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Five. (Full marks 20 points) Through the point $M(2,-1)$, draw four chords $P_{i} Q_{i}(i=1,2,3,4)$ of the parabola $y^{2}=x$, and the y-coordinates of points $P_{1}, P_{2}, P_{3}, P_{4}$ form an arithmetic sequence. Find whether: $\frac{P_{1} M}{M Q_{1}}-\frac{P_{2} M}{M Q_{2}}>\frac{P_{3} M}{M Q_{4}}-\frac{P_{4} M}{M... | Let $P_{i}\left(y_{i}^{2}, y_{i}\right), Q_{i}\left(y^{\prime 2}, y_{i}^{\prime}\right), \lambda_{i}=\frac{P_{i} M}{M Q_{i}}\left(\lambda_{i} \neq\right.$ $-1), i=1,2,3,4$. By the section formula, we get
$$
\left\{\begin{array}{l}
\frac{y_{i}^{2}+\lambda_{i} y^{\prime 2}}{1+\lambda_{i}}=2, \\
\frac{y_{i}+\lambda_{i} y_... | \frac{P_{1} M}{M Q_{1}}-\frac{P_{2} M}{M Q_{2}}>\frac{P_{3} M}{M Q_{3}}-\frac{P_{4} M}{M Q_{4}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,808 |
Let $a$ and $b$ be integers. Find all integer solutions that satisfy $(a, b) + 9[a, b] + 9(a + b) = 7ab$.
The text translated to English is as follows:
Let $a$ and $b$ be integers. Find all integer solutions that satisfy $(a, b) + 9[a, b] + 9(a + b) = 7ab$.
The problem is to find all pairs of integers $(a, b)$ that ... | Let $(a, b)=k$, then $a=k x, b=k y$, where $x, y \in N$, and $(x, y)=1$, then $[a, b]=k x y$. Substituting into the original equation, we get
$$
1+9 x y+9(x+y)=7 k x y .
$$
Solve the equation (1) to determine the range of values for the parameter $k$.
From (1) $7 k=\frac{1}{x y}+9\left(\frac{1}{x}+\frac{1}{y}\right)+9... | (x, y) = (2, 19), (19, 2), (1, 1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,810 |
List all positive integers $a$ such that the quadratic equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one integer root.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Transform the original equation into a linear equation in terms of $a$: $(x+2)^{2} a=2(x+6)$.
Therefore, $x+2 \neq 0$.
$\therefore a=\frac{2(x+6)}{(x+2)^{2}}$. $(a$ is a positive integer)
(*)
$$
\therefore \frac{2(x+6)}{(x+2)^{2}} \geqslant 1 \text {. }
$$
Solving this, we get $-4 \leqslant x \leqslant 2$. T... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,811 |
In $\triangle ABC$, the area is $1, DE \parallel AB$, connect $BD$, let the largest area among $\triangle DCE$, $\triangle ABD$, and $\triangle BDE$ be $y$. Find the minimum value of $y$.
The area of $\triangle ABC$ is $1, DE \parallel AB$, connect $BD$, and let the largest area among $\triangle DCE$, $\triangle ABD$,... | Let $\frac{C D}{A C}=x$. Since $\triangle C D E \sim \triangle A B C$, we have $S_{\triangle C D E}=x^{2}$, $\frac{A D}{A C}=1-x$. Therefore, $S_{\triangle A B D}=1-x$, $S_{\triangle B D E}=1-x^{2}-(1-x)=x-x^{2}=x(1-x)$, where $0<x<1$.
From $x(1-x)<1-x$, we know that $S_{\triangle B D E}$ cannot be the largest. When $... | \frac{3-\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,814 |
Prove that $a_{n}=72 n+63-2^{n+6}+$ $(-7)^{n}(n \in N)$ is a multiple of 576. | Proof: From $(x-1)^{2}(x-2)(x+7)=x^{4}+$ $3 x^{3}-23 x^{2}+33 x-14$, it is easy to verify that $a_{n+4}+3 a_{n+3}-$ $23 a_{n+2}+33 a_{n+1}-14 a_{n}=0(n \geqslant 0)$ (assuming $y^{0}=1$ for any $y$).
Also, $a_{0}=0, a_{1}=0, a_{2}=0, a_{3}=-576$ are all multiples of 576, so by the recurrence relation, $a_{4}=-3 a_{3}+... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,815 |
62 Given: In a non-obtuse $\triangle A B C$, the three sides are $a, b, c$, and the lengths of the medians to these sides are $m_{a}, m_{b}, m_{c}$. $R$ and $r$ are the circumradius and inradius, respectively. Prove:
$$
\frac{a^{2}}{m_{a}}+\frac{b^{2}}{m_{b}}+\frac{c^{2}}{m_{c}} \geqslant 8 R-4 r .
$$ | Proof: $\because b^{2}+c^{2}-a^{2}=2 b c \cos A$
$$
\begin{array}{c}
\leqslant\left(b^{2}+c^{2}\right) \cos A . \\
\therefore a^{2} \geqslant\left(b^{2}+c^{2}\right)(1-\cos A) \\
=\left(b^{2}+c^{2}\right) \cdot 2 \sin ^{2} \frac{A}{2} \\
\therefore 2\left(b^{2}+c^{2}\right) \leqslant a^{2} \csc ^{2} \frac{A}{2} \\
=a^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,816 |
Example 13 Find all values of $a$ such that the roots of the equation $a^{2} x^{2}+a x+1-7 a^{2}=0$ are integers.
(1992, Shanghai Junior High School Mathematics Competition) | Solution: Transform the original equation into a quadratic equation in $a$:
$$
\begin{array}{l}
\left(x^{2}-7\right) a^{2}+x a+1=0 . \\
\because \Delta=x^{2}-4\left(x^{2}-7\right)=28-3 x^{2} \geqslant 0, \\
\therefore x^{2} \leqslant \frac{28}{3} .
\end{array}
$$
Since $x$ is an integer,
$$
\therefore x^{2}=0,1,4,9 \t... | \frac{11}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,817 |
1. Given the equation $\left(a^{2}-1\right) x^{2}-2(5 a+1) x+24=0$ has two distinct negative integer roots. Then the integer value of $a$ is $\qquad$ .
（The 1st Zu Chongzhi Cup Junior High School Mathematics Competition） | (Solution: $a=-2$ ) | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,818 |
2. Given that $k$ is an integer, and the equation $\left(k^{2}-1\right) x^{2}-$ $3(3 k-1) x+18=0$ has two distinct positive integer roots. Then $k=$ $\qquad$
（4th Hope Forest Junior High School Mathematics Competition） | (Solution: $k=2$ ) | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,819 |
3. $m$ is an integer not greater than 4, and the equation $x^{2}-2(2 m-3) - x+4 m^{2}-14 m+8=0$ has two integer roots. Find these two integer roots. | (Answer: $m=0$ 时, $x_{1}=-2, x_{2}=-4 ; m=4$ 时, $x_{1}$ $=2, x_{2}=8$ )
(Answer: $m=0$ when, $x_{1}=-2, x_{2}=-4 ; m=4$ when, $x_{1}$ $=2, x_{2}=8$ ) | m=0 \text{ 时, } x_{1}=-2, x_{2}=-4; m=4 \text{ 时, } x_{1}=2, x_{2}=8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,820 |
4. For the equation $\left(1+m^{2}\right) x^{2}-2 m(1+n) x+m^{2}+n^{2}=$ 0 to have integer roots, find the conditions that integers $m, n$ must satisfy, and determine these two integer roots. | (Answer: $m$ is an integer and $m^{2}=n, x_{1}=x_{2}=m$ ) | m^2 = n, x_1 = x_2 = m | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,821 |
5. Find all integer values of $a$ such that the equation $(a+1) x^{2}-\left(a^{2}+1\right) x+2 a^{3}-6$ $=0$ has integer roots.
(1996, Huanggang Region, Hubei Province Junior High School Mathematics Competition) | (Tip: Discuss separately for $a=-1, a \neq-1$, $a=-1,0,1$) | -1, 0, 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,822 |
6. 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 $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$.
(1991, Nanchang City Junior High School Mathematics Competition) | (Given $a>0$, we know $0>x_{1}>x_{2}$, the result is -1) | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,823 |
Example 3 Let $m$ be an integer, and $4<m<40$, and the equation $x^{2}-2(2 m-3) x+4 m^{2}-14 m+8=0$ has two integer roots. Find the value of $m$ and the roots of the equation. | Solution: From $\Delta=[-2(2 m-3)]^{2}-4\left(4 m^{2}-\right.$ $14 m+8)=4(2 m+1)$ and $4<m<40$,
we have $9<2 m+1<81$.
Since $2 m+1$ is an odd number,
thus $2 m+1=5^{2}$ or $7^{2}$.
Then $m=12$ or 24.
When $m=12$, $x^{2}-42 x+416=0$, $x_{1}=26, x_{2}=16$;
When $m=24$, $x^{2}-90 x+1976=0$, $x_{1}=52, x_{2}=38$. | m=12, x_{1}=26, x_{2}=16 \text{ or } m=24, x_{1}=52, x_{2}=38 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,825 |
13. Find all integers $a$ such that the quadratic equation $a x^{2}+2 a x+a-9=0$ has at least one integer root. | (Hint: $a=\frac{9}{(x+1)^{2}} \geqslant 1, -4 \leqslant x \leqslant 2$. Discuss the values of $x$ respectively, we can get $a=1,9$) | a=1,9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,831 |
14. Find all positive numbers $a$ such that the quadratic equation $\left(a^{2}+1\right) x^{2}+2 a x+\left(a^{2}-1\right)=0$ has both roots as integers. | $\begin{array}{l}\quad \text { (Hint: }\left(x^{2}+1\right) a^{2}+2 x a+\left(x^{2}-1\right)=0, \Delta=4 x^{2} \\ -4\left(x^{2}+1\right)\left(x^{2}-1\right) \geqslant 0, x^{4}-x^{2}-1 \leqslant 0,0 \leqslant x^{2} \leqslant \\ \frac{1+\sqrt{5}}{2}<2,-\sqrt{2}<x<\sqrt{2}, \therefore x=-1,0,1 \cdot a \\ =1)\end{array}$ | a=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,832 |
Example $1 \quad \cdots$ Family I: A factory has $n(n \geqslant 3)$ jobs, numbered from 1 to $n$. $n$ job seekers are numbered from 1 to $n$ in the order of their increasing ability. A job seeker $i$ can take on job $j$ if and only if $i \geqslant j$.
Job seekers arrive in a random order. Each person is hired for the ... | Proof: Arrange the job seekers who find jobs in a sequence according to their arrival order. Let the set
$A=$ \{sequences containing job seeker $n$\},
$B=\{$ sequences containing job seeker $n-1$\},
$A_{1}=A \backslash(A \cap B), B_{1}=B \backslash(A \cap B)$.
Establish the mapping $f: A_{1} \rightarrow B_{1}$ as swapp... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,833 |
Example 2 A glasses workshop in a factory has received a batch of tasks, requiring the processing of 6000 type $A$ parts and 2000 type $B$ parts. This workshop has 214 workers, each of whom can process 3 type $B$ parts in the time it takes to process 5 type $A$ parts. These people are divided into two groups, both work... | Solution: Let the number of people assigned to produce $\mathrm{C} A$ type parts be $x$, then the number of people assigned to produce $B$ type parts is $214-x$. Suppose the respective rates are 5 and 3. The time to complete the entire task is
$$
t(x)=\max \left\{\frac{6000}{5 x}, \frac{2000}{3(214-x)}\right\},(x \in N... | 137 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,834 |
Example 3 In a pharmacy, there are several kinds of medicine, some of which are potent. The pharmacist uses these medicines to prepare 68 prescriptions $j$, each prescription containing exactly 5 kinds of medicine, with at least one being potent, but ensuring that any 3 selected medicines are precisely contained in one... | Solution: Let there be $n$ types of medicines, then there are $C_{n}^{3}$ "medicine groups". In this mixture, there are $C_{5}^{3}$ "medicine groups", leading to the equation $C_{n}^{3}=68 C_{5}^{3} \Rightarrow n=17$.
Select any one medicine $a$, there should be $\frac{C_{16}^{2}}{C_{1}^{2}}=20$ prescriptions containi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,835 |
Example 4 Find all integer pairs $(x, y)$ that satisfy the equation $y^{4}+2 x^{4}+1=4 x^{2} y$. | Solution: Transform the original equation into
$$
\begin{array}{l}
2 x^{4}-4 y x^{2}+\left(y^{4}+1\right)=0 . \\
\text { Then } \Delta=(-4 y)^{2}-8\left(y^{4}+1\right) \\
=-8\left(y^{2}-1\right)^{2} \geqslant 0, \\
\left(y^{2}-1\right)^{2} \leqslant 0 .
\end{array}
$$
Thus, $y^{2}-1=0$, i.e., $y=1, \cdots 1$.
When $y=... | (1, 1), (-1, 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,836 |
Example 4 Suppose 10 people each carry a container of the same size to fill with water from 10 identical taps. Let the time it takes to fill the $i$-th container $(i=1,2, \cdots, 10)$ from a tap be $T_{i}$ minutes. How should the 10 people be arranged to minimize the total waiting time? What is this minimum total waiti... | Solution: Let $T_{1}S_{\text {min }}$.
Similarly, in other cases where the number of people at the two taps is unequal, the total time spent is greater than $S_{\min }$.
Therefore, the result of (1) is the answer. | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,837 |
Example 5 The mathematical model of the population dynamics of four types of chicks, where the young can become mature within the same year, estimates the number of a female population after 5 years.
(1) Each female breeds once a year;
(2) Each female breeds only once in the first year of development, and does not bree... | Solution: Let $a_{n}$ represent the number of female birds in the $n$th generation.
(1) According to the problem, $a_{0}=1, a_{n}=2 a_{n-1}, n \geqslant 1$.
It is easy to get $a_{n}=2^{n} a_{0}=2^{n}$.
Therefore, $a_{5}=2^{5}=32$.
(2) According to the problem, $a_{0}=1, a_{n}=a_{n-1}+1$.
Thus, $a_{5}=6$.
(3) Let $b_{... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,838 |
Example 6 The main control panel of a chemical plant is 1 meter wide, the bottom edge of the panel is 2 meters above the ground, and the eyes of the on-duty personnel are 1.2 meters above the ground when sitting. Where should the on-duty personnel sit to see the panel most clearly? And find the maximum angle of view at... | Solution: As shown in Figure 1, the distance from the person's leg to the ground $A E=1.2$, the height of the clock face $B C=1$, and the distance from the bottom of the clock face to the ground $C F=2$. Draw $A D \parallel E F$ intersecting $B F$ at $D$.
Then $D F=1.2, C D=2-1.2=0.8$.
Let $\angle B A D=\alpha, \angl... | \operatorname{arctg} \frac{5}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,839 |
Example 7 A factory produces two types of products, A and B. The production of each ton of product requires electricity (kWh), coal (tons), labor (people), and output value (thousand yuan) as shown in the table below:
\begin{tabular}{|c|c|c|c|c|}
\hline Type & Electricity & Coal & Labor & Output Value \\
\hline A & 2 &... | Solution: Let the production of products A and B be $x, y$ tons, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
2 x+8 y \leqslant 160, \\
3 x+5 y \leqslant 150, \\
5 x+2 y \leqslant 200, \\
x \geqslant 0, \\
y \geqslant 0 .
\end{array}\right.
$$
In the Cartesian coordinate system, draw the ... | 336 \frac{16}{19} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,840 |
Example 8 There are 6 bus stops on a road. A bus is traveling from the first stop to the sixth stop. Passengers can freely get on and off at each stop along the way, but the bus can carry at most 5 passengers at any time. Prove: Among these stops, there must be two pairs (4 different) of stops $X_{1}, Y_{1}$ and $X_{2}... | Proof: Let $a_{ij} (1 \leqslant i < j \leqslant 6)$ represent the number of passengers who board at station $i$ and alight at station $j$. Then, $0 \leqslant a_{ij} \leqslant 5$.
Assume that at each station, the staff must fill out a registration card for the number of passengers boarding and alighting, as shown in th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,841 |
Example 9 Suppose there are $n$ gas stations on a circular highway, each storing a certain number of barrels of gasoline (some stations may store none). The total amount of gasoline stored at the $n$ stations is sufficient for a car to travel around the highway once. Now, let a car that initially has no fuel start from... | Proof: Let these $n$ stations be $A_{1}$, $A_{2}$, ..., $A_{n}$ in sequence (counterclockwise), and each station has oil (otherwise, the station can be canceled).
When $n=1$, the conclusion is obviously true.
Assume that when $n=k$, the conclusion holds.
When $n=k+1$, by the given condition, the total oil is sufficient... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,842 |
For example, $10 n(n>3)$ table tennis players have played several singles matches. After the matches, any two players have played against a set of opponents that are not exactly the same. Try to prove: it is always possible to remove one player, so that among the remaining players, any two players still have not played... | Proof: Let $M_{x}$ denote the set of opponents of $x$, and assume $A$ is the player who has played the most games (if there is a tie, choose any one).
If the proposition does not hold, then there exist $B, C$ such that after removing $A$, the opponent sets of $B, C$ are the same (of course, at this time, $B, C$ have n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,843 |
2. A stationery store that operates both wholesale and retail has the following rule: If you buy 51 pencils or more (including 51), you pay the wholesale price; if you buy 50 pencils or fewer (including 50), you pay the retail price. The wholesale price for 60 pencils is 1 yuan cheaper than the retail price for 60 penc... | (Tip: Let the whole class have $x$ students, the equation is $\frac{60 m}{x}-\frac{60 m}{x+10}$ $=1(40<x \leqslant 50)$, solving it yields $x=50)$ | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,844 |
4. A workshop produces no more than 20 TVs on the first day, and the production of each subsequent day increases by no more than 20 TVs compared to the previous day. After several days, the production on a certain day reaches 1991 TVs. Find the minimum value of the total production from the first day.
将上面的文本翻译成英文，请保留源... | (Tip: Let the production on the $i$-th day be $a_{i}$ (units), $i=1,2, \cdots, n$, and $a_{n}=1991$. Let the total production over $n$ days be $S_{n}$. Let $x_{i}=a_{i}-$ $a_{i-1}(i \geqslant 1)$, and agree that $a_{0}=0$, then $x_{i} \leqslant 20$. Thus, the original problem is transformed into finding the minimum val... | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 709,845 |
5. There are $n$ cities, and it is known that any 2 cities are connected by one of three means of transportation: bus, train, or airplane. Among these cities, all three means of transportation are used, but each city uses at most two of these means, and any 3 cities are connected by at least two means of transportation... | ( Hint: Use points to represent cities, connect every two points with a line, and color them in red, white, and blue to represent three different modes of transportation. Use the coloring method to solve, $n_{\max }=4$. )
Translate the above text into English, please retain the original text's line breaks and format... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,846 |
Example 5 If $k$ is a positive integer, and the two roots of the linear equation $(k-1) x^{2}-p x+k=0$ are both positive integers, then the value of $k^{p k}\left(p^{p}+k^{k}\right)+(p+k)$ is $\qquad$ . | Solution: From $x_{1} x_{2}=\frac{k}{k-1}$, we know $\frac{k}{k-1}$ is a positive integer.
$$
\begin{array}{l}
\therefore k-1=1, k=2 . \\
\text { So } x_{2}=2 \text {. } \\
\therefore x_{1}=1, x_{2}=2 \text {. }
\end{array}
$$
From $x_{1}+x_{2}=\frac{p}{k-1}$, we get $p=3$.
Therefore, the original expression $=2^{6}\l... | 1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,847 |
6. There are two flour mills supplying flour to three residential areas. The first flour mill produces 60 tons per month, and the second flour mill produces 100 tons per month. The first residential area requires 45 tons per month, the second residential area requires 75 tons per month, and the third residential area r... | (1960, Shanghai City Mathematics Recontest)(Hint: Let the flour transported by the first flour mill to the first and second residential areas be $x, y$ tons, respectively. Thus, $s=15 x+6 y+840$, where $0 \leqslant x \leqslant 45,0 \leqslant y \leqslant 60 . x+y$ $\geqslant 20$. When $x=20, y=0$, $S_{\text {min }}=960$... | 960 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,848 |
Example 1 Let the sequence be $u_{0}=2, u_{1}=\frac{5}{2}, \cdots, u_{n+1}=$ $u_{n}\left(u_{n-1}^{2}-2\right)-u_{1}$.
Prove: $\left[u_{n}\right]=2^{\frac{1}{3}\left\{2^{n}-(-1)^{n}\right\}}$.
Here, $[x]$ denotes the greatest integer not exceeding $x$. | Analysis: From the conclusion to be proved, we know that $u_{n}$ should be in the form of $2^{b} n+\alpha$. Here, $b_{n}=\frac{1}{3}\left\{2^{n}-(-1)^{n}\right\}, 0 \leqslant \alpha<1$. To determine the form of $\alpha$, we can take $n=1,2,3, \cdots$, and calculate the first few terms of the sequence:
$$
\begin{array}{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,849 |
Example 2 Let $v_{1}=2, v_{2}=2+1=3, v_{3}=2 \times 3$
$$
\begin{array}{c}
+1=7, \cdots, v_{n}=v_{1} v_{2} \cdots v_{n-1}+1, \text { and } \\
S=\frac{1}{v_{1}}+\frac{1}{v_{2}}+\cdots+\frac{1}{v_{n}} .
\end{array}
$$
Try to prove or disprove the following conclusion: for any positive integer $n, S_{n}<1$, and for any s... | Analysis: First, find some special values of $S_{n}$, and identify the pattern.
$$
\begin{array}{l}
S_{1}=\frac{1}{2}=1-\frac{1}{2}=1-\frac{1}{v_{2}-1}, \\
S_{2}=-\frac{1}{2}+\frac{1}{3}=1-\frac{1}{6}=1-\frac{1}{v_{3}-1}, \\
S_{3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=1-\frac{1}{42}=1-\frac{1}{v_{4}-1}, \\
\quad\left(v_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,850 |
Example 3. Given: $a_{1}=1, a_{2}=2$, $a_{n+2}=\left\{\begin{array}{l}5 a_{n+1}-3 a_{n}, \text { when } a_{n} a_{n+1} \text { is even; } \\ a_{n+1}-a_{n}, \text { when } a_{n} a_{n+1} \text { is odd. }\end{array}\right.$ Prove: for all $n \in N, a_{n} \neq 0$. (1989, National High School Mathematics Contest) | Analysis: If it is difficult to find the starting point for solving the problem, you can first calculate the first few terms of the sequence:
$$
\begin{array}{l}
a_{1}=1, \\
a_{2}=2, \\
a_{3}=5 a_{2}-3 a_{1}=7, \\
a_{4}=5 a_{3}-3 a_{2}=29, \\
a_{5}=a_{4}-a_{3}=29-7=22, \\
a_{6}=5 a_{5}-3 a_{4}=23, \\
\cdots \cdots .
\e... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,851 |
Example 4 Given the numbers $1,2,2^{2}, \cdots, 2^{n-1}$. For any permutation $\sigma=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of them, define $S_{1}(\sigma)=$ $x_{1}, S_{2}(\sigma)=x_{1}+x_{2}, \cdots, S_{n}(\sigma)=x_{1}+x_{2}+\cdots+$ $x_{n}$. Also let $Q(\sigma)=S_{1}(\sigma) S_{2}(\sigma) \cdots S_{n}(\sigma)$, f... | Analysis: Generalize, first calculate the value of $S\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\sum \frac{1}{Q(\sigma)}$ for any $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$.
When $n=1$, $S\left(a_{1}\right)=\frac{1}{a_{1}}$.
When $n=2$,
$$
\begin{aligned}
S\left(a_{1}, a_{2}\right) & =\frac{1}{a_{1}\left(a_{1}+a_... | 2^{-\frac{n(n-1)}{2}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,852 |
Example 6 Let $u_{1}, u_{2}, u_{3}, \cdots$ be an integer sequence, satisfying $u_{1}=39, u_{2}=45, u_{n+2}=u_{n+1}^{2}-u_{n}$. Prove: 1986 can divide the holeless polynomial of this sequence. | Analysis: Since $u_{3}=u_{2}^{2}-u_{1}=45^{2}-39=1986$, $\therefore 1986 \mid u_{3}$.
If we can prove that there are infinitely many terms congruent to $u_{3}$, the problem is solved. Therefore, we need to study the sequence of remainders of $u_{n}$.
Let $v_{n} \equiv u_{n}(\bmod 1986)$.
Then $0 \leqslant v_{n}<1986$.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,854 |
Example 1 A class has 6 lessons scheduled on Monday: Chinese, Math, History, PE, Music, and Geography.
(1) If PE cannot be scheduled as the first lesson in the morning, History cannot be the second lesson in the morning, Music cannot be the third lesson in the morning, Chinese cannot be the fourth lesson in the morning... | (1) Essentially, it is the Bernoulli-Euler wrong envelope problem when $n=6$, so the number of different ways to arrange the schedule is
$$
\begin{aligned}
u_{5} & =6!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}\right) \\
& =265 \text { (ways). }
\end{aligned}
$$
(2) Essentially, it is the Ber... | 265, 44, 309, 362, 97, 18, 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,855 |
Example 2. Rearrange $1,2,3,4,5,6,7,8$.
(1) such that the even numbers are in their original positions, and the odd numbers are not in their original positions, how many different arrangements are there?
(2) such that the prime numbers are in the odd positions, and the composite numbers are in the even positions, how m... | Solution: (1) In essence, it is the Bernoulli-Euler letter misplacement problem for $n=4$, the number of different arrangements is
$$
u_{1}=9 \text { (ways). }
$$
(2) The different arrangements of even numbers in odd positions and odd numbers in even positions are 4! ways each. By the multiplication principle, the numb... | 225 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,856 |
Example: $3 n$ people attend a party, each bringing a hat and an umbrella. After the party, each person randomly takes a hat and an umbrella.
(1) How many possibilities are there such that no one gets back their original items?
(2) How many possibilities are there such that someone gets back their original items?
(3) H... | (1) $\boldsymbol{n}$ people first take hats, by the problem, none of them take the right one, there are $u_{n}$ ways; then they take umbrellas, by the problem, none of them take the right one again, there are also $u_{n}$ ways. By the multiplication principle, the total number of ways is
$$
u_{n}^{2}=\left[n!\sum_{i=2}... | n!\quad(n-1)!\left[\sum_{i=2}^{n-1} \frac{(-1)^{i}}{i!}\right] \cdot \left[\sum_{i=2}^{n-1} \frac{(2 n+1)(-1)^{i}}{i!}+\frac{2(-1)^{n}}{(n-1)!}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,857 |
Example 6 The roots of the equation $x^{2}+a x+1=b$ are natural numbers. Prove: $a^{2}+b^{2}$ is a composite number.
(20th Annual Math Competition) | Proof: The original equation is rearranged as
$$
\begin{array}{l}
x^{2}+a x+(1-b)=0, \\
\text { then, } x_{1}+x_{2}=-a, x_{1} x_{2}=1-b . \\
\begin{array}{l}
a^{2}+b^{2}= {\left[-\left(x_{1}+x_{2}\right)\right]^{2}+\left(1-x_{1} x_{2}\right)^{2} } \\
= x_{1}^{2}+2 x_{1} x_{2}+x_{2}^{2}+1-2 x_{1} x_{2} \\
\quad+x_{1}^{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,858 |
Example 1 For any $a>1, b>1$,
we have $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$.
(1992, 26th Commonwealth Mathematics Olympiad (10th grade)) | To prove for $a:=1+t_{1}, b:=1+t_{2}$, where $t_{1}, t_{2}>0$. The inequality to be proven becomes
$$
\begin{array}{l}
\frac{\left(1+t_{1}\right)^{2}}{t_{2}}+\frac{\left(1+t_{2}\right)^{2}}{t_{1}} \geqslant 8 . \\
\text { The left side of the above equation } \geqslant \frac{2\left(1+t_{1}\right)\left(1+t_{2}\right)}{\... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false | 709,859 |
Example 2 Let $x, y, z > 0$ be real numbers. Prove the inequality
$$
\begin{array}{l}
x(x-z)^{2}+y(y-z)^{2} \\
\geqslant(x-z)(y-z)(x+y-z),
\end{array}
$$
and determine when equality holds.
(1992, Canadian Mathematical Olympiad) | Prove: The Schur's inequality is equivalent to
$$
\begin{array}{l}
\quad x^{3}+y^{3}+z^{3}+3 x y z-x^{2} z-y^{2} z-y^{2} x \\
\quad-z^{2} x-z^{2} y \geqslant 0, \\
\text { i.e., } x(x-y)(x-z)+y(y-x)(y-z) \\
+z(z-x)(z-y) \geqslant 0 .
\end{array}
$$
(1) is symmetric with respect to $x, y, z$. Without loss of generality,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,860 |
Example 3 Let $x_{k}>0(k=1,2, \cdots, n), \mathrm{H} x_{1}+$ $x_{2}+\cdots+x_{n}=1$. Prove:
$$
\frac{x_{1}^{2}}{x_{1}+x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n-1}+x_{n}}+\frac{x_{n}^{2}}{x_{n}+x_{1}} \geqslant \frac{1}{2} \text {. }
$$
(1990, All-Soviet Union Mathematical Olympiad for High School Students) | Prove: Let $x_{k}=\frac{x_{k}+x_{k+1}}{2}+t_{k}, k=1,2, \cdots, n$. And $x_{n+1}=x_{1}$, where $t_{1}+t_{2}+\cdots+t_{n}=0$.
$$
\begin{aligned}
\text { LHS }= & \frac{\left[\frac{1}{2}\left(x_{1}+x_{2}\right)+t_{1}\right]^{2}}{x_{1}+x_{2}} \\
& +\frac{\left[\frac{1}{2}\left(x_{2}+x_{3}\right)+t_{2}\right]^{2}}{x_{2}+x_... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 709,861 |
For example, $4 n \in N, a, b \in R, x_{a}, x_{1}, \cdots, x_{n}$ are real variables, and $\sum_{i=0}^{n} x_{i}=a, \sum_{i=1}^{n} x_{i}^{2}=b$. Try to determine the range of $x_{0}$.
(30th IMO Preliminary Problem) | Let $x_{i}=\frac{a-x_{0}}{n}+t_{i}, i=1,2, \cdots, n$. Where $t_{1}+t_{2}+\cdots+t_{n}=0$. Then
$$
\begin{aligned}
x_{0}^{2}-b= & -\sum_{i=1}^{n} x_{i}^{2}=-\sum_{i=1}^{n}\left(\frac{a-x_{0}}{n}+t_{i}\right)^{2} \\
= & -\sum_{i=1}^{n}\left(\frac{a-x_{0}}{n}\right)^{2} \\
& -2 \cdot \frac{a-x_{0}}{n} \cdot \sum_{i=1}^{n... | \frac{a-\sqrt{n^{2} b-n a^{2}+n b}}{n+1} \leqslant x_{0} \leqslant \frac{a+\sqrt{n^{2} b-n a^{2}+n b}}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,862 |
Example 6 How many ordered quadruples of integers $(a$, $b, c, d)$ satisfy $0<a<b<c<d<500, a+d=b$ $+c$ and $bc-ad=93$?
(11th American Invitational Mathematics Examination) | Solution: Let $b=a+t, t \in N$, then $c=d-t$. Substituting into $b c-a d=93$, we get $(d-a-t) t=3 \times 31$.
When $t=1$, we have $b=a+1, c=a+93, d=$ $a+94$. Since $a+94<500$, we have $0<a<406$. Thus, there are 405 sets of $(a, b, c, d)$.
When $t=3$, we have $b=a+3, c=a+31, d=$ $a+34$. In this case, $0<a<466$, so the... | 870 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,864 |
Let $a, b, c$ be positive real numbers, determine all triples $\left\{\begin{array}{l}x+y \div z=a+b+c, \\ 4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c .\end{array}\right.$ of $x, y, z$.
(36th IMO Shortlist) | It is not difficult to verify that $x=\frac{b+c}{2}, y=\frac{c+a}{2}, z=\frac{a+b}{2}$ is a solution. Below, we prove that this is the only solution.
$$
\text{Let } x=\frac{b+c}{2}+t_{1}, y=\frac{c+a}{2}+t_{2}, z=\frac{a+b}{2}+
$$
$t_{3}$, substituting into the original system of equations yields
$$
\left\{\begin{array... | x=\frac{b+c}{2}, y=\frac{c+a}{2}, z=\frac{a+b}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,865 |
Example 8 Find all positive integers $a, b, c$ such that $1<a<b<c$, and $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.
(33rd IMO Problem) | Let $a=1+t_{1}, b=1+t_{2}, c=1+t_{3}$, then $1 \leqslant t_{1}<t_{2}<t_{3}$. Also,
$$
\begin{aligned}
2 & \leqslant \frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)-1}{t_{1} t_{2} t_{3}} \\
& <\frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)}{t_{1} t_{2} t_{3}} \\
& =\left(1+\frac{1}... | (3,5,15) \text{ and } (2,4,8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,866 |
Lemma Let $a^{\prime}, b^{\prime}, c^{\prime}$ be the side lengths that can form a triangle $C^{\prime}$, and let $A . C C$ be a triangle with $c, b, c$ as its three sides.
(1) If at least one of $A+A^{\prime}, B+B^{\prime}, C+C^{\prime}$ is less than $180^{\circ}$, without loss of generality, assume $A+A^{\prime} \geq... | Proof: Since $F, B, F, C$ are concyclic, by Ptolemy's theorem, we have
$$
P C \cdot F B + P B \cdot F C = B C \cdot F P \text{.}
$$
It is easy to see that $\triangle B C P \sim \triangle A^{\prime} B^{\prime} C^{\prime}$, thus
$$
\begin{array}{l}
\frac{P C}{b^{\prime}} = \frac{P B}{c^{\prime}} = \frac{B C}{a^{\prime}}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,867 |
Example 7 The roots of the equation $x^{2}+p x+q=0$ are both positive integers, and $p+q=1992$. Then the ratio of the larger root to the smaller root is $\qquad$. | Given: $\because x_{1}+x_{2}=-p, x_{1} x_{2}=q$,
$$
\begin{array}{l}
\therefore x_{1} x_{2}-x_{1}-x_{2}=q+p=1992, \\
\left(x_{1}-1\right)\left(x_{2}-1\right)=1993 .
\end{array}
$$
$\because 1993$ is a prime number,
$$
\therefore\left\{\begin{array}{l}
x_{1}-1=1, \\
x_{2}-1=1993 .
\end{array}\right.
$$
Solving, we get ... | 997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,869 |
Text $[2]$, $[3]$ introduced Leibniz's formula and relevant knowledge about the nine-point circle. Based on this, this article will provide an algebraic proof of Feuerbach's formula, to make up for the lengthy geometric proof.
Assume $\triangle A B C$ is a right triangle, but not an isosceles triangle. Then the radius... | $$
\begin{array}{c}
\text { Proof: As shown in the figure, } \\
A N=p-a,\left(p=\frac{1}{2}(a\right. \\
+b+c)) \\
\therefore I A^{2}=A N^{2}+N I^{2} \\
=(p-a)^{2}+r^{2} . \\
\text { Therefore, } I A^{2}+I B^{2}+I C^{2} \\
=(p-a)^{2}+r^{2}+(p-b)^{2}+r^{2}+(p-c)^{2} \\
\quad+r^{2} \\
=(p-a)^{2}+(p-b)^{2}+(p-c)^{2}+3 r^{2... | I P=\frac{R}{2}-r | Geometry | proof | Yes | Yes | cn_contest | false | 709,870 |
Question: There are 12 points on a circle, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Convex polygons formed by these points are classified as follows: polygons whose vertices include both the red and blue points are called bicolored polygons; polygons that include... | Simplified: From the problem, we know that a bicolored polygon must include two points of different colors, while a colorless polygon must not include any of these points. Since all 12 points are on the circumference of the circle, the number of bicolored polygons is $\sum_{i=1}^{10} C_{10}^{i}$, and the number of colo... | 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,871 |
Let $4, b, c$ be the side lengths of $\triangle ABC$, and $S$ be its area. It is shown in [1] that for non-obtuse triangles,
$$
\sum \frac{1}{a^{2}} \geqslant \frac{5}{4 S} \text {. }
$$
Equality holds if and only if $\triangle ABC$ is an isosceles right triangle.
Let $R, r, s$ denote the circumradius, inradius, and ... | Prove that from
$$
\left(\sum_{a b)^{2}}=\sum a^{2} b^{2}+2 a b c \sum_{a}\right.
$$
and
$$
\sum a b=s^{2}+4 R r+r^{2}, a b c=4 R r s
$$
the following identity holds:
$$
\sum \frac{1}{a^{2}}=\frac{s^{4}+\left(2 r^{2}-8 R r\right) s^{2}+\left(4 R r+r^{2}\right)^{2}}{16 R^{2} r^{2} s^{2}} .
$$
Inequality (2) is equiva... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,872 |
Lemma. The function $n==\sin ^{n} x+\cos ^{n} x(n \in N$, $n \geqslant 2, x \in\left(0, \frac{\pi}{2}\right))$ has a minimum value of $2^{\frac{2-n}{2}}$.
Note: When $n=2$, $u \equiv 1=2^{\frac{2-2}{2}}$; when $n \geqslant 3$, it is known from Example 2 in [1] that the minimum value of $u$ is $2^{\frac{2-n}{2}}$, henc... | Proof: By the lemma, we have
$$
\begin{aligned}
u_{n} & =\cos ^{n} \frac{A}{2}+\sin ^{n} \frac{A+C}{2}+\sin ^{n} \frac{C}{2} \\
& >\cos ^{n} \frac{A}{2}+\sin ^{n} \frac{A}{2} \geqslant 2^{\frac{2-n}{2}} .
\end{aligned}
$$
Also, \( u_{n} < 3 \), hence \( 2^{\frac{2-n}{2}} < u_{n} < 3 \).
Notice that when \( \angle C \r... | 2^{\frac{2-n}{2}} < u_{n} < 3 | Inequalities | proof | Yes | Yes | cn_contest | false | 709,873 |
Theorem. Let the side lengths of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$; $\omega_{a}, \omega_{b}, \omega_{c}$ and $\omega_{a}^{\prime}, \omega_{b}^{\prime}, \omega_{c}^{\prime}$ be the angle bisectors of the corresponding sides, respectiv... | Proof: By the formula for the length of the angle bisector in a triangle and the arithmetic mean inequality, we get
$$
\omega_{a}=\frac{2 a b}{b+c} \cos \frac{A}{2} \leqslant \sqrt{b c} \cos \frac{A}{2}.
$$
Similarly, $\omega_{a}^{\prime} \leqslant \sqrt{b^{\prime} c^{\prime}} \cos \frac{A^{\prime}}{2}$. Thus, $\omega... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,874 |
Let $\triangle A B C$ have sides $a, b, c$, and the radii of the excircles be $r_{a}, r_{b}, r_{c}$. Then
$$
\frac{a}{r_{a}}+\frac{b}{r_{b}}+\frac{c}{r_{c}} \geqslant 2 \sqrt{3} .
$$
The author provides the following strengthened inequality for this proposition:
Let $\triangle A B C$ have sides $a, b, c$, and let $r, ... | Proof: $\because a=r\left(\operatorname{ctg} \frac{J}{2}+\operatorname{ctg} \frac{C}{2}\right)$,
$$
\begin{array}{l}
2==r\left(\operatorname{ctg} \frac{A}{2}+\operatorname{ctg} \frac{C}{2}\right), \\
c=r\left(\operatorname{ctg} \frac{A}{2}+\operatorname{ctg} \frac{B}{2}\right), \\
r_{a}=r \operatorname{ctg} \frac{B}{2}... | \sqrt{16-\frac{8 r}{R}} | Inequalities | proof | Yes | Yes | cn_contest | false | 709,875 |
Let $\triangle A B C$ have sides $B C=a, C A=b, A B=c$, as shown in Figure 1. When $\angle A, \angle B, \angle C$ are all less than $120^{\circ}$, $F$ is the Fermat point of $\triangle A B C$, $F A=u, F B=v, F C=w$. Then
$$
u+v+w=\frac{\sqrt{2}}{2} \sqrt{a^{2}+b^{2}+c^{2}+4 \sqrt{3} S} .
$$ | Proof: As shown in Figure 2, construct a regular $\triangle MNL$, and let $P$ be a point inside it such that $PM = a$, $PN = b$, and $PL = c$.
Rotate $\triangle PML$ around $M$ by $60^\circ$ clockwise to get $\triangle P'MN$.
Then $\triangle PMP'$ is an equilateral triangle with side length $a$, and $\triangle PP'N$ is... | u + v + w = \frac{\sqrt{2}}{2} \sqrt{a^2 + b^2 + c^2 + 4\sqrt{3} S} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,877 |
In a non-obtuse $\triangle A B C$, $A B>A C$, $\angle B=45^{\circ}$, $O$ and $I$ are the circumcenter and incenter of $\triangle A B C$, respectively, and $\sqrt{2} \theta I=A B-A C$. Find $\sin A$. | Solution 1: From the known conditions and Euler's formula, we have
$$
\left(\frac{c-b}{\sqrt{2}}\right)^{2}=O r^{2}-R^{2}-2 R
$$
From the known geometric relationship, we get
$$
r=\frac{c+a-b}{2} \operatorname{tg} \frac{b}{2}=\frac{c+a-b}{2} \operatorname{tg} \frac{\pi}{8}
$$
$$
=\frac{\sqrt{2}-1}{2}(c+a-b) .
$$
From... | \sin A=\frac{\sqrt{2}}{2} \text{ or } \sin A=\sqrt{\sqrt{2}-\frac{1}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,879 |
Example 8 Find the integer $k$ such that the quadratic equation
$$
k x^{2}+(k+1) x+(k-1)=0
$$
has integer roots.
(1993, 5th Ancestor's Cup Junior Mathematics Contest) | $$
\begin{array}{l}
x_{1}+x_{2}=-\frac{k+1}{k}=-1-\frac{1}{k}, \\
x_{1} x_{2}=\frac{k-1}{k}=1-\frac{1}{k} .
\end{array}
$$
( $k$ is a real number, at this point we cannot conclude $k=1,-1$ )
If $x_{1} x_{2}-x_{1}-x_{2}=2$,
then $\left(x_{1}-1\right)\left(x_{2}-1\right)=3$,
thus $\left\{\begin{array}{l}x_{1}-1=1, \\ x_{... | k=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,880 |
For a given large number $x$, does there exist $2n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, $b_{1}, b_{2}, \cdots, b_{n}$, that simultaneously satisfy the following two conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{z}+\cdots+b_{n}$;
(2) $n-1>\sum_{i=1}^{n} \frac{a_{i}-b_{i}}{a_{i}+b_{i}}>n-1-\fra... | Solution 1: There exist $2n$ numbers that meet the proposition requirements. Let $a_{i} = 2Mi, b_{i} = 2i$ (for $i=1,2,3, \cdots, n-1$; $M$ is a positive integer greater than or equal to $8000n$), $a_{n} = (M-1)^{2}n(n-1), b_{n} = M(M-1)n(n-1)$.
Clearly, the above $2n$ numbers are all distinct, and $a_{1} + a_{2} + \c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,881 |
Find all natural numbers $n$ greater than 3 such that $1+C_{n}^{1}+C_{n}^{2}+C_{3}^{3}$ divides $2^{2000}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | For $n > 3$, we only need to consider which natural numbers $n$ can make $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}=2^{k}(k \in \mathbb{N})$. After calculation, we get
$$
\begin{array}{l}
1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} \\
=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} \\
=\frac{(n+1)\left(n^{2}-n+6\right)}{6},
\end{array}
$$
whic... | n=7 \text{ and } n=23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,883 |
How many pairs of positive integers $x, y, x \leqslant y$, satisfy $(x, y)=5!$ and $[x, y]=50!$? | 1. Let $p_{1}, p_{2}, \cdots, p_{12}$ represent the 12 prime numbers from 7 to 47 in increasing order.
Then, $5!=2^{3} \cdot 3^{2} \cdot 5^{1} \cdot p^{3} \cdot f_{2}^{\prime \prime} \cdot \cdots \cdot f_{12}^{\prime}$.
For 50! $=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{*_{3}} \cdot \hat{o}_{1}^{b_{1}}=p_{2}^{b_{2}} \cdot \c... | 2^{14} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,886 |
3. Prove:
$$
\frac{1}{1999}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \cdots \cdot \frac{1997}{1998}<\frac{1}{44} .
$$ | 3. Let $p=\frac{1}{2} \cdot \frac{3}{4} \cdot \cdots \cdot \frac{1997}{1998}$.
Since $2\frac{1}{3}, \frac{3}{4}>\frac{3}{5}, \cdots, \frac{1997}{1998}>\frac{1997}{1999}$.
Therefore, $p>\frac{1}{3} \cdot \frac{3}{5} \cdots \cdots \cdot \frac{1997}{1999}=\frac{1}{1999}$.
Also, since $1 \cdot 3<2 \cdot 2,3 \cdot 5<4 \cdo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,888 |
4. Point $O$ is a point inside parallelogram $A B C D$ such that $\angle A O B + \angle C O D = 180^{\circ}$. Prove: $\angle O B C = \angle O D C$. | 4. Proof 1: Consider the translation that maps point $D$ to point $\Lambda$. This translation also maps point $O$ to $\vec{E} \vec{A}$. Since $\overrightarrow{C B} = \overrightarrow{D A}$, it also maps point $C$ to point $B$.
Translation preserves angles, so $\angle A O^{\prime} B = \angle D O C = 180^{\circ} - \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,889 |
5. Try to simplify $\sum_{k=0}^{n} \frac{(-1)^{k} C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$ into the form $\frac{p(n)}{q(n)}$, where $p(n)$ and $q(n)$ are two polynomials with integer coefficients. | 5. Consider the following:
$$
\begin{array}{l}
k^{3}+9 k^{2}+26 k+24=(k+2)(k+3)(k+4) . \\
\text { i. } S(n)=\sum_{k=0}^{n} \frac{(-1)^{k} C_{k}^{k}}{k^{3}+9 k^{2}+26 k+24} .
\end{array}
$$
Then \( S(n)=\sum_{k=1}^{n} \frac{(-1)^{4} n!}{k!(n-k)!(k+2)(k+3)(k+4)} \)
$$
\begin{aligned}
= & \sum_{k=0}^{n}\left[\frac{(-1)^{... | \frac{1}{2(n+3)(n+4)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,890 |
Example 9 If the discriminant of $x^{2}+2 k x+2 t-1=0$ is to be determined ().
(A) is an integer
(B) is an integer, but it cannot be determined whether it is odd or even
(C) is an odd number
(D) is an even number | Solution: \| $\alpha+\beta=-2k, \beta=-2k-\alpha$ are integers, and $\alpha \beta=2t-1$ is odd,
$\therefore \alpha, \beta$ are both odd.
Therefore, the answer is (C). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,891 |
1. Among the following questions (or say problems)
(1) The equation $x^{2}+\frac{1}{x-1}=1+\frac{1}{x-1}$ is a one-variable quadratic equation;
(3) Extend line $A B$ to $C$;
(1) Simplifying $a \sqrt{-\frac{1}{a}}$ yields $\sqrt{-a}$. The number of correct statements is ( .
(A) 1
(B) 2
(C) 3
(D) 4 | $-.1 .(\mathrm{C})$.
$x^{2}+\frac{1}{x-1}=1+\frac{1}{x-1}$ is a fractional equation; an equilateral triangle is of course an isosceles triangle; a ray can only be extended in the reverse direction; for $a \sqrt{-\frac{1}{a}}$ to be meaningful, then $a<0$, thus, $a \sqrt{-\frac{1}{a}}=$ $-\sqrt{-a}$. Therefore, the answ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,892 |
2. If $x+y+z \neq 0, a=\frac{x}{y+z}, b=\frac{y}{x+z}$, $c=\frac{z}{x+y}$, then $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=(\quad)$.
(A) 0
(B) 1
(C) $a+b+c$
(D) Uncertain | 2. (B).
$$
\frac{a}{a+1}=\frac{\frac{x}{y+z}}{\frac{x+y+z}{y+z}}=\frac{x}{x+y+z} \text {. }
$$
Therefore,
$$
\frac{b}{b+1}=\frac{y}{x+y+z}, \frac{c}{c+1}=\frac{z}{x+y+z} .
$$
$\therefore$ Adding them together gives the result. Hence, the answer is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,893 |
$G, A G>B G, C D=16, A E \perp C D$ at $E, B F \perp$ $C D$ at $F$. Then $A E-B F=$ ( ).
(A) 6
(B) 8
(C) 10
(I) 12 | 3. (D) .
Chord distance $O H=\sqrt{10^{2}-8^{2}}=6$. From $O H / / A E / / B F$ we get
$$
\frac{A E}{O H}=\frac{10+O G}{O G}, \frac{B F}{O H}=\frac{10-O G}{O G} .
$$
Subtracting the two equations yields
$$
A E-B F=2 O H=12 \text {. Therefore, the answer is (D). }
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,894 |
4. The equation $\left|x^{4}-4\right|-a-1=0$ has 4 solutions
(A) $a=0$
(B) $-1<a \leqslant 0$
(C) $-1<a<3$
(D) $a \geq 3$ | 4. (C).
From $\left|x^{4}-4\right|=a+1$, we have $a+1 \geqslant 0$, which means $a \geqslant-1$. At this point, $x^{4}=4 \pm(a+1)$. For the equation to have 4 real roots, we also need $a+1 \neq 0$ and $4-(a+1)>0$, which means $-1<a<3$. Therefore, the answer is (C). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,895 |
5. Cut a wire of length $100 \mathrm{~cm}$ into $n$ small segments $(n \geqslant 3)$, each segment no shorter than $10 \mathrm{~cm}$. If for any such cutting method, there always exist 3 small segments that can form a triangle, then the minimum value of $n$ is ().
(A) 3
(B) 4
(C) 5
(D) 6 | 5. (C).
If a rod is divided into 4 segments, it may not form a quadrilateral, for example, with lengths $10, 17, 28$, 45. Consider the case of dividing it into 5 segments, assuming the lengths are $a, b, c, d, e$, and $10 \leqslant a \leqslant b \leqslant c \leqslant d \leqslant e$. If no segment can form a polygon, t... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 709,896 |
6. In an acute triangle $\triangle ABC$ with sides $a, b, c$, the distances from the circumcenter to the sides are $m, n, p$. Then, $m: n: p=(\quad)$.
(A) $\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$
(B) $a: b: c$
(C) $\cos A: \cos B: \cos C$
(I) $\sin A: \sin B: \sin C$ | 6. (C).
As shown in Figure 1, let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l}
m=\sqrt{R^{2}-\left(\frac{a}{2}\right)^{2}} \\
=\frac{1}{2} \sqrt{4 R^{2}-a^{2}} \\
=\frac{1}{2} \sqrt{\frac{a^{2}}{\sin ^{2} A}-a^{2}} \\
=\frac{a}{2} \cdot \frac{\cos A}{\sin A}=R \cos A .
\end{array}
$$
Similarly... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,897 |
1. Let $x, y$ be numbers, $x^{2}+x y+y^{2}=3$. Then, the value range of $x^{2}-x y+y^{2}$ is $\qquad$ | ii. $1.1 \leqslant x^{2}-x y+y^{2} \leqslant 9$.
Let $x^{2}-x y+y^{2}=k$.
Also, $x^{2}+x y+y^{2}=3$.
From ((2) - (1)), we get $x y=\frac{3-k}{2}$.
From (2), we have $(x+y)^{2}=3+x y=\frac{9-k}{2}$.
Since $(x+y)^{2} \geqslant 0$,
Therefore, $k \leqslant 9$. At this point, $x+y= \pm \sqrt{\frac{9-k}{2}}$.
Therefore, $x$ ... | 1 \leqslant x^{2}-x y+y^{2} \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,898 |
3. If $a \geqslant 1$, then, when $\sqrt{a-\sqrt{a-x}}=x$, the solution is
保留源文本的换行和格式，直接输出翻译结果如下：
3. If $a \geqslant 1$, then, when $\sqrt{a-\sqrt{a-x}}=x$, the solution is | 3. $\frac{\sqrt{4 a-3}-1}{2}$. Therefore, we get
$$
a^{2}-\left(2 x^{2}+1\right) a+\left(x^{4}-x\right)=0 .
$$
Solving, we get $a=\frac{\left(2 x^{2}+1\right) \pm(2 x+1)}{2}$,
which means $a=x^{2}+x+1$ or $a=x^{2}-x$.
From $a=x^{2}+x+1$ we get
$$
x=\frac{-1 \pm \sqrt{4 a-3}}{2}(\because a \geqslant 1, \therefore 4 a-3... | \frac{\sqrt{4 a-3}-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,900 |
Example 10 The equation $x^{2}+p x+1997=0$ has exactly two integer roots $x_{1} 、 x_{2}$. Then the value of $\frac{p}{\left(x_{1}+1\right)\left(x_{2}+1\right)}$ is
(A) 1
(B) -1
(C) $-\frac{1}{2}$
(D) $\frac{1}{2}$ | Solution: Given $|1 x_{1} x_{2}=1997$, and 1997 is a prime number, we know
$$
\begin{array}{l}
x_{1}=1, x_{2}=1997, \\
p=-\left(x_{1}+x_{2}\right)=-1998 . \\
\text { The original expression }=\frac{-1998}{2 \times 1998}=-\frac{1}{2} .
\end{array}
$$
Therefore, the answer is (C). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,902 |
i. (25 points) Given circle $\odot O$ with radius $R$, two chords $A B \perp C D$ intersect at $M$. Prove: $A B^{2}+(C M-D M)^{2}$ is a constant. | As shown in Figure 3, draw the diameter $A E$, and connect $B E$, $D E$, and $B C$. It is easy to see that $B E \parallel C D$, thus $B C = D E$.
$$
\text{Also, } A M \cdot B M = C M \text{.}
$$
$D M$,
$$
\text{Then } \begin{aligned}
& A B^{2} + (C M - D M)^{2} \\
= & (A M + B M)^{2} + (C M \\
& - D M)^{2} \\
= & A M^{... | 4R^2 | Geometry | proof | Yes | Yes | cn_contest | false | 709,903 |
II. (25 points) Let real numbers $a, b, c, m$ satisfy the condition $\frac{a}{m+2}+\frac{b}{m+1}+\frac{c}{m}=0$, and $a \geq 0, m>0$. Prove that the quadratic equation $a x^{2}+b x+c=0$ has a root $x_{0}$, such that $0<x_{0}<1$. | (1) When $a=0$, if $b \neq 0$, then $x_{0}=-\frac{c}{b}=$ $\frac{m}{m+1}$, satisfying $0 < x_{0} < 1$;
When $a>0$, let $f(x)=a x^{2}+b x+c$, it is easy to find that
$$
f\left(\frac{m}{m+1}\right)=-\frac{a m}{(m+1)(m+2)}<0,
$$
If $c>0$, then
$$
f(0)=c>0,
$$
Thus, there must be a root satisfying
$$
0 < x_{0} < \frac{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,904 |
Three. (30 points) If the three sides of $\triangle A B C$ are $a$, $b$, and $c$, and $a^{2}+b^{2}+c^{2}<6$. Prove: $\triangle A B C$ can be covered by a unit circle.
| Three, let $a \geqslant b \geqslant c$, then
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2} \\
= & a^{2}+\frac{1}{2}(b+c)^{2} \\
> & a^{2}+\frac{1}{2} a^{2} \\
= & \frac{3}{2} a^{2},
\end{aligned}
$$
so $a\sqrt{3}$, by the median formula
$$
\begin{aligned}
2 m_{u}^{2} & =b^{2}+c^{2}-\frac{1}{2} a^{2} \\
& <6-\frac{3}{2} a^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,905 |
One, Select (6 points per question, total 36 points)
1. The number of solutions to the equation $\left(\frac{3}{19}\right)^{x}+\left(\frac{5}{19}\right)^{x}+\left(\frac{11}{19}\right)^{x}=$ $2 \sqrt{x-1}$ is (
(A) 0
(B) 1
(C) 2
(D) Infinite | $-1 .(B)$
\| $x-1 \geqslant 0$, so $x \geq 1$.
$$
\because f(x)=\left(\frac{3}{19}\right)^{x}+\left(\frac{5}{19}\right)^{x}+\left(\frac{11}{19}\right)^{x} \text { in }[1,+\infty) \mathrm{E} .
$$
is a decreasing function,
$\therefore$ when $x=1 \|_{1}^{+}, f(x)_{\max }{ }^{2} 1$.
X $g(x)=2 \sqrt{x \cdots 1}$ in $[1,+\i... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,906 |
3. Given $\frac{\cos \alpha \cos \frac{\beta}{2}}{\cos \left(\alpha-\frac{\beta}{2}\right)}+\frac{\cos \beta \cos \frac{\alpha}{2}}{\cos \left(\beta-\frac{\alpha}{2}\right)}=1$. Then the value of $\cos \alpha+\cos \beta$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) $\sqrt{2}$
(D) $\frac{\sqrt{2}}{2}$ | 3. (A).
Let's set
$$
\begin{array}{l}
\frac{\cos \left(x+\frac{\beta}{2}\right)+\cos \left(\alpha-\frac{\beta}{2}\right)}{\cos \left(\alpha-\frac{\beta}{2}\right)} \\
+\frac{\cos \left(\beta+\frac{\alpha}{2}\right)+\cos \left(\beta-\frac{\alpha}{2}\right)}{\cos \left(\beta-\frac{\alpha}{2}\right)}=2, \\
\frac{\cos \le... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,908 |
4. Let $w=\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}$. Then the expanded form of $(x-w)(x$ $\left.-w^{3}\right)\left(x-w^{7}\right)\left(x-w^{9}\right)$ is ( ).
(A) $x^{4}+x^{3}+x^{2}+x+1$
(B) $x^{4}+x^{3}+x+1$
(C) $x^{4}-x^{4}+x^{2}-x+1$
(D) $x^{4}-x^{3}-x+1$ | 4. (C).
$\because w=\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}$,
$\therefore(x-w)\left(x-w^{2}\right) \cdots\left(x-w^{3}\right)\left(x-w^{10}\right)$
$$
=x^{10}-1 \text {. }
$$
$$
\begin{array}{l}
\therefore\left(x-w^{2}\right)\left(x-w^{2}\right)(x-w)\left(x-w^{4}\right)\left(x-w^{10}\right) \\
\quad=x^{5}-1 .
\en... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,909 |
5. For the circle $x^{2}+y^{2}-5 x=0$, passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$, if $d \in\left(\frac{1}{6}, \frac{1}{3}\right]$, then the set of values for $n$ is ( ).
(A) $\{4,5,6\}$
(B) $\{6,7,8,9\}$
(C) $\{3,4,5\}$
(D) $\{3,4,5,6\}$ | 5. (D).
It is easy to know that the longest chord passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$ is the diameter, whose length is 5, and the shortest chord perpendicular to this diameter has a length of 4. Let the lengths of $n$ chords forming an arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$.... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,910 |
Example 1 Find the real solutions of the equation $x+x y+y^{2}-3 x-3 y$ ~ $3=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Transform the original equation into
$$
x^{2}+(y-3) x+\left(y^{2}-3 y+3\right)=0 \text {. }
$$
Since the required $x$ is a real number, then $\Delta \geqslant 0$, i.e., $(y-3)^{2}-4\left(y^{2}-3 y+3\right) \geqslant 0$.
Rearranging, we get $(y-1)^{2} \leqslant 0$.
Since the required $y$ is also a real number... | x=1, y=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,913 |
Example 2 If the equation $(x+a)(x+b)+(x+b)(x+c)+(x+c)(x+a)=0$ (where $a, b, c$ are all positive numbers) has two equal real roots, prove: the segments of lengths $a, b, c$ can form a triangle, and indicate the characteristics of the triangle.
$(1996$, Anhui Province Partial Areas Junior High School Mathematics League) | Proof: Given the equation transformed as
$$
\begin{array}{l}
3 x^{2}+2(a+b+c) x+(a b+b c+c a) \\
=0 .
\end{array}
$$
Since the equation has two equal real roots, then
$$
\begin{aligned}
\Delta & =[2(a+b+c)]^{2}-12(a b+b c+c a) \\
& =2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\
& =0 . \\
\therefore & (a-b)^{2}=(b-c)^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,914 |
Example 10 Given that $\triangle ABC$ intersects with a line $PQ$ parallel to $AC$, and the area of $\triangle APQ$ equals a constant $k^2$. What is the relationship between $\iota^2$ and the area $S$ of $\triangle ABC$ for the problem to have a solution? How many solutions are there?
(1987, Shanghai Junior High School... | Solution: Let $\frac{B P}{A B}=x$, then
$S_{\triangle B P Q}=x^{2} S, S_{\triangle A B Q}=x S$.
According to the problem, we have $x S-x^{2} S=k^{2}$,
which is $S x^{2}-S x+k^{2}=0$.
The condition for this equation to have real solutions is
$\Delta=S^{2}-4 S k^{2} \geqslant 0$, that is, $S \geqslant 4 k^{2}$.
Moreover,... | S \geqslant 4 k^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,915 |
3. The sum of the first 1997 terms of the sequence $1,1,2,1,2,3,1,2,3,4,1,2 \cdots$ is $\qquad$ . | 3.42654 .
Sequence grouping: Let $a_{1997}$ be in the $(k+1)$-th group. (1), (1,2), (1,2,3), ..., (1,2,...,k), (1,2,..., $a_{1997}$, ..., k+1). From the sum of terms, we have
$$
\left\{\begin{array}{l}
1+2+\cdots+k=\frac{k(k+1)}{2}<1997, \\
1+2+\cdots+(k+1)=\frac{(k+1)(k+2)}{2} \geqslant 1997 .
\end{array}\right.
$$
... | 42654 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,916 |
4. For a regular tetrahedron $V-ABC$ with base edge length $a$ and side edge length $b$, let $M$ be a point on the height $VO$ such that $\frac{VM}{MO}=\frac{b}{a}$. A plane passing through $M$ is parallel to the side edge $VA$ and the base edge $BC$. The area of the section formed by this plane cutting the regular tet... | 4. $\frac{2}{9} \cdot \frac{a b^{2}(3 a+b)}{(a+b)^{2}}$.
Make a section $E F G H$, it is easy to know it is a rectangle. Let it intersect the base at $N$, and intersect $V D$ at $P$. Since
$$
\begin{array}{l}
A D=\frac{\sqrt{3}}{2} a, A O= \\
\frac{\sqrt{3}}{3} a, O D=\frac{\sqrt{3}}{6} a, \frac{A N}{N D} \\
=\frac{V ... | \frac{2}{9} \cdot \frac{a b^{2}(3 a+b)}{(a+b)^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,917 |
5. Product $\prod_{k=1}^{7}\left(1+2 \cos \frac{2 k \pi}{7}\right)=$ $\qquad$ | 5.3 .
$$
\begin{array}{l}
\text { Let } \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} \text {. Then } \omega^{7}=1 . \\
\omega^{k}=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}, \\
\omega^{-k}=\cos \frac{2 k \pi}{7}-i \sin \frac{2 k \pi}{7}, \\
\therefore \omega^{k}+\omega^{-k}=2 \cos \frac{2 k \pi}{7} . \\
\tex... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,918 |
$\begin{array}{l}\text { 6. Given } P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+ \\ a_{3} x^{2}+a_{4} x+a_{5} \text {, and when } k=1,2,3,4 \text {, } P(k) \\ =k \cdot 1997 \text {. Then } P(10)-P(-5)=\end{array}$ | 6.75315 .
Let $Q(x)=P(x)-1997 x$. Then for $k=1,2,3,4$, $Q(k)=P(k)-1997 k=0$. Hence, $1,2,3,4$ are roots of $Q(x)=0$. Since $Q(x)$ is a fifth-degree polynomial, we can assume
$$
Q(x)=(x-1)(x-2)(x-3)(x-4)(x-r) \text {. }
$$
Therefore,
$$
\begin{array}{l}
P(10)= Q(10)+1997 \cdot 10 \\
= 9 \cdot 8 \cdot 7 \cdot 6(10-r)+... | 75315 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,919 |
Three, the sequence $\left\{x_{n}\right\}: x_{1}=1, x_{n+1}=3 x_{n}^{3}+2 x_{n}^{2} + x_{n}$; the sequence $\left\{y_{n}\right\}: y_{n}=\frac{1}{1+2 x_{n}+3 x_{n}^{2}}$; the sequence $\left\{z_{n}\right\}: z_{n}=\frac{2+3 x_{n}}{1+2 x_{n}+3 x_{n}^{2}}$. If the product of the first $n$ terms of $\left\{y_{n}\right\}$ is... | $$
\text { Three, since } \begin{aligned}
y & =\frac{1}{1+2 x_{n}+3 x_{n}^{2}}=\frac{x_{n}}{x_{n}+2 x_{n}^{2}+3 x_{n}^{3}} \\
& =\frac{x_{n}}{x_{n+1}},
\end{aligned}
$$
then, $P=y_{1} \cdot y_{2} \cdots y_{n}=\frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}} \cdots \frac{x_{n}}{x_{n+1}}=\frac{1}{x_{n+1}}$,
$$
\begin{align... | S+P=1 | Algebra | proof | Yes | Yes | cn_contest | false | 709,920 |
Four, on the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$ there are 16 points, sequentially $P_{1}, P_{2}, \cdots, P_{16}, F$ is the left focus, and the angles between each adjacent pair of points and $F$ are equal $\left(\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{16} F P_{1}\right)$. Let the dis... | $$
\begin{array}{l}
a=5, b \\
=4, c=3 . \text { Let } \angle X F P_{1} \\
=\alpha \cdot, \angle P_{1} F P_{2}= \\
\angle P_{2} F P_{3}=\cdots= \\
\angle P_{16} F P_{1}=\frac{\pi}{8} . \\
F M=d:-P_{i} F \cos \left[(i-1) \frac{\pi}{8}+\alpha\right] \\
\quad=\frac{a^{2}}{c}-c=\frac{16}{3} .
\end{array}
$$
By the definiti... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,921 |
Prove: There exists a multiple of 1997 that has no more than 11 digits, and none of its digits are $2,3,4,5,6,7$. | $-、 \because 1997 \leqslant 2^{11}(=2048)$, construct the set $A=$ $\left\{a_{1} a_{2} \cdots a_{11}\right\}$, where $a_{i}=0$ or $1(i=1,2, \cdots, 11)$.
Each digit in the numbers in $A$ has two possibilities $(0,1)$, and by the multiplication principle, the number of elements in set $A$ is $|A|=2^{11}>1997$. Therefore... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,923 |
II. The side length of square $ABCD$ is 1. With vertex $C$ as the center and $CB$ as the radius, draw a circle. On $AB$, take $BP = \frac{1}{5}$. Draw a tangent line from $P$ to the circle, intersecting $AD$ at $Q$, with the point of tangency being $E$. Connect $CE$ to intersect the diagonal $BD$ at $K$. Connect $AK$ t... | Given that points $C, D, Q, E$ are concyclic, then $\angle 3=\angle 1$.
Given that points $C, E, P, B$ are concyclic, then $\angle 4=\angle 2$.
$$
\begin{array}{l}
\because \frac{D K}{K B}=\frac{S_{\triangle D C K}}{S_{\triangle K C B}} \\
=\frac{\sin \angle 1}{\sin \angle 2}=\frac{\sin \angle 3}{\sin \angle 4} . \\
\f... | \frac{13}{30} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,924 |
Three, Proof: In any group of $2 n+2$ people, there exist two people $A$ and $B$ such that among the remaining $2 n$ people, at least $n$ of them either all know $A$ and $B$; or all do not know $A$ and $B$.
| Three, consider any person $M$, who either knows or does not know the other $2n+1$ people. Suppose $M$ knows $k$ of them and does not know the other $2n+1-k$ people. The total number of "pairs" among these $2n+1$ people is $C_{2n+1}^{2} = n(2n+1)$. Pairs where $M$ either knows both or neither are called "Class A pairs,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,925 |
Example 1 Prove: There uniquely exists a triangle with side lengths being consecutive integers and one angle being twice another angle.
(10th INO) | Proof: Let $\triangle ABC, a=n-1, b=n, c=n+1$ ($n$ is a natural number), and the three angles are $\alpha, 2\alpha, \pi - 3\alpha$. Then
$$
\begin{array}{l}
\frac{\sin (\pi-3 \alpha)}{\sin \alpha}=\frac{\sin 3 \alpha}{\sin \alpha}=\frac{3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha} \\
=3-4 \sin ^{2} \alpha=4 \cos ^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,926 |
Initial 63. Without solving the equation, construct a quadratic equation in $y$ such that its two roots are the $t$-th powers of the roots of the equation $x^{2}-x-1=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: From the problem, we have $x^{2}=x+1$. For convenience, we first calculate $x^{4}$ and $x^{3}$.
$$
\begin{aligned}
x^{4} & =\left(x^{2}\right)^{2}=(x+1)^{2}=x^{2}+2 x+1 \\
& =(x+1)+2 x+1=3 x+2, \\
x^{3} & =x^{2} \cdot x=(x+1) \cdot x=x^{2}+x \\
& =(x+1)+x=2 x+1, \\
x^{7} & =x^{4} \cdot x^{3}=(3 x+2)(2 x+1) \\... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,927 |
Thorough 64. On a plane, there are 103 points, among which no three points are collinear, and no four points are concyclic. Then there exists at least one circle passing through three of these points, such that 50 points are inside the circle, and 50 points are outside the circle. | Solution: Given that no three of the 103 points are collinear, we can find two points $B_{1}$ and $B_{2}$ such that the remaining points are on the same side of the line $B_{1} B_{2}$.
Since no four points are concyclic, for the other 101 points, the angles $\angle B_{1} A_{i} B_{2}$ (where $A_{i}$ is one of the 101 p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,928 |
63. Let $a_{1}=1997^{1997^{1997}}{ }^{.197}$ (1997 sevens in total), the sum of the digits in the decimal representation of $a_{1}$ is $a_{2}$, the sum of the digits of $a_{2}$ is $a_{3}$, and so on. Find $a_{2000}$. | Solution: Let $x_{n}=10000^{10000^{10000}}$ (with $n$ 10000s), $n=1,2, \cdots$, then for $n \geqslant 2$, we have
$$
\begin{array}{l}
x_{n}=10000^{x_{n-1}}=10^{4 x_{n-1}} . \\
\therefore a_{1}<x_{1997}=10^{4 x_{1996}}, \\
a_{2} \leqslant 9 \times 4 x_{1996}<100 x_{1996} \\
=100 \cdot 10^{4 x 1995}=10^{4 x_{1995}+2}, \... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,929 |
64. The general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=\frac{\cos 3^{n} x+3 \cos 3^{n-1} x}{3^{n-1} \sin 3^{n} x} \text {. Find } S_{n}
$$ | $$
\text { Sol: } \begin{array}{l}
\frac{2(\cos 3 \alpha+3 \cos \alpha)}{\sin 3 \alpha} \\
=\frac{3(2 \cos \alpha+\cos 3 \alpha)-\cos 3 \alpha}{\sin 3 \alpha} \\
=\frac{3\left(4 \cos ^{2} \alpha-1\right) \cos \alpha}{\sin \alpha\left(3 \sin ^{2} \alpha\right)}-\operatorname{ctg} 3 \alpha \\
=3 \operatorname{ctg} \alph... | S_{n}=\frac{3 \operatorname{ctg} x}{2} \cdots \frac{\operatorname{ctg}^{n} x}{2 \cdot 3^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,930 |
For example, $2 M$ is a point on the side $AB$ of the cyclic quadrilateral $ABCD$, $MP \perp BC$, $MQ \perp CD$, $MR \perp AD$, $PR$ intersects $MQ$ at $N$. Prove: $PN \cdot MA = RN \cdot MB$.
---
This is the translated text, maintaining the original format and line breaks. | Proof: Let $\angle A=\angle P M Q=\alpha$,
$$
\angle B=\angle Q M R=\beta, \angle R N M=\gamma \text {. }
$$
In $\triangle P M N$ and $\triangle P M B$,
$$
\begin{array}{l}
\frac{P N}{P M}=\frac{\sin \alpha}{\sin \gamma}, \frac{P M}{B M}=\sin \beta, \\
\therefore \frac{P N}{B M}=\frac{\sin \alpha \cdot \sin \beta}{\si... | PN \cdot MA = RN \cdot MB | Geometry | proof | Yes | Yes | cn_contest | false | 709,931 |
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