problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3. Let $a, b, c$ be 3 distinct real numbers, and $P(x)$ be a polynomial with real coefficients. It is known that
(1) $P(x)$ divided by $(x-a)$ gives a remainder of $a$;
(2) $P(x)$ divided by $(x-b)$ gives a remainder of $b$,
(3) $P(x)$ divided by $(x-c)$ gives a remainder of $c$.
Find the remainder when $P(x)$ is divid... | Given that the remainder of $P(x)$ divided by $(x-a)$ is $a$, we can assume
$$
P(x)=(x-a) Q_{1}(x)+a .
$$
Thus,
$$
P(b)=(b-a) Q_{1}(b)+a=b .
$$
So $Q_{1}(b)=1, Q_{1}(x)=(x-b) Q_{2}(x)+1$. Substituting into (1) gives
$$
P(x)=(x-a)(x-b) Q_{2}(x)+x .
$$
From $P(c)=(c-a)(c-b) Q_{2}(c)+c=c$, we get
$$
Q_{2}(c)=0, Q_{2}(x... | x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,476 |
5. (France) Let $P$ be a point inside $\triangle A B C$. Prove that $\angle P A B, \angle P B C, \angle P C A$ is at least one less than or equal to $30^{\circ}$. | Let $\alpha=\angle P A B, \beta=\angle P B C$, $\gamma=\angle P C A$. Then,
the distance from $P$ to $A B$
$$
\begin{array}{c}
=P A \sin \alpha \\
=P B \sin (B-\beta),
\end{array}
$$
the distance from $P$ to $B C$
$$
\begin{array}{l}
=P B \sin \beta \\
=P C \sin (C-\gamma), \\
=P A \sin (A-\alpha). \\
\end{array}
$$
th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,478 |
5. Prove: For each integer $x, x^{2}+5 x+16$ cannot be divisible by 169. | Proof by contradiction. If there exists an integer $x$, such that $169 \mid x^{2}+5 x+16$,
$$
169 \mid(x-4)^{2}+13 x \text{. }
$$
So $13 \mid(x-4)^{2}$. Since 13 is a prime number, $13 \mid(x-4)$. Thus $169 \mid(x-4)^{2}$. Combining with (1), we get $13 \mid x$. This contradicts $13 \mid(x-4)$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,479 |
6. Several balls are distributed among $2 n+1$ bags. If any one of the bags is taken away, the remaining $2 n$ bags can always be divided into two groups, each containing $n$ bags, and the number of balls in these two groups is equal. Prove: the number of balls in each bag is the same. | Prove: Let the numbers $a_{1}, a_{2}, \cdots, a_{2n+1}$ represent the number of balls in each of the $2n+1$ bags. Clearly, $a_{1}, a_{2}, \cdots, a_{2n+1}$ are non-negative integers. Without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2n+1}$. Thus, the problem is transformed into: th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,480 |
3. It is known that a parliament has 30 members, where any two are either friends or enemies, and each member has exactly 6 enemies. Any three members form a committee. Find the total number of such committees where the three members are either all friends or all enemies. | 3. Let the set of all three-member committees that satisfy the requirements of the problem be denoted as $X$, and the number of elements in set $X$ be denoted as $x$. The number of other three-member committees is denoted as $y$, thus
$$
x+y=C_{30}^{s}=4060 \text {. }
$$
For any senator $a$, we denote the set of commi... | 1990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,481 |
6. Two people, A and B, play the following game: A selects three non-zero numbers, and then B arranges these three numbers to replace the three asterisks in the expression $* x^{2}+* x+*$. If the resulting quadratic trinomial has two distinct rational roots, A is considered the winner. Prove that A always has a way to ... | 6. Party A only needs to take three numbers as $1, 2, -3$, and they will win for sure. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,484 |
7. Find the maximum value of the expression
$$
|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots \mid-x_{1000}
$$
where $x_{1} ， x_{2}, \cdots, x_{1000}$ are different natural numbers from 1 to 1990. | 7. Noting that when $x \geqslant 0, y \geqslant 0$, the following formulas hold:
$$
\begin{array}{l}
|x-y| \leqslant \max \{x, y\}, \\
\max \{\max \{x, y\}, z\}=\max \{x, y, z\},
\end{array}
$$
we can directly obtain
$$
\begin{array}{l}
\left.|| \cdots|| x_{1}-x_{2}\left|-x_{3}\right|-\cdots \mid-x_{\mathrm{n}}\right\... | 1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,485 |
8. Use lines parallel to the sides of an equilateral triangle to divide it into \( n^{2} \) smaller equilateral triangles, each with a side length of 1. | 8. The total number of vertices of the small triangles on the left is
$$
1+2+\cdots+(n+1)=\frac{1}{2}(n+1)(n+2) \text {, }
$$
This means that the broken line consists of $\frac{1}{2}\left(n^{2}+3 n\right)$ small line segments. If we color all the small triangles that are similar to the original triangle and have the s... | \frac{1}{2}(n^2 + 3n) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,486 |
1. Is it possible to color each small square in a $1990 \times 1990$ grid with one of two colors, black or white, such that two squares symmetric about the center of the grid are colored differently, and in each row and each column of the grid, the number of black and white squares are equal? | 1. Assume that a $1990 \times 1990$ square grid has been colored to meet the requirements of the problem. Write +1 on black cells and -1 on white cells, then divide the square grid into four smaller squares of size $995 \times 995$ using a cross line through the center of the square and parallel to the sides of the squ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,487 |
6. (Netherlands) Given a real number $a>1$, construct a bounded infinite sequence $x_{0}, x_{1}, x_{2}, \ldots$, such that for every pair of distinct non-negative integers $i, j$ we have
$$
\left\lvert\, \begin{array}{ll}
x_{i} & x_{j}|\cdot| i-\left.j\right|^{a} \geqslant 1 \\
\end{array}\right.
$$ | For any non-negative integer $n$, let its decimal representation be
$$
\begin{aligned}
n & =b_{0}+b_{1} \times 10+b_{2} \times 10^{2}+\ldots \\
& +b_{\mathrm{k}} \times 10^{\mathrm{b}},
\end{aligned}
$$
where $b_{0}, b_{1}, \ldots, b_{x} \in\{0,1,2, \ldots, 9\}$. Let
$$
\begin{aligned}
y_{\mathrm{n}} & =b_{0}+b_{1} \t... | not found | Inequalities | proof | Yes | Yes | cn_contest | false | 705,489 |
3. On the side $AB$ of the convex quadrilateral $ABCD$, take a point $E$ different from $A$ and $B$, and the line segment $AC$ intersects $DE$ at point $F$. Prove that the circumcircles of $\triangle ABC$, $\triangle CDF$, and $\triangle BDE$ are concurrent. | 3. Examine two possible scenarios:
1) The circumcircles of $\triangle A B C$ and $\triangle C D F$ intersect at another point $K$ besides $C$ (see the figure below). Clearly, it is only necessary to prove that points $B, D, E$, and $K$ are concyclic.
If point $K$ coincides with point $B$ or $D$, the conclusion is obvio... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,490 |
4. There are two fleas at the two endpoints of the line segment $[0, 1]$. Some points are marked within the line segment. Each flea can jump over the marked points such that the positions before and after the jump are symmetric about the marked point, and they must not jump out of the range of the segment $[0, 1]$. Eac... | 4. The segment [0, 1] is divided into smaller segments by certain fixed points, with lengths $\frac{17}{23}$ and $\frac{19}{23}$. Therefore, the two fleas cannot land on the same segment after each taking one step (see the following supplement). From this, we can conclude that the minimum number of steps required must ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,491 |
5 . Find the integer solution to the equation
$$
\left[\frac{x}{1!}\right]+\left[\begin{array}{c}
x \\
2!
\end{array}\right]+\cdots+\left[\frac{x}{10!}\right]=1001
$$ | 5. From the old equation, we know that $x$ is a positive integer not exceeding 1001, so $x<6$!. Therefore, the last five terms on the left side of the equation can be removed. Each positive integer $x<6$! can be uniquely expressed as
$$
x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+e,
$$
where $a$, $b$, $c$, $d$, $e$ ... | 584 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,492 |
7. In a convex polygon, draw all diagonals, and color each side and each diagonal with one of $k$ colors, so that there is no monochromatic closed broken line with the vertices of the polygon as its vertices. How many vertices can such a polygon have at most? | 7. Let the number of vertices of a polygon be $n$, then an $n$-sided polygon has $C_{\mathrm{n}}^{2}=\frac{1}{2} n(n-1)$ edges and diagonals. If there is a segment of the same color, then by a graph theory theorem, there must be a graph, i.e., there must be a closed broken line, which contradicts the given condition. T... | n \leqslant 2 k | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,494 |
8. In the plane, a point $A_{0}$ and $n$ vectors $\vec{a}_{1}$, $\overrightarrow{a_{2}}, \cdots, \vec{a}_{\mathrm{n}}$ are given, such that $\vec{a}_{1}+\cdots+\vec{a}_{\mathrm{n}}=\stackrel{\rightharpoonup}{0}$. Each permutation $\overrightarrow{a_{1_{1}}}, \ldots, \overrightarrow{a_{1_{n}}}$ of these vectors defines ... | 8. Since $\overrightarrow{a_{1}}+\overrightarrow{a_{2}}+\cdots+\overrightarrow{a_{\mathrm{n}}}=\overrightarrow{0}$, they can be appropriately placed so that their endpoints form a convex polygon $A_{0} A_{1} \cdots A_{\mathrm{n}-1}$. Let $\triangle A_{1} A_{1} A_{k}(0 \leqslant i<j<k \leqslant n-1)$ be the triangle wit... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,495 |
3. Given the quadratic trinomial $f(x)=a x^{2}+b x+c$
all coefficients are positive, and $a+b+c=1$.
Prove that for any positive numbers $x_{1}, x_{2}, \cdots, x_{\mathrm{n}}$, as long as $x_{1} x_{2} \cdots x_{\mathrm{n}}=1$, then
$$
f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{\mathrm{n}}\right) \geqslant... | 3. Note, for any $x, y > 0$, we have
$$
f(x) f(y) \geqslant [f(\sqrt{x y})]^{2}.
$$
In fact, if we denote $\sqrt{x y} = z$, then we have
$$
\begin{aligned}
& f(x) f(y) - f^{2}(z) \\
= & a^{2}(x^{2} y^{2} - z^{4}) + b^{2}(x y - z^{2}) \\
+ & c^{2}(1 - 1) + a b(x^{2} y + x y^{2} - 2 z^{3}) \\
+ & a c(x^{2} + y^{2} - 2 z... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,496 |
4. Given a cube with an edge length of 100, which is composed of one million unit cubes, these small cubes form the framework of the large cube. The three edges extending from a vertex of a unit cube, which are mutually perpendicular, are called a frame. How can the entire framework be divided into two groups of frames... | 4. Introduce a Cartesian coordinate system in space so that all the vertices of the large cube have coordinates of either 0 or 100. Mark certain nodes (vertices of unit cubes) in the framework so that each large edge (of length 100) of the framework has exactly one marked node. For example, all points whose coordinates... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,497 |
6. Let $d$ be the minimum distance between opposite edges of any tetrahedron, and $h$ be the minimum height of the tetrahedron. Prove that $2 d > h$.
untranslated text remains the same as the source text. | 6. For definiteness, let $h$ be the height of tetrahedron $ABCD$ drawn from point $A$, and $d$ be the distance between edges $AB$ and $CD$. Draw a line $l \parallel CD$ through $B$, and a plane perpendicular to edge $CD$ through $A$, intersecting line $l$ and $CD$ at points $E$ and $F$ (upper right figure). Thus, the h... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,499 |
7. Write the equation $x^{3}+\ldots x^{2}+\ldots x$ $+\ldots=0$ on the blackboard. Two people play the following game: Player A chooses a number, and Player B arbitrarily fills this number into one of the three blanks in the equation; then Player A chooses another number, and Player B fills it into one of the remaining... | Place 0 in the last position, then the left side of the equation becomes the polynomial $x^{3}+a x^{2}+b x$. Then, Party A can take the numbers 2 and -3 in sequence, and depending on where Party B places 2, the polynomial can be factored as follows:
$$
\begin{array}{l}
x^{3}+2 x^{2}-3 x=x(x+3)(x-1), \\
x^{3}-3 x^{2}+2 ... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,501 |
2. (3rd Latin American Mathematical Competition) $a$, $b$, $c$, $d$, $p$, $q$ are natural numbers, and satisfy
$$
a d-b c=1, \frac{a}{b}>\frac{p}{q}>\frac{c}{d} \text {. }
$$
Prove:
(1) $q \geqslant b+d$,
(2) if $q=b+d$, then $p=a+c$. | $$
\begin{aligned}
& \text { (1) } \frac{1}{b d}=\frac{a d-b c}{b d} \frac{b}{d} \\
= & \left(\frac{a}{b}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{c}{d}\right) \\
\geqslant & \frac{1}{b q}+\frac{1}{q d} \\
= & \frac{b+d}{b q d} .
\end{aligned}
$$
Thus, $q \geqslant b+d$.
(2) If $q=b+d$, from (1) it must be that
$$
\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,504 |
3. (3rd Latin American Mathematical Competition) Consider expressions of the form $x+y t+z t^{2}\left(x, y, z \in Q, t^{2}=2\right)$. Prove: If $x+y t+z t^{2} \neq 0$, then there exist $u$, $v, w \in Q$, such that
$$
\left(x+y t+z t^{2}\right)\left(u+\nu t+w t^{2}\right)=1
$$ | Prove that since $t$ is an irrational number, when
$$
x+y t+z t^{2}=(x+2 z)+y t \neq 0,
$$
then, $(x+2 z)-y t \neq 0$. Therefore,
$$
\begin{array}{l}
(x+2 z)^{2}-y^{2} t^{2} \\
=(x+2 z)^{2}-2 y^{2} \neq 0 .
\end{array}
$$
Take $\quad v=\frac{-y}{(x+2 z)^{2}-2 y^{2}}$,
then $v$ is a rational number. Take
$$
u+w t^{2}=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,505 |
4. (19th Austrian Mathematical Competition) Find all real-coefficient polynomials with the following property:
$$
\begin{aligned}
P(x) & =x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x \\
& +a_{0}
\end{aligned}
$$
For any root $a$ of $P(x)$ (real or complex), $\frac{1}{a}$ and $1-a$ are also roots of $P(x)$. | If $a$ is any root of $P(x)$, then $\frac{1}{a}$ and $1-a$ are also roots of $P(x)$. Therefore,
$$
1-\frac{1}{a}, \frac{1}{1-a}, 1-\frac{1}{1-a}=\frac{a}{a-1}
$$
are also roots of $P(x)$.
Thus, $a \neq 0,1$.
Since $P(x)$ is a 5th-degree polynomial, the six roots $a, \frac{1}{a}, 1-a, 1-\frac{1}{a}, \frac{1}{1-a}, \fra... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,506 |
5. (16th All-Russian Mathematical Competition) On the sides $AB$ and $BC$ of the convex quadrilateral $ABCD$, points $E$ and $F$ are taken such that segments $DE$ and $DF$ trisect the diagonal $AC$. It is known that the areas of $\triangle ADE$ and $\triangle CDF$ are each equal to $\frac{1}{4}$ of the area of quadrila... | Prove that $E, F$ are the midpoints of $AB, BC$ respectively.
Let $DE, DF$ intersect $AC$ at $P, Q$.
From $AP=QC$ we know
$S_{\triangle P D}=S_{\triangle C Z D}$.
Then by $S_{\triangle D B}=S \triangle C D F$
we get
$S_{\triangle P E}=S_{\triangle C Q F}$.
Thus, $E, F$ are equidistant from $AC$,
$EF \| AC$. By
$$
\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,507 |
6. (3rd Latin American Mathematical Olympiad) In $\triangle ABC$, the lengths of the three sides are $a, b, c$. Each side is divided into $n$ equal segments. Let the sum of the squares of the distances from each vertex to the $n-1$ division points on its opposite side be $s$.
Prove that $\overline{a^{2}+b^{2}+c^{2}}$ i... | Proof: Let $A_{1}$ be the $i$-th division point on side $BC$ starting from $B$. By the cosine rule,
$$
\begin{aligned}
c^{2}= & A A_{1}^{2}+\left(\frac{i}{n} a\right)^{2}-2 A A_{1} \\
& \cdot \frac{i}{n} \cos \theta, \\
b^{2}= & A A_{1}^{2}+\left(\frac{n-i}{n} a\right)^{2}-2 A A_{1} \frac{n-i}{n} \\
& \cdot \cos (\pi-\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,508 |
7. (19th Austrian Mathematics Competition) In $\triangle ABC$, the angle bisectors of $\angle B, \angle C$ intersect the opposite sides at $B', C'$ respectively. Prove that the line $B'C'$ intersects the incircle of $\triangle ABC$.
保留源文本的换行和格式，翻译结果如下：
7. (19th Austrian Mathematics Competition) In $\triangle ABC$, th... | Proof: Let $I$ be the incenter of $\triangle ABC$, and draw $ID \perp BC$ at $D$, and $IE \perp B'C'$ at $E$. Let $ID = r$, and $IE = h$. Then,
$$
\begin{array}{l}
\angle B'C'C + \angle C'B'B \\
=\frac{1}{2}(\angle B + \angle C).
\end{array}
$$
Assume $\angle C'B'B \leqslant \frac{1}{2} \angle B$, then
$$
\begin{align... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,509 |
8. (16th All-Russian Mathematical Competition) In space, given 6 different points, it is known that a sphere can be constructed through any 5 of these points. Can we conclude that a sphere can be constructed through all 6 points?
保留源文本的换行和格式，翻译结果如下：
8. (16th All-Russian Mathematical Competition) In space, given 6 dif... | First, prove that all 6 points lie on the same sphere.
(1) If it is known that 4 of the 6 points are not coplanar, then there exists a unique sphere \( S \) passing through these points. Therefore, by the condition (any 5 of these 6 points lie on some sphere), the remaining 2 points are also on this sphere;
(2) If it i... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,510 |
1. Given $2 n$ points in the plane, where no three points are collinear, and $n$ points are colored red, $n$ points are colored blue. Prove: It is always possible to find $n$ line segments without common points, such that the two endpoints of each segment have different colors. | 1. Since there are only a finite number of ways to pair red and blue points one-to-one, among the sums of the lengths of the $n$ segments obtained by each pairing method, there must be a minimum. The pairing method at this time is the one we seek. If not, suppose $AC$ and $BD$ intersect, with $A, B$ being red points an... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,511 |
9. (16th All-Russian Mathematical Competition) For any pyramid with a convex quadrilateral base, can a plane be used to cut it such that the cross-section is always a parallelogram.
Preserve the original text's line breaks and format, and output the translation result directly. | Consider any pyramid $S-ABCD$ with a convex quadrilateral base $ABCD$. The intersection line of plane $SAD$ and plane $SBC$ is $a$, and the intersection line of plane $SAB$ and plane $SDC$ is $b$.
Let the plane determined by $a$ and $b$ be $\alpha$, the base plane be $\gamma$, and the intersection line of $a$ and $\ga... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,512 |
10. (19th Austrian Mathematical Competition) In the region $\{(x, y) \mid x, y>0, x y=1\}$, find the range of the function
$$
f(x, y)=\frac{x+y}{[x][y]+[x]+[y]+1}
$$
where $[\alpha]$ denotes the integer part of $\alpha$. | By symmetry, let $x \geqslant 1$.
(1) When $x=1, y=1$, $f(1, 1)=\frac{1}{2}$.
(2) When $x>1$, let $x=n+\alpha$ $(n \geqslant 1, n \in \mathbb{N}, 0<\alpha<1)$. If $f(x, y)$ is an increasing function when $x>1$, then
$$
f(x, y) \in\left(\frac{n+1}{n+1}, \frac{n+1+\frac{1}{n+1}}{n+1}\right).
$$
Let $a_{n}=\frac{n+\frac{... | \left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,513 |
```
11. (16th All-Russian Mathematics Competition) Replace the “*” in the following division problem with appropriate digits
$$
\begin{array}{l}
\text { * } 8 \text { * } \\
* * \sqrt{* * * * *} \\
-) * * * * * * \\
\frac{-) * *}{* *} \\
\frac{-) * * *}{0} \\
\end{array}
$$
``` | From the fact that the divisor multiplied by 8 yields a two-digit number, we know the divisor is no greater than 12. Furthermore, since the divisor multiplied by the first or last digit of the quotient results in a three-digit number, we know the divisor is no less than 12.
Therefore, the divisor is 12.
And since 12 mu... | 11868 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,514 |
13. (19th Austrian Mathematical Competition) Find $N=$ $19^{88}-1$ for all divisors $d=2^{2} \cdot 3^{b}$ (where $a, b$ are natural numbers) and determine the sum of these divisors $d$. | $$
\begin{aligned}
N & =(20-1)^{88}-1 \\
& =(1-4 \times 5)^{88}-1 \\
& =-C_{88}^{1} 4 \times 5+C_{88}^{2} 4^{2} \times 5^{2} \\
& -C_{88}^{8} 4^{8} \times 5^{3}+\cdots \\
& -C_{88}^{87} 4^{87} \times 5^{87}+C_{88}^{88} 4^{88} \times 5^{88} \\
& =-2^{5} \times 55+2^{8} M \\
& =2^{5}(-55+2 M),
\end{aligned}
$$
where $M$... | 744 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,516 |
14. (16th All-Russian Mathematical Competition) A digit in a decimal number is called periodically repeating if the positions of this digit after the decimal point, from left to right, form an arithmetic sequence.
Prove: Any irrational number in the interval $(0,1)$ has at least one digit that is not periodically repe... | Prove that the conclusion is equivalent to: If a decimal number $A \in(0$, 1 ), and each of its digits is periodically repeated, then $A$ is a rational number.
Let the different digits used in $A$ be $i_{1}, i_{2}, \cdots$, $i_{k}$, and the repetition periods of these digits be $T_{1}, T_{2} ， \cdots$, $T_{x}$.
The l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,517 |
15. (16th All-Russian Mathematics Competition) A desert is shaped like a half-plane, which is divided into many small squares of size $1 \times 1$. In the desert, 15 squares away from the boundary, there is a robot with energy $E=59$. The "energy consumption" of each small square is a natural number not greater than 5,... | Solve As shown in the figure, use routes with arrows to represent five different stubborn movement routes of the robot, * indicates the position of the robot.
The total energy consumption of the small squares passed by these five routes does not exceed $3 \times 88 + 2 \times 10 + 2 \times 5 = 294$ (the above routes p... | 59 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,518 |
16. (3rd Latin American Mathematical Olympiad) Consider a set formed by $n(n \geqslant 3)$ natural numbers (where no three form an arithmetic progression). Prove that among all such sets, there exists one whose sum of the reciprocals of its elements is maximum. | Let $M=\left\{1,2,3, \cdots, n+C_{\mathrm{n}}^{2}\right\}$.
For each set $A=\left\{a_{1}, a_{2}, \cdots, a_{0}\right\}$, denote $S_{\Delta}=\sum_{1=1}^{n} \quad \frac{1}{a_{i}}$.
Since for any two numbers $a, b$, there are at most three numbers $2a - b, \frac{a+b}{2}, 2b - a$ that form an arithmetic sequence with $a, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,519 |
17. (16th All-Russian Mathematics Competition) 100 athletes participate in a running competition. It is known that among any 12 of them, 2 can be found who are acquaintances. Prove that no matter how the athletes are assigned numbers (not necessarily from 1 to 100), 2 acquainted athletes can always be found whose numbe... | Let $N_{1}$ represent the number of numbers starting with the digit $i$ $(i=1,2, \cdots, 9)$.
This problem is equivalent to proving that there exists some $k$ for which the inequality $N_{k} \geqslant 12$ holds.
We use proof by contradiction.
Assume the conclusion does not hold, i.e., for each $k, N_{8} \leqslant 11$,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,520 |
18. (16th All-Russian Mathematics Competition) In the school football championship, each team must play a match against every other team. Each match awards 2 points to the winning team, 1 point to each team in the event of a draw, and 0 points to the losing team. It is known that one team has the highest score, but it ... | Let the team with the highest score, denoted as team $A$, be the champion. Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must win at least $n + 1$ matches, meaning their score is no less than $2(n + 1)$ points. ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,521 |
19. (16th All-Russian Mathematics Competition) Can 8 non-overlapping squares with side length 1 be placed inside a circle with radius 2? | Prove that the circumradius of isosceles trapezoid $ABCD$ is less than 2, and that points $E, F$ lie inside this circle.
Let the center of the circle be $O$, and $K$, $L$, $M$ be the midpoints of $AD$, $BC$, $EF$ respectively. Let $OK = x$, then $LO = 3 - x$.
$$
\begin{array}{l}
AO^2 = AK^2 + KO^2 = \frac{9}{4} + x^2,... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,523 |
Example 6. The figure formed by the diagonals of squares with side length 1 is called a cross. Prove: In a circle with a radius of 100, the number of non-overlapping crosses that can be placed does not exceed 80000. | Prove that for each cross, consider a circle with the center of the line segment as the center and the half of the cross diagonal (diagonal length... half) $\frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{1}{2 \sqrt{2}}$ as the radius.
Then, in the figure, the intersection $O_{1} O_{2} \leqslant r_{1}+r_{2}=2 r$ $=\frac{1}... | 80000 | Geometry | proof | Yes | Yes | cn_contest | false | 705,524 |
Example 7. Prove: Any $n(n \geqslant 2)$ points in a plane can always be covered by some non-intersecting circles, the sum of the diameters of these circles is less than $n-a+1$, and the distance between any two circles is greater than $a$, where $0<a<1$. (First Zhejiang Provincial Higher Normal Colleges Mathematics Le... | Prove (1) Given $n$ points as centers, and $r=\frac{1}{2}+\frac{1}{2 n}$ as the radius, draw $n$ circles. The sum of the diameters of these $n$ circles is $n \cdot 2 r=n+1$.
(2) First adjustment: Make the circles covering the $n$ points non-intersecting, and ensure that the sum of their diameters does not increase. The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,525 |
Example 8. In a convex $n(n \geq 4)$-sided polygon, if any three of all the diagonals do not intersect at the same point, into how many parts do they divide the $n$-sided polygon? | Method One: Draw the diagonals in sequence. When a new diagonal is drawn, if the number of intersection points with the already drawn diagonals is $m$, then this new diagonal increases the number of segments the $n$-sided polygon is divided into by $m+1$. According to this incremental pattern, the number of segments th... | C_{\mathrm{n}-1}^{2}+C_{4}^{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,526 |
Question: There is an $n \times n (n>4)$ sheet of blank grid paper, each cell filled with 1 or $-1$; the product of $n$ numbers that are in different rows and different columns is called a basic term. Try to prove: for each grid table filled in the above manner, the sum of the basic terms is divisible by 4.
Translate ... | Prove: Let $a_{i_{1}}(i, j=1,2, \ldots, n)$ represent the element in the $i$-th row and $j$-th column of the table, then each basic term is $a_{11_{1}} a_{2} 1_{2} \cdots a_{n 1_{n}}\left(i_{1}, i_{2}, \cdots, i_{\mathrm{D}}\right)$, where $(i_{1}, i_{2}, \cdots, i_{\mathrm{D}})$ is a permutation of $(1,2, \cdots, n)$.... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,527 |
Question: Try to find three consecutive integers greater than 7, each divisible by 7, 8, and 9 respectively. | To generalize: The proposition can be extended as follows:
Proposition If three consecutive natural numbers $x-2, x-1$, $x$ are divisible by natural numbers $a-2, a-1, a$ respectively, where $x > a > 2$, then $x = a + k a(a-1)(a-2) \quad(k \in \mathbb{N})$.
Proof Since $a \mid x$, $(a-1) \mid x-1 = (a-1)(1 + k a(a-2))$... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,528 |
Example 1. Solve the equation $8 x^{3}+8 \sqrt{3} x^{2}+6 x+\sqrt{3}$ $-1=0$ | Let $\sqrt{3}=t$, the original equation becomes
$$
2 x t^{2}+\left(8 x^{2}+1\right) t+8 x^{3}-1=0 \text {. }
$$
Solving the quadratic equation for $t$ yields
$$
t_{1}=-2 x+1, t_{2}=\frac{4 x^{2}+2 x+1}{-2 x} .
$$
Given $t=\sqrt{3}$, the original equation's roots are
$$
\begin{array}{l}
x_{1}=\frac{1-\sqrt{3}}{2}, \\
... | x_{1}=\frac{1-\sqrt{3}}{2}, x_{2}, 3=-\frac{1+\sqrt{3} \pm \sqrt[4]{12}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,529 |
Example 2. Solve the inequality
$$
\sqrt{x^{2}+6 x+10}+\sqrt{x^{2}-6 x+10}>10 .
$$ | Solve the inequality which can be transformed into
$$
\sqrt{(x+3)^{2}+1}+\sqrt{(x-3)^{2}+1}>10 \text {. }
$$
Let $1=y^{2}$, then we have
$$
\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}>10 \text {. }
$$
This precisely indicates that the sum of the distances from the moving point $(x, y)$ to the points $(3,0)$ and $(-... | x>\frac{5}{4} \sqrt{15} \text { or } x<-\frac{5}{4} \sqrt{15} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 705,530 |
For example, find the angle $\theta$ formed by $B D^{\prime}$ and $D C^{\prime}$ in the cube on the right.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Let the lower base $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be $\gamma$, then
$$
\begin{array}{c}
\sin \alpha=-\overline{1} \overline{\sqrt{3}}, \cos \alpha=\sqrt{\frac{2}{3}}, \\
\sin \beta=\cos \beta=\frac{\sqrt{2}}{2}, \\
\cos \varphi=\cos \angle B^{\prime} D^{\prime} C^{\prime}=\cos \frac{\pi}{4}=\sqrt{2} .
\e... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,531 |
Example 2. As shown in Figure 2, with the sides $AB, AC$ of $\triangle ABC$ extended outward to form squares $ABEF, ACGH$, the extension of the altitude $AD$ from $BC$ intersects $FH$ at $M$. Prove: $FM = MH, \quad$ and $AM = \frac{1}{2} BC$. | Proof: Let $\angle ABC = \alpha, \angle BCA = \beta$. From the given conditions, it is easy to see that $\angle FAM = \alpha, \angle MAH = \beta$. Then, by the Angle Bisector Theorem, we have
$$
\begin{aligned}
\frac{FM}{MH} & =\frac{AF \cdot \sin \alpha}{AH \cdot \sin \beta}=\frac{AB \cdot \sin \alpha}{AC \cdot \sin \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,534 |
Example 11. As shown in Figure 11, $AB = CD = 1, \angle ABC$ $= 90^{\circ}, \angle CBD = 30^{\circ}$. Find $AC$. (1986, Canadian Competition) | Let $A C=x$. By the Angle Bisector Theorem, we have
$$
\frac{A D}{D C}=\frac{A B \cdot \sin 120^{\circ}}{B C \cdot \sin 30^{\circ}},
$$
which means $\frac{x+1}{1}=\frac{1 \cdot \frac{\sqrt{3}}{2}}{B C \cdot \frac{1}{2}}$.
Therefore, $B C=\frac{\sqrt{3}}{x+1}$.
Applying the Pythagorean Theorem in $\mathrm{Rt} \triangle... | \sqrt[3]{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,535 |
Example 1. Let $k$ be a real number, discuss the real roots of the equation
$$
\left|x^{2}-1\right|-x-k=0
$$ | Solve by discussing the relationship between the intersection points of images and the solutions of equations. Let $y=f_{1}(x)=\left|x^{2}-1\right|$ and $y=f_{2}(x)=x+k$. Sketch the graphs of these two functions, and note that $y=x+k$ is a set of lines parallel to $y=x$. Thus, we can obtain a partition of the set of re... | 0, 1, 2, 3, 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,536 |
Example 2. Can a partition $\left(S_{1}, S_{2}, S_{3}, S_{4}\right)$ of the set $X=\{1,2,3, \cdots$, 1989\} be given such that the sums of the numbers in $S_{1}, S_{2}, S_{3}$, and $S_{4}$ form an increasing arithmetic sequence with a common difference of 10? | Let the sum of the numbers in $S_{1}$ be $a$, then the sums of the numbers in $S_{2}, S_{3}, S_{4}$ are $a+10, a+20, a+30$ respectively. Thus, the total sum $=4a+60$ is an even number, but $1+2+\cdots+1989=\frac{1989 \times 1990}{2}$ is an odd number. This means the requirement of the problem cannot be achieved. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,537 |
Example 3. Let $S$ denote the set of all composite numbers not exceeding 79.
(1) Try to prove that $S$ can be divided into three subsets, such that the elements in each subset form an arithmetic sequence;
(2) Discuss whether $S$ can be divided into two such subsets. | (1) It is only necessary to construct a specific partition of $S$. Note that each element in $S$ can be written as $(2 r+1) \cdot(2 r+2 s+1)$, where $r \geqslant 1, s \geqslant 0$. Therefore, the first term of each arithmetic sequence is $(2 r+1)^{2}$, and the common difference is $2(2 r+1)$. Taking $r=1,2,3$, we have
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,538 |
Example 4. Prove: The set $\{1,2, \cdots, 1989\}$ can be partitioned into 117 mutually disjoint subsets $A_{i}(i=1,2$, ..., 117), such that
(1) each $A_{i}$ contains 17 elements;
(2) the sum of the elements in each $A_{i}$ is the same.
(30th IMO). | Solve the general problem: The set $\{1,2, \cdots, m n\}$ (where $m, n$ are positive integers, $n>1$) can be divided into $m$ mutually disjoint subsets $A_{1}(i=1,2, \cdots, m)$, such that
(1) Each $A_{1}$ contains $n$ elements;
(2) The sum of the elements in each $A_{1}$ is the same.
When $n=2 k$, arrange $1,2, \cdots... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,539 |
Example 5. Let $r, s, n$ be natural numbers, and $r+s$ $=n$. Prove: $A=\left\{\left[\frac{n}{r}\right],\left[\frac{2 n}{r}\right], \cdots\right.$, $\left.\left[\frac{(r-1) n}{r}\right]\right\}$ and $B=\left\{\left[\begin{array}{c}n \\ s\end{array}\right],\left[\frac{2 n}{s}\right]\right.$, $\left.\cdots,\left[\frac{(s-... | Given that the number of elements in $A, B, N$ are $r-1, s-1$, and $n-2$ respectively, therefore, $A-\mathrm{j} B$ forming a partition of $N$ is equivalent to $A \cap B=\emptyset$.
Necessity. If $r$ and $s$ have a common divisor $d$ with $n$, then by $n=r+s$, $d$ is also a common divisor of $r$ and $s$. Let $r=r^{\pri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,540 |
Example 6. Let $m \geqslant 3$ be a fixed natural number. Prove that the smallest natural number $r(m)$ equal to $m^{2}-m-1$:
For any partition $A$ and $B$ of the set $\{1,2, \cdots, r(m)\}$, the equation $x_{1}+x_{2}+\cdots+x_{m-1}=x_{\mathrm{m}}$ must have a solution in one of them. | First, we prove that $r(m) \geqslant m^{2}-m-1$. For the set
$$
M=\left\{1,2, \cdots, m^{2}-m-2\right\},
$$
consider one of its partitions
$$
A=\left\{1,2, \cdots, m-2,(m-1)^{2},\right.
$$
$\left.(m-1)^{2}+1, \cdots, m^{2}-m-2\right\}$ and $B=\{m-1$, $\left.m, m+1, \cdots,(m-1)^{2}-1\right\}$.
We will prove that the e... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,541 |
Let $a$ and $b$ be natural numbers, and $(ab + 1)$ divides $a^2 + b^2$. Prove that:
$$
\frac{a^2 + b^2}{ab + 1}
$$
is the square of some integer. (29th IMO Problem) | Let $k=\frac{a^{2}+b^{2}}{a b+1}$,
i.e., $a^{2}+b^{2}=k(a b+1)$.
If we relax the restrictions and allow $b$ to be any non-negative integer, then for the special case of $b=0$, it is clear that $k=a^{2}$. This special case provides inspiration for the general case
$$
a \geqslant b>0,
$$
We should try to use a des... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,543 |
Example 3. Prove that in a coordinate plane, it is impossible to construct: a closed broken line with rational points as vertices, having an odd number of sides, each of length 1. (31st IMO Preliminary Problem, provided by the Soviet Union) | Analysis The original solution involves a lot of calculations, and the idea is clear (please refer to the "Middle School Mathematics" magazine, 1990, for a more natural solution). To prove that such a closed broken line does not exist, it is natural to use proof by contradiction ("assume it exists, derive a contradicti... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,544 |
Shape, through point $M$ on $AB$ draw $MP$, $MQ$, $MR$ perpendicular to $BC$, $CD$, $AD$ respectively. Connect $P$, $R$ and intersect $MQ$ at $N$. Prove:
$\frac{PN}{NR}=\frac{BM}{MA}$ . (1983, Fujian Province Junior Competition) | Proof: Let $\angle P M N = \alpha, \angle N M R = \beta$. From the given conditions, $\alpha$ is supplementary to $\angle C$, and $\angle C$ is supplementary to $\angle A$, thus $\alpha = \angle A$. Similarly, $\mathscr{L} = \angle B$. It is also easy to see that $B M - \sin \angle B = P M$, and $M A \cdot \sin \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,545 |
Example 2. Prove that when $n$ is a positive integer, the smallest integer greater than $(3+$ $\sqrt{5})^{2 n}$ is divisible by $2^{n}$.
(1987. Weizhou High School Competition) | Let $\alpha=3+\sqrt{5}, \beta=3-\sqrt{5}$. Then $\alpha, \beta$ are the roots of $x^{2}-6 x+4=0$, and we have
$$
0<3-\sqrt{5}<1 \text {. }
$$
By Theorem 1, $\alpha^{2 n}+\beta^{2 n}$ is the smallest integer greater than $\alpha^{2 n}$.
$$
\begin{aligned}
& \alpha^{2 n}+\beta^{2 n} \\
= & (3+\sqrt{5})^{2 n}+(3-\sqrt{5}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,547 |
Example 3. Prove that the smallest integer greater than $(\sqrt{3}+1)^{2a}$ is divisible by $2^{n+1}$. (6th U.S. College Mathematics Competition)
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\alpha^{2 u 1}+\beta^{2 n}=(1+\sqrt{3})^{2 n}+(1-\sqrt{3})^{2 u} \\
=(4+2 \sqrt{3})^{n}+(4-2 \sqrt{3})^{n} \\
= 2^{n}\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right] \\
= 2^{n}\left[2 \cdot(2+\sqrt{3})^{2-1}\right. \\
+\sqrt{3}(2+\sqrt{3})^{n-1}+2(2-\sqrt{3})^{\bar{a}} \\
\left.-\sqrt{3}(2-\sqrt{3})^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,548 |
Example 4. Let $a$ be a positive integer, $\lambda_{1}, \lambda_{2}\left(\lambda_{1}>\lambda_{2}\right)$ be the two roots of the equation $x^{2}-a x-1=0$.
Let $f_{n}=\frac{1}{\sqrt{a^{2}+4}}\left(\lambda_{1}^{\mathrm{n}}-\lambda_{2}^{\mathrm{n}}\right)(n=1,2,3$,
$\cdots$, then all $f_{\mathrm{n}}$ are positive integers... | Prove II Dingyue 2 Know
$$
\lambda_{1}^{\mathrm{n}}-\lambda_{2}^{\mathrm{n}}=\sqrt{\Delta} N_{\mathrm{n}}=\sqrt{a^{2}+4} N \mathrm{n}_{0}
$$
$N_{\mathrm{a}}$ is an integer. Also, $-a1 j . \\
\because \lambda_{1} \lambda_{2}=-1, \\
\therefore f_{t}^{f_{t}}=\frac{f_{\frac{1}{s}}}{f_{s}}=\frac{\left(\lambda_{1}^{s}\right)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,549 |
Example 1. In a right-angled triangle $ABC$, a square $MNQR$ is inscribed. Try to prove that the area of this square does not exceed half the area of $\triangle ABC$. (1978, Guangdong Competition) | Prove as follows:
1. Draw $C C_{1} \perp A B$, with $C_{1}$ as the foot of the perpendicular, and set $A B=c$, $M N=x$, $C C_{1}=h$. From $\triangle C N Q \sim \triangle C A B$, we have
$$
\frac{x}{c}=\frac{h-x}{h} \Rightarrow \frac{x}{c}+\frac{x}{h}=1.
$$
Also, $S_{\mathrm{MNQR}}=x^{2} \Rightarrow \frac{x}{c} \cdot \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,550 |
Example 2. As shown in Figure 2, the area of trapezoid $ABCD$ is $S$,
$$
\begin{array}{l}
AB \| CD, AB=b, \\
CD=a(a<b),
\end{array}
$$
Diagonals $AC$ and $BD$ intersect
at $O$, and the area of $\triangle BOC$ is $\frac{2 S}{9}$. Find $-\frac{a}{b}$.
(1988, Jiangsu Junior
Competition) | Given $\because A B \| C D, \therefore \triangle A O B \backsim \triangle C O D$,
$$
\therefore \frac{O D}{O C}=\frac{O B}{O A} \text {. }
$$
Also, $\frac{S \triangle \frac{C O D}{}}{2 S / 9}=\frac{O D}{O B}, \frac{2 S / 9}{S \triangle A O B}=\frac{O C}{O A}$,
$\therefore \frac{S \triangle C O D}{2 S / 9}=\frac{2 S / ... | -\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,551 |
Example 1. Draw the graph of $||x|-1|+||y|-1|=2$, and find the area of the enclosed region.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve: By substituting $-y$ and $-x$ for $y$ and $x$ respectively, the equation remains unchanged. Therefore, the graph is symmetric about the $x$-axis and $y$-axis. Thus, it is only necessary to draw:
$$
|x-1|+|y-1|=2(x \geqslant 0, y \geqslant 0) \text { I }
$$
For equation (1), let $x-1=x^{\prime}, y-1=y^{\prime}$,... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,554 |
Example 1. Find the equation of the tangent line at point $P\left(x_{1}, y_{1}\right)$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. | First, the ellipse transforms into a circle $x^{\prime 2}+y^{\prime 2}=a^{2}$ under the transformation $x=x^{\prime}, y=\frac{b}{a} y^{\prime}$, and point $P$ transforms into $P^{\prime}\left(x_{1}, \frac{a}{b} y_{1}\right)$. The equation of the tangent line to the circle at $P^{\prime}$ is
$$
y^{\prime}-\frac{a}{b} y_... | \frac{x_{1} x}{a}+\frac{y_{1} y}{b}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,558 |
Example 2. Find the area of the smallest circumscribed parallelogram around the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. | Let the points of tangency of an external quadrilateral with the circle $x^{\prime 2}+y^{\prime 2}=a^{2}$ be $R(\cos \alpha, a \sin \alpha)$, $R^{\prime}(a \cos \beta, a \sin \beta)$, $Q(-a \cos \alpha$, $-a \sin \alpha)$, and $Q^{\prime}(-a \cos \beta, -a \sin \beta)$. Then the area of the external parallelogram is
$$... | 4ab | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,559 |
Example 1. (1987 Math Summer Camp) Solve the equation
$$
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}
$$ | Let $1+x=u$, then
$x=\sqrt{1+\sqrt{1}+\sqrt{u}}$ which means $\sqrt{u}=\sqrt{1+x}$
$=\sqrt{1+\sqrt{1+\sqrt{u}}}$, the structure of the equation is similar to the original, so
$x=\sqrt{u}$, that is $x=\sqrt{1+x}$, solving this gives $x=\frac{1+\sqrt{5}}{2}$. | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,560 |
Example 2. (1978 Canadian Mathematical Competition) Determine the maximum real number $z$, where $(x, y$ are also real numbers )
$$
\left\{\begin{array}{l}
x+y+z=5, \\
x y+y z+z x=3 .
\end{array}\right.
$$ | Solving, we get $x+y=5-z, xy=3-(5-z)z$. By the inverse of Vieta's formulas, we obtain the equation for $t$
$$
t^{2}-(5-z) t+3-(5-z) z=0
$$
For real roots, we have $\Delta \geqslant 0$, solving which gives $-1 \leqslant z \leqslant \frac{13}{3}$,
$$
z_{\max }=\frac{13}{3}
$$ | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,561 |
7. Let the area of parallelogram $ABCD$ be $S$, points $E, F$ are on $AB, AD$ respectively, and $EB=\frac{2}{m} AB$, $AF=\frac{3}{n} AD$ (where $m, n$ are natural numbers greater than 4). $BF$ intersects $CE$ at $G$. Find:
(1) The area of $\triangle BEG$,
(2) $EG: GC=$ ? , $BG: GF=$ ?
(1989, Hefei City Junior High Scho... | $\begin{array}{c}\text { (Answer: } S \triangle \mathbf{B \mathbf { G }}=\frac{6 S}{m(m n+6)}, \\ E G: G C=6: m n, B G: G F=2 n:[n \\ \cdot(m-2)+6])\end{array}$ | S \triangle \mathbf{B \mathbf { G }}=\frac{6 S}{m(m n+6)}, \\ E G: G C=6: m n, B G: G F=2 n:[n \cdot(m-2)+6] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,562 |
Let the real-coefficient polynomial be
$$
f(x)=x^{\mathrm{n}}+a_{1} x^{\mathrm{n}-1}+\cdots+a_{\mathrm{n}}
$$
with roots being real numbers $b_{1}, b_{2}, \ldots, b_{n}$, where $n \geqslant 2$.
Prove: For $x>\max \left(b_{1}, \ldots, b_{\mathrm{n}}\right)$,
$$
f(x+1) \geqslant \frac{2 n^{2}}{\frac{1}{x-b_{1}}+\frac{1}... | For $n \geqslant 2$, we have
$$
n^{2}-2 n(n-1) \leqslant 0 \text {. }
$$
For any $t>0$, we have
$$
\begin{array}{c}
n(n-1) \\
2
\end{array} t^{2}-n t+1 \geqslant 0 .
$$
From this, we get $\left.(1+t)^{n} \geqslant 1+n t+\frac{n(n}{2} 1\right) t^{2} \geqslant 2 n t$.
$$
\text { When } x>\max \left(b_{1}, b_{2}, \cdots... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,563 |
For $i=1,2, \cdots, 1991$, take $n_{1}$ points on a circle, and label each point with a number $i$, with only one number per point. The requirement is to draw a set of chords such that:
(1) No two chords have a common point;
(2) The numbers at the endpoints of each chord are different.
To ensure that for all possible l... | If, according to the problem, all the points with labels can be connected by chords, since any two labeled points correspond to and only to one chord, $n_{1}+n_{0}+\cdots 1 n_{1991}$ must be even. For any $i \in \{1,2, \cdots, 1991\}$, among the $n$ points with the label $i$, each point must correspond to one of the ch... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,564 |
Three, given any 5 points on a plane, where no three points are collinear and no four points are concyclic. If a circle passes through three of these points, and the other two points are respectively inside and outside the circle, then it is called a "good circle".
Let the number of good circles be $n$, find all possib... | Three, among: 5 points, take any two points $A, B$ and draw the line through $A$, $B$. If the other three points $C, D, E$ are on the same side of line $A B$, then consider $\angle A C B, \angle A D B, \angle A E B$. Without loss of generality, if $\angle A C B < 180^{\circ}$, then circle $A D B$ is the unique good cir... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,565 |
Example 5. For an isosceles trapezoid $\triangle B C D$ with a base angle of $67.5^{\circ}$, a circle is drawn with the leg $B C$ as the diameter, intersecting the lower base $A B$ at $E$. 11. It is exactly tangent to the leg $A D$ at $M$. Find $B E: A E$. (1981, Beijing High School Competition) | As shown in Figure 5, let the center be \( O \), and \( OE, OA \). Then, by the Angle Bisector Theorem, we have
\[
\begin{aligned}
& \frac{B E}{A E} \\
= & \frac{O B \cdot \sin \angle B O E}{O A \cdot \sin \angle E O A}
\end{aligned}
\]
Since \( O E = O B \) and \( \angle E B O = 67.5^{\circ} \), it follows that \( \a... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,567 |
Let the function $f$ be defined for non-negative integers and satisfy the conditions:
$$
\begin{array}{l}
f(0)=0, f(1)=1, \\
f(n+2)=23 f(n+1)+f(n), \\
n=0,1,2, \cdots .
\end{array}
$$
Prove: For any $m \in \mathbb{N}$, there exists $d \in \mathbb{N}$, such that
$$
m|f(f(n)) \Leftrightarrow d| n .
$$ | Five, prove for any $m^{-} N$, for any $n \in N$, there is $m^{\prime} f(n)$.
In fact, we do not need to assume $m>1$. Let $(n)$ be the remainder of $f(n)$ divided by $m$, and let $g(n) \in\{0,1,2, \cdots, m-1\}$, with $g(0)=0, g(1)$ $=1$,
$$
\begin{array}{r}
g(n+2) \equiv 23 g(n+1)+g(n)(\bmod m), \\
n=0,1,2, \cdots .... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,568 |
Six, color all the edges of a convex polyhedron red, and the corners. The number of singular face angles at a vertex $A$ is called the singularity degree of that vertex, denoted as $S_{A}$. Prove: There always exist two vertices $B$ and $C$, such that $S_{B} + S_{C} \leqslant 4$. | Six, mark the red edges of a convex polyhedron with the number 1, and the yellow edges with the number 0. Define the degree of any dihedral angle as the sum of the numbers marked on its two sides, modulo 2, resulting in a remainder of 0 or 1. Thus, a dihedral angle is a singular dihedral angle if and only if its degree... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,569 |
Example 6. Given $\triangle A B C$ is constant, $E, F$ are the trisection points of $B C$, $M$ is the midpoint of $A C$, $B M$ intersects $A E, A F$ at $G, H$ respectively. Find $B C: C H: H M$. (1982 Tianjin Junior High School Competition) | Let $ABM$
$$
=\sim, \angle M B C=\beta \text {. }
$$
Since $M$ is the midpoint of $AC$,
by the Angle Bisector Theorem, we have
thus $\frac{\sin \alpha}{\sin \beta}=\frac{BC}{AB}$.
By the Angle Bisector Theorem again, we get
$$
\begin{aligned}
\frac{AG}{CE} & =\frac{AB \cdot \sin \alpha}{BE \cdot \sin \beta}=\frac{AB}... | 5: 3: 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,570 |
Example 7. In $\triangle A B C$, $D, E$ are points on $B C$, $C A$ respectively, and $B D: D C=m: 1, C E: E A$ $=n: 1, A D$ intersects $B E$ at $F$. Then the area of $\triangle A B F$ is $\qquad$ times the area of $\triangle A B C$. | Let $\angle E B C$
$$
=\alpha, \angle A B E=P \text {. }
$$
By the Angle Bisector Theorem,
$$
n=\frac{C E}{E A}=\frac{B C \cdot \sin \alpha}{A B \cdot \sin \beta} .
$$
Therefore, $\sin \alpha / \sin \beta=n A B / B C$.
Thus, $\frac{D F}{F A}=\frac{B D \cdot \sin \alpha}{A B \cdot \sin \beta}=\frac{B D}{A B} \cdot \fr... | \frac{m}{m n+m+1} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,571 |
Example 8. In $\triangle ABC$, $D$ is the midpoint of $AC$.
(1) If $BC$ is trisected, with division points from $C$ being $C_{1}, C_{2}$, and $C_{1}A, C_{2}A$ intersect $BD$ at $D_{1}$, $D_{2}$. Find $S \wedge N_{1} D_{2}: S \triangle A C_{1} C_{2}$;
(2) If $BC$ is divided into four equal parts, with division points fr... | To solve, we only need (3).
It is easy to see that $B C_{2}={ }_{n}^{n \cdots 2} B C, B C_{1}=\frac{n-1}{n} B C$.
Let $\angle A B D=\alpha$,
$\angle D B C=\beta,$
Since $D$ is the point dividing $A C$ internally, by the angle bisector theorem, we can get
$\sin \alpha / \sin \beta=$
$B C / A B$, thus
$$
\begin{array}{l}... | \frac{n^2}{2(n-1)(2n-1)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,572 |
Example 10. The cosine value of the angle between the median on one leg of an isosceles right triangle and the hypotenuse is $\qquad$
(1986, Jilin Province Eight
Cities Junior High School Competition) | Given $\triangle A B C$ is an isosceles right triangle with $A B$ as the base, $A D$ is the median of side $B C$, and $\angle D A B = \alpha$.
Then, for the angle bisector, we have
$$
\begin{array}{l}
D B = \frac{A D \cdot \sin \alpha}{A C \cdot \cos \angle C A B} \\
= A D \cdot \sin \alpha \\
A C \cdot \sin 45^{\circ}... | \frac{\sqrt{10}}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,574 |
Example 2. In quadrilateral
$$
A B C D \text {, }
$$
(1) If the diagonals $A C = B D$, prove that the perimeter of the inscribed quadrilateral, whose sides are parallel to the two diagonals, is a constant;
(2) If $A C: B D = m: n (m, n$ are constants, $m \neq n$), does the conclusion in (1) still hold? Please provide a... | Proof (1) As shown in Figure 2, it is easy to see that quadrilateral $E F G H$ is a parallelogram. Let $A C=B D=1, H E=x$, $E F=y, A E=a, B E=b$, then
$$
\begin{array}{l}
x=\frac{a}{a+b}, \\
y=\frac{b}{a+b} .
\end{array}
$$
Thus, $x+y=\frac{a+b}{a+b}$
$$
=1 \text {, }
$$
which means $2(x+y)=2$.
Therefore, the perimet... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,576 |
Example 1. (1989, American Mathematical Competition) For every positive integer $n$, let $S_{4}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$, $T_{n}=S_{1}+S_{2}+\cdots+S_{n}, U_{a}=\frac{1}{2} T_{1}+\frac{1}{3} T_{2}+\cdots+\frac{1}{n+1} T_{n}$. Try to find integers $0<a, b, c, d<1000000$, such that $T_{1088}=a \cdot ... | Let $S_{0}=0$, then by Abel's identity, we have
$$
\begin{aligned}
T_{\mathrm{n}} & =\sum_{\mathrm{i}=0}^{\mathrm{n}} S_{1}=\sum_{k=0}^{\mathrm{n}-1}\left(\sum_{\mathrm{i}=0}^{\mathrm{k}} 1\right) \\
& \cdot\left(S_{\mathrm{k}}-S_{k+1}\right)+\left(\sum_{i=0}^{\mathrm{n}} 1\right) S_{\mathrm{n}} \\
& =\sum_{\mathbf{k}=... | a=1989, b=1989, c=1990, d=3978 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,578 |
Example 2. $(1978, \mathrm{IMO})$ Let $\left\{a_{\mathrm{k}}\right\}(k=1,2$, for all positive integers $n$, we have
$$
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{1}{k} .
$$ | Proof: From the condition, it is clear that $\sum_{1=1}^{k} a_{1} \geqslant \sum_{1=1}^{k}$.
$$
\text { Also, } \frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}>0(k=1,2, \cdots, n), \text { J }
$$
By repeatedly applying Abel's transformation, we get
$$
\begin{array}{l}
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}}=\sum_{k=1}^{n-1}\left(\sum... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,579 |
For a positive integer x, we have
$$
\frac{2}{3} n \sqrt{n}<\sum_{i=1}^{n} \sqrt{i}<\frac{4 n}{6}+3 \sqrt{n} .
$$ | $$
\frac{2 n+1}{3} \sqrt{n} \leqslant \sum_{i=1}^{n} \sqrt{i} \leqslant \frac{4 n+3}{6} \sqrt{n}-\frac{1}{6} .
$$
In fact, when $n=1$, equation (14) obviously holds. Hereafter, we assume $n \geqslant 2$.
Let $S_{n}=\sum_{i=1}^{n} \sqrt{i}$, by Abel's identity, we have
$$
\begin{aligned}
S_{n}= & \sum_{k=1}^{n-1}\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,584 |
Example 1. $O$ is the origin of the complex plane, $Z_{1}$ and $Z_{2}$ are moving points, $\theta$ and $-\theta\left(0<\theta<\frac{\pi}{2}\right)$;
( 2 ) The area of $\triangle O Z_{1} Z_{2}$ is a constant $S$. Find the minimum multiple of the modulus of the complex number corresponding to the centroid of $\triangle O... | Let the coordinates of $Z_{1}, Z_{2}, Z$ be $\left(\gamma_{1} \cos \theta, \gamma_{1} \sin \theta\right)$, $\left(\gamma_{2} \cos (-\theta), \gamma_{2} \sin (-\theta)\right)$, $(x, y)$, and $Z$ be the centroid of $\triangle O Z_{1} Z_{2}$,
$$
x=\frac{1}{3}\left(\gamma_{1}+\gamma_{2}\right) \cos \theta, y=\frac{1}{3}\le... | \frac{2}{3} \operatorname{ctg} \theta \sqrt{S} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,585 |
1 (Philippines) Let $P$ be any point inside $\triangle ABC$, $P_{1}$ and $P_{2}$ are the feet of the perpendiculars from $P$ to sides $AC$ and $BC$, respectively, and $Q_{1}$ and $Q_{2}$ are the feet of the perpendiculars from $C$ to $AP$ and $BP$, respectively. Prove that the three lines $Q_{1} P_{2}$, $Q_{2} P_{1}$, ... | It is known that $C, P_{1}, Q_{2}, P, Q_{1}, P_{2}$ lie on a circle. Since $C P_{1}$ intersects $P Q_{1}$ at point $A$, and $Q_{2} P$ intersects $P_{2} C$ at point $B$, applying Pascal's theorem, we know that the intersection of $Q_{1} P_{2}$ and $Q_{2} P_{1}$ lies on the line $A B$. That is, the lines $Q_{1} P_{2}$, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,588 |
2. (Japan) Given an acute triangle $ABC$. Let $M$ be the midpoint of side $BC$, and take a point $P$ on segment $AM$ such that $PM = BM$. Let $H$ be the foot of the perpendicular from $P$ to $BC$. Draw the perpendicular from $H$ to $PB$ intersecting $AB$ at $Q$, and draw the perpendicular from $H$ to $PC$ intersecting ... | If $A B$
$=A C$, then $M$ coincides with $H$, and the conclusion to be proved
is obviously true. Without loss of generality, assume
$A B>A C$, then $H$ is
on $M C$, it suffices to prove $\angle R H C=\angle R Q H$.
Since $P M=B M=C M$, $\angle B P C$ is a right angle, thus $\angle Q H R$ is also a right angle. Therefor... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,589 |
5. (Spain) In $\triangle A B C$, it is known that $\angle A = 60^{\circ}$. Draw a line through the incenter $I$ of the triangle parallel to $A C$ intersecting $A B$ at $F$. Take a point $P$ on $B C$ such that $3 B P = B C$. Prove that $\angle B F P = \frac{1}{2} \angle B$. | Suppose $B C=3$, and the inradius of $\triangle A B C$ is $r$. By the Law of Sines, we know that
$$
\begin{aligned}
A B & =3 \frac{\sin C}{\sin A}=2 \sqrt{3} \sin (B+A) \\
& =\sqrt{B} \sin B+3 \cos E .
\end{aligned}
$$
Since: $r=4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}$, where $R$ is the circ... | \alpha=\frac{B}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 705,591 |
9. (France) In the plane, a set $E$ of 1991 points is given. Each point in $E$ is connected by a line segment to at least 1593 other points in $E$. Prove that there exist six points in $E$ such that every pair of them is connected by a line segment.
Does there exist a set of 1991 points such that each point is connect... | Proof: Let $A_{1}$ and $A_{2}$ be two points in $E$. In $E \backslash\left\{A_{1}, A_{2}\right\}$, the points not connected to $A_{1}$ are at most $1989-1592=397$, and similarly, the points not connected to $A_{2}$ are also at most 397. Therefore, in $E \backslash\left\{A_{1}, A_{2}\right\}$, the points connected to bo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,594 |
An isosceles right triangle. Now fix $\triangle A B C$, and rotate $\triangle A D E$ around point $A$ on the plane. Try to prove: no matter where $\triangle A D E$ rotates to, there must be a point $M$ on line segment $E C$ such that $\triangle B M D$ is an isosceles right triangle. (1987, National Mathematics Competit... | Prove: As shown in the figure, extend $C B$ to $G$, such that $B G = B C$. Extend $E D$ to $N$, such that $D N = E D$. Then $\triangle A C G$ and $\triangle A E N$ are both isosceles right triangles. By rotating $\triangle A E G$ around vertex $A$ by $90^{\circ}$, it can coincide with $\triangle A N C$, thus $N C = E G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,598 |
17. (Hong Kong) Find all positive integer solutions \(x, y, z\) for the equation \(3^{x} + 4^{y} = 5^{x}\). | Solving: $x=y=z=2$ is a solution. The following proves that this is the only solution. Let positive integers $x, y, z$ satisfy $3 j$ product, then
$$
1 \equiv 5^{\circ}(\bmod 3)=2^{\circ}(\bmod 3),
$$
thus $z$ is even. Let $z=2 z_{1}$, then the original equation becomes
$$
\left(5^{x}+2^{y}\right)\left(5^{x}-2^{y}\rig... | x=2, y=2, z=2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,600 |
19. (Ail Lamb) Let $a$ be a rational number and $0<a<1$. If $\cos (3 \pi a)+2 \cos (2 \pi a)=0$, prove that $a=\frac{2}{3}$. | Let $x=\cos (\pi a)$, then the equation becomes
$$
4 x^{3}+4 x^{2}-3 x-2=0 \text {, }
$$
which can be factored as $(2 x+1)\left(2 x^{2}+x-2\right)=0$.
If $x=\cos (\pi a)=-\frac{1}{2}$, then $a=\frac{2}{3}$. If $x$ satisfies $2 x^{2}+x-2=0$, then $\cos (\pi)$ $=\frac{\sqrt{17}-1}{4}$, and it can be shown that $a$ is no... | a=\frac{2}{3} | Algebra | proof | Yes | Yes | cn_contest | false | 705,602 |
20. (Iran) Let $\alpha$ be the positive root of the equation $x^{2}=1991 x+1$. For any natural numbers $m$, $n$ define
$$
m * n=m n+[\alpha m][\alpha n] .
$$
Prove that for all natural numbers $p, q, r$ we have
$$
(p * q) * r=p *(q * r) .
$$ | Let $k=1991$, then $\alpha > 1991$ and $\alpha(\alpha-k) = 1$. For any natural numbers $p, q$, we have
$$
\begin{aligned}
\alpha(p * q) & =\alpha p q+\alpha[\alpha p][\alpha q]=\frac{1}{\alpha}\left(\alpha^{2} p q+\alpha^{2}[\alpha p][\alpha q]\right) \\
& =\frac{1}{\alpha}\left(\alpha^{2} p q+[\alpha p][\alpha q]+k \a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,603 |
21. (Xiang Zhe) Let $f(x)$ be a 1991st degree polynomial with integer coefficients, and let $g(x) = f^2(x) - 9$. Prove that $g(x)$ cannot have more than 1991 distinct integer roots. | Assume that the number of distinct integer roots of $g(x)$ is $\geqslant 1992$. From $g(x)=(f(x)-3)(f(x)+3)$, it is easy to see that $|x-y| \geqslant 7$. | $f(x)-f(y)=6$ easily leads to $x-y \mid 6$, a contradiction. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,604 |
22. (USA) Let real numbers $a, b, c$ be such that there is exactly one square with all its vertices on the cubic curve $y=x^{3}+a x^{2}+b x+c$. Prove that the side length of this square is $\sqrt[1]{72}$. | To prove that for the transformation $\left(x^{\prime}=x+\frac{a}{3}, y^{\prime}=y-d\right)$, the curve $\Gamma_{1}: y=x^{3}+b x$ is symmetric with respect to the origin. The center of a square with vertices on $\Gamma_{1}$ is the origin; otherwise, rotating this square $180^{\circ}$ around the origin would yield anoth... | \sqrt{72} | Geometry | proof | Yes | Yes | cn_contest | false | 705,605 |
25. (USA) Let $0 \leqslant x_{1} \leqslant 1, i=1,2, \cdots$, $n, n>2$. Prove that there exists $1 \leqslant i \leqslant n-1$ such that
$$
x_{1}\left(1-x_{1+1}\right) \geqslant \frac{1}{4} x_{1}\left(1-x_{n}\right) .
$$ | Let $x_{1}=a=\max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, $x_{1}=b: x \min \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$.
If $x_{2} \leqslant \frac{1+b}{2}$, then $x_{1}\left(1-x_{2}\right) \geqslant x_{1}$
- $\left(1-\frac{1+b}{2}\right)=\frac{1}{2} x_{1}(1-b)$, obviously when $i=1$
H(1) holds. If $x_{2}>\frac{1+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,608 |
Example 5. As shown in Figure 5, the endpoints of a fixed-length chord $P Q$ (length less than the diameter) move along the semicircular arc $\widehat{A B}$. Try to prove: regardless of the position of $P Q$, the perpendiculars from $P$ and $Q$ to $A B$ intersect $A B$ at points $P^{\prime}$ and $Q^{\prime}$, and the t... | Prove that connecting $O M, O P, O Q$, then $O M \perp P Q$. Thus, $O, M, P, P^{\prime}$ are concyclic, $\angle M P O = \angle M P^{\prime} O$.
Similarly, $\angle M Q O = \angle M Q^{\prime} O$.
Therefore, $\triangle M P^{\prime} Q^{\prime} \sim \triangle O P Q$.
However, regardless of the position to which $P Q$ slide... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,609 |
29. (Finland) We call a subset $S$ of the set of real numbers $\mathbb{R}$ super-invariant if for any $a>0, x_{0} \in \mathbb{R}$, there exists $b \in \mathbb{R}$ such that
$$
\begin{array}{ll}
\left\{x_{0}+a\left(x-x_{0}\right),\right. & x \in S\} \\
=\{y+b, y \in S\} .
\end{array}
$$
Find all super-invariant sets. | Let's solve the problem. $S$ is a super-invariant set $\Leftrightarrow$ the complement of $S$ is a super-invariant set. It is easy to see that the following three types of sets are super-invariant sets:
(1) $R$ and its complement $\varnothing_{3}$
(2) Single-point set $\{x\}$ and its complement $\{y \in R; y \neq x\}$
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,612 |
30. (Bulgaria) Two students, A and B, play the following game: they each write a positive integer on a piece of paper, and both know that one of the numbers is the sum of A and B's numbers. Then the referee asks A: "Do you know what the other student wrote?" If A answers no, the referee asks B the same question. If B a... | Let the numbers written by $A$ and $B$ be $a$ and $b$ respectively. The numbers written on the blackboard are $x$ and $y$, and without loss of generality, assume $0 < x < y$, thus $a + b = y$. In this case, $A$'s response should also be “knows”. Therefore, (i) holds. Similarly, (ii) can be proven to hold.
For (i) and ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 705,613 |
1. Let the equation be
$$
x^{n}+a_{0-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0
$$
where all coefficients are real, and satisfy:
$$
0<a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n-1} \leqslant 1 .
$$
Given that $\lambda$ is a complex root of this equation, and satisfies the condition $|\lambda| \geqslant 1$. Prove: $\... | Proof 1: Given the condition $\lambda \neq 0,1$. Consider
$$
\begin{aligned}
0 & =(\lambda-1)\left(\lambda^{\mathrm{u}}+a_{\mathrm{n}-1} \lambda^{\mathrm{n}-1}+\cdots+a_{1} \lambda\right. \\
& \left.+a_{0}\right) \\
= & \lambda^{\mathrm{n}+1}+\left(a_{\mathrm{n}-1}-1\right) \lambda^{\mathrm{n}}+\left(a_{\mathrm{n}-2}-a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,614 |
3. Draw a $9 \times 9$ grid on a plane, and arbitrarily fill each small square with +1 or -1. The following method of changing the numbers is called making one move: for any small square, take the product of the numbers in all the small squares that share a common edge with this square (excluding the square itself). Th... | Method 1 It is easy to verify that the following three number tables remain unchanged under the aforementioned transformation.
The cells without numbers are all filled with +1.
Fill the middle row and middle column of the $9 \times 9$ grid with +1, and then fill the remaining four $4 \times 4$ sub-grids symmetrically w... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,616 |
4. Convex quadrilateral $ABCD$ is inscribed in circle $O$, diagonals $AC \leqslant BD$ intersect at $P$. The circumcircles of $\triangle ABP$ and $\triangle CDP$ intersect at $P$ and another point $Q$, and $O, P, Q$ are not collinear. Prove: $\angle OQP=90^{\circ}$.
(Cui Wei provided the problem) | Proof 1 As shown,
Connect $A O, A Q$,
$$
\begin{array}{l}
D O, D Q . \\
=\angle D Q P \\
+\angle P Q A \\
-\angle D C P \\
+\angle A B P \\
=2 \angle A B D, \\
\angle A O D=2 \angle A B D,
\end{array}
$$
Thus, $\angle D Q A=\angle A O D$.
Therefore, $A, Q, Q, D$ are concyclic.
$$
\begin{array}{l}
\text { Also, } \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,617 |
6. Given the integer sequence $\left\{a_{0}, a_{1}, a_{2}, \cdots\right\}$ satisfies:
(1) $a_{n+1}=3 a_{n}-3 a_{n-1}+a_{n-2}$, $n=2,3, \cdots$,
(2) $2 a_{1}=a_{0}+a_{2}-23$
(3) For any natural number $m$, in the sequence $\left\{a_{0}\right.$, $\left.a_{1}, a_{2}, \cdots\right\}$, there must be $m$ consecutive terms $a... | Proof $\mid 11(1)$ shows that for any $n \geq 2$,
$$
a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right)+a_{n-1}-a_{n-2} \text {. }
$$
Let $d_{n}=a_{n}-a_{n-1}, n=1,2, \cdots$, then
$$
d_{n+1}-d_{n}=d_{n}-d_{n-1}=\cdots=d_{2}-d_{1} .
$$
From (2), we get $d_{2}-d_{1}=a_{2}-2 a_{1}+a_{0}=2$. Therefore,
$$
a_{n}=a_{0}+\sum_{x=1}^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,619 |
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