problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Three, in a certain sports competition, every two players play one match, and each match must have a winner. The competition is used to determine outstanding players. Player $A$ is determined to be an outstanding player under the condition that: for any other player $B$, either $A$ beats $B$; or there exists a player $... | First, we prove that in any competition, there must exist an excellent player. In fact, the player with the most net wins, let's denote this player as $A$, is an excellent player. This is because, if $A$ is not an excellent player, then there must be at least one player $B$ who defeats $A$, and $B$ also defeats all the... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 703,104 |
In a regular triangle with an area of 1, place five points arbitrarily. Try to prove: within this regular triangle, it is always possible to draw three regular triangles to cover these five points. The sides of these three regular triangles are parallel to the sides of the original triangle, and the sum of their areas ... | Proof: Since any five points already placed are inside this triangle, the distance from any of these points to the three sides is greater than zero, so there must exist an equilateral triangle with sides parallel to the three sides of this triangle and with an area less than 1, such that these five points are still ins... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,105 |
Let $A_{1} A_{2} A_{3} A_{4}$ be a tetrahedron, $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$ be the spheres with centers at $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ respectively, and they are mutually tangent. If there exists a point $O$, from which a sphere with radius $r$ can be drawn to be tangent to $S_{1}$, $S_{2}$, $S_{3}$, $S_{... | 1 It is easy to prove that $S_{1}, S_{2}, S_{3}, S_{4}$ are only pairwise externally tangent. Let the sphere with center at $O$ and radius $R$ be tangent to $A_{1} A_{2}, A_{2} A_{2}, A_{3} A_{1}$ at $P_{1}, P_{2}, P_{3}$ (Figure 6). From the congruence of the right triangles $\triangle A_{1} O P_{3} \cong$ right $\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,106 |
$\therefore 、 m$ distinct positive even numbers and $n$ distinct positive odd numbers have a total sum of 1987, for all such $m$ and $n$, what is the maximum value of $3 m+4 n$? Please prove your conclusion.
The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 1987, for all such $m$ a... | The maximum value of $3 m+4 n$ is 221. The proof is as follows: Let $a_{1}+\cdots+a_{n}+b_{1}+\cdots+b_{n}=1987$. Here, $a_{i} (1 \leqslant i \leqslant m)$ are distinct positive even numbers, and $b_{j} (1 \leqslant j \leqslant n)$ are distinct positive odd numbers. Clearly, $u$ is always an odd number, and
$$
\begin{a... | 221 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,107 |
1. Given the quadratic equation in $x$, $x^{2}-(2 m+1) x$ $+m+2=0$ has two integer solutions, find the value of $m$ and the two integer solutions of the equation. | 1. Answer: $m=2$, solutions 1, 4; $m=-2$, solutions -3, 0. Hint: Use the relationship between roots and coefficients, then eliminate $m$. | m=2, \text{solutions } 1, 4; m=-2, \text{solutions } -3, 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,108 |
3 . Find the integers $x, y$ that satisfy the equation $2 x-3 y=17$, and for which $x^{2}+y^{2}$ is minimized, as well as the minimum value of $x^{2}+y^{2}$. | 3. Answer: $x=4, y=-3, x^{2}+y^{2}$ has a minimum value of 25.
Hint: The general solution of $2 x-3 y=17$ is $x=3 t+7$, $y=2 t-1$. Substituting into $x^{2}+y^{2}$, we get $x^{2}+y^{2}$ $=13\left(t+\frac{19}{13}\right)^{2}+\frac{289}{13}$. The integer $t$ closest to $t=-\frac{19}{13}$ is -1. | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,110 |
4. Find the positive integer solutions for the equation $x+y+z=19$, 1.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The note above is not part of the translation but is provided to clarify the instruction. Here is the... | Eliminate $x$, we get $9z = 4(19 - y), (9, 4) = 1, z$ is a multiple of 4. From the second equation $95 - 10z = x + 5y \geqslant 6, \quad z \leqslant 8.9, \quad z = 4, 8$. | null | Number Theory | other | Yes | Yes | cn_contest | false | 703,111 |
5 . An integer when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 4, find this integer. | 5. Answer: $105 k+53$ ( $k$ is an integer $)$.
| 105k+53 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,112 |
6 . Find the integer values of $m, n$ that satisfy $m^{4}-n^{4}=369$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6 . Answer: $(m, n)=(5,4),(5,-4)$, $(-5,4),(-5,-4)$.
Hint: $\left(m^{2}+n^{2}\right)\left(m^{2}-n^{2}\right)=369$, from this, we have $\left(m^{2}+n^{2}, m^{2}-n^{2}\right)=(369,1)$, $(123,3),(41,9)$. Since $m, n$ are integers, we have only
$$
\left(m^{2}+n^{2}, m^{2}-n^{2}\right)=(41,9),
$$
i.e., $\left\{\begin{array... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,113 |
7. $a, b, c$ are natural numbers. (1) Find the solutions $(a, b)$ that satisfy $a>b$ and $(a+b)^{2}=a^{3}+b^{3}$, (2) Find the solutions $(a, b, c)$ that satisfy $a>b>c$ and $(a+b+c)^{2}=a^{3}+b^{3}+c^{3}$. | 7. (1) Answer: $(a, b)=(2,1)$.
Hint: $a^{3}+b^{3}-(a+b)^{2}=(a+b)$
$[a(a-b-1)+b(b-1)] \geqslant 0$, thus $a^{3}$
$+b^{3} \geqslant(a+b)^{2}$, equality holds if and only if $a-b-1=0$,
$b-1=0$, so $(a, b)=(2,1)$.
(2) Answer: $(a, b, c)=(3,2,1)$.
Hint: $a^{3}+b^{3}+c^{3}-(a+b+c)^{2}$ $=a\left[(a-3)^{2}+5(a-b-1)+3(b-c-\rig... | (a, b, c)=(3,2,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,114 |
8. (1) Given that $a, b, c$ are real numbers, and $a>0$,
$$
b>0, c \neq 0, \sqrt{a+b}=\sqrt{a+c}
$$
$+\sqrt{b+c}$. Prove: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$.
(2) Find the integers $a, b$ that satisfy $\sqrt{a+b}=\sqrt{a-3}+\sqrt{b-3}$. | 8. (1) Hint: Squaring both sides twice, simplify to get $a b+b c+c a=0$, dividing both sides by $a b c$ completes the proof.
(2) Answer: $(a, b)=(4,12),(6,6)$, $(12,4)$.
Hint: In (1), let $c=-3$, which is (2), use the result from (1). | (a, b)=(4,12),(6,6),(12,4) | Algebra | proof | Yes | Yes | cn_contest | false | 703,115 |
Side 3. Find the general integer solution of the equation $41 x+177 y=4$.
untranslated text retained as requested:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Solution:
$3\left|\begin{array}{rr}41 & 177 \\ 39 & 164 \\ \hline 2 & 13 \\ 2 & 12 \\ \hline 0 & 1\end{array}\right| 6$
It is clear that $177=4 \times 41+13,13=177-4 \times 41$,
$$
\begin{array}{ll}
41=3 \times 13+2, & 2=41-3 \times 13, \\
13=6 \times 2+1, & 1=13-6 \times 2 .
\end{array}
$$
To express 1 using 41 and ... | x=-328+177 t, \\ y=76-41 t | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,116 |
9. Given that $x, y, z$ are integers from 1 to 9, and $x>y>z$, find the integer solutions $(x, y, z)$ for the equation $3(x-y)=4(y-z)$. | 9. Answer: $(x, y, z)=(8,4,1),(9,5,2)$
Hint $\because(3,4)=1, \quad \therefore x-y(>0)$ is a multiple of 4. Similarly, $y-z(>0)$ is a multiple of 3.
$$
\begin{array}{l}
\text { Also } \because 9 \geqslant x>y>z \geqslant 1, \quad \therefore x-y=4, \\
y-z=6,3 .
\end{array}
$$ | (8,4,1),(9,5,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,117 |
10. Let $ab$ be a two-digit number in decimal notation, and let $ba$ be the two-digit number formed by swapping the unit and tens digits, the product of these two two-digit numbers is 4930, find the values of $a$ and $b$. | 10. Answer: $(a, b)=(5,8),(8,5)$. Hint: $(10 a+b)(10 b+a)=4930$. | (a, b)=(5,8),(8,5) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,118 |
Example 1. On a plane, there are fixed points $A, B$ and any four points $P_{1}, P_{2}, P_{3}, P_{4}$. Prove: Among these four points, there must be two points $P_{i}, P_{j},(i, j=1,2,3,4, i \neq j)$, such that $\left|\sin A P_{i} B-\sin A P_{j} B\right| \leqslant \frac{1}{3}$. | [Hint] Divide $[0,1]$ into three equal subintervals. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,119 |
Example 2. Prove that there exist integers $a, b$, $c$, not all zero, and each with absolute value less than 1,000,000, such that $|a + b \sqrt{2} + c \sqrt{3}| < 10^{-11}$.
(1980 Putnam Competition Problem) | 〔Prompt〕Let $S=\{r+s \sqrt{2}+t \sqrt{3} \mid \cdot$ $\left.r, s, i<10^{\circ}, r, s, t \in \bar{Z}^{-}\right\}$, then the elements of $S$ are $10^{18}$.
Let $d=(1+\sqrt{2}+\sqrt{3}) \cdot 10^{\circ}$, divide the interval $[0, d)$ into $10^{18}-1$ subintervals, each with a length of $\frac{d}{10^{18}-1}$: thus, there ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,120 |
Example 3. For any irrational number $a$ and positive integer $Q$, there must exist a rational number $\frac{n}{m}$ (where $0 < m \leqslant Q$), such that $\left\lvert\, \alpha-\frac{n}{m} \right\rvert < \frac{1}{m Q}$. | 〔Hint〕Assume $0<a<1$, divide the interval $(0,1)$ into $Q$ equal parts, and consider $Q+1$ fractional parts
$$
\{0\},\{a\},\{2 a\}, \cdots,\{Q a\},
$$
then there must be two numbers, say $\{k a\},\{h a\}$, in the same interval. Let $k \leqslant h$, then $0 \leqslant\{k a\}-\{h a\}<\frac{1}{Q}$.
$$
\text { Let } \begin{... | \left|\alpha-\frac{n}{m}\right|<\frac{1}{m Q} | Number Theory | proof | Yes | Yes | cn_contest | false | 703,121 |
Example 4. Among 18 people, there must be four people who either all know each other or are all strangers to each other. | This problem is equivalent to the 18th order two-color complete graph け, there must be a monochromatic quadrilateral.
[Hint] Consider the 17 lines drawn from a point $O$, then at least 9 lines are of the same color, for example, 9 red lines, denoted as $O V_{1}, O V_{2}, \cdots, O V_{8}$.
If there are at least four re... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,123 |
Example 1. A road has 6 bus stops. Now a bus is traveling from $a_{1}$ to $a_{8}$, and passengers can freely get on and off at each stop along the way, but this bus can carry at most 5 passengers at any time. Prove: Among these 6 stops, there must be two pairs (four different) stops $A_{1}, B_{1}; A_{2}, B_{23}$ such t... | [Hint] Let $C_{kl}$ denote the number of passengers getting on at station $a_k$ and getting off at station $a_l$. Then all $C_l$ can be listed in a table:
$$
\begin{array}{cc:ccc}
C_{12} & C_{13} & C_{14} & C_{15} & C_{18} \\
C_{23} & C_{24} & C_{25} & C_{26} \\
& C_{34} & C_{25} & C_{28} \\
\hdashline & & C_{45} & C_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,124 |
Side 3. In a 7-vertex, 2-colored complete graph, there are at least four monochromatic triangles. Generally, in an $n$-vertex, 2-colored complete graph, there are $f_n$ monochromatic triangles, then $f_n$ satisfies
$$
f_{n} \geqslant\left(\begin{array}{l}
\frac{2}{3} m(m-1)(m-2), \text { when } n=2 m; \\
\frac{2}{3} m(... | This can be obtained by finding the extremum of $f_{n}=C_{n}^{3}-\frac{1}{2} \sum_{i=1}^{n} x_{i}\left(n-1-x_{i}\right)$. | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,125 |
Example 3. There are 650 points inside a circle $C$ with a radius of 16. Prove: there exists an annulus with an inner radius of 2 and an outer radius of 3, which covers at least ten of these points. | [Hint] Draw a concentric circle $K$ of circle $C$ with a radius of 19, the area of $K$ is $19^{2} \pi$.
With the given points $\alpha_{i}(i=1,2, \cdots, 650)$ as centers, draw annuli with inner radius 2 and outer radius 3, each with an area of $\left(3^{2}-2^{2}\right) \pi=5 \pi$.
The total area of these 650 annuli is... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,126 |
Example 4. Find the general solution of the equation $12 x+8 y+36 z=100$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: For the equation $12 x+8 y+36 z=100$, dividing each term by the greatest common divisor 4 of the coefficients, we get
$$
3 x+2 y+9 z=25 \text {. }
$$
Divide equation (2) into two equations
$$
\begin{array}{l}
3 x+2 y=u \\
u+9 z=25 .
\end{array}
$$
For equation (3), it is not difficult to see that
$$
3 \cdot... | \left\{\begin{array}{l}
x=-2+2 t_{1}+9 t_{2}, \\
y=2-3 t_{1}-9 t_{2}, \\
z=3-t_{2} .
\end{array}\right.
} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,127 |
Example 1. Distribute 1600 peanuts among 100 monkeys, prove that no matter how they are distributed, at least 4 monkeys will get the same number of peanuts.
(Beijing 1962 Competition Question) | [Hint] If 3 get 0 peanuts, 3 get 1 peanut, $\cdots$, 3 get 32 peanuts, 1 gets 33 peanuts, then the minimum number of peanuts required is
$$
3 \times(1+2+\cdots+32)+33=1617>1600
$$ | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,130 |
Example 2. Color each point on the plane in black or white arbitrarily. Prove that there must exist an equilateral triangle with side length 1 or $\sqrt{3}$, whose three vertices are of the same color.
(1st National Mathematical Winter Camp (1986) Problem) | 〔Prompt〕First prove that there must exist a line segment of length 2 with endpoints of different colors in the plane, then construct a regular hexagon using this line segment as one of its diagonals. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,132 |
Example 3. There are $2 n$ people $(n>1)$ at a party, each of whom knows several of the other attendees. Prove that it is possible to select 4 people from them to sit in a circle, such that each person is seated next to their acquaintances. | [Hint] From $2 n$ people, two people $A, B$ sit opposite each other: leaving $2 n-2$ people. Since $A$, $B$ together know $2 a$ people, then among the $2 a-2$ people, there are at least 2 people who know both $A$ and $B$, let them be $C, D$, and have $C, D$ sit between $A, B$. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,133 |
Example 2 Factorize: $f(x)=x^{8}+x^{4}+1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve $f(x)$ has no linear factors (why?) Let's examine the quadratic factors. This polynomial is of degree 8, and using method 1 would be quite difficult. We adopt method 2. Suppose $g(x)$ is a quadratic factor. Then $g(0)$ divides $f(0)$ $=1, g( \pm 1)$ divides $f( \pm 1)=3$. From $g(1)$ $+g(-1)=2(1+g(0))$, we know o... | f(x)=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)\left(x^{4}-x^{2}+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,135 |
Factorize the expression
$$
x^{3}+x^{2}-a^{2} x+\left(a^{2}+a\right) \text {. }
$$ | Let $f(x)=x^{3}+x^{2}-a^{2} x+\left(a^{2}+a\right)$. This is a cubic polynomial, which needs to be factored into linear factors. Suppose $g(x)=x+p$ is such a factor (where $p$ may depend on $a$), then $g(0)=p, f(0)=a^{2}+a$, so $p$ must divide $a^{2}+a=a(a+1)$, and $g(1)=1+p$ must divide $f(1)=2+a$. Therefore, the poss... | (x+a+1)\left(x^{2}-a x+a\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,136 |
Example 4. Factorize $x y z\left(x^{3}+y^{3}+z^{3}\right)-x^{3} y^{3}$ $-y^{3} z^{3}-z^{3} x^{3}$. | Solve: The original expression can be seen as a polynomial in $x$
$$
\begin{aligned}
f(x)= & x^{4} y z-x^{3}\left(y^{3}+z^{3}\right)-x y z\left(y^{3}\right. \\
& \left.+z^{3}\right)-y^{3} z^{3} .
\end{aligned}
$$
If $f(x)$ has a linear factor $a x+b$, then $a$ divides $y z$, and $b$ divides $y^{3} z^{3}$. Therefore, w... | f(x)=\left(x^{2}-y z\right)\left(y^{2}-z x\right)\left(z^{2}-x y\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,137 |
Example 5. Find two sets of smaller positive integer solutions for the system of equations
$$
\left\{\begin{array}{l}
2 x+3 y-4 z=8, \\
5 x-7 y+6 z=7
\end{array}\right.
$$ | Eliminate $z$: (1) $\times 3+$ (2) $\times 2$, we get
$$
16 x-5 y=38 \text{. }
$$
Solve equation (3) for $y$, we get
$$
y=\frac{9 x-38}{5}=3 x-7+\frac{-3}{5}.
$$
Since $x, y$ in (4) are integers, we can let $\frac{x-3}{5}=k$ (an integer), i.e., $\square$
$$
x=5 k+3 \text{. }
$$
Substitute (5) into (4), we get
$$
y=16... | (3,2,1) \text{ and } (13,34,30) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,138 |
Example 5. Factorize: $a^{4}+b^{4}+c^{4}+1-2\left(a^{2}\right.$ $\left.+b^{2}+c^{2}+a^{2} b^{2}+c^{2} a^{2}\right)+8 a b c$. | Solve, treating as a polynomial in $a$:
$$
\begin{array}{l}
f(a)=a^{4}-2\left(b^{2}+c^{2}+1\right) a^{2}+8 b c a \\
+\left(b^{4}+c^{4}+1-2 b^{2}-2 c^{2}-2 b^{2} c^{2}\right),
\end{array}
$$
where the last term, treated as a polynomial in $b$, is $F(b)=b^{4}$ $-2\left(1+c^{2}\right) b^{2}+\left(c^{4}+1-2 c^{2}\right) \... | (a-b+c-1)(a+b-c-1)(a+b+c+1)(a-b-c+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,139 |
Example 1. ( $i M O \cdot 24$) Given that $A$ is an intersection point of two unequal circles $O_{1}$ and $O_{2}$, the external common tangent $P_{1} P_{2}$ touches the circles at $P_{1}$ and $P_{2}$, another external common tangent touches the circles at $Q_{1}$ and $Q_{2}$, and $M_{1}$ and $M_{2}$ are the midpoints o... | Extend the common chord $AB$, intersecting $P_{1} P_{2}$ at $T$. From
$$
T P_{1}^{2}=T A \times T B=T P_{2}^{2},
$$
it follows that $T$ is the midpoint of $P_{1} P_{2}$.
By symmetry (the entire figure is symmetric about $\mathrm{O}_{1} \mathrm{O}_{2}$), $M_{1}$ and $M_{2}$ both lie on the line segment $O_{1} O_{2}$, a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,140 |
Example 2. (IHO.19.1) Inside a known square $ABCD$, construct equilateral triangles $ABK, CDU, BCL, DAN$. Prove that the midpoints of the four segments $KL, LM, MN$, and $NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon. | When $B, C, D, A$ are in place, $K$ rotates to $L, M, N, K$, so it is called shape $K L \| N$ square, with $O$ as the center, and $D: \triangle A N$.
Let $D$ intersect $A N$ at: $P_{2}$, since $D M$ is the height of the equilateral triangle $D A N$, so $P_{2}$ is the midpoint of $A N$. Similarly, the intersection poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,141 |
Example 3. (IMO.17.3) On the sides of any $\triangle ABC$, construct $\triangle BPC$, $\triangle CQA$, and $\triangle ARB$ outwardly, such that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, and $\angle ABR = \angle BAR = 15^{\circ}$. Prove:
(1) $\angle QRP = 90^{\circ}$, (2) $QR = RP$. | Translate the given text into English, maintaining the original line breaks and formatting:
Rotate point $B$ around $R$ by $90^{\circ}$ to point $D$ (we hope to prove: $\triangle R B P \cong \triangle R D Q$, i.e., $P$ rotates around $R$ by $90^{\circ}$ and intersects at $Q$). Clearly, $\angle A R D=\angle A R B-90^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,142 |
4. (IMO.1.5) On the line segment $AB$, take a point $M$, and construct squares $\triangle MCD$ and $MBEF$ on the same side of $AB$ with $AM$ and $MB$ as sides, respectively. The circumcircles of these two squares are $\odot P$ and $\odot Q$, and they intersect at $M$ and another point $N$. The line $AF$ intersects $BC$... | If $\triangle A M F$ rotates $90^{\circ}$ around point $M$, then $A$ becomes $C$, $F$ becomes $B$, and thus $A F$ becomes $B C$. This indicates that $B C$ is perpendicular to $A F$, hence $\angle A N^{\prime} C=90^{\circ}$. Therefore, $N^{\prime}$ lies on the circle, and thus $N^{\prime}$ coincides with $N$.
Since $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,143 |
Example 5. (IMO .22.5) Three congruent circles have a common point $O$, and are all inside a given triangle $ABC$, with each circle touching two sides of $\triangle ABC$. Prove that the incenter, circumcenter, and point $O$ of the triangle are collinear.
The translation maintains the original text's line breaks and fo... | Let the centers of three circles be $A', B', C'$. Since $AB$ is the common tangent of $\bigcirc A'$ and $\bigcirc B'$, the distances from $A'$ and $B'$ to $AB$ are equal, both being the radius, hence $A'B' \parallel AB$. Similarly, $B'C' \parallel BC$, $C'A' \parallel CA$.
Since $AB$ and $AC$ are tangents to $\bigcirc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,144 |
The 27th International Mathematical Olympiad problem 1 can be stated as follows:
Let $d$ be a positive integer not equal to $2, 32n^2 - 40n + 13, 32n^2 - 24n + 5$ (where $n$ is a positive integer). Prove:
In the set $\left\{2, 32n^2 - 40n + 13, 32n^2 - 24n + 5\right\}$, there exist two distinct elements $a, b$ such t... | To facilitate the discussion, let's first prove two lemmas.
Lemma 1: If and only if $d=2 k^{2}-2 k+1$ (where $k$ is a positive integer), then $2 d-1$ is a perfect square.
$$
\text{Proof: Let } d=2 k^{2}-2 k+1 \text{ (where } k \text{ is a positive integer)}
$$
then,
$$
2 d-1=4 k^{2}-4 k+1=(2 k-1)^{2},
$$
hence, $2 d-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,145 |
In the candidate problems of the 22nd International Mathematical Olympiad held in the United States in 1981, there was such a problem: "Given a square $\triangle B C D$, for which positive integers $n$ can this square always be divided into $n$ non-overlapping smaller squares (these smaller squares can be of different ... | (Divide and Increase Equally)
Divide the adjacent sides $AB, AD$ of a square into $n$ equal parts. $1^{\circ}$ When $n=1$, no division is needed, and it remains the original square.
$2^{\circ}$ When $n \geqslant 2$, the division points are $B_{1}, B_{2}, \cdots, B_{n-1}; D_{1}, D_{2}, \cdots, D_{n-1}$. As shown in Figu... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,146 |
Example 1. Indicate one method for each of the following to divide a square with a side length of 1986 mm:
(1) Divide it into 1986 smaller squares with integer side lengths;
(2) Divide it into 1986 smaller squares with not all side lengths being integers. | (1) $\because 1986=2 \times 993$, divide the two sides $A B$ and $A D$ of the square into 993 equal parts (each part being $2 \text{ mm}$). As shown in Figure 5, using the discussion from Proof 1, we get: $2 \times 993-1=1985$ small squares with a side length of $2 \text{ mm}$, and one square $F C E E_{1}$ with a side ... | 1986 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,147 |
Given an equilateral triangle $A B C$, prove that when $n$ is a natural number not equal to $2, 3, 5$, the equilateral triangle $A B C$ can be divided into $n$ smaller equilateral triangles (these smaller equilateral triangles can be of different sizes). | Proof 1 (Equal Division of Side Length Method)
$1^{\circ}$ Divide the sides $AB, AC$ of the equilateral triangle into $n$ equal parts. When $n=1$, it is the original triangle. When $n \geq 2$, the division points are: $B_{1}, B_{2}, \cdots, B_{n-1}; C_{1}, C_{2}, \cdots, C_{n-1}$. Draw $C_{1} D \parallel AB$ intersecti... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,148 |
Example 6. Given a three-digit integer that is a multiple of 5, the sum of its digits is 20, and the sum of one digit and the hundreds digit is a multiple of 3, find this integer. | Solution: Let the required three-digit integer be
$$
N=100 x+10 y+z \text {. }
$$
Since $N$ is a multiple of 5, it must be that $z=0$ or $z=5$. It is also given that the sum of the digits of $N$ is 20, so
$$
x+y+z=20 \text { . }
$$
If $z=0$, then $x+y=20$.
Given $1 \leqslant x \leqslant 9,0 \leqslant y \leqslant 9$, ... | 785 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,149 |
Example 3. Prove that an equilateral triangle with side length $a$ can definitely be divided into $n^{2}$ congruent smaller equilateral triangles. $(n$ is a natural number) | Proof: As shown in Figure 11, divide the two sides $AB$ and $AC$ of an equilateral triangle into $n$ equal parts, obtaining the points $B_{1}, B_{2}, \cdots, B_{n-1}$ and $C_{1}, C_{2}, \cdots, C_{n-1}$. Connect $B_{1} C_{i} [i=1,2, \cdots, (n-1)]$ and so on. It is known that $\left|B_{i} C_{i}\right| = \frac{i}{n} a$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,150 |
In junior high school plane geometry, there is such a problem: "There is a point $P$ inside the acute angle $\angle X O Y$. Try to find points $Q$ and $R$ on $O X$ and $O Y$ respectively, such that the perimeter of $\triangle P Q R$ is the shortest." This problem can be solved using geometric construction methods. | Proof: Let $Q^{\prime} \backslash R^{\prime}$ be any points on $O X$ and $O Y$ respectively, such that at least one of $Q^{\prime}=Q$ and $R^{\prime}=R$ does not hold. Connect $P Q$, $P R$, $P Q^{\prime}$, $P R^{\prime}$, and $Q^{\prime} R^{\prime}$ to form $\triangle P Q R$ and $\angle P Q^{\prime} R^{\prime}$. Then c... | 2 a \sin \angle X O Y | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,151 |
Example 1. Let $\sim$, $w$, $\lambda$ be complex numbers, and $|\lambda| \neq 1$, solve the equation $\bar{z} - \lambda z = w$ for $z$. | Analysis: We consider the equation $\bar{z}-\lambda z=w$ as a linear equation in two variables $z$ and $\bar{z}$. Therefore, to find $z$, we must construct a new equation involving $z$ and $\bar{z}$.
Solution $\because \bar{z}-\lambda z=w, \therefore(\bar{z}-\lambda z)=\bar{w}$, i.e., $z-\bar{\lambda} \bar{z}=\bar{w}$... | z=\frac{\bar{\lambda} w+w}{1-|\lambda|^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,152 |
Example 2. Prove: Any function $f(x)$ in the symmetric interval $(-l, l)$ can be expressed as the sum of an even function and an odd function. | Analysis: Let $f(1)=G(x)+H(x)$ (where $G(x)$ is an even function, and $H(x)$ is an odd function). We can view this equation as a linear equation in two variables, $G(x)$ and $H(x)$. To find $G(x)$ and $H(x)$, we only need to construct a linear equation involving $G(x)$ and $H(x)$.
Proof: Let $f(x)=G(x)+H(x)$ (where $G... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,153 |
Example 4. Prove: $\sqrt{3}-\sqrt{2}$ is an irrational number. | Proof: Let $x = \sqrt{3} - \sqrt{2}$, then $x + \sqrt{2} = \sqrt{3}$,
$$
(x + \sqrt{2})^2 = 3, \quad x^2 + 2 \sqrt{2} x + 2 = 3,
$$
which simplifies to $1 - x^2 = 2 \sqrt{2} x$.
$$
(1 - x^2)^2 = (2 \sqrt{2} x)^2,
$$
simplifying further, we get $x^4 - 10 x^2 + 1 = 0$.
The possible rational roots of this equation are $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,155 |
Given point $P$ is on the circumcircle of equilateral triangle $ABC$, prove:
(1) $AP = BP + CP$;
(2) $AP^2 = PB - PC + AB^2$. | In $\triangle A P C \|$, by the cosine theorem, we get $A P^{2}+P C^{2}-2 A P \cdot P C \cos 60^{\circ}=A C^{2}$, thus $P C^{2}-A P \cdot P C+A P^{2}-A B^{2}=0$.
Similarly, in $\triangle A P B \|$, $P B^{2}-A P \cdot P B + A P^{2}-A B^{2}=0$.
By the definition of roots, $P B, P C$ are the roots of the equation $x^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,156 |
Example 7. If $a+b+c=1, \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, then at least one of $a, b, c$ is equal to 1. | From the known, $a+b+c=1$, $abc=ab+ac+bc=t$. By Vieta's theorem, $a, b, c$ satisfy the equation $x^{3}-x^{2}+tx-t=0$ as its three roots. Observing, we find that $x=1$ is a root of this equation,
$\therefore a, b, c$ at least one of them equals 1, | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,158 |
Example 8. In $\triangle A B C$, it is known that (1) $\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C = -\frac{1}{6}$,
(2) $\operatorname{tg}^{3} A + \operatorname{tg}^{3} B + \operatorname{tg}^{3} C = -\frac{181}{216}$. Find the sizes of $A, B, C$. | $$
\begin{array}{c}
\text { Sol } \because A, B, C \text { are the three interior angles of } \triangle A B C, \\
\therefore \operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C=\operatorname{tg} A \operatorname{tg} B \operatorname{tg} C . \\
\because \operatorname{tg}^{3} A+\operatorname{tg}^{3} B+\operatorname... | \frac{3 \pi}{4}, \operatorname{arctg} \frac{1}{2}, \operatorname{arctg} \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,159 |
1. Find all integer solutions to the equation:
$$
x^{2}+x y+2 y^{2}=29
$$ | Solving equation (1) for $x$, we get
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$
You get
$$
x=-3 \pm \sqrt{\frac{116-7 y^{2}}{2}} .
$$
Since $x$ is a real number, it must be that
$$
\begin{array}{l}
116-7 y^{2} \geqslant 0, \\
1 \quad \left\lvert\, \leqslant \sqrt{\frac{116}{7}}=4.07 \cdots .\right.
\end{arr... | (1,-4), (3,-4), (-1,4), (-3,4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,160 |
Example 9. Given $(c-a)^{2}-4(a-b)(b-c)$ $=0$. Prove that $a, b, c$ form an arithmetic sequence. | Prove that because the discriminant $\Delta=0$ of the equation $(a-b) x^{2}+(c-a) x$ $+(b-c)=0$, its roots are equal.
$$
=0, \quad \because(a-b) \times 1^{2}+(c-a) \times 1+(b-c)
$$
$\therefore$ the two roots $x_{1}=x_{2}=1$.
By Vieta's formulas, we get $a-c=1, b-c=a-b$, i.e., $a, b, c$ form an arithmetic sequence. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,161 |
Example 10. Find the value of $\cos \frac{2 \pi}{7}+\cos -\frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$. | Let $7 \theta=2 n \pi(n=1,2,3)$, then $4 \theta$ $=2 n \pi-3 \theta, \cos 4 \theta=\cos 3 \theta, 2 \cos ^{2} 2 \theta-1$ $=4 \cos ^{3} \theta-3 \cos \theta, 2\left(2 \cos ^{2} \theta-1\right)^{2}-1$ $=4 \cos ^{3} \theta-3 \cos \theta$. Simplifying, we get $8 \cos ^{4} \theta-4 \cos ^{3} \theta$ $-8 \cos ^{2} \theta+3 ... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,162 |
Example 11. Prove: $\cos \frac{2 \pi}{2 n+1}+\cos \frac{4 \pi}{2 n+1}+$ $\cdots+\cos \frac{2 n \pi}{2 n+1}=-\frac{1}{2}$. | $$
\begin{array}{c}
\text { Analysis: The original expression is transformed to } \\
1+2\left(\cos \frac{2 \pi}{2 n+1}+\cos \frac{4 \pi}{2 n+1}+\cdots\right. \\
\left.+\cos \frac{2 n \pi}{2 n+1}\right)=0 . \\
\because \quad \cos \frac{2 \pi}{2 n+1}=\cos \frac{4 n \pi}{2 n+1}, \cos \\
\frac{4 \pi}{2 n+1}=\cos \frac{(4 n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,163 |
Example 12. Prove $\left|\begin{array}{lll}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|=0$. | Prove that the determinant of the system of homogeneous equations
$$
\text { coefficient }\left\{\begin{array}{l}
x \sin \alpha+y \cos \alpha+z \sin (\alpha+\delta)=0, \\
x \sin \beta+y \cos \beta+z \sin (\beta+\delta)=0, \\
x \sin \gamma+y \cos \gamma+z \sin (\gamma+\delta)=0 .
\end{array}\right.
$$
$\because$ It has ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,164 |
4. As shown in Figure 4: $A B C D$ and $C E F G$ are both squares, then among the following equations:
(1) $\angle 1+\angle 8=130^{\circ}$;
(2) $\angle 2+\angle 8=180^{\circ}$;
(3) $\angle 1+\angle 5=180^{\circ}$;
(4) $\angle 3+\angle 5=180^{\circ}$, the incorrect one(s) is/are ( ).
(A) only (1); (B) only (2);
(C) only... | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,168 |
5. In Figure 5, $ABCD$ is a square, $DE = EC, \angle CDE = 66^{\circ}$, the following relationships:
(1) $\angle 1: \angle 4 = 4: 1$;
(2) $\angle 1: \angle 5 = 1: 1$;
(3) $(\angle 1 + \angle 2) : (\angle 3 + \angle 4) = 5: 3$
(A) All correct;
(B) Only (1);
(C) Only (2),
(D) Only (1) and (2),
(E) Only (2) and (3). | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,169 |
6. As shown in Figure 6, $A B \perp B C, E D \perp B C$ $\angle E A B=\angle E B A$, among the following equations:
(1) $A E=C E$;
(2) $E D=\frac{1}{2} A B$;
(3) $2 B E^{2}=A B^{2}+B C^{2}$;
(4) $\Delta C^{2}=4\left(E D^{2}+B D^{2}\right)$
which always hold true are ( ).
(A) All hold true;
(B) Only (1) and (2;
(C) Onl... | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,170 |
Example 2. Find the integer solutions of the equation
$$
2 x y+2 x z-x y z=4 z
$$
for the unknowns $x, y, z$ satisfying the conditions $x \geq 1, y=3, z=3$. | Factorizing the left side of equation (1), we get
$$
x(2 y+2 z-y z)=4 z .
$$
Since $x \geqslant 1, z \geqslant 3$, it follows that $2 y+2 z-y z>0$, i.e., $\square$
$$
y z-2 y-2 z<0 \text {. }
$$
(3) can be transformed into $(y-2)(z-2)<4$.
Since $y \geqslant 3, z \geqslant 3$, we know that $y-2 \geqslant 1, z-2 \geqsl... | (x, y, z)=(4,3,3),(6,4,3),(8,3,4),(12,5,3),(20,3,5) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,171 |
8. In circle $8, P Q // S R, Q S$ bisects $\angle P Q R, R P$ bisects $\angle Q R S$: Among the following conclusions:
(1) $P R$ bisects $\angle S P Q$;
(2) $P R = S Q$;
(3) The area of $P Q R S$
$$
=\frac{1}{2} P P \cdot S Q
$$
which of the following is incorrect?
( )
(A) All are incorrect;
(B) Only (1);
(C) Only (2)... | $\mathrm{C}$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,173 |
9. As shown in Figure 9, $ABCD$ is a square: $AE \parallel BF$, $CF \parallel BD$. Among the following statements:
(1) $\triangle AED \cong \triangle BFC$;
(2) $EF \perp BC_{3}$
(3) $EF = AB$ are correct ( ).
(A) All correct;
(B) (1) and (2);
(C) (2) and (3);
(D) Only (3);
(E) Only (1).
保留源文本的换行和格式如下：
9. As shown in ... | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,174 |
11. As shown in the figure, $\angle A=\angle R, \angle B=\angle Q$, $\angle C=\angle P, P Q=6, A B=4.5, A C=3$, $B C=4$, among the following three relations:
(1) $\frac{S_{\triangle A B O}}{S_{\triangle P Q R}}=\frac{4}{9}$;
(2) $R P=4.5$;
(3) $Q R=6.5$ are incorrect ( ).
(A) All are incorrect;
(B) Only (1);
(C) Only (... | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,176 |
12. In Rt $\triangle A B C$, the right-angle side $A B=6$, $A C=8$, the following three conclusions:
(1) The distance from the circumcenter to $A B$ is 1;
(2) The distance from the orthocenter to $B C$ is 4.8;
(3) The distance from the incenter to the circumcenter is $1 \frac{2}{3}$ ( ).
(A) Only (1);
(B) (1) and (2);
... | $\mathrm{D}$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,177 |
13. Given an equilateral triangle $A B C$ with $B C$ as the diameter, a circle is drawn intersecting $A B$ and $A C$ at $D$ and $E$ respectively. If $B C=a$, which of the following three statements are correct:
(1) $C D=\frac{1}{2} \sqrt{3} a$;
(2) $\overparen{B D}=\overparen{D E}=\overparen{E C}$,
(3) $S_{\triangle A ... | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,178 |
14. As shown in Figure 12, $\triangle A B C$ is inscribed in $\odot O, A E$ bisects $\angle B A C, A D \_B C$, among the following four conclusions:
(1) $O E \perp B C$;
(2) $O E$ bisects $B C$;
(3) $A E$ bisects $\angle D A O$;
(4) $\frac{A B}{A \bar{D}}=\frac{2 A O}{A C}$
which of the following are always true?
( ).... | $\mathrm{E}$ | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,179 |
16. As shown in Figure 14,
$\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ intersect
at points $A$ and $B$, $\bigcirc \mathrm{O}_{2}$
passes through $\odot O_{1}$, and $A C$ and $A D$
are diameters of $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$
respectively. Then the following conclusions:
(1) $C, B, D$ are col... | $\mathrm{E}$ | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,181 |
19. As shown in Figure 16, the radius of $\odot O$ is $1, PO=2$, and $A, B$ are the points of tangency. Among the following three conclusions:
(1) $PA=\sqrt{3}$, (2) $\angle APB=60^{\circ}$,
(3) The area of the shaded region is $\sqrt{3}-\frac{1}{3} \pi$. The correct one(s) is/are ( ).
(A) Only (1);
(B) Only (2);
(C) (... | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,185 |
2. $\odot A$ and $\odot B$ intersect at $P$ and $Q$, $X P Y \parallel A B$, $M$ and $N$ are the midpoints of $X P$ and $P Y$ respectively (see Figure 17). Among the following statements:
(1) $A M \perp P X$;
(2) $A B N M$ is a parallelogram;
(3) $X Y = 2 A B$.
Which of the following is correct? ( ).
(A) Only (1),
(B) ... | $\mathrm{E}$ | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,186 |
$\begin{array}{c}-2^{2}+4^{2}+6^{2}+\cdots+(2 n)^{2} \\ =\frac{2}{3} n(n+1)(2 n+1) .\end{array}$
The translation is as follows:
$\begin{array}{c}-2^{2}+4^{2}+6^{2}+\cdots+(2 n)^{2} \\ =\frac{2}{3} n(n+1)(2 n+1) .\end{array}$
This equation represents the sum of the squares of even numbers from $-2^2$ to $(2n)^2$, whi... | \begin{array}{l}\text { Prove: } 2^{2}+4^{2}+6^{2}+\cdots+(2 n)^{2} \\ =2^{2} \cdot 1^{2}+2^{2} \cdot 2^{2}+2^{2} \cdot 3^{2}+\cdots \\ +2^{2} \cdot n^{2} \\ =2^{2}\left(1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right) \\ =4 \cdot \frac{1}{6} n(n+1)(2 n+1) \\ =\frac{2}{3} n(n+1)(2 n+1) \text {. } \\\end{array} | \frac{2}{3} n(n+1)(2 n+1) | Algebra | proof | Yes | Yes | cn_contest | false | 703,187 |
\begin{aligned}=1^{2} & +3^{2}+5^{2}+\cdots+(2 n-1)^{2} \\ & =\frac{1}{3} n\left(4 n^{2}-1\right)\end{aligned} | Prove: $\begin{aligned} & 1^{2}+3^{2}+5^{2}+\cdots+(2 n-1)^{2} \\ = & {\left[1^{2}+2^{2}+\cdots+n^{2}+\cdots+(2 n)^{2}\right] } \\ & -\left[2^{2}+4^{2}+\cdots+(2 n)^{2}\right] \\ = & \frac{1}{6} \cdot 2 n(2 n+1)(2 \cdot 2 n+1) \\ & -\frac{2}{3} n(n+1)(2 n+1) \\ = & \frac{1}{3} n\left(4 n^{2}-1\right) .\end{aligned}$ | \frac{1}{3} n\left(4 n^{2}-1\right) | Algebra | proof | Yes | Yes | cn_contest | false | 703,188 |
\begin{aligned} \text { } & 1^{3}+2^{3}+3^{3}+\cdots+n^{3} \\ & =\left[\frac{n(n+1)}{2}\right]^{2} .\end{aligned} | Prove: Transforming $(n+1)^{4}=n^{4}+4 n^{3}+6 n^{2}+4 n+1$, we get
$$
(n+1)^{4}-n^{4}=4 n^{3}+6 n^{2}+4 n+1 \text {. }
$$
When $n$ takes $1,2,3, \cdots, n$, we get
$$
\begin{array}{l}
2^{4}-1^{4}=4 \cdot 1^{3}+6 \cdot 1^{2}+4 \cdot 1+1 \\
3^{4}-2^{4}=4 \cdot 2^{3}+6 \cdot 2^{2}+4 \cdot 2+1 \\
4^{4}-3^{4}=4 \cdot 3^{3... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,189 |
$\begin{array}{l}\text { Four, } 2^{3}+4^{3}+6^{3}+\cdots+(2 n)^{3} \\ =2 n^{2}(n+1)^{2} \text {. } \\\end{array}$ | Prove: $\begin{aligned} & 2^{3}+4^{3}+6^{3}+\cdots+(2 n)^{3} \\ = & 2^{3} \cdot 1^{3}+2^{3} \cdot 2^{3}+2^{3} \cdot 3^{3}+\cdots \\ & +2^{3} \cdot n^{3} \\ = & 2^{3}\left(1^{3}+2^{3}+3^{3}+\cdots+n^{3}\right) \\ = & 8 \cdot\left[\frac{n(n+1)}{2}\right]^{2} \\ = & 2 n^{2}(n+1)^{2} .\end{aligned}$
Note: There is a corre... | 2 n^{2}(n+1)^{2} | Algebra | proof | Yes | Yes | cn_contest | false | 703,190 |
$\begin{array}{l}\text { Five, } 1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3} \\ =n^{2}\left(2 n^{2}-1\right) \text {. } \\\end{array}$ | $$
\begin{aligned}
: & 1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3} \\
= & {\left[1^{3}+2^{3}+\cdots+n^{3}+\cdots+(2 n)^{3}\right] } \\
& -\left[2^{3}+4^{3}+\cdots+(2 n)^{3}\right] \\
= & {\left[\frac{2 n(2 n+1)}{2}\right]^{2}-2 n^{2}(n+1)^{2} } \\
= & n^{2}\left(2 n^{2}-1\right) .
\end{aligned}
$$ | n^{2}\left(2 n^{2}-1\right) | Algebra | proof | Yes | Yes | cn_contest | false | 703,191 |
$\begin{array}{l}\text { Six, } 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1) \\ =\frac{1}{3} n(n+1)(n+2)\end{array}$ | Proof: [Method 1]
$$
\because a_{n}=n(n+1)=n^{2}+n .
$$
$\therefore$ When $n$ takes $1,2,3, \cdots, n$, we get
$$
\begin{array}{l}
1 \cdot 2=1^{2}+1, \\
2 \cdot 3=2^{2}+2, \\
3 \cdot 4=3^{2}+3, \\
\cdots \cdots \\
n(n+1)=n^{2}+n .
\end{array}
$$
Adding the $n$ equations, we get
$$
\begin{array}{l}
1 \cdot 2+2 \cdot 3... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,192 |
Example 1. Find the solution $(x, y)$ that satisfies the equation $\frac{1}{x}-\frac{1}{y}=\frac{1}{12}$, and makes $y$ the largest positive integer. | Solving the original equation for $y$, we get
$$
y=\frac{12 x}{12-x}=-12+\frac{144}{12-x}.
$$
From this equation, it is clear that for $y$ to take its maximum value, $12-x$ must be positive and as small as possible. Also, $12-x$ should be a divisor of 144, so $12-x=1, x=11$, at which point $y=132$. Therefore, the posi... | (11, 132) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,193 |
\begin{aligned} \text { t,} & 1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+\cdots+n(n+1) \\ & \cdot(n+2) \\ & =\frac{1}{4} n(n+1)(n+2)(n+3)\end{aligned} | Using the general formula $a_{n}=n(n+1)(n+2)=n^{3}+3 n^{2}+2 n$, we can prove by analogy with formula six. | \frac{1}{4} n(n+1)(n+2)(n+3) | Algebra | proof | Yes | Yes | cn_contest | false | 703,194 |
\begin{array}{l}\text { Eight, } 1 \cdot 3 \cdot 5+3 \cdot 5 \cdot 7+\cdots+(2 n-1)(2 n \\ \quad+1)(2 n+3)=n\left(2 n^{3}+8 n^{2}+7 n-2\right) .\end{array} | Using the general formula $a_{n}=(2 n-1)(2 n+1)(2 n$ $+3)=8 n^{3}+12 n^{2}-2 n-3$, we can prove by analogy with formula six. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,195 |
\begin{aligned} \text { Nine. } & \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} \\ & =1-\frac{1}{n+1} .\end{aligned} | Proof: $\because a_{n}=\frac{1}{n(n+1)}=\frac{1}{n}$
$$
\begin{aligned}
& -\frac{1}{n+1} \\
\therefore \quad & \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} \\
& =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}\right. \\
& \left.-\frac{1}... | 1-\frac{1}{n+1} | Algebra | proof | Yes | Yes | cn_contest | false | 703,196 |
\begin{aligned}+ & \frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\cdots \\ & +\frac{1}{n(n+1)(n+2)} \\ & =\frac{1}{2}\left[\frac{1}{1 \cdot 2}-\frac{1}{(n+1)(n+2)}\right]\end{aligned} | \begin{array}{l}\text { Proof: } \because a_{n}=\frac{1}{n(n+1)(n+2)} \\ =\frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right] \text {, } \\ \therefore \quad \frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\cdots \\ +\frac{1}{n(n+1)(n+2)} \\ =\frac{1}{2}\left(\frac{1}{1 \cdo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,197 |
$$
\begin{array}{l}
\text { Given: Point } P\left(x_{0}, y_{0}\right) \text {, line } l: A x+B y \\
+C=0 \quad\left(A^{2}+B^{2} \neq 0\right) .
\end{array}
$$
Find: The distance from point $P$ to line $l$. | Solution: Let $H(x, y)$ be any point on the line $l$.
$\because$ In the line equation $A x+B y+C=0$,
$\therefore A, B$ at least one is not zero. Without loss of generality, let $B \neq 0$,
$$
y=-\frac{A x+C}{B}.
$$
$\therefore$ The coordinates of point $H$ are $\left(x,-\frac{A x+C}{B}\right)$.
The distance from point ... | \frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,198 |
Example. The infinite sequence
$$
a_{1}, a_{2}, a_{3}, \ldots
$$
has the relation
$$
a_{n}=4+\frac{1}{3} u_{n-1}(n=2,3,1, \cdots)
$$
Find $\lim _{n \rightarrow \infty} a_{n}$. | $$
\begin{array}{l}
a_{2}=4+\frac{1}{3} a_{1}, \\
a_{3}=4+\frac{1}{3} a_{2}=4+\frac{1}{3}\left(4+\frac{1}{3} a_{1}\right) \\
=4+4 \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{4}=4+\frac{1}{3} a_{3} \\
=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{n}=4+4 \cdo... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,202 |
Lake 1. Given the arithmetic sequence with the first term $a_{1}=3, a_{n+1}=a_{n}+5$, find up to the 5th term. | Solution: From the recursive formula $a_{2}=a_{1}+5=3+5=8$,
$$
\begin{array}{l}
a_{3}=a_{2}+5=8+5=13, \\
a_{4}=a_{3}+5=13+5=18, \\
a_{5}=a_{4}+5=18+5=23 .
\end{array}
$$
From the recursive formula to find the general term, there are commonly four forms:
I $\cdot a_{n+1}=a_{n}+d$,
Arithmetic sequence type. | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,203 |
Example 1. Find the smallest positive integer $n$ that makes $\frac{n-13}{5 n+6}$ a non-zero reducible fraction. | Solution: Since $(1,5)=1,|1 \times 6-(-13) \times 5|$ $=71 \neq(1,5),(1,71)=1$, the fraction can be simplified by 71. From $\left.n-13=71 m_{1}, n=71 m_{1}+13\left(m_{1} \in Z\right)\right)$, and $n-13>0$, so we take $m_{1}=1$, obtaining the smallest $n=84$. | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,204 |
Example 9. In the figure, a circle intersects $\triangle A B C$ at $B C$, $C A$, $A B$ respectively at $A_{1}$ and $A_{2}, B_{1}$ and $B_{2}, C_{1}$ and $C_{2}$.
If the perpendiculars from points $A_{1}, B_{1}, C_{1}$ to $B C, A C$, $A B$ respectively intersect at one point, then the perpendiculars from $A_{2}, B_{2},... | Draw perpendiculars to sides $BC, AC, AB$ through $A_{1}, B_{1}, C_{1}$ respectively, denoted as $l_{1}, l_{2}, l_{3}$, and $l_{1}, l_{2}, l_{3}$ intersect at a point $P$. Draw perpendiculars to sides $BC, AC, AB$ through $A_{2}, B_{2}, C_{2}$ respectively, denoted as $l_{1}^{\prime}, l_{2}^{\prime}, l_{3}^{\prime}$, t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,206 |
Example 11. Prove: The orthocenter $H$, the centroid, and the circumcenter $O$ of a triangle are collinear, and $O M: M H=1: 2$.
untranslated text remains the same as it seems there is a typo or missing information in the original text. The term "重心足外心" is not clear, it might be intended to say "the centroid and the ... | Let $A_{1}=D_{1}, C_{1}$ be the midpoints of sides $BC$, $AC$, and $AB$ of $\triangle ABC$, respectively, and let $M$ be the centroid. Then,
$$
\begin{array}{l}
A_{1} \xrightarrow{H(M,-2)} \rightarrow A, B_{1} \xrightarrow{H(M,-2)} \rightarrow B, \\
C_{1} \xrightarrow{H(M,-2)} \rightarrow C,
\end{array}
$$
The median o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,208 |
Example 12. Three equal circles have a common point $O$, and are all inside a given triangle, with each circle tangent to two sides of the triangle.
Try to prove: the incenter, circumcenter, and $O$ of this triangle are collinear. | Proof: Let the centers of two equal circles be $A_{1}, B_{1}, C_{1}$, $A B / / A_{1} B_{1}, \quad B C / / B_{1} C_{1}, \quad C A / / C_{1} A_{1}$,
$\therefore \triangle A B C$ is similar to $\triangle A_{1} B_{1} C_{1}$, $A A_{1}$, $B B_{1}, C C_{1}$ are respectively the angle bisectors of $\triangle A B C$, therefore ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,209 |
Example 13. Circle $S$ is tangent to the equal legs $AB$ and $BC$ of isosceles $\triangle ABC$ at points $P$ and $K$, respectively, and is also externally tangent to the incircle of $\triangle ABC$. Prove that the midpoint $O_{1}$ of segment $PK$ and the incenter of $\triangle ABC$ coincide. | Proof: Let the center of circle $S$ be $O$.
Draw a tangent line through $D$ intersecting the extensions of $BA, BC$ at $A^{\prime}, C^{\prime}$ respectively, then $P K / / A C / / A^{\prime} C^{\prime}$.
Let $k=\frac{B A}{B A^{\prime}}$, then
$$
\begin{array}{c}
\triangle B A^{\prime} C^{\prime} H(B, k) \rightarrow \tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,210 |
Example 1: Prove: Using 15 rectangular tiles of size $4 \times 1$ and 1 rectangular tile of size $2 \times 2$, it is impossible to perfectly cover an $8 \times 8$ rectangular floor. 〔Second National Junior High School Mathematics Correspondence Contest〕 | Proof 1: As shown in Figure 2,
using a coloring pattern where the diagonals parallel to the secondary diagonal and spaced two cells apart are of the same color, and using two colors to color the entire grid.
Obviously, (1) there are 32 black cells and 32 white cells on the grid;
(2) no matter whether a 4 × 1 rectangul... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,211 |
Example 3. A convex $n$-sided polygon is divided into triangles by some diagonals, satisfying the following conditions: $1^{\circ}$ The number of diagonals emanating from each vertex is even; $2^{\circ}$ No two diagonals have any common points other than the vertices. Prove: Then $n$ is a multiple of 3. | Ze, Mao Xian, we point out, it can be proved by mathematical induction: “If a plane figure $F$ is divided into $r$ parts by a straight line, then these parts can be colored with two colors, so that no two adjacent parts are colored the same color.”
Because for a known $n$-sided polygon divided by some diagonals into s... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,212 |
Example 4. In space, there are six points, any three of which form the vertices of a scalene triangle. Prove that one of these triangles has its longest side simultaneously as the longest side of another triangle. | Proof Let $P_{1}, P_{2}, \cdots, P_{8}$ be six known points in space. In each triangle $P_{j} P_{j} P_{k}$, we color the shortest side red, and the rest of the sides remain uncolored.
From each point $P$, the 5 segments connecting it to the other known points are still uncolored.
(1) If at least three of the five segm... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,213 |
Example 5. Let $a_{1}, a_{2}, a_{3}, \cdots$ be a non-decreasing sequence of positive integers. For $m=1$, define $b_{m}=\min \{n: a_{n} \geqslant m\}$, i.e., $b_{m}$ is the smallest $n$ such that $a_{n} \geqslant m$. If $a_{q}=p$, where $p$ and $q$ are integers, prove:
$$
a_{1}+a_{2}+\cdots+a_{q}+b_{1}+b_{2}+\cdots+b_... | Analysis: Let's take the case of $a_{3}=5$, and $a_{1}=1, a_{2}=3, a_{3}=5$ as an example for analysis.
At this time,
$\because a_{1} \geqslant 1, \quad \therefore b_{1}=1$;
$\because a_{1}<2, a_{2} \geqslant 2, \therefore l_{2}=2$; similarly, $b_{3}=2$;
$\because a_{2}<4, a_{3} \geqslant 4, \therefore b_{4}=3$; simila... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,214 |
Example 1. The four vertices of a convex quadrilateral $P Q R S$ lie on the four sides of a square $A B C D$ with side length $a$. Prove that the perimeter of $P Q R S$ is not less than $2 \sqrt{2} a$ | $$
\begin{array}{l}
A B C D \xrightarrow{S(B C)} A_{1} B C D_{1} \text {, } \\
P Q R S \xrightarrow{S(B C)} P_{1} Q R_{1} S_{1}, \\
A_{1} B C D_{1} \xrightarrow{S\left(C D_{1}\right)} A_{2} B_{2} C D_{1} \text {, } \\
P_{1} Q R_{1} S_{1} \xrightarrow{S\left(C D_{1}\right)} P_{2} Q_{2} R_{1} S_{2} \text {, } \\
A_{2} B_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,216 |
Example 3, Solve the equation: $\frac{4 x-5}{2 x-3}-\frac{15 x-23}{3 x-4}$
$$
=\frac{12 x-17}{6 x-10}-\frac{20 x-29}{4 x-5}
$$ | Solution: By the division algorithm, we get
$$
\begin{aligned}
& \left(2+\frac{1}{2 x-3}\right)-\left(5-\frac{3}{3 x-4}\right) \\
= & \left(2+\frac{3}{6 x-10}\right)-\left(5-\frac{4}{4 x-5}\right) .
\end{aligned}
$$
Simplifying:
$$
\frac{1}{2 x-3}+\frac{3}{3 x-4}=\frac{3}{3 x-10}+\frac{4}{4 x-5} .
$$
Solving it yield... | x=-\frac{10}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,217 |
Example 4. Find the extremum of the function $f(x)=\frac{x^{2}+x-6}{x^{2}+x+1}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | \begin{array}{l}\text { Solution: } f(x)=1-\frac{7}{x^{2}+x+1} \\ =1-\frac{1}{\left(x+\frac{1}{2}\right)^{2}+3}, \text { the minimum value is }-\frac{25}{3} .\end{array} | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,218 |
Example 1. Given $a^{2}+a-1=0$, find the value of $a^{3}+2 a^{2}+$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Applying long division, we get
$$
a^{3}+2 a^{2}+3=(a+1)\left(a^{2}+a-1\right)+4=4 .
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,219 |
For example, $2, n$ are natural numbers, prove that $3^{2 n+2}-8 n-9$ is a multiple of 4.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: When $n=1$, the proposition is obviously true. Applying long division:
$$
\begin{aligned}
& 3^{2}(k+1)+2-8(k+1)-9 \\
= & 3^{2 k+4}-8 k-17 \\
= & 9\left(3^{2 k+2}-8 k-9\right)+64(k+1) .
\end{aligned}
$$
Therefore, if the proposition holds for $n=k$, then it also holds for $n=k+1$. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,220 |
Example 2. Consider a parallelogram $O A B C$, where point $O$ is fixed at the origin, point $A$ moves on the line segment $x=1$, $y=t \quad(-1 \leqslant t \leqslant 1)$, and point $C$ moves on the unit circle. Find the equation of the trajectory of point $B$. | Solution 1: Let the complex numbers corresponding to points $A, B, C$ be $1+t i(-1 \leqslant t \leqslant 1), x+y i, \cos \theta +i \sin \theta . \quad(0 \leqslant 0 \leqslant 2 \pi)$
$\because O A B C$ is a parallelogram,
$$
\begin{array}{c}
\therefore \overrightarrow{O B}=\overrightarrow{O A}+ \\
\overrightarrow{A B}=... | (x-1)^{2}+(y-t)^{2}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,222 |
Example 3. Let $w, z$ be complex numbers, and the equation of $w$ is $w^{2}+z \cdot w$ $+z i=0$ which always has a real root $\alpha$ ( $\alpha$ is a real parameter). Try to find the locus of point $Z$.
Translate the above text into English, please keep the original text's line breaks and format, and output the transl... | Given that the equation of $w$ is $w^{2}+z \cdot w+z \boldsymbol{i}=0$ always has a real root $\ell$, hence
$$
a^{2}+r_{i} z+i=0 \text {. }
$$
Let $z=x+y i(x, y$ be real numbers $)$,
we get: $\left(a^{2}+a x-y\right)+i(a y+x)=0$.
From the equality of complex numbers, we have: $\left\{\begin{array}{l}x^{2}+a x-y=0, \\ ... | x^{2}=\frac{y^{3}}{1-y} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,223 |
3. As shown in the figure, $AB$ is the diameter of a semicircle, $O$ is the center of the circle, $CD \perp AB, DE \perp OC$. If the lengths of $AD, BD$, and $CD$ are all rational numbers, then
Proposition
甲、The length of $OE$ is a rational number;
乙、The length of $DE$ is a rational number;
丙、The lengths of all line ... | $3 .(D)$; | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,226 |
Example 2. Through the midpoint $A$ of the chord $BC$ of $\odot O$, draw two chords $PQ$, $FS$. Connect $PS$, $FQ$ intersecting $BC$ at $M$, $N$ respectively.
Prove: $AM = AN$.
保留了原文的换行和格式，以下是翻译结果：
```
Example 2. Through the midpoint $A$ of the chord $BC$ of $\odot O$, draw two chords $PQ$, $FS$. Connect $PS$, $FQ$ in... | $\begin{array}{l}\text { Prove } S \xrightarrow{S(O A)} K \text {, } \\ K \in \odot O \text {, } \\ \xrightarrow{S(O A)} A, \\ \therefore S A=A K \text {. } \\ \text { Also } A B \xrightarrow{S(O A)} A C \text {, } \\ \therefore \angle B A S=\angle C A K \text {. } \\ \text { But } \angle 3=\angle 1 \text {, } \\ \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,227 |
5. Given a quadrilateral $A B C D$ with a point $E$ inside it, connecting $A E, B E, C E, D E$ divides the quadrilateral $A B C D$ into four triangles of equal area. Then the proposition
甲、 $A B C D$ is a convex quadrilateral;
乙、 $E$ is the midpoint of diagonal $A C$ or diagonal $B D$;
丙、 $A B C D$ is a parallelogram
... | $5 .(B)$; | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,229 |
For what values of $a, b$ does the equation
$$
\begin{array}{l}
x^{2}+2(1+a) x+\left(3 a^{2}+4 a b+4 b^{2}\right. \\
+2)=0
\end{array}
$$
have real roots? | $$
\begin{aligned}
\Delta= & 4\left[(1+a)^{2}-\left(3 a^{2}+4 a b+4 b^{2}+2\right)\right] \\
= & 4\left(-1+2 a-2 a^{2}-4 a b-4 b^{2}\right) \\
= & -4\left[\left(1-2 a+a^{2}\right)\right. \\
& \left.+\left(a^{2}+4 a b+4 b^{2}\right)\right] \\
& -4\left[(1-a)^{2}+(a+2 b)^{2}\right] \geqslant 0 . \\
\because & -4\left[(1-... | a=1, b=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,236 |
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