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Example 6. In square $A B C D$, $E$ is on $B C$. $B E=2, C E=1$,
$P$ is on $B D$, then what is the minimum value of the sum of the lengths of $P E$ and $P C$?
(Fifth Junior Mathematics Correspondence Competition) | Solve in $\triangle B C P$, $B E: C E=2$,
$$
\therefore P E^{2}=\frac{1}{3} P B^{2}+\frac{2}{3} P C^{2}-2 \text {. }
$$
In $\triangle B C D$, $B C=C D=3$,
$$
\begin{aligned}
B D= & \sqrt{2} B C=3 \sqrt{2}, \\
\therefore P C^{2} & =B C^{2}-B P \cdot P D \\
& =9-B P(3 \sqrt{2}-B P) \\
& =P B^{2}-3 \sqrt{2} P B+9,
\end{a... | \sqrt{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,202 |
1. If two real-coefficient quadratic equations in $x$, $x^{2}+x+a=0$ and $x^{2}+a x+1=0$, have at least one common real root, then $a=$ $\qquad$ | 1. $-2 ;$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,205 |
3. If $x, y$ are real numbers, and $\log _{12}\left(x^{2}+3\right)-$ $\log _{12} x y+\log _{12}\left(y^{2}+3\right)=1$, then the ordered pair $(x, y)=$ | $3 .(\sqrt{3}, \sqrt{3})$ or
$$
(-\sqrt{3},-\sqrt{3})_{3}
$$ | (\sqrt{3}, \sqrt{3}) \text{ or } (-\sqrt{3}, -\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,207 |
Example 7. As shown in the figure, $D$ is a point on side $AC$ of $\triangle ABC$, and $AD$
$$
\begin{array}{l}
: DC=2 \\
: 1, \angle C \\
=45^{\circ}, \\
\angle ADB
\end{array}
$$
$=60^{\circ}$. Prove that $AB$ is the tangent to the circumcircle of $\triangle BCD$. (1987 National Junior High School Mathematics Competi... | Prove $\because A D: D C=2: 1$, according to Corollary 5, we have $B D^{2}=\frac{1}{3} A B^{2}+\frac{2}{3} B C^{2}-\frac{2}{9} A C^{2}$. Also, $\because \angle C=45^{\circ}, \angle A D B=60^{\circ}$. By the Law of Sines, we get $\frac{B D}{\sin \angle C}=\frac{B C}{\sin \angle B D C}$, i.e., $\frac{B D}{\sqrt{2} / 2}=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,213 |
II. Find all integer values of $a$ for which the equation $(a+1) x^{2}-\left(a^{2}+1\right) x$ $+2 a^{3}-6=0$ has integer roots. | When $a=-1$, the original equation has integer roots.
When $a \neq-1$, $\triangle=-7 a^{4}-8 a^{3}+2 a^{2}+24 a+25$. We have
$$
\left\{\begin{array}{c}
\triangle \leqslant-84 a+24 a+25<0, \text { when } a \geqslant 2 \\
\triangle \leqslant-10|a|^{2}-24|a|+25<0, \\
\text { when } a \leqslant-2
\end{array}\right.
$$
The... | -1,0,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,216 |
In $\triangle A B C$, the opposite sides of $A, B, C$ are $a, b, c$, respectively, and $a > b > c$. $A S, A S^{\prime}$ are the angle bisector and external angle bisector of $\angle A$, $B T, B T^{\prime}$ are the angle bisector and external angle bisector of $\angle B$, $C U, C U^{\prime}$ are the angle bisector and e... | Three, Proof: In a triangle, since $A S, A S^{\prime}$, are the angle bisector and the external angle bisector of $\angle A$,
$$
\begin{array}{l}
\therefore \frac{C S}{S B}=\frac{b}{c}=\frac{C S^{\prime}}{S^{\prime} B}, \\
\therefore \frac{C S}{B C}=\frac{b}{b+c}, \frac{C S^{\prime}}{B C}=\frac{b}{b-c},
\end{array}
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,217 |
Four, there are 1990 points distributed on a straight line. We will mark the midpoints of all possible line segments with these points as endpoints. Try to find the minimum number of distinct midpoints that can be obtained.
untranslated text:
直线上外布着1990个点, 我们来标出以这些点为端点的一切可能的线段的 中点. 试求至少可以得出多少个互不重合的中点.
translated te... | Four, Solution: Let the two points that are farthest apart be denoted as $A, B$. Consider the 1988 line segments formed by $A$ and the other 1988 points excluding $B$. The midpoints of these 1988 line segments are all distinct, and their distances to point $A$ are less than $\frac{1}{2} A B$.
Similarly, for point $B$,... | 3977 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,218 |
Example 8. In $\triangle A B C$, $A B>A C, A E$ bisects $\angle A$, and intersects $B C$ at $E$. There is a point $S$ on $B C$ such that $B S=E C$. Prove: $A S^{2}-A E^{2}=(A B-$ $A C)^{2}$. (1979 Jiangsu Province Mathematics Competition) | $$
\begin{array}{l}
A E^{2}=A B \\
\cdot A C-B E \\
\cdot E C, \\
A S^{2}=A B^{2} \\
- \frac{S C}{B C}+A C^{2} \\
\cdot \frac{B S}{B C}-B S \cdot S C . \\
\therefore A S^{2}-A E^{2}=A B^{2} \cdot \frac{S C}{B C}+A C^{2} \cdot \frac{B S}{B C} \\
-A B \cdot A C+B E \cdot E C-B S \cdot S C . \quad(1) \\
\because B S=E C, ... | A S^{2}-A E^{2}=(A B-A C)^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 705,224 |
3. As stated, for square $A B C D$ with side length $a$, $E$ is a point on $D C$, and the length of $D E$ is $b$. The median of $A E$ intersects $A D, A E, B C$ at points $P, M, Q$ respectively. Then $P M: M Q=$ | 3. $b:(2 a-b)$ | b:(2 a-b) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,228 |
4. Let $A$ be a three-digit number, and $B$ be a two-digit number, and $A: B=3: 1$. If placing $B$ to the left of $A$ forms a five-digit number $C$, and placing $A$ to the left of $B$ also forms a five-digit number $D$, and $D$ is 40014 less than $C$, then, the last digit of the number $D^{\mathrm{A}}$ is $\qquad$ | 4. 3.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,229 |
Given $A B C D$ is a square, $M, N$ are on $A B, B C$ sides respectively, and $B M=B N$, also $B P \perp M C$ at $P$.
Prove: $P D \perp P N$. | Given $\triangle B P M \sim \triangle C P B$, it can be deduced that: $\triangle P C D \sim \triangle P B N$, thus $\angle N P D=\angle B P C$ $=90^{\circ}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,232 |
Three, as shown in the figure, divide the regular hexagon into $3 n^{2}$ small equilateral triangles, and label these small triangles with numbers $1,2, \cdots, m$, such that triangles with adjacent numbers have adjacent sides.
(1) When $n=4$, please provide a labeling method according to the above requirements, so tha... | Three, (1) Numbered as shown in the figure.
(2) Color each small triangle with one of two colors, black or white, so that triangles with adjacent sides are colored differently. The difference in the number of triangles of the two colors is $n$, and among the triangles with adjacent numbers, the difference in their numb... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,234 |
Example 2. A chord $CD$ of fixed length less than the diameter has its endpoints sliding on the arc $\overparen{AB}$. Try to prove that regardless of the position of $CD$, the triangle formed by the projections $E, F$ of $C, D$ on the diameter $AB$ and the midpoint $P$ of $CD$ is always a similar isosceles triangle. | Let's choose
$C D \| A B$
this special
position to explore
(Figure 2).
It is easy to see that at this time,
$\triangle P E F$ is
an isosceles triangle,
and
$\triangle P E F$
$\sim \triangle O C D$.
When the chord $C D$ moves,
which relationships remain unchanged? See (Figure 3):
$O P \perp C D ; \triangle O C D$ shape ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,246 |
Three, if the two roots $\alpha, \beta$ of the equation $x^{2}-3 x+1=0$ are also roots of the equation $x^{8}-p x^{2}+q=0$, where $p, q$ are integers, find the values of $p, q$.
---
The translation is provided as requested, maintaining the original text's line breaks and format. | Three, Hint: Using Vieta's formulas, we get $p=48$, $q=7$. $\quad$. | p=48, q=7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,250 |
Four, try to find such a positive integer $a$, so that the quadratic equation
$$
a x^{2}+2(2 a-1) x+4(a-3)=0
$$
has at least one integer root. | Given $-4 \leqslant x \leqslant 2$, by testing integer values of $x$, we can obtain $a=1, 3, 6, 10$. | a=1, 3, 6, 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,251 |
Five, as shown in the figure, $P, Q$ are points on the sides $AB, BC$ of the square $ABCD$, respectively, and $BP = BQ$. A perpendicular line is drawn from point $B$ to $PC$, with the foot of the perpendicular being $H$. Prove: $DH \perp HQ$.
保留源文本的换行和格式，直接输出翻译结果如下：
Five, as shown in the figure, $P, Q$ are points on ... | Five, Hint: From $\triangle B C P \backsim \triangle H C B$, we can deduce
$$
\begin{array}{l}
\triangle H B Q \sim \triangle H C D . \\
\text { Hence } \angle D H Q=\angle C H B \\
=90^{\circ} .
\end{array}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,252 |
Example 2. For $\triangle A B C=$ with sides $a, b, c$, construct squares outward on each side, with areas sequentially $S_{a}, S_{b}, S_{c}$. If $a+b+c=18$, find the minimum value of $S_{\mathrm{a}}+S_{\mathrm{b}}+S_{\mathrm{c}}$. | Given $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ $\geqslant 0$, which means $2\left(a^{2}+b^{2}+c^{2}\right)=2(a b+b c+c a)$. Therefore, $3\left(a^{2}+b^{2}+c^{2}\right) \geqslant a^{2}+b^{2}+c^{2}+2 a b$ $+2 a c+2 b c=(a+b+c)^{2}$. Thus, $S_{1}+S_{v}+S_{r} \geqslant \frac{18^{2}}{3}=108$.
When $a=b=c$, the equality holds, so th... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,257 |
2. The teacher distributed a stack of books among A, B, C, L, E. $\frac{1}{4}$ of the books were given to $B$, then $\frac{1}{3}$ of the remaining books were given to $C$, and the rest were split equally between $D$ and $E$. If $E$ received 6 books, then the teacher originally had $\qquad$ books. | 2. 48 | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,261 |
$=$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $3-\sqrt{5}$ | 3-\sqrt{5} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,262 |
Example 3. $A, B$ are two fixed points, try to find point $C$ on the known line $l$ ($A, B$ are on the same side of $l$) such that $A C^{2}+B C^{2}$ is minimized. | Take the midpoint $M$ of $A B$, and take any point $X$ on $l$, connect $M X$, then in $\triangle A X B$, $X M$ is a median. By the median length formula, we have
$$
\begin{array}{l}
4 X_{M^{2}}=2\left(A X^{2}+B X^{2}\right)-A B^{2}, \\
\text { hence } A X^{2}+B X^{2}=\frac{4 X M^{2}+A B^{2}}{2} .
\end{array}
$$
The ri... | M C \perp l | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,268 |
Five, arrange $p$ boys and $q$ girls around a circle $(p \geqslant 0, q \geqslant 0, p+q \geqslant 2)$, and let the number of groups of adjacent boys be $a$, and the number of groups of adjacent girls be $b$.
Prove: $a-b=p-q$. | Five, Proof
1. If $p+q \leqslant 3$ and $p=0$ or $q=0$, the proposition is obviously true.
2. If $p+q \geqslant 4, p q \neq 0$, swapping a pair of adjacent boys and girls does not change the value of $a-b$. Repeated swaps can lead to a special arrangement of the children, thus $a-b=p-q$. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,273 |
Example 1. (1983 Shanghai Mathematics Competition) For a natural number $n$, consider the quadratic equation $x^{2}+(2 n+1) x+n^{2}=0$, with its two roots being $\alpha_{\mathrm{n}}, \beta_{\mathrm{n}}$. Find the value of the following expression:
$$
\begin{array}{l}
\frac{1}{\left(\alpha_{3}+1\right)\left(\beta_{3}+1... | \[
\begin{array}{l}
\text { Solve } \alpha_{\mathrm{n}}+\beta_{\mathrm{n}}=(2 n+1), \alpha_{\mathrm{n}} \beta_{\mathrm{n}}=n^{2}, \\
\text { then }\left(\alpha_{n}+1\right)\left(\beta_{n}+1\right)=n(n-2). \\
\text { The original expression }=\frac{1}{3 \times 1}+\frac{1}{4 \times 2}+\cdots+\frac{1}{20 \times 18} \\
=\f... | \frac{531}{760} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,279 |
Three, given three numbers $89, 12, 3$. Perform the following operations: take any two of these numbers, find their sum and divide by $\sqrt{2}$, and simultaneously find their difference and divide by $\sqrt{2}$. Can we obtain the three numbers $90, 10, 14$ after several such operations? Prove your conclusion. | Three, Prompt: Cannot: Keep the sum of the squares of three numbers plus one unchanged in each step of the operation. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,285 |
Five, there are 1990 matches on the table, and two children, A and B, take turns to pick 1, 2, or 3 matches each time. Whoever picks the last match wins. A goes first. Which child will win? How should he play this game? | Five, Hint: A wins. As long as he can make the remaining number of matches a multiple of 4 each time he picks them up. | A | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,287 |
Six, 1990 points are arranged from left to right on a straight line: $P_{1}, P_{2}, \cdots, P_{1990}$, and it is known that point $P_{k}$ is the $k$-th equal division point closest to $P_{k+1}$ on the line segment $P_{k-1} P_{k+1}$ $(2 \leqslant k \leqslant 1989)$, for example, $P_{5}$ is the 5th equal division point c... | Six, Hint: $P_{k} P_{k+1}=\frac{1}{k+1} P_{k-1} P_{k}(k$ $\geqslant 2$ ). | \frac{1}{l}=1988 \cdot 1987 \cdot \cdots \cdot 3 \cdot 2 | Algebra | proof | Yes | Yes | cn_contest | false | 705,288 |
$$
\text { 1. Given } x^{2}-3 y^{2}=2 x y, \quad x>0, y>0 \text {, }
$$
then the value of $\frac{x+2 y}{x-y}$ is | 1. $\frac{5}{2} ;$ | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,294 |
Given any six points on a plane, with no three points collinear. Prove: There always exist three points such that the triangle formed by these three points has an angle not exceeding $30^{\circ}$. If $30^{\circ}$ is changed to $29^{\circ}$, does the conclusion still hold? | Five, Prompt:
1. There exists a line $L$ passing through two points, such that all other points are on the same side of $L$. Divide into two cases to prove the proposition: $\angle B A F \leqslant 120^{\circ}$ or $\angle B A F > 120^{\circ}$.
2. Changing to $29^{\circ}$ does not hold. A counterexample is the six vertic... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,300 |
Example 3. (Zu Chongzhi Cup Mathematics Competition) The sum of all digits of the natural numbers $1, 2, 3, \ldots, 9999$ is $\qquad$ | Solve $0+9999, 1+9998, \cdots, 4999+5000$
all sum to 9999 without any carry, so the sum of digits is more:
$$
4 \times 9 \times 5000 \div 2-5=89995 \text {. }
$$
This is called the method of integer value combination. | 89995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,301 |
Six, in all integers that start and end with 1 and alternate between 1 and 0 (i.e., $101$, $10101$, $1010101$, etc.), how many of them are prime numbers? | Six, Proof that there is only one prime number 101.
If $n \geqslant 2$, then $A=10^{2 n}+10^{2 n-2}+\cdots+10^{2}$
$$
+1=\frac{\left(10^{\mathrm{n}+1}-1\right)\left(10^{\mathrm{n}+1}+1\right)}{99} \text {. }
$$
When $n=2 m+1$, $\frac{10^{2 m+2}-1}{99}=10^{2 \mathrm{~m}}$ $+\cdots+10^{2}+1, A$ is a composite number;
Wh... | 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,302 |
3. Let $[x]$ denote the greatest integer not exceeding $x$. If
$$
\begin{array}{l}
f=[1 \mathrm{~g} 1]+[1 \mathrm{~g} 2]+[1 \mathrm{~g} 3]+\cdots+[1 \mathrm{~g} 1989] \\
+[1 \mathrm{~g} 1990], \text { then } f=
\end{array}
$$ | 3. $4863 ;$ | 4863 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,305 |
2. Let $y=|x-a|+|x-15|+|x-a-15| (0<a<15)$, if $a \leqslant x \leqslant 15$, then the minimum value of $y$ is ( ).
( A ) 30 .
(B) 25 .
( C ) 15.
(D) 7.5 . | $\begin{array}{l}2 \\ C\end{array}$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,307 |
4. A fraction with a numerator of 1 and a denominator that is a natural number greater than 1 is called a unit fraction. If $\frac{1}{8}$ is expressed as the sum of two different unit fractions, then the number of all possible combinations is ( ).
(A) 1 combination.
(B) 2 combinations.
(C) 3 combinations.
(D) 4 combina... | $\begin{array}{c}4 \\ - \\ C\end{array}$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,309 |
Three, let rational numbers $a, b$ satisfy the equation $a^{5}+b^{5}=2 a^{2} \cdot b^{2}$, prove that 1-ab is the square of a rational number. | Three, Proof: If $a \cdot b=0$, the conclusion holds.
If $a \neq 0$, let $b=k a$, then $a=\frac{2 k^{2}}{1+k^{2}}$, $1-a b=1-\frac{4 k^{5}}{\left(1+k^{5}\right)^{2}}=\left(\frac{1-k^{5}}{1+k^{5}}\right)^{2}$. The proposition holds. | \left(\frac{1-k^{5}}{1+k^{5}}\right)^{2} | Algebra | proof | Yes | Yes | cn_contest | false | 705,310 |
Four, as shown in the figure, the radius of $\odot O$ is $R$, $AB$ and $CD$ are two diameters of $\odot O$, and $\widehat{AC}=60^{\circ}$. Take any point $P$ on $\widehat{CB}$, and let $PA$, $PD$ intersect $CD$, $AB$ at $E$, $F$. Let $m=AE \cdot AP + DF \cdot DP$. Student A says: “$m$ is a variable.” Student B says: “$... | Connect $A C, B D$. From the fact that $O, E, P$, $F$ are concyclic, we get:
$$
\begin{array}{l}
m=A O \cdot A F+D O \cdot D E \\
=R(R+O E+R+O F) \\
=R(2 R+O F+C E) \\
=3 R^{2} \text { (constant). }
\end{array}
$$ | 3R^2 | Geometry | proof | Yes | Yes | cn_contest | false | 705,311 |
Example 3. Given an $8 \times 8$ chessboard, change the color of each square within every $2 \times 2$ internal square to the opposite color. In the end, can there be exactly one white square left on the board? | In an $8 \times 8$ chessboard, there are 32 black squares and 32 white squares. Assign the value 1 to each black square and the value -1 to each white square. Initially, there are 32 ones and 32 negative ones, and their product $S_{0}=1^{32} \cdot(-1)^{32}=1$.
Each operation changes the color of $2 \times 2=4$ squares... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,312 |
Example 4. (Shanghai Mathematics, 1987)
$$
\log _{3}\left[(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right) \cdot \ldots\right.
$$
- $\left.\left(3^{84}+1\right)+\frac{1}{2}\right]+\log _{3} 2$ The value is ( ).
(A) 32.
( B) 64.
(C) 128.
(D) None of the above. | $$
\begin{aligned}
\text { Original expression }= & \log _{3}(3-1)(3+1)\left(3^{2}+1\right) \\
& \cdot \cdots \cdot\left(3^{84}+1\right) \\
= & \log _{3} 3^{128}=128 .
\end{aligned}
$$
Expressing "3-1" as 2 in the second line is a kind of simplification. This can be referred to as the method of adding factors. | 128 | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,313 |
Five, Given: The two roots of the equation $x^{2}+85 x+1=0$ are $m, n$, and the two roots of the equation $x^{2}+96 x+1=0$ are $p, q$. Find: $(m+p) \cdot(n+p) \cdot(m-q) \cdot(n-q)$. | Five, Hint: Contrary to what it seems. 1791.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,314 |
Six, point $P$ is inside
$\square A B C D$.
$P A, P B, P C, P D$
divide the quadrilateral $ABCD$ into four
triangles, their areas are $a, a r, a r^{2}, a r^{3}(a>0, r>0)$. Determine the position of point $P$. Please explain your reasoning. | Six, Hint: Using $S \triangle P^{A} A_{R}+S \triangle P C$ $=S \triangle \mathrm{PBC}+S \triangle \mathrm{PDA}$, we can get $r=1$, so point $P$ is the intersection of $A C$ and $B D$.
| r=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,315 |
1. Factorize: $4 x^{3}-31 x+15=$ | 1. $(2 x-1)(2 x-5)(x+3)$ | (2 x-1)(2 x-5)(x+3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,324 |
4. In $\triangle A B C$, $\angle C=90^{\circ}, A C=\frac{1}{2} B C$
$\angle A$ bisector $A D$ intersects $C B$ at $D$, and a point $G$ is taken on $D B$ such that $D G=C D$, then $C G: G B=$ $\qquad$ | 4. $\frac{\sqrt{ } \overline{5}+1}{2}$. | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,328 |
1. Two cars start from the same location at the same time, driving in the same direction at the same speed. Each car can carry a maximum of 24 barrels of gasoline. They cannot use any other fuel during the journey. Each barrel of gasoline can make a car travel 60 kilometers. Both cars must return to the starting point,... | If so, A gives B $(24-2 x)$ barrels of gasoline. B continues to move forward, carrying $(24-2 x)+(24-x)=48-3 x$ barrels of gasoline. According to the problem, we should have
$$
48-3 x \leqslant 24 \text {. That is, } x \geqslant 8 \text {. }
$$
After A and B separate, the distance B continues to travel is
$$
\begin{ar... | 1920 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,344 |
3. There is a sequence of numbers which are $1, 5, 11, 19, 29$, $A, 55$, where $A=$ $\qquad$ . | 3. 41; | 41 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,346 |
3. Find the positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$. | 3. Solution Since $x, y, z$ are positive integers and
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6} \leqslant 1 .
$$
Assume without loss of generality that $1<x \leqslant y \leqslant z$, then $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}$.
Thus, $\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslan... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,347 |
2. If $\sqrt{a-1}+(a b-2)^{2}=0$, then $\frac{1}{a b}$
$$
+\frac{1}{(a+1)(b+1)}+\cdots+\frac{1}{(a+1990)(b+1990)}
$$ | 2. $\frac{1991}{1992}$; | \frac{1991}{1992} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,354 |
3. Given $a, b, c$ satisfy $a+b+c=0, a b c$ $=8$, then the range of values for $c$ is $\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | $\begin{array}{l}3 . c>0 \text { or } \\ c \geqslant 2 \sqrt[3]{4} ;\end{array}$ | c \geqslant 2 \sqrt[ | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,355 |
1. From the natural numbers 1, 2, 3, ..., 354, choose 178 numbers. Prove: among them, there must be two numbers whose difference is 177. | Three, 1. Proof: Divide the natural numbers $1, 2, 3$, …, 354 into the following 177 groups:
$$
(1,178),(2,179),(3,180),
$$
$\cdots, \quad(177,354)$. Each pair of numbers in a group has a difference of 177. From the given 354 numbers, if any 178 numbers are chosen, by the pigeonhole principle, there must be two numbers... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,359 |
$A B C D, A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, H square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ has its vertex $A^{\prime}$ at the center of square $A B C D$. When square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ rotates around $A^{\prime}$, the area of the overlapping part of the two squares is alwa... | 2. Prove that when the sides of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are parallel to the corresponding sides of $A B C D$ (as shown by the dashed lines), it is easy to see that the overlapping area is $\frac{1}{4}$ of the square's area. When rotated to the position of the solid square $A^{\prime} B^{\prime} C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,360 |
3. Among all possible four-digit numbers formed using the digits $1,9,9,0$, for each such four-digit number and a natural number $n$, their sum when divided by 7 does not leave a remainder of 1. List all such natural numbers $n$ in descending order.
$$
n_{1}<n_{2}<n_{3}<n_{4}<\cdots \cdots,
$$
Find: the value of $n_{1... | 3. Solution
$1,0,9,0$ digits can form the following four-digit numbers: 1099, 1909, 1990, 9019, 9091, 9109, $9190,9901,9910$ for a total of nine. The remainders when these nine numbers are divided by 7 are $0,5,2,3,5,2,6$, 3 , 5 . Since $n$ and their sum cannot be divisible by 7 with a remainder of 1, $n$ cannot have a... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,361 |
5. Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}$ $=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$. Then $x-a-b-c=$ | 5. 0 ;
The above text has been translated into English, maintaining the original text's line breaks and format. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,362 |
Three, A and B start from locations $A, B$ respectively, heading towards each other. They meet at point $C$ on the way, after which A takes 5 hours to reach $B$, and B takes $3 \frac{1}{5}$ hours to reach $A$; it is known that A walks 1 kilometer less per hour than B. Find the distance between $A$ and $B$. | Three, Hint: Let $A C=S_{1}, C B=S_{2}$, the time required for both to meet is /small hour, then we get $t=4$ hours. From the problem, we solve to get $S_{1}=16, S_{2}=20$. Therefore, $A B=35$ km. | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,377 |
Example 1. In $\triangle A B C$, $A B=A C=2, B C$ side has 100 different points $P_{1}, P_{2}, \cdots, P_{1} 00$. Let $m_{1}=A P_{1}^{2}+B P_{1} \cdot P_{1} C(i=1,2, \cdots, 100)$, then $m_{1}+m_{2}+\cdots+m_{100}$ equals what? (1990 National Junior High School Mathematics League) | Solve $\because A B$ $=A C$, according to Corollary 1 we have
$$
\begin{array}{c}
A P_{1}^{2}=A B^{2} \\
-B P_{\mathrm{i}} \cdot P_{1} C, \\
\therefore A P_{1}^{2} \\
+B P_{1} \cdot P_{1} C \\
=A B^{2}, \\
\therefore m_{\mathrm{i}}=A P_{1}^{2}+B P_{1} \cdot P_{\mathrm{i}} C \\
=A B^{2}=2^{2}=4,
\end{array}
$$
that is
... | 400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,378 |
Observe several sets of numbers: $3,4,5 ; 5,12,13$; $7,24,25 ; \cdots$. (1) If $a, b, c$ represent the first, second, and third numbers in each set, respectively, write the equation that $a$, $b$, $c$ satisfy; (2) Indicate how to derive the next two numbers from the first number $a$ (where $a$ is an odd number), and pr... | (1) $\boldsymbol{a}^{\mathbf{2}}+b^{2}=c^{2}$;
(2) Let $\boldsymbol{a}=2 n+1$, then
$$
\begin{array}{l}
b=n(2 n+1)+n, \quad b=\frac{1}{2}\left(a^{2}-1\right) ; \\
c=n(2 n+1)+n+1, \quad c=\frac{1}{2}\left(a^{2}+1\right) .
\end{array}
$$
(3) Proof omitted. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,379 |
Example 2. In $\triangle A B C$, the angle bisector of $\angle C$ intersects $A B$ at $D$. Prove: $C D<\sqrt{ } C \overline{A \cdot C B}$. (Hungarian Middle School Mathematics Competition) | Prove that $CD$ bisects $\angle C$, according to Corollary 3 we know
$$
\begin{aligned}
& C D^{2} \\
= & C A \cdot C B \\
- & A D \cdot D B
\end{aligned}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,389 |
$\{x\}=x-[x]$. Then the equation $x+2\{x\}=3[x]$ has a solution of $\qquad$ .
Let's translate the solution part as well:
To solve the equation $x + 2\{x\} = 3[x]$, we start by expressing $x$ in terms of its integer and fractional parts:
\[ x = [x] + \{x\} \]
Substituting this into the equation, we get:
\[ [x] + \{x\... | 3. $\frac{5}{3}$ | 0 \text{ or } \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,390 |
5. Given a convex quadrilateral $ABCD$ with area $S$, $O$ is a point outside the quadrilateral, $M_{1}, M_{2}, M_{3}, M_{4}$ are the centroids of $\triangle OAB$, $\triangle OBC$, $\triangle OCD$, $\triangle ODA$ respectively. Then the area of quadrilateral $M_{1} M_{2} M_{3} M_{4}$ is $\qquad$ | 5. $\frac{2}{9} S$.
untranslated text remains in its original format and position. | \frac{2}{9} S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,392 |
Three, let $x<0$ and $x-\frac{1}{x}=\sqrt{5}$. Find the value of the rational expression $\frac{x^{10}+x^{0}+x^{4}+1}{x^{16}+x^{8}+x^{2}+1}$. | Three, Hint: From $x-\frac{1}{x}=\sqrt{5}, x<0$, deduce that $x+\frac{1}{x}=-3$. Then transform the required expression into the form $x^{\mathrm{n}}+\frac{1}{x^{\mathrm{n}}}$. The value obtained is $\frac{42}{47}$. | \frac{42}{47} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,393 |
$$
\begin{array}{l}
\text { Four, Prove: } 1991^{1992}+1993^{1994} \\
+1995^{1996}+1997^{1998}+1999^{2000}
\end{array}
$$
is divisible by 5. | Four, Hint: By discussing the regularity of the powers of the unit digits, we know that the unit digits of the terms in the sum are $1, 9, 5, 9, 1$, and their sum is 25. Therefore, the sum can be divisible by 5. | 25 | Number Theory | proof | Yes | Yes | cn_contest | false | 705,394 |
Six, In pentagon $A B C D E$, $\angle A B C$ $=\angle A E D$ $=90^{\circ}, \angle B A C$ $=\angle E A D, M$ is the midpoint of $C D$.
Prove:
$M B=M E$ | Six, Proof: Take the midpoints $P, Q$ of $A C, A D$, and connect the segments as shown in the figure.
$$
\begin{array}{c}
\because U^{\prime}, M, Q \text{ are the midpoints of the sides of } \angle A C D^{-}. \\
\therefore A P=M Q, \\
A Q=P M . \\
\therefore B P=M Q, \\
P M=Q E . \\
\text { Also } \because 1=2 \angle ;... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,396 |
Given $\triangle A B C$ has a perimeter of 20, area of $01 \sqrt{3}$, and $\angle A=60^{\circ}$. Find $\sin A: \sin B: \sin C$. | Seven, Hint: By the cosine theorem, we get $(b+c)^{2}-a^{2}$ $=120$. Then, we get the system of equations $\left\{\begin{aligned} b+c & =13, \\ b c & =40 \text {. Thus, we have }\end{aligned}\right.$ $\sin A: \sin B: \sin C=7: 8: 5$ | 7: 8: 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,397 |
7. Let $b=1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-\cdots$ $-1988^{2}+1989^{2}$. When $b$ is divided by 1991, the remainder is $(.$.
(A) 0 .
(B) 1 .
(C) 2 .
(D) 4 . | $\begin{array}{c}7 \\ B\end{array}$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,405 |
Example 3. In $\triangle A B C$, $D$ is a point on side $B C$. It is known that $A B=13, A D=12, A C=15, B D=5$. What is $D C$? (3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | According to Stewart's Theorem, we have
$$
A D^{2}=A B^{2} \cdot \frac{C D}{B C}+A C^{2} \cdot \frac{B D}{B C}-B D \cdot D C \text {. }
$$
Let $D C=x$,
then $B C=5+x$.
Substituting the known data
into the above equation, we get
$$
\begin{array}{l}
12^{2}=13^{2} \\
\cdot \frac{x}{5+x}+15^{2} \cdot \frac{5}{5+x}-5 x,
\e... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,411 |
Three, given that $a$ is an integer, the equation $x^{2}+(2 a+1) x$ $+a^{2}=0$ has integer roots $x_{1}, x_{2}, x_{1}>x_{2}$. Try to find the value of $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$. | Three, Solution: From the problem, we know that the discriminant $4a + 1$ is a perfect square, so $a \geqslant 0$. Since $4a + 1$ is odd, we can set $(2k + 1)^2 = 4a + 1$, solving for $a = k(k + 1)$.
$$
x_{1,2} = \frac{1}{2} \left[ \left( -2k^2 - 2k - 1 \right) \pm \sqrt{(2k + 1)^2} \right].
$$
When $k \geqslant 0$, $... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,412 |
Four, let $B D, C E$ be the angle bisectors of $\angle B$, $\angle C$ in $\triangle A B C$, and $\angle A E D: \angle D E C = \angle A D E: \angle E D B$. Prove that $\triangle A B C$ is an isosceles triangle. | Four, Hint: Let $\angle A E D: \angle D E C$ $=\angle A D E: \angle E D B=k$, solve to get $k=2$. Then prove $\triangle D C E \cong \triangle E B D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,413 |
Five, there is a large vat of water and 5 measuring cups, with capacities of $1$, $5$, $25$, $125$, $625$ (liters). Try to prove: for any integer $a$ not exceeding 1562, it is possible to use these 5 measuring cups, each at most twice, to pour $a$ liters of water into an empty bucket.
(Explanation: Pouring a cup of wat... | Consider the polynomial
$$
625 b+125 c+25 d+5 e+f \text {, }
$$
where each letter represents the number of times a measuring cup with a capacity corresponding to its coefficient (in milliliters) is used (if the letter is positive, it indicates water is poured from the water jug into the bucket; if the letter is negati... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,414 |
Example 4. As shown in the figure, points $A, B, C, D$ lie on the same circle, and $BC=DC=4, AE=6$. The lengths of segments $BE$ and $DE$ are both positive integers. What is the length of $BD$? (1988
National Junior High School Mathematics Competition) | Solve in $\triangle B C D$, $B C=D C$, H
Inference 1 gives
$$
C E^{2}=B C^{2}-B E \cdot D E,
$$
$\because A, B, C, D$ are concyclic,
$$
\begin{array}{l}
\therefore B E \cdot D E=A E \cdot C E, \\
\therefore C E^{2}=B C^{2}-A E \cdot C E=4^{2}-6 \times C E,
\end{array}
$$
which is $C E^{2}+6 \cdot C E-16=0$.
Solving gi... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,422 |
3. Factorize: $a^{3}+2 a^{2}-12 a+15$ $=$ $\qquad$ - If $a$ is a certain natural number, and the above expression represents a prime number, then this prime number is | 3. $\left(a^{2}-3 a+3\right)(a+5), 7$; | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,423 |
5. In $\triangle A B C$, $A B>A C$, the altitude $A D$ and the median $A E$ as well as the angle bisector $A F$ of $\angle A$ divide $\angle A$ into four equal parts, then $\angle A=$ $\qquad$ degrees,
$\angle B=$ $\qquad$ degrees, $\angle C=$ $\qquad$ degrees. | $5.90^{\circ}, 22.5^{\circ}, 67.5^{\circ}$ | 90^{\circ}, 22.5^{\circ}, 67.5^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,425 |
Three, real numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a^{2}-b c-6 a+3=0, \\ b^{2}+c^{2}+b c-2 a-1=0 .\end{array}\right.$ Find the range of values for $a$. | Three、 $\frac{10-2 \sqrt{19}}{3} \leqslant a \leqslant \frac{10+2 \sqrt{19}}{3}$. | \frac{10-2 \sqrt{19}}{3} \leqslant a \leqslant \frac{10+2 \sqrt{19}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,427 |
Let $\triangle A B C$ have three sides $a, b, c$. Prove: (1) If $\angle A=2 \angle B$, then $a^{2}=b(b+c)$; ( 2 ) If $\angle A=3 \angle B$, then $c^{2}=-\frac{1}{b}(a-b)\left(a^{2}-b^{2}\right)$. | (1) As shown in the figure, draw
the angle bisector of
$\angle A$ intersecting
$B C$ at $D$. It is easy to see that
$\triangle A B C \sim D A C$.
Therefore, we have
$$
\frac{c}{A \bar{D}}=\frac{a}{b}=\frac{b}{a-A D} .
$$
Eliminating $A D$, we get $a^{2}=b(b+c)$.
$$
\text { ( } 2 \text { ) As }
$$
shown in the figure,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,428 |
Five or six football teams hold a round-robin tournament, with rankings determined by points. Each pair of teams plays one match, with the winner getting two points, a draw giving each team 1 point, and the loser getting 0 points. If two (or more) teams have the same points, the ranking is decided by a draw. Questions:... | Five, (1)15 games, 30 points; (2)24 points, 15 points; (3)4 points, 9 points; (4)5 points, 10 points. | 5, 15, 30, 24, 15, 4, 9, 5, 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,429 |
1. If $m$ is a perfect square, then the smallest perfect square greater than $m$ is | 1. $m+2 \sqrt{m}+1$; | m+2 \sqrt{m}+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,439 |
Three, let $\triangle A B C$ be an acute triangle, with its three sides satisfying the inequality $A B<A C<B C$. If point $I$ is the incenter of $\triangle A B C$, and point $O$ is the circumcenter of $\triangle A B C$. Prove:
$$
\angle O B C<\angle I B C, \angle O C B<\angle I C B .
$$ | $$
\begin{aligned}
& \text { i) } \angle O B C \\
= & \angle O C B=90^{\circ} \\
- & \angle A .
\end{aligned}
$$
ii) $I B C$
$$
=\frac{1}{2} \angle B, \angle I C B=\frac{1}{2} \angle C \text {. }
$$
iii) $\angle C<\angle B<\angle A$.
Combining the above, we get $90^{\circ}-\angle A=\frac{1}{2} \angle C + (\angle B$
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,443 |
1. (Soviet) Given $\triangle A B C$, let $I$ be its incenter, and the internal angle bisectors of $\angle A, \angle B, \angle C$ intersect the opposite sides at $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Prove:
$$
\frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}} \le... | Let $B C=a, C A=b, A B=c$, it is easy to prove:
$$
\begin{array}{l}
\frac{A I}{A A^{\prime}}=\frac{b+c}{a+b+c}, \frac{B I}{B B^{\prime}}=\frac{a+c}{a+b+c}, \\
C I I^{\prime}=\frac{a+b}{a+b+c} .
\end{array}
$$
By the AM-GM inequality, we have
$$
\begin{array}{l}
A I \cdot B I \cdot C I \\
A A^{\prime} \cdot B B^{\prime... | \frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}} \leqslant \frac{27}{8} | Geometry | proof | Yes | Yes | cn_contest | false | 705,444 |
2. (Romania) Let $n$ be a positive integer, and let $a_{1}, a_{2}, \cdots, a_{\mathrm{k}}$ be all the natural numbers less than $n$ and coprime with $n$. If
$$
a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{k}-a_{\mathrm{k}-1}>0,
$$
prove that $n$ is either a prime or a positive integer power of 2. | Obviously, $a_{1}=1$.
By $(n-1, n)=1$, thus $a_{\mathrm{k}}=n-1$.
Let $d=a_{2}-a_{1}>0$.
When $a_{2}=2$, $d=1$, thus, $k=n-1$ and $a_{1}=i(i=1,2, \cdots, n-1)$. Therefore, $n$ is coprime with all natural numbers less than $n$. This implies that $n$ is a prime number.
When $a_{2}=3$, $d=2$, thus $n$ is coprime with all... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,445 |
4. Let $a_{1}=2, a_{2}=7$. For $n \geqslant 2$, let $a_{n+1}$ be an integer, and determined by
$$
-\frac{1}{2}<a_{n+1}-\frac{a_{n}^{2}}{a_{n-1}} \leqslant \frac{1}{2}
$$
Prove: For all $n \geqslant 2, a_{n}$ is odd. | 4. It is known that $a_{1}=2, a_{2}=7, a_{3}=25, a_{4}=89$,
‥ We will use induction to prove
$$
a_{n}=3 a_{n-1}-2 a_{n-2} \text {. }
$$
For $n=3$, 4, equation (1) holds. Assume (1) holds for:
$k(>2)$, then
$$
\begin{array}{l}
\frac{a_{\mathrm{k}}^{2}}{a_{k-1}}=\frac{a_{\mathrm{k}}\left(3 a_{k-1}-2 a_{k-2}\right)}{a_{k... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,446 |
5. In isosceles $\triangle A B C$, $\angle A=100^{\circ}, A B$ $=A C, \angle B$'s bisector intersects $A C$ at $D$. Prove: $B D+A D=B C$. | 5. Take point $E$ on $BC$ such that $CE = AD$. Then
$$
\frac{CE}{CD} = -\frac{AD}{CD} = \frac{BA}{BC}.
$$
Also, $\angle DCE = \angle ABC = 40^{\circ}$, so,
$$
\begin{array}{l}
\triangle CED \sim \triangle BAC, \\
\angle CDE = \angle BCA = 40^{\circ}.
\end{array}
$$
Thus, $\angle BDE = 80^{\circ}, \angle BED = 80^{\ci... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,447 |
$6 . I$ is the incenter of $\triangle A B C$, and the extensions of $A I, B I, C I$ intersect the circumcircle of $\triangle A B C$ at $D, E, F$ respectively. Prove: $E F \perp A D$. | $$
\text { 6. } \begin{aligned}
& \angle A D F + \angle D F E \\
= & \angle A D F + \angle D F C + \angle C F E \\
= & \angle \frac{C}{2} + \angle \frac{A}{2} + \angle \frac{B}{2} = \frac{\pi}{2},
\end{aligned}
$$
Branch $A D \perp E F$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,448 |
7. For all positive integers $n$, prove:
$$
\begin{aligned}
1! & (2 n-1)!-2!(2 n-2)!+\cdots \\
& -(2 n-2)!2!+(2 n-1)!1! \\
= & \frac{(2 n)!}{n+1} .
\end{aligned}
$$ | 7. To prove
$$
\begin{array}{l}
\frac{1}{C_{2 n}^{1}}-\frac{1}{C_{2 n}^{2}}+\cdots-\frac{1}{C_{2 n}^{2 n-2}}+\frac{1}{C_{2 n}^{2 n-1}} \\
=\frac{1}{n+1} .
\end{array}
$$
Because
$$
\begin{array}{l}
-\frac{1}{C_{2 n+1}^{\mathrm{k}}}+\frac{1}{C_{2 \mathrm{n}+1}^{\mathrm{b}+1}} \\
=\frac{2 n-k+1}{(2 n+1) C_{2 n}^{\mathrm... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,449 |
9. $O$ is the center of a circle, $AC$ is a fixed chord of circle $O$, but not a diameter. $BD$ is a moving chord perpendicular to $OC$, with $D$ on the minor arc $AC$. If $BD$ intersects $AC$ at $E$, find the locus of the circumcenter of $\triangle ABE$.
Translate the above text into English, please keep the original... | 9. Let the circumcenter of $\triangle A B E$ be $O_{1}$, and connect $O_{1} O$, $O_{1} A$, $O_{1} E$, $O A$, $O D$, $A D$, as shown in the figure.
Then $\angle A O D=$
$2 \angle B=$
$\angle A O_{1} E$, so
$\triangle A O D C S$
$\triangle A O_{1} E$,
$$
\frac{A D}{A E}=\frac{A O}{A O_{1}},
$$
$\angle D A O=\angle E A O_... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,451 |
10. A hostess is waiting for 7 or 11 children to arrive, and she has prepared 77 marbles as gifts. She puts these marbles into $n$ bags so that each child (whether 7 or 11) can receive several bags of marbles, and the 77 marbles are evenly distributed among these children. Find the minimum value of $n$. | 10. Since these $n$ bags of sand can be divided into 7 portions, each containing 11 marbles, represented by 7 vertices $x_{1}, x_{2}, \cdots, x_{7}$. These $n$ bags of marbles can also be divided into 11 portions, each containing 7 marbles, represented by vertices $y_{1}, y_{2}, \cdots, y_{11}$. If a certain bag of mar... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,452 |
11. (1) Let $a, b, c$ be real numbers. Prove:
$$
\begin{array}{l}
\min \left[(b-c)^{2},(c-a)^{2},(a-b)^{2}\right] \\
\leqslant \frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}
$$
(2) If the number $k$ is such that
$$
\begin{array}{l}
\min \left[(a-b)^{2},(a-c)^{2},(a-d)^{2} .\right. \\
\left.\quad(b-c)^{2},(b-d)... | 11. Generally, we have
$$
\min _{1 \leqslant \mathrm{i}<\mathrm{j} \leqslant \mathrm{n}}\left(a_{\mathrm{i}}-a_{1}\right)^{2} \leqslant \frac{12}{n\left(n^{2}-1\right)} \sum_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}^{2} \text {. }
$$
Assume without loss of generality that $a_{1} \geqslant a_{2} \geqslant \cdots \geqs... | \frac{1}{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 705,453 |
12. $A, B$ are two given points, $C$ moves on a circle with center $A$, the angle bisector of $\triangle A B C$ at angle $A$ intersects side $B C$ at point $P$, find the locus of $P$. | 12. Taking $A$ as the origin and $A B$ as the $x$-axis to establish a rectangular coordinate system. Suppose $C$ moves on the unit circle, the coordinates of point $B$ are $(a, 0)$, the coordinates of point $C$ are $\left(x_{0}, y_{0}\right)$, and the coordinates of point $P$ are $(x, y)$, then
$$
\left\{\begin{array}{... | x^{2}+y^{2}-\frac{2 a}{1+a} x=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,454 |
13. Place several points on the unit sphere such that the distance between any two points is (1) at least $\sqrt{2}$; (2) greater than $\sqrt{2}$. Determine the maximum number of points and prove your conclusion. | 13. (1) The maximum number of points is 6. If $A$ is one of these points, let's assume $A$ is at the North Pole, then the remaining points must all be in the Southern Hemisphere (including the equator). If there is only one point $B$ at the South Pole, then the rest of the points are all on the equator, in which case t... | 6 | Geometry | proof | Yes | Yes | cn_contest | false | 705,455 |
3. (China) Let $S=\{1,2,3, \ldots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers. | Let $A_{1}=\{S \mapsto$ all natural numbers divisible by $i\}, i=2,3,5,7$. Let $A=A_{2} \cup A_{3} \cup A \cup \cup A_{7}$. Using the principle of inclusion-exclusion, it is easy to calculate that the number of elements in $A$ is 216. Since any 5 numbers chosen from $A$ must have at least two numbers in the same $A_i$,... | 217 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,456 |
14. Given that $x, y$ are positive real numbers, and $x+y=1$, $m, n$ are positive integers greater than 1. Prove:
$$
\left(1-x^{\mathrm{m}}\right)^{\mathrm{n}}+\left(1-y^{\mathrm{n}}\right)^{\mathrm{m}}<1 .
$$ | 14. First, prove the following fact:
If $a, b, c, d \in R^{+}, a+b=c+d$, $|a-b|>|c-d|$, and $n(>1)$ is a positive integer, then $a^{\mathrm{n}}+b^{\mathrm{n}}>c^{\mathrm{n}}+d^{\mathrm{n}}$.
The proof is left to the reader. Now, we proceed to prove the original problem:
The original inequality is equivalent to
$$
\beg... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,457 |
16. Given length $a$, angle $\theta$, and ratio $k>1$. Construct a triangle $ABC$ such that $BC=a, \angle C - \angle B = \theta, \quad \frac{AB}{AC}=k_{0}$ | 16. Let $BC=a$, draw $\angle BCD=\theta$ through point $C$, and $CD=\frac{1}{k} a$, then connect $BD$. Draw $\angle DCA = \angle DBC$ through point $C$, so that $CA$ intersects the extension of $BD$ at $A$, then $\triangle ABC$ is the desired triangle. | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,458 |
17. Each small square of an $n \times n$ chessboard is colored with one of $m$ colors, and no two rows can be colored exactly the same. Then, all the small squares in a certain column are repainted white. Prove: It is possible to choose such a column so that after repainting, no two rows are exactly the same. | 17. By contradiction. Otherwise, coloring any column white would result in two rows becoming identical. Use $n$ points $v_{1}, v_{2}, \cdots, v_{\mathrm{n}}$ to represent the $1, 2, \cdots, n$ rows. After coloring the $i$-th column white $(i=1,2, \cdots, n)$, connect a line between the points corresponding to the two i... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,459 |
18. Let the closed interval $[0,1]$ be denoted as $I$. Suppose the function $f: I \rightarrow I$ is a monotonic continuous function, and $f(0)=0, f(1)=1$. Prove: The graph of $f$ can be covered by $n$ rectangles of area $\frac{1}{n^{2}}$.
Translate the above text into English, please retain the original text's line br... | 18. Since $f(1)>f(0)$, $f(x)$ is monotonically increasing on $I$. Let $x_{0} \in[0,1)$, then $\left(f(x)-f\left(x_{0}\right)\right)(x-x_{0})$ is monotonically increasing and continuous on $\left[x_{0}, 1\right]$. Also, the graph of $f(x)$ on $\left[x_{0}, x_{1}\right]$ is covered by the rectangle with vertices $\left(x... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,460 |
19. A straightedge with parallel sides can be used to draw two parallel lines, the distance between which is exactly the width of the ruler. Let $A B C D E F G H I$ be a regular nonagon.
(1) Given vertices $A, B, C$ and $D$, use this straightedge to construct the remaining vertices of the regular nonagon.
(2) If any fo... | 19. First, for any angle, a person can use a ruler with parallel sides to construct its angle bisector. As shown in the figure, given $\angle A O B$, use a ruler with parallel sides to draw lines parallel to the two sides of $\angle A O B$, and let them intersect at point $T$. Connect $O T$, then $O T$ is the angle bis... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,461 |
20. Let $x$ be an $n$-digit number. Is there always a non-negative integer $y \leqslant 9$ and $z$ such that $10^{\mathrm{n}+1} z+10 x+y$ is a perfect square? | 20. Not necessarily.
For example, when $x=111$, there do not exist non-negative integers $y \leqslant 9$ and $z$ such that $10^{4} z+1110+y$ is a perfect square.
If otherwise, let $10^{4} z+1110+y=k^{2}$. Since the square of an odd number is congruent to 1 modulo 8, and the square of an even number is congruent to 0 ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,462 |
21. Let the weight of the counterfeit coin be $a$, and the weight of the genuine coin be $b$ $(a \neq b)$. There are two piles of three coins each, and it is known that each pile contains exactly one counterfeit coin. How many times at least must a precise scale (not a balance) be used to find these two counterfeit coi... | 21. (1) At least 3 weighings are required.
First, take one coin from each of the two piles, (1) if the weight is $2a$, then the real coins have been found; (2) if the weight is $2b$, then take one coin from the remaining two in each pile and weigh them separately, and the two fake coins can be found; (3) if the weight... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,463 |
22. The diameter of a figure refers to the maximum distance between any two points in the figure. Given an equilateral triangle with a side length of 1, indicate how to cut it into two parts with a straight line so that when these two parts are reassembled into a figure, the figure has the maximum diameter.
(1) If the ... | 22. (1) The maximum value of the diameter is $\frac{\sqrt{13}}{2}$.
If the cutting line passes through one vertex and does not pass through the midpoint of the opposite side, there are the following three scenarios:
Since the diameter is generated by the distance between two endpoints, among the three figures, only $... | \frac{\sqrt{13}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,464 |
23. Let $n \geqslant 1$, for $2 n+1$ positive numbers $x_{1}, x_{2}$, $\cdots, x_{2 n+1}$, prove or disprove
\[
\begin{array}{l}
\frac{x_{1} x_{2}}{x_{3}}+\frac{x_{2} x_{3}}{x_{1}}+\cdots+\frac{x_{2 n+1} x_{1}}{x_{2}} \\
\geqslant x_{1}+x_{2}+\cdots+x_{2 n+1} .
\end{array}
\]
Equality holds if and only if all $x$ are ... | 23. When $n=1$, the inequality holds.
$$
\begin{aligned}
\because \quad & \frac{x_{1} x_{2}}{x_{3}}+\frac{x_{2} x_{3}}{x_{1}} \geqslant 2 x_{2}, \\
& \frac{x_{2} x_{3}}{x_{1}}+\frac{x_{3} x_{1}}{x_{2}} \geqslant 2 x_{3}, \\
& \frac{x_{3} x_{1}}{x_{2}}+\frac{x_{1} x_{2}}{x_{3}} \geqslant 2 x_{1}, \\
\therefore \quad & 2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,465 |
24. A tetrahedron has four faces that are all congruent triangles. If $\alpha$ is the angle between a pair of opposite edges, prove:
$$
\cos \alpha=\left|\frac{\sin (\angle B-\angle C)}{\sin (\angle B+\angle C)}\right|,
$$
where $\angle B, \angle C$ are the angles between the other two edges and the edge in the face c... | 24. Since the four faces of this tetrahedron are congruent triangles, it is easy to see that the opposite edges of this tetrahedron are equal. By constructing a plane through each edge parallel to the opposite edge, a parallelepiped is formed. Each face of this parallelepiped has diagonals equal to the opposite edges o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,466 |
4. (USA) Let $G$ be a connected graph with $k$ edges. Prove that the edges of $G$ can be labeled with $1, 2, \cdots, k$ such that for every vertex that belongs to two or more edges, the greatest common divisor of the labels of the edges incident to that vertex is 1. | Proof: By the connectivity of $G$, every vertex in $G$ belongs to at least one edge. Take any vertex $v_{0}$ and start walking along the edges of $G$, passing through each edge at most once, but allowing multiple passes through each vertex. Suppose after passing through $t_{1}$ edges, it is no longer possible to contin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,467 |
25. Determine the number of non-congruent triangles with positive integer side lengths and a perimeter of $n$.
Try to determine the number of non-congruent triangles with positive integer side lengths and a perimeter of $n$. | 25. The side lengths of a triangle with integer sides and perimeter $n$
Let the side lengths be represented by $x, y, z$, then
$x+y+z=n$,
$x>0, y>0, 2x>y$.
The integer solutions for $x$ are given by $y=n-2x$, thus
$\frac{n}{4}<x<\frac{n}{2}$.
Therefore, the integer solutions for $x$ are $\left[\frac{n-1}{2}\right]-\... | T=\begin{cases}
N\left(\frac{n^{2}+6n}{48}\right) & \text{if } n \text{ is odd} \\
N\left(\frac{n^{2}}{48}\right) & \text{if } n \text{ is even}
\end{cases} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,468 |
26. The circumcenter of $\triangle A B C$ is $O, A B=A C, D$ is the midpoint of $A B$, and $E$ is the centroid of $\triangle A C D$. Prove: $O E \perp C D$. | 26. Let $F$ be the midpoint of $A C$, then $E$ is on $D F$, and $\frac{D E}{D F}=\frac{2}{3}$.
Let $G$ be the midpoint of $B C$,
$A G$ intersects $C D$ at $H$, then
$O$ is on $A G$, and $H$ is
the centroid. Connect $F G$ intersecting $C D$
at $I$, then $I$ is the midpoint of $D C$, so
$$
\frac{D H}{D I}=\frac{2}{3... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,469 |
29. $A B$ is the diameter of a fixed circle with center $O$. The center of another variable circle lies on the line $A B$, it passes through point $O$ and intersects the fixed circle at two points. The common tangent of the two circles touches the variable circle at point $P$. Find the locus of point $P$. | 29. Let the center of the moving circle be $O^{\prime}$, and the common tangent line touch circle $O$ at
$$
\begin{array}{l}
\text { point } C, \text { then } O C \perp C P, O^{\prime} P \perp C P, \\
\text { so } \angle C O P \\
= \angle O P O^{\prime} . \\
\text { Also, since } \angle O P O^{\prime} \\
= \angle O^{\... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,472 |
1. A cube with an edge length of 3 is composed of 27 unit cubes. How many lines pass through the centers of 3 unit cubes? How many lines pass through the centers of 2 unit cubes? | There are 49 lines passing through the centers of 3 unit cubes; there are
$$
C_{27}^{2}-3 \times 49=204 \text { (lines). }
$$ | 204 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,474 |
2. Given $\triangle A B C$, the angle bisector of $\angle B$ intersects side $A C$ at $P$, and the angle bisector of $\angle A$ intersects side $B C$ at $Q$. If the circle passing through points $P, Q, C$ also passes through the incenter $R$ of $\triangle A B C$, and $P Q=1$, find the lengths of $P R$ and $R Q$. | Extend $C R$ and let it intersect side $A B$ at $T$, then
$$
\begin{aligned}
\angle P R Q= & \angle P R C+\angle C R Q \\
= & \angle B R T+\angle A R T \\
& =\frac{1}{2} \angle B+\frac{1}{2} \angle C+\frac{1}{2} \angle A \\
& +\frac{1}{2} \angle C \\
& =\frac{1}{2}(\angle A+\angle B)+\angle C
\end{aligned}
$$
Given th... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,475 |
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