end{aligned}
$$
Figure 2 is a construction that achieves 9. | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,962 |
The 1996 Canadian Mathematical Olympiad, Problem 2 is: Find all real solutions to the system of equations
$$
\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y, \\
\frac{4 y^{2}}{1+4 y^{2}}=z, \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.
$$
The 1996, Issue 5 provided one solution method. Below, we present two cl... | Solution 1: From the problem, we know $x \geqslant 0, y \geqslant 0, z \geqslant 0$. Obviously, $x=0, y=0, z=0$ is one set of solutions to the system of equations.
Assume $x, y, z > 0$.
(1) $\times$ (2) $\times$ (3) gives
$$
\frac{64 x y z}{\left(1+4 x^{2}\right)\left(1+4 y^{2}\right)\left(1+4 z^{2}\right)}=1.
$$
By t... | (x, y, z)=(0,0,0), (x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,963 |
W. Janous conjectures:
Let $x, y, z$ be positive numbers, then
$$
\frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z} \geqslant 0.
$$
We will generalize this to the proposition: Let $x, y, z \in R^{+}, \alpha, \beta \in R, \alpha \beta>0$,
then
$$
\frac{y^{\alpha}-x^{\alpha}}{(z+x)^{\beta}}+\frac{... | Proof: Since (2) is homogeneous with respect to $x, y, z$, we may assume $x+y+z=1$, then (2) transforms into
$$
\begin{array}{l}
\frac{x^{\alpha}}{(1-x)^{\beta}}+\frac{y^{\alpha}}{(1-y)^{\beta}}+\frac{z^{\alpha}}{(1-z)^{\beta}} \\
\geqslant \frac{z^{\alpha}}{(1-x)^{\beta}}+\frac{x^{\alpha}}{(1-y)^{\beta}}+\frac{y^{\alp... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,964 |
Three, Seeking Training: Find all natural numbers $n$ such that the numbers
$$
1,2, \cdots, 3 n
$$
can be arranged in a table
$$
\begin{array}{llll}
a_{1} & a_{2} & \cdots & a_{n} \\
b_{1} & b_{2} & \cdots & b_{n} \\
c_{1} & c_{2} & \cdots & c_{n}
\end{array}
$$
satisfying the following two conditions:
$$
\text { (1)... | ```
Let the set of natural numbers $n$ that satisfy the two conditions be denoted as $S$. Suppose $n \in S$, from conditions (1) and (2) we can deduce: there exist natural numbers $s$ and $t$, such that
$\qquad$
```
i.e., $3n+1=4s$,
$n(3n+1)=12t$.
Therefore, $n \equiv 1 \pmod{4}$,
$n \equiv 0 \pmod{3}$.
Thus, $n$ must ... | 12k + 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,968 |
Four, quadrilateral $A B C D$ is inscribed in a circle, the extensions of its sides $A B$ and $D C$ intersect at point $P, A D$ and $B C$ intersect at point $Q$, from $Q$ draw two tangents $Q E$ and $Q F$ to the circle, with the points of tangency being $E$ and $F$ respectively.
Prove: $P, E, F$ are collinear. | Connect $P Q$,
and take a point
$M$ on $P Q$ such that
$B, C, M, P$
are concyclic. Then
$$
\begin{array}{c}
Q E^{2}=Q M \\
Q P=Q C \cdot Q B \text {, (1) } \\
\text { and } \angle P M C \\
=\angle P D Q .
\end{array}
$$
and $\angle P M C$
Thus, $C, D, Q$,
$M$ are concyclic. Therefore,
$$
P M \cdot P Q=P C \cdot P D \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,969 |
Five, let $A=\{1,2,3, \cdots, 17\}$. For any function $f: A \rightarrow A$, denote
$$
f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)
$$
$(k \in \mathbb{N})$. Find the natural number $M$ such that:
$$
\begin{array}{l}
\text { (1) When } m<M, 1 \leqslant i \leqslant 16, \text { we have } \\
f^{[m]}(i+1)-f^{[m]}(i... | Five, the required $M_{0}=8$.
First, prove $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$,
then $3 i-2 \equiv 3 j-2(\bmod 17)$,
we have $i \equiv j(\bmod 17)$,
$$
\therefore i=j \text {. }
$$
The mapping $f$ is a mappi... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,970 |
Example 6 Solve the equation
\[
\begin{array}{l}
a^{4} \cdot \frac{(x-b)(x-c)}{(a-b)(a-c)}+b^{4} \cdot \frac{(x-c)(x-a)}{(b-c)} \\
+\epsilon^{4} \cdot \frac{(x-a)(x-b)}{(c-a)(c-b)}=x^{4} .
\end{array}
\] | Solution: It is easy to know from the definition that $x_{1}=a$ is a root of the original equation. Since $a, b, c$ in the original equation are cyclically symmetric, by the property of cyclically symmetric expressions, $x_{2}=b$ and $x_{3}=c$ are also roots of the original equation. Let $x_{4}$ be another root of the ... | x_{1}=a, x_{2}=b, x_{3}=c, x_{4}=-(a+b+c) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,971 |
Six, let the sequence of non-negative numbers $a_{1}, a_{2}, \cdots$ satisfy the condition
$$
a_{n+m} \leqslant a_{n}+a_{m}, m, n \in N \text {. }
$$
Prove: For any $n \geqslant m$ we have
$$
a_{n} \leqslant m a_{1}+\left(\frac{n}{m}-i\right) a_{n n} .
$$ | Proof 1: By hypothesis, we have
$$
0 \leqslant a_{n} \leqslant a_{n-1}+a_{1} \leqslant \cdots \leqslant n a_{1} .
$$
If $n>m$, then by hypothesis, we know
$$
\begin{aligned}
\frac{a_{n}}{n}-\frac{a_{m}}{m} & \leqslant \frac{a_{n-m}+a_{m}}{n}-\frac{a_{m}}{m} \\
& =\frac{m a_{n-m}-(n-m) a_{m}}{n m} \\
& =\frac{n-m}{n}\l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,972 |
1. Given the universal set $I=\{1,2,3,4,5\}, A \cap B$ $=\{2\}, \bar{A} \cap B=\{1,4\}$. Then $\bar{B}$ equals: ( ).
(A) $\{3\}$
(B) $\{5\}$
(C) $\{1,2,4\}$
(D) $\{3,5\}$ | 1. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 708,973 |
4. The expression for $i \sin \frac{7 \pi}{5}$ is ().
(A) $\cos \frac{7 \pi}{5}+i \sin \frac{7 \pi}{5}$
(B) $\sin \frac{7 \pi}{5}\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$
(C) $\sin \frac{2 \pi}{5}\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$
(D) $\sin \frac{2 \pi}{5}\left(\cos \frac{3 \pi}{2}+i \sin... | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,976 |
Example 7. Solve the equation
$$
\begin{array}{l}
\sqrt{x-1}+\sqrt{2 x-3}+\sqrt{3 x-5} \\
+\sqrt{4 x-7}=5 x-6 .
\end{array}
$$
(First Yangtze Cup Correspondence Competition for Junior High School Students) | Solution: By applying the method of completing the square, the original equation can be transformed into
$$
\begin{array}{l}
(\sqrt{x-1}-1)^{2}+(\sqrt{2 x-3}-1)^{2} \\
+(\sqrt{3 x-5}-1)^{2}+(\sqrt{4 x-7}-1)^{2}=0 . \\
\therefore \sqrt{x-1}-1=\sqrt{2 x-3}-1 \\
\quad=\sqrt{3 x-5}-1=\sqrt{4 x-7}-1=0 .
\end{array}
$$
Solv... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,982 |
12. The condition for the function $f(x)=3 \sin (2 x+\theta)$ to be an even function is (
(A) $\theta=k \pi+\frac{\pi}{2}(k \in Z)$
(B) $\theta=k \pi+\pi(k \in Z)$
(C) $\theta=2 k \pi+\frac{\pi}{2}(k \in Z)$
(D) $\theta=2 k \pi+\pi(k \in Z)$ | 12. A | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,985 |
18. A moving circle is externally tangent to $(x+2)^{2}+y^{2}=4$ and tangent to the line $x=2$. Then, the equation of the locus of the center of the moving circle is | $\begin{array}{l}\text { 18. } y^{2}+12 x-12 \\ =0\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 18. } y^{2}+12 x-12 \\ =0\end{array}$
Since the text is a mathematical equation, it remains the same in English. | y^{2}+12 x-12 = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,991 |
19. The solution set of the inequality $\sqrt{1-x}>-x$ is | 19. $\left\{x \left\lvert\, \frac{-1-\sqrt{5}}{2}<x \leqslant 1\right.\right\}$ | \left\{x \left\lvert\, \frac{-1-\sqrt{5}}{2}<x \leqslant 1\right.\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,992 |
Example 8 Solve the equation
$$
\begin{array}{l}
\quad \frac{b+c}{x-a}+\frac{c+a}{x-b}+\frac{a+b}{x-c}=\frac{3(a+b+c)}{x} . \\
(a>0, b>0, c>0)
\end{array}
$$ | $$
\begin{array}{l}
\left(\frac{b+c}{x-a}-\frac{a+b+c}{x}\right)+\left(\frac{c+a}{x-b}-\frac{a+b+c}{x}\right) \\
+\left(\frac{a+b}{x-c}-\frac{a+b+c}{x}\right)=0 . \\
\because \frac{b+c}{x-a}-\frac{a+b+c}{x} \\
\quad=\left(\frac{b+c}{x-a}-1\right)+\left(1-\frac{a+b+c}{x}\right) \\
\quad=\frac{a+b+c-x}{x-a}-\frac{a+b+c}{... | x_{1}=a+b+c, \quad x_{2,3}=\frac{(a b+b c+c a) \pm \sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}-a b c(a+b+c)}}{a+b+c} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,993 |
21. On a sphere with a diameter of 1, there are three points $A$, $B$, and $C$. If the spherical distances between $A$ and $B$, and between $A$ and $C$ are both $\frac{\pi}{3}$,
and the spherical distance between $B$ and $C$ is $\frac{\pi}{2}$, then the distance of the plane section $A B C$ through the center of the s... | 21. $\frac{\sqrt{2}}{2}$ | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,995 |
22. (10 points) Given $t>0$,
$$
\begin{array}{l}
f(t)=\sqrt{t}+\frac{1}{\sqrt{t}}+\sqrt{t+\frac{1}{t}+1}, \\
g(t)=\sqrt{t}+\frac{1}{\sqrt{t}}-\sqrt{t+\frac{1}{t}+1} .
\end{array}
$$
Find the minimum value of $f(t)$ and the maximum value of $g(t)$. | Three, 22. From the inequality $x+\frac{1}{x} \geqslant 2(x>0)$, we can get $f(t) \geqslant 2+\sqrt{3}$ (with equality if and only if $t=1$).
$\therefore$ The minimum value of $f(t)$ is $2+\sqrt{3}$.
Also, $\because g(t)=\frac{1}{f(t)}$,
$\therefore$ The maximum value of $g(t)$ is $\frac{1}{2+\sqrt{3}}$, which is $2-\s... | 2-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,996 |
23. (8 points) Given $\sin \left(x+20^{\circ}\right)=\cos (x$ $+10^{\circ})+\cos \left(x-10^{\circ}\right)$. Find $\operatorname{tg} x$. | 23. The given conditions can be transformed into
$$
\begin{array}{l}
\sin x \cos 20^{\circ}+\cos x \sin 20^{\circ}=2 \cos x \cos 10^{\circ}, \\
\sin x \cos 20^{\circ}=\cos x \cdot\left(2 \cos 10^{\circ}-\sin 20^{\circ}\right) . \\
\begin{aligned}
\therefore \operatorname{tg} x & =\frac{2 \cos 10^{\circ}-\sin 20^{\circ}... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,997 |
24. (15 points) Given the parabola $y^{2}=\operatorname{tar}(0<a<1)$ with focus $\vec{F}$, a semicircle is constructed above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two distinct points $M$ and $N$. Let $P$ be the midpoint of segment $M N$.
(1) Find the value of $|M F|+|N F|$... | 24. (1) From the given, we have $F(a, 0)$, and the semicircle is
$$
[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0) \text {. }
$$
Substituting $y^{2}=4 a x$ into the equation, we get
$$
x^{2}-2(4-a) x+a^{2}+8 a=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then, by the definition of the parabola... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,998 |
3. In a convex quadrilateral $ABCD$, $\angle ABC = 60^{\circ}, \angle BAD = \angle BCD = 90^{\circ} \cdot AB = 2, CD = 1$, the diagonals $AC$ and $BD$ intersect at point $O$, as shown in the right figure. Then $\sin \angle AOB =$ $\qquad$ | 3. $\frac{15+6 \sqrt{3}}{26}$.
It is known that points $A$, $B$, $C$, and $D$ are concyclic.
Extend $BA$ and $CD$ to intersect at $P$. Then,
$$
\angle ADP = \angle ABC = 60^{\circ}.
$$
Let $AD = x$,
then $AP = \sqrt{3} x$, $DP = 2x$.
By the secant theorem, we have
$$
\begin{array}{l}
(2 + \sqrt{3} x) \cdot \sqrt{3} x... | \frac{15 + 6 \sqrt{3}}{26} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,001 |
4. The maximum value $M(a)$ of the function $f(x)=\left|x^{2}-a\right|$ on the interval $[-1,1]$ has its minimum value equal to $\qquad$ . | 4. $\frac{1}{2}$.
Since $f(x)=\left|x^{2}-a\right|$ is an even function,
$\therefore M(a)$ is the maximum value of $f(x)$ in the interval $0 \leqslant x \leqslant 1$.
When $a \leqslant 0$, $f(x)=x^{2}-a$, it is easy to know that $M(a)=1-a$.
When $a>0$, from the graph we can see that.
If $\sqrt{2 a} \geqslant 1$, then ... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,002 |
5. In $\triangle A B C$, $A C=2, B C=3, \angle A C B$ $=90^{\circ}, C D$ is the angle bisector of $\angle A C B$, and the right triangle is folded along $C D$ to form a right dihedral angle $A^{\prime}-C D-$
$B$. Then the distance between $A^{\prime}$ and $B$ is . $\qquad$ | 5. $\sqrt{7}$.
Since $A^{\prime}-C D-B$
is a right dihedral angle, i.e., plane $A^{\prime} C D$
$\perp$ plane $B C D$. We have $\angle A^{\{\prime}} C D$
$$
=45^{\circ}, \angle B C D=45^{\circ} \text {. }
$$
From $\cos \angle A^{\prime} C D=\cos \angle A^{\prime} C D \cdot \cos \angle B C D$,
we get $\cos \angle ... | \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,003 |
Example 9 Solve the equation
\[
\begin{array}{l}
\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}+4 \sqrt{x-2}+\sqrt{y-1}-28 \\
=0 .
\end{array}
\]
(1988, Xuzhou City Junior High School Mathematics Competition) | Solution: The original equation can be reorganized as
$$
\begin{array}{l}
\left(4 \sqrt{x-2}+\frac{36}{\sqrt{x-2}}\right) \\
+\left(\sqrt{y-1}+\frac{4}{\sqrt{y-1}}\right)=28 .
\end{array}
$$
We know an important inequality: if $a \geqslant 0$, $b \geqslant 0$, then $a+b \geqslant 2 \sqrt{a b}$. Equality holds if and o... | x=11, y=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,004 |
II. (15 points) The function $f(k)$ is defined on $N$ and takes values in $N$, being a strictly increasing function (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$), and satisfies the condition $f(f(k))=... | For $k \in N, f(f(k))=3_{i}^{2}$.
(1)
$\therefore f[f(f(\hat{i})]:=\hat{j}(3 k)$
Also, $f[f(f(i))]=3 f(k)$,
$\therefore f(3 k)=3 f(k)$.
If $f(1)=1$, substituting into (1) gives $f(1)=3$, a contradiction.
$\therefore f(1)=a>1$, but $f(f(1))=f(a)=3$.
By the strict monotonicity of $f(k)$, i.e., $1<a \Rightarrow f(1)<f(a)=... | 197 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,005 |
Three. (15 points) Each point in the plane is colored either red or blue. Prove: There exists a triangle with two interior angles of $\frac{2 \pi}{7}$ and $\frac{4 \pi}{7}$, and the length of the side between these angles is 1996, such that all three vertices are the same color. | In a two-colored plane, draw any regular heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ with a side length of 1996. Among these seven vertices, there must be four points of the same color, and among these four points of the same color, there must be two adjacent points. For definiteness, let these two points be $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,006 |
Four. (15 points) The base of the pyramid $S-ABCD$ is a rectangle $ABCD$ with center $O$, where $AB=4$, $AD=12$, $SA=3$, $SB=5$, $SO=7$. A section of the pyramid is made through vertex $S$, the center $O$ of the base, and a point $N$ on edge $BC$. For what value of $BN$ is the area of the resulting section $\triangle S... | $$
\begin{array}{l}
S . A^{2}+12^{2}=S B^{2}, S A^{2}+A 0^{2}=S C^{2}, \\
\angle S A B=90^{\circ}, \angle S A O-90^{\circ} .
\end{array}
$$
$\therefore S A \perp$ plane $A B C D . C B \perp A B, D A \perp A B \Rightarrow C B \perp$ plane $S A B, D A \perp$ plane $S A B, \angle N B S=90^{\circ}, \angle S A M=90^{\circ}$... | \frac{42 \sqrt{13}}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,007 |
Five. (15 points) Proposition (*): Let $a, b, c$ be non-negative real numbers. If $a^{4}+b^{4}+c^{4} \leqslant 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)$, then $a^{2}+b^{2}+c^{2} \leqslant 2(a b+b c+c a)$.
(1) Prove that proposition (*) is correct;
(2) Write the converse of proposition (*), and determine whethe... | (1) Let
$$
\begin{aligned}
D & =2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right) \\
& =4 a^{2} b^{2}-\left(a^{2}+b^{2}\right)^{2}+2 c^{2}\left(a^{2}+b^{2}\right)-c^{4} \\
& =(2 a b)^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2} .
\end{aligned}
$$
Since \( D \geqslant 0 \), then \((2 a b)^{2} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,008 |
1. Can four points on a plane be painted red and another four points be painted black, such that for any three points of the same color, there must exist a point of the opposite color, and these four points are precisely the vertices of a parallelogram? | 1. It can be done. The eight points shown below satisfy the conditions.
\begin{tabular}{llll}
$A$ & $B$ & $C$ & $D$ \\
$\bigcirc$ & $\bigcirc$ & & \\
$\bigcirc$ & & $\bigcirc$ & \\
$E$ & $F$ & $G$ & $H$
\end{tabular}
The parallelograms thus formed are: $A B E-F, A B G-H$,
$$
\begin{array}{l}
A E G-C, B E G-D, C D F-E,... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,009 |
2. Do there exist three distinct prime numbers $p$, $q$, $r$, such that $q r$ divides $p^{2}+d$, $r p$ divides $q^{2}+d$, $p q$ divides $r^{2}+d$, where
(1) $d=10$;
(2) $d=11$. | 2. (1) Let $p, q, r$ satisfy the conditions, assuming $p>q>r$. Since $r \neq 2$ (otherwise, $r^{2}+10=14=p q$ would contradict $q>2$), it follows that $q \geqslant r+2$. Because $3^{2}+10=19$ is a prime number, $r \neq 3, p \geqslant r$ $+6$ (since $p, q, r$ cannot be consecutive odd numbers, as one of three consecutiv... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,010 |
4. (1) A square is divided into several right-angled triangles with legs of 3 and 4. Prove: The total number of right-angled triangles is even.
(2) A rectangle is divided into several right-angled triangles with legs of 1 and 2. Prove: The total number of right-angled triangles is even. | 4. (1) The hypotenuse of a right-angled triangle is 5, and the sides of the triangle are all integers, so we can assume the side length of the square is $n \in N$. The area of the triangle is 6, i.e., $6\left|n^{2} .2\right| n, 3 \mid n$, let $n=6 m$, then $n^{2}=36 m^{2}$, the total number of triangles is $6 m^{2}$, w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,012 |
5. Does there exist a six-digit number $A$, such that the last six digits of any number in $A, 2A, 3A, \cdots, 500000A$ are not all the same? | 5. Let $A$ be a six-digit number. If $A$ is even, then $500000A$ ends with six zeros; if $A$ is divisible by 5, then $200000A$ ends with six zeros. Therefore, we can assume that the last digit of $A$ is $1, 3, 5, 7,$ or $9$. For any odd digit $d$, there exists a one-digit number $b$ such that the last digit of $Ab$ is ... | 888889 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,013 |
Fill in any 6 numbers from the first 36 negative natural numbers on the ticket. When the draw takes place, 6 numbers are crossed out from $1,2, \cdots, 36$. If all the numbers on the ticket are among the remaining 30 numbers, then the ticket wins. Prove:
(1) If 9 tickets are purchased, then it is possible to fill in th... | 6. (1) We fill out 9 lottery tickets as follows: (1-6), (1-3, $7-9$), (4-9), (10-15), (10-12, 16-18), (13-18), (19-24), (25-30), (31-36). To prevent the first three tickets from winning, at least two numbers from 1 to 9 must be crossed out; to prevent the next three tickets from winning, at least two numbers from 10 to... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,014 |
Example 10 Find $A^{2}$, where $A$ is the sum of the absolute values of all roots of the following equation:
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}
$$
(9th American Invitational Mathematics Examination) | Solution: According to the structural characteristics of the original equation, set
$$
\begin{array}{ll}
x=\sqrt{19}+\frac{91}{y}, & y=\sqrt{19}+\frac{91}{z}, \\
z=\sqrt{19}+\frac{91}{u}, & u=\sqrt{19}+\frac{91}{v}, \\
v=\sqrt{19}+\frac{91}{x} . &
\end{array}
$$
Assume $x>y$ and $x<y$ respectively, then
$$
\begin{arra... | 383 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,015 |
1. Can four vertices of a cube be painted red and the other four vertices black, so that any plane passing through three points of the same color must contain another point? | 1. It can be done. As shown in the figure, the corresponding planes are:
$$
\begin{array}{l}
A C D-B, A C G-E, A D G- \\
F, C D C-H, B E-A, B E H-C, \\
B F H-D \text { and } E F: T-G.
\end{array}
$$ | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,016 |
3. Let $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}, F^{\prime}$ be the midpoints of the six sides $AB, BC, CD, DE, EF, FA$ of hexagon $ABCDEF$. Try to express the area of hexagon $ABCDEF$ using the areas of $\triangle ABC^{\prime}, \triangle BCD^{\prime}, \triangle CDE^{\prime}, \triangle DEF^{\prime}, ... | 3. $A^{\prime}$ is the midpoint of $A B$, so $2 S_{\triangle E F A^{\prime}}=S_{\triangle E F A}+S_{\triangle E F B}$. Similarly, we have:
$$
\begin{array}{l}
2 S_{\triangle F A B^{\prime}}=S_{\triangle F A B}+S_{\triangle F A C}, \\
2 S_{\triangle A B C}=S_{\triangle A B C}+S_{\triangle A B D}, \\
2 S_{\triangle B C D... | \frac{2}{3} \left( S_{\triangle ABC^{\prime}} + S_{\triangle BCD^{\prime}} + S_{\triangle CDE^{\prime}} + S_{\triangle DEF^{\prime}} + S_{\triangle EFA^{\prime}} + S_{\triangle FAB^{\prime}} \right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,018 |
4. Prove: There does not exist a function $f(x)$ that satisfies for all real numbers $x$
$$
f[f(x)]=x^{2}-1996
$$ | 4. Let $g(x)=f[f(x)]=x^{2}-1996$, and let $a, b$ be the two real roots of $x^{2}-1996=x$. Then $a, b$ are fixed points of $g(x)$. Let $f(a)=p$, then $f[f(p)]=f[f(f(a))]=f(a)=p$, which means $p$ is also a fixed point of $g(x)$, so $p \in \{a, b\}$.
Similarly, $f(b) \in \{a, b\}$.
$$
\begin{array}{l}
\text { Let } h(x)=g... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,019 |
5. On a certain island's coast, there are several ports, and there are several towns inland. The roads on the island are all one-way, and two roads can only intersect at a port or a town. It is known that starting from any port or town and traveling along the roads cannot lead back to the starting point. For ports $i, ... | 5. (1) If $f_{13}=0$ or $f_{24}=0$, the proposition is obviously true. Therefore, we may assume $f_{13}>0, f_{24}>0$. Consider all pairs of routes $(P, Q)$, where $P$ is a route from 1 to 3, and $Q$ is a route from 2 to 4. Clearly, $P$ and $Q$ must intersect. Let $X$ be the first intersection point of $P$ and $Q$. Then... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,020 |
6. A math lottery is now issued, where 10 numbers are filled in on a lottery ticket from the first 100 natural numbers. During the draw, 10 numbers are crossed out from $1,2, \cdots, 100$. If the 10 numbers on the lottery ticket are all among the remaining 90 numbers, then the ticket wins.
Prove:
（1）If 13 lottery tick... | 6. Fill in 13 lottery tickets as follows: $(1-10),(1-5, 11-15),(16-15),(16-25)(16-20,26-30)$, $(21-30),(31-40),(41-50),(51-60), \times(61-$ $70),(71-80),(81-90),(91-100)$, to prevent the first three tickets from winning, two numbers within $1-15$ need to be crossed out; to prevent the next three tickets from winning, t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,021 |
1.6 Using 6 straight lines, a regular 12-sided polygon is divided into 12 congruent shapes, there are ( ) ways.
(A) 1 way
(B) 2 ways
(C) more than 2 ways
(D) infinitely many ways | -1. (D).
Obviously, $A_{1} A_{7}$, $A_{2} A_{8}$, $A_{3} A_{9}$, $A_{4} A_{10}$, $A_{5} A_{11}$, and $A_{6} A_{12}$ divide the regular dodecagon $A_{1} A_{2} A_{3} \cdots A_{10} A_{11} A_{12}$ into 12 congruent triangles. With the center $O$ of the regular dodecagon as the center, rotate $A_{1} A_{7}$, $A_{2} A_{8}$, $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,022 |
3. If $A, B, C, D$ are four distinct points on a plane, any circle passing through $A B$ and any circle passing through $C D$ either intersect or coincide, then these four points ( ).
(A) must be collinear
(B) must be concyclic
(C) must be collinear or concyclic
(D) none of the above answers is correct | 3. (C).
Let $l_{1}$ and $l_{2}$ be the perpendicular bisectors of $AB$ and $CD$, respectively. If $l_{1} \parallel l_{2}$ or $l_{1}$ coincides with $l_{2}$, then $AB$ and $CD$ are either parallel or collinear. Clearly, $AB \parallel CD$ does not meet the problem's requirements, so $A, B, C, D$ must be collinear, as sh... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,024 |
4. The solution situation of the equation $x^{2}-x=\frac{1}{x}$ is ( ).
(A) Only one positive root
(B) Only one negative root
(C) One positive root and one negative root
(D) No real roots | 4. (A).
Method 1. Let $y=x^{2}-x$ $=\left(x-\frac{1}{2}\right)-\frac{1}{4}, y=$ $\frac{1}{x}$, their graphs intersect at only one point in the first quadrant.
Method 2. If $x0$, so $x>0$, when $x \leqslant 1$, $x^{2}-x$ $\leqslant 0$, when $x \geqslant 2$, $x^{2}-x>1, \frac{1}{x}0$. Therefore, the equation $x^{2}-x-\... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,025 |
Example 1 If the real numbers $x, y, m$ satisfy the relation
$$
\begin{array}{l}
\sqrt{3 x+5 y-2-m}+\sqrt{2 x+3 y-m} \\
=\sqrt{x-199+y} \cdot \sqrt{199-x-y},
\end{array}
$$
try to determine the value of $m$.
(1994, Beijing Junior High School Mathematics Competition) | Analysis: An equation with one unknown cannot be solved by brute force, but only by intelligent methods. Observing the characteristics of the equation, all are square roots. By definition, the right side of the equation has
$$
\left\{\begin{array}{l}
x-199+y \geqslant 0, \\
199-x-y \geqslant 0,
\end{array}\right.
$$
w... | 201 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,026 |
Example 2 Use $[a]$ to denote the greatest integer not exceeding the real number $a$, and $\{a\}=a-[a]$ to denote the fractional part of $a$. Solve the equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]=\{x\}-1$.
(1991, Shanghai Junior High School Mathematics Competition) | Analysis: By definition, we have $0 \leqslant\{x\}<1$, which implies
$$
\begin{array}{l}
0 \leqslant\left[x^{3}\right]+\left[x^{2}\right]+[x]+1<1, \\
-1 \leqslant\left[x^{3}\right]+\left[x^{2}\right]+[x]<0 .
\end{array}
$$
Since $\left[x^{3}\right],\left[x^{2}\right],[x]$ are all integers, it follows that $\left[x^{3}... | x=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,027 |
Example 2 In $\triangle A B C$, $\angle C=30^{\circ}, O$ is the circumcenter, $I$ is the incenter, point $D$ on side $A C$ and point $E$ on side $B C$ are such that $A D=B E=A B$. Prove: $O I \perp D E$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 2 In $\triangle A B C$, $\angle C=30^{\circ}, O$ is the circumcenter, $I$ is the i... | Analysis: As shown in the figure, let the radius of the circumcircle be $R$, then the coordinates of $D$, $E$, $O$, and $I$ can be expressed using $R$, $\angle A$, and $\angle B$. However, $\angle A=150^{\circ}-\angle B$, so we can use $\angle B$ to express $k_{O I}$ and $k_{D E}$. Then prove that $k_{O I} \cdot k_{D E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,028 |
52. Given $a, b, c$ are any positive numbers, and
$$
0 \leqslant \lambda < \min \left\{\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\right\} .
$$
Prove: $\frac{a}{b+c-\lambda a}+\frac{b}{c+a-\lambda b}+\frac{c}{a+b-\lambda c} \geqslant \frac{3}{2-\lambda}$. | Proof: If $b+c-\lambda a, c+a-\lambda b, a+b-\lambda c$ and $2-\lambda$ are all positive numbers, then in the inequality concerning positive numbers $x, y, z$:
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}\right) \geqslant 9 \text{,}
$$
by taking $x=b+c-\lambda a, y=c+a-\lambda b, z=a+b-\lambda c$, we get:
$$
\b... | \frac{3}{2-\lambda} | Inequalities | proof | Yes | Yes | cn_contest | false | 709,029 |
Example 3 In $\triangle ABC$, it is known that $\angle A=60^{\circ}$. Through the incenter $I$ of the triangle, draw a line parallel to $AC$ intersecting $AB$ at $F$. On side $BC$: take point $P$ such that $3BP=BC$. Prove: $\angle BFP=\frac{1}{2} \angle B$ | Analysis: As $\operatorname{tg} \angle B F P=k_{P F}$, it is only necessary to express the coordinates of points $P$ and $F$ in terms of $\angle B$.
Proof: Clearly, $B(2 R \sin C, 0), C(R \sin B, \sqrt{3} R \sin B)$, and $3 B P=B C$,
$$
\begin{array}{l}
\therefore P\left(\frac{R}{3}(4 \sin C+\sin B), \frac{\sqrt{3}}{3... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,030 |
Example 4 As shown in the figure, let the circumradius of $\triangle ABC$ be $R$, the incenter be $I$, $\angle B=60^{\circ}, \angle A<\angle C$, and the external angle bisector of $\angle A$ intersects the circle $O$ at $E$. Prove:
$$
\text { (1) } IO=AE \text {; }
$$
(2) $2 R<IO+IA+IC<(1+\sqrt{3}) R$.
(1994, National ... | Analysis: As shown in the figure, extend $A I$ to intersect $\odot O$ at $F$, then $E F$ is a diameter. Therefore,
$$
\angle E=\angle B+\frac{1}{2} \angle A=60^{\circ}+\frac{1}{2} \angle A.
$$
Thus, the lengths of each side can be expressed using $R$ and $\angle A$.
Proof: (1) Clearly, $O(R \sin A, R \cos A)$, and $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,031 |
Example 6 A line $l$ does not intersect with a circle $\omega$ with center $O$, $E$ is a point on $l$, $O E \perp l, M$ is any point on $l$ different from $E$, from $M$ two tangents to circle $\omega$ are drawn, touching $\omega$ at points $A$ and $B, C$ is a point on $M A$ such that $E C$ $\perp M A, D$ is a point on ... | Analysis: As shown in the figure, let the radius of the circle be $r$, $OE = a$, $\angle OME = \alpha$, $\angle OMA = \theta$. It is given that $\frac{\sin \theta}{\sin \alpha} = \frac{r}{a}$. Express the equation of $CD$ in terms of $\alpha$ and $\theta$, set $x = 0$, and simplify $y_F$.
$$
\begin{array}{c}
\text{Pro... | \frac{a^2 - r^2}{2a} | Geometry | proof | Yes | Yes | cn_contest | false | 709,032 |
Example 8 As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are tangent to the three sides of $\triangle A B C$. $E, F, G, H$ are the points of tangency, and the extensions of $E G$ and $F H$ intersect at point $P$. Prove: the line $P A$ is perpendicular to $B C$.
$(1996$, National High School Mathematics Competi... | Analysis: As shown in the figure, it is only necessary to prove that $O A, E G, F H$ are concurrent, which means proving that the x-coordinate of the intersection point $P$ of $E G$ and $F H$ is 0. The equations of $E G$ and $F H$ depend on $\triangle A B C$, so we can introduce $\angle B$ and $\angle C$ as parameters.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,034 |
Example 1 As shown in Figure 1, let $P A_{1}, P A_{2}, P A_{3}$ intersect $B_{2} B_{3}, B_{3} B_{1}, B_{1} B_{2}$ at $C_{1}, C_{2}, C_{3}$ respectively. Prove:
(1) $\frac{A_{1} B_{3}}{B_{3} A_{2}}+\frac{A_{1} B_{2}}{B_{2} A_{3}}=\frac{A_{1} P}{P B_{1}}=2 \frac{A_{1} C_{1}}{C_{1} B_{1}}$;
(2) $\frac{A_{1} A_{2}}{B_{3} A... | Proof: From Figure 1 and the meaning of the area parameters, we have
$$
\begin{array}{l}
\frac{A_{1} B_{3}}{B_{3} A_{2}}+\frac{A_{1} B_{2}}{B_{2} A_{3}}=\frac{S_{\triangle P A_{3} A_{1}}}{S_{\triangle P A_{2} A_{3}}}+\frac{S_{\triangle P A_{1} A_{2}}}{S_{\triangle P A_{2} A_{3}}} \\
=\frac{\lambda_{2}}{\lambda_{1}}+\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,035 |
Example 2 As shown in Figure 2, the areas of $\triangle A_{1} B_{2} B_{3}$, $\triangle P B_{3} A_{2}$, and $\triangle P A_{3} B_{2}$ are $5, 8, 3$ respectively. Find the area of $\triangle A_{1} A_{2} A_{3}$. (Letters have been changed, the same below)
(1994, Japan Mathematical Olympiad Preliminary Problem) | $$
\begin{array}{c}
\frac{\lambda_{1}}{\mu_{1}}=\frac{\lambda_{0}}{v_{1}}=\frac{S_{\triangle A_{1} A_{2} A_{3}}}{S_{\triangle A_{1} B_{2} B_{3}}} \\
=\frac{\lambda_{1}+\mu_{1}+5+8+3}{5} . \\
\text { Also } \frac{8}{\lambda_{1}}=\frac{\mu_{1}}{3}, \\
\therefore \lambda_{1}=12, \mu_{1}=2, \\
S_{\triangle A_{1} A_{2} A_{3... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,036 |
Example 4 As shown in Figure $4, S_{\triangle P B_{2} A_{1}}=3, S_{\triangle P A_{1} A_{2}}=2$, $S_{\triangle P A_{2} B_{1}}=1$. Find $S_{\triangle A_{3} B_{2} B_{1}}$.
(1990, Shanghai Junior High School Mathematics Competition) | Solution: With the symbols having the same meanings as before, we have
$$
\begin{array}{l}
\lambda_{3}=S_{\triangle P A_{1} A_{2}}=2, \\
S_{\triangle P B_{2} A_{1}}=\frac{A_{1} B_{2}}{A_{1} A_{3}} \\
\cdot S_{\triangle P A_{3} A_{1}} \\
=\frac{\lambda_{3}}{\lambda_{3}+\lambda_{1}} \cdot \lambda_{2} \\
=3, \\
S_{\triang... | \frac{45}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,038 |
Example 3 Find all integer pairs $(x, y)$ that satisfy the equation $y^{4}+2 x^{4}+1=4 x^{2} y$.
(1995, Jiangsu Province Junior High School Mathematics Competition) | Analysis: The original equation can be transformed into $2 x^{4}-4 y x^{2}+\left(y^{4}+1\right)$ $=0, \Delta=16 y^{2}-8\left(y^{4}+1\right) \geqslant 0$, then $\left(y^{2}-1\right)^{2} \leqslant$
0. Therefore, $y^{2}-1=0, y= \pm 1$.
When $y=1$, $x= \pm 1$;
When $y=-1$, $x$ has no real solutions.
Thus, $(x, y)=( \pm 1,1... | (x, y)=(\pm 1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,039 |
Example 5 As shown in Figure $5, \triangle A_{1} A_{2} A_{3}$ is divided into six smaller triangles by three concurrent lines. If the three smaller triangles marked with $*$ have equal areas, prove: the areas of the six smaller triangles are all equal.
Translate the above text into English, please keep the original te... | Proof: Symbolic meanings as before, then
$$
\begin{array}{l}
S_{\triangle P B_{2} A_{1}}=\frac{B_{2} C_{1}}{B_{2} B_{3}} \\
\cdot S_{P B_{2} A_{1} B_{3}} \\
=\frac{\mu_{3}}{\mu_{2}+\mu_{3}}\left(\mu_{1}+v_{1}\right) \\
=\frac{\mu_{3} \mu_{3}}{\mu_{2}+\mu_{3}} \cdot \frac{\lambda_{1}+\lambda_{1}}{\lambda_{1}} \\
\text {... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,040 |
Example 6 As shown in Figure $6, A_{1} B_{1}$ is the altitude of the acute $\triangle A_{1} A_{2} A_{3}$, $A_{2} B_{2}, A_{3} B_{3}$ intersect $A_{1} B_{1}$ at $P$. Prove that: $\angle A_{1} B_{1} B_{2}=\angle A_{1} B_{1} B_{3}$.
(18th Putnam Mathematical Competition) | Prove: Connect $B_{2} B_{3}$ intersecting $A_{1} B_{1}$ at $C_{1}$, draw the perpendiculars $B_{2} B_{2}^{\prime}, B_{3} B_{3}^{\prime}$ to $A_{2} A_{3}$. Then
$$
\begin{array}{l}
\frac{B_{2} C_{1}}{C_{1} B_{3}}=\frac{\mu_{3}}{\mu_{2}} \\
=\frac{\lambda_{1} /\left(\lambda_{3}+\lambda_{1}\right)}{\lambda_{1} /\left(\lam... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,041 |
Example 7 As shown in Figure 1, find:
(1) $\frac{C_{1} B_{1}}{A_{1} B_{1}+A_{1} C_{1}}+\frac{C_{2} B_{2}}{A_{2} B_{2}+A_{2} C_{2}}$
$$
+\frac{C_{3} B_{3}}{A_{3} B_{3}+A_{3} C_{3}}=1 ;
$$
(2) $\frac{A_{1} C_{1}}{C_{1} B_{1}}+\frac{A_{2} C_{2}}{C_{2} B_{2}}+\frac{A_{3} C_{3}}{C_{3} B_{3}} \geqslant 3$
(Safta conjecture);... | Prove: (1) $\frac{C_{1} B_{1}}{A_{1} B_{1}+A_{1} C_{1}}+\frac{C_{2} B_{2}}{A_{2} B_{2}+A_{2} C_{2}}$
$$
\begin{array}{c}
+\frac{C_{3} B_{3}}{A_{3} B_{3}+A_{3} C_{3}}=\frac{1}{1+\frac{2 A_{1} C_{1}}{C_{1} B_{1}}}+\frac{1}{1+\frac{2 A_{2} C_{2}}{C_{2} B_{2}}} \\
+\frac{1}{1+\frac{2 A_{3} C_{3}}{C_{3} B_{3}}}=\frac{1}{1+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,042 |
The 34th IMO Shortlist Problem 2 (provided by Canada):
Let the circumradius of $\triangle A B C$ be $R=1$, the inradius be $r$, and the inradius of its pedal triangle $A^{\prime} B^{\prime} C^{\prime}$ be $\rho$. Prove that: $\rho \leqslant 1-\frac{1}{3}(1+r)^{2}$.
This article will prove that $\rho \leqslant r^{2}$. | Prove: As shown in the figure, the area of $\triangle ABC$ is denoted as $\triangle$. The circumradius $R=1$.
Construct the right triangle $\triangle BB'A$ such that
$$
\begin{aligned}
AB' & =c \cos A. \text{ Similarly, } \\
AC' & =b \cos A. \\
& \text{The area of } \triangle AB'C' \\
& =\frac{1}{2} AB' \cdot AC' \sin ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,043 |
Question: Let the function $f: N \rightarrow N$ satisfy the conditions $f(1)=1, f(2 n+1)=f(2 n)+1, f(2 n)$ $=3 f(n)$. Find the value set of $f$.
| Solution: We slightly modify the original problem. Let the function $g: N \rightarrow N$ satisfy
$$
\begin{array}{l}
g(1)=1, \\
g(2 n+1)=g(2 n)+1, \\
g(2 n)=2 g(n) .
\end{array}
$$
Using the first principle of mathematical induction, we can prove that this results in the identity mapping on $N$.
We already have $g(1)... | \left\{\sum_{i=0}^{r} a_{i} 3^{i} \mid a_{r}=1, a_{j}=0 \text { or } 1, j=0,1, \cdots, r-1, r \in N\right\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,044 |
Theorem 1. Let $I$ be the incenter of $\triangle A B C$, and $A I, B I, C I$ intersect the opposite sides at $D^{\prime}, E^{\prime}, F^{\prime}$, and the circumcircle at $D_{1}, E, F$. Then we have
$$
\frac{D D^{\prime}}{A D^{\prime}}+\frac{E E^{\prime}}{B E^{\prime}}+\frac{F F^{\prime}}{C F^{\prime}}=\frac{R}{r}-1,
$... | Proof: As shown in Figure 1, $O$ is the circumcenter, and $D$ is the midpoint of $BC$. Then $OD \perp BC$ and $BG = GC$. Let $OG = r_{a}$, and the area of $\triangle ABC$ be $\triangle$. Then
$$
\begin{array}{l}
\frac{D D^{\prime}}{A D^{\prime}}=\frac{S_{\triangle D B C}}{\triangle} \\
=\frac{a\left(R-r_{a}\right)}{2 \... | \frac{R}{r}-1 | Geometry | proof | Yes | Yes | cn_contest | false | 709,047 |
Theorem 2 Let $P$ be any point inside $\triangle ABC$, and let $AP$, $BP$, $CP$ extended meet the opposite sides at $X'$, $Y'$, $Z'$, and the circumcircle at $X$, $Y$, $Z$. Then
$$
\frac{X X^{\prime}}{A X^{\prime}}+\frac{Y Y^{\prime}}{B Y^{\prime}}+\frac{Z Z^{\prime}}{C Z^{\prime}} \leqslant \frac{R}{r}-1 \text {. }
$$ | Proof: As shown in Figure 2, let the angle bisector of $\angle A$ intersect $BC$ at $D'$ and the circumcircle at $D$. The tangent line through $D$ intersects the extension of $AX'$ at $X''$. Then $DX'' \parallel BC$, and $X'X'' \geqslant X'X$. Therefore,
$$
\begin{array}{l}
\frac{X'X}{AX'} \leqslant \frac{X'X''}{AX'} \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,048 |
Theorem Let $P$ be any point in the plane of $\triangle A B C$, and $I$ be the incenter. Then
$$
P I^{2}=\frac{a \cdot P A^{2}+b \cdot P B^{2}+c \cdot P C^{2}-a b c}{a+b+c},
$$
where $a=BC, b=CA, c=AB$. | Prove: As shown in the figure, $AD$ bisects $\angle A$, hence
$$
\begin{array}{l}
BD=\frac{ac}{b+c}, \\
CD=\frac{ab}{b+c}.
\end{array}
$$
Applying the Steiner theorem to $\triangle ABC$, we get
$$
PD^2=\frac{b}{b+c} PB^2 + \frac{c}{b+c} PC^2 - \frac{a^2 bc}{(b+c)^2}.
$$
Since $I$ is the incenter, we have
$$
\begin{al... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,049 |
Example 4 In the border desert area, patrol vehicles travel 200 kilometers per day, and each vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their tasks, and then return along the same route to the base. To allow 3 of them to patro... | Analysis: The key is to list the relationship formula for the distance traveled by the car, and then solve it by analyzing the characteristics of the formula.
Let 5 cars travel to point $B$ in $x$ days, and 3 cars continue to travel for $y$ days, with the distance $s=200(x+y)$, which is to find the maximum value of $s... | 1800 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 709,050 |
In $\triangle A B C$, if $A, B, C$ are its three interior angles, then
$$
\begin{array}{l}
\cos \frac{A}{k}+\cos \frac{B}{k}+\cos \frac{C}{k} \\
\leqslant 3 \cos \frac{\pi}{3 k} \cdot(k \in N)
\end{array}
$$ | Proof: Without loss of generality, let $C$ be an acute angle, then $\frac{A+B}{2 k}, \frac{C+\frac{\pi}{3}}{2 k}, \frac{\pi}{3 k}$ are all acute angles. Therefore,
$\cos \frac{A+B}{2 k}>0, \cos \frac{C+\frac{\pi}{3}}{2 k}>0, \cos \frac{\pi}{3 k}>0$.
$\therefore \cos \frac{A}{k}+\cos \frac{B}{k}+\cos \frac{C}{k}+\cos \f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,052 |
57.21. A student did not notice the multiplication sign written between two 7-digit numbers, mistaking it for a 14-digit number. This 14-digit number happens to be 3 times the original product. Try to find these 3 numbers. | 57. 21. The original expression is
$$
1666667 \times 3333334 .
$$ | 1666667 \times 3333334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,054 |
57. 22. The sequence $x_{n}$ is defined by the following rule:
$$
x_{n+1}=|1-| 1-2 x_{n}|| \text{, if } 0 \leqslant x_{1} \leqslant 1 \text{. }
$$
（1）Prove: The sequence becomes periodic from a certain term onward if and only if $x_{1}$ is a rational number. $\cdots)$ ? $\quad$ - | 57. 22. The recursive relation in the problem is given by
$$
x_{n+1}=\left\{\begin{array}{ll}
2 x_{n}, & \text { if } 0 \leqslant x_{n}<\frac{1}{2}, \\
2-2 x_{n}, & \text { if } \frac{1}{2} \leqslant x_{n} \leqslant 1 .
\end{array}\right.
$$
(1) If $x_{1}$ is a rational number, i.e., $x_{1}=\frac{p}{q}$, where $(p, q)=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,055 |
57. 23. 1994 A parliament has the property that each of its members has had a low degree of interaction with one of their colleagues. Prove: it is possible to form a committee of 665 members such that none of its members have the aforementioned relationship (i.e., no member has had a low degree of interaction with any ... | 57. 23. First, we prove a general conclusion: "In a group of $n$ people, if each person has harmed at most one companion, then when $3(m-1)3 k$, we have $\left|A_{3}\right|=3 k+1$ or $3 k+2$. At this point, we first remove the one or two people who have been harmed the most from $A_{3}$, and denote the remaining set as... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,056 |
57.25. Investigate the following statements about polygons (not necessarily convex).
(1) Is it... always possible to find a chord of the polygon that divides it into two parts of equal area?
(2) Prove: Any polygon can be divided by one of its chords into two parts, each with an area not less than $\frac{1}{3}$ of the p... | 57. 25. (1) Not necessarily. The polygon shown in Figure 2(a) cannot be divided into two equal-area parts by any of its chords. (b) and (c) illustrate the construction process of this polygon. Starting from four congruent equilateral triangles in (b), keeping the central one unchanged, and reflecting the surrounding th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,058 |
57. 26. Does there exist a real-coefficient polynomial $P(x)$: it has even coefficients, and for any $n>1, P^{n}(x)$ also has even coefficients. | 57. 26. Existence.
For example, the polynomial with integer coefficients
$$
\begin{aligned}
P(x) & =10\left(x^{3}+1\right)(x+1)-x^{2} \\
& =10 x^{4}+10 x^{3}-x^{2}+10 x+10
\end{aligned}
$$
has negative coefficients. However, the polynomial
$$
\begin{aligned}
P^{2}(x)= & x^{4}+100\left(x^{3}+1\right)^{2}(x+1)^{2} \\
&... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,059 |
Example: Does a right-angled triangle with a perimeter of 6 and an integer area exist? If not, please provide a proof; if it does, please prove how many there are.
(1994, National Junior High School Mathematics League) | Analysis: Let the two legs be $a$ and $b$, the hypotenuse be $c$, and the area be $S$ and an integer. Thus, we have
$$
\left\{\begin{array}{l}
a+b+c=6, \\
a^{2}+b^{2}=c^{2}, \\
S=\frac{1}{2} a b .
\end{array}\right.
$$
From this, we know that the key is to find the value of $S$, which will allow us to solve the system... | a=\frac{5-\sqrt{7}}{3}, b=\frac{5+\sqrt{7}}{3}, c=\frac{8}{3} | Number Theory | proof | Yes | Yes | cn_contest | false | 709,061 |
57. 32. Proof: For any natural number $k>1$, there exists a power of 2 such that at least half of the last $k$ digits are 9 (for example, for $k=2$, there is $2^{12}=4096$; for $k=3$, there is $2^{53}=\cdots 992$; etc.). | 57. 32. Let's prove: for any natural number $k$, in the last $k+2$ digits of $2^{10 \cdot 3^{k}+k+2}$, there are at least $\left[\frac{2}{3}(k+2)\right]$ nines.
By F $2^{10}=1024=10^{3}+24$. In the expansion of $\left(10^{3}+24\right)^{5^{k}}$, we have $10^{3 \cdot 5^{k}}+5^{k} \cdot 10^{3 \cdot 5^{k}-1} \cdot 24+\cdo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,064 |
58. 01. A person spent one coin to buy a loaf of bread and a bottle of kvass (a drink). When prices increased by $20 \%$, this coin was only enough to buy half a loaf of bread and a bottle of kvass. Try to determine, if prices increase by another $20 \%$, whether this coin would be enough to buy just a bottle of kvass. | 58.01. Ans: Enough. The prices of gas are $x^{1} \mathrm{j} y$, then
$$
\begin{array}{l}
x+y=1, \\
1.2(0.5 x+y)=1 .
\end{array}
$$
The price is
$$
\text { (1.2) } y=1.44 y \text {. }
$$
H J $1.44 y<1$, so enough to head - Stankebis. | 1.44y < 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,065 |
58. 02. Proof: All integers of the form 10017, 100117, 1001117, ... are divisible by 53. | 58.02. Knowing that the first number 10017 is divisible by 53, and the difference between adjacent numbers in the sequence
is also divisible by 53, so all numbers in the sequence are divisible by 53. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,066 |
58. 03- Given a convex quadrilateral $A B C D$ and a point $O$ inside the shape, $\left[\angle A O B=\angle C O D=120^{\circ}, A O=O B\right.$, $C O=O D$. Let $K, L, M$ be the midpoints of segments $A B, B C$, and $C D$, respectively. Prove:
(1) $K L=L M$;
(2) $\triangle K L M$ is an equilateral triangle. | 58. 03. Discuss $B()^{\prime}(C)$
Price small no difference as $N$ I. $P$.
$$
\begin{array}{c}
\because K N=\frac{1}{2} A O \\
=\frac{1}{2} B O=L P, \\
I N=\frac{1}{2} C O=\frac{1}{2} D O \\
=M P, \\
\quad \angle K N L=\angle A O C=120^{\circ}+\angle B O C=\angle B O D \\
=\angle L P M, \\
\therefore \triangle K N L \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,067 |
58. 04. To make a rectangular box that can hold no less than 1995 unit cubes, each face of the box needs to be covered with decorative fabric. Now, we have prepared: $a) 962 ; b$ ) $960 ; c) 958$ square units of decorative fabric. Is it enough? | 58. 04. The volume of a box with dimensions $11 \times 13 \times 14$ is 2002, which is sufficient to hold no less than 1995 unit cubes, and its surface area is $2 \times(11 \times 13+11 \times 14+13 \times 14)=$ | 958 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,068 |
58. 07. Prove: No matter how many 3s are added between the two 0s in the number 12008, the resulting number is always divisible by 19. | 58.07. We live
$$
\begin{array}{l}
1203 \underbrace{3 \cdots 308}_{n \uparrow}=\underset{n+1 \uparrow}{126 \cdots 6} 40-\underset{n+1 \uparrow}{63 \cdots} 32 \\
=20 \times 63 \cdots 32-63 \cdots 32 \\
n+1 \uparrow \quad n+1 \text { digits } \\
=19 \times \underset{n+1 \text { digits }}{63 \cdots 32} \\
\end{array}
$$
a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,071 |
Example 6 Given that $x, y, z$ are two non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $S=2x+y-z$, then what is the sum of the maximum and minimum values of $S$?
(1996, Tianjin City Junior High School Mathematics Competition) | Analysis: By representing the algebraic expression $S$ with one letter and determining the range of this letter, we can find the maximum or minimum value of $S$.
From the given, we solve $y=\frac{7-4 x}{3}, z=\frac{1-x}{3}$, and.
$$
\left\{\begin{array}{l}
\frac{7-4 x}{3} \geqslant 0, \\
\frac{1-x}{3} \geqslant 0, \\
x... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,072 |
58. 08. Given an equilateral $\triangle ABC$, for any point $P$ inside the shape, consider the intersection point $A'$ of line $AP$ and $BC$, and the intersection point $C'$ of line $CP$ and $BA$. Try to find the geometric position of point $P$ such that segments $AA'$ and $CC'$ are equal. | 58. 08. Take any point \( A' \) on \( BC \), and draw the line \( AA' \). The intersection point lies on the altitude of \( \triangle ABC \) from \( A \) to \( BC \). It is easy to see that every point on this altitude meets the requirements of the problem.
Now, let the intersection of \( AA' \) and \( CC'' \) be \( P... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,073 |
58. 09. For any natural number $k$, we call a rectangle of size $1 \times k$ a "strip". Try to determine for which natural numbers $n$, a rectangle of size $1995 \times n$ can be divided into distinct "strips"?
| 58. 09. Clearly, the sides of the "strips" are parallel to the sides of the given rectangle. Let's assume the side lengths of the rectangle are in centimeters. We will discuss the following two cases.
(1) $n<1995$. In this case, the maximum length of the "strips" that can be divided is $1995 \mathrm{~cm}$. However, the... | n \leqslant 998 \text{ or } n \geqslant 3991 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,074 |
58. 10. Let $a, b, c, d$ be natural numbers, and $a b$ $=c d$. Can $a+b+c+d$ be a prime number? | 58. 10. Answer: cannot be a prime number.
Solution 1: Since $a b=c d$, by the Fundamental Theorem of Arithmetic, there exist positive integers $c_{1}, c_{2}, d_{1}, d_{2}$, such that $c=c_{1} c_{2}, d=d_{1} d_{2}, a=c_{1} d_{1}, b=c_{2} d_{2}$, thus,
$$
a+b+c+d=\left(c_{1}+d_{2}\right)\left(d_{1}+c_{2}\right)
$$
is a... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,075 |
Shape. In each step, a triangle is divided into two triangles along the altitude to the hypotenuse. Prove: no matter how many steps of the above operation are performed, among the resulting triangles, two congruent triangles can always be found.
Every step involves dividing a triangle into two triangles along the alti... | 58. 11. Suppose it were not so. We consider the smallest $n$ with the following property: after $n$ operations, we can obtain pairwise non-congruent triangles from four congruent right triangles. Assume that the four given triangles are such.
It should be noted that the order of operations does not matter (i.e., the r... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,076 |
58. 12. A geological team collected 80 core samples during their survey, each with a different but known weight (registered on a list). After some time, the labels on the sample boxes became blurred, but only the custodian knew what they were. He can prove he is correct without opening the boxes, using only the list an... | 58. 12. $a$ ) The custodian can proceed as follows: for convenience, he adds one more box with a weight of 0, and then arranges the 81 boxes in ascending order of weight, numbering them from 1 to 81 (since he knows what each box contains, he can do this). Thus, he only needs to ensure that his numbering is correct. For... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 709,077 |
58. 13. Given the value of $\sin \alpha$. How many different values can $\left.a) \sin \frac{\alpha}{2}, b\right)$ $\sin \frac{\alpha}{3}$ have at most, respectively? | 58. 13. a) Let $\sin \alpha=x$, the $\alpha$ that satisfies this must!! make $\frac{\alpha}{2}$ correspond to 4 points on the unit circle:
$$
\begin{array}{l}
\frac{1}{2} \arcsin x, \frac{1}{2} \arcsin x+\pi, \\
\frac{1}{2}(\pi-\arcsin x), \frac{1}{2}(3 \pi-\arcsin x) .
\end{array}
$$
Thus, $\sin \frac{\alpha}{2}$ can... | 4, 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,078 |
58. 15. In trapezoid $ABCD$, the two diagonals intersect at point $K$. A circle is drawn with each of the trapezoid's legs as diameters (the centers of the circles are at the midpoints of the legs). It is known that point $K$ is outside both circles. Prove: The lengths of the tangents drawn from point $K$ to these two ... | 58. 15. Let the non-parallel sides of the trapezoid be \(AB^{Y_{j}} CD\) (see Fig. 5). Suppose 1': lines \(AC\) and \(BD\) intersect the circle at points \(M^{\digamma}\) and \(N\), respectively. The lengths of the tangents from the point of intersection of the secants to the points of tangency are given by \(KM \cdot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,079 |
58. 17. Integers $a, b, c$ make $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$ both integers. Prove: $|a|=|b|=|c|$. | 58. 17. Proof 1: Using proof by contradiction.
Assume the conclusion does not hold. If \(a, b, c\) have a common divisor \(d\), then we can consider \(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}\). Therefore, without loss of generality, we can assume that no number other than \(\pm 1\) can simultaneously divide \(a, b, c\).... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,080 |
58. 18. There is a board with several light bulbs, all of which are on. Now there are several buttons, pressing which can change the on/off state of all the bulbs connected to that button. It is known that for any group of bulbs, there is a button connected to an odd number of bulbs in that group. Prove: By pressing th... | 58. 18. First, note that the final result is independent of the order in which the buttons are pressed. Let the number of bulbs be $n$, and we will use induction on $n$. When $n=1$, the conclusion is obviously true. Assuming the conclusion holds for $n-1$ bulbs, we will consider the case of $n$ bulbs. By the induction ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,081 |
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7... | Analysis: Let $\frac{n^{2}-71}{7 n+55}=k$ ( $k$ is an integer), then $n^{2}$ $-7 k n-(55 k+71)=0$, and $\Delta=49 k^{2}+4$ (5jk $+71)=49 k^{2}+220 k+284$ should be a perfect square.
$$
\begin{array}{l}
\text { and }(7 k+15)^{?}=49 k^{2}+210 k+225 \\
<49 k^{2}+220 k+284 \\
<49 k^{2}+238 k+289 \\
=(7 k+17)^{2}, \\
\there... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,083 |
58. 20. Is it possible to color each edge of an $n$-gon with one of three colors so that every face has edges of all three different colors, and at every vertex, edges of different colors meet, if: $a$) $n=1995$; b) $n=1996$? | 58. 20. $a$ ) It is possible. Note that 1995 is a multiple of 3, so the bottom face can be alternately colored with three different colors: each edge is colored with the third color not present at its endpoints (the missing color at both endpoints is the same). It is easy to see that all three different colors appear o... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,084 |
58. 21. In $\triangle A B C$, $A A_{1}$ is a median, $A A_{2}$ is an angle bisector, $K$ is a point on $A A_{1}$ such that $K A_{2} / / A C$. Prove: $A A_{2} \perp K C$. | 58. 21. Let the vectors $\overline{A B}$ and $\overline{A C}$ be denoted as $\vec{b}$ and $\vec{c}$, respectively. Then, $\overrightarrow{A A_{1}}=\frac{1}{2}(\vec{b}+\vec{c})$,
$\overrightarrow{A A_{2}}=\frac{\vec{b} \cdot|\vec{c}|+\vec{c} \cdot|\vec{b}|}{2(|\vec{b}|+|\vec{c}|)}$,
$\frac{A K}{A A_{1}}=\frac{A_{2} C}{A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,085 |
58. 22. Divide the interval $[-1,1]$ into black intervals and white intervals so that the integral values of any $a$ ) linear function; $b$ ) quadratic trinomial over the black and white intervals are equal. | 58. 22. The elbow method proves that any interval can be divided into black and white intervals such that the integral values of any polynomial of degree not exceeding $n$ over the two different colored intervals are equal.
When $n=0$, it is sufficient to divide the interval into two equal parts.
Assuming it has been s... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 709,086 |
58. $24^{*}$. Prove that there exist infinitely many composite numbers $n$ such that $3^{n-1}-2^{n-1}$ is a multiple of $n$.
保留源文本的换行和格式，翻译结果如下：
58. $24^{*}$. Prove that there exist infinitely many composite numbers $n$ such that $3^{n-1}-2^{n-1}$ is a multiple of $n$. | 58. $24^{*}$. Let $n=3^{2^{2}}-2^{z^{2}}$, where $t \geqslant 2$, then it is clear that $n$ is a composite number. It is well known that as long as $x$ and $y$ are unequal integers, and $k$ is a natural number, then $x^{k}-y^{k}$ can be divided by $x-y$. Therefore, to prove that $3^{n-1}-2^{n-1}$ can be divided by $n$,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,088 |
1. If $x$ takes real values, then the minimum value of $\sqrt{4 x^{2}+4 x+1}+$ $\sqrt{4 x^{2}-7984 x+1996^{2}}$ is ( ).
(A) 0
(B) 1995
(C) 1996
(D) 1997 | $\begin{array}{l}\text { I. 1. (D). } \\ \text { Original expression }=|2 x+1|+|2 x-1996| . \\ \text { When } x<-\frac{1}{2} \text { or } x>998 . \\ \text { When } x<-\frac{1}{2} \text {, } \\ \text { the expression }=-(2 x+1)-(2 x-1996)>1997 . \\ \text { When } x>998 \text {, } \\ \text { the expression }=2 x+1+2 x-19... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,090 |
3. The equation $9 x^{2}+4 y^{2}-12 x-4 y+3=0$ has ( ) integer solutions.
(A) : None
(B) Some
(C) Too many
(D) Exactly one integer solution | 3. (A).
The original equation is transformed into $(3 x-2)^{2}+(2 y-1)^{2}=2$. Therefore, the system can be:
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 3 x - 2 = 1 , } \\
{ 2 y - 1 = 1 ; }
\end{array} \quad \left\{\begin{array}{l}
3 x-2=-1, \\
2 y-1=-1 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,092 |
Example 8 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
$(1996$, National Junior High School Mathematics Leagu... | $$
\begin{array}{l}
\text{Analysis: Let } x_{1} \text{ and } x_{2} \text{ be the x-coordinates of the points where the parabola intersects the x-axis, then } x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}. \text{ It is easy to know that } -\frac{b}{a}<0, \text{ and } -116 c(a-b+c). \\
\because a-b+c>0, c \geqslant 1... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,094 |
5. $P, Q$ are points on the legs $AB, AC$ of Rt $\triangle ABC$, respectively. $M$ is the midpoint of the hypotenuse $BC$, and $PM \perp MQ$. If $PB=x, QC=y$, then $PM^2 + QM^2 =$ ( ).
(A) $2xy$
(B) $x^2 + y^2$
(C) $(x+y)^2$
(D) $x^2 + xy + y^2$ | 5. (B).
As shown, draw: $M D \perp A B$ at $F$. $M E \perp A C$ at $E$, then $D, E$
are the feet of the perpendiculars from $A B, A C$. Let $A B=$
$c, A C=b$. Then $M D=\frac{b}{2}$.
$M E=\frac{c}{2}$. Thus,
$$
\begin{array}{l}
P M^{2}+Q M^{2}=M D^{2}+P D^{2}+M E^{2}+E Q^{2} \\
=\left(\frac{b}{2}\right)^{2}+\left(x... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,095 |
6. As shown in the figure, the graph of the function $y=-\frac{1}{2} x+2$ intersects the $y$-axis at $M$, and the $x$-axis at $N$. Points $A, B$ are on $MN$, and their projections on the $x$-axis are $A_{1}, B_{1}$ respectively. If $O A_{1}+O B_{1}>4$, then the relationship between the area $S_{1}$ of $\triangle O A_{1... | 6. (A).
Let $O A_{1}=x_{1}, O B_{1}=x_{2}$. Then $x_{1}0 .
\end{array}
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,096 |
1. Arrange the natural numbers $1,2,3, \cdots, 999$. 1.5 into the number $N=1234 \cdots 998999$. Then, the sum of the digits of $N$ is $\qquad$ | II. 1.13500.
Since $1.2, \cdots .999$ contains 999 numbers. |1. $1+998=2+997$ 999 , for sure, hence the sum of each pair is 27. There are 500 pairs. Therefore, the sum of the digits of $N$ is $500 \times 27=13500$. | 13500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,097 |
3. Natural numbers $m, n$ satisfy $8 m+10 n>9 m n$. Then
$$
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4}=
$$
$\qquad$ | 3. 2 .
It is known that $\frac{8}{n}+\frac{10}{m}>9$. It is easy to see that one of the two natural numbers $m, n$ must be 1 (otherwise $\frac{8}{m}+\frac{10}{m} \leqslant \frac{8}{2}+\frac{10}{2}=9$ which is a contradiction). Without loss of generality, let $m=1$, then
$$
\begin{array}{l}
m^{2}+n^{2}-m^{2} n^{2}+m^{4... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,099 |
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