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If $f_{n}\left(a^{m}\right)$ represents the last $n$ digits of the natural number $a^{m}$, and $m$ is the smallest value such that the equation $f_{2000}\left(7^{m}\right)=1$ holds, prove that $m \leqslant 2^{1907} \times 5^{19948}$ (where $m \neq 0 \in \mathbf{N}$). | Proof: First, we prove two very interesting lemmas:
Lemma one: If $q \geqslant 3 \in \mathbf{N}$, then $2^{q+1} \mid\left(7^{2^{q-2}}-1\right)$.
Lemma two: If $q \geqslant 1 \in \mathbf{N}$, then $5^{q+1} \mid\left(7^{4 \times 5^{q-1}}-1\right)$.
First, we prove Lemma one by induction:
When $q=3$, it is easy to see tha... | m \leqslant 2^{1997} \times 5^{1998} | Number Theory | proof | Yes | Yes | cn_contest | false | 712,428 |
Let $a$ be a real constant such that the inequality $\frac{1}{\sqrt{x}+1} \geqslant a \sqrt{\frac{x}{x-1}}$ has non-zero real solutions for $x$. Prove that:
$$
a<\frac{2}{5} \text {. }
$$ | Proof: Obviously, when $a \leqslant 0$, the inequality has non-zero real solutions. Below, we only need to consider the case $a>0$.
Let $a>0$, and let $t=\sqrt{x}$. Since $x>0$, we have $x-1>0$, i.e., $t>1$. The original inequality becomes $\frac{1}{t+1} \geqslant a \cdot \frac{t}{\sqrt{t^{2}-1}}$, which is equivalent... | a<\frac{2}{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,429 |
Example 1 Choose any natural number as $a_{0}$, and then arbitrarily select $a_{1} \in \{a_{0}+54, a_{0}+77\}$, and so on, that is, after selecting $a_{k}$, select $a_{k+1} \in \{a_{k}+54, a_{k}+77\}$. Prove that in the sequence $a_{0}, a_{1}, a_{2}, \cdots$, there must be a term $a_{n}$ whose last two digits are the s... | Proof: Since 77 and 100 are coprime, let $k \cdot 77 \equiv b_{k} \pmod{100}$, then $B \equiv \{b_{k} \mid k=1,2, \cdots, 100\}$ is a complete residue system modulo 100, i.e., $B=\{0,1,2, \cdots, 99\}$. Arrange $b_{1}, b_{2}, \cdots, b_{100}$ in increasing order of their indices clockwise on a circle. Thus, the last tw... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,430 |
Let $2 x_{1}, x_{2}, \cdots, x_{n}$ be a sequence, with the general term $x_{n}$ $=\frac{n}{n+a}$, where $a$ is a natural number. Prove that for any natural number $n$, the term $x_{n}$ of the sequence can always be expressed as the product of two other terms.
(1986, United German Mathematics Competition) | Proof: The problem is to prove the existence of two natural numbers $u, v$ such that $x_{n}=x_{n} x_{v}$.
Let $\frac{n}{n+a}=\frac{u}{u+a}+\frac{v}{v+a}(u \leqslant v)$, then
$$
v=\frac{n u+n a}{u-n} \text{. }
$$
Since $n u+n a$ are positive numbers, it follows that $u>n$. Let $p=$ $u-n$, then $p \in \mathbf{N}$, subs... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,431 |
Example 3 Suppose an arithmetic sequence composed of positive integers has one term that is a perfect square. Prove: the sequence also has infinitely many terms that are perfect squares.
(1986, Hungarian Adani・Daniel Mathematics Competition) | Proof: Let the common difference of an arithmetic sequence be $d$, then $(n+k d)^{2}=n^{2}+\left(2 n k+k^{2} d\right) d$. From this, we can conclude that the statement holds. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,432 |
Example 4 Given a positive integer $m \geqslant 2$. Prove:
(1) There exist integers $x_{1}, x_{2}, \cdots, x_{2 m}$, such that
$$
x_{i} x_{m+i}=x_{i+1} x_{m+i-1}+1,1 \leqslant i \leqslant m \text {. }
$$
(2) For any integer sequence $x_{1}, x_{2}, \cdots, x_{2 m}$ satisfying condition (1), one can construct an integer ... | Proof: (1) If $x_{1}=x_{2}=\cdots=x_{m}=1$, from $x_{i} x_{m+i}=x_{i+1} x_{m+i-1}+1(1 \leqslant i \leqslant m)$, we have
$$
\begin{array}{l}
x_{m+i}=x_{m+i-1}+1(1 \leqslant i \leqslant m-1), \\
x_{2 m}=x_{m+1} x_{2 m-1}+1 .
\end{array}
$$
Thus, $x_{m+1}=x_{m}+1=2$. Therefore,
$$
\begin{array}{l}
x_{k+m}=k+1(1 \leqslan... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,433 |
Example $5 A_{1}, A_{2}, \cdots, A_{29}$ are 29 different positive integers. For $1 \leqslant i < j \leqslant 29$, the sum $A_{i} + A_{j}$ is divisible by 200.
(29th IMO Shortlist) | Proof: Suppose the elements in $A_{i}(1 \leqslant i \leqslant 29)$ are all no more than 1988, and the number of elements in each set $N_{i}$ (1988) $\geqslant \frac{1988}{\mathrm{e}}=731.3 \cdots$, i.e., $\left|A_{i}\right| \geqslant 732$. Without loss of generality, assume $\left|A_{i}\right|=732$ (otherwise, remove s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,434 |
Example 6 The first term is a positive integer, Tang constructs all the even terms, and Xia constructs the subsequent odd terms. Xiao Du's construction method is: subtract any digit of the previous term from the previous term; Xiao Xia's construction method is: subtract any one digit of the previous term from the previ... | Proof: If the first term is a single-digit number, then all subsequent terms are 0, which is obviously true.
If the first term is an $n$-digit number, we will prove that the sequence cannot carry over.
Assume it can carry over, then it must carry over at an odd term, let it be the $(2k+1)$-th term, then the $(2k)$-th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,435 |
Example 7 Let $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ denote two integer sequences as follows:
$$
x_{0}=1, x_{1}=1, x_{n+1}=x_{n}+2 x_{n-1}(n=1 \text {, }
$$
$2,3, \cdots)$,
$$
y_{0}=1, y_{2}=7, y_{n+1}=2 y_{n}+3 y_{n-1}(n=1 \text {, }
$$
$2,3, \cdots \downarrow)$
Therefore, the first few terms of the two seq... | Consider the sequence $\{\bmod 8\}$.
$$
\begin{array}{l}
x_{4}=5+2 \times 3=11 \equiv 3\{\bmod 8\}, \\
x_{5}=3+2 \times 5=13=5\{\bmod 8\}, \\
y_{2}=2 \times 7+3 \times 1=17 \equiv 1(\bmod 8\}, \\
y_{3}=2 \times 1+3 \times 7=23 \equiv 7\{\bmod 8\},
\end{array}
$$
We can归纳 get that these two sequences are periodic seque... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,436 |
Example 8 The Fibonacci sequence $\left\{f_{n}\right\}$ is defined as $f_{1}=1$, $f_{2}=1, f_{n}=f_{n-1}+f_{n-2}(n>2)$. Prove: There is a unique set of positive integers $a, b, m$, such that $0<a<m, 0<b<m$, and for all positive integers $n, f_{n}-a n b^{n}$ is divisible by $m$.
(1983, British Mathematical Olympiad) | Let $M(m)$ denote the multiples of $m$. From $f_{1}-a b=1-a b=M(m)$, we know that $m$ and $ab$ are coprime. Subtracting $f_{2}-2 a b^{2}=1-2 a b^{2}=M(m)$, we get
$$
2 a b^{2}-a b=a b(2 b-1)=M(m) \text {. }
$$
Since $ab$ and $m$ are coprime, it follows that
$$
2 b-1=M(m) \text {. }
$$
When $n>2$,
$$
\begin{array}{l}
... | a=2, b=3, m=5 | Number Theory | proof | Yes | Yes | cn_contest | false | 712,437 |
Example 3 On a circle with a circumference of $300 \mathrm{~cm}$, there are two balls, A and B, moving in uniform circular motion at different speeds. Ball A starts from point A and moves in a clockwise direction, while ball B starts from point B at the same time and moves in a counterclockwise direction. The two balls... | Solution: Let the length of $\overparen{A C P}$ be $x \mathrm{~cm}$, and the original speeds of A and B be $v_{1}$ and $v_{2}\left(v_{1} \neq v_{2}\right)$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{40}{v_{1}}=\frac{x-40}{v_{2}}, \\
\frac{300-20-(x-40)}{2 v_{1}}=\frac{x-40+20}{\fr... | 120 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,438 |
Example 9 The sequence $p_{1}, p_{2}, \cdots$ is defined as follows: $p_{1}=2$, and for $n \geqslant 2$, $p_{n}=\left(p_{1} p_{2} \cdots p_{n-1}+1\right)$'s largest prime factor. Prove: 5 is not a term in this sequence.
(1982, Australian Mathematical Olympiad) | Proof: It is evident that the elements in the sequence are distinct.
Assume 5 is a term in the sequence, initially the $k$-th term, i.e., $p_{k}=5$, and when $k>n$, $p_{n} \neq 5$. By definition: $p_{1}=2, p_{2}=2+1=3, p_{k}=\left(2 \times 3 \times \cdots p_{k-1}+1\right.$'s largest prime factor $)=5(k>2)$.
Clearly, $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,439 |
Example 10 Given $v_{0}=0, v_{1}=1, v_{n+1}=8 v_{n}-$ $w_{n-1}(n=1,2,3, \cdots)$. Is there any term in the sequence $\left\{v_{n}\right\}$ that can be divisible by 15? How many such terms are there? Prove your conclusion.
(1991, Zhejiang Province High School Mathematics Summer Camp) | Solution: From $v_{0}=0, v_{1}=1$ and $v_{n+1}=8 v_{n}-v_{n-1}$, we get $v_{n+1} \equiv 8 v_{n}-v_{n-1}(\bmod 15)$. According to this, we can obtain the following table:
Thus, $v_{n}(\bmod 15)$ is a sequence with a cycle of 30, where $v_{0}(\bmod 15)=0, v_{15}(\bmod 15)=$ 0. Therefore, there exist terms divisible by 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,440 |
Example 11 Does there exist a number $\alpha(0<\alpha<1)$, such that there is an infinite sequence of positive numbers $\left\{a_{n}\right\}$, satisfying
$$
1+a_{n+1} \leqslant a_{n}+\frac{\alpha}{n} \cdot a_{n}, n=1,2, \cdots \text {. }
$$
(29th IMO Shortlist) | Assume there exists $\alpha$ and $\left\{a_{n}\right\}$ satisfying the conditions, then we have
$$
a_{n}\left(1+\frac{1}{n}\right)>a_{n}\left(1+\frac{\alpha}{n}\right) \geqslant 1+a_{n+1} \text {. }
$$
This implies $a_{n}>\frac{n}{n+1}\left(1+a_{n+1}\right)$.
Thus, $a_{n}>n \times \frac{1}{n+1}$.
If there exists $k \l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,441 |
1. As shown in Figure 1, given the positions of real numbers $a$, $b$, and $c$ on the number line. Then the value of $|c-1|+|a-c|+|a-b|$ is $(\quad)$.
(A) $b-1$
(B) $2 a-b-1$
(C) $1+2 a-b-2 c$
(D) $1-2 c+b$ | 1.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text "1.D" seems to be a label or a choice in a list, and it doesn't require translation as it is already in a form that is commonly used in both Chinese... | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,442 |
3. If $x-\frac{1}{x}=1$, then the value of $x^{3}-\frac{1}{x^{3}}$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,444 |
4. As shown in Figure 2, in the right triangle $\triangle A B C$, $\angle C=90^{\circ}$, $\angle C A B=30^{\circ}$, $A D$ bisects $\angle C A B$. Then the value of $\frac{A B}{C D}-\frac{A C}{C D}$ is ( ).
(A) $\sqrt{3}$
(B) $\frac{\sqrt{3}}{3}$
(C) $3-\sqrt{3}$
(D) $6-2 \sqrt{3}$ | 4.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,445 |
6. In the plane of isosceles $\triangle A B C(A B=A C \neq B C)$, there is a point $P$ such that $\triangle P A B, \triangle P B C, \triangle P A C$ are all isosceles triangles. Then the number of points satisfying this condition is ( ).
(A) 1
(B) 3
(C) 6
(D) 7 | $6 . C$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,447 |
Example 4 As shown in Figure 2, there is a rectangular plot of land $A B C D$, and a rectangular flower bed $E F G H$ is to be built in the center, with its area being half of the area of this plot, and the width of the paths around the flower bed being equal. Now, without any measuring tools, only a sufficiently long ... | Solution: Let the width of the road be $x$, $AB=a$, $AD=b$. According to the problem, we have
$$
(a-2 x)(b-2 x)=\frac{1}{2} a b \text{. }
$$
Thus, $8 x^{2}-4(a+b) x+a b=0$.
Solving for $x$ gives $x=\frac{1}{4}\left[(a+b) \pm \sqrt{a^{2}+b^{2}}\right]$.
When the plus sign is taken,
$$
\begin{aligned}
2 x & =\frac{1}{2}... | x=\frac{a+b-\sqrt{a^{2}+b^{2}}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,449 |
8. If the solution to the equation $\frac{2 x+a}{x-2}=-1$ with respect to $x$ is positive, then the range of values for $a$ is $\qquad$ . | 8. $a<2$ and $a \neq-4$ | a<2 \text{ and } a \neq -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,450 |
11. In the Cartesian coordinate system: there are four points $A(-8,3)$, $B(-4,5)$, $C(0, n)$, $D(m, 0)$. When the perimeter of quadrilateral $ABCD$ is minimized, the value of $\frac{m}{n}$ is -. $\qquad$ | 11. $-\frac{3}{2}$ | -\frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,453 |
12. At 12 o'clock, the three hands of the clock coincide. After $x$ minutes, the second hand first bisects the acute angle between the minute and hour hands. Then the value of $x$ is $\qquad$ . . | 12. $\frac{1440}{1427}$ | \frac{1440}{1427} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,454 |
14. As shown in Figure 4, $ABCD - A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a rectangular prism, $A^{\prime} = 50 \mathrm{~cm}, AB$
$$
=40 \mathrm{~cm}, AD=30 \mathrm{~cm} \text {, }
$$
The top and bottom faces are each divided into
$3 \times 4$ small squares,
each with a side length of $10 \mathrm{~cm}$, result... | $14.10 \sqrt{17}$ | 10 \sqrt{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,456 |
15. As shown in Figure 5, given that $M$ and $N$ are points on side $BC$ of $\triangle ABC$, and satisfy $BM = MN = NC$. A line parallel to $AC$ intersects the extensions of $AB$, $AM$, and $AN$ at points $D$, $E$, and $F$ respectively. Prove that $EF = 3DE$. | Three, 15. As shown in Figure 7, draw lines through $N$ and $M$ parallel to $AC$ intersecting $AB$ at points $H$ and $G$, respectively. Line $NH$ intersects $AM$ at point $K$.
$$
\begin{array}{l}
\because B M=M N=N C, \\
\therefore B G=G H=H A .
\end{array}
$$
It is easy to see that $H K=\frac{1}{2} G M$,
$$
G M=\frac... | E F=3 D E | Geometry | proof | Yes | Yes | cn_contest | false | 712,457 |
16. Given the equations in $x$, $4 x^{2}-8 n x-3 n=2$ and $x^{2}-(n+3) x-2 n^{2}+2=0$. Does there exist a value of $n$ such that the square of the difference of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain the reason. | 16. $\Delta_{1}=(8 n+3)^{2}+23>0$.
Then $n$ is any real number, equation (1) always has real roots.
Let the two roots of the first equation be $\alpha, \beta$. Then
$$
\begin{array}{l}
\alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} . \\
\therefore(\alpha-\beta)^{2}=4 n^{2}+3 n+2 .
\end{array}
$$
From the second equa... | n=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,458 |
17. As shown in Figure 6, given an equilateral triangle $\triangle ABC$ inscribed in a circle, take a point $M$ on the minor arc $\overparen{AB}$, different from $A$ and $B$. Let the line $AC$ intersect $BM$ at point $K$, and the line $CB$ intersect $AM$ at point $N$. Prove: The product of segments $AK$ and $BN$ is ind... | $$
\begin{array}{l}
\text { 17. } \because \angle A M K=\angle C=\angle C A B=\angle K+\angle A B K, \\
\angle A M K=\angle N A B+\angle A B K, \\
\therefore \angle K=\angle B A M=\angle B A N .
\end{array}
$$
Similarly, $\angle A B K=\angle N$.
Thus, $\triangle A B K \sim \triangle B N A$, and we have $\frac{A B}{B N... | A K \cdot B N = A B^2 | Geometry | proof | Yes | Yes | cn_contest | false | 712,459 |
Example 5 A mall installs an escalator between the first and second floors, which travels upwards at a uniform speed. A boy and a girl start walking up the escalator to the second floor (the escalator itself is also moving). If the boy and the girl both move at a uniform speed, and the boy walks twice as many steps per... | Solution: (1) Let the girl's speed be $x$ steps/min, the escalator's speed be $y$ steps/min, and the stairs have $s$ steps, then the boy's speed is $2x$ steps/min. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} .
\end{array}\right.
$$
Dividing... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,460 |
18. A construction vehicle loads cement power poles from the warehouse and delivers them to the roadside 1000 meters away from the warehouse for installation. It is required to install a power pole every 100 meters along one side of the road. Given that the vehicle can carry a maximum of 4 poles per trip, and the task ... | 18. Obviously, the fewer the number of trips, the shorter the distance traveled, and the less the fuel cost. Therefore, transporting 18 power poles in 5 trips results in a shorter travel distance. There are two methods for these 5 trips:
(1) Four trips of 4 poles each, and one trip of 2 poles;
(2) Three trips of 4 pole... | 19mn | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,461 |
1. In an acute triangle $\triangle ABC$, the three altitudes $AD$, $BE$, and $CF$ intersect at $H$. Among the seven points $A$, $B$, $C$, $D$, $E$, $F$, and $H$, the number of groups of four points that can be concyclic is ( ).
(A) 4 groups
(B) 5 groups
(C) 6 groups
(D) 7 groups | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,462 |
$\sqrt{\sqrt{2001}-\sqrt{2000}}$. Then $a^{4}-b^{4}$ equals ( ).
(A) 2000
(B) 2001
(C) $\sqrt{2} 900$
(D) $\sqrt{2001}$ | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,464 |
6. As shown in Figure 2; two equal circles intersect at points $A$ and $B$. A line through $B$ intersects the two circles at $M$ and $N$. Tangents to the circles at $M$ and $N$ intersect at $C$. Then the quadrilateral $A M C N$ has the following relationship ( ) established.
(A) has an incircle but no circumcircle
(B) ... | 6.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,467 |
1. The equation $x^{2}-4 x+3 a^{2}-2=0$ has real roots in the interval $[-1,1]$. Then the range of real number $a$ is | 1. $-\frac{\sqrt{15}}{3} \leqslant a \leqslant \frac{\sqrt{15}}{3}$ | -\frac{\sqrt{15}}{3} \leqslant a \leqslant \frac{\sqrt{15}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,468 |
Example 6: On a page of a cloth store's ledger, ink has stained the page, as shown in the table below:
The number of meters of cloth sold is unclear, but it is remembered that it was sold in whole meters; in the amount column, only the three digits 7.28 are visible, but the first three digits are unclear. Please help ... | Solution: Let the quantity purchased be $x$ meters, and the first three digits of the amount be $y$. According to the problem, we have
$$
49.36 x=10 y+7.28 \text {. }
$$
Therefore, $y=\frac{617 x-91}{125}$
$$
\begin{aligned}
=5 x-1+ & \frac{2(17-4 x)}{125} . \\
\text { Let } \frac{17-4 x}{125}=t \Rightarrow x & =\frac... | x=98, y=483 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,471 |
II. (14 points) Let $a, b, c$ be distinct real numbers. Prove that:
$$
\frac{a^{4}}{(a-b)(a-c)}+\frac{b^{4}}{(b-c)(b-a)}+\frac{c^{4}}{(c-a)(c-b)}>0 .
$$ | Given that $a$, $b$, and $c$ are distinct real numbers,
$$
\begin{aligned}
\therefore \text { LHS } & =\frac{-a^{4}(b-c)-b^{4}(c-a)-c^{4}(a-b)}{(a-b)(b-c)(c-a)} \\
& =\frac{(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\right)}{(a-b)(b-c)(c-a)} \\
& =a^{2}+b^{2}+c^{2}+a b+b c+c a \\
& =\frac{a^{2}+2 a b+b^{2}+b^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,474 |
Three. (14 points). Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+$ $\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots+\sqrt{1+\frac{1}{1999^{2}}+\frac{1}{2000^{2}}}$. Find the greatest integer $[S]$ not exceeding $S$. | $\begin{array}{l}\equiv 、 \begin{aligned} \because & \sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\ & =\sqrt{\frac{n^{2}+2 n+1}{n^{2}}-\frac{2 n}{n^{2}}+\frac{1}{(n+1)^{2}}} \\ & = \\ & =\sqrt{\left(\frac{n+1}{n}\right)^{2}-2 \cdot \frac{n+1}{n} \cdot \frac{1}{n+1}+\left(\frac{1}{n+1}\right)^{2}} \\ & \left.=1+\frac{1... | 1999-\frac{1}{2000} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,475 |
Four. (15 points) As shown in Figure 3, $AB$ is the diameter of semicircle $O$, $C$ is a point on the semicircle, $CD \perp AB$ at $D$, $\odot O_{1}$ is tangent to $BD$ at point $E$, tangent to $CD$ at point $F$, and tangent to the semicircle at point $G$. Prove:
(1) $A$, $F$, $G$ are collinear;
(2) $AC=AE$. | (1) As shown in Figure 5, connect $A G$, $F G$, $O_{1} F$, and then connect $O O_{1}$, its extension must pass through point $G$.
$$
\begin{array}{l}
\because O_{1} F \perp C D, \\
\therefore \angle F O_{1} G=\angle A O G .
\end{array}
$$
$\therefore \triangle F O_{1} G \backsim \triangle A O G$. Thus, $\angle A G O=\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,476 |
Five. (15 points) As shown in Figure 4, $AOB$ is a quarter of a unit circle. The center of the semicircle $O_{1}$ is on $OA$ and is internally tangent to the arc $\overparen{AB}$ at point $A$. The center of the semicircle $O_{2}$ is on $OB$ and is internally tangent to the arc $\overparen{AB}$ at point $B$. The semicir... | (1) Let the radius of $\odot C_{1}$ be $R$, and the radius of $\mathrm{S} \mathrm{O}_{2}$ be $r$ (as shown in Figure 6).
$$
\begin{aligned}
y= & \frac{1}{2} \pi\left(R^{2}+r^{2}\right) \\
= & \frac{1}{2} \pi\left[(R+r)^{2}\right. \\
& -2 R r] .
\end{aligned}
$$
Connecting $O_{1} O_{2}$, this line must pass through the... | (3-2 \sqrt{2}) \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,477 |
1. Given $a$ is a real number. Then, the number of subsets of the set $M=\left\{x \mid x^{2}-3 x-a^{2}+2=0, x \in \mathbf{R}\right\}$ is ( ).
(A) 1
(B) 2
(C) 4
(D) Uncertain | $-1 .(\mathrm{C})$
From the discriminant $\Delta=1+4 a^{2}>0$ of the equation $x^{2}-3 x-a^{2}+2=0$, we know that the equation has two distinct real roots. Therefore, $M$ has 2 elements, and the set $M$ has $2^{2}=4$ subsets. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,478 |
3. Among the four functions $y=\sin |x|, y=\cos |x|, y=$ $|\cot x|, y=\lg |\sin x|$, the one that is periodic with $\pi$, monotonically increasing in $\left(0, \frac{\pi}{2}\right)$, and an even function is ( ).
(A) $y=\sin |x|$
(B) $y=\cos |x|$
(C) $y=|\cot x|$
(D) $y=\lg |\sin x|$ | 3.(D).
$y=\sin |x|$ is not a periodic function. $y=\cos |x|=\cos x$ has a period of $2 \pi$. $y=|\cot x|$ is monotonically decreasing on $\left(0, \frac{\pi}{2}\right)$. Only $y=\lg |\sin x|$ satisfies all the conditions. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,480 |
4. If $\triangle A B C$ has exactly one solution when $\angle A B C=60^{\circ}, A C=12, B C=k$, then the range of $k$ is ( ).
(A) $k=8 \sqrt{3}$
(B) $0<k \leqslant 12$
(C) $k \geqslant 12$
(D) $0<k \leqslant 12$ or $k=8 \sqrt{3}$ | 4. (D).
According to the problem, $\triangle A B C$ has two types, as shown in Figure 5.
It is easy to get $k=8 \sqrt{3}$ or $0<k \leqslant 12$.
This problem can also be solved using the special value method, eliminating (A), (B), and (C). | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,481 |
Example 7 In a school's donation activity for the Hope Project, the total donation amount from $m$ boys and 11 girls in Class A is equal to the total donation amount from 9 boys and $n$ girls in Class B, both being $(m n+9 m+11 n+145)$ yuan. It is known that each person's donation amount is the same, and it is an integ... | Solution: Let the donation amount per person be $x$ yuan. Since the total donation amount of the two classes is equal and the donation amount per person is the same, the number of people in the two classes is also equal, denoted as $y$. Then,
$$
y=m+11=n+9
$$
․ $m n+9 m+11 n+145$
$$
=(m+11)(n+9)+46=y^{2}+46 \text {. }
... | 25 \text{ or } 47 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,482 |
5. If the expansion of $\left(1+x+x^{2}\right)^{1000}$ is
$$
a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2000} x^{2000} \text {, }
$$
then the value of $a_{0}+a_{3}+a_{6}+a_{9}+\cdots+a_{1998}$ is ( ).
(A) $3^{333}$
(B) $3^{666}$
(C) $3^{999}$
(D) $3^{2001}$ | 5. (C).
Let $x=1$, we get
$$
3^{1000}=a_{0}+a_{1}+a_{2}+a_{3}+\cdots+a_{2000} ;
$$
Let $x=\omega$, we get
$$
0=a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3} \omega^{3}+\cdots+a_{2000} \omega^{2 \omega 00} ;
$$
(Where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$, then $\omega^{3}=1$ and $\omega^{2}+\omega+1=0$.)
Le... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,483 |
8. If the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3,3 z_{1}$ $2 z_{2}=\frac{3}{2}-\mathrm{i}$, then $z_{1} \cdot z_{2}=$ $\qquad$ | 8. $-\frac{30}{13}+\frac{72}{13} \mathrm{i}$.
From
$$
\begin{array}{l}
3 z_{1}-2 z_{2}=\frac{1}{3} z_{2} \cdot \bar{z}_{2} \cdot z_{1}-\frac{1}{2} z_{1} \cdot \bar{z}_{1} \cdot z_{2} \\
=\frac{1}{6} z_{1} \cdot z_{2}\left(2 \bar{z}_{2}-3 \bar{z}_{1}\right)
\end{array}
$$
we can get $z_{1} \cdot z_{2}=\frac{6\left(3 z... | -\frac{30}{13}+\frac{72}{13} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,485 |
12. In a regular hexagon divided into six areas for观赏 plants, 加园: $\mathrm{i}$, it is required that the same type of plant be in the same area, and adjacent areas must have different plants. There are 4 different types of plants available, resulting in $\qquad$ planting schemes.
Note: The term "观赏 plants" is directly ... | 12.732.
Consider planting the same type of plant at $A$, $C$, and $E$, there are $4 \times 3 \times 3 \times 3=108$ methods.
Consider planting two types of plants at $A$, $C$, and $E$, there are $3 \times 4 \times 3 \times 3 \times 2 \times 2=432$ methods.
Consider planting three types of plants at $A$, $C$, and $E$, ... | 732 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,486 |
13. Let $\left\{a_{n}\right\}$ be an arithmetic sequence, $\left\{b_{n}\right\}$ be a geometric sequence, and $b$ $=a_{1}^{2}, b_{2}=a_{2}^{2}, b_{3}=a_{3}^{2}\left(a_{1}<a_{2}\right)$, and $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\right.$ $\left.\cdots+b_{n}\right)=\sqrt{2}+1$. Find the first term and common dif... | Three, 13. Let the required common difference be $d$.
$$
\because a_{1}0 \text {. }
$$
From this, we get $a_{1}^{2}\left(a_{1}+2 d\right)^{2}=\left(a_{1}+d\right)^{4}$.
Simplifying, we get $2 a_{1}^{2}+4 a_{1} d+d^{2}=0$.
Solving, we get $d=(-2 \pm \sqrt{2}) a_{1}$.
Since $-2 \pm \sqrt{2}<0$, it follows that $a_{1}<0$... | a_{1}=-\sqrt{2}, d=2\sqrt{2}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,487 |
14. Let the curve $C_{1}: \frac{x^{2}}{a^{2}}+y^{2}=1$ (where $a$ is a positive constant) intersect with $C_{2}$ : $y^{2}=2(x+m)$ at only one point $P$ above the $x$-axis.
(1) Find the range of real numbers $m$ (expressed in terms of $a$);
(2) $O$ is the origin, and $C_{1}$ intersects the negative half of the $x$-axis ... | 14. (1) From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1, \\ y^{2}=2(x+m)\end{array}\right.$ eliminating $y$, we get
$$
x^{2}+2 a^{2} x+2 a^{2} m-a^{2}=0 \text {. }
$$
Let $f(x)=x^{2}+2 a^{2} x+2 a^{2} m-a^{2}$, problem (1) is transformed into the equation (1) having a unique solution or equal roots in $x \in(... | S_{\text {nux }}=a \sqrt{a-a^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,488 |
One, (50 points) As shown in Figure 3, in $\triangle A B C$, $O$ is the circumcenter, and the three altitudes $A D, B E, C F$ intersect at point $H$. Line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Prove:
(1) $O B \perp D F, O C \perp$
$D E$;
(2) $O H \perp M N$. | (1) $\because A, C, D, F$ are concyclic points,
$\therefore \angle B D F=\angle B A C$.
Also, $\because \angle O B C=\frac{1}{2}\left(180^{\circ}-\angle B O C\right)$
$$
=90^{\circ}-\angle B A C \text {, }
$$
$\therefore O B \perp D F$.
Similarly, $C C \perp D E$.
(2) $\because C F \perp M A$,
$$
\begin{array}{l}
\ther... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,491 |
Three. (50 points) Divide a rectangle with side lengths of positive integers $m$ and $n$ into several squares with side lengths of positive integers, where each square's sides are parallel to the corresponding sides of the rectangle. Find the minimum value of the sum of the side lengths of these squares.
---
Please n... | Three, let the minimum value sought be $f(m, n)$, it can be proven that
$$
f(m, n)=m+n-(m, n).
$$
where $(m, n)$ represents the greatest common divisor of $m$ and $n$.
In fact, without loss of generality, assume $m \geqslant n$.
(1) Induction on $m$. It can be proven that there exists a valid partition such that the s... | f(m, n)=m+n-(m, n) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,493 |
Given a convex quadrilateral $ABCD$ and its interior points $E$ and $F$, such that
\[
\begin{array}{l}
A E=B E, C E=D E, \angle A E B=\angle C E D ; \\
A F=D F, B F=C F, \angle A F D=\angle B F C .
\end{array}
\]
Prove: $\angle A F D + \angle A E B = \pi$.
(Xu Yichao provided) | Solution: As shown in Figure 1, we agree to denote the intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral as $G$, and let $\angle EAB = \angle ABE = \theta, \angle FAD = \angle ADF = \varphi$.
Since $\triangle AEC$ can coincide with $\triangle BDE$ through a rotation around point $E$, we have $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,494 |
For given positive integers $a, b, b > a > 1$, where $a$ does not divide $b$, and a given sequence of positive integers $\left\{b_{n}\right\}_{n=1}^{\infty}$, satisfying that for all positive integers $n$, $b_{n+1} \geqslant 2 b_{n}$. Does there always exist a sequence of positive integers $\left\{a_{n}\right\}_{n=1}^{... | Solution: The answer is affirmative. We construct using induction.
Take $a_{1}$ as a positive integer, such that $2 a_{1} \notin \left\{b_{n}\right\}_{n=1}^{\infty}, a_{1}>b-a$ (for example, if $b_{n_{0}}>b-a+1$, take $a_{1}^{*}=b_{n_{0}}-1$).
Assume we have taken $a_{1}, a_{2}, \cdots, a_{k}$ such that
$$
a_{i+1}-a_{i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,495 |
Given an integer $k > 1$, let $\mathbf{R}$ be the set of all real numbers. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x$ and $y$ in $\mathbf{R}$, we have
$$
f\left(x^{k}+f(y)\right)=y+(f(x))^{k} .
$$
(Huang Xuan-guo provided) | $$
\begin{array}{l}
\text { Solution: Let } x=0, t=(f(0))^{k} \text {, by (1) we have } \\
f(f(y))=y+t . \\
\text { and } \left.f\left(f\left(x^{k}+f(f(y))\right)\right)=f(f(y)+f(x))^{k}\right) \\
=f\left((f(x))^{k}+f(y)\right)=y+(f(f(x)))^{k} \\
=y+(x+t)^{k} . \\
\text { By (2) we have } \\
\quad f\left(f\left(x^{k}+f... | f(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,496 |
Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$, $x_{n+1}, x_{n+2}$ satisfy the condition
$$
0 < x_{1} < x_{2} < \cdots < x_{n} < x_{n+1} < x_{n+2}.
$$
Find the minimum value of
$$
\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left... | Solution: (I) Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n+1$. The expression in the problem can be written as
$$
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} i_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i=1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)}... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,497 |
Given an equilateral $\triangle ABC$, $D$ is any point on side $BC$. The circumcenter and incenter of $\triangle ABD$ are $O_{1}, I_{1}$, and the circumcenter and incenter of $\triangle ADC$ are $O_{2}, I_{2}$. The line $O_{1} I_{1}$ intersects $O_{2} I_{2}$ at $P$. Try to find: When point $D$ moves on side $BC$, the l... | $$
\begin{array}{l}
\text { Solution 1: As shown in Figure 2, draw auxiliary } \\
\text { lines. From } \\
\angle \mathrm{AO}_{2} \mathrm{D}=2 \angle \mathrm{C}= \\
A D^{2}-O D^{2}=A O^{2}, \\
\text { i.e., }(\sqrt{3} y)^{2}-x^{2}=(\sqrt{3})^{2} \text {, } \\
y^{2}-\frac{x^{2}}{3}=1,-1<x<1, y<0 . \\
\left(2+1-a+\sqrt{a... | y=-\frac{\sqrt{3}}{3} \sqrt{a^{2}+3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,498 |
1. Given that $p$, $q$, and $r$ are real numbers, and $r+q=6-4p+$ $3p^{2}$, $r-q=5-4p+p^{2}$. Then the relationship in size between $p$, $q$, and $r$ is ( ).
(A) $r<p<q$
(B) $q<r<p$
(C) $q<p<r$
(D) $p<q<r$ | $-1 .(\mathrm{D})$
Completing the square, we get $r-q=(p-2)^{2}+1 \geqslant 1>0$, which implies $q0$, we know $q>p$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,501 |
2. Let positive integers $x \neq y$, and satisfy $-\frac{1}{x}+\frac{1}{y}=\frac{2}{5}$. Then the units digit of $x^{y}+$ $y^{x}$ is ( ).
(A) 2
(B) 3
(C) 5
(D) 7 | 2. (A).
From $\frac{1}{x}+\frac{1}{y}=\frac{2}{5}$ we get $2 x y-5 x-5 y=0$. Then $4 x y-10 x-10 y+25=25$,
which simplifies to $(2 x-5)(2 y-5)=25$.
Since $x, y$ are positive integers, and $x \neq y$,
$$
\therefore\left\{\begin{array}{l}
2 x-5=1,5,25, \\
2 y-5=25,5,1 .
\end{array}\right.
$$
Solving, we get $\left\{\be... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,502 |
3. Given an isosceles trapezoid $A B C D$ with legs $A B=C D=a, A C \perp B D, \angle A B C=\alpha$. Then the area of this trapezoid is ( )
(A) $2 a \sin \alpha$
(B) $a^{2} \sin ^{2} a$
(C) $2 a \cos \alpha$
(D) $a^{2} \cos \alpha$ | 3. (B).
As shown in Figure 3, draw $A E \perp B C$, with $E$ as the foot of the perpendicular. Note that $\triangle B O C$ is an isosceles right triangle, so $\triangle A E C$ is also an isosceles right triangle, hence
$$
E C=A E=a \sin \alpha .
$$
$S_{\text {square } A \text { ABDC }}=a^{2} \sin ^{2} \alpha$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,503 |
4. A university has no more than 5000 registered students, of which $\frac{1}{3}$ are freshmen, $\frac{2}{7}$ are sophomores, $\frac{1}{5}$ are juniors, and the remainder are seniors. In the mathematics department's student list, freshmen account for $\frac{1}{40}$ of all freshmen in the university, sophomores account ... | 4. (C).
Let the total number of students in the school be $n$, and the number of students in the mathematics department be $c$. Then $\frac{1}{3} n$ are first-year students, $\frac{2}{7} n$ are second-year students, and $\frac{1}{5} n$ are third-year students. The number of first-year students in the mathematics depar... | 183 | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,504 |
5. A retired worker receives a pension each year, the amount of which is proportional to the square of the number of years he worked. If he had worked $a$ more years, his pension would be $p$ yuan more than it is now; if he had worked $b$ more years $(a \neq b, a 、 b$ $>0$ ), his pension would be $q$ yuan more than it ... | 5. (D).
Let $x$ be the pension, $y$ be the years of service, and $k$ be the proportionality constant. From the given information, we have
$$
\begin{array}{l}
x=k \sqrt{y}, x+p=k \sqrt{y+a}, \\
x+q=k \sqrt{y+b} .
\end{array}
$$
Thus, $x^{2}=k^{2} y,(x+p)^{2}=k^{2}(y+a)$,
$$
(x+q)^{2}=k^{2}(y+b) .
$$
Substituting $x^{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,505 |
6. As shown in Figure $1, A B$ is the diameter of semicircle $O$, $A C$ and $A D$ are both chords, $\angle C A D=\angle D A B$. Then the size relationship between $A C+A B$ and $2 A D$ is ( ).
(A) $A C+A B>2 A D$
(B) $A C+A B=2 A D$
(C) $A C+A B<2 A D$
(D) None of the above, cannot compare the sizes | 6. (C).
As shown in Figure 4, draw $OM \perp AC$ at $M$, intersecting $AD$ at $N$, and connect $OD$. It is easy to see that $OD \parallel AC$, $OD \perp OM$. Therefore, $AM < AN$, $OD < ND$, which means $\frac{1}{2} AC < AN$, $\frac{1}{2} AB < ND$. Hence, $AC + AB < 2AD$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,506 |
1. If $16^{9 m}=a, 4^{37 n}=\frac{1}{a}$, then $(36 m+74 n-1)^{2000}$ | Ni.1.1.
Since $a=16^{9 m}=\left(2^{4}\right)^{9 m}=2^{36 m}$, and $\frac{1}{a}=4^{37 n}=2^{74 n}$, then $1=a \times \frac{1}{a}=2^{36 m} \cdot 2^{74 n}=2^{36 m+74 n}$,
we have $36 m+74 n=0$.
Therefore, $(36 m+74 n-1)^{2000}=(-1)^{20000}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,507 |
In rectangle $A B C D$, $A B=C D=2, B C=A D=$ 8, point $O$ is 1 unit away from $A B$, $B C$, and $A D$, and $O$ is inside the rectangle. If the rectangle is rotated $45^{\circ}$ about $O$, then the area of the overlapping part of the two rectangles is $\qquad$ | $3.6 \sqrt{2}-4$.
As shown in Figure 5, the common part of the two figures after rotation is the hexagon $E F G H L S$, whose area can be obtained by $S_{\text {rhombus } A_{1} B_{1} S}-2 S_{\triangle M_{1} F G}$.
Connecting $O A_{1}$ intersects $A B$ at $K$, then
$$
\begin{array}{l}
O A_{1} \perp A B \text { and } F ... | 6 \sqrt{2}-4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,509 |
4. As shown in Figure 2, in
Rt $\triangle A B C$,
$\angle C=90^{\circ}$, point $M$
is the intersection of the three
medians of the triangle. Perpendiculars are drawn from $M$
to $A B$, $B C$, and $A C$,
with the feet of the perpendiculars being $D$, $E$, and $F$, respectively. If $A C=3$, $B C=12$, then the area of $\... | 4.4.
$$
\begin{array}{l}
AB = \sqrt{3^2 + 12^2} = \sqrt{153} = 3 \sqrt{17}, \\
S_{\triangle ABC} = \frac{1}{2} \times 3 \times 12 = 18.
\end{array}
$$
Draw a perpendicular from $C$ to $AB$, with the foot of the perpendicular at $H$, and let it intersect $EF$ at $G$. We have
$$
\begin{array}{l}
S_{\triangle ABC} = \fra... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,510 |
In the Cartesian coordinate system, the parabola $y=-\frac{4}{9} x^{2} +\frac{2}{9} m x+\frac{5}{9} m+\frac{4}{3}$ intersects the $x$-axis at points $A$ and $B$. It is known that point $A$ is on the negative half of the $x$-axis, and point $B$ is on the positive half of the $x$-axis, with the length of $BD$ being twice... | (1) Given $|O B|=2|O A|$, let $|O A|=u$. Then $|O B|=2 u (u>0)$.
Thus, $-u, 2 u$ are the roots of the equation $-\frac{4}{9} x^{2}+\frac{2}{9} m x+\frac{5}{9} m+ \frac{4}{3}=0$.
By Vieta's formulas, we have $\left\{\begin{array}{l}-u+2 u=\frac{1}{2} m, \\ -u \cdot 2 u=-\frac{5}{4} m-3 .\end{array}\right.$
Solving, we g... | P\left(1, \frac{3}{2}\right) \text{ or } P(1,-6) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,512 |
II. (25 points) From a point $P$ outside $\odot O$, draw two tangents $PA$ and $PB$ to $\odot O$, with $A$ and $B$ as the points of tangency. Take any point $C$ on the minor arc $\overparen{AB}$, and draw the tangent to $\odot O$ through point $C$, intersecting $PA$ and $PB$ at points $D$ and $E$, respectively. Let $AB... | $$
\begin{array}{l}
\text { II. As shown in Figure 8, connect } O A, O B, O C, \text { then } \\
O A \perp A P, O B \perp \\
B P, O C \perp D E, \\
\angle O A P=\angle O B P \\
=\angle O C D \\
=\angle O C E=90^{\circ}. \\
\text { Also, } P A = P I ; I A \\
=D C, E B=E C, \\
\text { then } \angle P A B=\angle P B A, \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,513 |
Three. (25 points) In a square array of 16 rows and 16 columns composed of red and blue dots, adjacent dots of the same color are connected by a line segment of the same color, and adjacent dots of different colors are connected by a yellow line segment. It is known that there are 133 red dots, of which 32 are on the b... | Three, each row has 15 line segments, 16 rows have a total of $15 \times 16=240$ horizontal line segments; 16 columns similarly have 240 vertical line segments; in total, there are 480 line segments. It is known that 195 of these line segments are yellow, so the remaining 284 line segments are red and blue.
There are ... | 134 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,514 |
3. Given a non-constant sequence $\left\{a_{i}\right\}$, satisfying $a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}$ $=0$, and $a_{i+1} \neq a_{i-1}, i=0,1,2, \cdots, n$. For a given natural number $n, a_{1}=a_{n+1}=1$, then $\sum_{i=0}^{n-1} a_{i}$ equals ( ).
(A) 2
(B) -1
(C) 1
(D) 0 | $2, \cdots, n$, so, $a_{i+1}+a_{i-1}=a_{i}, i=1,2, \cdots, n$. Thus we get
$$
\begin{array}{l}
a_{0}=a_{1}-a_{2}, \\
a_{1}=a_{2}-a_{3}, \\
\ldots \ldots .
\end{array}
$$
$$
\begin{array}{l}
a_{n-2}=a_{n-1}-a_{n}, \\
a_{n-1}=a_{n}-a_{n+1} .
\end{array}
$$
Therefore, $\sum_{i=0}^{n-1} a_{i}=a_{1}-a_{n+1}=0$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,517 |
4. Given that $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$ (where $a, b, c$ are real numbers), and $c$ is an imaginary number, $\frac{\alpha^{2}}{\beta}$ is a real number. Then the value of $\sum_{k=1}^{5985}\left(\frac{\alpha}{\beta}\right)^{k}$ is ( ).
(A) 1
(B) 2
(C) 0
(D) $\sqrt{3} \mathrm{i}$ | 4. (C).
From $\frac{\alpha^{2}}{\beta}$ being a real number, we get $\frac{\alpha^{3}}{\alpha \beta} \in \mathbf{R}$.
$$
\because \alpha \beta \in \mathbf{R}, \therefore \alpha^{3} \in \mathbf{R} \text {. }
$$
$\because \alpha, \beta$ are conjugates, $\therefore \beta^{3} \in \mathbf{R}$, and $\alpha^{3}-\beta^{3}=0$.... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,518 |
5. If $a+b+c=a b c, A=\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-a^{2}\right)\left(1-c^{2}\right)}{a c}+\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}$, then the value of $\mathrm{A}$ is
(A) 3
(B) -3
(C) 4
(D) -4 | $\begin{array}{l}\text { 5. (C). } \\ \text { Let } a=\tan \alpha, b=\tan \beta, c=\tan \gamma. \text { Then } \\ \tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \cdot \tan \beta \cdot \tan \gamma, \\ \text { and } A=4(\cot 2 \beta \cdot \cot 2 \gamma+\cos 2 \alpha \cdot \cot 2 \gamma \\ \quad+\cot 2 \alpha \cdot \cot 2... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,519 |
6. Let $x_{i} \in\{1,2, \cdots, n\}, i=1,2, \cdots, n$, satisfy $\sum_{i=1}^{n} x_{i}=\frac{n(n+1)}{2}, x_{1} x_{2} \cdots x_{n}=n$ !, the maximum number $n$ for which $x_{1}, x_{2}, \cdots, x_{n}$ must be a permutation of $1,2, \cdots, n$ is ( ).
(A) 4
(B) 6
(C) 8
(D) 9 | 6. (C).
When $n \geqslant 9$, $9, 4, 4$ can replace the factors 8, 6, 3 in $n$!, since $8+6+3=9+4+4$ and $8 \times 6 \times 3=9 \times 4 \times 4$. Thus, the sum $x_{1}+x_{2}+\cdots+x_{n}$ and the product $x_{1} x_{2} \cdots x_{n}$ remain unchanged. However, when $n=8$, it is known that $x_{1}, x_{2}, \cdots, x_{8}$ i... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 712,520 |
1. Let point $P$ be a point inside the convex polygon $A_{1} A_{2} \cdots A_{n}$, the distance from point $P$ to the line $A_{1} A_{2}$ is $h_{1}$, to the line $A_{2} A_{3}$ is $h_{2}, \cdots$, to the line $A_{n, 1} A_{n}$ is $h_{n-1}$, and to the line $A_{n} A_{1}$ is $h_{n}$. If there exists a point $P$ such that $\f... | 1. A polygon that can be inscribed in a circle.
Let the area of a convex polygon be $S$, then we have
$$
a_{1} h_{1}+a_{2} h_{2}+\cdots+a_{n} h_{n}=2 S \text {, }
$$
and $2 S\left(\frac{a_{1}}{h_{1}}+\frac{a_{2}}{h_{2}}+\cdots+\frac{a_{n}}{h_{n}}\right) \geqslant\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}$, with equalit... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,521 |
Example 1 Find the integer solutions of the equation $x y + x + y = 6$.
| Solution: Adding 1 to both sides of the equation, we get
$$
\begin{array}{l}
x y + x + y + 1 = 7 . \\
\text { Left side }=(x+1)(y+1), \\
\text { Right side }=1 \times 7=(-1) \times(-7) .
\end{array}
$$
Therefore, the integer solutions of the original equation are determined by the following system of equations: | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,522 |
Example 2 Find all integer solutions to the equation $\frac{x+y}{x^{2}-x y+y^{2}}=\frac{3}{7}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Taking $x$ as the main variable, rearrange the equation as
$$
3 x^{2}-(3 y+7) x+\left(3 y^{2}-7 y\right)=0 \text {. }
$$
Since $x$ is an integer, then
$$
\begin{array}{l}
\Delta=[-(3 y+7)]^{2}-4 \times 3\left(3 y^{2}-7 y\right) \geqslant 0 \\
\Rightarrow \frac{21-14 \sqrt{3}}{9} \leqslant y \leqslant \frac{2... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,523 |
Example 12 Find the value of $k$ that makes the roots of the equation $k x^{2}+(k+1)$ $x+(k-1)=0$ integers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Discuss in two cases, $k=0$ and $k \neq 0$.
When $k=0$, the given equation is $x-1=0$, which has the integer root $x=1$.
When $k \neq 0$, the given equation is a quadratic equation.
Let the two integer roots be $x_{1}$ and $x_{2}$, then we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{k+1}{k}=\cdots 1-\f... | k=0, k=-\frac{1}{7}, k=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,524 |
Example 13 Consider the equation $m^{2}+(k-4) m$ $+k=0$ with respect to $m$. Let the largest integer root of this equation be the diameter of $\odot O$. Let $P$ be a point outside $\odot O$, and draw the tangent $P A$ and the secant $P B C$ through $P$, as shown in Figure 1, with $A$ being the point of tangency. It is ... | Solution: Let the two roots of the equation be $m_{1}$ and $m_{2}$, then
$$
\left\{\begin{array}{l}
m_{1}+m_{2}=4-k, \\
m_{1} m_{2}=k .
\end{array}\right.
$$
Let $P A=x, P B=y, B C=z$, then $x, y, z$ are all positive integers.
By the secant-tangent theorem, we have
$$
P A^{2}=P B \cdot P C=P B(P C+3 C),
$$
which mean... | P A=2, P B=1, P C=4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,525 |
Example 14 Given the equations in $x$: $4 x^{2}-8 n x- 3 n=2$ and $x^{2}-(n+3) x-2 n^{2}+2=0$. Does there exist a value of $n$ such that the square of the difference of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain the reason.... | Solution: From $\Delta_{1}=(-8 n)^{2}-4 \times 4 \times(-3 n-2)$ $=(8 n+3)^{2}+23>0$, we know that for any real number $n$, equation (1) always has real roots.
Let the two roots of the first equation be $\alpha, \beta$. Then
$$
\alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} \text {. }
$$
Thus, $(\alpha-\beta)^{2}=(\a... | n=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,526 |
Example: 11984 points are distributed on the circumference of a circle, each marked with +1 or -1. If starting from a certain point and moving in any direction around the circle to any point, the sum of all the numbers passed is positive, then that point is called a "good point." Prove: if the points marked with -1 are... | Analysis: The relationship between 661 and 1984 is unclear, but the problem implies that when the number of points labeled -1 is sufficiently small, a "good point" must exist. We can use the relationship between 1984 and 661 to consider proving a generalized conclusion: in $3 n+1$ points, if there are $n$ points labele... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,527 |
Example 2: Does there exist 1000000 consecutive positive integers, each of which has a repeated prime factor, i.e., can be divided by the square of some prime number? | Analysis: If we only consider 1000000, our thinking will be blocked. We consider proving a generalized proposition: there exist $n$ consecutive positive integers, each of which has a repeated prime factor.
When $n=1$, we just need to take the square of any prime (such as $4, 9$, etc.), then the proposition holds.
Ass... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,528 |
Example 3 Let $n \geqslant 2$, in an $n \times n$ (i.e., $n$ rows and $n$ columns) number table, if no two rows are exactly the same, prove that it is always possible to delete one column so that in the remaining number table, no two rows are exactly the same. | Analysis: It is difficult to prove this problem directly using mathematical induction, mainly because it is hard to transition from $n \leqslant k$ to $n = k+1$. However, if we relax the conditions of the proposition and convert it into the following generalized problem, we can achieve unexpected results.
In an $m \ti... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,529 |
Example 4 Given $a_{1}=1, a_{2}=2$, $a_{n+2}=\left\{\begin{array}{ll}5 a_{n+1}-3 a_{n}, & a_{n} a_{n+1} \\ a_{n+1}-a_{n}, & a_{n} a_{n+1}\end{array}\right.$ is odd; prove that for all $n \in \mathbf{N}, a_{n} \neq 0$. (1988, National High School Mathematics Competition) | Analysis: Direct proof is not easy by hand, so we use the idea of specialization to examine the characteristics of the terms of the sequence.
From the given conditions, we can deduce that the first 9 terms of the sequence $\left\{a_{n}\right\}$ are $1,2,7,29,22,23,49,26,-17$. It can be observed that:
(1) The first 9 t... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,530 |
The second question: Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$. Find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$. | Solution: The conclusion is: the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1, and the maximum value is $\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}$.
$$
\text { By } \begin{aligned}
1 & =\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j<n} \sqrt{\frac{k}{j}} x_{k} x_{j} \\
& =\left(\sum_{i=1}^{n} x_{i}\right)^{2}-2 ... | \sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,533 |
Example 4 Given the equation $a^{2} x^{2}-\left(3 a^{2}-8 a\right) x+$ $2 a^{2}-13 a+15=0$ (where $a$ is a non-negative integer) has at least one integer root. Then, $a=$ $\qquad$ | Solution: Clearly, $a \neq 0$. Therefore, the original equation is a quadratic equation in $x$.
$$
\begin{aligned}
\Delta & =\left[-\left(3 a^{2}-8 a\right)\right]^{2}-4 a^{2}\left(2 a^{2}-13 a+15\right) \\
& =[a(a+2)]^{2}
\end{aligned}
$$
is a perfect square.
Thus, $x=\frac{\left(3 a^{2}-8 a\right) \pm a(a+2)}{2 a^{2... | 1,3,5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,534 |
Third question: Divide a rectangle with side lengths of positive integers $m$ and $n$ into several squares with side lengths of positive integers, where each square's sides are parallel to the corresponding sides of the rectangle. Try to find the minimum value of the sum of the side lengths of these squares.
---
Tran... | Solution: The minimum value of the sum of the side lengths of the squares $\Sigma$ is $m+n-d$, where $d$ is the greatest common divisor of $m$ and $n$, i.e., $d=(m, n)$. Below is the proof, divided into two steps:
The first step is to prove $\Sigma \geqslant m+n-d$. Use mathematical induction on the larger side length... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,535 |
Example 1 Let the sequence $\left\{a_{n}\right\}$ satisfy the following conditions:
$$
a_{1}=5, a_{n+1}=\frac{a_{n}^{2}+8 a_{n}+16}{4 a_{n}}(n=1,2, \cdots) \text {. }
$$
Prove: $a_{n} \geqslant 4(n=1,2, \cdots)$. | Analysis: Given $a_{1}=5$ and the recursive formula, we can deduce that $a_{n}>0$. The recursive formula can be written as
$$
a_{n+1}=\frac{1}{4}\left(a_{n}+\frac{16}{a_{n}}+8\right) \text {. }
$$
Since $a_{n}>0, \frac{16}{a_{n}}>0$,
we have $a_{n+1}=\frac{1}{4}\left(a_{n}+\frac{16}{a_{n}}+8\right)$
$$
\geqslant \frac... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,536 |
Example 2 The sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=1, F_{2}=$ 2, for any $n \in \mathbf{N}$, there is $F_{n+2}=F_{n+1}+F_{n}$. Prove: for any $n \in \mathbf{N}$, we have
$$
\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}} .
$$ | Analysis: Let $F_{0}=F_{2}-F_{1}=1$.
Then, $1=\frac{F_{k}}{F_{k+1}}+\frac{F_{k-1}}{F_{k+1}}, k=1,2, \cdots, n$.
Thus, $1=\frac{1}{n} \sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\frac{1}{n} \sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}}$
$$
\begin{array}{l}
\geqslant \sqrt[n]{\frac{F_{1}}{F_{2}} \cdot \frac{F_{2}}{F_{3}} \cdots \cdots... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,537 |
Example 3 Define the sequence $\left\{a_{n}\right\}$ as follows:
$$
a_{1}=\frac{1}{2}, 2 k a_{k}=(2 k-3) a_{k-1} .(k \geqslant 2)
$$
Prove: For all $n \in \mathbf{N}$, we have $\sum_{k=1}^{n} a_{k}<1$. | Analysis: The transformation of the recurrence relation can be achieved through the method of partial fraction decomposition to form a "consecutive difference" form.
From the given, we have $2(k+1) a_{k+1}=(2 k-1) a_{k}$, which leads to $-a_{k}=2\left[(k+1) a_{k+1}-k a_{k}\right]$.
$$
\begin{aligned}
-\sum_{k=1}^{n} a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,538 |
Example 4 Let $x_{0}=5, x_{n+1}=x_{n}+\frac{1}{x_{n}}(n=0$, $1,2, \cdots)$. Prove: $45<x_{1000}<45.1$. | Analysis: From the recurrence relation, we have $x_{n+1}^{2}=x_{n}^{2}+\frac{1}{x_{n}^{2}}+2$, $x_{n}>0$ and $x_{n+1}>x_{n}$. Therefore,
$$
\begin{aligned}
x_{n+1}^{2} & =x_{0}^{2}+\sum_{k=0}^{n}\left(x_{k+1}^{2}-x_{k}^{2}\right) \\
& =x_{0}^{2}+\sum_{k=0}^{n}\left(\frac{1}{x_{k}^{2}}+2\right) \\
& =25+2(n+1)+\sum_{k=0... | 45<x_{1000}<45.1 | Algebra | proof | Yes | Yes | cn_contest | false | 712,539 |
Example 5 Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=1, a_{n} a_{n+1}=$ $n+1(n \in \mathbf{N})$. Prove:
$$
\sum_{k=1}^{n} \frac{1}{a_{k}} \geqslant 2(\sqrt{n+1}-1) .
$$ | Analysis: Given $a_{2}=2, a_{n}>0$. When $n=1$, $\frac{1}{a_{1}}=1>2(\sqrt{2}-1)$, the proposition holds.
When $n \geqslant 2$, from $a_{n} a_{n+1}=n+1$, we get $a_{n} a_{n}=n$.
Therefore, $a_{n}\left(a_{n+1}-a_{n-1}\right)=1$.
Hence, $\frac{1}{a_{n}}=a_{n+1}-a_{n-1}$.
Thus, $\sum_{k=1}^{n} \frac{1}{a_{k}}=\frac{1}{a_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,540 |
Example 6 Let $n(n \geqslant 2)$ be an integer. Prove:
$$
\sum_{k=1}^{n-1} \frac{n}{n-k} \cdot \frac{1}{2^{k}}<4 .
$$ | Analysis: Let $a_{n}=\sum_{k=1}^{n-1} \frac{n}{n-k} \cdot \frac{1}{2^{k-1}}$. Below, we establish a recurrence relation for $a_{n}$ and then prove the conclusion.
$$
\begin{aligned}
a_{n+1}= & \sum_{k=1}^{n} \frac{n+1}{n+1-k} \cdot \frac{1}{2^{k-1}} \\
= & \frac{n+1}{n}+\frac{n+1}{n-1} \cdot \frac{1}{2}+\cdots+\frac{n+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,541 |
Example 7 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=$ $a_{n}+\frac{1}{a_{n}}$. Prove: $14<a_{100}<18$. | Analysis: When $k>1$, $a_{k}^{2}=a_{k-1}^{2}+2+\frac{1}{a_{k-1}^{2}}$, and $a_{k}>1$. Thus, $a_{k-1}^{2}+2<a_{k}^{2} \leqslant a_{k-1}^{2}+3$, $k=2,3, \cdots, n$. The equality holds only when $k=2$. Noting that $a_{1}=1$, summing up the above inequality, we get $2 n-1<a_{n}^{2}<3 n-2$. Therefore, $\sqrt{2 n-1}<a_{n}<\s... | 14<a_{100}<18 | Algebra | proof | Yes | Yes | cn_contest | false | 712,542 |
Example 8 Prove: $16<\sum_{k=1}^{80} \frac{1}{\sqrt{k}}<17$. | Analysis: From $\sqrt{k-1}<\sqrt{k}<\sqrt{k+1}$, we can obtain
$$
\frac{2}{\sqrt{k+1}+\sqrt{k}}<\frac{1}{\sqrt{k}}<\frac{2}{\sqrt{k}+\sqrt{k-1}},
$$
which means $2(\sqrt{k+1}-\sqrt{k})<\frac{1}{\sqrt{k}}<2(\sqrt{k}-\sqrt{k-1})$.
Summing up, we get
$$
\begin{array}{l}
2 \sum_{k=2}^{80}(\sqrt{k+1}-\sqrt{k})=2(\sqrt{81}-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,543 |
Example 9 Let $a_{0}=1, a_{n}=\frac{\sqrt{1+a_{n-1}^{2}}-1}{a_{n-1}^{2}}$ $(n=1,2, \cdots)$. Prove: $a_{n}>\frac{\pi}{2^{n+2}}$. | Analysis: Clearly, for $n \in \mathbf{N}$, we have $a_{n}>0$. According to the structural characteristics of the recurrence relation, let $a_{n} = \tan \alpha_{n}, \alpha_{n} \in \left(0, \frac{\pi}{2}\right)$.
From the given conditions, we have $a_{n} = \tan \alpha_{n} = \tan \frac{\alpha_{n-1}}{2}$.
Since $a_{0} = 1 ... | a_{n} > \frac{\pi}{2^{n+2}} | Algebra | proof | Yes | Yes | cn_contest | false | 712,544 |
Example 5 Let $m \in \mathbf{Z}$, and $4<m<40$, the equation $x^{2}-2(2 m-3) x+4 m^{2}-14 m+8=0$ has two integer roots. Find the value of $m$ and the roots of the equation. | Solution: Since the equation has integer roots, then
$$
\begin{aligned}
\Delta & =[-2(2 m-3)]^{2}-4\left(4 m^{2}-14 m+8\right) \\
& =4(2 m+1)
\end{aligned}
$$
is a perfect square.
Thus, $2 m+1$ is a perfect square.
Also, since $m \in \mathbf{Z}$ and $4<m<40$,
it follows that $2 m+1$ is a perfect square only when $m=12... | m=12, x_1=16, x_2=26; m=24, x_3=38, x_4=52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,545 |
Example 10 Let $n, k \in \mathbf{N}$. Prove:
$$
\sum_{k=1}^{n} \frac{1}{(k+1) \sqrt{k}}<2\left(1-\frac{1}{\sqrt{n+1}}\right) .
$$ | Analysis: To prove $a_{1}+a_{2}+\cdots+a_{n}) b_{n}$, it is sufficient to prove $a_{1}) b_{1}$, and $a_{k}) b_{k}-b_{k-1}(k \geqslant 2)$.
Let $a_{n}=\frac{1}{(n+1) \sqrt{n}}, b_{n}=2\left(1-\frac{1}{\sqrt{n+1}}\right)$.
Then $b_{1}=2\left(1-\frac{1}{\sqrt{2}}\right)=2-\sqrt{2}>\frac{1}{2}=a_{1}$.
For $k \geqslant 2$,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,546 |
Example 12 The sequence $\left\{a_{n}\right\}$ is a sequence of non-negative real numbers, and satisfies $a_{k}-2 a_{k+1}+a_{k+2} \geqslant 0, \sum_{i=1}^{k} a_{i} \leqslant 1, k=1,2, \cdots$. Prove: $0 \leqslant a_{k}-a_{k+1}<\frac{2}{k^{2}}(k=1,2, \cdots)$. | Analysis: According to the condition $\sum_{i=1}^{k} a_{i} \leqslant 1$, if there is some $a_{k} < a_{k+1}$, then $a_{k+1} \leqslant a_{k} - a_{k+1} + a_{k+2} < a_{k+2}$. Thus, starting from $a_{k}$, the sequence $\left\{a_{n}\right\}$ is monotonically increasing, and the sum $a_{1} + a_{2} + \cdots + a_{n}$ will tend ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,548 |
Example 13 Let $a_{1}, a_{2}, \cdots, a_{n}$ be a sequence of distinct natural numbers. Prove that: $\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{1}{k}$. | \begin{array}{l}\text { Analysis: Let } S_{k}=a_{1}+a_{2}+\cdots+a_{k} \\ \geqslant 1+2+\cdots+k=\frac{1}{2} k(b+\cdots) \text {, } \\ b_{k}=\frac{1}{k^{2}} \\ \text { then } \sum_{k=1}^{n} \frac{a_{k}}{k^{2}}=\sum_{k=1}^{n} a_{k} b_{k} \\ =S_{n} b_{n}+\sum_{k=1}^{n-1} S_{k}\left(b_{k}-b_{k+1}\right) \\ \geqslant \frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,549 |
1. Given that the average of the data $x_{1}, x_{2}, x_{3}$ is $a$, and the average of $y_{1}, y_{2}, y_{3}$ is $b$. Then the average of the data $2 x_{1}+3 y_{1}, 2 x_{2}+3 y_{2}, 2 x_{3}+$ $3 y_{3}$ is ( ).
(A) $2 a+3 b$
(B) $\frac{2}{3} a+b$
(C) $6 a+9 b$
(D) $2 a+b$ | $-1 . \mathrm{A}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,550 |
2. A home appliance mall once sold two air conditioners of different brands, one made a profit of $12\%$, and the other incurred a loss of $12\%$, and the total selling price of the two air conditioners was 30 thousand. Then, the profit of the mall in this transaction is: .
(A) Neither loss nor gain
(B) Loss of 90 yuan... | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,551 |
3. As shown in Figure 1, in quadrilateral
$A B C D$, one pair of opposite sides $A B$ $=C D$, the other pair of opposite sides $A D \neq$ $B C$, take the midpoints $M 、 N$ of $A D 、 B C$ respectively, and connect $M N$. Then the relationship between $A B$ and $M N$ is ( ).
(A) $A B=M N$
(B) $A B>M N$
(C) $A B<M N$
(D)... | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,552 |
4. Let the line $n x+(n+1) y=\sqrt{2}$ (where $n$ is a natural number) form a triangle with the coordinate axes, with the area of the triangle being $S_{n}(n=1,2, \cdots, 2000)$. Then the value of $S_{1}+S_{2}+\cdots+S_{2000}$ is ( ).
(A) 1
(B) $\frac{1999}{2000}$
(C) $\frac{2000}{2001}$
(D) $\frac{2001}{2002}$ | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,553 |
6. Person A and Person B start from point $A$ at the same time, traveling along the same road to point $B$, both using two different speeds $v_{1}$ and $v_{2}\left(v_{1}<\right.$ $v_{2}$ ). Person A uses speed $v_{1}$ for half of the distance and speed $v_{2}$ for the other half. Person B uses speed $v_{1}$ for half of... | $6 . B$ | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,555 |
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