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 |
|---|---|---|---|---|---|---|---|---|---|
Example 6. Using three different numbers from the set $S=\{1,2, \cdots, n\}(n>3)$ as side lengths, how many different triangles can be formed? | Let $T_{\mathrm{n}}$ be the number we are looking for, then $T_{\mathrm{n}}+{ }_{1}$ is the corresponding number for the set $S \cup\{n+1\}$. Consider $\Gamma_{n}+1-T_{n}$, it represents the number of triangles with one side length $n+1$ and the other two sides being different numbers from $S$. For each such triangle, ... | \frac{1}{24} n(n-2)(2 n-5), (n=2 r) \\ \frac{1}{24}(n-1)(n-3)(2 n-1), (n=2 r+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,998 |
9. In trapezoid $A B C D$, $A B \| C D$, and $A B$ $=92, B C=50, C D=19, A D=70$. A circle with center $P$ on $A B$ is tangent to sides $B C$ and $A D$. If $A P=\frac{m}{n}$, where $m, n$ are coprime positive integers. Find $m+n_{0} \quad$ | 9. As shown in the figure, extend $A D$ and $B C$ to intersect at point $Q$. Since the distances from $P$ to $A Q$ and $B Q$ are equal, $P$ lies on the angle bisector of $\angle A Q B$, thus we have
$$
\frac{A P}{B P}=\frac{A Q}{B Q}
$$
Since $A B \parallel C D$, we have
$$
\frac{A Q}{B Q}=\frac{A D}{B C}=\frac{7}{5}.... | 164 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,999 |
11. Lines $l_{1}, l_{2}$ both pass through the origin, and in the first quadrant, they form angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ with the positive $x$-axis, respectively. Define $R(l)$ as the line obtained by reflecting $l$ about $l_{1}$ and then reflecting the result about $l_{2}$. Let $R^{(1)}(l)=R(l), R^{... | A line passing through the origin and forming angles $\theta_{0}, \theta, \lambda$ with the positive x-axis. The symmetric line $\lambda^{\prime}$ about $\lambda$ forms an angle with the x-axis of $\theta_{0} + (\theta_{0} - \theta) = 2 \theta_{0} - \theta$.
Therefore, the symmetric line $\lambda_{1}$ about $l_{1}$ pa... | 945 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,001 |
12. In a game, two players take turns to "eat squares" from a $5 \times 7$ grid chessboard. To "eat a square," a player selects an uneaten square and moves the piece to that square, then all the squares in the quadrant formed (along the left edge of the square upwards, and along the bottom edge of the square to the rig... | 12. In this game, the height of the table formed by the un-eaten squares from left to right is not constant (as shown in the right figure). It is easy to prove that this situation is both sufficient and necessary. Moreover, each shape can be completely described by a 12-step zigzag line. This zigzag line starts from th... | 792 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,002 |
13. In $\triangle A B C$, it is known that $A B=9, B C: C A$ $=40: 41$. Find the maximum value of the area of $\triangle A B C$. | 13. Let $AB = c$, $AC = br$, $BC = ar$, $(a < b)$. Establish a Cartesian coordinate system with point $A$ as the origin and the line $AB$ as the x-axis (as shown in the figure), set $A(0,0)$, $B(c, 0)$, $C(x, y)$.
From $\frac{BC}{AC} = \frac{a}{b}$, we get
$$
\frac{\sqrt{(x-c)^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} = \fra... | 820 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,003 |
14. In $\triangle A B C$, points $A^{\prime}, B^{\prime}, C^{\prime}$ are on sides $B C, A C, A B$ respectively, $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect at point $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \frac... | 14. As shown in the figure, since $\triangle A O B$ and $\triangle A^{\prime} O B$ have the same height, therefore,
$$
\begin{aligned}
\frac{A O}{O A^{\prime}} & =\frac{S_{\triangle A O B}}{S_{\triangle A^{\prime} O B}} \\
& =\frac{S_{\triangle C O A}}{S_{\triangle C O A^{\prime}}}+S_{\triangle C O A} \\
& =\frac{S_{\t... | 94 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,004 |
15. If there exists a positive integer $m$ such that $m!$ ends with exactly $n$ zeros, then the positive integer $n$ is called a "factorial tail number." How many non-"factorial tail number" positive integers are there less than 1992? | 15. Let $f(m)$ be the number of trailing zeros in the number $m!$. Clearly, $f(m)$ is a non-decreasing function of $m$, and when $m$ is a multiple of 5, we have
$$
\begin{aligned}
f(m) & =f(m+1)=f(m+2) \\
& =f(m+3)=f(m+4) \\
& 4 \times 1991=7964. \text{ Using (2), it is easy to get }
$$
$$
f(7965)=1988, f(7975)=1991 .
... | 396 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,005 |
3. On a plane, there are 4 straight lines, any two of which intersect, but no three are concurrent. Thus, on each line, there are 3 intersection points, which divide the line into two segments, resulting in a total of 8 segments. Can the lengths of these 8 segments be equal to
(a) $1,2,3,4,5,6,7,8$ ?
(b) distinct natur... | 3. (a) No. Otherwise, by the triangle inequality, a line segment of length 1 cannot form a triangle with the other 7 line segments. Therefore, as shown in Figure 4, the length of $BC$ or $CD$ must be 1. Without loss of generality, let $BC=1$. Since $BF$ and $CF$ are both integers, they must be equal. Applying the cosin... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,008 |
Example 7. Let the sum of the lengths of all diagonals of a convex polygon with $n(n>3)$ vertices be $d$, and the perimeter be $p$. Prove: $n-3<\frac{2 d}{p}<\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right] -2$, where $[x]$ denotes the greatest integer not exceeding $x$. (Problem from the 25th IMO) | Proof: Let the vertices of a convex polygon be $A_{1}, A_{1}, \cdots, A_{2}$. For convenience, we agree that $A_{1}+\mathrm{R}$ is $A_{1}$. Now we compare the diagonals with those sides not exceeding half the perimeter. Clearly, we have
$$
\begin{array}{l}
A_{1} A_{2}n-3$.
As shown in the right figure, by
$$
\begin{arr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,009 |
4. A lottery ticket has 50 spaces arranged in sequence, and each participant should fill in the numbers 1 to 50 without repetition on their ticket, with the order determined by themselves. The host also fills in these 50 numbers on a ticket as the answer key. If a participant's number in any space matches the number in... | 4. 26 tickets. To ensure a win, for example, 26 tickets can be filled out as follows:
$$
\begin{array}{l}
1,2,3, \cdots, 25,26,27, \cdots, 50 \\
2,3,4, \cdots, 26,1,27, \cdots, 50 \\
3,4,5, \cdots, 1,2,27, \cdots, 50 \\
\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
25,26,1, \cdots, 23,24,27, \cdots, 50 \\
26,1,2, \c... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,010 |
7. An investigator is drafting an outline for questioning a witness who is defending a public crime. He intends to ask only such questions that can be answered with a "yes" or "no" (of course, whether to ask a certain question depends on the witness's answer to the previous question). The investigator believes that all... | 7. Denote the best outline as $S$. First, ask 13 questions according to outline $S$, then
Figure 6
ask a verification question: "Do you agree with the previous set of questions?" If the answer is "No," it indicates that there is no untruth, so continue asking questions according to outline $S$, still in groups, with t... | 105 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,012 |
8. A square is divided into a $5 \times 5$ grid. In one of the cells, a “-..” sign is filled, and in the other cells, a “+” sign is filled. Each time, you are allowed to select a square, which contains more than one cell, and has cell lines as boundaries, and then simultaneously change all the signs within that square.... | 8. In Figure 7, any square composed of more than one cell contains an even number of shaded cells. Therefore, if the initial negative sign is located in a shaded cell, then after each step of changing the sign, there will be an odd number of negative signs in such cells, and thus at least one will remain. By rotating F... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,013 |
3. On the blackboard, there are natural numbers $1 ; 2,2,3, \cdots$, $n$, where $n \geqslant 3$. Each time, it is allowed to erase any two numbers $p$ and $q$, and replace them with $p+q$ and $|p-q|$. After such repeated modifications, all the numbers on the blackboard become $k$. What values can $k$ possibly be? | Natural numbers.
After each step, the blackboard will write non-negative integers. If the sum and difference of two non-negative integers can be divided by a small odd number $d$, then these two numbers themselves can also be divided by $d$. Therefore, if the number $k$ can divide the odd number $d>1$, then the number... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,015 |
7. A blackboard of size $n \times n$ is divided into $n^{2}$ squares of $1 \times 1$. The intersection of $k$ rows and $l$ columns of these squares forms a $k \times 1$ submatrix of the blackboard; and the semi-perimeter of this submatrix is $k+l$. It is known that if the semi-perimeter of each submatrix covering the m... | 7. Use induction. When $n=1$ and 2, the proposition is obviously true. Suppose $n>2$, and for $m\frac{1}{2} n$
cases, the proposition holds. Now assume that $(i, j)$ and $(j, i)$ are not covered, and we delete the $i$-th row and the $j$-th row, as well as the $i$-th column and the $j$-th column. Thus, the remaining pa... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,018 |
8. Let $a_{1}, a_{2}, \cdots, a_{100}, b_{1}, b_{2}, \cdots$, $b_{100}$ be distinct real numbers, and fill them into a $100 \times 100$ grid as follows: in the cell at the intersection of the $i$-th row and the $i$-th column, fill in $a_{1}+b_{1}$. It is known that the product of the numbers in any column is 1. Prove t... | 8. Consider the polynomial
$$
\phi(x)=\left(a_{1}+x\right)\left(a_{2}+x\right) \cdots\left(a_{100}+x\right)
$$
-1, it is known that $b_{1}, b_{2}, \cdots, b_{100}$ are its roots. In this polynomial, the coefficient of $x^{100}$ is 1, hence we have
$$
p(x)=\left(x-b_{1}\right)\left(x-b_{2}\right) \cdots\left(x-b_{101}\r... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,019 |
For the 20th IMO problem:
In $\triangle A B C$, $A B$
$=A C$, there is a circle that is internally tangent to the circumcircle of $\triangle A B C$, and is tangent to $A B, A C$ at $P, Q$ respectively. Prove that the midpoint of the line segment $PQ$ is the incenter of $\triangle A B C$. | In fact, to obtain the conclusion of this problem, it is not necessary to treat $\triangle ABC$ as an "isosceles" triangle. That is, the conclusion can also hold under the "general" conditions of $\triangle ABC$. It is evident that "$AB = AC$" is an unnecessary condition. However, the existence of such an unnecessary c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,020 |
1. Given a natural number $a_{0}$, construct the sequence of numbers $a_{\mathrm{n}}, n=1,2, \cdots$ according to the following rules. If the last digit of $a_{n-1}$ is less than or equal to 5, then it is discarded to obtain $a_{\mathrm{n}}$ (at this point, it may be that nothing remains, and the sequence process ends)... | 1. Impossible. In fact, if the last digit $\alpha$ of the number $a_{n-1}$ is not greater than 5, then
$$
10 a_{n}=a_{n-1}-\alpha \leqslant a_{n-15}
$$
Wang then has $a_{n}=10 a_{n-1}-a_{n-1}$, the last digit of which is not greater than 5. Therefore, $10a_{n+1} \leqslant a_{n}=9 a_{n-10}$. Thus, in any case, either $... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,021 |
3. Take points $A, B, C, D, E$ on a sphere such that segments $A B, C D$ intersect at point $F$, and points $A, C$ and $F$ are equidistant from point $E$. Prove that line $B D$ is perpendicular to $E F$. | 3. Let ray $E F$ intersect the sphere at point $G$, and let $X Y$ be either of the segments $A B$ or $C D$. Prove that point $Y$ lies in the plane passing through the midpoint of segment $E G$ and is perpendicular to it. In fact, points $Z, E, Y, G$ lie on a line. And $\triangle Z E F \backsim \triangle G Y F$, because... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,023 |
5. On the sides $AB, AD$ of the square $ABCD$, take points $K, N$ respectively, such that $AK \cdot AN = 2 BK \cdot DN$. If the line segments $CK, CN$ intersect the diagonal $BD$ at points $L, M$ respectively, prove that points $K, L, M, N$ and $A$ lie on the same circle. | $\angle B K C+\angle D N C=\frac{3 \pi}{4}$ (Figure 12), prove that points $K$, $L, M, A$ lie on the same circle, and similarly, $A$, $N$, $M$, $L$ lie on the same circle. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,025 |
6. On the planet, there are 100 mutually hostile countries. To maintain peace, they decide to form several alliances, with the requirement that each alliance includes no more than 50 countries, and any two countries must be in at least one alliance. How many alliances are needed at a minimum?
(a) What is the minimum nu... | 6. 6 alliances. Each country should join no less than 3 alliances, so the number of alliances is at least 6.
(a) Only by dividing 100 countries into 4 groups, each with 25 countries, and then through combinations, we get $C{ }_{4}^{2}=6$ alliances.
(b) Only by dividing 100 countries into 10 groups $a_{1}, a_{2}, a_{8},... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,026 |
1. Let the points dividing the sides $BC, CA, AB$ of $\triangle ABC$ internally in the ratio $t(1-t)$ be $P, Q, R$, respectively. The area of the triangle formed by the segments $AP, BQ, CR$ is $K$, and the area of $\triangle ABC$ is $L$. Find $\frac{K}{L}$ (expressed in terms of $t$). | 1. As shown in the figure, regarding line $AB$, take $C^{\prime}$ on the same side as $C$, such that $A R C C^{\prime}$ forms a parallelogram. In $\triangle C C^{\prime} Q$ and $\triangle A B C$,
$$
\begin{array}{l}
\because C Q=t b, C C^{\prime}=t c, \\
\angle Q C C^{\prime}=\angle C A B, \\
\therefore \triangle C C^{... | t^2 - t + 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,029 |
2. Let $N$ be the set of all positive integers, and the mappings $p, q$ from $N$ to $N$ are defined as follows:
$$
\begin{array}{ll}
p(1)=2, & p(2)=3, \\
p(3)=4, & p(4)=1,
\end{array}
$$
When $n \geqslant 5$, $p(n)=n$.
$q(1)=3, q(2)=4$,
$q(3)=2, q(4)=1$,
When $n \geqslant 5$, $q(n)=n$.
(1) Try to find a mapping $f$, s... | 2. (1) Represent $n \rightarrow p(n)+2$, then it is:
Therefore, the following expression is valid: $f(1)=2, f(2)$ $=4, f(3)=7, f(4)=5, f(5)=3, n \geqslant 6$ when, $f(\pi)=\left\{\begin{array}{ll}n+3, & n \text { is even, } \\ n-1, & n \text { is odd. }\end{array}\right.$
(2) According to the problem, $n \rightarrow q... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,030 |
3. Let $A$ be a positive integer with 1.6 digits. Prove: It is possible to properly select several adjacent digits from $A$, such that the product of these digits is a perfect square. For example, if one of the digits of $A$ is 4, then just taking this digit is sufficient.
保留源文本的换行和格式，直接输出翻译结果。 | 3. When one of the digits of $A$ is 0, then regardless of the number of digits taken, including that digit, their product is 0 (a square number), so we assume that none of the digits are 0. Since each number between 1 and 9 can be uniquely represented in the form $2^{a} 3^{b} 5^{c} 7^{d}$, the product of the first to t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,032 |
4. Divide a rectangle with a length of 10 and a width of 1.4 into 140 small squares with a side length of 1. Then, as shown, color the small squares in alternating black and white. For each square, fill in 0 or 1, so that the number of 1s in each row and each column is odd. Prove: The number of 1s in the black squares ... | 4. Let the numbers (0 or 1) filled into the small squares be represented as a $1.0 \times 14$ matrix $A=\left(a_{1}\right)$, and let
$i$ be odd, $f$ be odd, then the sum of $a_{1}$ is $P$,
$i$ be even, $j$ be even, then the sum of $a_{1}$ is $Q$,
$i$ be even, $j$ be odd, then the sum of $a_{13}$ is $S$.
Then $P+Q=$ th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,033 |
Ming: In the circle with a diameter whose endpoints are some points of $A$, there exists such a circle (including its circumference) that contains at least $\left[\frac{n}{3}\right]$ points of $A$. Here, $[x]$ denotes the greatest integer not exceeding $x$. | 5. Let $C$ be the smallest circle containing all points of $A$ (including the circumference and interior). If $C$ contains only two points of $A$ on its circumference, then these two points are the endpoints of the diameter of $C$ (otherwise, it would contradict the minimality of $C$), and $C$ is the desired circle. If... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,034 |
In the acute triangle $\triangle A B C$, $A B=c, A C=1$, $\angle A=60^{\circ}$, and the circumradius $R \leqslant 1$ of $\triangle A B C$, then ).
(A) $\frac{1}{2}<c \leqslant 1$.
(B) $1<c \leqslant \sqrt{3}$.
(C) $\sqrt{3}<c \leqslant 2$.
( D) ) $c=2$. | Solve As shown in Figure 5, construct $C D \perp A B \exists^{\circ} D$, $B E \perp A C$ at $E$. Since $A B C$ is an acute triangle, $D$ and $E$ are located on $A B$ and $A C$ respectively. Therefore,
$$
\begin{array}{l}
c>A D=A C \cos 60^{\circ} \\
=\frac{1}{2}, \\
c=2 A E<2 A C=2 .
\end{array}
$$
Therefore, the answ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,035 |
$1 \cdot 1$ Menelaus' Theorem A line intersects the three sides $B C, C A$,
$A B$ of $\triangle A B C$ or their extensions at
$D, E, F$, then
$$
\begin{array}{l}
\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B} \\
=1
\end{array}
$$ | Proof 1 (Parallel Line Proportion Method)
As shown in Figure 1, draw $C K \| A B$, intersecting $F D$ at $K$, then
$$
\frac{B D}{D C}=\frac{B F}{C K}, \quad \frac{C E}{E A}=\frac{C K}{A F} .
$$
Multiplying the above two equations, we get
$$
\frac{B D}{D C} \cdot \frac{C E}{E A}=\frac{B F}{F A}, \frac{B D}{D C} \cdot \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,037 |
1.2 Menelaus' Inverse Theorem In $\triangle A B C$, take points $D, E, F$ on the sides $B C$, $C A$, $A B$ or their extensions, respectively. If $\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1$, then points $D, E$, $F$ are collinear. | Prove: Draw a line through $D, E$ intersecting $A B$ (or its extension) at $F_{1}$. By Menelaus' theorem, we get
$$
\frac{B D}{D C} \cdot \frac{C E}{E} A \cdot \frac{A F_{1}}{F_{1} B}=1 \text {. }
$$
From the given, $\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1$.
Therefore, $\frac{A F_{1}}{F_{1} B}=\f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,038 |
$A C, A D, B C, B D, C D$ are six edges, only one of which is longer than 1. Find the maximum possible sum of the lengths of these six edges. (1935, 14th American High School Mathematics Olympiad) | Without loss of generality, assume the length of $AD$ is greater than 1, and the lengths of the other five sides are no more than 1. Suppose the lengths of the other five sides are fixed. To maximize the sum of the six side lengths, $AD$ should be as long as possible. Since the other five side lengths are fixed, i.e., ... | 5 + \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,039 |
Example 3. Given: A regular tetrahedron $ABCD$ with each edge of length 1. Try to find the point $S$ whose sum of distances to the four vertices is minimized. | Solution: Obviously, the sum of distances from any point outside the regular tetrahedron to the vertices of the regular tetrahedron $ABCD$ cannot reach the minimum. Therefore, we only need to consider any point $P$ inside the regular tetrahedron.
As shown in the figure, let $E, F$ be the midpoints of $AB, CD$ respecti... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,040 |
Example 4. Given $0<p \leqslant a, b, c, d, e \leqslant q$. Prove:
$$
\begin{array}{l}
(a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right) \\
\quad \leqslant 25+6\left(\sqrt{\frac{q}{p}}-\sqrt{\frac{p}{q}}\right)^{2} .
\end{array}
$$
(1977, 6th American High School Mathematics Olympiad) | Proof: First, if at least one number (let's assume it is $a$) is not $p$ or $q$, temporarily keep $b, c, d, e$ unchanged. Set
$$
\begin{aligned}
w= & (a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right. \\
& \left.+\frac{1}{e}\right), m=b+c+d+e, \\
n= & \frac{1}{b}+\frac{1}{c}+\frac{1}{d} .
\end{alig... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,041 |
Example 6. $m+n$ points divide the circumference of a circle into $m+n$ arcs. Among these, $m$ points are denoted by $A$, and the remaining $n$ points are denoted by $B$. If the two endpoints of an arc are both $A$, then this arc is marked with the number 2; if the two endpoints of an arc are both $B$, then this arc is... | Prove that by fixing the labels of all $m+n-2$ points except for the endpoints of one arc, consider the change in the product caused by swapping these two endpoints. Obviously, if the labels of the endpoints of this arc are the same, the product does not change after swapping. If the labels of the two endpoints of this... | 2^{n-m} | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,042 |
Example 7. (Schwarz Problem) In an acute $\triangle ABC$, construct an inscribed $\triangle DEF$ such that the perimeter of $\triangle DEF$ is minimized.
In an acute $\triangle ABC$, construct an inscribed $\triangle DEF$ so that the perimeter of $\triangle DEF$ is the smallest. | Solve for the perimeter of $\triangle D E F$ which is $=D E+E F+F D$ containing three variables. Let's assume $D, E$ are known points, thus $D E$ is a known length. Adjust the values of $D F, E F$, i.e., adjust the position of $F$ on $A B$, to minimize $E F+F D$.
Reflect $D$ over $A B$ to $D^{\prime}$, connect $D^{\pri... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,043 |
Example 4. Try to decompose 1988 into 8 positive integers $a_{1}$, $a_{2}$, ..., $a_{8}$, so that $a_{1}!a_{4}! \cdots a_{8}!$ is minimized. | $$
\begin{array}{l}
\quad \text { Solving } 1988=8 \times 248+4 . \\
\quad \text { Let } a_{1}=a_{2}=a_{3}=a_{4}=248, a_{5}=a_{6}=a_{7} \\
=a_{8}=249 \text {. }
\end{array}
$$
This makes $a_{1}!\cdot a_{2}!\cdot \cdots \cdot a_{8}!$ the smallest.
| a_{1}=a_{2}=a_{3}=a_{4}=248, a_{5}=a_{6}=a_{7}=a_{8}=249 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,047 |
For every $A \sim S$, let
$$
\beta_{\mathbf{A}}=\left\{\begin{array}{ll}
\sum_{\mathrm{a} \in \mathrm{A}}(-1)^{\mathrm{a}} a, & A \neq \varnothing, \\
0, & A=\varnothing .
\end{array}\right.
$$
Find $\sum_{\mathrm{A} \subset \mathrm{S}} \beta_{\mathrm{A}}$. | Take any $i \in S$, then the number of $A$ such that $i \in A \subset S$ is $2^{\mathrm{n}}-1$, so the contribution of $i$ to the expression $\sum_{\mathrm{A} \subset \mathrm{S}} \beta_{\mathrm{A}}$ is $(-1)^{1} \cdot i \cdot 2^{n-1}$. Therefore, we have
$$
\begin{array}{l}
\sum_{A \subset S} \beta_{1}=2^{n-1}\left(-1+... | (-1)^{n}\left[\frac{n+1}{2}\right] 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,051 |
For every $A \subset S$, let
$$
S_{\mathrm{A}}=\left\{\begin{array}{ll}
(-)^{\mid \mathrm{A}} \mid \sum_{\mathbf{a} \in \mathrm{A}} a, & A \neq \varnothing, \\
0, & A=\varnothing .
\end{array}\right.
$$
Find $\sum_{\mathrm{A} \subset \mathrm{S}} S_{\mathrm{A}}$. | Solve: There are $2^{n-1}$ sets $A$ satisfying $n \in A \subset S$, and there are also $2^{n-1}$ sets $B$ satisfying $n \notin B \subset S$. Note that $B$ covers all subsets of $\{1,2, \cdots, n-1\}$, and each $A$ satisfying $n \in A \subset S$ can be obtained by adding the element $n$ to some $B$. For each $B \subset ... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,054 |
Question 5 Let the set $S=\{1,2, \cdots, 1000\}$. Now for any non-empty subset $A$ of $S$, let $\alpha_{\mathrm{A}}$ denote the sum of the largest and smallest numbers in $A$, then, the arithmetic mean of all such $\alpha_{\Delta}$ is $\qquad$ | The answer is 1001.
Let $S=\{1,2, \cdots, n\}, n \in N$. Let $m_{A}, M_{A}$ represent the minimum and maximum elements of a non-empty subset $A$ of $S$, respectively. Then the arithmetic mean of all $\alpha_{A}$ is
$$
\frac{1}{2^{n}-1} \sum_{\substack{A \subset S \\ A \neq \varnothing}}\left(m_{A}+M_{A}\right).
$$
Sin... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,055 |
Question 6 Let $A \subset S$ and $A \neq \varnothing$, we call
$$
D_{\mathbf{A}}=\frac{1}{|A|} \sum_{\mathbf{a} \in \mathrm{A}} a_{\mathbf{.}}
$$
the average value of $A$. Find the arithmetic mean of all $D_{\mathrm{A}}$. | $$
\begin{array}{ll}
\text { Let } P(S)=\{A \mid A \subset S\} . \\
& R_{\mathrm{E}}=\{A \subset S|| A \mid=k\}, k=1,2 \\
\cdots, n . \\
\text { Then, } \quad R_{1} \cup R_{2} \cup \cdots \cup R_{\mathrm{n}}=P(S)-\{\varnothing\}, \\
& R_{1} \cap R_{1}=\varnothing, 1 \leqslant i<1 \leqslant n .
\end{array}
$$
Then, $\q... | \frac{n+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,056 |
Question 7 Calculate $\omega_{\mathrm{D}}=\sum_{\substack{\mathrm{A} \subset \mathrm{S} \\ \mathrm{A} \neq \varnothing}}\left(\prod_{\mathrm{a} \in \mathrm{A}} a\right)$. | Let $n \in A \subset S$ and $|A| \geqslant 2$, define $f(A)=A - \{n\}$, then $f(A) \neq A$ and $f(A) \subset \{1,2, \cdots, n-1\} - S$. Notice that
\[
\prod_{a \in A} a + \prod_{a \in f(A)} a = n \prod_{a \in f(A)} a + \prod_{a \in f(A)} a = (n+1) \prod_{a \in f(A)} a
\]
only the singleton subsets $\{n\}$ of $S$ do not... | \omega_{n} = (n+1)! - 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,057 |
1. (Soviet Union) Given $\triangle A B C$, let $I$ be its incenter, and the internal angle bisectors of angles $A, B, C$ intersect the opposite sides at $A^{\prime}, B^{\prime}, C^{\prime}$, respectively. Prove that:
$$
\frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}} \leqs... | Given $\triangle A B C$ with the half-angles of its interior angles denoted as $\alpha, \beta, \gamma$, and the inradius as $r$.
Then $\alpha+\beta+\gamma=\frac{\pi}{2}$,
$$
\begin{aligned}
A I & =\frac{r}{\sin \alpha}, \\
I A^{\prime} & =\frac{r}{\sin (\alpha+2 \beta)}
\end{aligned}
$$
It is easy to obtain, $\frac{A ... | \frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}} \leqslant \frac{8}{27} | Inequalities | proof | Yes | Yes | cn_contest | false | 706,059 |
Example 2. Given that $D, F$ are points on the sides $A B$, $A C$ of $\triangle A B C$ respectively, and $A D: D B=C F: F A=2: 3$. Connect $D F$ to intersect the extension of side $B C$ at point $E$. Then, $E F: F D=$ $\qquad$
(3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | Given Figure 3, line $D F E$ intersects $\triangle A B C$, we have
$$
\begin{aligned}
& \frac{C F}{F A} \cdot \frac{A D}{D B} \cdot \frac{B E}{E C} \\
= & 1 . \\
& \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{B E}{E C}=1, \\
& \frac{B E}{E C}=\frac{9}{4}, \frac{B C}{C E}=\frac{5}{4} .
\end{aligned}
$$
Line $A F C$ inters... | 2: 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,060 |
$\begin{array}{l}\text { (France) Let } P \\ \text { be a point in } \triangle A B C. \text { Prove } \\ \text { that among } \angle P A B, \angle P B C, \\ \angle P C A \text { at least one } \\ \text { is less than or equal to } 30^{\circ} .\end{array}$ | Draw perpendiculars from $P$ to each side, with the feet of the perpendiculars being $D$, $E$, $F$. By the Erdős-Mordell theorem, we have
$$
P A+P B+P C \geqslant 2(P D+P E+P F) \text {. }
$$
Then, at least one of the inequalities $P A \geqslant 2 P F, P B \geqslant 2 P D, P C \geqslant 2 P E$ must hold. Otherwise, it... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,061 |
Theorem: The trisectors of each interior angle of $\triangle A B C$ intersect at $P, Q, R$ (as shown in the figure), then $\triangle P Q R$ is an equilateral triangle.
untranslated part:
the text "如图" means "as shown in the figure" and "正二促形" should be "equilateral triangle". The term "每巫个" seems to be a typo and sh... | Proof: Let $\angle A = 3 \alpha, \angle B = 3 \beta$, $\angle C = 3 \gamma$, and assume $\alpha + \beta + \gamma = 60^{\circ}$. Then, for sides $a, b, c$,
$$
\begin{array}{c}
=\frac{4 \sin \left(60^{\circ}-\beta\right) \sin \rho \sin \left(60^{\circ}+\beta\right) \cdot \sin \gamma \sin \left(60^{\circ}-\gamma\right)}{4... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,062 |
Theorem In a convex quadrilateral $A_{1} A_{2} A_{3} A_{4}$, the circumradii of $\triangle A_{2} A_{3} A_{4}$, $\triangle A_{3} A_{4} A_{1}$, $\triangle A_{4} A_{1} A_{2}$, and $\triangle A_{1} A_{2} A_{3}$ are $R_{1}$, $R_{2}$, $R_{3}$, and $R_{4}$, respectively. Then
$$
\begin{array}{l}
\left(R_{1} R_{2}+R_{3} R_{4}\... | Proof: Let the diagonals intersect at $O$. By Steiner's theorem, we have
$$
\begin{aligned}
& A_{1} O^{2} \cdot A_{2} A_{4} \\
= & A_{1} A_{4}^{2} \cdot A_{2} O \\
+ & A_{1} A_{2}^{2} \cdot O A_{4} - A_{2} O \cdot O A_{4} \cdot A_{2} A_{4}.
\end{aligned}
$$
Rearranging and multiplying both sides by $\frac{A_{2} A_{4} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,063 |
Theorem Let $P$ be a point on the circumcircle of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$
$$
\begin{array}{l}
\quad \frac{P A_{1}+P A_{1+2 j}}{P A_{1+1}}=2 \cos \frac{j \cdot 180^{\circ}}{n} . \\
(i, j=1,2, \cdots, i+2 j \leqslant n)
\end{array}
$$ | Proof As shown in the figure, extend $P A_{1+2}$ to $K$ such that $A_{1+21} K=P A_{1}$. Connect $A_{1} A_{1+j}, A_{1+j}$ $A_{1+2j}, A_{1+j} K$, then $=A_{1+1} A_{1+21}$. Therefore, $A_{1} A_{1+1}=A_{1+1} A_{1+21}$. Also, $\angle A_{i+j} A_{i+2} K=\angle A_{1+1} A_{1} P$ (the exterior angle of a cyclic polygon equals th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,064 |
In plane geometry, there is a famous Euler's theorem about the nine-point circle, which states that the midpoints of the three sides of a triangle, the feet of the three altitudes, and the midpoints of the three segments connecting the orthocenter to the vertices, these nine points are concyclic, and this circle is cal... | Prove that, as shown in the figure, $E, F, G, H, I, J$ are the midpoints of $A_{1} A_{4}, A_{1} A_{3}, A_{1} A_{2}, A_{2} A_{4}$, $A_{3} A_{4}, A_{2} A_{3}$ respectively. Let $O$ be the intersection of $\odot E I F$ and $\odot G F J$. Prove that point $O$ lies on $\odot E H G$.
The following reasoning will use the fact... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,066 |
Consider a rectangle with side lengths of $m \times n$ and a rectangle with side lengths of $k \times 1$. By dividing them into several squares with side lengths of 1, we can appropriately select several identical small rectangles to exactly cover the large rectangle. A natural question is, what kind of relationship mu... | A $5 \times 1$ small rectangle can cover.
(2) A $10 \times 10$ square cannot be covered by 25 $4 \times 1$ small rectangles. This is similar to Example 2, where the cells of the square are colored with four colors. The result is that the number of cells of each color is $26, 25, 24, 25$, which means,
2. If $k \mid m$, ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,067 |
16. In a disorderly boxing tournament, many matches were held, all completely at random. After the tournament, the organizers decided to orderly divide the boxers into $n>1$ teams in such a way that for each boxer, at most $\frac{1}{n}$ of the boxers in their team have fought with them. Is such a partition always possi... | 16. Such a partition is always possible. For any partition of all boxers, specify using... reed pipes to indicate a pair of boxers who have fought in the same team. Since the number of possible partitions of boxers is finite, there must be a partition with the minimum number of reed pipes. We assert that this partition... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,069 |
17. If $n$ is a positive integer, find the formula for the sum $\binom{n+1}{1}$
$$
+\binom{n+2}{2} 2^{2}+\binom{n+3}{3} 3^{2}+\cdots+\binom{2 n}{n} n^{2}
$$ | 17. Consider a chess club with $2 n+1$ players, ranked from 1 to $2 n$ +1. Select $n+1$ players to form a team for an upcoming competition with a competitive club. Additionally, choose one captain and one coach from the remaining $n$ players. The same player can hold both positions, but each captain's rank must be at l... | \frac{n(n+1)^{3}\binom{2 n+1}{n+1}}{(n+2)(n+3)} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,070 |
Example 3. Let $D$ be the midpoint of the hypotenuse $BC$ of the isosceles right triangle $ABC$, and let $E$ be on $AB$ such that $AE: EB = 2: 1$. Prove that $CE \perp AD$.
(1978, Wuhu City Junior High School Mathematics Competition) | Proof: Let $A C$ $=B C=x$, then
$$
\begin{array}{c}
A B:=\sqrt{2} x, \\
A B=\frac{3}{3} \sqrt{2} x .
\end{array}
$$
In $\triangle A E C$
$$
\begin{aligned}
C E^{2} & =A C^{2}+A E^{2}-2 A C \cdot A E \cdot \cos 45^{\circ} \\
& =\frac{5}{9} x^{2} .
\end{aligned}
$$
Thus, $C E=\frac{\sqrt{5}}{3} x$.
Line $A K D$ interse... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,071 |
18. $O$ is the circumcenter of $\triangle A D$, $X, Y$ are points on $A C, A B$, respectively, such that $B X$ and $C Y$ intersect at $O$. Also, $\angle B A C=\angle A Y X=\angle X Y C$. Try to find their common value. | 18. It is easy to prove that $O$ is inside $\triangle ABC$. Connect $AO$ and extend it to intersect $BC$ at $Z$. By Ceva's Theorem,
$$
\frac{A Y \cdot B Z \cdot C X}{Y B \cdot Z C \cdot X A}=1.
$$
Since $YX$ bisects $\angle AYC$, we have $\frac{A Y}{Y C}=\frac{A X}{X C}$. Therefore, $\frac{B Y}{Y C}=\frac{B Z}{Z C}$, ... | 50^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,072 |
19. For $0<x<\frac{\pi}{2}$, let $\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}$ take all real numbers except those in the open interval $(a, b)$. Find $a+b$.
---
Note: The function $\operatorname{ctg} x$ is the cotangent function, often denoted as $\cot x$ in English. | 19. Notice that $\operatorname{ctg} 3 x=\frac{\operatorname{ctg}^{3} x-3 \operatorname{ctg} x}{3 \operatorname{ctg}^{2} x-1}$. Let $y=\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}, t=\operatorname{ctg}^{2} x$, then
$$
3 t^{2}-(y+1)t+3 y=0 \text {. }
$$
Thus this equation has real roots, hence
$$
\Delta=(y+1)... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,073 |
20. In $\triangle A B C$, $A B=c, B C=a, C A$ $=b$. Where $a, b, c$ are positive integers, and $\angle A=2$ ( $\angle B$ $-\angle C)$, find all such triples $(a, b, c)$. | 20. If the angle bisector of $\angle B A C$ intersects $B C$ at $D$, then $\angle A D B:=\frac{\angle A}{2}+\angle C=\angle B$, hence $A D=c$. From $A$ draw a line intersecting $B C$ at $E$. Let $B E - D E = x$, then $C D = a - 2x$. Therefore,
$$
b^{2}-(a-x)^{2}=c^{2}-x^{2},
$$
we get $x=\frac{\left(c^{2}+a^{2}-b^{2}\... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,074 |
21. Let when $0 \leqslant x \leqslant 1$, there exists a positive number $q$ such that $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\mathrm{t}}}{q}$ holds, find the smallest positive number $t$ that makes the above inequality true. For this smallest $t$ value, what is the minimum value of $q$ that makes the above inequa... | 21. Since $y=\sqrt{x}$ is strictly concave, for $0<x<1$, we have
$$
q \geqslant 2-(\sqrt{1+x}+\sqrt{1-x}).
$$
After two constant transformations, we get
$$
\begin{array}{l}
q \geqslant x^{1-2}\left(1+\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \\
\cdot\left(1+\sqrt{1-x^{2}}\right).
\end{array}
$$
If $t<2$, then as $x$ app... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,075 |
22. Let $G$ be a connected graph, and for each vertex $V$, associate a non-negative integer $f(V)$, satisfying:
(1) If $V$ and $W$ are adjacent, then
$|f(V)-f(W)| \leqslant 1$.
(2) If $f(V)>0$, then $V$ is adjacent to at least one vertex $W$ such that $f(W)<f(V)$.
(3) There is exactly one vertex $U$ such that $f(U)=0$.... | 22. If $S_{\mathbf{K}}$ denotes the set of all vertices $V$ such that the shortest path connecting $U$ and $V$ has $k$ edges. Since the graph is connected, every vertex belongs to some $S_{\mathrm{k}}$. If $f(V)=k$, then assert that $V$ belongs to $S_{x}$. Clearly, $S_{0}=\{U\}$, and by (3), $U$ is the only vertex sati... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,076 |
24. $f(x)$ is a strictly increasing concave function defined on $0 \leqslant x \leqslant 1$, and satisfies $f(0)=0, f(1)=1$. If $g(x)$ is its quasi-inverse function, prove: when $0 \leqslant x \leqslant 1$, $f(x) g(x) \leqslant x^{2}$. | 24. Let $0<b \leqslant 1$ and $a=f(b)$, then $(b, a)$ lies on the line segment connecting $(0,0)$ and $(1,1)$ or above it. Thus, $a \geqslant b$. Let $c=g(b)$, then $b=f(c)$, and as stated above, $0<c \leqslant b$, so $(c, b)$ lies on the line segment connecting $(0,0)$ and $(b, a)$ or above it. Therefore, we have
$$
b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,078 |
26. Find all positive integers $m, n$ and prime $p$ such that $\binom{m}{2}-1=p^{\mathrm{n}}$ holds. | 26. There is $\frac{(m+1)(m-2)}{2}=p^{n}$. If $m$ is an odd number, then for some integer $t, 0 \leqslant t \leqslant n$, we have $\frac{m+1}{2}=p^{\mathrm{t}}$ and $m-2=p^{n-t}$, thus $p^{n-t}+3=2 p^{t}$. Clearly, $t>0$.
When $n-t>0$, then it must be that $p=3$, hence $3^{n-t-1}+1=2 \cdot 3^{t-1}$.
From $n-t-1=0$, we ... | (m, n, p)= (3,1,2),(4,1,5),(5,2,3),(8,3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,080 |
27. $n$ is a positive integer, $X_{\mathrm{n}}$ represents the set of all $n$-tuples of non-negative integers $\left(x_{1}, x_{2}, \cdots, x_{\mathrm{n}}\right)$, where $\left(x_{1}, x_{2}, \cdots, x_{\mathrm{n}}\right)$ satisfies the equation $x_{1}+x_{2}+\cdots+x_{\mathrm{n}}$ $=n$. And $Y_{\mathrm{n}}$ represents th... | 27. Consider a special set $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ in $X_{n}$. Suppose it is compatible with $\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ in $Y_{n}$. For $1 \leqslant i \leqslant n$, let $z_{1}=y_{1}-x_{1} \geqslant 0$, because
$$
\boldsymbol{z}_{1}+\boldsymbol{z}_{2}+\cdots+\boldsymbol{z}_{n}=2 n-n=n,... | \left|X_{n}\right|^{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,081 |
Example 4. In $\triangle A B C$, $\angle A C B=90^{\circ}, C H$ $\perp A B, A D=D C, C E$ bisects $\angle A C H, D E$ intersects the extension of $C H$ at $F$. Prove: $B F \| C E$.
(1987, Sichuan Province Junior High School League) | Proof as shown in the figure:
5, it is easy to prove
$$
\begin{array}{c}
B E=B C, \\
\frac{A E}{E H}=\frac{C A}{C H} \cdot(1)
\end{array}
$$
From $\frac{B C}{B H}=\frac{C A}{C H}$ and
( 1 ) we get
$$
\frac{A E}{E H}=\frac{B B}{B H}=\frac{B H+E H}{B H}=1+\frac{E H}{B H}
$$
Then $\frac{E H}{B H}=\frac{A E}{E H}-1$.
In ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,082 |
29. If $a, b, c$ are positive real numbers, solve the equation
$$
\begin{array}{l}
\frac{x^{8}+a^{8}}{(x+a)^{3}}+\frac{x^{3}+b^{3}}{(x+b)^{3}}+\frac{x^{3}+c^{3}}{(x+c)^{8}} \\
=\frac{3}{2}-\frac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)} .
\end{array}
$$ | 29. Notice that
$$
\left(x^{3}+a^{3}\right)=(x+a)\left((x+a)^{2}-3 a x\right),
$$
Therefore, the given equation can be rewritten as:
$$
\begin{array}{l}
\frac{a x}{(x+a)^{2}}+\frac{b x}{(x+b)^{2}}+\frac{c x}{(x+c)^{2}} \\
=\frac{x^{8}+(a b+b c+c a) x}{(x+a)(x+b)(x+c)} .
\end{array}
$$
Thus, one root of the equation i... | x=0, \sqrt{a b}+\sqrt{b c}+\sqrt{c a}, \sqrt{a b}-\sqrt{b c}-\sqrt{c a}, -\sqrt{a b}+\sqrt{b c}-\sqrt{c a}, -\sqrt{a b}-\sqrt{b c}+\sqrt{c a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,084 |
30. $P$ is a fixed point outside a known fixed straight line $l$. Consider all lines $m$ passing through $P$ and not coplanar with $l$. Take point $L$ on $l$ and point $M$ on $m$, such that $LM$ is perpendicular to both $l$ and $m$. Find the locus of point $M$. | 30. If $\alpha$ is the plane containing $P$ and perpendicular to $l$, it intersects $l$ at $L^{\prime}$. Any line $m$ passing through $P$ and obliquely intersecting $l$ lies in the unique plane $\beta$ perpendicular to $\alpha$, and $\beta$ intersects $\alpha$ at the line $m^{\prime}$. Let $M^{\prime}$ be the foot of t... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,085 |
Test Question $A-1$. How many evil numbers have a 10-adic representation that is exactly 0,1 alternating and ends with 1? | Let $P_{\mathrm{k}}=1010 \cdots 101$ ( $k$ zeros) be a prime number in decimal notation, $k \geqslant 1$. Clearly, $P_{1}=101$ is a prime number. When $k>2$, if $k$ is odd, then 101 clearly divides $P_{\times 3}$. If $k$ is even, then
$$
\begin{array}{l}
11 P_{k}=11 \cdots 1(2 k+2 \text { ones}) \stackrel{\text { denot... | 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,086 |
Test Question $A-3$. Prove: If the complex number $z$ satisfies
$$
11 z^{n}+10 i z^{9}+10 i z-11=0 \text {. }
$$
then $|z|=1$. | Proof: Let $z=a+bi, a, b \in \mathbb{R}$, then
$$
\left|z^{0}\right|=\left|\frac{11-10 i z}{11 z+10 i}\right|=\frac{f(a, b)}{g(a, b)},
$$
where $f(a, b)=|11-10 i z|$
$$
\begin{array}{l}
=\sqrt{11^{2}+220 b+10^{2}}\left(a^{2}+b^{2}\right), \\
g(a, b)=|11 z+10 i| \\
=\sqrt{11^{2}\left(a^{2}+b^{2}\right)+220 b+10^{2}} .
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,087 |
Test Question $A-5$. Let $G$ be a regular $(2 m+1)$-sided polygon inscribed in the unit circle ($m$ is a positive integer). Prove: there exists a positive constant $A$ (depending only on $m$), such that for any point $P$ inside $G$, there exist two distinct points $V_{1}, V_{2}$, satisfying
$$
\left|P V_{1}-P V_{2}\rig... | Prove that the diameter of $G$ (the maximum distance between any two points in $G$) is $2d$, where $d = \cos \frac{\pi}{2(2m+1)}$. For any given point $P$ in $G$, the value of $PV_1$ for any $2m+1$ points $V_1, V_2, \cdots, V_{2m+1}$ in $G$ falls within the interval $[0, 2d]$. Dividing this interval into $2m$ equal par... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,088 |
Test $B-5$. Let $T$ be the inscribed trapezoid $ABCD$ (counterclockwise) in the unit circle $O$, $ABCDI, AB=s_{1}$, $CD=s_{2}, OE=d, E$ is the intersection point of the heights. When $d \neq 0$, determine the minimum upper bound of $\frac{s_{1}-s_{2}}{d}$. If the minimum upper bound can be achieved, determine all such ... | Solve: As shown in the figure, establish a rectangular coordinate system with $O$ as the origin, $AB \perp x$-axis, and the coordinates of $E$ are $(d, 0)$. The equation of $BD$ can be set as $x - d = k y$, where $k^{-1}$ is the slope of line $BD$. Since $B$ and $D$ are on the unit circle, we have
$$
(d + k y)^{2} + y^... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,089 |
Example 5. As shown in Figure 6, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $AB$, $F$ is the midpoint of $BC$, $AF$ and $DE$ intersect at $I$, and $BD$ and $AF$ intersect at $H$. Find the area of quadrilateral $BEIH$. | Connect $AC$ intersecting $BD$ at $O$,
It is easy to prove $S_{\triangle BCD}=4, S_{\triangle BOD}$ $=S_{\triangle BOC}=S_{\triangle BFB}=1$.
Line $AHF$ intersects Rt $\triangle BOC$, by Menelaus' theorem we get
$$
\frac{BF}{FC} \cdot \frac{CA}{AO} \cdot \frac{OH}{HB}=1 \text {. }
$$
We can get $BH=\frac{2}{3} BO$.
S... | \frac{7}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,090 |
Example 1. Find the minimum value of the quotient of a three-digit number divided by the sum of its digits, and write down this three-digit number.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let the three-digit number be $100x + 10y + z$, and the quotient of the number and the sum of its digits be $w$, where $1 \leqslant x \leqslant 9, 0 \leqslant y, z \leqslant 9$.
$$
w = \frac{100x + 10y + z}{x + y + z} = 1 + \frac{9(11x + y)}{x + y + z}.
$$
In equation (1), if $x$ and $y$ are fixed and $z$ varies, then... | 10 \frac{9}{19} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,091 |
Example 2. In a given acute angle $A O B$, there is a fixed circle $C$. Try to find a point $P, Q, R$ on $\odot C, O A, O B$ respectively, such that $P Q+Q R+R P$ is minimized. | Analysis: If point $P$ is determined on the circle, then the method to determine the positions of $Q, R$ should be considered. As shown in the figure, construct the symmetric points $P_{1}, P_{2}$ of point $P$ with respect to $O A, O B$. Connect $P_{1} P_{2}$, intersecting $O A$, $O B$ at $Q, R$ respectively. According... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,092 |
Example 1. Given an equilateral $\triangle ABC$ with side length 4, points $D, E$, and $F$ are on $BC, CA$, and $AB$ respectively, and $|AE|=|BF|=|CD|=1$. Connect $AD, BE, CF$ to form $\triangle RQS$. Point $P$ moves within $\triangle RQS$ and on its boundary. The distances from $P$ to the three sides of $\triangle ABC... | Proof: Let $P$ be any point inside or on the sides of a regular $\triangle ABC$. The sum of the distances from $P$ to each side, $x+y+z$, is a constant. Consider a line through $P$ parallel to $BC$, intersecting $AB$, $BE$, $AD$, and $AC$ at $U$, $V$, $M$, and $N$, respectively. When point $P$ moves along $UN$, the val... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,093 |
Example 1. Let $a$ be a digit from 1 to 9. Discuss the number of perfect squares in the sequence $a, \overline{a a}, \bar{a} \overline{a a}, \overline{a a a a}, \cdots$. | Analysis: For $a$ equal to $2,3,7,8$, $a$ equal to $1,5,9$, and $a$ equal to $4,6$, respectively. It is not difficult to see that only $1,4,9$ are perfect squares. | 1,4,9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,094 |
Example 2. Is the sum of the squares of five consecutive natural numbers a square number?
Will the above text be translated into English, please keep the original text's line breaks and format, and output the translation result directly. | Let the sum of the squares of five consecutive natural numbers be
$$
\begin{aligned}
A= & (n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2} \\
& +(n+2)^{2} .
\end{aligned}
$$
Simplifying, we get $A=5\left(n^{2}+2\right)$. Clearly, $A$ is a multiple of 5. However, since the units digit of $n^{2}+2$ will never be 0 or 5, $A$ is not a... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,095 |
Example 11. Let $a$ and $b$ be any integers. Prove that the equations
$$
x^{2}+10 a x+5 b+3=0
$$
and $x^{2}+10 a x+5 b-3=0$
have no integer solutions. | To prove that it is only necessary to verify that the discriminants of the two equations are not perfect squares.
$$
\begin{array}{l}
\Delta_{1}=\left(25 a^{2}-5 b-3\right) \times 4, \\
\Delta_{2}=\left(25 a^{2}-5 b+3\right) \times 4 .
\end{array}
$$
Notice that the last digit of $25 a^{2}-5 b=5\left(5 a^{2}-b\right)$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,096 |
Example 12. Discuss whether it is possible to find natural numbers $m$ and $n$ such that
$$
A=3^{m}+3^{n}+1
$$
becomes a perfect square. | To facilitate the discussion, we introduce a notation. Let $a$ be an integer and $m$ a natural number. If there exist integers $q$ and $r$ such that $a = m q + r$, then $r$ is called the remainder of $m$ dividing $a$, denoted as $a \equiv r(m)$. If there is also $b \equiv s(m)$, then
$$
a + b \equiv r + s(m), \quad a \... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,097 |
Example 13. If $x, y$ are natural numbers such that $x^{2}+y^{2}-x$ is divisible by $2 x y$, then $x$ is a perfect square. | Proof: Let $x^{2}+y^{2}-x \doteq 2 k x y$, where $k$ is an integer. Then the quadratic equation in $y$
$$
y^{2}-2 k x y+\left(x^{2}-x\right)=0
$$
has roots that are all integers. Consequently, the discriminant $\Delta=4 x\left[\left(k^{2}-1\right) x+\right.$ 1] must be a perfect square. However, since $x$ and $\left(k... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,098 |
Example 1. Let $G$ be any convex shape tightly sandwiched between two parallel lines $L_{1}$ and $L_{2}$, with its boundary $C$ intersecting both $L_{1}$ and $L_{2}$. A line $L$ parallel to $L_{1}$ divides $G$ into two parts, $A$ and $B$, as shown in the figure, and the distances between $L$ and $L_{1}$, and $L$ and $L... | Let $M$ be one of the intersection points of $G$ and $L_{2}$, and the intersection of $L$ and $G$ must be a line segment, which we can denote as $N P$. To maximize $\frac{S_{A}}{S_{B}}$, $S_{A}$ should be as large as possible, and $S_{B}$ should be as small as possible.
Connect $M N, M P$. Clearly, $\triangle M N P$ i... | \frac{a^{2}+2 a b}{b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,099 |
Example 2. Let $G$ be a convex shape in the plane, $G$ contains the point $(0, 0)$ but no other lattice points. Also, suppose $G$ is distributed with equal area in all four quadrants. Prove: the area of $G$ $\leqslant 4$.
| Analyzing, we only need to prove that the area of $G$ in a certain quadrant is no greater than 1.
$G$ does not contain any lattice points except the origin, and the lattice points closest to the origin are: $(1,0),(0,1),(-1,0)$, and $(0,-1)$.
Obviously, through the point $(1,0)$, we can draw a supporting line of $G$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,100 |
Example 3. If the area of a figure $G$ is greater than or equal to $\frac{\pi}{4}$, then $G$ contains at least one pair of points that are 1 unit apart. | Let the diameter of $G$ be $AB=2d$, and let $d<\frac{1}{2}$. Taking the midpoint of $AB$ as the origin and $AB$ as the positive direction of the $x$-axis, we establish a Cartesian coordinate system. Clearly, the convex shape $G$ is sandwiched between the lines $x=-d$ and $x=d$. The intersection of the line $x=a(-d \leq... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,101 |
Example 4. $n(\geqslant 3)$ parallel lines $M_{1}, M_{2}, \cdots, M_{n}$, if for any three of them a line can be drawn intersecting them, then a line can be drawn intersecting all $n$ line segments. | Analyze the establishment of a Cartesian coordinate system. Without loss of generality, assume that the known $n$ line segments are parallel to the $y$-axis, so $M_{i}$ consists of points $(x, y)$: $x=c_{i}, a_{i} \leqslant y \leqslant b_{i}$, where $a_{i}, b_{i}, c_{i}$ are constants, $i=1,2, \cdots, n$.
The necessary... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,102 |
Example 5. Given 5 points on a plane, any three of which can form the vertices of a triangle. Prove: there exist at least 3 non-overlapping triangles, whose vertices are among the given points.
The text above is translated into English, preserving the original text's line breaks and format. | The convex hull $M$ of five points on a plane can only be one of three cases:
(1) $M$ is a convex pentagon $A B C D E$.
The sum of the interior angles of a convex pentagon is $3 \pi=4 \times \frac{\pi}{2}+\pi$. Therefore, there are at least two obtuse angles among the interior angles.
If two adjacent interior angles $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,103 |
Example 7. Let $A$ be a non-empty finite set of points in three-dimensional space, and connect any two points within $A$ with a line segment (a point $x$ can be considered as a degenerate line segment from $x$ to $x$). The set of all points on these line segments is denoted as $S(A)$. For a given non-empty finite point... | Let $M$ be the convex hull of $A_{0}$. It is easy to see that
$$
A_{0} \subseteq A_{1} \subseteq A_{2} \subseteq \cdots \subseteq M.
$$
To prove the original proposition, it suffices to show that: $M \subseteq A_{2}$.
From the construction method of $S(A)$, we know:
$$
\begin{array}{l}
A_{1}=\left\{Y \mid \overrightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,104 |
Example 1. (Menelaus' Theorem) If a straight line intersects the three sides (or their extensions) of triangle $ABC$, then among the three pairs of segments $A' B$ and $A' C, B' C$ and $B' A, C' A$ and $C' B$, the following relationship holds:
$$
\begin{aligned}
& A' C \cdot B' A \cdot C' B \\
= & A' B \cdot B' C \cdot... | Proof 1: Draw a line $AX // C'A'$, intersecting the extension of $BC$ at $X$. Then we have $\angle BAC': BC' = XA': BA'$, $AB': CB' = XA': CA'$.
Eliminating $XA'$ from these two equations yields equation (1).
Proof 2: Utilize the principle that the ratio of the areas of two triangles with one equal or supplementary ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,105 |
Example 3. Try to find the four-digit number $\overline{x x y y}$, such that it is a perfect square. | Solution: Since the required square number $\overline{x x y y}=11 \cdot(100 x + y)$, therefore, $100 x+y$ can be divided by 11, i.e., $11 | (x + y)$. Noting that $x$ and $y$ are both digits, thus $x+y=11$. After verification, only $x=7, y=4$ satisfies the condition, hence the required four-digit square number is 7744. | 7744 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,106 |
Example 5. Five monkeys found Wang Yiguan's peaches. How can they be fairly divided when one monkey secretly gets up, eats one peach, and the remaining peaches are exactly divided into five equal parts, after which it hides its share and goes back to sleep. The second monkey gets up, also eats one peach, and the remain... | This problem has many solutions, for example, it can first be transformed into an indeterminate equation, and then its positive integer solutions can be sought. However, there is a simpler solution: Imagine that when Sun Wukong, the Great Sage Equal to Heaven, found his subordinates facing such a difficult problem, he ... | 3121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,109 |
Example 1. The numbers 1447, 1005, and 1231 have something in common. That is, each number is a four-digit number starting with $\mathrm{i}$, and in each four-digit number, exactly two digits are the same. How many such four-digit numbers are there? (1st AIME) | Solve for the four-digit numbers that must contain one 1 or two 1s.
(i) The case with two $\hat{\mathrm{i}} 1$s
Choose two numbers from the remaining 9 digits (excluding 1), which has $C_{9}^{2}$ ways. Then, form any permutation of a three-digit number with these two numbers and one 1, which has $P_{3}^{3}$ ways. Thus,... | 432 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,110 |
Example 2. Use 6 white beads, 8 blue beads, and 1 red bead to form a string. How many different ways can this be done?
保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be part of the translation. Here is the corrected version:
Example 2. Use 6 white beads, 8 blue beads... | This is a circular permutation problem. If we fix the red bead, it turns into a linear permutation problem. By the principle of the whole, apart from the red bead, the remaining 14 beads have $N_{1}=\frac{P_{14}^{14}}{P_{6}^{6} \cdot P_{8}^{8}}=3003$ ways to be strung.
Below, we discuss two cases based on whether the ... | 1519 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,111 |
List 3. From $1,2,3, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways are there to select the numbers that meet the above requirements? (1389, National High School League) | Solution: Obviously, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+(14-$ $\left.a_{3}\right)=14$, where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-$ $a_{3} \geqslant 0$.
Transform the above equation to
$$
\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,112 |
Example 4.8 girls and 25 boys form a circle, with at least two boys standing between any two girls. How many different arrangements are there? (Arrangements that can be made to coincide by rotating the circle are considered the same) (1990, National High School Competition) | Co-cultivation $\frac{25!16!}{9!}$ different permutation methods.
Solution reference: "Shou Deng Mathematics Initial 1901 Issue 2, Another Solution to the First Problem of the National Commercial Middle School Competition." There is a unique creative ability.
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。
Co-cultivation $\frac{25... | \frac{25!16!}{9!} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,113 |
List 5. Arrange 8 cards $A A B B C D E F$ in a row, the number of arrangements where the same letter cards are allowed to be adjacent is how many?
Arrange 8 cards $A A B B C D E F$ in a row, with the same letter cards allowed to be adjacent. How many such arrangements are there? | Let the number of ways to arrange 8 cards in a row be $S$, then $|S|=\frac{P_{8}^{8}}{P_{2}^{2} \times \bar{P}_{2}^{2}}$; let the situation where the two cards with the letter $\Lambda$ are adjacent be $A_{1}$, and the situation where the two cards with the letter $B$ are adjacent be $A_{2}$, then $\left|A_{1}\right|:=... | 5760 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,114 |
Example 6. A person wrote 6 letters to 6 different people and prepared 6 envelopes with the recipients' addresses written on them. How many ways are there to place the letters into the envelopes so that no letter matches the recipient on the envelope? (Polish Competition Question) | Let's denote the set of all ways to place 6 letters into 5 different envelopes (i, $2, \cdots, 6$) such that one letter is left out, and the remaining letters are placed into the 5 envelopes in any order, as $\Lambda_{i}$. Then, $\left|A_{i}\right|=P_{5}^{5}=5!$.
Similarly,
$$
\begin{array}{l}
\left|A_{i} \cap A_{j}\ri... | 265 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,115 |
Example 4. It is known that $A$ is a hundred-digit number, 99 of whose digits are 5. Can $A$ be a perfect square? | Assume $A$ is a perfect square, then there are two cases: Case $1^{\circ}$ The unit digit of $A$ is 5, so its tens digit must be 2, i.e., $A=\underbrace{55 \cdots 5}_{98 \uparrow} 25$. For a perfect square with a unit digit of 5, its tens digit should be even, which is a contradiction.
Case $2^{\circ}$ The unit digit ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,117 |
. (iMO-32) Let $P$ be a point inside $\triangle A B C$. Prove that $\angle P A B, \angle P B C, \angle P C A$ at least one does not exceed $30^{\circ}$. | Let $\alpha=\angle P A B, \beta=\angle P B C, \gamma=\angle P C A$. Then, by Proposition 1 and the AM-GM inequality, we have
$$
\sin \alpha \cdot \sin \beta \cdot \sin \gamma \leqslant \frac{P F}{A P} \cdot \frac{P D}{B P} \cdot \frac{P E}{C P}=k .
$$
Let $\lambda=\frac{P D}{A D}, \mu=\frac{P E}{B E}, u=\frac{P F}{C E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,120 |
Example 2. (7th American Mathematical Competition) In the scalene $\triangle ABC$, two altitudes are 4 and 12. If the third altitude is also an integer, then it is most likely to be ( ).
(A) 4.
(B) 5.
(C) 6.
(D) 7.
(E) An answer different from (A)-(D).
Solve the system of inequalities
Translate the above text into ... | $$
\left\{\begin{array}{l}
\frac{1}{4}<\frac{1}{12}+\frac{1}{h} \\
\frac{1}{h}<\frac{1}{4}+\frac{1}{12}
\end{array}\right.
$$
We get $3<h<6$. It is clear that $h_{\max }=5$ (since $h$ is a positive integer). | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,122 |
$\mathbf{A}-\mathbf{1}$. Given a $2 \times 3$ rectangle, with vertices at $(0,0),(2,0),(0,3)$, and $(2,3)$. Rotate it $90^{\circ}$ clockwise around the point $(2,0)$, then $90^{\circ}$ clockwise around the point $(5,0)$, then $90^{\circ}$ clockwise around the point $(7,0)$, and finally $90^{\circ}$ clockwise around the... | As shown in the figure, the area in question includes four right triangles of $1 \times 1$ with an area of $\frac{1}{2}$ each, four right triangles of $1 \times 2$ with an area of 1 each, two quarter-circles with an area of $\frac{\pi}{4}(\sqrt{2})^{2}=\frac{\pi}{2}$ each, and two quarter-circles with an area of $\frac... | \frac{7}{2} \pi+6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,123 |
$\mathbf{A}-2$. Let $A, B$ be different $n \times n$ real matrices. If $A^{3}=B^{3}, A^{2} B=B^{2} A$. Is $A^{2}+B^{2}$ invertible? | Solution: No. Otherwise, $A-B=\left(A^{2}+B^{2}\right)^{-1}\left(A^{2}+B^{2}\right)(A-B)=\left(A^{2}+B^{2}\right)^{-1}\left(A^{3}+B^{2} A-A^{2} B-B^{3}\right)=\left(A^{2}+B^{2}\right)^{-1} \cdot 0=0$, so $A=B$, a contradiction. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,124 |
A-5. Find the maximum value of the following expression:
$$
\int_{0}^{y} \sqrt{x^{4}+\left(y-y^{2}\right)^{2}} d x \text {, }
$$
where $0 \leqslant y \leqslant \mathrm{i}$. | The maximum value is $\frac{1}{3}$, which is obtained when $y=1$. In fact, when $y=1$, $\int_{0}^{1} \sqrt{x^{4}} d x=\frac{1}{3}$. On the other hand, $\int_{0}^{y} \sqrt{x^{3}+\left(y-y^{2}\right)^{2}} d x \leqslant \int_{0}^{y}\left(x^{2}+\left(y-y^{2}\right)\right) d x$ $=y^{2}-\frac{2}{3} y^{3}$. Let $g(y)=y^{2}-\f... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 706,127 |
Example 5. If $a$ is a perfect square, then the number of positive divisors of $a$ must be odd. Conversely, if the number of positive divisors of a natural number $a$ is odd, then $a$ is a perfect square. | Analysis For natural number $a$, 1 and $a$ are both its positive divisors, and the rest of the positive divisors (if any) lie between 1 and $a$. Arrange the positive divisors of $a$ in ascending order:
$$
\begin{aligned}
1 & =a_{1}<a_{2}<\cdots<a_{i}<a_{i+1}<\cdots<a_{m-1}<a_{m} \\
& =a .
\end{aligned}
$$
Notice that ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,128 |
A-6. Let $A(n)$ denote the number of multiplicative partitions $a_{1}+a_{2}+\cdots+a_{r}$ of positive integers whose sum is $n$. These multiplicative partitions satisfy: $a_{1}>a_{2}+a_{3}, a_{2}>a_{3}+a_{4}, \cdots, a_{r-2}>a_{r-1}+a_{r}$, $a_{r-1}>a_{r .}$
$B(n)$ denotes the number of additive partitions $b_{1}+b_{2}... | Proof. Given $a_{1}+a_{2}+\cdots+a_{r}=n$, $a_{1}>a_{2}+a_{3}, \cdots, a_{r-1}>a_{r}$.
Then
$$
\begin{aligned}
n & =a_{r}+a_{r-1}+\cdots+a_{1} \\
& =a_{r}\left(g_{r}-g_{r-1}-g_{r-2}\right) \\
& +a_{r-1}\left(g_{r-1}-g_{r-2}-g_{r-3}\right) \\
& +\cdots+a_{2}\left(g_{2}-g_{1}\right)+a_{1} g_{1} \\
& =a_{r} g_{r}+\left(a_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,129 |
$\mathbf{B}-1$. For any integer $n \geqslant 0$, let $S(n)=n-$ $m^{2}$, where $m$ is the largest integer such that $m^{2} \leqslant n$. The sequence $\left(a_{k}\right)_{k=0}^{\infty}$ is defined as follows: $a_{0}=A, a_{k+1}=a_{k}+S\left(a_{k}\right), k \geqslant$ 0 . For which positive integers $A$ does this sequence... | If $A$ is a perfect square, the sequence will eventually be a constant, as it always equals $A$. Clearly, if the sequence does not contain any perfect squares, it will diverge to infinity. Furthermore, if $a_{n}$ is not a perfect square, and $a_{n+1}=(r+1)^{2}$, if $a_{n} \geqslant r^{2}$, then
$$
\begin{array}{l}
a_{n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,130 |
$\mathbb{E}-2$. Let $f, g$ be non-constant differentiable real-valued functions on $R$, and assume that for any pair of real numbers $x, y$, we have
$$
\begin{array}{l}
f(x+y)=f(x) f(y)-g(x) g(y), \\
g(x+y)=f(x) g(y)+g(x) f(y) .
\end{array}
$$
If $f^{\prime}(0)=0$, prove that for any $x,(f(x))^{2}+$
$$
(g(x))^{2}=1
$$ | Solve: Differentiate the two equations with respect to $y$, we get
$$
\begin{array}{l}
f^{\prime}(x+y)=f(x) f^{\prime}(y)-g(x) g^{\prime}(y), \\
g^{\prime}(x+y)=f(x) g^{\prime}(y)+g(x) f^{\prime}(y) .
\end{array}
$$
Let $y=0$, we get
$$
\begin{array}{l}
f^{\prime}(x)=-g^{\prime}(0) g(x), \\
g^{\prime}(x)=g^{\prime}(0)... | (f(x))^{2}+(g(x))^{2}=1 | Algebra | proof | Yes | Yes | cn_contest | false | 706,131 |
$\mathbf{B}-3$. Does there exist a real number $L$ such that: if $m$, $n$ are integers greater than $L$, then an $m \times n$ rectangle can be expressed as the union of several $4 \times 6$ and $5 \times 7$ rectangles, and any two of these smaller rectangles intersect at most along their boundaries? | There exists.
Proposition If $a, b$ are positive integers, then there exists a number $L_{0}$, such that any multiple of $(a, b)$ (the greatest common divisor of $a$ and $b$) greater than $L_{0}$ can be expressed as $r a + s b$, where $r, s$ are non-negative integers.
Proof of the proposition First, assume $(a, b) = 1... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,132 |
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