problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Three. (25 points) Person A and Person B went to the same mall to buy a type of small handkerchief, each costing 0.50 yuan. The mall stipulates that a 20% discount is available for purchasing no less than 10 small handkerchiefs. As a result, A spent 4 yuan more than B, and it is known that the amount A spent does not e... | Three, let A and B buy $x$ and $y$ pieces of small handkerchiefs, respectively.
(1) When $x<10, y<10$, we have
$$
0.5 x - 0.5 y = 4 \text{, }
$$
which simplifies to $x = 8 + y$.
Thus, only $x = 9, y = 1$ satisfies the equation.
(2) When $x \geqslant 10, y < 10$, we have
$$
0.5 \times (1 - 20\%) x - 0.5 y = 4 \text{, }... | x = 9, 15, 20; y = 1, 4, 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,031 |
1. Given sets
$$
\begin{array}{l}
A=\left\{(x, y)\left\{\frac{y-3}{x}-2=a+1\right\},\right. \\
B=\left\{(x, y) \mid\left(a^{2}-1\right) x+(a-1) y=15\right\} .
\end{array}
$$
If $A \cap B=\varnothing$, then all possible values of $a$ are ( )
(A) $-1,1$
(B) $-1, \frac{1}{2}$
(C) $\pm 1, \frac{1}{2}$
(D) $\pm 1,-4, \frac... | - (D).
From the problem, when $a=1$, $B=\varnothing$, then we have $A \cap B=\varnothing$.
When $a=-1$, $a+1=-(a+1)=0$, and $A=$ $\{(x, 3) \mid x \neq 2, x \in \mathbf{R}\}, B=\left\{\left.\left(x,-\frac{15}{2}\right) \right\rvert\, x \in \mathbf{R}\right\}$, then we have $A \cap B=\varnothing$.
When $a, t \pm 1$, we ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,032 |
Example 3 Prove: By connecting any given 12 different positive integers with appropriate operation symbols (such as $+、-、 \times 、 \div$), the result obtained is always a multiple of 20790. | Analysis: First decompose 20790 to get
$$
20790=2 \times 3 \times 5 \times 7 \times 9 \times 11 \text {, }
$$
which is a product of 6 integers. We can also think that: in any given $i$ +1 numbers, there must be 2 numbers whose difference is a multiple of $i$.
Let the 12 positive integers be $a_{i}(i=1,2, \cdots, 12)$.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,033 |
2. Given the equation in $x$
$x^{2}-2 a x+a^{2}-4 a=0$ (where $a$ is a real number)
has at least one complex root with a modulus of 3. Then the possible values of $a$ are ( ).
(A) 1,9
(B) $-1,9,2-\sqrt{13}$
(C) $\pm 1,9,2+\sqrt{13}$
(D) $1,9,2-\sqrt{13}$ | 2. (D).
When a complex root is a real number, it is 3, then we have $3^{2}-2 a \times$ $3+a^{2}-4 a=0$, which is $a^{2}-10 a+9=0$, yielding $a=1,9$ (at this point $\left.\Delta=4 a^{2}-4\left(a^{2}-4 a\right)>0\right)$.
When the complex root is an imaginary number, we have $\Delta=4 a^{2}-4\left(a^{2}-4 a\right)$ $<0... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,034 |
3. Given $\triangle A B C, O$ is a point inside $\triangle A B C$, $\angle A O B=\angle B O C=\angle C O A=\frac{2 \pi}{3}$. Then the largest number $m$ such that
$$
A B+B C+C A \geqslant m(A O+B O+C O)
$$
holds is $(\quad$.
(A) 2
(B) $\frac{5}{3}$
(C) $\sqrt{3}$
(D) $\frac{\sqrt{3}}{2}$ | 3. (C).
As shown in Figure 2, draw $OD \perp BC$ intersecting $BC$ at point $D$, and $OE$ as the angle bisector of $\angle BOC$. Let $OB=x$, $OC=y$, $OA=z$. Then, by $\angle BOC=\frac{2\pi}{3}$ and the cosine rule, we have
$$
\begin{array}{l}
BC=\sqrt{x^{2}+xy+y^{2}}. \\
\text{And } S_{\triangle BOC}=\frac{1}{2} BC \c... | \sqrt{3} | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,035 |
4. Given $f(x)=a \sin x+b \sqrt[3]{x}+c \ln (x$ $\left.+\sqrt{x^{2}+1}\right)+4(a, b, c$ are real numbers $)$, and $f\left(\lg \log _{3} 10\right)=5$. Then the value of $f(\lg \lg 3)$ is ( ).
(A) -5
(B) -3
(C) 3
(D) varies with $a, b, c$ | 4. (C).
Since the function $y=\ln \left(x+\sqrt{x^{2}+1}\right)$ is an odd function, $g(x)=a \sin x+b \sqrt[3]{x}+c \ln \left(x+\sqrt{x^{2}+1}\right)$ is also an odd function, which means $f(x)-4$ is an odd function.
$$
\begin{aligned}
\text { Also, } 5 & =f(\lg \lg 10) \\
& =f\left(\lg \lg ^{-1} 3\right)=f(-\lg \lg 3... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,036 |
5. The last two digits of the integer $\left[\frac{10^{1995}}{10^{95}+3}\right]$ are ( ).
(A) 10
(B) 01
(C) 00
(D) 20 | 5. (C).
$$
\begin{array}{l}
\because A=\frac{10^{1995}}{10^{95}+3}=\left(10^{95 \times 20}-3 \times 10^{95 \times 19}+\cdots\right. \\
\left.-3^{19} \times 10^{95}+3^{20}\right)-\frac{3^{21}}{10^{95}+3}, \\
\begin{aligned}
\therefore[A]= & 10^{1900}-3 \times 10^{1805}+\cdots-3^{19} \times 10^{95} \\
& \quad+3^{20}-1 .
... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,037 |
6. A circle with radius $x$ is tangent to (at least one side) of a square with a side length of 1, and rolls inside it. After rolling around the square once, let the area of the region not covered by $\odot O$ be $S$. The value of $x$ that minimizes $S$ is $(\quad)$.
(A) $4-\pi$
(B) $\frac{1}{4}$
(C) $\frac{4}{20-\pi}$... | 6. (C).
When $0 \leqslant x < \frac{1}{4}$, the area of the 4 right angles and the small central square region is $S$, then we have
$$
S=(20-\pi)\left(x-\frac{4}{20-\pi}\right)^{2}+\frac{4-\pi}{20-\pi} .
$$
When $\frac{1}{4} \leqslant x \leqslant \frac{1}{2}$, the area of the 4 corner regions is $S$, then we have
$$
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,038 |
1. The set of values for the real number $a$ such that the equation $\cos 2 x+\sin x=a$ has real solutions is $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be part of the translation. Here is the correct translation:
1. The set of values for t... | $$
=1 .\left[-2, \frac{9}{8}\right] \text {. }
$$
Since $a=-2\left(\sin x-\frac{1}{4}\right)^{2}+\frac{9}{8}, -1 \leqslant \sin x \leqslant$ 1, to make the equation have real solutions, the maximum value of $a$ can be $\frac{9}{8}$ (which can be achieved when $\sin x=\frac{1}{4}$), and the minimum value of $a$ is obta... | \left[-2, \frac{9}{8}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,039 |
2. Let the set $M=\{x \mid 0 \leqslant x \leqslant 11, x \in \mathbf{Z}\}$, $F=\{(a, b, c, d) \mid a, b, c, d \in M\}$, and the mapping $f: F \rightarrow \mathbf{Z}$, such that $(a, b, c, d) \xrightarrow{f} a b - c d$. Given that $(u, v, x, y) \xrightarrow{f} 39$, $(u, y, x, v) \xrightarrow{f} 66$. Then the values of $... | 2. 1,9,8,6.
From the problem, we know
$$
\begin{array}{l}
u v - x y = 39, \\
u y - x v = 66,
\end{array}
$$
where \( u, v, x, y \) are non-negative integers and do not exceed 11. Therefore,
$$
\begin{array}{l}
(y + v)(u - x) = 105. \\
(y - v)(u + x) = 27.
\end{array}
$$
Given \( 0 \leqslant y \leqslant 11, 0 \leqslan... | x = 1, y = 9, u = 8, v = 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,040 |
3. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $x_{1999}-x_{601}=$ $\qquad$ . | 3.0 .
From $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$ we get $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$.
Let $x_{n}=\tan \alpha_{n}$, then
$$
x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) \text {. }
$$
Therefore, $x_{n+6}=x_{n}$,
which means the sequence $\left... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,041 |
4. If the surface area and volume of a circular cone are divided into upper and lower parts by a plane parallel to the base in the ratio $k$, then the minimum value of $c$ that makes $k c>1$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | 4.7.
As shown in Figure 3, let $C O_{1}=r_{1}, A O=r_{2}$, the surface area of the smaller cone is $S_{1}$. The lateral surface area of the frustum is $S_{2}$, then
$$
\frac{S_{2}+S_{\text {base }}}{S_{1}}=\frac{1}{k},
$$
which means
$$
\frac{S_{1}}{S_{\text {lateral }}+S_{\text {base }}}=\frac{k}{k+1} \text {. }
$$
... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,042 |
5. For sets $A$ and $B$, their union $A \cup B=\left\{a_{1}, a_{2}\right.$, $\left.\cdots, a_{n}\right\}$, when $A \neq B$, $(A, B)$ and $(B, A)$ are considered different pairs. Then the number of such pairs $(A, B)$ is $\qquad$ | $5.3^{n}$.
First consider $A, A$ has $k$ elements $(k=0,1,2, \cdots, n)$, there are $C_{i}^{k}$ $A$s. For each $A$, since $(A \cup B) \backslash A \subset B$, $B$ has $2^{k}$ possibilities for each element in $A$ to exist or not. Thus, according to the problem, the number of pairs $(A, B)$ that satisfy the requirement ... | 3^n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,043 |
For example, 41987 can be written as a three-digit number $x y z$ in base $b$. If $x+y+z=1+9+8+7$, try to determine all possible values of $x$, $y$, $z$, and $b$.
$(1987$, Canadian Mathematics Competition) | Analysis: It is easy to know that $x b^{2}+y b+z=1987, x+y+z=25$, thus,
$$
x\left(b^{2}-1\right)+y(b-1)=1962 \text {, }
$$
which means
$$
\begin{array}{l}
(b-1)[(b+1) x+y]=1962 \\
=2 \times 3^{2} \times 109 .
\end{array}
$$
From $b>10$ we know $b-1>9$;
From $1962 \geqslant b^{2}-1$ we know $b \leqslant \sqrt{1563}<45... | x=5, y=9, z=11, b=19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,044 |
6.12 friends have a weekly dinner together, each week they are divided into three groups, each group 4 people, and different groups sit at different tables. If it is required that any two of these friends sit at the same table at least once, then at least how many weeks are needed. | 6.5.
First, for any individual, sitting with 3 different people each week, it would take at least 4 weeks.
Second, there are $C_{12}^{2}=66$ pairs among 12 people. Each table has $\mathrm{C}_{4}^{2}=6$ pairs, so in the first week, $3 \times 6=18$ pairs get to know each other.
Since 4 people sit at 3 tables, after th... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,045 |
Three. (20 points) Given a cyclic hexagon $A B C D E F$ with sides satisfying the relation $A B=C D=E F=r, r$ being the radius of the circle, and $G, H, K$ are the midpoints of sides $B C, D E, F A$ respectively. Try to explain whether $\triangle G H K$ is an equilateral triangle. | Three, using the conclusion: If the three points $A, B, C$ of $\triangle ABC$ correspond to the complex numbers $a, b, c$ respectively, then the sufficient and necessary condition for $\triangle ABC$ to be an equilateral triangle is $a + w b + w^2 c = 0$, where $w = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$.
From the proble... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,046 |
Four. (20 points) Transporting goods from a riverside city $A$ to another location $B$, where the nearest point $C$ on the riverbank to $B$ is $b$ kilometers away, and the distance from $A$ to $C$ along the river is $a$ kilometers. If the waterway transportation cost is $\frac{1}{n}$ of the highway transportation cost ... | As shown in Figure 4, suppose a road is built from point $B$ to point $D$, i.e., $BD$ is the road, and then from point $D$ to point $A$ is changed to water transport, then we have
$$
\begin{array}{l}
A D=a-b \tan \varphi, \\
C D=b \tan \varphi, \\
B D=\frac{b}{\cos \varphi} .
\end{array}
$$
Thus, according to the prob... | S=k b \sqrt{n^{2}-1}+k a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,047 |
Five. (20 points) Given the function $f(x)=c x^{4}+b x^{3}+c x^{2}+d x$, satisfying:
(i) $a, b, c, a$ are all non-zero;
(ii) For any $x \in\{-2,-1,0,1,2\}$, $f(x)$ is an integer;
(iii) $f(1)=1, f(5)=70$.
Determine whether, for each integer $x$, $f(x)$ is an integer. | (1) Let $x= \pm 1$ to get $2 a+2 c, 2 b+2 d$ are integers; let $x= \pm 2$ to get $32 a+8 c, 16 b+4 d$ are integers. Therefore, $24 a, 12 b, 24 c, 12 d$ are all integers. Thus, we can let
$$
a=\frac{\alpha}{24}, b=\frac{\beta}{12}, c=\frac{\gamma}{24}, d=\frac{\delta}{12},
$$
where $\alpha, \beta, \gamma, \delta$ are p... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,048 |
一、(50 分) Let $K$ be the incenter of $\triangle A B C$, and points $C_{1}, B_{1}$ be the midpoints of sides $A B, A C$ respectively. Line $A C$ intersects $C_{1} K$ at point $B_{2}$, and line $A B$ intersects $B_{1} K$ at point $C_{2}$. If the area of $\triangle A B_{2} C_{2}$ is equal to the area of $\triangle A B C$, ... | Let $a=BC, b=CA, c=AB, b^{*}=AC_{2}, c^{*}=AB_{2}, s=\frac{1}{2}(a+b+\epsilon)$. Suppose the inradius of $\triangle ABC$ is $r$.
$\because S_{\triangle AC_{1}B_{2}}=\frac{1}{2} AC_{1} \cdot AB_{2} \sin A$,
$S_{\triangle ABC}=\frac{1}{2} AC \cdot AB \sin A$,
$\therefore$ From $C_{1}$ being the midpoint of $AB$ and $S_{\... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,049 |
(50 points) Given a natural number $n \geqslant 5$. Try to find:
(1). In the $n$-element set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, how many different numbers are produced by $a_{i}+a_{j}$ $(1<i<j \leqslant n)$ at least;
(2) Determine all $n$-element sets that achieve the above minimum value. | (1) Suppose $a_{1}<a_{2}<\cdots<a_{n}$, then it is clear that at least the following $2 n-3$ different numbers are generated:
$$
\begin{array}{l}
a_{1}+a_{2}<a_{1}+a_{3}<\cdots<a_{1}+a_{n}<a_{2}+a_{n} \\
<\cdots<a_{n-1}+a_{n} .
\end{array}
$$
(2) Clearly, a non-constant arithmetic sequence satisfies the requirement.
C... | 2n-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,050 |
Three. (50 points) Given a conic section $\Gamma$, with foci $F_{1}$ and $F_{2}$, and directrices $l_{1}$ and $l_{2}$, $P$ is a point on the conic section $\Gamma$. Draw a line through point $P$ parallel to the axis of symmetry of the conic section, intersecting $l_{1}$ and $l_{2}$ at points $M$ and $N$, respectively. ... | Three, (1) As shown in Figure 5, let the curve $\Gamma$ be an ellipse, with the equation of the ellipse being $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Then $F_{1}(a e, 0), F_{2}(-a e, 0), 0 \\
1, b=\sqrt{e^{2}-1} a,
\end{array}$
$P(a \sec \theta, b \tan \theta), M\left(\frac{a}{e}, b \tan \theta\right)$. By symmetr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,051 |
In $\triangle A B C$; $A B=A C, C D$ is the altitude, $D M \perp B C, D N \perp A C$, $M, N$ are the feet of the perpendiculars. When $C D=D M+D N$, find $\frac{A B}{B C}$ (do not use trigonometric functions). | Solution: It is obvious that $\angle A \neq 90^{\circ}$. We discuss in two cases:
(1) As shown in Figure 1, when $\angle A$ is an acute angle, let $BC=a, AB=$
$AC=b$. We have
$$
\begin{array}{l}
b \cdot CD=2 S_{\triangle ABC} \\
=2 S_{\triangle DBC}+2 S_{\triangle ACC} \\
=a \cdot DM+b \cdot DN \\
\Rightarrow b(CD-DN) ... | \frac{1+\sqrt{17}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,052 |
Given $a, b, c$ are Pythagorean numbers, and $a<b<c, p$ is an odd prime. Then the Pythagorean triplet has the property $a^{2}=(b+c) p$ if and only if $b+p=c$.
| Proof: Sufficiency.
If $b+p=c$, and the Pythagorean triple can generally be expressed as
$$
\left(u^{2}-v^{2}\right) d, 2 u v d,\left(u^{2}+v^{2}\right) d,
$$
where $u, v, d \in \mathbf{N}, u>v$, and $u$ and $v$ are coprime. Then
$$
c=\left(u^{2}+v^{2}\right) d.
$$
If $b=\left(u^{2}-v^{2}\right) d$, then $2 v^{2} d=p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,053 |
For $\triangle A B C$ with side lengths $a, b, c, n \geqslant 0$. Prove: $a^{n} \cos A+b^{n} \cos B+c^{n} \cos C \leqslant \frac{1}{2}\left(a^{n}+b^{n}+c^{n}\right)$. | Proof: According to symmetry, without loss of generality, let $a \geqslant b \geqslant c$, then $a^{n} \geqslant b^{n} \geqslant c^{n}$, and $\cos A \leqslant \cos B \leqslant \cos C$. Therefore,
$$
\left(a^{n}-b^{n}\right)(\cos A-\cos B) \leqslant 0,
$$
which means $a^{n} \cos A+b^{n} \cos B \leqslant a^{n} \cos B+b^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,054 |
Example 5 Arrange all powers of 3 and the sums of distinct powers of 3 in an increasing sequence:
$$
1,3,4,9,10,12,13, \cdots \text {. }
$$
Find the 100th term of this sequence. | Analysis: Write the known sequence in the form of the sum of powers of 3:
$$
\begin{array}{l}
a_{1}=3^{0}, a_{2}=3^{1}, a_{3}=3^{1}+3^{0}, \\
a_{4}=3^{2}, a_{5}=3^{2}+3^{0}, a_{6}=3^{2}+3^{1}, \\
a_{7}=3^{2}+3^{1}+3^{0}, \cdots
\end{array}
$$
It is easy to find that the terms correspond exactly to the binary represent... | 981 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,055 |
Find the smallest positive integer $n$, such that every $n$-element subset of $S=\{1,2, \cdots, 150\}$ contains 4 pairwise coprime numbers (it is known that $S$ contains a total of 35 prime numbers). | Solution: Consider the number of multiples of 2, 3, or 5 in $S$. We have
$$
\begin{array}{l}
{\left[\frac{150}{2}\right]+\left[\frac{150}{3}\right]+\left[\frac{150}{5}\right]-\left[\frac{150}{2 \times 3}\right]-\left[\frac{150}{2 \times 5}\right]} \\
-\left[\frac{150}{3 \times 5}\right] +\left[\frac{150}{2 \times 3 \ti... | 111 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,056 |
Example 6 The sequence $\left\{x_{n}\right\}$ is defined as follows:
$x_{1}=1$, for $k>0$,
$x_{2 k}=\left\{\begin{array}{l}2 x_{k}, \text { if } k \text { is even, } \\ 2 x_{k}+1, \text { if } k \text { is odd; }\end{array}\right.$
$x_{2 k+1}=\left\{\begin{array}{l}2 x_{k}, \text { if } k \text { is odd, } \\ 2 x_{k}+1... | Prove: Represent all numbers in binary, let $n=$ $\left(\overline{a_{m} a_{m-1} \cdots a_{1} a_{0}}\right)_{2}$, and use mathematical induction to prove: $x_{n}$ $=\left(\overline{b_{m} b_{m-1} \cdots b_{1} b_{0}}\right)_{2}$, where, $b_{m}=a_{m}=i, b_{i} \equiv s$ $a_{i}+a_{i+1}(\bmod 2), i=0,1,2, \cdots, m-1$.
When ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,057 |
Example 1 Let $n \in \mathbf{N}$. Prove: $1^{2001}+2^{2001}+$ $3^{2001}+\cdots+n^{2001}$ cannot be divisible by $n+2$. | Proof: Let $S_{n}=1^{2001}+2^{2001}+\cdots+n^{2001}$,
$$
\begin{array}{l}
\text { then } S_{n}= {\left[n^{2001}+(n-1)^{2001}+\cdots+2^{2001}\right] } \\
+1^{2001}, \\
2 S_{n}=2+\left(2^{2001}+n^{2001}\right)+\left[3^{2001}+(n-\right. \\
\left.1)^{2001}\right]+\cdots+\left(n^{2001}+2^{2001}\right) .
\end{array}
$$
Whe... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,060 |
Example 2 Let $a, b$ be positive real numbers, $\frac{1}{a}+\frac{1}{b}=1$. Try to prove: for every natural number $n$, we have
$$
(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1} \text {. }
$$
(1988, National High School Mathematics Competition) | Proof: From $\frac{1}{a}+\frac{1}{b}=1$ we get $a b=a+b$.
Since $a>0, b>0$, we have $a+b \geqslant 2 \sqrt{a b}$, which implies $a b \geqslant 2 \sqrt{a b}, a b \geqslant 4$.
$$
\text { Let } \begin{aligned}
S_{n}= & (a+b)^{n}-a^{n}-b^{n} \\
= & \mathrm{C}_{n}^{1} a^{n-1} b+\mathrm{C}_{n}^{2} a^{n-2} b^{2}+\cdots \\
& ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,061 |
Example 3 Suppose there are $n$ balls, and two people, A and B, play a game of taking balls in turns, each time one can take 1 or 2 balls at will, but not zero, and the one who takes the last ball loses. It is stipulated that A takes first. What are the winning and losing situations for A and B? | Solution: Since the loser ends up with only one ball, in the second-to-last turn, the number of balls must be 3 or 2. Therefore, as long as the number of balls left is 4, the first player will lose.
Therefore, when $n=3k+1(k \in \mathbf{N})$, A will lose. At this point, B can adopt the following strategy: if A takes 1... | n=3k+1 \text{ (A loses)}, n=3k \text{ or } 3k+2 \text{ (B loses)} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,062 |
Example 3 In $\triangle ABC$, $BD$ and $CE$ are angle bisectors, and $P$ is any point on $ED$. Perpendiculars are drawn from $P$ to $AC$, $AB$, and $BC$, with feet at $M$, $N$, and $Q$ respectively. Prove:
$$
P M + P N = P Q \text{. }
$$ | Proof: As shown in Figure 3, draw a line through point $P$ parallel to $AB$ intersecting $BD$ at $F$, and draw a line through point $F$ parallel to $BC$ intersecting $PQ$ and $AC$ at $K$ and $G$ respectively, then connect $PG$.
Since $BD$ bisects $\angle ABC$, the distances from point $F$ to $AB$ and $BC$ are equal. T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,063 |
Example 4 The sports meet lasted for $n$ days $(n>1)$, and $m$ medals were awarded. On the first day, 1 medal plus $\frac{1}{7}$ of the remaining $m-1$ medals were awarded, on the second day, 2 medals plus $\frac{1}{7}$ of the remaining medals were awarded, and so on, until on the $n$th day, the remaining medals were a... | Solution: Let the number of medals remaining after $k$ days of the sports meet be $x_{k}$ $(k \in \mathbf{N}, k>1)$, then the number of medals issued on the $k$-th day is
$$
\begin{array}{l}
k+\frac{1}{7}\left(x_{k-1}-k\right)=\frac{1}{7} x_{k-1}+\frac{5}{7} k, \\
x_{k}=x_{k-1}-\left(\frac{1}{7} x_{k-1}+\frac{6}{7} k\r... | 6, 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,064 |
Example 5. Does there exist a prime number that remains prime when 16 and 20 are added to it? If so, can the number of such primes be determined? | Solution: Testing with prime numbers $2, 3, 5, 7, 11, 13 \cdots$, we find that 3 meets the requirement. Below, we use the elimination method to prove that apart from 3, no other prime number satisfies the requirement.
Divide the natural numbers into three categories: $3n, 3n+1, 3n+2$ $(n \in \mathbb{N})$,
$\because 2n... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,065 |
Example 6 Find the integer points on the curve $4 x=y^{2}+5 y+2(0 \leqslant y \leqslant$ 50 ). | Solution: The problem is to find the coordinates of integer points on the curve.
Obviously, $x=\frac{y^{2}+5 y+2}{4}(0 \leqslant y \leqslant 50)$.
Divide the integer $y$ into four categories: $4 n, 4 n+1, 4 n+2$ $(n=0,1,2, \cdots, 12), 4 n+3(n=0,1,2, \cdots$, 11), and use the elimination method for discussion.
When $y=... | \left(4 n^{2}+7 n+2,4 n+1\right) \text{ and } \left(4 n^{2}+9 n+4,4 n+2\right)(n=0,1,2, \cdots, 12) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,066 |
Example 7 What is the largest even number that cannot be written as the sum of two odd composite numbers?
The largest even number that cannot be written as the sum of two odd composite numbers is 38.
Note: The original text did not provide the answer, this is based on known mathematical results. However, if you need... | Solution: First, prove that even numbers $n$ greater than 38 can all be written as the sum of two odd composite numbers.
Obviously, if the unit digit of $n$ is 0, then $n=5k+15$; if the unit digit of $n$ is 2, then $n=5k+27$; if the unit digit of $n$ is 4, then $n=5k+9$; if the unit digit of $n$ is 6, then $n=5k+21$; ... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,067 |
Example 1: Do there exist constants $a$, $b$, $c$ such that the equation
$$
\begin{array}{l}
1 \cdot 2^{2}+2 \cdot 3^{2}+\cdots+n(n+1)^{2} \\
=\frac{n(n+1)}{12}\left(a n^{2}+b n+c\right)
\end{array}
$$
holds for all natural numbers $n$? | Strategy: First, use the method of undetermined coefficients to find the values of $a$, $b$, and $c$, then use mathematical induction to prove that the equation holds for all natural numbers $n$.
Solution: Let $n=1, n=2, n=3$, we get
$$
\left\{\begin{array}{l}
\frac{1}{6}(a+b+c)=4, \\
\frac{1}{2}(4a+2b+c)=22, \\
9a+3b... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,068 |
Example 2 Given the equation of curve $c$ is
$$
(t+1) x^{2}+y^{2}-2\left(a^{2}+2 a t\right) x+3 a t+b=0 \text {. }
$$
For any real number $t$, curve $c$ always passes through the fixed point $P(1,0)$. Find the constants $a$ and $b$. | Strategy: Substitute $P(1,0)$ into the curve equation, isolate the parameter $t$, and then set the coefficients of the parameter to zero to determine $a$ and $b$.
Solution: Substituting $P(1,0)$ into the curve equation, we get
$$
t+1-2\left(a^{2}+2 a t\right)+3 a t+b=0 \text {. }
$$
Rearranging, we have
$$
\begin{arr... | a=1, b=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,069 |
Example 3 The parabolic equation $y^{2}=p(x+1)(p>$ $0)$, the line $x+y=t$ intersects the $x$-axis at a point to the right of the parabola's directrix. Prove: the line always intersects the parabola at two points. | Strategy: After combining two equations, prove that the discriminant $\Delta>0$ of the resulting equation.
Solution: The directrix of the parabola $l: x=-1-\frac{p}{4}$.
From the intersection point $(t, 0)$ of the line $x+y=t$ with the $x$-axis being to the right of the directrix $l$, we get
$$
t>-1-\frac{p}{4},
$$
wh... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,070 |
Example 4 Given $\cdots 1 \leqslant a \leqslant 1$, the inequality $\left(\frac{1}{2}\right)^{x^{2}+a x}<\left(\frac{1}{2}\right)^{2 x+a-1}$ always holds. Find the range of values for $x$. | Strategy: Convert the original inequality into an equivalent polynomial inequality, treating $a$ as the main variable, and express it as a linear inequality in $a$. Set up a linear function $f(a)$, and use the monotonicity of $f(a)$ to find the range of $x$.
Solution: The original inequality is transformed into
$$
x^{... | x>2 \text{ or } x<0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,071 |
Example 5 If the line $y=m x+b$ intersects the hyperbola $(x-1)^{2}-a^{2} y^{2}=a^{2}$ for any real number $m$, find the conditions that the real numbers $a$ and $b$ must satisfy. | Strategy: After combining the two equations, obtain a quadratic inequality about $m$ from $\Delta_{1} \geqslant 0$, isolate the parameter $m$, and then discuss to find the relationship between $a$ and $b$.
Solution: From $\left\{\begin{array}{l}y=m x+b, \\ (x-1)^{2}-a^{2} y^{2}=a^{2}\end{array}\right.$ we get
$$
\begin... | a^{2} b^{2}+a^{2}-1 \leqslant 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,072 |
Example 4 Let $M_{1}$ and $M_{2}$ be points on side $BC$ of $\triangle ABC$ such that $BM_{1} = CM_{2}$. Draw any line intersecting $AB$, $AC$, $AM_{1}$, and $AM_{2}$ at $P$, $Q$, $N_{1}$, and $N_{2}$, respectively. Prove that: $\frac{AB}{AP} + \frac{AC}{AQ} = \frac{AM_{1}}{AN_{1}} + \frac{AM_{2}}{AN_{2}}$. | Proof: As shown in Figure 4, if $P Q \parallel B C$, the conclusion is easily proven.
If $P Q$ is not parallel to $B C$, let $P Q$ intersect line $B C$ at $D$. Draw a line through point $A$ parallel to $P Q$ intersecting line $B C$ at $E$.
Given $B M_{1}=C M_{2}$, we know $B E+C E=M_{1} E + M_{2} E$. It is easy to se... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,074 |
Example 7 Given $0<a<1$,
$$
\begin{array}{l}
f(x)=\log _{a}(x+1), \\
g(x)=2 \log _{a}(2 x+t),
\end{array}
$$
for $x \in[0,1]$, $f(x) \geqslant g(x)$ always holds. Find the range of the real number $t$. | $$
\begin{array}{l}
\text { Solution: From } f(x) \geqslant g(x) \text { we get } \\
\log _{a}(x+1) \geqslant 2 \log _{a}(2 x+t) \\
\Leftrightarrow\left\{\begin{array}{l}
x+1>0, \\
2 x+t>0, \\
x+1 \leqslant(2 x+t)^{2}
\end{array}\right. \\
\Leftrightarrow t \geqslant \sqrt{x+1}-2 x .
\end{array}
$$
Let $u=\sqrt{x+1}$,... | t \geqslant 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,075 |
Example 8 Suppose for all real numbers $x$, the inequality
$$
\begin{array}{l}
x^{2} \log _{2} \frac{4(a+1)}{a}+2 x \log _{2} \frac{2 a}{a+1} \\
+\log _{2} \frac{(a+1)^{2}}{4 a^{2}}>0
\end{array}
$$
always holds. Find the range of real numbers $a$. | Strategy: Treat $x$ as the main variable and $a$ as the main parameter, transforming the inequality into $\log _{2} \frac{a+1}{2 a}>-\frac{3 x^{2}}{x^{2}-2 x+2}$. Since $-\frac{3 x^{2}}{x^{2}-2 x+2}=-\frac{3 x^{2}}{(x-1)^{2}+1} \leqslant 0$, it follows that $\log _{2} \frac{a+1}{2 a}>0$. Solving this inequality will gi... | 0<a<1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,076 |
Example 9 Given that the function $f(x)$ defined on $\mathbf{R}$ is an odd function, and is increasing on $[0,+\infty)$. For any $\theta \in \mathbf{R}$, find the real number $m$ such that $f(\cos 2 \theta-3)+f(4 m-$ $2 m \cos \theta)>0$ always holds. | Strategy: Utilize the properties of the function, isolate the parameter $m$, perform reasonable substitution, and determine the range of $m$ through extremum discussion.
Solution: The problem is equivalent to
$\cos 2 \theta-3>2 m \cos \theta-4 m$.
Since $2-\cos \theta \in[1,3]$, then
$$
\begin{array}{l}
m>\frac{2-\cos ... | m>4-2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,077 |
Example 10 Given $a>0$ and $a \neq 1$, the sequence $\left\{a_{n}\right\}$ is a geometric sequence with the first term $a$ and common ratio also $a$, let $b_{n}=$ $a_{n} \lg a_{n}(n \in \mathbf{N})$. If every term in the sequence $\left\{b_{n}\right\}$ is always less than the term that follows, find the range of values... | Strategy: Isolate the parameter $k$ from $b_{k+1}-b_{k}>0$, and determine the range of $a$ through extremum calculation.
Solution: $\because a_{n}=a \cdot a^{n-1}=a^{n}$,
$$
\therefore b_{n}=a_{n} \lg a_{n}=n a^{n} \lg a \text {. }
$$
From $b_{k+1}-b_{k}=a^{k}[k(a-1)+a] \lg a>0$,
we get
$$
[k(a-1)+a] \lg a>0 .
$$
Th... | 0<a<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,078 |
Example 11 Given the curve $y=x^{2}+(2 m-1) x+$ $m^{2}-2 m+1(m \in \mathbf{R})$. Does there exist a line $l$ with a defined slope that intersects the curve at only one point? If it exists, find this line; if not, explain the reason. | To solve the problem of converting the intersection issue into an equation issue, since $\Delta=0$, isolate the parameter $m$ and determine the values of $k$ and $b$.
Solution: Assume there exists a line $l$ with a defined slope that meets the requirements. Let the equation of line $l$ be $y=k x+b$.
$$
\text { From }\... | y=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,079 |
Example 12 Given $f(\theta)=\sin ^{2} \theta+\sin ^{2}(\theta+\alpha)$ $+\sin ^{2}(\theta+\beta)$, where $\alpha, \beta$ are constants satisfying $0 \leqslant \alpha \leqslant \beta \leqslant \pi$. For what values of $\alpha, \beta$ is $f(\theta)$ a constant value? | To make the value of $f(\theta)$ not change with $\theta$, i.e., the function $f(\theta)$ is a constant function, we can assign special values to the independent variable to explore.
$$
\begin{array}{l}
\text { Solution: Let } \theta=0, \frac{\pi}{6}, \frac{\pi}{2} \text {, we get } \\
f(0)=\sin ^{2} \alpha+\sin ^{2} \... | \alpha=\frac{\pi}{3}, \beta=\frac{2 \pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,080 |
Example 1 Given the parabola $y=x^{2}$, draw a line with a slope of 1 through the origin intersecting the parabola at $P_{1}$, draw a line with a slope of $\frac{1}{2}$ through $P_{1}$ intersecting the parabola at $P_{2}, \cdots \cdots$, generally, draw a line with a slope of $2^{-n}$ through the point $P_{n}$ intersec... | Given the conditions, we have
$$
\left\{\begin{array}{l}
\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}=2^{-n}, \\
y_{n}=x_{n}^{2} .
\end{array}\right.
$$
Substituting equation (2) into equation (1) and rearranging, we get
$$
x_{n+1}+x_{n}=2^{-n} \text {, }
$$
which can be rewritten as
$$
(-1)^{n+1} x_{n+1}=(-1)^{n} x_{n}-\left(... | \left(\frac{2}{3}, \frac{4}{9}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,081 |
Example 2 Given $a_{1}=1, a_{2}=\frac{5}{2}, a_{n+1}=$ $\frac{a_{n}+b_{n}}{2}, b_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$. Prove:
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .
$$ | Proof: From the characteristic of the simultaneous recurrence relations, we know that $\left\{a_{n}\right\}$ is an arithmetic mean sequence, and $\left\{b_{n}\right\}$ is a harmonic mean sequence. By considering the inequality of means, we can insert the geometric mean between $\left\{a_{n}\right\}$ and $\left\{b_{n}\r... | 2 | Algebra | proof | Yes | Yes | cn_contest | false | 712,082 |
Example 3 Given the sequences of positive integers $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy the recurrence relation $(1+\sqrt{3})^{2 n}=a_{n}+\sqrt{3} b_{n}$. Prove: $2^{n} \mid a_{n}$, $2^{n} \mid b_{n}$. | Prove: According to the binomial theorem, we have
$$
\begin{array}{l}
(1+\sqrt{3})^{2 n} \\
=\sum_{k=0}^{n} \mathrm{C}_{2 n}^{2 k} 3^{k}+\sum_{k=1}^{n} \mathrm{C}_{2 n}^{2 k-1}(\sqrt{3})^{2 k-1} \\
=\sum_{k=0}^{n} \mathrm{C}_{2 n}^{2 k} 3^{k}+\sqrt{3} \sum_{k=1}^{n} \mathrm{C}_{2 n}^{2 k-1} 3^{k-1} .
\end{array}
$$
No... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,083 |
Example 4 Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{n}=$ $a_{n-1} b_{n}, b_{n}=\frac{b_{n-1}}{1-a_{n-1}^{2}}(n \geqslant 2), a_{1}=p, b_{1}=$ $q$, and $p, q>0, p+q=1$. Find the general terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$. | Solution: When $n=1$, $a_{1}=p, b_{1}=q$.
When $n=2$,
$$
\begin{array}{l}
b_{2}=\frac{q}{1-p^{2}}=\frac{1-p}{1-p^{2}}=\frac{1}{1+p}, \\
a_{2}=a_{1} b_{2}=\frac{p}{1+p} .
\end{array}
$$
When $n=3$,
$$
\begin{array}{c}
b_{3}=\frac{\frac{q+p}{q+2 p}}{1-\frac{p^{2}}{(1+p)^{2}}}=\frac{1+p}{1+2 p}, \\
a_{3}=a_{2} b_{3}=\fra... | a_{n}=\frac{p}{1+(n-1) p}, b_{n}=\frac{1+(n-2) p}{1+(n-1) p} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,084 |
For example, $5 A D$ is the altitude of $\triangle A B C$, $K$ is a point on $A D$, $B K$ intersects $A C$ at $E$, and $C K$ intersects $A B$ at $F$. Prove: $\angle F D A=\angle E D A$. | Proof: As shown in Figure 5, draw a line through point $A$ parallel to $BC$, intersecting lines $DE$, $DF$, $BE$, and $CF$ at points $Q$, $P$, $N$, and $M$ respectively. Clearly,
$$
\frac{BD}{AN}=\frac{KD}{KA}=\frac{DC}{AM} \text{. }
$$
Thus, $BD \cdot AM = DC \cdot AN$.
From $\frac{AP}{BD}=\frac{AF}{FB}=\frac{AM}{BC}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,085 |
Example 1 Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)(n$ is an integer). Then, the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is ( ).
(A) 1991
(B) -1991
(C) $(-1)^{n} 1991$
(D) $(-1)^{n} 1991^{-1}$
(1991, National Junior High School Mathematics League) | Solution: From the given, let
$$
1991^{\frac{1}{n}}=x+t, -1991^{-\frac{1}{n}}=x-t \text {. }
$$
Then $-1=x^{2}-t^{2}$, which means $t^{2}=1+x^{2}$.
Also, $x+t>x-t$, so $t>0$.
Therefore, $t=\sqrt{1+x^{2}}$.
$$
\begin{array}{l}
\therefore\left(x-\sqrt{1+x^{2}}\right)^{n}=(x-t)^{n} \\
\quad=\left(-1991^{-\frac{1}{n}}\rig... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,086 |
Example 2 Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+\frac{9}{x}+\frac{4}{y}=10, \\
\left(x^{2}+9\right)\left(y^{2}+4\right)=24 x y .
\end{array}\right.
$$
(1993, Taiyuan Junior High School Mathematics Competition) | The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
x+\frac{9}{x}+y+\frac{4}{y}=10, \\
\left(x+\frac{9}{x}\right)\left(y+\frac{4}{y}\right)=24 .
\end{array}\right.
$$
From (1), let \( x+\frac{9}{x}=5+t \) and \( y+\frac{4}{y}=5-t \). Then from (2), we get
$$
(5+t)(5-t)=24,
$$
which sim... | x=3, y=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,087 |
Example 4 Given real numbers $x, y, z$ satisfy $x=6-y$, $z^{2}=xy-9$. Prove: $x=y$.
(1983, Tianjin Junior High School Mathematics Competition) | Proof: From $x=6-y$ we get $x+y=6$. Let $x=3+t, y=3-t$, then from the known equation we have
$$
\begin{array}{l}
z^{2}=(3+t)(3-t)-9, \\
z^{2}+t^{2}=0 . \\
\therefore t=z=0 .
\end{array}
$$
That is
$$
\begin{array}{l}
z^{2}+t^{2}=0 \\
\therefore t=z=0 .
\end{array}
$$
Therefore, $x=y$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,089 |
Example 5 Given $x+y+z=1$. Prove:
$$
x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3} \text {. }
$$
(Second Mathematical Competition of Ordzhonikidze, USSR) | Proof: From the given, we have \(x + y = 1 - z\).
Let \(x = \frac{1 - z}{2} + t, y = \frac{1 - z}{2} - t\). Then
\[
\begin{array}{l}
x^{2} + y^{2} + z^{2} \\
= \left(\frac{1 - z}{2} + t\right)^{2} + \left(\frac{1 - z}{2} - t\right)^{2} + z^{2} \\
= \frac{(1 - z)^{2}}{2} + z^{2} + 2 t^{2} \\
= \frac{3}{2}\left(z - \frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,090 |
Example 6 Let real numbers $a$, $b$, $c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0, \\
b^{2}+c^{2}+b c-6 a+6=0 .
\end{array}\right.
$$
Find the range of values for $a$.
(1995, Jilin Province Junior High School Mathematics Competition) | $$
\begin{array}{l}
\text { Solution: Adding (1) and (2) gives } \\
b^{2}+c^{2}=-a^{2}+14 a-13 \text {. } \\
\text { From (1) we get } b^{2} c^{2}=\left(a^{2}-8 a+7\right)^{2} \text {. } \\
\text { (4)(3) Let } b^{2}=\frac{-a^{2}+14 a-13}{2}+t \text {, } \\
c^{2}=\frac{-a^{2}+14 a-13}{2}-t, \\
\end{array}
$$
Substitut... | 1 \leqslant a \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,091 |
Example 7 Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}$ $=5$, let $s=x^{2}+y^{2}$. Then $\frac{1}{s_{\text {max }}}+\frac{1}{s_{\text {min }}}=$ $\qquad$ (1993, National High School Mathematics Competition) | Given:
$$
25 x^{2} y^{2}=\left[5-4\left(x^{2}+y^{2}\right)\right]^{2} \text {. }
$$
Let $s=x^{2}+y^{2}$, and set $x^{2}=\frac{s}{2}+t, y^{2}=\frac{s}{2}-t$. Substituting these into the equation and rearranging, we get:
$$
\frac{25}{4} s^{2}-(5-4 s)^{2}=25 t^{2} \geqslant 0 \text {, }
$$
which simplifies to:
$$
\left(... | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,092 |
Example 8 Given that $a, b, c$ are all real numbers, $a+b+c=0, abc=1$. Prove: $a, b, c$ at least one is greater than $\frac{3}{2}$.
(1991 "Shuguang Cup" Junior High School Mathematics Competition) | Proof: Given $a+b+c=0, abc=1$, we know that at least one of $a, b, c$ is positive. Without loss of generality, let $c>0$.
Let $a=-\frac{c}{2}+t, b=-\frac{c}{2}-t$. Then $\left(-\frac{c}{2}+t\right)\left(-\frac{c}{2}-t\right) c=1$,
i.e.,
$$
\begin{array}{l}
\frac{c^{2}}{4}-t^{2}=\frac{1}{c} . \\
\therefore \frac{c^{2}}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,093 |
Example 9 As shown in Figure
1, in $\square A B C D$, $\angle B A D$ is an obtuse angle, and $A C^{2} \cdot B D^{2}$ $=A B^{4}+A D^{4}$. Find the degree measure of $\angle B A D$.
(1986, Chengdu Junior High School Mathematics Competition) | Solution: By the properties of a parallelogram, we have
$$
\begin{array}{l}
A C^{2}+B D^{2}=2\left(A B^{2}+A D^{2}\right) \\
\text { Let } A C^{2}=A B^{2}+A D^{2}+t, \\
B D^{2}=A B^{2}+A D^{2}-t,
\end{array}
$$
Then, from the given equation, we get
$$
\begin{array}{l}
\left(A B^{2}+A D^{2}+t\right)\left(A B^{2}+A D^{2... | 135^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,094 |
For example, the area of $\triangle ABC$ is three times the area of its inscribed rectangle $PQRS$, and the values of side $BC$ and altitude $AU$ are both rational numbers. Under what conditions is the perimeter of rectangle $PQRS$ a rational number? Under what conditions is it an irrational number?
(1985, Guangzhou an... | Solution: As shown in Figure 2, let
$$
\begin{array}{c}
BC=a, AD=h, \\
PQ=x, PS=y. \text{ Then } \\
\frac{h-y}{h}=\frac{x}{a},
\end{array}
$$
Also, $xy=\frac{1}{3}$
$$
\cdot \frac{1}{2} a h=\frac{1}{6} a h \text{. }
$$
From (1), let $ay=\frac{ah}{2}+t, hx=\frac{ah}{2}-t$. Then
$$
x=\frac{a}{2}-\frac{t}{h}, y=\frac{h}... | a+h \pm \frac{\sqrt{3}}{3}(h-a) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,095 |
Example 6 In $\triangle A B C$, $A D$ is the median on side $B C$, point $M$ is on side $A B$, point $N$ is on side $A C$, and $\angle M D N=90^{\circ}$. If $B M^{2}+C N^{2}=D M^{2}+$ $D N^{2}$, prove that $A D^{2}=\frac{1}{4}\left(A B^{2}+A C^{2}\right)$.
(1997, Beijing Junior High School Mathematics Competition) | Proof: As shown in Figure 6, draw a line through point $B$ parallel to $AC$ intersecting the extension of $ND$ at $E$. Connect $ME$.
From $BD = DC$, we know $ED = DN$. Therefore, $\triangle BED \cong \triangle CND$.
Thus, $BE = NC$.
Clearly, $MD$ is the perpendicular bisector of $EN$. Hence,
$$
EM = MN.
$$
From $BM^2... | AD^2 = \frac{1}{4} (AB^2 + AC^2) | Geometry | proof | Yes | Yes | cn_contest | false | 712,096 |
Let $n$ be a fixed integer, $n \geqslant 2$.
a) Determine the smallest constant $c$ such that the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}
$$
holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n} \geqsla... | Solution: a) When non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ are not all 0, let
$$
\begin{array}{l}
x= \frac{\sum x_{i} x_{j}}{\left(\sum_{i=1}^{n} x_{i}\right)^{2}}, \\
y=\frac{\sum x_{i} x_{j} x_{k}\left(x_{i}+x_{j}+x_{k}\right)}{\left(\sum_{i=1}^{n} x_{i}\right)^{4}} . \\
\because \sum x_{i} x_{j}\left... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,097 |
Let $a, b, c$ be positive numbers, and $\sum$ denote the cyclic sum over the relevant quantities. Then for $n \geqslant 1$ we have
$$
\begin{array}{l}
\sum \frac{a^{n}}{b^{n}+c^{n}} \geqslant \sum \frac{a^{n-1}}{b^{n-1}+c^{n-1}}, \\
\sum \frac{a^{n+1}}{b^{n}+c^{n}} \geqslant \sum \frac{a^{n}}{b^{n-1}+c^{n-1}} .
\end{ar... | Proof: (1) Without loss of generality, let $a \geqslant b \geqslant c$, then it is easy to prove
$$
\begin{array}{l}
a^{n} b^{n-1}+a^{n} c^{n-1}-a^{n-1} b^{n}-a^{n-1} c^{n} \geqslant 0, \\
a^{n} c^{n-1}-a^{n-1} c^{n}+b^{n} c^{n-1}-b^{n-1} c^{n} \geqslant 0,
\end{array}
$$
and
$$
\begin{array}{l}
+\frac{b^{n} a^{n-1}+b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,098 |
Let $a_{i}, x_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n), \alpha \geqslant 1$. Then
$$
\begin{array}{l}
\quad\left(\sum_{i=1}^{n} a_{i}\right)^{\alpha} \sum_{i=1}^{n} x_{i}^{\alpha} \geqslant \sum_{j=1}^{n}\left(\sum_{i=1}^{n} a_{i} x_{i+j}\right)^{n} . \\
\left(x_{n+i}=x_{i}\right)
\end{array}
$$ | Prove: From $\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} b_{i}=\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i} b_{i+j}$ $\left(b_{n+i}=b_{i}\right)$, we get
$$
\sum_{i=1}^{n} b_{i}=\sum_{j=1}^{n} \frac{\sum_{i=1}^{n} a_{i} b_{i+j}}{\sum_{i=1}^{n} a_{i}}
$$
Let $b_{i}=x_{i}^{a}$. We get
$$
\sum_{i=1}^{n} x_{i}^{a}=\sum_{j=1}^{n} \frac{\su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,099 |
In optimization methods, there is an important inequality:
Let $a_{i}>0(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} a_{i}=1$, also $0<\lambda_{1} \leqslant \lambda_{2} \leqslant \cdots \leqslant \lambda_{n}$. Then
$$
\left(\sum_{i=1}^{n} \lambda_{i} a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{\lambda_{i}}\right) \leqslan... | Prove: For any real number $\mu>0$, there is always
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \lambda_{i} a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{\lambda_{i}}\right) \\
=\left(\sum_{i=1}^{n} \frac{\lambda_{i} a_{i}}{\mu}\right)\left(\sum_{i=1}^{n} \frac{\lambda_{i}}{\lambda_{i}}\right) \\
\leqslant=\frac{1}{4}\lef... | \left(\sum_{i=1}^{n} \lambda_{i} a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{\lambda_{i}}\right) \leqslant \frac{\left(\lambda_{1}+\lambda_{n}\right)^{2}}{4 \lambda_{1} \lambda_{n}} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,100 |
1. $a, b, c$ are rational numbers, and the equation $a+b \sqrt{2}+c \sqrt{3}=\sqrt{5+2 \sqrt{6}}$ holds. Then the value of $2 a+995 b+1001 c$ is ( ).
(A) 1999
(B) 2000
(C) 2001
(D) cannot be determined | $-1 .(\mathrm{B})$
$5+2 \sqrt{6}=(\sqrt{3}+\sqrt{2})^{2}$. Therefore, $a=0, b=1, c=1$. | 2001 | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,102 |
2. If $a b \neq 1$, and $5 a^{2}+2001 a+9=0$ and $9 b^{2}+2001 b+5=0$, then the value of $\frac{a}{b}$ is ( ).
(A) $\frac{9}{5}$
(B) $\frac{5}{9}$
(C) $-\frac{2001}{5}$
(D) $-\frac{2001}{9}$ | 2. (A).
$5 a^{2}+2001 a+9=0$,
$9 b^{2}+2001 b+5=0$ (obviously $b=0$ is not a solution of the equation)
$$
\Rightarrow 5 \cdot \frac{1}{b^{2}}+2001 \cdot \frac{1}{b}+9=0 \text {. }
$$
Therefore, $a$ and $\frac{1}{b}$ are both roots of the equation $5 x^{2}+2001 x+9=0$, but $a \neq \frac{1}{b}$. Given $\Delta>0$, that i... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,103 |
3. Given in $\triangle A B C$, $\angle A C B=90^{\circ}$, $\angle A B C=15^{\circ}, B C=1$. Then the length of $A C$ is ( ).
(A) $2+\sqrt{3}$
(B) $2-\sqrt{3}$
(C) 0.3
(D) $\sqrt{3}-\sqrt{2}$ | 3. (B).
As shown in Figure 4, construct $\angle E A \dot{D}=15^{\circ}$ intersecting $12 \mathrm{C}$ at $D$. Then $A D=$ $B D, \angle A D C=30^{\circ}$.
Let $A C=x$, then
$$
\begin{array}{l}
C D=\sqrt{3} x, \\
A D=2 x .
\end{array}
$$
Thus, $(2+\sqrt{3}) x=1$. Solving for $x$ yields $x=2-\sqrt{3}$. | 2-\sqrt{3} | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,104 |
4. As shown in Figure 1, in $\triangle A B C$, $D$ is a point on side $A C$. Among the following four situations, the one where $\triangle A B D \backsim \triangle A C B$ does not necessarily hold is ( ).
(A) $A D \cdot B C=A B \cdot B D$
(B) $A B^{2}=A D \cdot A C$
(C) $\angle A B D=\angle A C B$
(D) $A B \cdot B C=A ... | 4. (D).
It is obvious that (B) and (C) must hold. For (A), draw $B E \perp A C$ at $E$, and draw $D F \perp A B$ at $F$, then,
$$
D F=A D \sin A, B E=A B \sin A .
$$
From $A D \cdot B C=A B \cdot B D$, we get $D F \cdot B C=B E \cdot B D$.
Therefore, Rt $\triangle B D F \sim R t \triangle C B E$.
Thus, $\angle A B D=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,105 |
Example 7 As shown in Figure 7, $AB$ is the diameter of a semicircle, and $D$ is a point on $AB$. Points $E$ and $F$ are taken on the semicircle such that $EA=DA, FB=DB$. A perpendicular line is drawn from $D$ to $AB$, intersecting the semicircle at $C$. Prove that $CD$ bisects $EF$.
保留源文本的换行和格式，直接输出翻译结果。 | Proof: In Figure 7, draw perpendiculars from points $E$ and $F$ to $AB$, with $G$ and $H$ as the feet of the perpendiculars. Connect $FA$ and $EB$. It is easy to see that
\[
\begin{array}{l}
DB^{2} = FB^{2} = AB \cdot HB, \\
AD^{2} = AE^{2} = AG \cdot AB.
\end{array}
\]
Subtracting the two equations, we get
\[
DB^{2} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,107 |
6. A mall offers discounts to customers, stipulating:
(1) If a single purchase does not exceed 200 yuan, no discount is given;
(2) If a single purchase exceeds 200 yuan but does not exceed 500 yuan, a 10% discount is applied to the marked price;
(3) If a single purchase exceeds 500 yuan, the first 500 yuan is discounte... | 6. (C)
Obviously, 168 is less than $200 \times 0.9=180$, so it has not been discounted; 423 is less than $500 \times 0.9=450$, and greater than 200, so this is the price after a 10% discount; together, it is $168+423 \div 0.9=638>$ 500, according to (3), the amount to be paid is
$$
500 \times 0.9+138 \times 0.8=560.4 ... | 560.4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,108 |
1. Given point $P$ in the Cartesian coordinate system has coordinates $(0,1)$, $O$ is the origin, $\angle Q P O=150^{\circ}$, and the distance from $P$ to $Q$ is 2. Then the coordinates of $Q$ are $\qquad$. | 1. $Q( \pm 1,1+\sqrt{3})$ | Q( \pm 1,1+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,109 |
3. Given that $x$ and $y$ are positive integers, and $xy + x + y$ $=23, x^2y + xy^2=120$. Then $x^2 + y^2=$ $\qquad$ | 3.34 .
Let $x+y=s, xy=t$. Then $s+t=23, st=120$. We get $s=8, t=15$ or $s=15, t=8$ (discard). $x^{2}+y^{2}=(x+y)^{2}-2xy=s^{2}-2t=34$. | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,111 |
4. A positive integer, if added to 100 and 168 respectively, can result in two perfect squares. This positive integer is $\qquad$ | 4. $n=156$.
Let this number be $n$, and $n+168=a^{2}, n+100=b^{2}$. Then $a^{2}-b^{2}=68=2^{2} \times 17$,
which means $(a+b)(a-b)=2^{2} \times 17$.
However, $a+b$ and $a-b$ have the same parity,
so $a+b=34, a-b=2$.
Thus, $a=18$, and hence $n=156$. | 156 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,112 |
One. (Full marks 20 points) In the Cartesian coordinate system, there are three points $A(0,1), B(1,3), C(2,6)$. It is known that the points on the line $y=a x$ $+b$ with x-coordinates $0, 1, 2$ are $D, E, F$ respectively. Try to find the values of $a, b$ that minimize $A D^{2}+B E^{2}+C F^{2}$. | $F(2,2 a+b)$. From the graph, we can see
$$
\begin{array}{l}
A D^{2}+B E^{2}+C F^{2} \\
=(b-1)^{2}+(a+b-3)^{2}+(2 a+b-6)^{2} \\
= 5 a^{2}+6 a b+3 b^{2}-30 a-20 b+46 \\
= 5 a^{2}+(6 b-30) a+3 b^{2}-20 b+46 \\
= 5\left(a+\frac{3}{5} b-3\right)^{2}-5\left(\frac{3}{5} b-3\right)^{2}+3 b^{2}-20 b \\
+46 \\
= 5\left(a+\frac... | a=\frac{5}{2}, b=\frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,113 |
(1) Prove: If $x$ takes any integer value, the quadratic function $y=a x^{2}+b x+c$ always takes integer values, then $2a$, $a-b$, and $c$ are all integers.
(2) Write the converse of the above proposition, determine its truth, and prove your conclusion. | (1) If $x$ takes integer values, the quadratic function $y=a x^{2}+b x+c$ always takes integer values, then
When $x=0$, $y_{0}=a \cdot 0^{2}+b \cdot 0+c$ is an integer, so $c$ is an integer.
When $x=-1$,
$y_{-1}=a \cdot(-1)^{2}+b \cdot(-1)+c$ is an integer, thus $a - b = y_{-1} - y_{0}$ is an integer.
When $x=-2$,
$y_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,114 |
As shown in Figure $3, D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $F$ is a point on the extension of $BC$. $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the radiu... | Three, (1) Two circles are externally tangent.
Draw the tangent line $l$ of $\odot A B D$, then $\angle 1=\angle B$.
$$
\begin{array}{l}
\because \angle 3=\angle B+\angle C, \\
\therefore \angle 3=\angle 1+\angle C . \\
\because \angle 1+\angle 2=\angle 3=\angle 1+\angle C, \\
\therefore \angle 2=\angle C .
\end{array}... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,115 |
1. Given that $m$ and $n$ are integers, the equation
$$
x^{2}+(n-2) \sqrt{n-1} x+m+18=0
$$
has two distinct real roots, and the equation
$$
x^{2}-(n-6) \sqrt{n-1} x+m-37=0
$$
has two equal real roots. Find the minimum value of $n$, and explain the reasoning. | 1. $\left\{\begin{array}{l}n \geqslant 1, \\ (n-2)^{2}(n-1)-4(m+18)>0, \\ (n-6)^{2}(n-1)-4(m-37)=0 .\end{array}\right.$
(2) - (3), and rearrange to get
$$
(n-4)(n-1)>27.5 \text {, }
$$
then $n \geqslant 8$.
When $n=8$, $m=44$, which satisfies the given conditions. Therefore, the minimum value of $x$ is 8. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,116 |
2. Given that $M$ and $N$ are on the sides $DA$ and $AB$ of square $ABCD$ respectively, and $AM = AN$. A perpendicular line is drawn from $A$ to $BM$, with the foot of the perpendicular being $P$. Prove that $\angle APN = \angle BNC$.
| 2. As shown in Figure 3, from the given information
$$
\begin{array}{l}
\frac{A P}{A N}=\frac{A P}{A M}=\frac{B P}{A B} \\
=\frac{B P}{B C},
\end{array}
$$
$$
\text { and } \angle P A N=\angle P B C \text {, }
$$
then $\triangle P A N \sim \triangle P C B$.
Thus, $\angle P N A=\angle P C B$.
Therefore, points $P, N, B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,117 |
4. A water company charges for water usage as follows: for each household, the first 5 tons of water used in a month are charged at 0.85 yuan per ton, and any amount exceeding 5 tons is charged at a higher fixed rate. It is known that in July this year, the ratio of water usage between the Zhang family and the Li famil... | 4.1 .15
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "4.1 .15" does not contain any non-English content, the translation is the same:
4.1 .15 | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,123 |
9. The real number $x$ that satisfies the equation
$$
\begin{array}{l}
\sqrt{2 x^{2}+x+5} \cdot \sqrt{x^{2}+x+1} \\
=\sqrt{x^{2}-3 x+13}
\end{array}
$$
is | 9. $\frac{-3 \pm \sqrt{17}}{4}$ | \frac{-3 \pm \sqrt{17}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,128 |
Example 9 As shown in Figure $10, \odot O$ is the excircle of $\triangle A B C$ opposite to side $B C$, and $D, E, F$ are the points of tangency of $\odot O$ with $B C, C A$, and $A B$, respectively. If $O D$ intersects $E F$ at $K$, prove that $A K$ bisects $B C$. | Proof: As shown in Figure 10, draw a line through point $K$ parallel to $BC$ intersecting lines $AB$ and $AC$ at points $Q$ and $P$ respectively, and connect $OP$, $OQ$, $OE$, and $OF$.
Since $OD \perp BC$, it follows that $OK \perp PQ$.
Since $OF \perp AB$, it follows that points $O$, $K$, $F$, and $Q$ are concyclic. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,129 |
1. When $0<x \leqslant 1$, which of the following inequalities is correct? ( ).
(A) $\frac{\sin x}{x}<\left(\frac{\sin x}{x}\right)^{2} \leqslant \frac{\sin x^{2}}{x^{2}}$
(B) $\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin x}{x} \leqslant \frac{\sin x^{2}}{x^{2}}$
(C) $\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin \frac{... | -1 (B).
Because $0\left(\frac{\sin x}{x}\right)^{2}$. Also, since $f(x) = \frac{\sin x}{x}$ is a decreasing function on $\left(0, \frac{\pi}{2}\right)$, and $0 < x^{2} \leqslant x \leqslant 1 < \frac{\pi}{2}$, it follows that $\frac{\sin x^{2}}{x^{2}} \geqslant \frac{\sin x}{x}$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,131 |
2. Inside the circle $x^{2}+y^{2}-5 x=0$, there are three chords passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$ whose lengths form a geometric sequence. Then the range of the common ratio is ( ).
(A) $\left[\frac{\sqrt{2}}{\sqrt[4]{5}}, \frac{2}{\sqrt{5}}\right]$
(B) $\left[\sqrt[3]{\frac{4}{5}}, \fr... | 2. (C).
The equation $x^{2}+y^{2}-5 x:=0$ has a diameter of 5, so the maximum length of the chord passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$ is 5. The chord passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$ and perpendicular to the diameter through that point is the shortest, wit... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,132 |
3. Given $a, b \in \mathbf{R}^{+}, m, n \in \mathbf{R}, m^{2} n^{2}>$ $a^{2} m^{2}+b^{2} n^{2}$, let $M=\sqrt{m^{2}+n^{2}}, N=a+$ $b$. Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M<N$
(C) $M=N$
(D) Cannot be determined | 3. (A).
Obviously, $m^{2} n^{2} \neq 0$. Dividing both sides of the inequality by $m^{2} n^{2}$, we get $\frac{a^{2}}{n^{2}}+\frac{b^{2}}{m^{2}}\left(m^{2}+n^{2}\right)\left(\frac{a^{2}}{n^{2}}+\frac{b^{2}}{m^{2}}\right) \\
=a^{2}+b^{2}+\frac{m^{2}}{n^{2}} a^{2}+\frac{n^{2}}{m^{2}} b^{2} \\
\geqslant a^{2}+b^{2}+2 a b... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,133 |
5. If the complex number $z$ satisfies $|z|=1, A(-1,0)$, and $B(0,-1)$ are two points on the complex plane, then the function $f(z) = |(z+1)(\bar{z}-\mathrm{i})|$ reaches its maximum value when the figure formed by points $Z, A, B$ on the complex plane is ( ).
(A) Equilateral triangle
(B) Isosceles triangle
(C) Right t... | 5. (B).
Let $z=\cos \theta+\mathrm{i} \sin \theta$, then
$$
\begin{aligned}
& (z+1)(\vec{z}-\mathrm{i}) \\
= & (1+\cos \theta+\mathrm{i} \sin \theta)[\cos \theta-(1+\sin \theta) \mathrm{i}] \\
= & \cos \theta(1+\cos \theta)+\sin \theta(1+\sin \theta) \\
& +\mathrm{i}[\sin \theta \cdot \cos \theta-(1+\cos \theta)(1+\si... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,134 |
6. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, O$ is the center of the base $A B C D$, and $M, N$ are the midpoints of edges $C C_{1}$ and $A_{1} D_{1}$, respectively. Then the volume of the tetrahedron $O-$ $M N B_{1}$ is ( ).
(A) $\frac{1}{6}$
(B) $\frac{5}{48}$
(C) $\frac{1}{8}$
(D) $\frac{7}... | 6. (D):
As shown in Figure 3, draw a line through $O$ parallel to $AB$, intersecting edges $AD$ and $BC$ at points $E$ and $F$ respectively. Connect $BE$, and take the midpoint of $BF$ as $Q$. It is easy to see that $OQ \parallel BE \parallel B_{1}N$. Therefore, $OQ \parallel$ plane $MNB_{1}$. Hence, $V_{O M N B_{1}} ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,135 |
4. When $a>1$, if the inequality
$$
\begin{array}{l}
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n} \\
>\frac{7}{12}\left(\log _{a+1} x-\log _{a} x+1\right)
\end{array}
$$
holds for all positive integers $n$ not less than 2, then the range of $x$ is ( ).
(A) $21$ | 4. (D).
Let $f^{\prime}(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}$. Then
$$
\begin{aligned}
f(n+1) & =f(n)+\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1} \\
& =f(n)+\frac{1}{2 n+1}-\frac{1}{2 n+2} \\
& >f(n) .
\end{aligned}
$$
Therefore, $f(n)$ is an increasing function, so $f(n) \geqslant f(2)$.
Given that ... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,136 |
1. Let $x^{2}+y^{2}-2 x-2 y+1=0(x, y \in$
$\mathbf{R})$. Then the minimum value of $F(x, y)=\frac{x+1}{y}$ is $\qquad$ . | II. $1 . \frac{3}{4}$.
The original expression is transformed into $(x-1)^{2}+(y-1)^{2}=1$. Let $x-1=\cos \theta, y-1=\sin \theta$, then $x=1+\cos \theta, y=1+\sin \theta$. Therefore, $\frac{x+1}{y}=\frac{2+\cos \theta}{1+\sin \theta}$, let this be $\mu$. Then $\mu+\mu \sin \theta=2+\cos \theta$, which means $\mu \sin ... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,137 |
2. Let $\left|z_{1}\right|=\left|z_{2}\right|=a(a \neq 0)$, and $z_{1}+z_{2}$ $=m+m \mathrm{i}$, where $m$ is a non-zero real number. Then the value of $z_{1}^{3} \cdot z_{2}^{3}$ is $\qquad$ . | 2. $-a^{6} \mathrm{i}$.
Let $z_{1}=a(\cos \alpha+i \sin \alpha), z_{2}=(\cos \beta+i \sin \beta)$.
Then according to the given conditions, we have
$$
\left\{\begin{array}{l}
\cos \alpha+\cos \beta=\frac{m}{a}, \\
\sin \alpha+\sin \beta=\frac{m}{a},
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
2 \cos \... | -a^{6} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,138 |
3. Given $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a 、 b \in \mathbf{C}$. Then $a+a^{2}+a^{3}+\cdots+a^{2000}+b+b^{2}$ $+b^{3}+\cdots+b^{2000}=$ | 3. -2 .
From the known $f(-x)=f(x), g(-x)=-g(x)$ we get
$$
\begin{array}{l}
\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b x \\
=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x . \\
2^{-x}+\frac{a+b}{2^{-x}}=-\left(2^{x}+\frac{a+b}{2^{x}}\right) .
\end{array}
$$
Simplifying, from equation (1) we g... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,139 |
Example 1 Find the smallest positive integer $n$ that has exactly 144 different positive divisors, and among them, there are 10 consecutive integers.
(26th IMO Shortlist) | Analysis: According to the problem, $n$ is a multiple of the least common multiple of 10 consecutive integers, and thus must be divisible by $2, 3, \cdots, 10$. Since $8=2^{3} ; 9=3^{2}, 10=2 \times 5$, its standard factorization must contain at least the factors $2^{3} \times 3^{2} \times 5 \times 7$. Therefore, we ca... | 110880 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,140 |
4. In the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=1, a_{n+1}>a_{n}$, and $a_{n+1}^{2}+a_{n}^{2}+1=2\left(a_{n+1} a_{n}+a_{n+1}+a_{n}\right)$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n a_{n}}=$ $\qquad$ . | 4. $\frac{1}{3}$.
From the given, we have
$$
\begin{array}{l}
1+a_{n+1}^{2}+a_{n}^{2}+2 a_{n+1} a_{n}-2 a_{n+1}-2 a_{n} \\
=4 a_{n+1} a_{n},
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
\left(a_{n+1}+a_{n}-1\right)^{2}=4 a_{n} a_{n+1} . \\
\because a_{1}=1, a_{n+1}>a_{n}, \\
\therefore a_{n+1}+a_{n}-1>0 .
\... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,141 |
6. As shown in Figure 1, if
the three sides of $\triangle A B C$ are
$n+x, n+2 x, n$
$+3 x$, and the height $A D$ from $B C$ is $n$, where $n$
is a positive integer, and $0<x \leqslant$
1. Then the number of triangles that satisfy the above conditions is $\qquad$. | 6.12 .
Let $a=n+x, b=n+2 x, c=n+3 x, s=\frac{1}{2}(a+b+c)$. By Heron's formula and the general area formula, we have
$$
\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2} n(n+2 x) .
$$
Simplifying, we get $12 x=n$.
And $0<x=\frac{n}{12} \leqslant 1$,
so $0<n \leqslant 12$,
which means $n$ has exactly 12 possibilities (correspondin... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,143 |
Three. (Full marks 20 points) Let $y^{2}=2^{n+1} x(n=1,2$, $3, \cdots, 2000)$ be a family of parabolas. Draw a line with an inclination angle of $45^{\circ}$ through the focus of each parabola, and intersect the parabola to get the chord $A_{n} B_{n}(n=1,2, \cdots, 2000)$. By rotating each chord $90^{\circ}$ around its... | Three, the equation of the $n$-th parabola is $y^{2}=2^{n+1} x=2 \times 2^{n} x$, hence its focus coordinates are $\left(2^{n-1}, 0\right)$. The equation of the line passing through this focus with an inclination angle of $45^{\circ}$ is
$$
y=x-2^{n-1} \text {. }
$$
By solving the system of equations
$$
\left\{\begin{... | \frac{16}{3} \pi\left(2^{4000}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,144 |
Four, (Full marks 20 points) In $\triangle A B C$, if $\frac{\cos A}{\sin B}+$ $\frac{\cos B}{\sin A}=2$, and the perimeter of $\triangle A B C$ is 12. Find the maximum possible value of its area. | $$
\begin{array}{l}
\text { From the given, } \\
\sin A \cdot \cos A+\sin B \cdot \cos B=2 \sin A \cdot \sin B, \\
\text { i.e., } \sin A(\cos A-\sin B)+\sin B(\cos B-\sin A) \\
=0 \text {. } \\
\sin A\left[\sin \left(90^{\circ}-A\right)-\sin B\right] \\
+\sin B\left[\sin \left(90^{\circ}-B\right)-\sin A\right] \\
=2 \... | 36(3-2\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,145 |
Five. (Full marks 20 points) Given the function $f(x)=a x^{2} + b x + c$ whose graph passes through the point $(-1,0)$. Does there exist constants $a, b, c$ such that the inequality $x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$? If they exist, find the values of $a, b, c$; ... | Five, assuming there exist $a, b, c$ that meet the conditions.
$\because f(x)$'s graph passes through $(-1,0)$,
$\therefore f(-1)=0$, i.e., $a-b+c=0$.
Also, $\because x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$, let $x=1$, then
$$
\begin{array}{l}
1 \leqslant a+b+c \leqsla... | a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,146 |
1. After cutting off a triangle from a convex $n$-sided cardboard, the remaining polygon has an interior angle sum of $2160^{\circ}$. Then the value of $n$ ( ).
(A) can only be 12
(B) can only be 13
(C) can only be 14
(D) None of the above is correct | -1 (D).
The number of sides of a polygon with an interior angle sum of $2160^{\circ}$ is
$$
2160^{\circ} \div 180^{\circ}+2=14 \text { (sides). }
$$
(1) As shown in Figure 2, when cutting $\triangle A B C$ through two vertices $A$ and $C$ of an $n$-sided polygon, $n=14-1+2=15$.
(2) As shown in Figure 2, when cutting $\... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,147 |
2. For any natural numbers $n, m$ can divide $1992^{n}-999 n-1$. Then the maximum value of $m$ is ( ).
(A) 9
(B) 27
(C) 37
(D) 999 | 2. (D).
$\because 1999^{n}-999 n-1 \equiv 1^{n}-0-1 \equiv 0(\bmod 999)$,
$\therefore 999$ can divide $1999^{n}-999 n-1$.
$$
\therefore m_{\max } \geqslant 999 \text {. }
$$
When $n=1$, $1999^{n}-999 n-1=999$.
Then $m_{\text {max }} \leqslant 999$. Therefore, $m_{\text {max }}=999$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,148 |
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