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8. The speed of the yacht in still water is $x$ km/h, and the speed of the current is $y$ km/h. Pleasant Goat drives the yacht from downstream location $\mathrm{A}$ to upstream location $\mathrm{B}$, and then immediately returns to downstream $\mathrm{A}$. The time it takes for the yacht to travel from $\mathrm{A}$ to ... | $3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) Given the function
$$
f(x)=a \cos x+b \cos 2 x+c \cos 3 x,
$$
and $f(x) \geqslant-1$ always holds. Find the maximum value of $a-b+c$. | Let $x=\pi$. Then
$$
f(\pi)=-a+b-c \geqslant-1 \Rightarrow a-b+c \leqslant 1 \text {. }
$$
Thus, the maximum value of $a-b+c$ does not exceed 1.
Let $a=1, b=0, c=0$, at this point,
$$
f(x)=\cos x \geqslant-1
$$
always holds.
Therefore, the maximum value of $a-b+c$ is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Among all lines passing through the point $(\sqrt{1994}, 0)$, the number of lines that pass through two different rational points is | 3. 1 line.
Obviously, the line $l$ that meets the conditions is not parallel to the $y$-axis, so we can set its equation as $y=k(x-\sqrt{1994})$. Since $l$ passes through two rational points, let's assume they are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$. If $k$ is a rational number, $y_{1}=k\left(x_{1}-\... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 2 The number of sets $A$ that satisfy $\{a, b\} \subseteq A \varsubsetneqq \{a, b, c, d\}$ is $\qquad$.
| The number of sets $A$ that satisfy the condition is $2^{2}-1=3$.
Key to the solution The essence of this example is to find the number of proper subsets of the set $\{c, d\}$; generally, if $n, k \in \mathbf{N}$, $k<n$, then the number of sets $A$ that satisfy $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\} \subseteq A \v... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. (5 points) Use small cubes with edge length $m$ to form a large cube with edge length 12. Now paint the surface (6 faces) of the large cube red. If the number of small cubes with only one red face is equal to the number of small cubes with only two red faces, then $m=$ $\qquad$ . | 【Solution】Solution: According to the problem, there are $12 \div m$ small cubes on each edge of the large cube. Let $12 \div m=n$, which means there are $n$ small cubes on each edge of the large cube,
$$
\begin{array}{c}
6(n-2)^{2}=12(n-2) \\
(n-2)^{2}=2(n-2) \\
n-2=2 \\
n=4
\end{array}
$$
Since $12 \div m=4$
Therefor... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. Four people, A, B, C, and D, are competing in a table tennis tournament, where each pair of players will play one match. In the end, A defeated D, and A, B, and C won the same number of matches. How many matches did D win? | 10. 【Solution】Since a total of six matches were played, and "Jia, Yi, and Bing won the same number of matches," they either each won one match or each won two matches. If Jia, Yi, and Bing each won one match, Ding should have won three matches, but since Ding has already lost to Jia, it is impossible for him to win thr... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On the three sides of $\triangle A B C$, take points $P_{1}, P_{2}, \cdots, P_{6}, \cdots$, such that $P_{1}, P_{4}, P_{7}, \cdots$ are on $A C$, $P_{2}, P_{5}, P_{8}, \cdots$ are on $A B$, and $P_{3}, P_{6}, P_{9}, \cdots$ are on $B C$, and $A P_{1}=A P_{2}, B P_{2}=B P_{3}, C P_{3}=C P_{4}, A P_{4}=A P_{5}, B P_{5... | 6. A
Establish the complex plane, and let the letter at each point represent the complex number of that point. We have $P_{2}-A=e^{A_{1}}\left(P_{1}-\right.$ A), $P_{5}-A=e^{A i}\left(P_{4}-A\right)$. Subtracting these, we get $P_{5}-P_{2}=e^{A 4}\left(P_{4}-P_{1}\right)$. Similarly, $P_{6}-P_{3}=e^{R_{1}}$ $\left(P_{... | 0 | Geometry | MCQ | Yes | Yes | olympiads | false |
19.4.3 ** For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by $q_{i}=\left(q_{i-1}-1\right)^{3}+3, i=1,2, \cdots, n$. If each $q_{i}(i=1,2, \cdots, n)$ is a power of a prime, find the largest possible value of $n$. | Since $m^{3}-m=(m-1) m(m+1)$ is divisible by 3, we have $m^{3} \equiv m(\bmod 3)$. Therefore, $q_{i}-\left(q_{i-1}-1\right)^{3}+3 \equiv q_{i-1}-1(\bmod 3)$. Thus, one of $q_{1}, q_{2}, q_{3}$ must be divisible by 3, making it a power of 3. However, $\left(q_{i}-1\right)^{3}+3$ is a power of 3 only when $q_{i}=1$, so $... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 73$ an item is worth $C$ dollars, and it is sold for 100 dollars at a loss of $x \%$ of the selling price, then it is sold again at a profit of $x \%$ of the new selling price $S$. If the difference between $S$ and $C$ is $1 \frac{1}{9}$ dollars, then $x$ is
(A) undetermined.
(B) $\frac{80}{9}$.
(C) 10.
(D) $\... | [Solution] Let $C=100+100 \cdot x \% =100+x$, and $S=C+1 \frac{1}{9}=(100+x)+1 \frac{1}{9}=101 \frac{1}{9}+x$, also
$$
S=100+\frac{x}{100}\left(101 \frac{1}{9}+x\right) \text {. }
$$
From the above two equations, we have $\quad \frac{x^{2}}{100}+\frac{x}{90}-\frac{10}{9}=0$, solving this gives $x=10$ (the negative val... | 10 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 6 Given that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$, find the sum of the coefficients of this polynomial.
(2005 Western Olympiad Problem) | Solve for the sum of coefficients in the expansion of $\alpha^{k}+\beta^{k}$, given $\alpha+\beta=1, \alpha \beta=1$, denoted as $S_{k}$. Then $\alpha, \beta$ are the roots of the equation $x^{2}-x+1=0$. Solving this, we get
$$
\alpha=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}, \beta=\cos \frac{\pi}{3}-i \sin \frac{\pi}{3... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8 Let the function $f(x)=x^{3}+3 x^{2}+6 x+14$, and $f(a)=1, f(b)=19$, then $a$ $b=(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -2 | 8 From
$$
\begin{aligned}
f(x) & =x^{3}+3 x^{2}+6 x+14 \\
& =(x+1)^{3}+3(x+1)+10,
\end{aligned}
$$
let $g(y)=y^{3}+3 y$, then $g(y)$ is an odd function and monotonically increasing. And
$$
\begin{array}{l}
f(a)=(a+1)^{3}+3(a+1)+10=1, \\
f(b)=(b+1)^{3}+3(b+1)+10=19,
\end{array}
$$
so $g(a+1)=-9, g(b+1)=9, g(-b-1)=-9$,... | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. (1990 National High School Mathematics Competition) The number of elements in the set $\left\{(x, y) \left\lvert\, \lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y\right.\right\}$ is ( ).
A. 0
B. 1
C. 2
D. More than 2 | 5. Choose B. Reason: From $\lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg$,
we get $\left\{\begin{array}{l}x>0, \\ y>0, \\ x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}=x y .\end{array}\right.$
Therefore, the problem of the number of elements in the set of points is equivalent to the number of ordered pairs $(x,... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. In a sequence of numbers $x_{1}, x_{2}, \cdots$, it is known that $x_{1}=1$, and for $k \geqslant 2$,
$$
x_{k}=x_{k-1}+1-4\left(\left[\frac{k-1}{4}\right]-\left[\frac{k-2}{4}\right]\right) \text {. }
$$
Then $x_{2010}=(\quad$.
(A) 1
(B) 2
(C) 3
(D) 4 | Hint From the given, we have
$$
\begin{array}{l}
x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=1, \\
x_{6}=2, x_{7}=3, x_{8}=4, \cdots \cdots
\end{array}
$$
Since $2010=4 \times 502+2$, therefore,
$$
x_{2010}=x_{2}=2 \text{. }
$$
So the answer is (B). | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 5 A regular tetrahedron and a regular square pyramid have congruent lateral faces, so their lateral faces can coincide with each other. How many lateral faces does the polyhedron formed by combining them have?
(19th All-Russian Competition Problem) | As shown in Figure 23-12, the given quadrilateral pyramid is $P-ABCD$, and $l$ is the intersection line of faces $PAD$ and $PBC$. Point $E$ is on $l$ such that $EP=AD=BC$, then $PAEB$ is a tetrahedron with all faces congruent to the side faces of the quadrilateral pyramid.
In fact, $EP \parallel BC$ and $EP$ is equal ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. Person A participated in five tests, each with a full score of 100 points. A's score in each test is a natural number between 0 and 100, inclusive of 100. It is known that A scored the same in the first four tests and scored higher in the last test. If A's average score over the five tests is 82 points, then the nu... | 13. A.
Let the scores of the first four tests be $x$, and the score of the last test be $y$.
Then $4x + y = 410$. Clearly, $y$ is even.
Given $x < 82$.
If $y = 100$, then $x \notin \mathbf{Z}_{+}$. Therefore, the maximum value of $y$ is 98. Thus, the minimum value of $x$ is 78.
Upon verification, $x$ can be $78, 79, ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 8 (The 1st China Southeast Mathematical Olympiad) Given the inequality
$$
\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6
$$
holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$, find the range of values for $a$. | Let $\sin \theta + \cos \theta = x$, then $\cos \left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} x$, $\sin 2\theta = x^2 - 1$, and $x \in [1, \sqrt{2}]$.
Thus, the original inequality can be transformed into $(2a + 3)x + \frac{6}{x} - 2(x^2 - 1) > 0$.
Rearranging gives $(2x - 3)\left(x + \frac{2}{x} - a\right) ... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
13・11 Let $f(x)=|x-2|+|x-4|-|2 x-6|$, where $2 \leqslant x$ $\leqslant 8$, then the sum of the maximum and minimum values of $f(x)$ is
(A) 1 .
(B) 2 .
(C) 4 .
(D) 6 .
(E) None of the above values.
(33rd American High School Mathematics Examination, 1982) | [Solution] When $2 \leqslant x \leqslant 3$, $|x-2|=x-2,|x-4|=-(x-4)$,
$$
|2 x-6|=-(2 x-6),
$$
so at this time
$$
\begin{aligned}
f(x) & =(x-2)-(x-4)+(2 x-6) \\
& =-4+2 x,
\end{aligned}
$$
Similarly, when $3 \leqslant x \leqslant 4$,
$$
f(x)=(x-2)-(x-4)-(2 x-6)=8-2 x,
$$
When $4 \leqslant x \leqslant 8$,
$$
f(x)=(x-... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
2・76 Let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}, a_{2}, \cdots, a_{n}$ have the following property: for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
2. 77 In a $... | [Solution] The minimum value of $n$ is 8.
First, we prove that each number in $s$ appears at least twice in the sequence $a_{1}, a_{2}, \cdots, a_{n}$.
In fact, if a number in $s$ appears only once in this sequence, since there are 3 subsets containing this number, but in the sequence, there are at most two pairs of a... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) There are two docks, $A$ and $B$, on a river, with $A$ upstream and $B$ downstream. Two people, A and B, start from $A$ and $B$ respectively at the same time, rowing towards each other, and meet after 4 hours. If A and B start from $A$ and $B$ respectively at the same time, rowing in the same direction, ... | 【Analysis】In the problem of boats traveling in a stream, the speed at which two boats meet is the sum of their speeds, and the speed at which one boat catches up with the other is the difference in their speeds.
When two boats travel towards each other, the distance they cover is the distance between docks A and B. Wh... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
ii. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square of a natural number. | Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$
Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.
By verifying $n=2,4,6,8$ one by one, we find that $... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Taking the vertex $C(0,1)$ of the ellipse $x^{2}+a^{2} y^{2}=a^{2}(a>1)$ as the right-angle vertex, construct an inscribed isosceles right-angled triangle $A B C$ in this ellipse. The maximum number of such triangles that can be constructed is $\qquad$. | 10. 3
Obviously, such a triangle must exist. Without loss of generality, assume points $A$ and $B$ are located on the left and right sides of the $y$-axis, respectively. Let the slope of $CA$ be $k (k > 0)$, and the equation of the line $CA$ be $y = kx + 1$. Substituting this into the ellipse equation yields $\left(a^... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1 ;
(B) 2 ;
(C) $\pi$;
(D) 4 . | 4. (B)
For expressions containing absolute values, discuss cases to remove the absolute value symbols, resulting in four line segments.
$$
\begin{array}{c}
\left\{\begin{array}{l}
x-1 \geqslant 0 \\
y-1 \geqslant 0 \\
x+y \geqslant 3,
\end{array}\right. \\
\left\{\begin{array} { l }
{ x - 1 \leqslant 0 } \\
{ y - 1 \... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 6 As shown in Figure 7, the side length of square $A B C D$ is $1, P$ is any point on side $B C$ (it can coincide with points $B$ or $C$), and perpendiculars are drawn from points $B, C, D$ to the ray $A P$. The feet of the perpendiculars are $B^{\prime}, C^{\prime}, D^{\prime}$, respectively. Find the maximum ... | Prove that $S_{\triangle A P D}=\frac{1}{2} S_{A B C D}$,
$$
\begin{array}{c}
S_{\triangle A B P}+S_{\triangle A C P}=S_{\triangle A B C}=\frac{1}{2} S_{A B C D} \\
S_{\triangle A B P}+S_{\triangle \triangle A P}+S_{\triangle A P D}=S_{A B C D}=1 \text {, i.e., } \\
\frac{1}{2} A P \cdot B B^{\prime}+\frac{1}{2} A P \c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. If the real number $x$ satisfies $\arcsin x>\arccos x$, then the range of the expression
$$
f(x)=\sqrt{2 x^{2}-x+3}+2^{\sqrt{x^{2}-x}}
$$
is $\qquad$ | 3. $\{3\}$.
If $-1 \leqslant x \leqslant 0$, then $\arcsin x \in\left[-\frac{\pi}{2}, 0\right], \arccos x \in\left[\frac{\pi}{2}, \pi\right]$. At this time, $\arcsin x\sin (\arccos x) \\
\Rightarrow x>\sqrt{1-x^{2}} \Rightarrow|x|>\frac{\sqrt{2}}{2} \text {. } \\
\text { Also } 0<x \leqslant 1 \Rightarrow \frac{\sqrt{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(7) Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{20} x^{20}$, then $a_{0}+$ $a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=$ $\qquad$ | (7) -9 Hint: Let $x=0$, we get $a_{0}=1$. Differentiate both sides
$$
\begin{array}{c}
10\left(1+x-x^{2}\right)^{9}(1-2 x)=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1, a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=-10 \text {, thus } \\
a_{0}+a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=-9 .
\end{ar... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3、The first digit after the decimal point of $(1+\sqrt{2})^{2017}$ is
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
3、The first digit after the decimal point of $(1+\sqrt{2})^{2017}$ is | 3.0
Analysis: We consider the conjugate $(\sqrt{2}-1)^{2017}$. It is easy to know that $(1+\sqrt{2})^{2017}+(1-\sqrt{2})^{2017} \in Z$, and $0<(\sqrt{2}-1)^{2017}<1$. Since $(\sqrt{2}-1)^{4}<0.5^{4}=0.0625$, it follows that $(\sqrt{2}-1)^{2017}<0.0625$, meaning the first digit after the decimal point of $(\sqrt{2}-1)^... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k$ appears consecutively $k$ times, $k=1,3,5, \cdots$. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c$ $+d=$ | 6. 3 Detailed Explanation: From $x_{k}^{2}+1=x_{k}^{2}+2=\cdots=x_{(k+1)^{2}}=2k+1$, that is, when $k^{2}+1 \leqslant n \leqslant (k+1)^{2}$, $x_{n}=2k+1$.
$k=[\sqrt{n-1}]$, so $x_{n}=2[\sqrt{n-1}]+1$, thus, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.9 Define $\left\{a_{n}\right\}_{n=1}^{\infty}$ as follows:
$a_{1}=1989^{1989}$, and $a_{n}(n>1)$ is equal to the sum of the digits of $a_{n-1}$, what is $a_{5}$?
(21st Canadian Mathematics Competition, 1989) | [Solution] Since $a_{1}=1989^{1989}<10000^{1989}$,
and $10000^{1989}$ has $4 \cdot 1989+1=7957$ digits.
Therefore, $a_{1}$ has no more than 7957 digits.
By the problem statement, $a_{2}$ is the sum of the digits of $a_{1}$, then
$$
a_{2}<10 \cdot 8000=80000 .
$$
So $a_{2}$ is at most a 5-digit number, hence
$$
\begin{... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (1994 National High School Competition Question) Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
A. 5
B. 6
C. 7
D. 8 | 1. Choose C.
Simplified solution: The original recurrence relation can be transformed into $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right)$ (where $x=1$ is the root of the equation $3 x+x=4$). Let $b_{n}=a_{n}-1$, then $b_{n+1}=-\frac{1}{3} b_{n}, b_{1}=\frac{1}{3} b_{n}, b_{1}=a_{1}-1=8$, so $b_{n}$ is a geometric seq... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Given the sets
$$
\begin{array}{c}
M=\{x, x y, \lg (x y)\} \\
N=\{0,|x|, y\},
\end{array}
$$
and
and $M=N$, then
$\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\frac{1}{y^{3}}\right)+\cdots+\left(x^{2001}+\frac{1}{y^{2001}}\right)$ is equal to . $\qquad$ | 1. From the equality of sets, we know that the elements of the two sets are the same. Thus, there must be an element 0 in $M$. Also, by the definition of logarithms, $x y \neq 0$, so $x, y$ are not zero. Therefore, $\lg (x y)=0$, $x y=1$.
$$
\therefore \quad M=\{x, 1,0\}, N=\left\{0,|x|, \frac{1}{x}\right\} \text {. }
... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the maximum value of the natural number $a$ such that the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3 n+1}>2 a-5$ holds for all natural numbers $n$.
untranslated text remains unchanged. | 4. Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3 n+1}\left(n \in \mathbf{N}^{+}\right)$
$$
f(n+1)-f(n)=\frac{1}{3 n+2}+\frac{1}{3 n+4}-\frac{1}{3 n+3}=\frac{2}{3(n+1)(3 n+2)(3 n+4)}>0 \text {, }
$$
Therefore, $f(n+1)>f(n)$, so $f(n)$ is an increasing function on $\mathbf{N}^{+}$, and $f(n)_{\min }=f(1)=\frac... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
19.33 A convex $n$-sided polygon has a total of 27 diagonals, then the number of sides of this convex $n$-sided polygon is
(A) 11.
(B) 10.
(C) 9.
(D) 8.
(E) 7.
(Wuhan, Hubei Province, China Junior High School Mathematics Competition, 1985) | [Solution] The number of diagonals in a convex $n$-sided polygon is $\frac{1}{2} n(n-3)$. According to the problem, we have
$$
\frac{1}{2} n(n-3)=27 \text {, }
$$
which simplifies to
$$
n^{2}-3 n-54=0 \text {. }
$$
Solving this, we get $n=9$ or -3 (discard the negative solution).
Therefore, the answer is (C). | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
21. The number of three-digit positive integers that leave a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11 is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 21. E.
Notice that, $11-7=9-5=6-2=4$.
Then the number that meets the requirement is 4 less than the common multiple of 11, 9, and 6. The least common multiple of these numbers is $11 \times 3^{2} \times 2=198$, so the number that meets the requirement is $198k-4\left(k \in \mathbf{Z}_{+}\right)$.
Since this number is... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
8. For a plane $\alpha$, if the distances from all vertices of a polyhedron to the plane $\alpha$ are equal, then the plane $\alpha$ is called a "median plane" of the polyhedron.
(1) How many distinct median planes does a tetrahedron have?
(2) How many distinct median planes does a parallelepiped have?
(3) Given four n... | 8. (1) Answer: 7.
Consider the tetrahedron as $ABCD$.
If all four vertices are on the same side of plane $\alpha$, then these four vertices must lie on a plane parallel to plane $\alpha$, which does not meet the condition.
We only consider the following two scenarios.
(i) Three vertices on one side of plane $\alpha$ a... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given the curve $y=x^{3}-x$, draw a tangent line to the curve from a point $A(t, 0)$ on the $x$-axis, then the maximum number of tangent lines is $\qquad$. | 4. 3 Detailed Explanation: Let the point of tangency be $\left(x_{0}, x_{0}^{3}-x_{0}\right)$, then the slope of the tangent line is $k=3 x_{0}^{2}-1$, and the tangent line is $y-\left(x_{0}^{3}-x_{0}\right)=\left(3 x_{0}^{2}-1\right)\left(x-x_{0}\right)$, which simplifies to $y=\left(3 x_{0}^{2}-1\right) x-2 x_{0}^{3}... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
5. $[x]$ represents the greatest integer not exceeding the real number $x$, for example $[3]=3$, $[2.7]=2,[-2.2]=-3$, then the last two digits of $\left[\frac{10^{93}}{10^{31}+3}\right]$ are . $\qquad$ | 5. 08 Detailed Explanation: Since $\frac{10^{93}}{10^{31}+3}=\frac{\left(10^{31}\right)^{3}+3^{3}}{10^{31}+3}-\frac{3^{3}}{10^{31}+3}=10^{62}-3 \times 10^{31}+9-\frac{3^{3}}{10^{31}+3}$, therefore $10^{62}-3 \times 10^{31}+8<\frac{10^{93}}{10^{31}+3}<10^{62}-3 \times 10^{31}+9$, hence $\left[\frac{10^{93}}{10^{31}+3}\r... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Given real numbers $x, y, z$ satisfy $x+y+z \neq 0$, and $(x+y+z)^{2}=3+3(x y+y z+z x)$, then $\frac{x^{3}+y^{3}+z^{3}-3 x y z}{x+y+z}=$ | $3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16-14 As shown in the figure, point $B$ is on line segment $A C$, $M$ is the midpoint of line segment $A B$, $N$ is the midpoint of line segment $A C$, $P$ is the midpoint of $N A$, and $Q$ is the midpoint of $M A$.
Then the ratio $M N: P Q$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(2nd "Hope Cup" National Mathematics Invit... | [Solution] $\because M N=A N-A M$, then $P Q=\frac{A N}{2}-\frac{A M}{2}$. $\therefore \quad M N: P Q=(A N-A M): \frac{1}{2}(A N-A M)=2: 1$. Therefore, the answer is (B). | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
A car has a fuel tank capacity of 50 liters, and it departs from Shanghai to Harbin, which is 2560 kilometers away, with a full tank of fuel. It is known that the car consumes 8 liters of fuel for every 100 kilometers driven, and to ensure driving safety, at least 6 liters of fuel should be kept in the tank. Therefore,... | 【Analysis and Solution】
From Shanghai to Harbin, a total of $2560 \div 100 \times 8=204 \frac{4}{5}$ liters of fuel is consumed;
After refueling, the maximum fuel consumption before needing to refuel again is $50-6=44$ liters; $204 \frac{4}{5} \div 44=4 \frac{36}{55}$, using the ceiling method, we get 5 times;
But thes... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 64$ As shown in the figure, a beam of light emitted from point $A$ propagates in a plane, reflecting between the lines $AD$ and $CD$ several times before hitting point $B$ (point $B$ may be on $AD$ or $CD$) perpendicularly, and then reflecting back to point $A$ along the same path (at each reflection point, a... | [Solution] Let $\angle D A R_{1}=\theta$, and let $\theta_{i}$ be the angle of incidence (which is also the angle of reflection) at the $i$-th reflection point.
For $\triangle A R_{1} D$ and successively for $\triangle R_{i-1} R_{i} D, 2 \leqslant i \leqslant n$, up to $\triangle R_{n} B D$, using the exterior angle t... | 10 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Define a sequence: if $a_{n}=112$, then $a_{n+1}=2$ 112, which means reading the digits of $a_{n}$ from left to right as two 1s and one 2 to form $a_{n+1}=2112$. Following this rule, we get
$$
a_{n+2}=122112, a_{n+3}=11222112 .
$$
Given $a_{1}=13255569$, and following this rule to find $a_{100}$. Then, from $a_{1}$... | $-1 . B$.
According to the operation principle, if the numbers $4, 7, 8$ are to appear, they must appear in the number (let's assume this number is $a_{i}(i=3,4$, $\cdots, 100$, and $i$ is the smallest)) with the previous number $a_{i-1}$ having a continuous sequence of 4 or 7 or 8 numbers, i.e., $a_{i-1}$ has a part l... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
(2) The number of roots of the equation $10 \sin \left(x+\frac{\pi}{6}\right)=x$ is $(\quad)$.
(A) 8
(B) 7
(C) 6
(D) 5 | (2) By the properties of the sine function, draw the graphs of $y=10 \sin \left(x+\frac{\pi}{6}\right)$ and $y=x$ simultaneously, noting that
$$
3 \pi-\frac{\pi}{6}<10<3 \pi+\frac{\pi}{2}
$$
and
$$
\left|-\frac{\pi}{6}-3 \pi\right|<10<\left|-\frac{\pi}{6}-3 \pi-\frac{\pi}{2}\right|,
$$
we can see that the two curves ... | 7 | Calculus | MCQ | Yes | Yes | olympiads | false |
8.9 (1) Two people, A and B, are playing a game: A writes two rows, each containing 10 numbers, such that they satisfy the following rule: if $b$ is below $a$, and $d$ is below $c$, then $a+d=b+c$. Knowing this rule, B wants to determine all the numbers written. B can ask A questions like, "What is the number in the th... | [Solution] It is easy to see that, knowing all 10 numbers in one row and one number in another row, it is possible to reconstruct all other numbers. If only 10 numbers are known, then either each column contains exactly one known number, or there is a column where both numbers are unknown. In the former case, any numbe... | 11 | Other | math-word-problem | Yes | Yes | olympiads | false |
4. Given that $x, y$ are both positive numbers, and $\sqrt{x}(\sqrt{x}-2 \sqrt{y})=\sqrt{y}(6 \sqrt{y}-\sqrt{x})$, then $\frac{x+2 \sqrt{x y}+5 y}{2 x-3 \sqrt{x y}+y}=$ | $2$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. As shown in the figure, in the rectangular plate $ABCD$, $AB=3, BC=2$, with the edge $AD$ standing vertically on the ground. There is another vertical rod $MN$ on the ground, such that $NE$ is the perpendicular bisector of $AB$, with $E$ being the foot of the perpendicular, $NE=3$, $MN=5$. There is a point light so... | 10. 8 Extend $M D, M C$ to meet the ground at points $P, Q$, connect $N P, N Q, P Q$, extend $N E$ to meet $P Q$ at $F$, connect $M F$ to meet $C D$ at point $G$,
then $G E=2$, and $A B=3, B C=2$,
so $\frac{M N}{G E}=\frac{N F}{E F}$, i.e., $\frac{5}{2}=\frac{E F+3}{E F}$, thus $E F=2$,
similarly, we get $P Q=5$,
there... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
50. The weight of Big Bad Wolf is $2 \mathrm{~kg}$ less than 3 times the weight of Happy Sheep. Then 9 times the weight of Happy Sheep is $\qquad$ kg more than 3 times the weight of Big Bad Wolf. | answer: 6 | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1's common left and right vertices, $P$ and $Q$ are moving points on the hyperbola and ellipse, respectively, different from $A$ and $B$, and satisfy: $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\widehat{A Q}+\overrightarrow{B Q})(\lambda \in \mathbf{R}$ and $|\lambda|>1)$. Let the slopes of the lines $A P$, $B ... | (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and we know $A(-a, 0), B(a, 0)$, then we have
$$
\begin{array}{c}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \frac{x_{1}}{y_{1}}, \\
k_{3}+k_{4}=\frac{y_{2}}{x_{2}+a}+\frac{y_{2}}{x... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Let the graph of the parabola $y=a x^{2}+b x+c$ pass through the points $A(-1,-3) 、 B(4,2) 、 C(0,2)$. Let $P$ be a moving point on the axis of symmetry of the parabola. If $P A+P C$ takes the minimum value when the coordinates of point $P$ are $(m, n)$, then $n=$ $\qquad$ | 8.0.
From the condition, we know that the axis of symmetry of the parabola $l: x=2$.
Thus, $m=2$.
Let $l_{A B}: y=k x+b$. Then
$$
\left\{\begin{array} { l }
{ - 3 = - k + b , } \\
{ 2 = 4 k + b }
\end{array} \Rightarrow \left\{\begin{array}{l}
k=1, \\
b=-2,
\end{array}\right.\right.
$$
Thus, $l_{A B}: y=x-2$.
By sym... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. There are 35 books in total on a bookshelf, including storybooks, science books, picture books, and dictionaries, with each type of book having a different number of copies. Among them, storybooks and science books total 17 copies, and science books and picture books total 16 copies. One type of book has 9 copies, s... | 【Analysis】There are 18 books in total of albums and dictionaries, so it cannot be albums and dictionaries. If there are 9 storybooks, then there would be 8 science books, which conflicts with 8 albums; If there are 9 science books, 8 storybooks, 7 albums, and 11 dictionaries, it meets the requirements, so it is the sci... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$30.11 \times 12 \times 13 \times 14 \times 15 \times 16 \times 17 \times 18$, the unit digit of the product is | Answer: 0 | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Peppa Pig is fetching water from a well. At the beginning, the bucket is just on the water surface. If Peppa turns the handle 12 times, the bucket is still 2 meters away from the well's mouth; if Peppa turns the handle 16 times, the bucket just reaches the well's mouth. Then the distance from the water surface to th... | $8$ | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 9】Nine beauties, some are always sincere and tell the truth as angels, the rest are always flattering and tell lies as devils. The first beauty says: “There is exactly 1 devil among us”; the second beauty says: “There are exactly 2 angels among us”; the third beauty says: “There are exactly 3 devils among us”... | Analysis:
"Among us, there is exactly 1 devil," "Among us, there are exactly 3 devils," "Among us, there are exactly 5 devils," "Among us, there are exactly 7 devils," "Among us, there are exactly 9 devils" — among these 5 statements, at most 1 can be true; similarly, "Among us, there are exactly 2 angels," "Among us, ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. A gardener is preparing to plant a row of 20 trees, with two types of trees available: maple trees or sycamore trees. The number of trees between any two maple trees (not including these two maple trees) cannot be equal to 3. How many maple trees can there be at most among the 20 trees? $\qquad$ | 【Answer】 12
【Solution】In any continuous eight trees, once a maple tree is planted, it means that another position can only plant a sycamore tree. We use the following diagram to illustrate, using — to represent a maple tree, and ○ to represent a sycamore tree. Once the second position is planted with a maple tree, posi... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4} \cos x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4020}-a_{4027}-a_{4028}
$$
the value is $\qquad$ . | 6.2.
$$
\text { Let } x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right) \text {. }
$$
Then $x^{2}+x+2=1$.
So $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 37$ Three non-zero numbers $a, b, c$ form an arithmetic sequence. When $a$ is increased by 1 or $c$ is increased by 2, these three numbers form a geometric sequence. Then $b$ equals
(A) 16.
(B) 14.
(C) 12.
(D) 10.
(E) 8.
(14th American High School Mathematics Examination, 1963) | [Solution] Given $2 b=a+c$, and $b^{2}=c(a+1)$, also $b^{2}=a(c+2)$. $\therefore c=2 a$, then $a=\frac{2}{3} b$.
$\therefore b^{2}=\frac{8}{9} b^{2}+\frac{4}{3} b$, solving this gives $b=12$. Therefore, the answer is $(C)$. | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
Rixita $\star \star$ Find the unit digit of $\left[\frac{10^{20000}}{10^{100}+3}\right]$. | Since $\frac{3^{200}}{10^{100}+3}<1, I=\frac{10^{20000}-3^{200}}{10^{100}+3}$ is an integer, so $\left[\frac{10^{20000}}{10^{100}+3}\right]=I$. Also, $I \equiv \frac{-3^{200}}{3} \equiv-3^{199} \equiv-3^{3} \cdot\left(3^{4}\right)^{49} \equiv-27 \equiv 3(\bmod 10)$, hence the unit digit of $\left[\frac{10^{20} 000}{10^... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given $x \in\left(0, \frac{\pi}{2}\right)$, then the integer part of the number $m=3^{\cos ^{2} x}+3^{\sin ^{5} x}$ is
A. 2
B. 3
C. 4
D. Cannot be determined | 3. $\mathrm{B}$
Let $t=\sin ^{2} x$, then $0<t<1$, so $m=3^{t}+3^{1-t}=3^{t}+\frac{3}{3^{t}}, 3^{t} \in(1,3)$, hence $2 \sqrt{3} \leqslant m<4$. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
10. Let $a, b, c \in \mathbf{R}$, and there exist $\alpha, \beta, \gamma \in\{-1,1\}$, such that $a \alpha + b \beta + c \gamma = 0$, find the minimum value of $f=\left(\frac{a^{3}+b^{3}+c^{3}}{a b c}\right)^{2}$. | 10. Solution: $\because \quad \alpha, \beta, \gamma \in\{-1,1\}, \therefore$ by the pigeonhole principle, it is known that among the three real numbers $\alpha, \beta, \gamma$, at least two are equal. (1) All three real numbers are equal, $\alpha=\beta=\gamma \in\{-1,1\}$, then it must be that $a+b+c=0$, thus $f=\left(... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. (6 points) December 31, 2013, was a Tuesday, so, June 1, 2014, is $\qquad$. (Answer with a number: Monday is represented by 1, Tuesday by 2, Wednesday by 3, Thursday by 4, Friday by 5, Saturday by 6, and Sunday by 7.) | 【Analysis】First, calculate the number of days from December 31 to June 1, then determine how many weeks and extra days are in this period, and finally judge based on the remainder.
【Solution】Solution: $2014 \div 4=503 \cdots 2$
So this year is a common year, February has 28 days,
$1, 3, 5$ are large months with 31 day... | 7 | Other | math-word-problem | Yes | Yes | olympiads | false |
9. Let $[x]$ denote the greatest integer not exceeding the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is . $\qquad$ | Solving
Since $\lg ^{2} x-[\lg x]-2=0 \Rightarrow \lg ^{2} x=[\lg x]+2 \geqslant 0 \Rightarrow[\lg x] \geqslant-2$.
When $[\lg x]=-2$, $\lg ^{2} x=0$, no solution; When $[\lg x]=-1$, $\lg ^{2} x=1 \Rightarrow x=\frac{1}{10}$;
When $[\lg x]=0$, $\lg ^{2} x=2$, no solution; When $[\lg x]=1$, $\lg ^{2} x=3 \Rightarrow \lg... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 22 (Problem 21 of the 1989 Canadian Mathematics Competition) Define $\left\{a_{n}\right\}_{n=1}^{\infty}$ as follows: $a_{1}=1989^{1989}$, and $a_{n} (n>1)$ is equal to the sum of the digits of $a_{n-1}$, what is $a_{5}$? | Since $a_{1}=1989^{1989}<10000^{1989}$,
and $10000^{1989}$ has $4 \cdot 1989+1=7957$ digits.
Therefore, $a_{1}$ has no more than 7957 digits.
By the problem, $a_{2}$ is the sum of the digits of $a_{1}$, then
$$
a_{2}<10 \cdot 8000=80000 \text {. }
$$
So $a_{2}$ is at most a 5-digit number, hence
$$
a_{3} \leqslant 7+4... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) The figure is a quadrilateral composed of 18 small equilateral triangles of the same size. Some adjacent small equilateral triangles can form several larger equilateral triangles. Therefore, the number of large and small equilateral triangles containing the “**” mark in the figure is .
Preserve the ori... | 5. (3 points) The figure is a quadrilateral composed of 18 small equilateral triangles of the same size. Some adjacent small equilateral triangles can form several larger equilateral triangles. Therefore, the number of large and small equilateral triangles containing “*” in the figure is 6.
【Solution】Solution: Observin... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$, $f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, then $g(2002)=$ $\qquad$ . | $$
\begin{array}{l}
f(x)+5 \leqslant f(x+5) \leqslant f(x+4)+\leqslant f(x+3)+2 \leqslant f(x+2)+3 \\
\leqslant f(x+1)+4 \leqslant f(x)+5 \Rightarrow f(x+1)=f(x)+1 \Rightarrow f(2002)=2002 .
\end{array}
$$
So $g(2002)=f(2002)+1-2002=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given real numbers $x, y, z$ satisfy
$$
x^{2}+2 y^{2}+3 z^{2}=24 \text {. }
$$
Then the minimum value of $x+2 y+3 z$ is $\qquad$ | 4. -12 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x+2 y+3 z)^{2} \\
=(1 \times x+\sqrt{2} \times \sqrt{2} y+\sqrt{3} \times \sqrt{3} z)^{2} \\
\leqslant\left[1^{2}+(\sqrt{2})^{2}+(\sqrt{3})^{2}\right]\left(x^{2}+2 y^{2}+3 z^{2}\right)
\end{array}
$$
$$
=144 \text {. }
$$
Therefore, $x+2 y+3 z \geqsl... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
23. Find the smallest positive integer $n(\geqslant 3)$, such that in any set of $n$ points in the plane with no three points collinear, there must be three points that are the vertices of a non-isosceles triangle.
(2005 China National Training Team Problem) | 23. First, when 6 points in a plane are the 5 vertices of a regular pentagon and its center, all triangles formed by these 6 points are isosceles triangles, so the required positive integer \( n \geqslant 7 \). Next, if there exist 7 points in a plane (where no three points are collinear) such that all triangles formed... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 1.3 Divide 4 people into two groups, with at least 1 person in each group, and find the number of different grouping methods. | Solution: Let the required number be $N$. Using Jia, Yi, Bing, and Ding to represent the 4 people, the $N$ grouping methods that satisfy the problem can be divided into the following two categories:
(1) Grouping methods where one group contains only 1 person.
Since the person in the 1-person group can be any one of Jia... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(2) Consider a tangent line to the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$, which intersects the two symmetry axes of the ellipse at points $A$ and $B$, respectively. Then the minimum length of the line segment $AB$ is | (2) 8 Hint: Let the point of tangency be $P(5 \cos \theta, 3 \sin \theta)$, then the equation of the tangent line to the ellipse at point $P$ is $\frac{\cos \theta}{5} x+\frac{\sin \theta}{3} y=1$, which intersects the $x$-axis at $A\left(\frac{5}{\cos \theta}, 0\right)$ and the $y$-axis at $B\left(0, \frac{3}{\sin \th... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Let $f(x)=|x-1|+|2 x-1|+|3 x-1|$, where $x \in \mathbf{R}$. Then the range of $f(x)$ is $\qquad$ | $-1 .[1,+\infty)$.
Notice,
$$
f(x)=|x-1|+2\left|x-\frac{1}{2}\right|+3\left|x-\frac{1}{3}\right| \text {. }
$$
When $\frac{1}{3} \leqslant x \leqslant \frac{1}{2}$, $f(x)$ reaches its minimum value.
We can take $x=\frac{1}{3}, f_{\text {min }}=f\left(\frac{1}{3}\right)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Four. (20 points) In the sequence $\left\{a_{n}\right\}$, $a_{1}, a_{2}$ are given non-zero integers, $a_{n+2}=\left|a_{n+1}-a_{n}\right|$.
(1) If $a_{16}=4, a_{17}=1$, find $a_{2018}$;
(2) Prove: From $\left\{a_{n}\right\}$, it is always possible to select infinitely many terms to form two different constant subsequen... | (1) It is easy to see,
$$
\begin{array}{l}
a_{16}=4, a_{17}=1, a_{18}=3, a_{19}=2, a_{20}=1, \\
a_{21}=1, a_{22}=0, a_{23}=1, a_{24}=1, a_{25}=0, \\
\cdots \cdots
\end{array}
$$
Notice that, starting from the 20th term, every three consecutive terms periodically take the values $1, 1, 0$.
$$
\begin{array}{l}
\text { A... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Define the length of the interval $\left[x_{1}, x_{2}\right]$ as $x_{2}-x_{1}$. If the domain of the function $y=\left|\log _{2} x\right|$ is $[a, b]$, and the range is $[0,2]$, then the difference between the maximum and minimum values of the length of the interval $[a, b]$ is $\qquad$ . | When the interval $[a, b]$ is $\left[\frac{1}{4}, 4\right]$, the maximum length is $\frac{15}{4}$; when the interval $[a, b]$ is $\left[\frac{1}{4}, 1\right]$, the minimum length is $\frac{3}{4}$. Therefore, the difference between the maximum and minimum lengths of the interval $[a, b]$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=\frac{3}{2}, \\
a_{n}^{2}-a_{n} a_{n+1}-a_{n} a_{n-1}+a_{n+1} a_{n-1}+ \\
2 a_{n}-a_{n+1}-a_{n-1}=0(n \geqslant 2) .
\end{array}
$$
Prove: $a_{2016}>6$. | Note that,
$$
\begin{array}{l}
\left(a_{n}-a_{n+1}\right)\left(a_{n}-a_{n-1}\right)+ \\
\left(a_{n}-a_{n+1}\right)+\left(a_{n}-a_{n-1}\right)=0 .
\end{array}
$$
Let $b_{n}=a_{n}-a_{n-1}(n \geqslant 2)$. Then equation (1) becomes
$$
-b_{n+1} b_{n}-b_{n+1}+b_{n}=0 \text {. }
$$
Assume $b_{n+1}=0$. Then $b_{n}=0$.
Furth... | 6 | Algebra | proof | Yes | Yes | olympiads | false |
(5) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with the left vertex $A_{1}$ and the right focus $F_{2}$. Point $P$ is a moving point on the ellipse. When $\overrightarrow{P A_{1}} \cdot \overrightarrow{P F_{2}}$ takes the minimum value, the value of $\left|\overrightarrow{P A_{1}}+\overrightarrow{P F_{2}}\ri... | (5) From the conditions, we know $a=2, b=\sqrt{3}$, so $c=1$, hence $A_{1}(-2,0) 、 F_{2}(1,0)$. Let $P(x, y)$, then
$$
\overrightarrow{P A_{1}}=(-2-x,-y), \overrightarrow{P F_{2}}=(1-x,-y) .
$$
Thus,
$$
\begin{aligned}
\overrightarrow{P A_{1}} \cdot \overrightarrow{P F_{2}} & =(x+2)(x-1)+y^{2} \\
& =x^{2}+x-2+3\left(1... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. In $\triangle A B C$, it is known that $\sin A=10 \sin B \sin C, \cos A=10 \cos B \cos C$, then the value of $\tan A$ is $\qquad$ | Answer 11.
Solution: Since $\sin A-\cos A=10(\sin B \sin C-\cos B \cos C)=-10 \cos (B+C)=10 \cos A$, we have $\sin A=11 \cos A$, thus $\tan A=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, a line $l$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=$ $\lambda$, and there are exactly 3 such lines $l$, then $\lambda=$ $\qquad$. | 2. 4 .
First, note the following conclusion: A chord passing through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ and intersecting the right branch at two points attains its minimum length $\frac{2 b^{2}}{a}=4$ if and only if the chord is perpendicular to the $x$-axis. (In fact, the polar equation of thi... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Multiply the consecutive natural numbers from 1 to 25, which is $1 \times 2 \times 3 \times \ldots \times 25$, denoted as 25! (read as 25 factorial). When 25! is divided by 3, it is clear that 25! is divisible by 3, resulting in a quotient: then divide this quotient by 3, ..., and continue dividing by 3 until the qu... | $10$ | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For example, in the sequence $x_{1}, x_{2}, \cdots, x_{n}, \cdots$, the sum of any three consecutive terms is 20, and $x_{1}=9, x_{12}=7$. Find the value of $x_{2000}$. | Given that $x_{n+1}+x_{n+2}+x_{n+3}=20$
$$
x_{n}+x_{n+1}+x_{n+2}=20
$$
Subtracting the two equations, we get $x_{n+3}=x_{n}(n \geqslant 1)$. Therefore, $\left\{x_{n}\right\}$ is a periodic sequence with a period of 3, so $x_{4}=x_{1}=9, x_{12}=$ $x_{3}=7$. Since $x_{1}+x_{2}+x_{3}=20$, we have $x_{2}=4$. Hence, $x_{20... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (8 points) A large cube with an edge length of 5 is cut into 125 smaller cubes with an edge length of 1. The total surface area of these smaller cubes is $\qquad$ times the surface area of the original large cube. | 2. (8 points) A large cube with an edge length of 5 is cut into 125 smaller cubes with an edge length of 1. The total surface area of these smaller cubes is $\qquad$ times the surface area of the original large cube.
【Solution】Solution: According to the analysis, the original cube has 6 faces. Each cut adds 2 faces. T... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. The sequence $\left\{x_{n}\right\}$ is defined as follows: $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$. Then the integer part of the following sum $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}$ is . $\qquad$ | 11. From $x_{k+1}=x_{k}^{2}+x_{k} \Rightarrow \frac{1}{x_{k+1}}=\frac{1}{x_{k}\left(1+x_{k}\right)}=\frac{1}{x_{k}}-\frac{1}{1+x_{k}}$, we get
$$
\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{101}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\cdots+\frac{1}{x_{100}}-\frac{1}{x_{101}}=\frac{1}{x_{1}}-\frac{1}{x_{101}}=2-... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given that for any real numbers $x, y$, the function $f(x)$ satisfies $f(x)+f(y)=f(x+y)+xy$.
If $f(1)=m$, then the number of positive integer pairs $(m, n)$ that satisfy $f(n)=2019$ is $\qquad$. | 7.8 .
Let $y=1$, we get
$$
\begin{array}{l}
f(x)+f(1)=f(x+1)+x \\
\Rightarrow f(x+1)-f(x)=m-x .
\end{array}
$$
Then $f(x)=f(1)+\sum_{k=1}^{x-1}(f(k+1)-f(k))$
$$
=m x-\frac{1}{2} x(x-1) \text {. }
$$
Also, $2019=f(n)=m n-\frac{1}{2} n(n-1)$, so
$$
n(2 m+1-n)=2 \times 3 \times 673 \text {. }
$$
When $n$ is even, $2 m... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
II. (50 points) Given a sequence $\left\{a_{n}\right\}, n \in \mathbf{N}_{+}$, where $a_{1}=1, a_{n}=a_{1} a_{2} \cdots a_{n-1}+1, n \geqslant 2$. Find the smallest real number $M$, such that for any $m \in \mathbf{N}_{+}, \sum_{n=1}^{m} \frac{1}{a_{n}}<M$ holds. | II. Solution: From $a_{n}=a_{1} a_{2} \cdots a_{n-1}+1$, we get $\frac{1}{a_{1} a_{2} \cdots a_{n-1}}=\frac{1}{a_{n}}+\frac{1}{a_{1} a_{2} \cdots a_{n}}$, i.e., $\frac{1}{a_{n}}=\frac{1}{a_{1} a_{2} \cdots a_{n-1}}-\frac{1}{a_{1} a_{2} \cdots a_{n}}$.
For $m \geqslant 2$, substituting $n=2,3, \cdots, m$ into the above ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7 If for any real number $x$, the function
$$
f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4
$$
is always a non-negative real number, then the minimum value of the real number $a$ is . $\qquad$ | (7) - 2 Hint: From the conditions, we have $\left\{\begin{array}{l}f(0)=-|1+a|+2 \geqslant 0, \\ f(1)=-|a|+2 \geqslant 0,\end{array}\right.$ Solving this, we get $-2 \leqslant a \leqslant 1$.
When $a=-2$,
$$
f(x)=x^{2}-2 x-|x+1|-|x-2|+4,
$$
i.e.,
$$
f(x)=\left\{\begin{array}{ll}
x^{2}+3, & x2 .
\end{array}\right.
$$
... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(22) For the sequence $\left\{a_{n}\right\}$, we define $\left\{\Delta a_{n}\right\}$ as the first-order difference sequence of $\left\{a_{n}\right\}$, where $\Delta a_{n} = a_{n+1} - a_{n} \left(n \in \mathbf{N}^{*}\right)$. For a positive integer $k$, we define $\left\{\Delta^{k} a_{n}\right\}$ as the $k$-th order di... | (22) (1) Since $\Delta^{2} a_{n}-\Delta a_{n+1}+a_{n}=-2^{n}$, and
$$
\Delta^{2} a_{n}=\Delta a_{n+1}-\Delta a_{n},
$$
we can get
$$
\Delta a_{n}-a_{n}=2^{n}.
$$
Therefore,
$$
a_{n+1}-2 a_{n}=2^{n}, \frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{1}{2},
$$
so $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic s... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) The large square grid shown is composed of 81 small squares, each 1 square centimeter in area, and $B, C$ are two grid points. If you are asked to mark another grid point $A$ such that the area of $\triangle A B C$ is exactly 3 square centimeters, then the number of such $A$ points is ( ).
A. 6
B. 5
C. 8... | 【Analysis】According to the fact that the distance between two parallel lines is equal everywhere, it is only necessary to find a point on each side of $B C$ that meets the area of $3 \, \text{cm}^2$, and then draw parallel lines to find all the points.
【Solution】Solution: As shown in the figure, find points $A$ and $D... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
$3 \cdot 1$ If the numbers $p, p+10, p+14$ are all prime, find $p$.
(Kiev Mathematical Olympiad, 1981) | [Solution] If $p \equiv 1(\bmod 3)$, then
$$
p+14 \equiv 15 \equiv 0 \quad(\bmod 3) .
$$
In this case, $p+14$ is not a prime number.
If $p \equiv 2(\bmod 3)$, then
$$
p+10 \equiv 12 \equiv 0 \quad(\bmod 3) .
$$
In this case, $p+10$ is not a prime number.
Therefore,
$$
p \equiv 0 \quad(\bmod 3) .
$$
Since $p$ is a pr... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 9 Let $S=\{1,2,3,4\}, n$ terms of the sequence $a_{1}$, $a_{2}, \cdots, a_{n}$ have the following property: for any non-empty subset $B$ of $S$ (the number of elements in set $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of the num... | Solution: For each $i \in S$, it can form a binary subset with each of the other 3 elements in $S$, meaning there are 3 binary subsets containing $i$. If $i$ appears only once in the sequence, then the number of adjacent pairs containing $i$ is at most two. Therefore, $i$ must appear at least twice in the sequence. Sin... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Given a complex number $z$ satisfying $|z|=1$. Then
$$
f(z)=\left|z^{5}+\bar{z}^{3}+6 z\right|-2\left|z^{4}+1\right|
$$
the minimum value of $f(z)$ is $\qquad$ | 8. 3 .
Let $x=z^{2}+\bar{z}^{2} \in[-2,2]$. Then
$$
\begin{array}{l}
f(z)=\left|x^{2}+4\right|-2|x| \\
=(|x|-1)^{2}+3 \in[3,4] .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.22 Given that $p$ and $q$ are rational numbers, and $x=\frac{\sqrt{5}-1}{2}$ satisfies $x^{3}+p x+q=0$. Then the value of $p+q$ is
(A) -1.
(B) 1.
(C) -3.
(D) 3.
(Anhui Province, China Junior High School Mathematics Competition, 1997) | [Solution] Substituting $x=\frac{\sqrt{5}-1}{2}$ into $x^{3}+p x+q=0$, simplifying yields $4 \sqrt{5}(2+p)+4(2 q-p-4)=0$. Since $p, q$ are rational numbers, we have
$$
\left\{\begin{array}{l}
2+p=0, \\
2 q-p-4=0 .
\end{array}\right.
$$
Solving these equations gives $p=-2, q=1$. Therefore, $p+q=-1$. Hence, the answer i... | -1 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. (3 points) In the two-digit addition operation shown in the figure, it is known that $A+B+C+D=22$, then $X+Y=(\quad)$
$$
\begin{array}{r}
A B \\
+\quad C D \\
\hline X Y 9
\end{array}
$$
A. 2
B. 4
C. 7
D. 13 | 【Analysis】According to the unit digit of the sum is 9, we can get: $B+D=9$, then $A+C=22-9=13$, so we can get $x=1, y=3$, according to this, we can find the value of $x+y$.
【Solution】Solution: According to the analysis of the problem, we can get: $B+D=9$, then $A+C=22-9=13$, so we can get $x=1, y=3$, then $x+y=1+3=4$.... | 4 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
B. Let $a=\sqrt[3]{2}+1$. Then $\frac{a^{3}+a^{4}}{a^{3}+1}=(\quad$.
(A) 3
(B) 4
(C) $\sqrt[3]{2}+2$
(D) $\sqrt[3]{2}+\sqrt[3]{4}$ | B. A.
Notice,
$$
\begin{array}{l}
(a-1)^{3}=(\sqrt[3]{2})^{3}=2 \\
\Rightarrow a^{3}=2+3 a^{2}-3 a+1=3\left(a^{2}-a+1\right) \\
\Rightarrow \frac{a^{3}+a^{4}}{a^{3}+1}=\frac{a^{3}(1+a)}{(a+1)\left(a^{2}-a+1\right)} \\
\quad=\frac{a^{3}}{a^{2}-a+1}=3 .
\end{array}
$$ | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Planar vectors $\vec{u}=(x, y), \vec{b}=\left(x^{2}, y^{2}\right), \vec{c}=(1,1), \vec{a}=\left(\frac{1}{4}, \frac{1}{9}\right)$, if $\vec{a} \cdot \vec{c}$ $=1, \vec{b} \cdot \vec{d}=1$, then the number of such vectors $\vec{a}$ is ( ).
A. 1
B. 2
C. More than 2
D. Does not exist | 4. B
From $\vec{a} \cdot \vec{c}=1 \Rightarrow x+y=1, \vec{b} \cdot \vec{d}=1 \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$, it can be seen from the graph that there are two such $\vec{a}$ | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. (32nd USAMO Problem) Let $a, b, c$ be positive real numbers. Prove:
$$
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8 .
$$ | 12. Let $s=a+b+c, f(t)=\frac{(t+s)^{2}}{2 t^{2}+(s-t)^{2}}, t \in[0, s)$.
Since $f(t)=\frac{1}{3}+\frac{2}{3} \times \frac{4 s t+s^{2}}{3\left(t-\frac{s}{3}\right)^{2}+\frac{2}{3} s^{2}} \leqslant \frac{1}{3}+\frac{4 s+s^{2}}{s^{2}}=4\left(\frac{1}{3}+\frac{t}{s}\right)$, thus $f(a)+f(b)+f(c) \leqslant 4\left(\frac{1}... | 8 | Inequalities | proof | Yes | Yes | olympiads | false |
7. (5 points)
Grandma's eyesight is poor, and she accidentally put soy sauce in the vinegar bottle, put cooking wine in the soy sauce bottle, put oil in the cooking wine bottle, and put vinegar in the oil bottle. Xiao Fei used an empty bottle to help Grandma put these seasonings back into the correct bottles. Xiao Fei ... | $5$ | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. (6 points) Define a new operation “*”: $a^{*} b=\left\{\begin{array}{l}\mathrm{a} \text { (if } \mathrm{a}>\mathrm{b}) \\ 1 \text { (if } \mathrm{a}=\mathrm{b}) \\ \mathrm{b} \text { (if } \mathrm{a}<\mathrm{b})\end{array}\right.$ For example, $3.5 * 2=3.5,1 * 1.2=1.2,7 * 7=1$, then $\frac{1.1 * \frac{7}{3}-\frac{1}... | 6. (6 points) Define a new operation “*”: $a^{*} b=\left\{\begin{array}{l}\mathrm{a}(\text { if } \mathrm{a}>\mathrm{b}) \\ 1 \text { (if } \mathrm{a}=\mathrm{b}) \\ \mathrm{b}(\text { if } \mathrm{a}<\mathrm{b})\end{array}\right.$
For example, $3.5 * 2=3.5,1 * 1.2=1.2,7 * 7=1$, then $\frac{1.1 * \frac{7}{3}-\frac{1}{3... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7 \cdot 20$ Given that all vertices of a broken line are located on the surface of a cube with an edge length of 2, each edge of the broken line is 3 units long, and the two endpoints are precisely the two farthest vertices of the cube. How many edges does this broken line have at minimum?
The original text's line br... | [Solution] Let $P, Q, R$ be the midpoints of the three edges extending from point $B_{3}$. Clearly, $A_{1} P=A_{1} Q=A_{1} R=3$. Thus, the set of points on the surface of the cube that are 3 units away from $A_{1}$ consists of three arcs passing through $P, Q$, and $R$ on the three sides with $B_{3}$ as the vertex (see... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
17. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and satisfies the following conditions: (1) For any $x, y \in \mathbf{R}$, $f(x+y)=f(x)+f(y)$; (2) When $x>0$, $f(x)<0$, and $f(1)=-2$. Find the maximum and minimum values of $f(x)$ on the interval $[-2,3]$. | 17. Since $f(x)$ is an odd function on $\mathbf{R}$, we first consider the case where $x \geqslant 0$. Let $x_{1}>x_{2} \geqslant 0$, then $f\left(x_{1}\right)=$ $f\left(x_{1}-x_{2}+x_{2}\right)=f\left(x_{1}-x_{2}\right)+f\left(x_{2}\right)$. Since $x_{1}-x_{2}>0$, we have $f\left(x_{1}-x_{2}\right)<0$, thus $f\left(x_... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7 As shown in the figure, in the right triangle $\triangle ABC$, it is known that $BC=a$. If the line segment $PQ$ of length $2a$ has point $A$ as its midpoint, for what value of the angle between $\overrightarrow{P Q}$ and $\overrightarrow{B C}$ is the value of $\overrightarrow{B P} \cdot \overrightarrow{C Q}$... | Analyze: Combining the diagram, according to the rules of vector operations, transform the problem into solving using function knowledge.
Solution: Since $\overrightarrow{A B} \perp \overrightarrow{A C}$,
it follows that $\overrightarrow{A B} \cdot \overrightarrow{A C}=0$.
Since $\overrightarrow{A P}=-\overrightarrow{... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
19. $\overline{P Q Q P Q Q}, \overline{P Q P Q P Q}, \overline{Q P Q P Q P}, \overline{P P P P P P}, \overline{P P P Q Q Q}$ are five six-digit numbers, where the same letters represent the same digits, and different letters represent different digits. Regardless of what $P$ and $Q$ are, among these five six-digit numb... | $4$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 7 Given $p, q, \frac{2q-1}{p}, \frac{2p-1}{q} \in \mathbf{Z}$, and $p>1, q>1$. Find the value of $p+q$. | Solution: Clearly, $p \neq q$.
(1) If $p>q$, then
$$
\begin{array}{l}
\frac{2 q-1}{p}<\frac{2 p-1}{p}<2 \Rightarrow \frac{2 q-1}{p}=1 \\
\Rightarrow p=2 q-1 .
\end{array}
$$
Thus $\frac{2 p-1}{q}=\frac{2(2 q-1)-1}{q}=4-\frac{3}{q} \in \mathbf{Z}$
$$
\Rightarrow q=3 \Rightarrow p=5 \Rightarrow p+q=8 \text {. }
$$
(2) I... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Three. (50 points) Given the set
$$
M=\{1,2, \cdots, 2020\} \text {. }
$$
Now, each number in $M$ is colored one of three colors: red, yellow, or blue, and each color is used at least once. Let
$$
\begin{aligned}
S_{1}= & \left\{(x, y, z) \in M^{3} \mid x, y, z\right. \text { are the same color, } \\
& 2020 \mid(x+y+z... | Let $T=\left\{\omega, \omega^{2}, \cdots, \omega^{n}\right\}$, where $\omega$ is an $n$-th primitive root of unity, and $\omega^{n}=1$ (with $n=2020$). Let the red set, yellow set, and blue set be $I_{1}$, $I_{2}$, and $I_{3}$, respectively. Then,
$$
\begin{array}{l}
\left|S_{1}\right|=\frac{1}{n} \sum_{x \in T}\left(\... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(1) The minimum value of the function $f(x)=\frac{5-4 x+x^{2}}{2-x}$ on $(-\infty, 2)$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | (1) When $x<2$, therefore
$$
\begin{aligned}
f(x) & =\frac{1+\left(4-4 x+x^{2}\right)}{2-x} \\
& =\frac{1}{2-x}+(2-x) \\
& \geqslant 2 \cdot \sqrt{\frac{1}{2-x} \cdot(2-x)} \\
& =2,
\end{aligned}
$$
Equality holds if and only if $\frac{1}{2-x}=2-x$. This equation has a solution $x=1 \in(-\infty, 2)$, so the minimum va... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
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