problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 8. (IMO-7-6)
In the plane, given $n$ (≥3) points, where the maximum distance between any two points is $d$. The line segment connecting two points with this distance $d$ is called the diameter of this set of points. Prove: the number of diameters is at most $n$. | Proof by contradiction.
Suppose there are $n$ points, with more than $n$ diameters. If there is a point from which fewer than two diameters emanate, we remove this point, leaving the remaining $n-1$ points with at least $n$ diameters, clearly $n-1 \geqslant 3$. Therefore, we may assume that at least two diameters emana... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,525 |
Example 9. (IMO-21-2)
A prism has pentagons $A_{1} A_{2} A_{3} A_{4} A_{5}$ and $B_{1} B_{2} B_{3} B_{4} B_{5}$ as its top and bottom bases. Each side of these polygons and each segment $A_{i} B_{j}(i, j=1,2, 3,4,5)$ is painted either red or green. Every triangle with a vertex at one of the prism's vertices and with si... | First, we prove that the five edges of the upper base have the same color.
If the five edges of the upper base do not have the same color, then there must be two adjacent edges with different colors. Let's assume $A_{1} A_{2}$.
From $A_{1}$, draw five lines $A_{1} B_{1}, A_{1} B_{2}, A_{1} B_{3}, A_{1} B_{4}, A_{1} B... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,526 |
Example 10. (IMO-6-4)
17 scientists, each of whom communicates with all the others. In their communications, they only discuss three topics, and any two scientists only discuss one topic. Prove: there are at least three scientists, whose mutual communications discuss the same topic. | Prove that scientists are represented by points $A_{0}, A_{1}, \cdots A_{18}$. An edge is drawn between every two points. If the discussion is about the first topic, the corresponding edge is painted red; if it is about the second topic, it is painted yellow; if it is about the third topic, it is painted blue.
From $A... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,527 |
Example 11. (IMO-24-4)
Let $A B C$ be an equilateral triangle, and let $E$ be the set of points consisting of the points on the three segments $B C$, $C A$, $A B$ (including $A, B, C$). If $E$ is divided into two subsets, is it always true that one of the subsets contains the vertices of a right triangle? Prove your co... | Prove that the problem can be reduced to the following form: color the points in $E$ red and blue, and prove: there must exist a right-angled triangle with all three vertices of the same color.
On the sides $A B, B C, C A$, take points $P, Q, R$ respectively, such that $A P: P B=B Q: Q C=C R: R A=2$, then $P Q \perp A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,528 |
Example 5. The first hundred natural numbers
From the set
$$
\{1,2,3,4, \cdots, 99,100\}
$$
, randomly select 51 numbers, prove: there must be two numbers, one of which can divide the other. | After repeatedly factoring out the number “2”, any original number can ultimately be expressed as: an odd number $\times 2^{i}\left(\|_{1} \prod l=1,2, \cdots\right)$, and the odd number will not exceed half of the original number. For example:
$$
\begin{array}{l}
16=8 \times 2=4 \times 2^{2}=1 \times 2^{4} ; \\
24=12 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,529 |
Example 12. $(I M O-10-4)$
Prove: For any tetrahedron, there is always a vertex from which the three edges emanating can form a triangle. | Prove that if the longest edge is $AB$, then $AD + AC > AB$, and $BD + BC > AB$, at least one of them must hold. Otherwise,
$$
AD + BD + AC + BC \leq 2AB,
$$
which is impossible. Without loss of generality, assume $AD + AC > AB$. Clearly, we also have: $AC + AB > AD$, $AD + AB > AC$. These are the necessary and suffici... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,530 |
Example 13. (IMO-14-6)
Given four non-coincident parallel planes, prove that there exists a regular tetrahedron, with exactly one vertex on each plane. | Proof Let the four given mutually parallel planes be $\alpha, \beta, \gamma, \delta$. Among them, $\beta, \gamma, \delta$ are on the same side of $\alpha$, and their distances from $a$ are $b, c, d, b0$, respectively. Then they intersect $\alpha, \beta, \gamma$ in three parallel lines $n_{1}$, $n$, $l_{1}$, $m_{1}$, $l... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,531 |
Example 1. As shown in Figure 6, given the dihedral angle $P-A C-Q$
$$
\begin{array}{l}
=120^{\circ}, \angle B A C=60^{\circ}, \angle D C A=45^{\circ} . \\
B \in P, D \in Q .
\end{array}
$$
Find the angle formed by the skew lines $A B$ and $C D$. | Solve: Translate line $AB$ within plane $P$ so that points $A$ and $C$ coincide. This results in three rays originating from point $C$. Let the angle between rays $AB$ and $CD$ be $\theta$, the angle between $AB$ and $CA$ be $\theta_{1}=120^{\circ}$, and the angle between $CD$ and $CA$ be $\theta_{2}=45^{\circ}$. By th... | \arccos \left(\frac{2 \sqrt{2}+\sqrt{6}}{8}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,532 |
Example 2. Given a rectangular cuboid $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with length, width, and height being $a, b, c(a>b)$ respectively. Find the angle formed by $A C^{\prime}$ and $B D$. (81 College Entrance Examination for Science Mathematics Supplementary Paper, Question 7) | Solve the dihedral angle $C^{\prime}-A C-D=90^{\circ}$.
Let the angle between $A C^{\prime}$ and $B D$ be $\theta$. By the corollary of the three-line theorem,
$$
\begin{array}{l}
\cos \theta=\cos \theta_{1} \cos \theta_{2} \\
=\cos \theta_{1} \cos 2 \theta_{3} \\
=\cos \theta_{1}\left(2 \cos ^{2} \theta_{3}-1\right) \... | \theta=\arccos \frac{a^{2}-b^{2}}{\sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{a^{2}+b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,533 |
Example 4. It is known that there are two places $A$ and $B$ on the Earth's surface, located at latitudes $60^{\circ}$ N and $45^{\circ}$ N, respectively, with a longitudinal difference of $60^{\circ}$: Find the spherical distance between $A$ and $B$. (Figure 9) | Solve As shown in Figure 9, the essence of this problem is to find $\angle A O B$.
Let $\angle A O B=\theta, \angle A O N=\theta_{1}, \angle B O N=\theta_{2}$
$\alpha=$ dihedral angle $A-S N-B=60^{\circ}$.
We have $\cos \theta=\cos \theta_{1} \cos \theta_{2}$
$$
\begin{aligned}
& +\sin \theta_{1} \sin \theta_{2} \cos \... | \arccos \left(\frac{2 \sqrt{6}+\sqrt{2}}{8}\right) R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,535 |
Example 6, Fold a rectangle $A B C D$ with length and width of 4 and 3 respectively along the diagonal $A C$ to form a right dihedral angle, find the dihedral angle $D-A B-C$. | Let the dihedral angle $D-A B-C=a$, $\angle D A C=\theta, \angle D A B=\theta_{1}, \angle B A C=\theta_{2}$.
By the corollary of the three-line theorem,
$$
\begin{aligned}
\cos \angle D A B & =\cos \angle D A C \cos \angle B A C \\
& =\frac{3}{5} \times \frac{4}{5}=\frac{12}{25},
\end{aligned}
$$
i.e., $\cos \theta_{1... | \arccos \frac{9}{\sqrt{481}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,537 |
Example 8. In the rectangular cuboid $A B C D-A_{1} B_{1} C \cdot D_{1}$ (Fig. 15), $A B=2, A A_{1}=A D=1$. Find the angle between line $A B$ and plane $A B_{1} C$.
| Let the angle between $AB$ and the plane $AB_{1}C$ be $a$.
$$
\angle B A B_{1}=\angle B A C=\arccos \frac{2}{\sqrt{5}} \cdot
$$
By the corollary of formula 1,
$$
\begin{aligned}
\cos \alpha=\frac{\cos \angle B A B_{1}}{\cos \frac{\angle B_{1} A C}{2}} & =\frac{\frac{2}{\sqrt{5}}}{\frac{3}{\sqrt{10}}} \\
& =\frac{2 \sq... | \frac{2 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,539 |
Column 6. Card Games Without Draws
Lanlan and Dongdong often play an intellectual card game together. Before the game starts, each of them prepares five blank pieces of paper and writes a positive integer on each card according to their own wishes. Then, they exchange the five cards that have been written with numbers.... | Prove that from the above $n$ numbers, $n$ partial sums can be made.
$$
\begin{array}{l}
s_{1}=a_{1}, \\
s_{2}=a_{1}+a_{2}, \\
s_{3}=a_{1}+a_{2}+a_{3}, \\
\cdots \cdots \\
s_{\mathrm{n}}=a_{1}+a_{2}+\cdots+a_{\mathrm{n}} .
\end{array}
$$
If at least one of the $n$ integers $s_{1}, s_{2}, \cdots, s_{\text {n }}$ is a m... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,540 |
Example 9. Given the dihedral angle $C-A B=D=120^{\circ}$, $\angle C B A=60^{\circ}, \angle D A B=30^{\circ}, A B=\sqrt{37} \mathrm{~cm}$ (Figure 17). Find the distance $d$ between the skew lines $B C, A D$. | Let the angle between $BC$ and $AD$ be $\theta$. By the triple product formula,
$$
\begin{aligned}
\cos \theta= & \cos 60^{\circ} \cos 150^{\circ} \\
& +\sin 60^{\circ} \sin 150^{\circ} \cos 120^{\circ} \\
= & -\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{8}=-\frac{3 \sqrt{3}}{8} .
\end{aligned}
$$
Thus, $\sin \theta=\frac{\sqr... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,541 |
Example 10. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $B D=10 \mathrm{~cm}, \angle D_{1} B D=30^{\circ}, \angle A O D=60^{\circ}$. Find the distance between $A C$ and $B D_{1}$ (Figure 18). | Let the angle between $B D_{1}$ and $O A$ be $\theta$.
The dihedral angle $D_{1}-B D-A=90^{\circ}$.
By the corollary of the three-line theorem,
$$
\begin{aligned}
\cos \theta & =\cos 30^{\circ} \cos 60^{\circ} \\
& =\frac{\sqrt{3}}{4}, \\
\sin \theta & =\frac{\sqrt{13}}{4} . \\
\text { Therefore, } d & =\frac{5 \cdot ... | \frac{5 \sqrt{39}}{13} \text{ cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,542 |
Given: $O A, O C$ are the oblique line and the projection of $O A$ on plane $\alpha$, respectively. $B D \subset \alpha, B D \perp O C$, $B$ is the foot of the perpendicular. (Figure 19)
Prove: $B D \perp O A$. | Proof: Let the angle between $O A$ and $B D$ be $\theta$.
The dihedral angle $A-O C-D=90^{\circ}$. By the corollary of the Three-Line Theorem,
$$
\begin{aligned}
\cos \theta & =\cos \angle A O C \cdot \cos \angle C B D \\
& =\cos \angle A O C \cdot \cos 90^{\circ} \\
& =0 . \\
\text { Hence } \theta & =90^{\circ},
\end... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,543 |
On the coordinate plane, points with both coordinates being integers are called integer points. Try to prove: There exists a collection of concentric circles such that:
(1) Each integer point lies on one of the circles in this collection.
(2) Each circle in this collection has exactly one integer point on it. | Consider the point $P\left(\sqrt{2}, \frac{1}{3}\right)$. Let the integer points $(a, b)$ and $(c, d)$ be equidistant from point $P$, then
$$
\begin{array}{l}
\quad(a-\sqrt{2})^{2} + \left(d-\frac{1}{3}\right)^{2} = (c-\sqrt{2})^{2} + \left(d-\frac{1}{3}\right)^{2}
\end{array}
$$
which simplifies to $2(c-a) \sqrt{2}=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,544 |
Make a set of concentric circles on the plane such that:
(1) Every lattice point on the plane must lie on one of the circumferences of the concentric circles;
(2) Each circumference contains exactly two lattice points. Please prove the correctness of your construction. | Let's consider the point $O^{\prime}\left(\sqrt{2}, \frac{1}{2}\right)$.
First, we prove that for any integer point $A(a, b)$, there must exist a unique integer point $B(c, d)$ such that $\left|O^{\prime} A\right|=\left|O^{\prime} B\right|$. Indeed, if $\left|O^{\prime} A\right|=\left|O^{\prime} B\right|$, then $(a-\sq... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,545 |
Question: Let $[x]$ be the greatest integer not exceeding $x$. When $n$ is a natural number, prove that: $[\sqrt{n}+\sqrt{n+1}]=$ $[\sqrt{4 n+2}]$.
(Putnam Mathematical Competition, 1948) | 1. Construct the inequality $\sqrt{4 n+1}<\sqrt{n} +\sqrt{n+1}<\sqrt{4 n+2}$.
Proof: When $n$ is a natural number,
$$
4 n^{2}<4 n^{2}+4 n<4 n^{2}+4 n+1
$$
is obviously true.
Taking the square root on both sides, we get $2 n<2 \sqrt{n}(n+1)<2 n+1$. Therefore, $2 n+(2 n+1)<n+2 \sqrt{n(n+1)}+(n+1)<(2 n+1)+(2 n+1)$. Hence... | [\sqrt{n}+\sqrt{n+1}]=[\sqrt{4 n+2}] | Number Theory | proof | Yes | Yes | cn_contest | false | 703,546 |
Example 1. In a mathematics competition: three math teachers made the following guesses about the results of the final for the participating students:
A: Xiao Wang first place, Xiao Liu second place;
B: Xiao Wang second place, Xiao Luo third place;
C: Xiao Li second place, Xiao Luo fourth place.
As a result, each of t... | Consider a $4 \times 4$ grid array (Figure 1). For convenience, let's denote Xiao Wang, Xiao Liu, Xiao Luo, and Xiao Li as $A_{1}, A_{2}, A_{3}, A_{4}$ respectively.
If a teacher estimates that $A_{i}$ is in the $j$th place, a circle is drawn in the cell $a_{1 j}$. Thus, circles should be drawn in the cells $a_{11}, a_... | Xiao Wang is first, Xiao Li is second, Xiao Luo is third, and Xiao Liu is fourth. | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,547 |
Example 2. Find a three-digit number such that the ratio of the number to the sum of its digits is minimized.
untranslated text remains unchanged:
例2. 求一个三位数, 便它与它的各位数字之和的比为最小.
However, for a proper translation, it should be:
Example 2. Find a three-digit number such that the ratio of the number to the sum of its d... | Solve: Arrange the ratios of all three-digit numbers to the sum of their respective digits in a $9 \times 100$ grid (Table 1).
We examine the numbers in each row and select the smaller ones. By the inequality $\frac{a}{b}>\frac{a+1}{b+1}(a>b)$, from left to right, every 10 numbers form a group, and the last number in ... | 159 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,548 |
Example 3. A factory weaves two-color cloth from six different colors of yarn, with each color being paired with at least three other colors. Prove: It is possible to find three different two-color cloths that include all six colors. | Prove that by establishing a $6 \times 6$ grid, using six colors numbered from 1 to 6, and using cell $a_{ij}$ to represent a two-color fabric of colors $i$ and $j$, the grid contains all possible color combinations. If the factory produces $a_{15}$, then mark a solid dot in cell $a_{61}$; otherwise, mark an "×". Note ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,549 |
Example 4. Let $S_{n}^{(k)}=1^{x}+2^{x}+3^{x}+\cdots+n^{x}$.
Prove: $S_{n}^{(k+1)}+S_{n}^{(k)}+\cdots+S_{2}^{(k)}$ $+S_{1}^{(k)}=n S_{n}^{(k)}$ | ```
Prove that filling $1^{n}, 2^{n}, 3^{n}, \cdots, n^{n}$ into each row of an $n \times n$ grid (Figure 5), and denoting the number in cell $a_{i,}$ as $a_{i,}$, we have $a_{i j}=j^{n}(i, j=1,2, \cdots, n)$. Thus,
\[
\begin{array}{l}
\sum_{j=1}^{n} a_{i j}=S_{n}^{(k)}, \\
\text { hence } \sum_{i=1}^{n} \sum_{j=1}^{n}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,550 |
Example 1. For the equation $(1984 x)^{2}-1983 \times 1985 x-1$ $=0$, the larger root is $r$, and for the equation $x^{2}+1983 x-1984=0$, the smaller root is $s$. What is $r-s$? (1984 Beijing Junior High School Mathematics Competition Question) | Solve $\because 1984^{2}+(-1983) \times 1985$ $+(-1)=0$, according to property 1, we can get the roots of the first equation $x_{1}=1, \quad x_{2}=-\frac{1}{1984^{2}}$.
Also $\because 1+1983-1984=0$, we can get the roots of the second equation $x_{1}^{\prime}=1, x_{2}^{\prime}=-1984$.
$$
\text { Therefore } r-s=1985 \... | 1985 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,552 |
Example 2. Given that 1 is a root of the equation $a x^{2}+b x+c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$. | Given that $1$ is a root of the equation $a x^{2}+b x+c=0$,
$$
\begin{array}{c}
\therefore a+b+c=0 . \\
\text { Also, } a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c) \\
.\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
From (1) and (2), we can deduce
$$
\begin{array}{c}
a^{3}+b^{3}+c^{3}=3 a b c . \\
\text { Therefore,... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,553 |
Example 4. If $A, B, C$ are the interior angles of a triangle, and the equation $(\sin B-\sin A) x^{2}+(\sin A-\sin C) x$ $+(\sin C-\sin B)=0$ has equal roots, prove that $B<60^{\circ}$ | Solve: Since $(\sin B-\sin A)+(\sin A-\sin C)$ $+(\sin C-\sin B)=0$, according to property 1, we get
$$
x_{1}=1, x_{2}=\frac{\sin C-\sin B}{\sin B-\sin A} \text {. }
$$
$\because$ The equation has equal roots,
$$
\therefore \frac{\sin C-\sin B}{\sin B-\sin A}=1 \text {. }
$$
Thus, $\sin B=\frac{1}{2}(\sin A+\sin C)$
$... | B<60^{\circ} | Algebra | proof | Yes | Yes | cn_contest | false | 703,555 |
Example 5. In $\triangle A B C$, if the equation $(a-b) x^{2}+(c-a) x+(b-c)=0$ has equal real roots, prove that $\operatorname{ctg} \frac{A}{2}, \operatorname{ctg} \frac{B}{2}, \operatorname{ctg} \frac{C}{2}$ form an arithmetic sequence. | Prove that from $(a-b)+(c-a)+(b-c)=0$, according to property 1, we can get $x_{1}=1, x_{2}=\frac{b-c}{a-b}$.
$\because$ The two roots of the equation are equal, $\therefore \frac{b-c}{a-b}=1$.
$$
\therefore a-b=b-c \text {. }
$$
Also, $\operatorname{ctg} \frac{A}{2}=\frac{s-a}{r}$,
$$
\operatorname{ctg} \frac{B}{2}=\f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,556 |
Example 1. Find the equation of the angle bisector of the two lines $l_{1}: \sqrt{3} x+y-3=0$ and $l_{2}$ : $4 x-3 y-4=0$. | The equation of the family of lines passing through the intersection of $l_{1}, l_{2}$ is
$$
\begin{array}{l}
\sqrt{3} x+y-3+2(4 x \\
-3 y-4)=0 .
\end{array}
$$
Let $M_{0}\left(0 x, y_{0}\right)$ be any point on the angle bisector of $l_{1}, l_{2}$ other than the intersection point, then the above equation becomes
$$... | (5 \sqrt{3}+8) x_{0}-y_{0}-23=0 \text { and } (5 \sqrt{3}-3) x_{0}+11 y_{0}-7=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,557 |
Example 2. Find the equation of the line that is parallel to the line $l: 4 x-3 y+8=0$ and forms a triangle with the coordinate axes with an area of 6 square units.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | The family of lines parallel to $l$ has the equation
$$
4 x-3 y+k=0 \text {, }
$$
which can be rewritten as $\frac{x}{-\frac{k}{4}}+\frac{y}{\frac{k}{3}}=1$.
The intercepts of this family of lines on the two axes are
$$
a=-\frac{k}{4}, \quad b=\frac{k}{3} .
$$
According to the problem, we have
$$
\begin{array}{l}
\fr... | 4 x-3 y \pm 12=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,558 |
Example 3. Prove: regardless of the value of $a$, the line
$$
(a+1) x+(-2 a-5) y-6=0
$$
must pass through a fixed point. | Prove that the known line equation can be transformed into
$$
a(x-2 y)+(x-5 y-6)=0 \text {. }
$$
Obviously, this line passes through the intersection of two fixed lines $x-2 y=0$ and the equation, regardless of the value of $a$:
$$
\left\{\begin{array}{l}
x - 2 y = 0, \\
x - 5 y - 6 = 0
\end{array} \text {, the soluti... | (-4,-2) | Algebra | proof | Yes | Yes | cn_contest | false | 703,559 |
Example 4. Find the equation of the circle passing through points $A(2,2), B(5,3), C(3$, $-1)$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | The family of circles passing through point $A$ is given by
$$
\begin{array}{c}
(x-2)^{2}+(y-2)^{2} \\
+\lambda(x-2)+\mu(y-2)=
\end{array}
$$
0. This family of circles also passes through points $B(5,3)$ and $C(3,-1)$, leading to
$$
\left\{\begin{array}{c}
(5-2)^{2}+(3-2)^{2}+\lambda(5-2) \\
+\mu(3-2)=0, \\
(3-2)^{2}+(... | x^{2}+y^{2}-8 x-2 y+12=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,560 |
Example 5. Given that the center of circle $C$ lies on $x-y-4=0$ and it passes through the intersection points of two circles $C_{1}: x^{2}+y^{2}-4 x-3=0$, $C_{2}: x^{2}+y^{2}-4 y-3=0$. Find the equation of the circle. | The equation of the family of circles passing through the intersection points of circles $C_{1}, C_{2}$ is
$$
\begin{aligned}
& x^{2}+y^{2}-4 x-3 \\
+ & \lambda\left(x^{2}+y^{2}-4 y-3\right) \\
= & 0 .
\end{aligned}
$$
That is,
$$
\begin{array}{l}
\quad(1+\lambda) x^{2}+(1 \\
+\lambda) y^{\prime 2}-4 x-4 \lambda y \\
... | x^{2}+y^{2}-6 x+2 y-3=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,561 |
Example 8. Let $n$ be an odd positive integer. Arrange the natural numbers $1, 2, \cdots, n$ in any order as $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$. Prove that $\left(1-a_{1}\right)\left(2-a_{2}\right) \cdots\left(n-a_{\mathrm{n}}\right)$ is always an even number. | Proof: Since $n$ is odd, we can set $n=2k+1$,
where $k$ is a non-negative integer. It is easy to see that among $1,2, \cdots, n$, there are $k+1$ odd numbers, so among $1,2, \cdots, n$ and $a_{1}, a_{2}, \cdots, a_{n}$, there are a total of $2(k+1)=2k+1+1=n+1$ odd numbers. Since the product $\left(1-a_{1}\right)\left(1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,562 |
Example 6. Try to find the equations of all ellipses with foci at $(\pm 3,0)$, and write down the conditions under which the equation can represent a hyperbola with the same foci as the ellipse. | The foci of the required ellipse lie on the $x$-axis and are symmetric with respect to the coordinate axes, so we can assume its equation to be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text {, where } a>b>0 \text {. }
$$
Since the foci are at $(\pm 3,0)$, we have
$$
a^{2}=b^{2}+9, \quad b^{2}=a^{2}-9 \text { . }
... | a>3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,563 |
Example 2. Solve the equation
$$
\frac{x-2}{x+2}-\frac{16}{x^{2}-4}=\frac{x+2}{x-2} .
$$ | Solution: Multiply both sides of the equation by $(x+2) \quad(x-2)$ to eliminate the denominators, we get
$$
(x-2)^{2}-16=(x+2)^{2} \text {. }
$$
Solving this polynomial equation, we get
$$
x=-2 \text {. }
$$
Verification: When $x=-2$, $(x+2) \cdot(x-2$ , $=0$, so -2 is an extraneous root, hence the original equation... | \text{no solution} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,564 |
1. For any given natural number $n$, if $n^{6}+3 a$ is a cube of a positive integer, where $a$ is a positive integer, then
(A) there are infinitely many such $a$,
(B) such $a$ exist, but only finitely many;
(C) such $a$ do not exist,
(D) none of the conclusions (A),
(B),
(C) are correct. | For any given natural number $n$, there must be $n^{6}+3 \cdot n^{2} \cdot 3 \cdot\left(n^{2}+3\right)+3^{3}=\left(n^{2}+3\right)^{3}$. Let $a=3 n^{2}\left(n^{2}+3\right)+3^{2}$, which makes $n^{6}+3 a$ a cube of a positive integer, so such an $a$ exists.
And it is easy to find that $\left(n^{2}+3 k\right)^{3}=n^{6}+3... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 703,565 |
3. In the Cartesian coordinate system, points whose both coordinates are rational numbers are called rational points. If $a$ is an irrational number, then among all lines passing through the point $(a, 0)$,
(A) there are infinitely many lines, each of which has at least two rational points:
(B) there are $n(2 \leqslant... | Solution: The $x$-axis passes through the point $(a, 0)$, and the $x$-axis contains at least two rational points, so option (D) should be excluded.
Assume there is another line with a slope of $k (k \neq 0)$ that contains at least two rational points, let's say $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 703,567 |
4. As shown in the figure, the vertex $B$ of $\triangle A B C$ is at the center of the unit circle, and $\angle A B C=2 \alpha\left(0<\alpha<\frac{\pi}{3}\right)$. Now, $\triangle A B C$ is rotated counterclockwise within the circle using the following method: First, rotate around $A$ so that $B$ lands on the circumfer... | Solve as shown in the figure,
First time $A \rightarrow A_{1}$, the distance traveled is 0,
Second time $A_{1} \rightarrow A_{2}$, the distance traveled is $A_{1} A_{2}$;
Third time $A_{2} \rightarrow A_{3}$, the distance traveled is $A_{2} A_{3}$. After three rounds, it returns to a similar initial position of $\trian... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,568 |
1. Given the sets $M=\{x, x y, \lg (x y)\}$ and $N=\{0,|x|, y\}$, and $M=N$. Then, $\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\right.$ $\left.\frac{1}{y^{3}}\right)+\cdots+\left(x^{2001}+\frac{1}{y^{2001}}\right)$ is equal to | Given $\because M=N$,
$\therefore 0 \in M$, and $x, y$ cannot both be 0, then
$\lg (x y)=0, x y=1$.
Thus, $1 \in N$. If $y=1$, then $x=1$, which contradicts the distinctness of the elements in the set. Therefore, $|x|=1$, and $x=-1$, so $y=-1$.
Therefore, $x+\frac{1}{y}=-2 \cdot x^{2}+\frac{1}{y^{2}}=2$, $x^{3}+\frac{... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,569 |
2. Given the sets $A=\{(x, y) \mid | x|+| y | = a, a>0\}, B=\{(x, y) \mid | x y |+1 = |x|+|y|\}$. If $A \cap B$ is the set of vertices of a regular octagon on the plane, then the value of $a$ is | Solve As shown in the figure, the coordinates of the two vertices of the regular octagon in the first quadrant should satisfy
$$
\left\{\begin{array}{l}
x+y=a, \\
x y=a-1 .
\end{array}\right.
$$
Solving, we get $\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=a-1\end{array}\right.$ or $\left\{\begin{array}{l}x_{2}=a-1, \\ y_... | 2+\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,570 |
Example 9. Prove: Among any 11 integers, there must be 6 numbers whose sum is divisible by 6. | Proof: First, we prove that among any given 5 integers, there must be 3 numbers whose sum is divisible by 3. The reason is as follows: Let $x, y, z, u, v$ be 5 integers. The remainders when divided by 3 can only be of three types. If these 5 numbers are distributed among these types such that no type is empty, we can a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,573 |
$\triangle A B C$ and $\triangle A D E$
are two non-congruent
right isosceles triangles. Given $\triangle A B C$, while $\triangle A D E$ is rotated around point $A$ in the plane. Prove: regardless of the position to which $\triangle A D E$ is rotated, there must be a point $M$ on segment $E C$, such that $\triangl... | Solution 4 As shown, place $\triangle A B C$ on the complex plane as 1, then $D\left(e^{i \theta}\right), E\left(\sqrt{2} e^{i(\theta+\pi)}\right)$. Using $D B$ as the hypotenuse, construct an isosceles right triangle $\triangle D M \dot{B}$ (with $D, M, B$ in clockwise order), then
$$
\begin{array}{l}
\text { - } \fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,574 |
The point is called an integer point. Prove: there exists a set of concentric circles such that: (1) each integer point is on one and only one circumference of this set;
(2) on each circumference of this set, there is one and only one integer point. | Proof II Assuming the center of concentric circles is $P(x, y)$, for any two integer points $A(a, b)$ and $B(c, d)$, it is not simultaneously true that $a=c$ and $b=d$.
$$
\begin{array}{c}
|P A|^{2}=(x-a)^{2}+(y-b)^{2} \\
=x^{2}+y^{2}+a^{2}+b^{2}-2 a x-2 b y . \\
|P B|^{2}=(x-c)^{2}+(y-d)^{2} \\
=x^{2}+y^{2}+c^{2}+d^{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,575 |
Three, in a singles table tennis tournament with $n$ (>3) players, after several matches, the opponents of any two players are exactly not the same. Try to prove: it is always possible to remove one player so that among the remaining players, the opponents of any two players are still not exactly the same.
| Prove: This is a pure logical reasoning problem, and it is easy to describe and prove using set notation with diagrams.
Suppose there are $n$ players $a_{1}, a_{2}, \cdots, a_{n}(n>3)$, and the sets of opponents they have played against are denoted as $A_{1}$, $A_{2}, \cdots, A_{n}$. Clearly, $a_{i} \notin A_{i}{ }^{\... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,576 |
Find all six-digit integers $\overline{a b c d e f}$:
where $a, b, c, d, e, f$ represent different digits, and $a \cdot e \neq 0$, | Let $\overline{a b c d}=x, \overline{e f}=y$.
According to the problem, we have
$$
3(100 x+y)=10000 y+x,
$$
which simplifies to $299 x=9997 y$.
Thus, we have
$$
\frac{x}{769}=\frac{y}{23}=k, k \in N .
$$
Therefore, $x=769 k, y=23 k$.
However, $a \neq 0$, and $y$ is a two-digit integer, so $k$ can only be 2, 3, or 4.
... | 153846, 229769, 307692 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,577 |
Three, given $x, y \in N$, find the largest $y$ value such that there exists a unique $x$ value satisfying the following inequality:
$$
\frac{9}{17}<\frac{x}{x+y}<\frac{8}{15} \text {. }
$$ | Solve the original system of inequalities $\Longleftrightarrow\left\{\begin{array}{l}8 x-9 y>0, \\ 7 x-8 y>0\end{array}\right.$ $\longleftrightarrow \frac{9 y}{8}<x < \frac{8 y}{7}$.
For the interval $\left(\frac{9 y}{\circ}, \frac{8 y}{7}\right)$ to contain a unique integer, the necessary and sufficient condition is e... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 703,579 |
1. Let $\mathrm{f}$ be a function satisfying the following conditions:
(i) If $x>y$, and $f(y)-y \geqslant v \geqslant f(x)-x$, then for some number $z$ between $x$ and $y$, $f(z)=v+z$;
(ii) The equation $\mathrm{f}(\mathrm{x})=0$ has at least one solution, and among the solutions, there is one that is not less than al... | 1. Solution
(1) From (ii), we know that $f(x)=0$ has at least one solution. Let's assume $u$ is a solution of $f(x)$.
(2) Assume $u>0$.
(3) From (i) and (iii), we know that there exists a number $z$ between 0 and $u$ such that $f(z)=z$.
(4) From (v), we get:
$$
\begin{array}{l}
0=f(z) \cdot f(u)=f(z \cdot f(u)+u \cdot... | 1988 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,580 |
2. Does there exist a binary quadratic polynomial $\mathrm{P}(\mathrm{x}, \mathrm{y})$, such that for a unique pair of non-negative integers $(k, m)$, every non-negative integer $\mathrm{n}$ equals $\mathrm{P}(\mathrm{k}, \mathrm{m})$)? | 2. Car Factory.
Proof Let $\mathrm{P}(\mathrm{k}, \mathrm{m})=\mathrm{k}+$ (number of points below the line $\mathrm{x}+\mathrm{y}=\mathrm{k}+\mathrm{m}$), calculate the number of points $(k, m)$, where $k$ and $m$ are integers. This number is $1+2+\cdots+(k+m)=\frac{1}{2}(k+m) \cdot(1+\mathrm{k}+\mathrm{m})$.
Theref... | P(x, y)=\left((x+y)^{2}+3 x+y\right) / 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,581 |
Example 10. The difference of four integers
Let $a, b, c, d$ be four arbitrarily given integers. Prove that the product of the following 6 differences
$$
b-a, c-a, d-a, c-b, d-b, d-c
$$
is always divisible by 12. | To prove that the product of these 6 differences, denoted as $p$, must and only needs to prove: 3 and 4 can both divide $p$.
The following is divided into two steps.
Step 1: Classify $a, b, c, d$ based on their remainders when divided by 3. There are only three such classes, so among $a, b, c, d$, at least two numbers... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,583 |
4. Find and prove the point \( \mathrm{P} \) inside an acute triangle \( \mathrm{ABC} \) such that for this triangle, \( \overline{\mathrm{BL}}^{2} + \overline{\mathrm{CM}}^{2} + \overline{\mathrm{AN}}^{2} \) is minimized, where \( \mathrm{L}, \mathrm{M}, \mathrm{N} \) are the feet of the perpendiculars from \( \mathrm... | 4. Proof Let $a=\overline{B C}, b=\overline{A C}, c=\overline{A B}$, $x=\overline{B L}, y=\overline{C M}, z=\overline{A N}$. By the Pythagorean theorem, we get
$$
\begin{aligned}
(a-x)^{2}+(b-y)^{2}+(c-z)^{2}= & x^{2}+y^{2} \\
& +z^{2} .
\end{aligned}
$$
If $\mathrm{P}$ is the intersection of the perpendicular bisecto... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,584 |
Example 1. A spider is at point $A$ of the rectangular prism $A C^{\prime}$, and a fly is at point $C^{\prime}$. How can the spider reach the fly by the shortest route? | As shown in the figure, $AB$
$$
=a, BC=3, CC^{\prime}=
$$
$c$, and without loss of generality, let $a>b>c$.
Then, there are three possible shortest routes from $A \rightarrow C^{\prime}$:
(1) From the front
$A B B^{\prime} A^{\prime} \rightarrow$ to the top $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (or bottom
to th... | \sqrt{a^{2}+b^{2}+c^{2}+2 b c} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,585 |
Example 2. As shown in the figure, eight identical squares are overlapped to form a square array (requiring that one diagonal of each of the eight squares coincides on the same straight line). How should they be combined to maximize the total number of squares formed? How many are there in total?
Translate the above t... | Solve as shown in the figure below, place eight squares on the eight equal division points of the first square.
(1) Top and bottom $\qquad$
The number of squares with side length “ $\frac{1}{8}$ ” is
$$
2 \times(1+2+3+4+5+6)=42 \text {, }
$$
The number of squares with side length “ $\frac{2}{8}$ ” is
$$
2 \times(1+2+3... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,586 |
Pan said to five students: “These five paper boxes each contain 1, 2, 3, 4, 5 marbles. Who among you can choose three boxes and guess how many marbles are in each one?”
Jia guessed: “The first box has 3 marbles, the fourth box has 5 marbles, and the fifth box has 1 marble.”
Yi guessed: “The first box has 4 marbles,... | Solution list as follows:
Each person only guessed one box correctly $\Rightarrow$ there is only one correct number in each row; each box was only guessed correctly by one person $\Rightarrow$ there is only one correct number in each column.
(1) The fourth column has 2 fives, which cannot both be correct $\Rightarrow$ ... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,589 |
Example 14. There are 10 pieces of paper, on each of which the positive integers from 1 to 10 are written. Then they are all folded and placed in a hat. Five people are then asked to each draw two pieces of paper (the pieces drawn by each person are not returned to the hat). Unfortunately, an error occurred in recordin... | (1) From $B=4 \Rightarrow B=1+3$,
(2) $C=7=\left\{\begin{array}{l}1+6, \\ 2+5, \\ 3+4,\end{array}\right.$
but 1, 3 are already determined, $\therefore C=2+5$;
(3) From $A=11=\left\{\begin{array}{l}4+7, \\ 3+8, \\ 2+9, \\ 1+10,\end{array}\right.$
and 1, 2, 3, 5 are already determined,
$$
\therefore A=4+7 \text {, }
$$... | A=4+7, B=1+3, C=2+5, D=10+6, E=8+9 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,590 |
Combination: Three-foot residence: Discussing Beijing, Baotou, Tianjin, the requirement is to find out their residence and occupation based on the following conditions: (1) Although Bai Ying's all relatives and close friends are in Beijing, he only occasionally visits the capital; (2) Among these three people, two of t... | From (1)(3) $\Rightarrow$ Bai Ying is not a tractor driver and does not live in Beijing. And “the tractor driver must live in Beijing”. Assuming Tang Li is the tractor driver and lives in Beijing, from (2) $\Rightarrow$ Bai Min and Bao Min must have the first letter of their residence and occupation in Chinese pinyin b... | Bai\ Ying\ is\ the\ broadcaster\ and\ lives\ in\ Baotou,\ Tang\ Li\ is\ the\ postman\ and\ lives\ in\ Tianjin,\ Bao\ Min\ is\ the\ tractor\ driver\ and\ lives\ in\ Beijing | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,591 |
Example 1. (IMO-1-1) Prove: For any natural number $n$, the fraction $\frac{21 n+4}{14 n+3}$ is irreducible. | To prove that the fraction mentioned cannot be simplified, it is actually to prove that the numerator and the denominator are coprime. Note that
$$
3(14 n+3)-2(21 n+4)=1 \text{. }
$$
In Bézout's identity, take
$$
a=14 n+3, b=21 n+4, x=3, y=-2
$$
and the result follows.
We can also use a modified proof method without ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,592 |
Example $2 . m$ boxes, each containing some balls. Let $n<m$ be a given natural number, and perform the following operation: choose $n$ of these boxes and place one ball in each of the chosen boxes. Prove:
If $(m, n)=1$, then it is possible to perform a finite number of operations to make the number of balls in all bo... | Proof: Since $m, n$ are coprime, according to Bézout's identity, there exist natural numbers $x, y$ such that
$$
x n - y m = 1 \text{. }
$$
This implies $\quad 2 = y m + 1 = y(m-1) + (y+1)$. This means that by adding $x n$ balls to the salt, we can increase each of $m-1$ boxes by $y$ balls, and one box by $y+1$ balls,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,593 |
Example 3. (IMO-27-1) Let $M=\{1,2, \cdots, n-1\}, n \geqslant 3, M$ be a set where each number is colored either red or black, such that
(i) for each $i \in M, i$ and $n-i$ are the same color;
(ii) for each $i \in M, i \neq k$, $i$ and $|k-i|$ are the same color, where $k$ is a fixed number in $M$ that is coprime with... | Proof: For the convenience of the following statement, we add the numbers $0$ and $n$ to $M$, and stipulate that they have the same color as $k$. It is easy to verify that properties (i) and (ii) still hold.
By Bézout's identity, there exist positive integers $x$ and $y$ such that
$$
x k - y n = 1.
$$
(Whenever the pro... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,594 |
List 4. (IMO-4-1) Find the smallest natural number $n$ whose decimal representation ends with 6, such that when the last digit 6 is deleted and written at the beginning of the remaining digits, it becomes four times $n$. | Solve for $n$ whose decimal representation ends with 6, meaning that it leaves a remainder of 6 when divided by 10. By the division algorithm, we have
$$
n=10 m+6 \text {. }
$$
where $m$ is a positive integer, and we assume it is an $l$-digit number. According to the problem,
$$
4(10 m+6)=6 \times 10^{l}+m .
$$
Thus,... | 153815 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,595 |
Example 5. (IMO-9-3) If $k, m, n$ are positive integers, and $m+k+1$ is a prime number greater than $n+1$, let $C_{s}=s(s+1)$, prove that the product
$$
\left(C_{\mathrm{m}+1}-C_{\mathrm{k}}\right)\left(C_{\mathrm{m}+2}-C_{\mathrm{k}}\right) \cdots\left(C_{\mathrm{m}+\mathrm{n}}-C_{\mathrm{k}}\right)
$$
is divisible b... | Proof: It is evident that $C_{1} \cdots C_{n}=(n+1)!n!$, and $n!$ must divide the product of any $n$ consecutive integers, thus
$$
\begin{array}{l}
n!\mid[(m+1-k)(m+2-k) \\
\cdots \cdot(m+n-k)] .
\end{array}
$$
Similarly,
$$
\begin{array}{c}
(n+1)!\mid[(m+k+1)(m+k+2) \\
\cdots \cdots \cdot(m+k+n+1)] .
\end{array}
$$
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,596 |
For example, in the figure, the side lengths of the equilateral triangle and the square are both 2, while the side lengths of the regular hexagon and the regular octagon are both 1. $O$ is the center of the regular polygons. Find the area of the shaded parts in each figure.
Translate the above text into English, pleas... | Solution (1) The area of the shaded part
$$
=\frac{1}{4} S_{\text {square }}=1 \text {, }
$$
(2) The area of the shaded part
$$
=\frac{1}{3} S_{\text {equilateral triangle }}=\frac{\sqrt{3}}{3} ;
$$
(3) The area of the shaded part
$$
=\frac{1}{3} S_{\text {regular hexagon }}=\frac{\sqrt{3}}{2} \text {; }
$$
(4) The are... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,597 |
Let $p$ be a prime number, $1 \leqslant k<p$, then $C_{\mathrm{P}}^{\mathrm{k}}$ is divisible by $p$.
| To prove that $C \underset{\mathrm{p}}{\mathrm{k}}$ is an integer by the meaning of combinations, we have:
\[ C_{p}^{\mathrm{k}}=\frac{p(p-1) \cdots(p-k+1)}{k!(p-k)!} \]
\[ = p \times \frac{(p-1) \cdots(p-k+1)}{k!(p-k)!}. \]
Since $1 \leqslant k < p$, it follows that $1 \leqslant p-k < p$. Therefore, $p \times k!$ and... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,598 |
Example 7. (IMO-21-1) Let $p, q$ be natural numbers such that
$\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{1318}+\frac{1}{1319}$. Prove that $p$ is divisible by 1979. | $$
\begin{array}{l}
\frac{\boldsymbol{p}}{\boldsymbol{q}}=1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{1318}+\frac{1}{1319} \\
=\left(1+\frac{1}{2}+\cdots+\frac{1}{1319}\right) \\
-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right) \\
=\left(1+\frac{1}{2}+\cdots+\frac{1}{1319}\right) \\
-\left(1+\frac{1}{2}+\cdot... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,599 |
Example 8. (IMO-12-4) Find all positive integers $n$ such that the set $\{n, n+1, n+2, n+3, n+4, n+5\}$ can be partitioned into two non-empty subsets with no common elements, such that the product of all elements in one subset is equal to the product of all elements in the other subset. | There is no solution to this problem, as proven below.
Assume $n$ has the stated property, then any prime divisor $p$ of $n, n+1, n+2$, $n+3, n+4, n+5$ must divide the product of the elements of each subset, and thus must divide at least two of these six numbers (by the property of prime numbers).
If it divides two in... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,600 |
Example 9. (IMO-11-1) Prove: There exist infinitely many natural numbers $\alpha$ with the property that for any natural number $n$, $z=n^{4}+\alpha$ is never a prime. | To prove: Given $n$, there must exist infinitely many $\boldsymbol{\alpha}$ such that $n^{4}+\boldsymbol{\alpha}$ is not a prime number. By the definition of prime numbers, this is equivalent to saying that there are infinitely many $a$ such that $n^{4}+\alpha$ can be factored into the product of two proper divisors (p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,601 |
Example 10. (IMO-6-1) a) Find all positive integers $n$ such that $2^{\mathrm{n}}-1$ is divisible by 7.
b) Prove that there is no positive integer $n$ such that $2^{n}+1$ is divisible by 7. | Solve $\alpha$) First note that for $k \geqslant 0, k$ an integer, $2^{3 k}-1$ is divisible by 7. Proof as follows:
Since $2^{3} \equiv 1 \pmod{7}$, it immediately follows that
$$
2^{3 k} \equiv 1 \pmod{7}, \text{ which means } 7 \mid 2^{3 k}-1 \text{. }
$$
(Or, using $2^{3 k}-1=\left(2^{3}\right)^{k}-1=\left(2^{3}-1\... | 3 \mid n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,602 |
12. (IMO-20-1) The last three digits of the numbers $1978^{n}$ and $1978^{m}$ are equal. Find the positive integers $n$ and $m$ such that $n+m$ is minimized, where $n>m \geqslant 1$. | From the given, we have
$$
\begin{array}{l}
1978^{m}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right) \\
=2^{m} \times 989^{m} \times\left(1978^{m}-1\right) \\
\equiv 0\left(\bmod 10^{3}\right) .
\end{array}
$$
Since
$$
10^{3}=2^{3} \times 5^{3},
$$
and $989^{m}$ and $1978^{\mathrm{m}}-1$ are both odd, it follows that $m \... | 106 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,604 |
Proposition 1 The necessary and sufficient condition for all pairs of opposite edges of a tetrahedron to be equal is that, for each vertex, the sum of the two adjacent angles is equal to $180^{\circ}$. | To prove sufficiency. If the sum of the three dihedral angles at each vertex of a tetrahedron is equal to $180^{\circ}$. We can flip $\triangle A B C, \triangle A C D, \triangle A B D$ outward and place them in the plane of the base $\triangle B C D$. Then $A_{1} 、 C 、 A_{2}$, $A_{2} 、 B 、 A_{3}$, $A_{3} 、 D 、 A_{2}$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,605 |
Example 1. In $\triangle A B C$, $\angle C=3 \angle A$, $a=27, c=48$. What is $b=$ ?
(A) 33 ;
(B) 35;
(C) 37 ;
(D) 39 ;
(E) The value of $b$ is not unique.
[36th American High School Mathematics Examination (February 26, 1985, Beijing), Question 28]
Your publication provided a trigonometric solution to this problem i... | Solve: In $\angle BCD$, construct $\angle BCD = \angle A$, then $\angle ACD = 2\angle A$.
Let $AD = x, CD = y$. Then
$$
\begin{array}{l}
x^{2} - y^{2} = yb, \\
\because \triangle ACB \sim \triangle CDB, \\
\therefore \frac{y}{b} = \frac{27}{48} = \frac{9}{16}, y = \frac{9}{16} b \\
\frac{48 - x}{27} = \frac{27}{48}, \\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,607 |
Example 4. The following figures are composed of congruent squares. Which one can form a cube?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: This question mainly tests the students' spatial imagination ability. It can let students practice how to unfold a cube into a plane figure to familiarize themselves with the relationship between the front, back, top, bottom, left, and right six faces of the cube. The answer to this question is (A). | A | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,608 |
In an equilateral triangle with an area of I, place five points arbitrarily. Try to prove: within this triangle, it is always possible to cover these five points with three smaller equilateral triangles, whose sides are parallel to the sides of the original triangle, and the sum of their areas does not exceed 0.64. | The probability does not exceed 0.64. In fact, this number is not the minimum value of the sum of the areas of the three equilateral triangles, meaning the sum of the areas of the three triangles covering five points can be further reduced.
Below we prove that the sum of the areas of these three triangles does not exc... | \left(\frac{10}{13}\right)^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 703,610 |
Example 1. As shown in the figure, in $\triangle P M N$, $C, B$ are points on $P$, $P M$, respectively. The extension of $C B$ intersects the extension of $N M$ at point $A$, and $P C = A M, P N = m, A B = n$. Find $M N: B C$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 1. As shown in the figure, in $\triangle P M N$, $C, B$ are... | Draw $CG // NA$ intersecting $PM$ at $G$.
$$
\begin{array}{l}
\frac{CG}{MN}=\frac{PC}{PN}, \\
\left.\begin{array}{l}
\frac{AM}{CG}=\frac{AB}{BC}
\end{array}\right\} \Rightarrow \begin{array}{l}
\frac{AM}{MN}=\frac{PC}{PN} \cdot \frac{AB}{BC} \\
PC=AM, PN=m, AB=n
\end{array} \\
\Rightarrow \frac{MN}{BC}=\frac{PN}{AB}=\f... | \frac{m}{n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,611 |
Example 2. As shown in the figure, line $A E B$ intersects the sides or extensions of $\triangle D C F$ at points $B, E, A$ and $\frac{B E}{F C}=\frac{A E}{C D}$. Prove that $A F=B D$.
保留源文本的换行和格式，直接输出翻译结果。 | Prove that through point $B$, $BG \parallel CA$ intersects $FD$ at $G$.
$$
\begin{array}{l}
\left.\begin{array}{l}
\frac{BG}{FC}=\frac{BD}{CD}, \\
\frac{AF}{BG}=\frac{AE}{BE}
\end{array}\right\} \Rightarrow \frac{AF}{FC}=\frac{BD}{CD} \cdot \frac{AE}{BE} \\
\Rightarrow AF \cdot \frac{BF}{FC}=\frac{AE}{CD} \cdot BD \Rig... | AF=BD | Geometry | proof | Yes | Yes | cn_contest | false | 703,612 |
Example 3. As shown in the figure, given: the oblique intersection of line $R T K$ with the extensions of the three sides of $\triangle S M N$ at points $R, T, K$. Prove: $\frac{K R}{R T} \cdot \frac{T S}{S N} \cdot \frac{N M}{M K}=1$. | Proof One: Draw $TG // KM$ intersecting $SR$ at $G$.
$$
\begin{array}{l}
\left.\begin{array}{l}
\frac{TG}{MK}=\frac{RT}{RK}, \\
\frac{MN}{TG}=\frac{SN}{ST}
\end{array}\right\} \Rightarrow \frac{NM}{MK}=\frac{RT}{RK} \cdot \frac{SN}{TS} \\
=\frac{KR}{RT} \cdot \frac{TS}{SN} \cdot \frac{NM}{MK}=1 \text{. } \\
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,613 |
Example 1. Prove that $4-\sqrt{5} \leqslant x+4+\sqrt{5-x^{2}}$ $\leqslant 4+\sqrt{10}$. Where $|x| \leqslant \sqrt{5}$. | Proof: Let $x=\sqrt{5} \sin \varphi$,
$$
\left(-\frac{\pi}{2} \leqslant \varphi \leqslant \frac{\pi}{2}\right. \text { ) }
$$
The original inequality can be transformed into
$$
\begin{array}{l}
4-\sqrt{5} \leqslant \sqrt{5} \sin \varphi+4+\sqrt{5} \cos \varphi \\
\leqslant 4+\sqrt{10}
\end{array}
$$
or $\quad-\frac{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,614 |
BSO. Proof, $\left(1+\frac{\alpha}{m}\right)^{m}n, 0<m<n$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: Let $f(x)=\ln (1+x),(x \geqslant 0)$, then $f^{\prime}(x)=\frac{1}{1+x}$.
Obviously, for $x \geqslant 0, f^{\prime}(x)$ is a strictly decreasing function.
Take $0<\frac{\alpha}{n}<\frac{\alpha}{m}$, by Rule $I$ we have
$$
\begin{aligned}
& \frac{a}{n} \ln \left(1+\frac{\alpha}{m}\right)+\left(\frac{a}{m}-\frac{\... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,616 |
Example 1. Prove: $N$ is equivalent to the extended set of natural numbers $N_{0}=\{0$, $1,2, \cdots\}$. | Proof: Let $f(n)=n-1, n \in N$, $f(n) \in N_{0}$, then $f$ is a one-to-one correspondence from $N$ to $N_{0}$: $f$ $N \longleftrightarrow N_{0}$, hence $N \sim N_{0}$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,617 |
Example 5. In the corner of the room, there are several identical cubes piled up. How many of the cubes are not visible? | In the second layer, there is 1 that is invisible,
in the third layer, there are $(1+2)$ that are invisible:
in the fourth layer, there are $(1+2+3)$ that are invisible, in the fifth layer, there are $(1+2+3+4)$ that are invisible, so the total number of invisible ones is $1+3+6+10=20$. | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,619 |
Example 3. Any two line segments (as sets of points) are equivalent.
保留源文本的换行和格式，翻译结果如下：
Example 3. Any two line segments (as sets of points) are equivalent. | Proof: Let $AB$ and $CD$ be two line segments. If they are not collinear, connect $AC$ and $BD$, and a one-to-one correspondence between points can be established as shown in Figure 2 a), b). If they are collinear, a one-to-one correspondence between $AB$ and $EF$, and $CD$ and $EF$ can be established as shown in c), t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,620 |
1. In the equation $\overline{x 5} \cdot \overline{3 y} \bar{z}=7850$, restore the digits $x, y, z$ | Since $\overline{x 5}$ is a factor of 7850, in $x=$ $0,1, \cdots, 9$, $x$ can only be 2. (If $x=0$, then $5 \cdot 400=2000<7850$, if $x=1, 3, 4, \cdots$,
9, then $\overline{x 5}$ cannot divide 7850.) Therefore, $\overline{3 y z}$ $=7850 \div 25=314$, which means $y=1, z=4$. | x=2, y=1, z=4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,622 |
2. For a regular $n$-sided polygon, construct $n$ squares outside the polygon, each sharing one side with the polygon. It is known that the $2n$ outer vertices of these $n$ squares form a regular $2n$-sided polygon. For what value of $n$ is this possible? | Solve As shown in Figure 1, $P, Q, R$ are three vertices of a regular $n$-sided polygon; $M, N, K, L$ are four vertices of a $2n$-sided polygon. According to the given conditions, we have $MN = NQ, QK = KL, NK = KL = MN$, which means $\triangle NQK$ is an equilateral triangle. Therefore, $\angle NQK = 60^{\circ}$. From... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,623 |
4. Figure 2, for an isosceles triangle $ABC (AJ=BC)$, draw a line $l$ through vertex $B$, such that $l \parallel AC$. Take a point $O$ on $l$ as the center to draw a circle, tangent to $AC$ at $D$, and intersecting $AB, BC$ at $E, F$ respectively. Prove: the length of $\overparen{EDF}$ is independent of the position of... | For any circle, its radius is always equal to the height of the triangle with vertex $B$. Let $D_{1}, E_{1}, F_{1}$ be the symmetric points of $D, E, F$ with respect to the line $l$, respectively. We have $\overparen{E D F}=E_{1} \overbrace{D_{1} F_{1}}, \angle E_{1} B O=\angle O B E$. Since $l \perp A C$, we have $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,624 |
3. Given two parallel lines, a circle is tangent to one line at $A$, and intersects the other line at $B, C$. Take a point $D$ on the circumference of the circle, different from $A, B, C$. Prove: The distances from $A$ to the lines $B D$ and $C D$ are equal. | To prove that the distances from $A$ to the lines $B D$ and $C D$ are equal, it is necessary to prove that $D A$ is the angle bisector of $\angle B D C$. Below, we discuss this in two cases:
(1) Suppose $D$ is on the arc $B a C$ (Figure 4). From $B c A = A b C$, we can deduce that $\angle \dot{B} D A = \angle C D A$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,628 |
4. The area of quadrilateral $A B C D$ is $S$. Prove: The area of the quadrilateral formed by the midpoints of $A C, A D, B C$ and $B D$ is less than $0.5 S$.
| Proof: Let $K, L, M, N$ be the midpoints of $AC, AD, BC, BD$ respectively; $P, Q$ are the midpoints of $AB$ and $CD$ (Figure 6). It is easy to see that quadrilaterals $PMQL$ and $KMLN$ are parallelograms, and $S_{PMQL}=0.5 S$. As long as we can prove that parallelogram $KMLN$ lies inside parallelogram $PMQL$, we can co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,629 |
5. On the blackboard are the numbers $1,2, \cdots, 1987$. Perform the following transformation: erase some of the numbers on the blackboard and add the remainder when the sum of the erased numbers is divided by 7. After several such transformations, only two numbers remain on the blackboard, one of which is 987. Find t... | Note the fact that with each transformation: the sum of all numbers on the blackboard, when divided by 7, leaves the same remainder. $1+2+\cdots+1987=1987 \cdot 7 \cdot 142$, which indicates that the original sum on the blackboard is divisible by 7, so the sum of the two final numbers can also be divided by 7. Among th... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,631 |
1. Find the smallest natural number such that when the last digit of this number is moved to the first position, the resulting number is 5 times the original number. | Let the required number be $a_{1} a_{2} \cdots a_{n-1} a_{n}$ - According to the given conditions, we have
$$
\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}=5 \cdot \overline{a_{1} a_{2} \cdots a_{n-1} a_{n}} \text {. }
$$
Then
$$
\begin{array}{l}
a_{n} \cdot 10^{n-1}+\overline{a_{1} a_{2} \cdots a_{n-1}} \\
=5\left(\ove... | 142857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,632 |
2. Find all solutions to the equation $x^{2}-8[x]+7=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $x$ be the root of the given equation, $n=\lceil x\rfloor$, then $x^{2}+7=8 n, n \geqslant 0, n \leqslant x4$. Therefore, $1 \leqslant n<2$ or $4<n \leqslant 7$, i.e., $n=1,5,6,7$. Accordingly, $x^{2}=1,33,41$, 49 - After inspection, $1, \sqrt{33}, \sqrt{41}, 7$ are roots of the equation, while $-1,-\sqrt{33},-\sqr... | null | Algebra | proof | Yes | Yes | cn_contest | false | 703,633 |
3. Two circles are externally tangent at point $D$. A straight line is tangent to one circle at $A$ and intersects the other circle at points $B$ and $C$. Prove: The distances from point $A$ to the lines $BD$ and $CD$ are equal. | $$
\begin{array}{c}
\because \angle B D A=\angle B D F+\angle F D A \\
=\angle D C B+\angle F A D=\angle A D E, \\
\therefore D A \text { is the bisector of } \angle B D E.
\end{array}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,634 |
5. At least how many circles with a radius of 1 are needed to cover a circle with a radius of 2. | Let $O$ be the center of a circle with radius 2, and let $A B C D E F$ be a regular hexagon inscribed in the circle (Figure 8). The side length of the hexagon is 2. Construct six circles with the sides of the hexagon as diameters. The intersection points of these six circles, other than $A, B, C, D, E, F$, are denoted ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,636 |
7. (a) Let $\mathrm{gcd}(\mathrm{m}, \mathrm{k})=1$, prove: there exist integers $a_{1}, a_{2}, \cdots, a_{m}$ and $b_{1}, b_{2}, \cdots, b_{k}$, such that each product $a_{i} b_{j}(i=1,2, \cdots, m ; j=1,2, \cdots, k)$ when divided by $\mathrm{m} \cdot \mathrm{k}$ yields distinct remainders.
(b) Let $\mathrm{gcd}(\mat... | 7. (c) Exam Street
$$
\begin{array}{l}
a_{i}=k, i, i=1,2, \cdots, m, \\
b_{j}=m j+1, j=1,2, \cdots, k .
\end{array}
$$
Assume $\mathrm{mk} \mid \mathrm{a}_{1} \mathrm{~b}_{j}-\mathrm{a}_{\mathrm{s}} \mathrm{b}_{\mathbf{t}}=\left(\mathrm{k}_{1}+1\right)(\mathrm{mj}+1)$
$$
\begin{array}{l}
-(k s+1)(m t+1)=k m(i j-s t) \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,639 |
Example 7. A pig sells for $3 \frac{1}{2}$ silver coins. A goat sells for $1 \frac{1}{3}$ silver coins, and a sheep sells for $\frac{1}{2}$ silver coins. Someone bought 100 animals with 100 silver coins. How many pigs, goats, and sheep were bought? | Let the number of pigs bought be $x$, the number of goats be $y$, and the number of sheep be $z$. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=100, \\
3 \frac{1}{2} x+1 \frac{1}{3} y+\frac{1}{2} z=100 .
\end{array}\right.
$$
(2) $\times 2$ - (1): $6 x+\frac{5}{3} y=100$,
$$
y=60-\frac{18}{5} x \te... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,641 |
Example 8. A merchant has a 40-pound weight, which broke into 4 pieces after falling to the ground. Later, it was found that the weight of each piece is an integer number of pounds, and these 4 pieces can be used to weigh any integer weight from 1 to 40 pounds. What are the weights of these 4 pieces of the weight? | To make two weights A and B measure the most weight, you can choose A=1, B=3, with which you can measure 1, 2, 3, 4 pounds of objects. If you choose a third weight C=2×4+1=9, then you can use A, B, C to measure objects from 1 to 13 (4+9=13) pounds. Finally, if you choose a fourth weight D=2×13+1=27, then you can use A,... | 1, 3, 9, 27 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,642 |
Example 9. Primary School One and Primary School Two have the same number of students participating in the Golden Cup Competition. The schools use cars to transport the students to the examination site. Primary School One uses cars that can seat 15 people each; Primary School Two uses cars that can seat 13 people each.... | Let's assume that at the beginning, both schools had $x$ participants. According to the problem, the first school sent $\frac{x}{15}=a \cdots 0$ ($a$ cars),
the second school sent $\frac{x}{13}=a \cdots 12(a+1$ cars $)$; the second time, the first school sent $\frac{x+1}{15}=a \cdots 1(a+1$ cars $)$,
the second schoo... | 184 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,643 |
In a hand clapping competition, the winner of each match gets 2 points, the loser gets 0 points; if it's a draw, each player gets 1 point. Now the scores are: $1979$, $1980$, $198$, $1985$. After verification, how many players participated in the competition? | Since $n(n-1)$ is the product of two consecutive natural numbers, it must be even, so 1979 and 1985 are impossible. And the last digit of $n(n-1)$, being the product of two consecutive natural numbers, can only be 0, 2, or 6, so 1984 is also impossible. The only possibility left is 1980. For $n(n-1)=1980$, solving give... | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,644 |
Example 1. If $a>1, b$ is a positive rational number, $a^{b}+a^{-0}$ $=2 \sqrt{2}$, find the value of $a^{b}-a^{-b}$. | Consider the following solution:
Let $a^{b}=x$, then $x+\frac{1}{x}=2 \sqrt{2}$.
Transform it into $x^{2}-2 \sqrt{2} x+1=0$.
Solving yields $x_{1}=\sqrt{2}+1, x_{2}=\sqrt{2}-1$.
$$
\text { When } \begin{aligned}
x & =\sqrt{2}+1, \\
& a^{b}-a^{-b} \\
& =\sqrt{2}+1-\frac{1}{\sqrt{2}+1} \\
& =\sqrt{2}+1-(\sqrt{2}-1) \\
& ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,645 |
Example 2. If $\lg ^{2} x \lg 10 x<0$, find the value of $\frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}}$. | From the known condition $\lg ^{2} x \lg 10 x<0$, it is not difficult to conclude that $\lg x \neq 0$ and $\lg 10 x<0$, which also implicitly indicates the existence of $\lg x$.
$$
\text { Hence } \begin{aligned}
& \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}} \\
= & \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+2 \lg x+\lg ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,646 |
Example 1. $(1 \mathrm{MO}-27-1)$ Let $d$ be a positive integer not equal to $2,5,13$. Prove that in the set $\{2,5,13, d\}$, there exist two distinct elements $a, b$ such that $a \dot{o}-1$ is not a perfect square. | Proof by contradiction. Assume the conclusion is not true, then
$$
\begin{array}{l}
2 d-1=x^{2} \\
5 d-1=y^{2} \\
13 d-1=z^{2}
\end{array}
$$
$x, y, z$ are all integers.
Clearly, from (1) we know that $x$ is odd, thus
$$
2 d=x^{2}+1 \equiv 2(\bmod 4),
$$
(we use congruence, which has been introduced in (I)) i.e., $d$ i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,648 |
Example 2. Prove: $x^{2}+y^{2}+z^{2}=7 k^{2 n}$ has no positive integer solutions. Here $k$ is a known positive integer. | Here are four unknowns $n, x, y, z$. We still use the method of proof by contradiction.
First, note a simple but important fact:
$$
x^{2} \equiv 0,1,4(\bmod 8).
$$
Because, if $x$ is odd, let $x=2 l+1$, then $x^{2}=4 l(l+1)+1$, but $l(l+1)$ is even, so $8 \mid 4 l(l+1)$, i.e., $x^{2} \equiv 1(\bmod 8)$; if $x$ is even... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,649 |
Example 3: Prove that $x^{2}+y^{2}=z+z^{5}$ has infinitely many positive integer solutions satisfying the condition: $(x, y)=1$.
| Proof: Our method is to directly find the required solution (note that it does not have to be all solutions). There is a key identity that readers should remember, as it is useful in many places:
$$
\begin{aligned}
\left(a^{2}\right. & \left.+b^{2}\right)\left(c^{2}+d^{2}\right)=(a b+c d)^{2} \\
& +(a c-b d)^{2} .
\end... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,650 |
Example 4. (1MO-25-6) Let $a, b, c, d$ be odd numbers, $0<a<b<c<d$, and $a d=b c$. Prove: If
$$
a+d=2^{k}, \quad \mathrm{~b}+c=2^{m},
$$
$k, m$ are integers, then $a=1$. | $$
\begin{array}{l}
\text { Prove } \quad H a+d=2^{k}, b+c=2^{m} \text { then } \\
d=2^{k}-a, c=2^{m}-b . \\
\text { Substitute into } a d=b c c^{\prime}, \text { we have } \\
a\left(2^{k}-a\right)=b\left(2^{m}-b\right), \\
\text { which is } b \cdot 2^{m}-a \cdot 2^{k}=b^{2}-a^{2} .
\end{array}
$$
Furthermore, we hav... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,651 |
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