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12. A pipe burst occurred in a low-lying area by the riverbank, and river water is continuously gushing out, assuming the water gushing out per minute is constant. If two water pumps are used to pump out the water, it takes 40 minutes to finish; if four water pumps are used, it takes 16 minutes to finish. If the water ... | 12.6 .
Let the amount of water that has already gushed out before the pumping starts be $u$ cubic meters, the rate of water gushing out per minute be $b$ cubic meters, and the amount of water each pump can extract per minute be $c$ cubic meters $(c \neq 0)$. From the given conditions, we have
$$
\left\{\begin{array}{l... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,881 |
14. As shown in Figure 6, it is known that quadrilateral $ABCD$ is inscribed in a circle $\odot O$ with a diameter of 3, diagonal $AC$ is the diameter, the intersection point of diagonals $AC$ and $BD$ is $P$, $AB=BD$, and $PC=0.6$. Find the perimeter of quadrilateral $ABCD$. | $1+$ Let the heart of the shadow be क”, so $\mathrm{BH} / \mathrm{C}$.
From this, $\triangle O P B C \triangle C P D, \frac{C D}{B O}=\frac{C P}{P O}$,
which means $\frac{C D}{1.5}=\frac{0.6}{1.5-0.6}$. Therefore, $C D=1$.
Thus, $A D=\sqrt{A C^{2}-C D^{2}}=\sqrt{9-1}=2 \sqrt{2}$.
Also, $O H=\frac{1}{2} C D=\frac{1}{2}... | 1+2\sqrt{2}+\sqrt{3}+\sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,883 |
15. Someone has written a program: starting from 1, alternately perform addition or multiplication (the first operation can be either addition or multiplication). Each time for addition, add 2 or 3 to the previous result; each time for multiplication, multiply the previous result by 2 or 3. For example, 30 can be obtai... | 15. (1) $1 \xrightarrow{\times 2} 2 \xrightarrow{+2} 4 \xrightarrow{\times 2} 8 \xrightarrow{+2} 10 \xrightarrow{\times 2} 20 \xrightarrow{+2}$
22. Or consider it in reverse:
$$
22 \xrightarrow{-2} 20 \stackrel{\div}{\longrightarrow} 10 \xrightarrow{-2} 8 \xrightarrow{\div 2} 4 \xrightarrow{-2} 2 \xrightarrow{\div 2} 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,884 |
Four, (Full marks 10 points, factorize $2 x^{3}-x^{2} z-4 x^{2} y+2 x y z+$ $2 x y^{2}-y^{2} z$) | Four,
$$
(x-y)^{2}(2 x-z)
$$ | (x-y)^{2}(2 x-z) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,888 |
Example 6 As shown in the figure, two rivers converge at point $O$, forming a $75^{\circ}$ angle. The traveler lives at point $P$, 200 meters from point $O$, and 100 meters from the riverbank $AO$. He wishes to go to any point $C$ on the riverbank $AO$ to enjoy the scenery, then turn to any point $D$ on the riverbank $... | Solution: Make the symmetric points of point $P$ about $AO$ and $BO$ as $P'$ and $P''$, respectively. Then, $PC + CD + DP = P'C + CD + DP''$ is the length of the broken line between $P'$ and $P''$. To minimize this, $P'P''$ must be a straight line segment.
From the given, we have $\angle AOP = 30^\circ$.
Also, $\angle... | 100(\sqrt{6} + \sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,891 |
Nine, (Full marks 10 points) Given in $\triangle A B C$, $\angle A=90^{\circ}, A B$ $=A C, D$ is the midpoint of $A C$, $A E \perp B D$ at $E$, and the extension of $A E$ intersects $B C$ at $F$. Prove: $\angle A D B=\angle C D F$.
---
The translation retains the original text's line breaks and formatting. | Nine, Prompt: Draw the angle bisector of $\angle A$, intersecting $BD$ at $G$. Prove that $\triangle A B G \cong \triangle A C F, \triangle A G D \cong \triangle C D F$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,894 |
One, (Full marks 10 points) As shown in the figure, there is a natural number on each face of the cube. It is known that the sum of the two numbers on opposite faces is equal. Given that the numbers opposite to $12$, $9$, and $3$ are $a$, $b$, and $c$ respectively, find the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$. | Final 1. From the problem, we have $13+a=9+b=3+c$, which gives $a-b=-4, b-c=-6, c-a=10$. Therefore,
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\
=\frac{1}{2}(16+36+100)=76 .
\end{array}
$$ | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,896 |
Three, (Full marks 10 points) The numbers A, B, and C are 312, $270$, and 211, respectively. When these three numbers are divided by a natural number $A$, the remainder of dividing A is twice the remainder of dividing B, and the remainder of dividing B is twice the remainder of dividing C. Find this natural number $A$.... | Three, from the problem, we have
$$
270 \times 2-312=228, \quad 211 \times 2-270=152 \text {. }
$$
Then we have
\begin{tabular}{|c|c|c|}
\hline \multicolumn{2}{|c|}{$2 \mid \quad 228$} & 152 \\
\hline $2 T$ & 114 & 76 \\
\hline 19 & 57 & 38 \\
\hline & 3 & 2 \\
\hline
\end{tabular}
Therefore, 4 could be $2, 4, 19, 38... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,898 |
Five. (Full marks 10 points) Given $\frac{a+b-c}{c}=\frac{a-b+c}{b}=$ $=\frac{a+b+c}{a}$. Find the value of $\frac{(a+b)(b+c)(c+a)}{a b c}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$.
Then
\[
\begin{array}{l}
a+b=(k+1) c, \\
b+c=(k+1) a, \\
c+a=(k+1) b . \\
(1)+(2)+(3) \text{ gives } \\
(k-1)(a+b+c)=0 .
\end{array}
\]
When $k=1$, the original expression $=8$;
When $k \neq 1$, $a+b+c=0$, the original expression $=-1$. | 8 \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,900 |
Six. (Full marks 10 points) As shown in the figure, given $A C=B C, \angle C=$ $90^{\circ},$ the angle bisector $A D$ of $\angle A$ intersects $B C$ at $D$, and a perpendicular $B E$ is drawn from $B$ to $A D$ at $E$. Prove:
$$
B E=\frac{1}{2} A D .
$$ | Six, Hint: Extend $A C$ and $B E$ to intersect at $F$. It can be deduced that $\triangle A C D \cong \triangle B C F$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,901 |
Seven, (Full marks 10 points) In $\triangle A B C$, $A D$ is the median on side $B C$, $E$ is a point on $A D$, and $B E=A C$. Extend $B E$ to intersect $A C$ at $F$. Prove: $A F=E F$. | Seven, Prompt: Extend $A D$ to $G$, such that $A D = D G$, and connect $B G$. Prove that $\triangle A D C \cong \triangle G D B$.
---
Note: The translation preserves the original text's line breaks and formatting. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,903 |
Eight, (Full marks 10 points) A student buys a certain type of pencil. When he buys $x$ pencils, he pays $y$ yuan ($x, y$ are both integers). The clerk says: "If you buy 10 more, I will charge you a total of 2 yuan. This way, for every 30 pencils you buy, you can save 2 yuan." Find $x$ and $y$.
---
The above text is ... | Eight, according to the salesperson, $y$ can only be 1 or 2.
(1) When $y=1$, according to the problem, the original price per item is $\frac{1}{x}$ yuan. Buying 10 more items saves $\frac{1}{15}$ yuan per item, so the current price per item is $\frac{1}{x}-\frac{1}{15}$ (yuan).
Let the total number of items after buyin... | x=5, y=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,904 |
-、(Full marks 10 points) There are two decks of playing cards, each deck arranged in such a way that the first two cards are the Big Joker and the Small Joker, followed by the four suits of Spades, Hearts, Diamonds, and Clubs, with each suit arranged in the order of $1,2,3$, $\cdots, J, Q, K$. Someone stacks the two de... | Ten, if there are only 64 cards in hand, and you discard according to this rule, then the last remaining card will be the 64th card.
Now, if you have 108 cards in hand, which is $108-64=44$ (cards) more, we only need to discard 44 cards according to the rule, and place the 88th card at the bottom of the remaining card... | 6 \text{ of Diamonds} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,906 |
3. Positive numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ simultaneously satisfy
$$
\begin{array}{l}
\frac{x_{2} x_{3} x_{4} x_{5} x_{6}}{x_{1}}=1, \frac{x_{1} x_{3} x_{4} x_{5} x_{6}}{x_{2}}=2, \\
\frac{x_{1} x_{2} x_{4} \cdot x_{5} \cdot x_{6}}{x_{5}}=3, \frac{x_{1} \cdot x_{2} x_{3} x_{5} x_{6}}{x_{4}}=4, \\
\f... | 3. $1+\sqrt{2}+\sqrt{3}+\frac{11 \sqrt{6}}{6}$ | 1+\sqrt{2}+\sqrt{3}+\frac{11 \sqrt{6}}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,909 |
5. If the centroid of $\triangle A B C$ is $G, A G=\sqrt{2}, B G=\sqrt{3}$, $C G=\sqrt{5}$, then the area of $\triangle A B C$ is $\qquad$ | 5. $\frac{3 \sqrt{6}}{2}$ | \frac{3 \sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,911 |
9. Two computer warehouses supply computers to three schools. Warehouse A has 12 units, Warehouse B has 20 units; School $A$ needs 9 units, School $B$ needs 15 units, School $C$ needs 8 units. It is known that the distances from Warehouse A to Schools $A, B, C$ are 10 km, 5 km, 6 km, respectively; the distances from Wa... | $9.0,4,8$ | 9.0,4,8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,932 |
II. As shown in the figure, in the acute triangle $\triangle ABC$, $PQRS$ is an inscribed rectangle of $\triangle ABC$, and $S_{\triangle MBC} = n \cdot S_{\text{PQRS}}$, where $n$ is a natural number not less than 3. Prove that $\frac{BS}{AB}$ is an irrational number. | $$
\begin{array}{l}
\text{Let } B C=a, \text{ the height from } A \text{ to } B C \text{ be } A D=h, P S=x, R S= \\
\because . \\
\because \triangle A S R \sim \triangle A B C, \\
\therefore \frac{h-x}{h}=\frac{y}{a}, \\
y=\frac{h-x}{h} \cdot a . \\
\because S_{\triangle \text{ ABx }}=n S_{\text{HSTRPRS }}, \\
\therefo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,934 |
5. Based on market research analysis, a home appliance manufacturing company has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be at least... | 5. Weekly
Producing 30 air conditioners, 270 color TVs, and 60 refrigerators will maximize the output value, with the highest output value being 1050 thousand yuan. | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,935 |
Three, write 10 different
natural numbers, such that each of them is a divisor of the sum of these 10 numbers (explain the reason why the 10 natural numbers written meet the conditions of the problem). | Three natural numbers $1,2,3$, each of them is a divisor of the sum of these 3 numbers. If there are already $k(=3)$ natural numbers $a_{1}$, $a_{2}, \cdots, a_{i}$, each of which is a divisor of the sum (denoted as $p$) of these $k$ numbers, then $k+1$ natural numbers $a_{1}, a_{2}, \cdots, a_{k}, p$, each of them is ... | 1,2,3,6,12,24,48,96,192,384 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,936 |
Four, let $x, y$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, and satisfy the condition: when $x>y$, $x-y \geqslant \frac{x}{15}$. Find the maximum number of these natural numbers $n$. | $$
\begin{array}{l}
\text { Let's assume } a_{1}=1, d_{4} \geqslant 2 ; \\
a_{5}=a_{4}+d_{4} \geqslant 6, d_{5} \geqslant \frac{6^{2}}{19-6}>2, d_{5} \geqslant 3 ; \\
a_{6} \geqslant 9, d_{6} \geqslant \frac{9^{2}}{19-9}>8, d_{6} \geqslant 9 ; \\
a_{7} \geqslant 18, d_{7} \geqslant \frac{18^{2}}{19-18}=324 ; \\
a_{8} \... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,937 |
5. Given that the lengths of the three sides of a triangle are all integers, the difference between the lengths of two of its sides is 5. If the perimeter of this triangle is odd, then the minimum value of the third side is ( ).
(A) 8
(B) 7
(C) 6
(D) 4 | $5 . C$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,942 |
Example 1 Suppose $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=$ 0 , and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is (1990, Hefei Junior High School Mathematics Competition) | Given $a^{2}+2 a-1=0$, we know $a \neq 0$.
$$
\therefore\left(\frac{1}{a}\right)^{2}-2\left(\frac{1}{a}\right)-1=0 \text {. }
$$
From $b^{4}-2 b^{2}-1=0$, we get
$$
\left(b^{2}\right)^{2}-2\left(b^{2}\right)-1=0 \text {. }
$$
From $1-a b^{2} \neq 0$, we know $\frac{1}{a} \neq b^{2}$.
From (1) and (2), we know $\frac{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,946 |
Example 2 As shown in Figure 1, $P$ is any point on the circumcircle of square $A B C D$
$A D$. Prove:
(1) $P A+P C=\sqrt{2} F P$;
(2) $P A \cdot P C=P B^{2}-A B^{2}$. | Prove: As shown in Figure 2, connect $AC$ and $BD$.
In $\triangle APB$,
$$
AB^{2}=PA^{2}+PB^{2}-2PA \cdot PB \cos 45^{\circ},
$$
which is $PA^{2}-(\sqrt{2} PB) PA+\left(PB^{2}-AB^{2}\right)=0$.
In $\triangle PBC$,
$$
BC^{2}=PC^{2}+PB^{2}-2PC \cdot PB \cos 45^{\circ},
$$
which is $PC^{2}-(\sqrt{2} PB) PC+\left(PB^{2}-... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,947 |
Example 2 Let $n=1999$. Find the value of the following expression:
$$
\frac{1}{2^{n}}\left(1-3 \mathrm{C}_{n}^{2}+3^{2} \mathrm{C}_{n}^{4}-3^{3} \mathrm{C}_{n}^{6}+\cdots-3^{999} \mathrm{C}_{n}^{1998}\right) \text {. }
$$ | $\begin{array}{l}\text { Solution: Original expression }=\frac{1}{2^{n+1}}\left[(1+\sqrt{3} i)^{n}+(1-\sqrt{3} i)^{n}\right] \\ =\frac{1}{2}\left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{n}+\left(\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^{n}\right) \\ =\frac{1}{2}\left[\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,948 |
Example 3 Find the value of $\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}$.
(23rd Putnam Mathematical Competition) | $$
\begin{array}{l}
\text { Solution: } \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k \cdot k \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k \cdot n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}[(k-1)+1] \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n} \mathrm{C}_{n-1}^{k-1}+n \sum_{k=2}^{n}(n-1) \mathrm{C}_{n-2}^{k-2} \\
=n ... | n(n+1) \cdot 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,949 |
Example 4 Solve the equation $w!=x!+y!+z!$. (1983, Canadian High School Mathematics Competition) | Solution: Without loss of generality, let $x \leqslant y \leqslant z$.
Obviously, $w \geqslant z+1$, then
$(z+1)! \leqslant w! = x! + y! + z!$
$\leqslant 3 \cdot z!$.
Thus, $(z+1) \leqslant 3$, i.e., $z \leqslant 2$.
When $z=0$ or $z=1$, the original equation has no solution. When $z=2$, the solution to the original eq... | x=y=z=2, w=3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,950 |
Example 5 Let $A B C D E F$ be a hexagon, and a frog starts at vertex $A$. It can jump to one of the two adjacent vertices at each step. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences are possible from... | Analysis: Grasping the possible scenarios of the frog's jumping is the key to solving the problem.
Solution: According to the conditions, the frog's jumping methods can only result in two situations:
(1) Jump 3 times to reach point $D$, with 2 ways of jumping.
(2) Stop after 5 jumps (the first 3 jumps do not reach poi... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,951 |
Example 6 As shown in the figure, the sides of rectangle $ABCD$ lie on the grid lines, and $AB$ is $k$ times $AD$ ($k$ is a positive integer). Consider all possible shortest paths from $A$ to $C$ along the grid lines. Prove: Among these paths, the number of paths containing $AB_1$ is $k$ times the number of paths conta... | Analysis: Let's assume that rectangle $A B C D$ contains $m \times n$ squares $(m=i n)$. Then the shortest path from $A$ to $C$ should be a permutation of elements! problem.
Solution: The shortest path containing $A B_{1}$, besides $A B_{1}$, should also include $m-1$ horizontal segments and $n$ vertical segments, thu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,952 |
Example: 78 girls and 25 boys stand in a circle, with at least two boys standing between any two girls. How many different arrangements are there?
$(1990$, National High School Mathematics Competition) | Solution: Arrangements that can coincide after rotation should be considered as the same arrangement. Therefore, we can consider fixing one girl $A$ and stretching the circle into a straight line in a clockwise direction. Let $a$ and $b$ represent girls and boys, respectively. First, choose 16 boys from 25 to form " $A... | \frac{25!16!}{9!} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,953 |
Example 8 There are 20 teams participating in the national football championship finals. To ensure that in any group of 3 teams, at least two teams have played against each other, how many matches must be played at minimum?
(1969, All-Soviet Union Mathematical Olympiad) | Analysis: The only possible competition schemes that meet the conditions are:
(1) No grouping, any two teams play a match.
(2) Divide into two groups, Group A and Group B, containing 1 team and 19 teams respectively, any two teams in Group B play a match.
(3) Divide into two groups, containing $k(2 \leqslant k \leqslan... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,954 |
Example 9 In a sequence of coin tosses, the number of times a tail is followed by a head (denoted as "tail-head"), a head is followed by a tail (denoted as "head-tail"), a head is followed by a head (denoted as "head-head"), and a tail is followed by a tail (denoted as "tail-tail") can be counted. How many different se... | Analysis: From "反" being 1 more than "正", we know that the sequence starts and ends with "反" and "正" respectively. According to the conditions, it is easy to know that the count of "正" equals $2 \times 2 + 3 + 4 = 11$, and the count of "反" is $5 \times 2 + 3 + 4 = 17$. Since the "反" and "正" at the beginning and end are... | 560 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,955 |
Example 10 Let \( A = \frac{1 \times 3 \times 5 \times \cdots \times (2k-1)}{2 \times 4 \times 6 \times \cdots \times 2k} \) (where \( k \) is a positive integer). Prove that in the sequence \( \{2^{n-1} A\} \), from a certain term onwards, all subsequent terms are integers.
(1934, Hungarian Mathematical Olympiad) | Analysis: It is easy to find that expression $A$ can be simplified using factorials.
Proof: Multiply the numerator and denominator of $A$ by $\prod_{j=1}^{k-1} 2 j$, then the numerator becomes $(2 k-1)!$, and the denominator becomes $2^{2 k-1} \cdot k! \cdot (k-1)!$. Thus,
$$
A=\frac{1}{2^{2 k-1}} \cdot \frac{(2 k-1)!}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,956 |
Example 11 Given the equation $\sum_{k=1}^{1950} x_{k}=2000$. Find:
(1) the number of solutions in natural numbers;
(2) the number of solutions in non-negative integers. | Solution: (1) Construct a line segment $AB$, and divide it into 2000 equal parts.
Let $A$ and $B$ be the 0th and 2000th positions, respectively. The other points are numbered from left to right as 1st, 2nd, $\cdots, 1999$th. It is stipulated that if $x_{1}$ is placed at the 2nd position, then $x_{1}=2$, and if $x_{2}$... | \mathrm{C}_{1999}^{1949}, \mathrm{C}_{3949}^{1949} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,957 |
For example, $3 a_{1}, a_{2}, \cdots, a_{100}, b_{1}, b_{2}, \cdots, b_{100}$ are distinct real numbers. They are filled into a $100 \times 100$ grid according to the following rule: in the cell at the intersection of the $i$-th row and the $j$-th column, the number $a_{i}+b_{j}$ is written. It is known that the produc... | Proof: $\because$ the product of any sequence of numbers equals 1,
$$
\begin{array}{l}
\quad \therefore\left(a_{1}+b_{j}\right)\left(a_{2}+b_{j}\right) \cdots\left(a_{100}+b_{j}\right)=1, \\
(j=1,2, \cdots, 100)
\end{array}
$$
i.e., $\left(a_{1}+b_{j}\right)\left(a_{2}+b_{j}\right) \cdots\left(a_{100}+b_{j}\right)-1=0... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,958 |
Example 12 Let $1 \leqslant r \leqslant n$, consider the set of numbers $\{1,2, \cdots, n\}$ and all its subsets containing $r$ elements. In each such subset, take the smallest number, and let $F(n, r)$ denote the arithmetic mean of these numbers. Prove: $F(n, r)=\frac{n+1}{r+1}$.
(22nd IMO) | Analysis: The number of subsets containing $r$ elements is $\mathrm{C}_{n}^{r}$. In any such subset, the smallest number $k$ does not exceed $n-r+1$, i.e., $1 \leqslant k \leqslant n-r+1$. From the $n-k$ numbers $k+1, k+2, \cdots, n$, choose $r-1$ distinct numbers $n_{1}, n_{2}, \cdots, n_{r-1}$. Then $\left\{k, n_{1},... | \frac{n+1}{r+1} | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,959 |
Example 1 As shown in Figure 1, in $\triangle A B C$, $A B=A C, \angle A=20^{\circ}, D$ is a point on $A C$, $\angle D B C=50^{\circ}, E$ is a point on $A B$, $\angle E C B=40^{\circ}$. Prove: $E D=E B$. | Proof: Given $\angle D B C=50^{\circ}$, we know
$$
\begin{array}{l}
\angle B D C=50^{\circ}=\angle D B C . \\
\therefore B C=C D .
\end{array}
$$
Given $\angle E C B=40^{\circ}$, we know $C E \perp B D$. Therefore, $E C$ is the perpendicular bisector of $B D$.
$$
\therefore E D=E B \text {. }
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,960 |
Example 2 As shown in Figure 2, in $\triangle A B C$, $A B=A C, \angle A=20^{\circ}, D$ is a point on $A C$, $\angle D B C=50^{\circ}, E$ is a point on $A B$, $\angle E C B=20^{\circ}$. Find the degree measure of $\angle B D E$. | In $\triangle B C D$, it is easy to see that $B C = D C$. In $\triangle B C E$, it is easy to see that $E C = B C$.
$$
\therefore C D = C E = C B \text {, }
$$
Thus, $C$ is the circumcenter of $\triangle B D E$.
$$
\therefore \angle B D E = \frac{1}{2} \angle B C E = 10^{\circ} \text {. }
$$ | 10^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,961 |
Example 3 In $\triangle A B C$, $A B=A C, \angle A=$ $20^{\circ}, D$ is a point on $A C$, $\angle D B C=60^{\circ}, E$ is a point on $A B$, $\angle E C B=50^{\circ}$. Find the degree measure of $\angle B D E$. | Solution: As shown in Figure 3, let $E^{\prime}$ be the symmetric point of $E$ with respect to $AC$. Then $E^{\prime}A$, $E^{\prime}B$, $E^{\prime}C$, $E^{\prime}D$, and $E^{\prime}E$ are formed. It is known that $\triangle E^{\prime}EC$ is an equilateral triangle, $BE^{\prime}$ is the perpendicular bisector of $EC$, a... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,962 |
Example 4 In $\triangle A B C$, $A B=A C, \angle A=$ $20^{\circ}, D$ is a point on $A C$, $\angle D B C=60^{\circ}, E$ is a point on $A B$, $\angle E C B=30^{\circ}$. Prove: $E C=E D$. | $$
\begin{array}{l}
\therefore \angle E G B=\angle E B G=20^{\circ}, \\
\text { and } \angle B E G=2 \angle B E C=140^{\circ} . \\
\angle \text { Therefore, } \angle F E G=40^{\circ}=\angle E F G . \\
\text { Thus, } E G=F G . \\
\therefore E G=G D, \angle B D E=\frac{1}{2} \angle E G B=10^{\circ} \\
\text { Thus, } \... | E C=E D | Geometry | proof | Yes | Yes | cn_contest | false | 710,963 |
Example 5 In $\triangle A B C$, $A B=A C, \angle A=$ $20^{\circ}, D$ is a point on $A C$, $\angle D B C=70^{\circ}, E$ is a point on $A B$, $\angle E C B=50^{\circ}$. Prove:
$$
A D=D E=E B=B C .
$$ | Proof: As shown in Figure 5, construct a regular $\triangle FBC$ inside $\triangle ABC$ with $BC$ as one side. Connect $FA$ and $FD$. It is known that $AF$ is the perpendicular bisector of $BC$, $\angle FAB = 10^\circ = \angle DBA$, and $\angle FBA = 20^\circ = \angle DAB$. Therefore, $\triangle DAB \cong \triangle FBA... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,964 |
Example 6 In $\triangle A B C=A B=A C, \angle A=$ $20^{\circ}, D$ is a point on $A C$, $\angle D B C=70^{\circ}, E$ is a point on $A B$, $\angle E C B=50^{\circ}$. Prove:
$$
B C \cdot B E=D B \cdot D E .
$$ | Proof: As shown in Figure 6, take point $F$ on $EC$, point $G$ on $AC$, and point $H$ on $AB$ such that
$$
A H = C G = B F = B C.
$$
Connect $BF$, $BG$, $HF$, and $HG$. It is known that $\triangle FBC$ is an equilateral triangle. From Example 5, we know $A H = A D = B C$, and $HF CG$ is a rhombus, so $H G \parallel E ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,965 |
Example 7 In $\triangle A B C$, $A B=A C, \angle A=$ $20^{\circ}$, take a point $D$ on side $A B$ such that $A D=B C$. Find the degree measure of $\angle B D C$. | Solution: As shown in Figure 7, construct a regular $\triangle E B C$ inside $\triangle A B C$ with $B C$ as a side, and connect $A E$. It is easy to see that $A E$ is the perpendicular bisector of $B C$, so
$$
\angle E A C=10^{\circ} \text {. }
$$
In $\triangle A D C$ and $\triangle C E A$,
$$
\begin{array}{c}
A D=B ... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,966 |
Example 8 In $\triangle A B C$, $E$ is the midpoint of $B C$, and $D$ is on $A C$. If $\angle B A C=60^{\circ}, \angle A C B=20^{\circ}$, $\angle D E C=80^{\circ}, S_{\triangle A B C}+2 S_{\triangle C D E}=\sqrt{3}$. Find the length of $A C$. | Solution: As shown in Figure 8, take points $F$ and $G$ on the extension of $AB$ such that $AF = AC$, $GF = AB$. Connect $FC$, $GC$. It is easy to see that $\triangle FAC$ is an equilateral triangle, $\triangle FCG \sim \triangle ABC$, $\triangle CGB \sim \triangle CDE$.
Since $E$ is the midpoint of $BC$, we have
$$
\... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,967 |
Example 9 In $\triangle A B C$, $\angle A=20^{\circ}, A B=$ $A C=a, B C=b$. Prove: $a^{3}+b^{3}=3 a^{2} b$.
| Proof: As shown in Figure 9, let $E$ be the reflection of $C$ over $AB$, and $F$ be the reflection of $B$ over $AC$. $EF$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. We have $BE^2 = BM \cdot BA$. Therefore,
$$
BM = \frac{b^2}{a}.
$$
Thus, $AM = AB - BM = \frac{a^2 - b^2}{a}$,
$$
\begin{aligned}
MN & = \frac... | a^3 + b^3 = 3a^2b | Geometry | proof | Yes | Yes | cn_contest | false | 710,968 |
Example 4 Given that $x, y, z$ are all real numbers, $a>0$ and satisfy the relations
$$
\begin{array}{l}
x+y+z=a, \\
x^{2}+y^{2}+z^{2}=\frac{1}{2} a^{2} .
\end{array}
$$
Find the range of values for $x, y, z$.
(1957, Beijing High School Mathematics Competition) | Solution: From (1) we get $x+y=a-z$.
$(3)^{2}-(2)$ and simplifying yields
$$
x y=\frac{1}{4} a^{2}-a z+z^{2} \text {. }
$$
From (3) and (4), we know that $x$ and $y$ are the two roots of the equation in $t$
$$
t^{2}-(a-z) t+\left(\frac{1}{4} a^{2}-a z+z^{2}\right)=0
$$
Therefore,
$$
\Delta=(a-z)^{2}-4\left(\frac{1}{... | 0 \leqslant x, y, z \leqslant \frac{2}{3} a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,969 |
Example 2 defines $T\left(a_{1}, a_{2}, \cdots, a_{2 n}\right)=\left(a_{n+1}\right.$, $\left.a_{1}, a_{n+2}, a_{2}, \cdots, a_{n-1}, a_{2 n}, a_{n}\right)$. Question: For which natural numbers $n$ is the above operation periodic? That is, after a finite number of operations, a previously encountered state reappears.
(2... | Let $A=\left(a_{1}, a_{2}, \cdots, a_{2 n}\right)$.
Consider the positions of each component $a_{i}$ in $T\left(a_{1}, a_{2}, \cdots, a_{2 n}\right)$. $a_{1}$ is the second component of $T(A)$, $a_{2}$ is the fourth component of $T(A)$, $\cdots, a_{n}$ is the $2 n$-th component of $T(A)$, and $a_{n+1}$ is the first com... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,971 |
Example 3 Let $A=\left(a_{n}, a_{n-1}, \cdots, a_{1}, a_{0}\right)_{10}$ be a decimal number, and define $f(A)=2^{n} a_{0}+2^{n-1} a_{1}+ \cdots+2 a_{n-1}+a_{n}$. Let $A_{1}=f(A), A_{i+1}=f\left(A_{i}\right), i$ be a natural number. Prove that there must exist a natural number $k$ such that $A_{k+1}=A_{k}$.
(1st Nation... | Proof: First, intuitively, when $A=\left(a_{n}, a_{n-1}, \cdots, a_{1}, a_{0}\right)_{10}$ is sufficiently large, there should be $f(A) < A$, i.e., $A$ must be at least a 3-digit number.
If $A$ is a one-digit number, i.e., $n=0$, let $A=a$, then $A_{1}=f(a)=2^{0} a=A$,
i.e., a one-digit number remains unchanged under t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,972 |
Example 4 Let $A=\left(a_{n}, a_{n-1}, \cdots, a_{1}, a_{0}\right)_{10}$ be a decimal number, and define $f(A)=a_{n}^{2}+a_{n-1}^{2}+\cdots+a_{0}^{2}$ $\left(a_{n} \neq 0\right)$. Prove that the operation $f$ must result in a cycle, and that there are only two distinct cycle diagrams:
(1),(4,1637,58,89,145,42,20) | Proof: Consider the difference:
$$
\begin{array}{l}
A-f(A) \\
=\left(10^{n}-a_{n}\right) a_{n}+\left(10^{n-1}-a_{n-1}\right) a_{n-1} \\
\quad+\cdots+\left(10-a_{1}\right) a_{1}+\left(1-a_{0}\right) a_{0} .
\end{array}
$$
Notice that
$$
\left(10^{n}-a_{n}\right) a_{n} \geqslant 10^{n}-a_{n} \geqslant 100-a_{n} \geqslan... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,973 |
Example 5 In a $1 \times n$ grid, each cell is arbitrarily filled with 1 or -1. For which $n$, can we always transform all numbers to 1 through "neighbor product transformation" (each cell is simultaneously replaced by the product of all its neighboring cells, i.e., cells sharing a common edge)? | For any sequence $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, construct a bi-infinite sequence with a period of $2(n+1)$:
$$
\cdots, a_{1}, a_{2}, \cdots, a_{2 n}, 1, a_{n}, a_{n-1}, \cdots, a_{1}, \cdots
$$
Call a certain $a_{1}$ the first cell, and the cell before the $i$-th cell is called the $(i-1)$-th cell, while ... | n=2^{r}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,974 |
Example 6 Arrange $1,2, \cdots, n$ in a clockwise circle. First, cross out 2, then continue to cross out every other number in a clockwise direction until only one number remains. Denote the last remaining number as $f(n)$. Find $f(n)$. | Solution: List the values of $f(1), f(2), \cdots$ as follows:
$\begin{array}{llllllllll}n: & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \cdots \\ f(n): 1 & 1 & 3 & 1 & 3 & 5 & 7 & 1 & 3 \cdots\end{array}$
From the above table, we can conjecture that the rule for the values of $f(n)$ is:
When $n=2^{k}$, $f(n)=f\left(2^{k}\right)... | 2 t+1=2\left(n-2^{k}\right)+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,975 |
Example 1 As shown in the figure, in $\triangle A B C$, $\angle A=\frac{5 \pi}{8}$, $\angle B=\frac{\pi}{8}$. Prove that its angle bisector $C F$, median $B E$, and altitude $A D$ are concurrent.
(1984, Greek Mathematical Olympiad) | Proof: $\because \angle C=\pi-\angle A-\angle B=\frac{\pi}{4}$,
$\therefore \triangle A D C$ is an isosceles right triangle.
Let $A D=1$, then $C D=1, A C=\sqrt{2}$.
$$
\begin{array}{l}
\because \angle B=\frac{\pi}{8}, A D=1, \\
\therefore B D=\operatorname{ctg} \frac{\pi}{8}, B C=1+\operatorname{ctg} \frac{\pi}{8} . \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,976 |
Example 2 As shown in the figure, on the sides $AB$ and $AC$ of $\triangle ABC$, construct $\triangle ABE$ and $\triangle ACF$ outside the triangle such that $\triangle ABE \backsim \triangle ACF$ and $\angle ABE = \angle ACF = 90^{\circ}$. Prove that the lines $BF$, $CE$, and the altitude $AH$ from $A$ to side $BC$ ar... | Analysis: The figure in this problem clearly contains the graph of Ceva's Theorem, the only thing missing is that the intersection points on $AB$ and $AC$ have not been named, which is easily done.
Proof: Let $AB \cap CE = D, AC \cap BF = G$.
$$
\begin{array}{l}
\because \frac{AD}{DB} = \frac{S_{\triangle ABC}}{S_{\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,977 |
Example 3 Let $P$ be any point inside $\triangle ABC$. Inside the shape, construct rays $AL$, $BM$, and $CN$ such that $\angle CAL = \angle PAB$, $\angle MBC = \angle PBA$, and $\angle NCA = \angle BCP$. Prove that $AL$, $BM$, and $CN$ are concurrent. | Analysis: If the rays $A L, B M, C N$ are drawn to intersect at a point, then the figure of Ceva's Theorem appears. If $A P, B P, C P$ are extended to intersect the opposite sides, then another standard figure of Ceva's Theorem is obtained. Therefore, it is advisable to use Ceva's Theorem to solve this problem.
Proof ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,978 |
Example 4 Quadrilateral $A B C D$ is inscribed
in a circle, its sides $A B$
and $D C$ extended meet
at point $P, A D$
and $B C$ extended meet
at point $Q$, through point
$Q$ draw two tangents to
the circle, the points of tangency are $E$ and $F$. Prove: $P 、 E 、 F$ are collinear.
(1997, China Mathematic... | Prove: Connect $E F$, intersecting $A D$ and $B C$ at points $M$ and $N$ respectively, and let $A C \cap B D = R$. To prove that $P, E, F$ are collinear, it is sufficient to prove that $P, M, N$ are collinear. For this, it is sufficient to prove that $P, R, M$ and $P, N, R$ are collinear. To prove that $P, R, M$ are co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,979 |
Example 5 Does a right-angled triangle with a perimeter of 6 and an integer area exist? If not, please provide a proof; if it exists, please specify how many there are?
(1904, National Junior High School Mathematics League) | Solution: Such a right-angled triangle exists, and there is exactly one.
Let the hypotenuse of this right-angled triangle be $c$, the two legs be $a, b (a \leqslant b)$, and the area be $S$. Then
$$
\begin{array}{l}
a+b+c=6, a^{2}+b^{2}=c^{2} . \\
\therefore a+b=6-c, a b=18-6 c .
\end{array}
$$
It is evident that $a,... | c=\frac{8}{3}, a=\frac{5-\sqrt{7}}{3}, b=\frac{5+\sqrt{7}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,980 |
Proposition 1 Let $P$ be the Brocard point of $\triangle ABC$ (as shown in the figure), the inradii and areas of $\triangle BPC$, $\triangle CPA$, and $\triangle APB$ are $r_a$, $r_b$, $r_c$, $\triangle_a$, $\triangle_b$, and $\triangle_c$, respectively. The inradius and area of $\triangle ABC$ are $r$ and $\triangle$,... | Proof: It is known that $P B+P C+a=\frac{2 \Delta}{r_{t}}$,
$$
\begin{array}{l}
P C+P A+b=\frac{2 \Delta_{b}}{r_{b}}, \\
P A+P B+c=\frac{2 \triangle_{c}}{r_{c}} . \\
\therefore 2(P A+P B+P C)+a+b+c \\
\quad=\frac{2 \Delta_{a}}{r_{a}}+\frac{2 \Delta_{b}}{r_{b}}+\frac{2 \Delta_{c}}{r_{c}} .
\end{array}
$$
Thus $P A+P B+... | \frac{a^2 b + b^2 c + c^2 a}{\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}} | Geometry | proof | Yes | Yes | cn_contest | false | 710,981 |
Proposition 2 Let $P$ be the Brocard point of the bicircular quadrilateral $ABCD$, and let the inradii and areas of $\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA$ be $r_{a}, r_{b}, r_{c}, r_{d}, \triangle_{a}, \triangle_{b}, \triangle_{c}, \triangle_{d}$, respectively. Let the inradius and area of quadril... | $$
\begin{array}{l}
P A+P B+a \\
=\frac{2 \triangle_{a}}{r_{a}}, \\
P B+P C+b \\
=\frac{2 \triangle_{b}}{r_{b}}, \\
P C+P D+c=\frac{2 \triangle_{c}}{r_{c}}, \\
\quad P D+P A+d=\frac{2 \triangle_{d}}{r_{d}} . \\
\therefore 2(P A+P B+P C+P D)+a+b+c+d \\
=\frac{2 \triangle_{a}}{r_{a}}+\frac{2 \triangle_{b}}{r_{b}}+\frac{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,982 |
Proposition Let $a, b, c$ be positive real numbers, and $abc=1$. Then for all non-zero integers $n$, the inequality holds:
$$
\sum \frac{ab}{a^n + b^n + ab} \leqslant 1.
$$
where $\sum$ denotes the cyclic sum over $a, b, c$.
When $n=5$, (1) is the 37th IMO Shortlist problem ${ }^{[1]}$. Therefore, (1) is an exponentia... | First, prove the following lemma.
Lemma 1 For $a, b, c$ satisfying the conditions of the theorem,
(i) $\sum a^{2} b \geqslant \sum a$;
(ii) $\sum a^{2} b \geqslant \sum a b$.
Proof: Since $a b c=1$, refer to the problem G66 in this journal's Mathematical Olympiad section to know that (i) holds.
For (ii), if $\sum a \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,983 |
Let $F$ be the Fermat point inside $\triangle ABC$, and extend $AF, BF, CF$ to intersect the opposite sides at $A', B', C'$, respectively. Let $AA' = x, BB' = y, CC' = z$.
In 1995, Wu Yuesheng obtained the following inequality: :1]
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geqslant \frac{3 \sqrt{3}}{p^3},
\]
where $p... | Proof: According to [2], we have
$$
\begin{array}{l}
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2} \\
=\frac{3\left(a^{2}+b^{2}+c^{2}+4 \sqrt{3} \triangle\right)}{8 \Delta^{2}} .
\end{array}
$$
By the identity in a triangle
$$
a^{2}+b^{2}+c^{2}=2\left(p^{2}-4 R r-r^{2}\right)
$$
and $\triangle=p r$ (where $p$ ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,984 |
1. Determine all finite point sets $S$ in the plane with at least three points that satisfy the following condition:
For any two distinct points $A$ and $B$ in $S$, the perpendicular bisector of segment $AB$ is a line of symmetry of $S$. | Solution: Let $G$ be the centroid of $S$. For any two points $A, B$ in $S$, let $r_{AB}$ be the reflection of $S$ about the perpendicular bisector of segment $AB$. Since $r_{AB}(S) = S$, it follows that $r_{AB}(G) = G$. This implies that the distance from each point in $S$ to $G$ is equal, hence all points in $S$ lie o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,985 |
2. Let $n$ be a fixed integer, $n \geqslant 2$.
a) Determine the smallest constant $c$ such that the inequality $\sum_{1 \leqslant i<j} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4}$ holds for all non-negative real numbers $x_{1}, \cdots, x_{n} \geqslant ... | Solution: Since the inequality is homogeneous and symmetric, we can assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant 0$ and $\sum_{i=1}^{n} x_{i}=1$. At this point, we only need to discuss
$$
F\left(x_{1}, \cdots, x_{n}\right)=\sum_{i} x_{i}^{3} x_{i+1} > 0.
$$
In other words, when adjusting $x$... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,986 |
3. Let $n$ be a fixed positive even number. Consider an $n \times n$ square board, which is divided into $n^{2}$ unit squares. Two distinct unit squares on the board are called adjacent if they share a common edge.
Mark $N$ unit squares on the board such that every unit square on the board (marked or unmarked) is adja... | Solution: Let $n=2k$. First, color the square board in a chessboard pattern with alternating black and white squares. Let $f(n)$ be the minimum value of $N$ sought, $f_{w}(n)$ be the minimum number of white squares that must be marked so that every black square has a marked white square adjacent to it. Similarly, defin... | k(k+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,987 |
4. Determine all pairs of positive integers $(n, p)$, satisfying:
' $p$ is a prime, $n \leqslant 2 p$, and $(p-1)^{n}+1$ is divisible by $n^{p-1}$. | Solution: Clearly, $(1, p)$ and $(2,2)$ satisfy the problem. Therefore, we consider the case where $n \geqslant 2, p \geqslant 3$ below.
Since $(p-1)^{n}+1$ is odd, $n$ must also be odd, hence $n<2 p$. Let $q$ be the smallest prime divisor of $n$, then $q \mid(p-1)^{n}+1$, so $(p-1)^{n} \equiv-1(\bmod q)$, and $(q, p-... | (2,2),(3,3),(1, p) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,988 |
5. Two circles $\Gamma_{1}$ and $\Gamma_{2}$ are contained within circle $\Gamma$, and are tangent to circle $\Gamma$ at two different points $M$ and $N$. $\Gamma_{1}$ passes through the center of $\Gamma_{2}$. The line through the two intersection points of $\Gamma_{1}$ and $\Gamma_{2}$ intersects $\Gamma$ at points $... | Proof: $C D$ is tangent to $\Gamma_{2}$.
Proof: First, we prove a lemma.
Lemma Given that circle $\Gamma_{1}$ is contained within circle $\Gamma$ and is tangent to $\Gamma$ at point $U$. A chord $P Q$ of $\Gamma$ is tangent to $\Gamma_{1}$ at point $V$. Let $W$ be the point on $\Gamma$ that is the antipodal point of $F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,989 |
Example 6 Given that $x, y$ are real numbers, and $x^{2}+x y+$ $y^{2}-2=0$. Then the range of $x^{2}-x y+y^{2}$ is $\qquad$ | (1996, Hubei Province Huanggang Region Junior High School Mathematics Competition)
Solution: Let $x^{2}-x y+y^{2}=k$ (i.e., introduce the parameter $k$), and combine with the given conditions to get
$$
x y=\frac{2-k}{2},(x+y)^{2}=\frac{6-k}{2} \text {. }
$$
$\therefore$ When $6-k \geqslant 0$, i.e., $k \leqslant 6$, we... | \frac{2}{3} \leqslant x^{2}-x y+y^{2} \leqslant 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,991 |
2. Given that $A B C D$ is a cyclic quadrilateral, $E$ and $F$ are points on $A B$ and $C D$ respectively, and satisfy $A E: E B = C F: F D$. Let $P$ be a point on segment $E F$ such that $P E: P F = A B: C D$. Prove: The ratio of the areas of $\triangle A P D$ and $\triangle B P C$ does not depend on the choice of $E$... | Prove: If
lines $A D, B C$
are not parallel (as shown
in the figure), let their
intersection point
be $S$. Since
$A B C D$ is a cyclic
quadrilateral, then $\triangle A S B \sim \triangle C S D$. Therefore, $\frac{A B}{A S}=\frac{C D}{C S}$.
Also, since $\frac{A E}{E B}=\frac{C F}{F D}$, and $\frac{A E}{A B}=\frac{C F}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,992 |
4. Let $M, N$ be two points inside $\triangle ABC$ such that $\angle MAB = \angle NAC$ and $\angle MBA = \angle NBC$. Prove:
$$
\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1.
$$ | Proof: Let $K$ be a point on ray $BN$, satisfying $\angle BCK = \angle BMA$. Since $\angle BMA > \angle ACB$, then $K$ is outside $\triangle ABC$. Also, because $\angle MBA = \angle CBK$, we have $\triangle ABM \sim \triangle KBC$.
Thus,
$$
\frac{AB}{BK} = \frac{BM}{BC} = \frac{AM}{CK}.
$$
From $\angle ABK = \angle MB... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,993 |
5. Given that the orthocenter of $\triangle ABC$ is $H$, the circumcenter is $O$, and the circumradius is $R$. Let the points $A, B, C$ have their respective reflections across the lines $BC, CA, AB$ be $D, E, F$. Prove: The points $D, E, F$ are collinear if and only if $OH=2R$. | Proof: Let the centroid of $\triangle ABC$ be $G$, and the midpoints of $BC$, $CA$, and $AB$ be $A'$, $B'$, and $C'$, respectively. Draw the perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$, respectively, and let their intersection points be $A''$, $B''$, and $C''$. Then, the centroid of $\triangle A''B''C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,994 |
7. Given $\triangle A B C$ satisfies $\angle A C B=2 \angle A B C$. Let $D$ be a point on side $B C$ such that $C D=2 B D$. Extend line segment $A D$ to $E$ so that $A D=D E$. Prove: $\angle E C B+180^{\circ}=2 \angle E B C$. | Prove: Let
the midpoint of
$CD$ be
$H$, then $ABEH$
is a parallelogram.
Extend $BC$ to
$G$, such that $OG=$
$CA$. Let $BD=$
$DH^{\top}=\mathrm{HC}=$
$\frac{2}{2}, CA=b$,
$AB=c, BE=AH=x, AD=DE=y, CE=z$.
$\because 2 \angle ABC=\angle ACB=\angle CGA+\angle CAG$
$=2 \angle CGA=2 \angle CAG$,
$\therefore \triangle ABG \sim ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,995 |
8. Let $\triangle A B C$ satisfy $\angle A=90^{\circ}, \angle B<\angle C$. Draw the tangent to the circumcircle $\omega$ of $\triangle A B C$ through $A$, intersecting line $B C$ at $D$. Let $E$ be the reflection of $A$ over line $B C$, and let $X$ be the foot of the perpendicular from $A$ to $B E$. Let $Y$ be the midp... | Proof: Let $GA$ be the diameter of the circumcircle of $\triangle ABC$, and $H$ be the intersection of $AE$ and $BD$. Since $\angle B < \angle C$, $B$ and $G$ are on the same side of $AE$.
Since $\angle AEG = 90^\circ = \angle AXB$, $\angle AGE = \angle ABE = \angle ABX$, then
$\triangle AGE \sim \triangle ABX$.
Thus,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,996 |
9. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and satisfy $a_{1}+a_{2}+\cdots+a_{n}<1$. Prove:
$$
\begin{array}{l}
\frac{a_{1} a_{2} \cdots a_{n}\left\{1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)\right\}}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\righ... | Proof: Let $a_{n+1}=1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$. Obviously, $a_{n+1}>0$. Thus, we obtain $n+1$ positive numbers whose sum is 1. Therefore, the inequality becomes
$$
\begin{array}{l}
n^{n+1} a_{1} a_{2} \cdots a_{n} a_{n+1} \\
\leqslant\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)\left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,997 |
10. Let $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers greater than or equal to 1. Prove:
$$
\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}+\cdots+\frac{1}{r_{n}+1} \geqslant \frac{n}{\sqrt{r_{1} r_{2} \cdots r_{n}}+1} \text {. }
$$ | Proof: When $n=1$, the inequality obviously holds. Below, we prove by mathematical induction that the inequality holds for $n=2^{k} (k=1,2, \cdots)$.
When $k=1$, we have
$$
\begin{array}{l}
\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}-\frac{2}{\sqrt{r_{1} r_{2}}+1} \\
=\left(\frac{\left.\sqrt{r_{1} r_{2}}-1\right)\left(\sqrt{r_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,998 |
11. Let $x, y, z$ be positive real numbers, and $xyz=1$. Prove:
$$
\begin{array}{l}
\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+c)} \\
+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4} .
\end{array}
$$ | Proof 1: The original inequality is equivalent to
$$
\begin{array}{l}
x^{4}+x^{3}+y^{4}+y^{3}+z^{4}+z^{3} \\
\geqslant \frac{3}{4}(x+1)(y+1)(z+1) .
\end{array}
$$
For any positive numbers \( u, v, w \), we have \( u^{3}+v^{3}+w^{3} \geqslant 3uvw \). We will prove the stronger inequality:
$$
\begin{array}{l}
x^{4}+x^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,999 |
12. Given integers $n \geqslant k \geqslant 0$, define the number $c(n, k)$:
$c(n, 0)=c(n, n)=1, \quad n \geqslant 0$ when,
$$
c(n+1, k)=2^{k} c(n, k)+c(n, k-1) \text {, }
$$
$n \geqslant k \geqslant 1$ when.
Prove: $c(n, k)=c(n, n-k)$ for all integers $n, k$ satisfying $n \geqslant k \geqslant 0$.
| Proof: For $m \geqslant 1$, let
$$
\begin{array}{l}
f(m)=\left(2^{1}-1\right)\left(2^{2}-1\right) \cdots\left(2^{m}-1\right), \\
f(0)=1 . \\
\text { Let } a(n, k)=\frac{f(n)}{f(k) f(n-k)} .
\end{array}
$$
Then for $n \geqslant 0$, we have $a(n, 0)=a(n, n)=1$. Also, since $f(m)\left(2^{m+1}-1\right)=f(m+1)$ holds for a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,000 |
15. Find all pairs of real numbers $(a, b)$ such that for all positive integers $n$, $a[m]=b[a n]$, where $[x]$ denotes the greatest integer not exceeding $x$.
保留源文本的换行和格式，翻译结果如下：
15. Find all pairs of real numbers $(a, b)$ such that for all positive integers $n$, $a[m]=b[a n]$, where $[x]$ denotes the greatest integ... | Solution 1: Obviously, all real number pairs $(a, b)$ satisfy the condition when $a b=0$, or $a=b$, or $a, b$ are both integers. Next, we will prove that these are the only solutions.
Assume $a b \neq 0, a \neq b$, and for $n=1,2, \cdots, a[b n]=$ $b[a n]$. We will prove that $a, b$ are both integers.
If $a$ is not an... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,001 |
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
(1994-1995, Chongqing and Four Other Cities Mathematics Competition) | Solution: Let $\frac{n^{2}-71}{7 n+55}=k$ (where $k$ is an integer),
then $n^{2}-7 k n-(55 k+71)=0$.
$$
\begin{aligned}
\Delta & =49 k^{2}+4(55 k+71) \\
& =49 k^{2}+220 k+284 .
\end{aligned}
$$
When $\Delta$ is a perfect square, the $n$ in equation (1) is an integer. And
$$
\begin{array}{l}
(7 k+15)^{2}<\Delta=49 k^{2... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,002 |
1. If $\frac{x y}{(z+x)(y+z)}+\frac{y z}{(x+y)(z+x)}$ $+\frac{z x}{(y+z)(x+y)}=1$, then the values of $x, y, z$ are ( ).
(A) all zero
(B) only two are zero
(C) only one is zero
(D) all are non-zero | $-1 .(\mathrm{C})$.
The original equation can be rewritten as $x y z=0$, then at least one of $x, y, z$ is zero. If two or three of $x, y, z$ are zero, the denominator in the original equation would be undefined. Therefore, only one of $x, y, z$ is zero. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,003 |
2. The integer solutions of the equation $\left(x^{2}+x-1\right)^{x+98}=1$ are ( ) in number.
(A) 2
(B) 3
(C) 4
(D) 5 | 2. (C).
From $\left\{\begin{array}{l}x+98=0, \\ x^{2}+x-1 \neq 0\end{array}\right.$ we get $x_{1}=-98$,
From $x^{2}+x-1=1$ we get $x_{2}=1, x_{3}=-2$,
From $\left\{\begin{array}{l}x^{2}+x-1=-1, \\ x+98 \text { is even }\end{array}\right.$ we get $x_{4}=0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,004 |
3. Let $n$ be a natural number. Then the divisibility of $n^{2}+n+2$ is ( ).
(A) Neither divisible by 2 nor by 5
(B) Divisible by 2, but not by 5
(C) Not divisible by 2, but divisible by 5
(D) Divisible by both 2 and 5 | 3. (B).
Since $n^{2}+n=n(n+1)$, and $n(n+1)$ is divisible by 2, therefore, $n^{2}+n+2$ is divisible by 2. Also, the unit digit of $n(n+1)$ can only be $0, 2, 6$, then the unit digit of $n^{2}+n+2$ can only be $2, 4, 8$, so it cannot be divisible by 5. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,005 |
4. The condition that can determine quadrilateral $ABCD$ is a rhombus is ( ).
(A) Diagonal $AC$ bisects diagonal $BD$, and $AC$ $\perp BD$
(B) Diagonal $AC$ bisects diagonal $BD$, and $\angle A$ $=\angle C$
(C) Diagonal $AC$ bisects diagonal $BD$, and bisects $\angle A, \angle C$
(D) Diagonal $AC$ bisects $\angle A, \a... | 4. (D) .
As shown in the figure (referred to as a pen shape),
$AC$ bisects $BD$, $AC \perp BD$, $AC$ also bisects $\angle A, \angle C$, so we can rule out (A) and (C); a parallelogram that is not a rhombus can rule out (B); according to the method of determining a rhombus, we can prove that (D) is correct. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,006 |
5. As shown in the figure, square $A B C D$ and square $A E F G$, connect $B E$, $C F$, $D G$. Then $B E: C F: D G$ equals ( ).
(A) $1: 1: 1$
(B) $1: \sqrt{2}: 1$
(C) $1: \sqrt{3}: 1$
(D) $1: 2: 1$ | 5. (B).
As shown in the figure,
$\triangle A B E \cong \triangle A D G$.
Connect $A C$ and $A F$.
$$
\begin{array}{l}
\because \angle C A F=45^{\circ}-\angle B A F \\
=\angle B A E, \\
\frac{A C}{A B}=\sqrt{2}, \frac{A F}{A E}=\sqrt{2}, \\
\therefore \frac{A C}{A B}=\frac{A F}{A E} .
\end{array}
$$
Therefore, $\tri... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,007 |
6. As shown in the figure, in two circles with $O$ as the center, $A$ is any point on the larger circle, and a secant line $A X Y$ is drawn through $A$ to the smaller circle. If $A X \cdot A Y = 3$, then the area of the annulus in the figure is ( .
(A) $3 \pi$
(B) $6 \pi$
(C) $9 \pi$
(D) cannot be determined | 6. (A).
As shown in the figure, draw the tangent $AB$ of the small circle through $A$, with the point of tangency being $B$, then
$$
AB^2 = AX \cdot AY = 3
$$
Connect $2U, \cup B$, then
$$
\begin{array}{c}
S_{\text {quadrilateral}} = \pi \cdot OA^2 - \pi \cdot OB^2 \\
= \pi \left(OA^2 - OB^2\right) = \pi \cdot AB^2 =... | 3\pi | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,008 |
1. Given that the real number $x$ satisfies $\sqrt{x^{2}-x^{3}}=x$. $\sqrt{1-x}$. Then the range of values for $x$ is | 2.15.
From $2 x^{2}-6 x+y^{2}=0$ we get
$$
y^{2}=-2 x^{2}+6 x \geqslant 0 \text {. }
$$
Solving this, we get $0 \leqslant x \leqslant 3$.
Let $z=x^{2}+y^{2}+2 x$. Then
$$
\begin{array}{l}
z=x^{2}-2 x^{2}+6 x+2 x \\
=-x^{2}+8 x \\
=-(x-4)^{2}+16 . \\
\because 0 \leqslant x \leqslant 3,
\end{array}
$$
$\therefore$ whe... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,009 |
3. As shown in the figure, from the right-angle vertex $C$ of the right triangle $\triangle A B C$, lines $C E$ and $C F$ are drawn to the trisection points of the hypotenuse $A B$. It is known that $C E=\sin \alpha, C F=\cos \alpha$ (where $\alpha$ is an acute angle). Then $A B=$ | 3. $\frac{3 \sqrt{5}}{5}$.
As shown in the figure, let $A C=b$, $B C=a$. Draw $E M \perp A C, F N \perp A C$, with the feet of the perpendiculars being $M, N$ respectively.
Since $B E=E F=F A$,
then
$$
\begin{array}{l}
E M=\frac{2}{3} a, F N=\frac{1}{3} a, \\
C M=\frac{1}{3} b, C N=\frac{2}{3} b .
\end{array}
$$
In ... | \frac{3 \sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,011 |
4. As shown in the figure, circle O is inscribed in $\triangle ABC$ touching sides $AC$, $BC$, and $AB$ at points $D$, $E$, and $F$ respectively, with $\angle C=90^{\circ}$. The area of $\triangle ABC$ is 6. Then $AF \cdot BF=$ $\qquad$ | 4.6 .
Let $A C=b, B C=a, A B=c, \odot O$ have a radius of $r$. Then
$$
\begin{array}{l}
r=\frac{a+b-c}{2} \cdot a^{2}+i^{2}=c^{2} . \\
\therefore A F \cdot B F=A C \cdot B E \\
=(b-r)(a-r) \\
=\frac{b+c-a}{2} \cdot \frac{a+c-b}{2} \\
=\frac{1}{4}\left[c^{2}-(a-b)^{2}\right] \\
=\frac{1}{2} a b=6 .
\end{array}
$$ | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,012 |
Example 8 Let $9 \cos A-3 \sin B+\tan C=0$, (1) $\sin ^{2} B-4 \cos A \tan C=0$.
Prove: $|\cos A| \leqslant \frac{1}{6}$. | Proof: When $\cos A=0$, the conclusion is obviously true.
When $\cos A \neq 0$, according to (2), we can construct the equation
$$
x^{2} \cos A - x \sin B + \operatorname{tg} C = 0.
$$
From (1), we know that 3 is a root of equation (3), and from (2), we know that equation (3) has two equal real roots.
Thus $3+3=\frac{... | |\cos A| \leqslant \frac{1}{6} | Algebra | proof | Yes | Yes | cn_contest | false | 711,013 |
One, (Full marks 20 points) As shown in the figure, $D$ is a point on side $B C$ of equilateral $\triangle A B C$. It is known that $B D = 1, C D = 2, C H \perp A D$ at point $H$, and $B H$ is connected. Prove:
$$
\angle B H D = 60^{\circ}
$$ | Through $A$ draw $A O \perp B C$ at
$O$.
$$
\begin{array}{l}
\because \angle A D O=\angle C D H, \\
\angle A O D=\angle C H D=
\end{array}
$$
$90^{\circ}$,
$\therefore \triangle A D O \sim \triangle C D H$.
Thus $\frac{A D}{C D}=\frac{O D}{D H}$.
$$
\because B D=1, C D=2 \text {, }
$$
$\therefore A B=A C=B C=3$,
$$
O D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,014 |
II. (Full marks 25 points) Find the real number $k$, such that the roots of the equation $k x^{2}+(k+1) x+(k-1)=0$ are all integers.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | When $k=0$, the original equation becomes $x-1=0 . x=1$ which satisfies the condition.
When $k \neq 0$, let the two integer roots of the original equation be $\alpha$ and $\beta$, and assume $\alpha \geqslant \beta$. By Vieta's formulas, we get
$$
\begin{array}{l}
\alpha+\beta=-\frac{k+1}{k}=-1-\frac{1}{k}, \\
\alpha ... | k=0,1,-\frac{1}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,015 |
Three. (Full marks 25 points) A natural number that can be expressed as the difference of the squares of two natural numbers is called a "smart number." For example, $16=5^{2}-3^{2}$ is a "smart number." Try to answer the following questions:
(1) Is 1998 a "smart number"? Explain your reasoning.
(2) Arranged in ascendi... | Three, (1) 1998 is not a "smart number". If 1998 were a "smart number", we could set \(1998 = m^2 - n^2\) (where \(m\) and \(n\) are natural numbers), then \((m+n)(m-n) = 2 \times 999\).
Since \(m+n\) and \(m-n\) have the same parity, \((m+n)(m-n)\) is either odd or divisible by 4, hence it cannot be 1998. Therefore, 1... | 2665 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,016 |
1. Given $f(x)=x^{2}+(\lg a+2) x+\lg b$, and $f(-1)=-2$, also $f(x) \geqslant 2 x$ for all $x \in R$. Then $a+b=(\quad)$.
(A) 100
(B) 110
(C) 120
(D) 130 | $-, 1 .(B)$
Since $f(-1)=-2$, then
$$
\begin{array}{l}
1-(\lg a+2)+\lg b=-2 \\
\Rightarrow \lg a=\lg b+1 \Rightarrow a=10 b .
\end{array}
$$
Since for all $x \in \mathbf{R}$, $f(x) \geqslant 2 x$,
$$
x^{2}+(\lg a) x+\lg b \geqslant 0 .
$$
Since the coefficient of $x^{2}$ is $1>0$, then
$$
... | 110 | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,017 |
2. If $m>n>1$, then it must be true that ( ).
(A) $\log _{m} n>\log _{n} m>\log _{n}^{1} m$
(B) $\log _{\frac{1}{n}} m>\log _{n} m>\log _{m} n$
(C) $\log _{n} m>\log _{m} n>\log _{\frac{1}{n}} m$
(D) $\log _{n} m>\log _{\frac{1}{n}} m>\log _{m} n$ | 2. (C).
Take $m=100, n=10$ and substitute into the option to verify. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,018 |
3. Given that 10 prime numbers form an arithmetic sequence. Then their common difference $d$ must satisfy ( ).
(A) $d$ is a multiple of 2, 3, 5, 7
(B) $d$ is a multiple of 11
(C) $d \leqslant 200$
(D) $d \leqslant 50$ | 3. $(\mathrm{A})$.
Obviously, 2 is not a number in this sequence. Next, we prove that $3 \mid d$. In $a, a+d, a+2d (a, d \in \mathbf{N})$, if $3 \nmid d$, then the remainders when these three numbers are divided by 3 must be different, otherwise $3 \mid d$. In this case, $a, a+d, a+2d$ must include a multiple of 3, wh... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,019 |
4. Let $A B$ be the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the major axis $A B$ is divided into 100 equal parts, and perpendicular lines are drawn through each division point to $A B$, intersecting the upper half of the ellipse at $P_{1}, P_{2}$, $P_{3}, \cdots, P_{99}$, and $F_{1}$ i... | 4. (A).
From the definition of an ellipse, we know
$$
\begin{array}{l}
\left|F_{1} P_{i}\right|+\left|F_{2} P_{i}\right|=2 a(i=1,2, \cdots, 99) . \\
\therefore \sum_{i=1}^{99}\left(\left|F_{1} P_{i}\right|+\left|F_{2} P_{i}\right|\right)=99 \times 2 a=198 a .
\end{array}
$$
According to the problem, $P_{1}, P_{2}, \c... | 101 a | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,020 |
5. In square $ABCD$, $M$ is the midpoint of $AD$, and $N$ is the midpoint of $AB$. Fold $\triangle CDM$ and $\triangle CBN$ along $CM$ and $CN$ respectively, so that $CB$ and $CD$ coincide. Let point $B$ coincide with point $D$ at point $P$, and the midpoint of $DM$ becomes the midpoint of $PM$, denoted as $T$. Then th... | $5 .(\mathrm{D})$.
In square $ABCD$, connect $AC$ and $MN$, and let their intersection be $S$. Since $M$ is the midpoint of $AD$ and $N$ is the midpoint of $AB$, by symmetry, $S$ is the midpoint of $MN$. Connect $TS$, from $\triangle MPN$ we get $TS \parallel PN$. Therefore, $\angle CTS$ is the angle formed by the skew... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,021 |
6. Let $N=69^{5}+5 \times 69^{4}+10 \times 69^{3}+10 \times$ $69^{2}+5 \times 69+1$. Then the number of positive integer divisors of $N$ is ( ) .
(A)3
(B) 216
(C) 69
(D) 5 | 6. (B).
Using the inverse of the binomial theorem, we get
$$
N=(69+1)^{5}=70^{5}=2^{5} \times 5^{5} \times 7^{5} \text {. }
$$
From this, we can see that the positive integer divisor $a$ of $N$ has the following form:
$$
\begin{array}{l}
a=2^{x} 5^{y} 7^{z}, \\
0 \leqslant x \leqslant 5,0 \leqslant y \leqslant 5,0 \l... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,022 |
1. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{5}+1999(x-1)=2519, \\
(y-1)^{5}+1999(y-1)=-2519 .
\end{array}\right.
$$
Then the value of $x+y$ is $\qquad$ | 1.2.
Let $f(t)=t^{5}+1999 t$. Then $f(t)$ is increasing on $R$, and $f(x-1)=-f(y-1)$. Since $f(y-1)=-f(1-y)$, then
$$
\begin{array}{l}
f(x-1)=f(1-y) . \\
\therefore x-1=1-y \Rightarrow x+y=2 .
\end{array}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,023 |
Example 9 Given the equation in $x$, $x^{3}-a x^{2}$ $-2 a x+a^{2}-1=0$, which has only one real root. Then the range of the real number $a$ is $\qquad$ . | Solving: Rearrange the original equation into a quadratic equation in terms of $a$:
$$
\begin{array}{l}
a^{2}-x(x+2) a+x^{3}-1=0, \\
(a-x+1)\left(a-x^{2}-x-1\right)=0 . \\
\therefore x=a+1 \text { or } x^{2}+x+1-a=0 .
\end{array}
$$
According to the problem, the equation $x^{2}+x+1-a=0$ has no real roots.
Thus, $\Delt... | a<\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,024 |
2. Given that all terms of the sequence $\left\{a_{n}\right\}$ are positive, and the sum of the first $n$ terms $S_{n}$ satisfies $6 S_{n}=a_{n}^{2}+3 a_{n}+2$. If $a_{2}$, $a_{4}$, and $a_{9}$ form a geometric sequence, then the general term of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$ | $2.3 n-2$
When $n=1$, we have
$6 a_{1}=a_{1}^{2}+3 a_{1}+2 \Rightarrow a_{1}=1$ or $a_{1}=2$.
When $n \geqslant 2$,
$$
\begin{array}{l}
6 S_{n}=a_{n}^{2}+3 a_{n}+2, \\
6 S_{n-1}=a_{n-1}^{2}+3 a_{n-1}+2,
\end{array}
$$
(1) - (2) gives
$$
\begin{array}{l}
\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-3\right)=0 . \\
\bec... | 3n-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,025 |
3. For the function $f^{\prime}(x):=$ ig $\left(\sin ^{6} x+\cos ^{6} x+\right.$ $a \sin x \cos x)$ to have the domain $\mathbb{R}$, the real number $a$ must satisfy the condition: | Therefore, it is only necessary to determine $a$ such that when $t \in\left[-\frac{1}{2}, \frac{1}{2}\right]$,
$$
-3 t^{2}+a t+1>0 \text {. }
$$
(i) When $t=0$, (1) always holds;
(ii) When $t>0$, (1) $\Leftrightarrow a>3 t-\frac{1}{t}$. Let $f(t)=3 t$ $-\frac{1}{t}$, at this time $f(t)$ is monotonically increasing on $... | -\frac{1}{2}<a<\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 711,026 |
4. Given four numbers: $3^{147}-5^{21}, 2^{2^{4}}+1,147^{4}-$ $121^{2}, 13^{4}-11$ where one of them is a prime number. Then this prime number is $\qquad$ . | $4 \cdot 2^{2^{4}}+1$.
$3^{147}-5^{21}=\left(3^{21}\right)^{7}-\left(5^{3}\right)^{7}$, it has a factor $3^{21}-5^{3}$, $147^{4}-121^{2}$ can be divided by $147^{2}-121$, $13^{4}$ has the last digit 1, so $13^{4}-11$ can be divided by 10. $\therefore 2^{2^{4}}+1$ is a prime number. | 2^{2^{4}}+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,027 |
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