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$16 \cdot 1$ Given $n$ points on a plane, it is known that $1, 2, 4, 8, 16, 32$ are all distances between some pairs of these points. Then, the minimum possible value of the number of points $n$ is
(A) 4 .
(B) 5 .
(C) 6 .
(D) 7 .
(China Junior High School Mathematics League, 1997) | ［Solution］Since no three of the above distances can form the three sides of the same triangle, the distribution of these points and distances can only be as shown in the right figure, in other words, there must be at least 7 points.
Therefore, the answer is $(D)$. | 7 | Geometry | MCQ | Yes | Yes | olympiads | false |
8 Arrange fifteen students numbered $1,2,3, \ldots, 15$ in a circle facing inward, in numerical order. The first time, the student numbered 1 turns around. The second time, the students numbered 2 and 3 turn around. The third time, the students numbered $4,5,6$ turn around, .... The 15th time, all students turn around.... | 【Answer】12
【Analysis】Key points: Reverse reasoning, parity
If all 15 rounds are completed, there will be a total of $1+2+3+\cdots+15=120$ turns, with each student turning $120 \div 15=8$ times.
In the 15th round, all students turn once. Since the 15th round did not occur, each student turns 7 times.
In the 14th round, ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
(1) Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$. Draw perpendiculars from point $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$, respectively. Then the value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$. | (1) -1 Hint: Method one Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$, then the equation of line $P A$ is $y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right)$, i.e., $y=-x+2 x_{0}+\frac{2}{x_{0}}$.
From $\left\{\begin{array}{l}y=x, \\ y=-x+2 x_{0}+\frac{2}{x_{0}}\end{array}\right.$, we get $A\left(x_{0}+\fra... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let $a, b, c, d$ be four distinct real numbers such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=$ $b d$. Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$. | 11. Let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we get $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$, so the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$. Let $x+... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 8 (2006 Thailand Mathematical Olympiad) 229 boys and 271 girls are evenly divided into 10 groups, and each group's students are marked with numbers from 1 to 50. Now, select 4 students (including an odd number of girls) who satisfy the property: they come from two groups, and among the 4 students, there are two... | Prove that a selection of 4 students with 2 pairs of the same number from two groups is called a "team". Let $S=\{\delta \mid \delta$ is a team $\}$,
$O=\{\delta \in S \mid \delta$ contains an odd number of girls $\}$,
$E=\{\delta \in S \mid \delta$ contains an even number of girls $\}$.
It suffices to prove: $|O|$ is ... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
Example 9 (1994 National High School League Question) Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$, and $\left\{\begin{array}{l}x^{3}+\sin x-2 a=0, \\ 4 y^{3}+\sin y \cdot \cos y+a=0,\end{array}\right.$ then $\cos (x+2 y)=$ $\qquad$ | Solution: Fill in 1. Reason: Consider the function $f(t)=t^{3}+\sin t$, which is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Let $f(x)=2a$, i.e., $x^{3}+\sin x-2a=0$, and $f(2y)=-2a$, i.e., $4y^{3}+\sin y \cdot \cos y+a=0$. From $f(x)=2a=f(-2y)$, we get $x=-2y$.
Since $2y \in\left[-\frac{\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6: On the table, there are 1989 same-sized circular paper pieces placed without overlapping. How many different colors are needed at least to color each circular paper piece so that, no matter how these paper pieces are placed, it is always possible to color them in such a way that any two touching circular pap... | The minimum number of different colors required is 4.
3 colors are not enough. As shown in the figure, if 3 colors are represented by the numbers 1, 2, and 3, circle $A_{1}$ is painted color 1, and $A_{3}$ is painted color 2, then $A_{2}, A_{5}, A_{11}$ can only be painted color 3, and $A_{4}$ can only be painted color... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$, we have $f(x+5) \geqslant f(x)+5$, $f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, then $g(2002)=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the... | 5. $g(2002)=1$ Hint: $f(x)=g(x)+x-1$, so $g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+$ $5, g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1$. That is, $g(x+5) \geqslant g(x), g(x+1) \leqslant g(x)$. Therefore, $g(x) \leqslant$ $g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x), g(x+1)=g(x), g(x)$ ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15 A finite set of points $M$ on the plane satisfies: for any two points $A, B$ in $M$, there must exist a third point $C$ such that $\triangle A B C$ is an equilateral triangle. Find the maximum number of elements in $M$. | 15 It is evident that the set of three vertices of an equilateral triangle satisfies the requirement, so
$$
\max |M| \geqslant 3 .
$$
Since $M$ is a finite set of points, among the equilateral triangles formed by three points in $M$, there must be a largest one. Let it be $\triangle A B C$, then all points in $M$ are ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$, respectively. Point $P$ is on the right branch of the hyperbola $C$, and the excenter of $\triangle P F_{1} F_{2}$ opposite to $\angle P F_{1} F_{2}$ is $I$. The line $P I$ intersects the $x$-axis at point $Q$... | 6. 4 .
Since $I F_{1}$ is the angle bisector of $\angle P F_{1} Q$, we have
$$
\frac{|P Q|}{|P I|}=1+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \text {. }
$$
Let $P\left(x_{0}, y_{0}\right)$. Then, $\left|P F_{1}\right|=\frac{3}{2} x_{0}+2$.
By the optical property of the hyperbola, the tangent line to the hyp... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Let the sum of the first 4 terms of an arithmetic sequence be 26, the sum of the last 4 terms be 110, and the sum of all terms be 187. This sequence has $\qquad$ terms. | \begin{array}{l}\text { 2. } n=11 . \\ \quad a_{1}+a_{2}+a_{3}+a_{4}=4 a_{1}+6 d=26, a_{n}+a_{n-1}+a_{n-2}+a_{n-3}=4 a_{n}-6 d=100, S_{n}= \\ \frac{n}{2}\left(a_{1}+a_{n}\right), a_{1}+a_{n}=34 \text {, so } \frac{n}{2} \cdot 34=187 \text {, i.e., } n=11 .\end{array} | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. The first term of the sequence $\left\{a_{n}\right\}$ is 2, the sum of the first $n$ terms is $S_{n}$, and $\left\{S_{n}\right\}$ is a geometric sequence with a common ratio of $\frac{1}{3}$.
(1) Find the general formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n} S_{n}$, find the sum of all terms... | 14. Solution: (1) $S_{1}=a_{1}=2, S_{n}=2\left(\frac{1}{3}\right)^{n-1}$,
$$
a_{n}=S_{n}-S_{n-1}=2\left(\frac{1}{3}\right)^{n-1}-2\left(\frac{1}{3}\right)^{n-2}=2\left(\frac{1}{3}\right)^{n-1}(1-3)=-4\left(\frac{1}{3}\right)^{n-1} \text {. }
$$
So $a_{n}=\left\{\begin{array}{ll}2 & (n=1) \\ -4\left(\frac{1}{3}\right)^... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=$ | 3. 4 .
$$
\begin{aligned}
& \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=\left(\tan 9^{\circ}+\frac{1}{\tan 9^{\circ}}\right)-\left(\tan 27^{\circ}+\frac{1}{\tan 27^{\circ}}\right) \\
= & \frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=\frac{2\left(\sin 54^{\circ}-\sin 18^{\circ}\right)}{\sin 18^{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. If $a>1, b>1$, and $\lg (a+b)=\lg a+\lg b$, then the value of $\lg (a-1)+\lg (b-1)$
A. equals $\lg 2$
B. equals 1
C. equals 0
D. is not a constant independent of $a$ and $b$ | $\lg (a+b)=\lg a+\lg b \Rightarrow a+b=a b$, thus $(a-1)(b-1)=a b-(a+b)+1=1$ $\Rightarrow \lg (a-1)+\lg (b-1)=0$. The derived conclusion is.
The translation is as follows:
$\lg (a+b)=\lg a+\lg b \Rightarrow a+b=a b$, thus $(a-1)(b-1)=a b-(a+b)+1=1$ $\Rightarrow \lg (a-1)+\lg (b-1)=0$. The derived conclusion is. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
10. (10th US Olympiad Question) In a certain county, every two districts are connected by exactly one of three modes of transportation: car, train, or airplane. It is known that all three modes of transportation are used in the county, but no district has all three modes. Furthermore, no three districts are pairwise co... | 10. There are at most 4 districts.
Let vertices represent districts, and edges represent means of transportation, colored red, blue, and yellow to represent three types of transportation, as shown in the figure. This meets the requirements of the problem, thus there exist four districts.
If there are more than 4 dist... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
15. Given $f(x)=x^{2}+c$, and $f(f(x))=f\left(x^{2}+1\right)$.
(1) Let $g(x)=f(f(x))$, find the analytical expression of the function $g(x)$;
(2) Let $\varphi(x)=g(x)-\lambda f(x)$, try to find the value of the real number $\lambda$ such that $\varphi(x)$ is a decreasing function on $(-\infty,-1]$ and an increasing fun... | 15. (1) $f(f(x))=f\left(x^{2}+c\right)=\left(x^{2}+c\right)^{2}+c, f\left(x^{2}+1\right)=\left(x^{2}+1\right)^{2}+c$, we get $c=1$, $g(x)=x^{4}+2 x^{2}+2$.
(2) $\varphi(x)=g(x)-\lambda f(x)=x^{4}+(2-\lambda) x^{2}+(2-\lambda)$, set $x_{1}1+1+2-\lambda=4-\lambda$, to make $\varphi(x)$ decreasing on $(-\infty, -1]$, it i... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. There are $\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}, \mathbf{E}$ five people, each of whom always tells lies or always tells the truth, and they all know each other's behavior. $\mathrm{A}$ says $\mathrm{B}$ is a liar, $\mathrm{B}$ says $\mathrm{C}$ is a liar, $\mathrm{C}$ says $\mathrm{D}$ is a liar, $\mathrm... | 【Analysis】If A tells the truth, from what A says, we know B is lying, from what B says, we know C is telling the truth, and so on, we find D is lying and E is telling the truth, which means 2 people are lying.
If A is lying, then B is telling the truth, C is lying, D is telling the truth, and E is lying, which means 3 ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.24 On a circle, there are several black and white chess pieces. Two people, A and B, play a game: A takes away all black pieces that have adjacent white pieces, then B takes away all white pieces that have adjacent black pieces. They take turns doing this until only one color of pieces remains.
(1) Suppose there are ... | [Solution] The right diagram shows 41 pieces, of which 29 are black and 12 are white. The numbers marked beside the pieces indicate how many steps before the end the piece was taken. Pieces without numbers should all be marked with 4, which is omitted in the diagram. It is easy to verify that after each of the two play... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. As shown in the figure, a rectangle is divided into two parts of exactly the same size and shape by a zigzag line, and these two parts can be combined to form a square. Given that the length of the rectangle is 27 cm, what is the width of the rectangle in cm? | Match the answer 12
Translate the text above into English, please retain the source text's line breaks and format, and output the translation result directly. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$31 \cdot 47$ If $m$ is a positive integer, and the line $13x + 11y = 700$ and $y = mx - 1$ intersect at an integral point, then $m$ can only be
(A) 4.
(B) 5.
(C) 6.
(D) 7.
(E) one of 4, 5, 6, 7.
(17th American High School Mathematics Examination, 1966) | [Solution] Substituting $y = mx - 1$ into $13x + 11y = 700$, we get
$$
x = \frac{711}{13 + 11m} = \frac{3^2 \cdot 79}{13 + 11m}.
$$
Since $x$ is an integer, $13 + 11m$ must be a factor of the numerator, which means $13 + 11m$ can only be one of the following numbers:
$$
1, 3, 3^2, 79, 3 \cdot 79, 3^2 \cdot 79.
$$
Let... | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
$31 \cdot 35$ satisfies $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ for the number of different integer pairs $(x, y)$ is
(A) 0.
(B) 1.
(C) 3.
(D) 4.
(E) 7.
(35th American High School Mathematics Examination, 1984) | [Solution 1] Given $1984=2^{6} \cdot 31$. Also, when $04$, $y<x$.
Therefore, the original equation has only three sets of solutions that satisfy the conditions.
Thus, the answer is $(C)$.
[Solution 2] Given $1984=2^{6} \cdot 31$, then
$$
\begin{aligned}
\sqrt{1984} & =8 \sqrt{31}=\sqrt{31}+7 \sqrt{31}=2 \sqrt{31}+6 \sq... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
10.115 Starting from any three-digit number $n$, we get a new number $f(n)$, which is the sum of the products of the three digits of $n$, taken two at a time, and the product of the three digits.
(1) When $n=625$, find $\frac{n}{f(n)}$.
(2) Find all three-digit numbers such that $\frac{n}{f(n)}=1$.
(UK Mathematical Oly... | [Solution] Let $n=\overline{a b c}=100 a+10 b+c$, where $a, b, c \in\{0,1,2, \cdots, 9\}$ and $a \geqslant 1$.
$$
f(n)=(a+b+c)+(a b+b c+c a)+a b c \text {. }
$$
(1) When $n=625$, $a=6, b=2, c=5$.
$$
\begin{array}{c}
f(625)=(6+2+5)+(12+10+30)+60=125, \\
\frac{625}{f(625)}=\frac{625}{125}=5 .
\end{array}
$$
(2) According... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Seven, $n$ teams are to hold a home-and-away double round-robin match (each 2 teams play 2 matches, each having 1 home match), each team can play multiple away matches within a week (from Sunday to Saturday). However, if a team has a home match in a certain week, no away matches for that team can be scheduled in that w... | (1) As shown in Table 1. The symbol “*” in the table indicates that the team has a home game in that week and cannot travel. It is easy to verify that, according to the schedule in the table, the six teams can complete the competition in four weeks.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline Team & Week 1 & Week 2 & W... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Question 11: Given the mapping f: $\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ is both injective and surjective, and satisfies $f(x)+f(f(x))=6$, then $f(1)=$ $\qquad$ _. | Question 11, Solution: Let $S=\{1,2,3,4,5\}$, for any $\mathrm{a} \in S$, according to the conditions:
$$
\left\{\begin{array}{c}
f(a)+f(f(a))=6 \\
f(f(a))+f(f(f(a)))=6
\end{array}\right.
$$
Thus, $\mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{a})))$.
Since $f$ is a bijection, $\{f(a) \mid a \in S\}=... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Find all values of $a$ such that the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy $\left(x_{1}-3\right)^{2}+$ $\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0$.
(1983 Austrian Olympiad Problem) | Let $y=x-3$, then $y_{1}=x_{1}-3, y_{2}=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$
(y+3)^{3}-6(y+3)^{2}+a(y+3)+a=y^{3}+3 y^{2}+(a-9) y+4 a-27
$$
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
y_{1}+y_{2}+y_{3}=-3, \\
y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}=a-9, \\
y_{1} y_{2} y_{3}=27-4 a .
\... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest positive integer $n$, such that the set $\{1,2,3, \cdots, 3 n-1,3 n\}$ can be divided into $n$ mutually disjoint triples $\{x, y, z\}$, where $x+y=3 z$. | 1. Let $n$ be a positive integer with the property mentioned in the problem, and let the $n$ triples be $\left\{x_{i}, y_{i}, z_{i}\right\}, i=1,2, \cdots, n$. Then, from $x_{i}+y_{i}=3 z_{i}$, we have $4 \sum_{i=1}^{n} z_{i}=\sum_{i=1}^{n}\left(x_{i}+y_{i}+z_{i}\right)=\sum_{j=1}^{3 n} j=\frac{3 n(3 n+1)}{2}$. Therefo... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In any city of a certain province, it is connected by railway to at most 3 other cities, and from any city to other cities, it passes through at most one city in between. How many cities can there be at most in the province? | 1. Arbitrarily select a city $v_{0}$, which is connected to at most 3 cities $v_{1}, v_{2}, v_{3}$. Each of $v_{1}, v_{2}, v_{3}$ is connected to at most two cities different from $v_{0}$, and there are no other cities (otherwise, from $v_{0}$ to that city, at least two cities would be passed through).
Thus, the total ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 27 (2004 National High School Competition Question) In the plane rectangular coordinate system $x O y$, two points $M(-1,2)$ and $N(1,4)$ are given, and point $P$ moves on the $x$-axis. When $\angle M P N$ takes the maximum value, the abscissa of point $P$ is $\qquad$ | Solution: The center of the circle passing through points $M$ and $N$ lies on the perpendicular bisector of segment $M N$, which is the line $y=3-x$. Let the center of the circle be $S$ $(a, 3-a)$, then the equation of circle $S$ is $(x-a)^{2}+(y-3+a)^{2}=2\left(1+a^{2}\right)$.
For a chord of fixed length, the inscri... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 8 Let $P(x)$ be a polynomial of degree $2n$, $P(0)=P(2)=\cdots=P(2n)=0, P(1)=$ $P(3)=\cdots=P(2n-1)=2, P(2n+1)=-30$, find $n$.
| Let $f(x) = P(x) - 1$, then $f(k) = (-1)^{k+1}, k = 0, 1, \cdots, 2n, f(2n+1) = -31$. Consider the Lagrange formula, and note the $k$-th term
$$
\begin{array}{l}
f_{k}(x) = \prod_{\substack{i \neq k \\
0 \leqslant i \leqslant 2n}} \frac{x-i}{k-i} \cdot f(k), \\
\therefore \quad f_{k}(2n+1) = \frac{(2n+1)!}{(2n+1-k)} \c... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. As shown in the figure, $P$ is a point inside rectangle $A B C D$, and the distances from $P$ to points $A, B, C$ are $3 \sqrt{2}, 4 \sqrt{2}$, $5 \sqrt{2}$, respectively. Then the distance from $P$ to point $D$ is $\qquad$. | $6$ | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. For an old clock, the minute hand and the hour hand overlap every 66 minutes of standard time. Therefore, in 24 hours of this old clock, compared to 24 hours of standard time, it ( ).
(A) is 12 minutes fast
(B) is 6 minutes fast
(C) is 6 minutes slow
(D) is 12 minutes slow | 【Answer】D
【Question Type】Clock Problem
【Analysis】The hour hand moves at a speed of 0.5 degrees per minute, and the minute hand moves at a speed of 6 degrees per minute. Every time the minute hand completes an extra full circle (360 degrees) compared to the hour hand, the two hands overlap once. This happens every $\fra... | 12 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
11. In a city football tournament, each team must play one match against every other team, with 3 points awarded for a win, 1 point for a draw, and 0 points for a loss. It is known that one team has the highest score, more than any other team, but this team has the fewest wins, fewer than any other team. How many teams... | 11. Let the team with the most points and the fewest wins be team $A$. If team $A$ wins $n$ games and draws $m$ games, then it scores $3n + m$ points. The other teams must win at least $n+1$ games, scoring at least $3(n+1)$ points. Therefore, $3(n+1) < 3n + m$, which implies $m \geq 4$. Team $A$ must draw at least 4 ga... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$29 \cdot 19$ has 10 numbers: $1983^{3}+3 \cdot 1983^{2}+2 \cdot 1983, 1984^{3}+3 \cdot 1984^{2}+2 \cdot 1984, \cdots, 1991^{3}+3 \cdot 1991^{2}+2 \cdot 1991, 1992^{3}+3 \cdot 1992^{2}+2 \cdot 1992$, the largest integer that can divide each of these 10 numbers is
(A) 2.
(B) 3.
(C) 6.
(D) 12.
(China Guangzhou, Wuhan, Fu... | [Solution] From $a^{3}+3 a^{2}+2 a=a\left(a^{2}+3 a+2\right)=a(a+1)(a+2)$, we know that $2 \mid\left(a^{3}+3 a^{2}+2 a\right)$, and $31\left(a^{3}+3 a^{2}+2 a\right)$, which means $6!\left(a^{3}+3 a^{2}+2 a\right)$.
Also, 12 Y $1985 \cdot 1986 \cdot 1987$, which means
$$
12 y\left(1985^{3}+3 \cdot 1985^{2}+2 \cdot 1985... | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Given that $x$ and $y$ are integers, and satisfy
$$
\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)=-\frac{2}{3}\left(\frac{1}{x^{4}}-\frac{1}{y^{4}}\right) \text {. }
$$
Then the number of possible values for $x+y$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4 | $-1 . C$.
From the given equation, we have
$$
\frac{x+y}{x y} \cdot \frac{x^{2}+y^{2}}{x^{2} y^{2}}=\frac{2}{3} \cdot \frac{x^{4}-y^{4}}{x^{4} y^{4}}
$$
Obviously, $x$ and $y$ are not equal to 0.
Thus, $x+y=0$ or $3 x y=2(x-y)$.
If $3 x y=2(x-y)$, then
$$
(3 x+2)(3 y-2)=-4 \text{. }
$$
Since $x, y$ are integers, solv... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
II. (25 points) As shown in Figure 2, given that $O$ is the circumcenter of $\triangle ABC$, $AB=AC$, and $D$ is a point on the circumcircle of $\triangle OBC$. A line perpendicular to $OD$ is drawn through point $A$, and the foot of the perpendicular is $H$. If $BD=7, DC=3$, find $AH$. | II. As shown in Figure 5, extend $BD$ and $OD$, intersecting $\odot O$ at points $N$ and $E$ respectively.
From the problem, we have
$$
\begin{array}{l}
\angle NDE \\
=\angle ODB \\
=\angle OCB \\
=\angle OBC \\
=\angle CDE.
\end{array}
$$
Thus, $DE$ is the external angle bisector of $\angle BDC$.
Since point $D$ lies... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. The product of the two ten-digit numbers 1111111111 and 9999999999 has how many odd digits? | 11. 【Solution】1111111111×9999999999
$$
\begin{array}{l}
=1111111111 \times(10000000000-1) \\
=11111111110000000000-1111111111 \\
=11111111108888888889
\end{array}
$$
Thus, there are 10 digits that are odd. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $P$ be a moving point on the surface of a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, such that the volume of the pyramid $P-B D D_{1} B_{1}$ is $\frac{1}{3}$. Then the length of the trajectory of point $P$ is $\qquad$ | 4.2.
Given $S_{\text {rectangle } B D D_{1} B_{1}}=\sqrt{2}, V_{\text {tetrahedron } P-B D D_{1} B_{1}}=\frac{1}{3}$, we know the distance $h$ from point $P$ to plane $B D D_{1} B_{1}$ is $h=\frac{\sqrt{2}}{2}$.
Since the distances from $A A_{1}$ and $C C_{1}$ to plane $B D D_{1} B_{1}$ are both $\frac{\sqrt{2}}{2}$,... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In the military drill formation performance, the students of a class happen to stand in a double-layer hollow square, with 9 students standing on each side of the outer layer. If the students of this class are to stand guard on a straight 250-meter long road, starting from one end and standing one person every 5 met... | answer: 5 | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 6 In $\triangle A B C$, find the value of $a^{3} \sin (B-C)+b^{3} \sin (C-A)+c^{3} \sin (A-B)$. | First, we prove a lemma:
Lemma: In $\triangle ABC$, we have
$$
\frac{a^{2}-b^{2}}{c^{2}}=\frac{\sin (A-B)}{\sin C}, \frac{c^{2}-a^{2}}{b^{2}}=\frac{\sin (C-A)}{\sin B}, \frac{b^{2}-c^{2}}{a^{2}}=\frac{\sin (B-C)}{\sin A}.
$$
In fact, it is sufficient to prove one of them, let's prove $\frac{a^{2}-b^{2}}{c^{2}}=\frac{\... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. Given the functions $f(x)=x-\ln x-2, g(x)=x \ln x+x$.
(1) Prove that $f(x)$ has a zero in the interval $(3,4)$;
(2) If $k \in \mathbf{Z}$, and $g(x)>k(x-1)$ for all $x>1$, find the maximum value of $k$. | (1) $f^{\prime}(x)=1-\frac{1}{x}=\frac{x-1}{x}$, then $f(x)$ is monotonically increasing on $(1,+\infty)$,
and $f(3)=1-\ln 30$, so $f(x)$ has a zero point in $(3,4)$.
(2) Since $x>1$, then $k<\frac{x \ln x+x}{x-1}$, let $\varphi(x)=\frac{x \ln x+x}{x-1}$,
$$
\varphi^{\prime}(x)=\frac{(2+\ln x)(x-1)-(x \ln x+x)}{(x-1)^... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
$5 \cdot 71$ A city has $n$ high schools, the $i$-th high school sends $C_{i}$ students to the stadium to watch a ball game. It is known that $0 \leqslant C_{i} \leqslant 39, i=1,2, \cdots, n, C_{1}+C_{2}+\cdots+C_{n}=1990$, each row of seats in the stand has 199 seats. It is required that students from the same school... | [Solution 1] Since $C_{i} \leqslant 39$, each row can accommodate at least 161 people. Therefore, with 13 rows, at least $161 \times 13 = 2093$ people can be seated, which is certainly enough for all 1990 students.
Next, we use the extremal principle to prove that 12 rows are sufficient. Since $C_{1}$, $C_{2}, \cdots,... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 8 If the function $y=f(x+1)$ is an even function, is the graph of the function $y=f(x)$ a curve with axis symmetry? If so, write the equation of the axis of symmetry; if not, explain the reason. | Let $g(x)=f(1+x)$, then by the function $y=g(x)$ being an even function, we get $f(1+x)=f(1-x)$, so, the graph of the function $y=f(x)$ is a curve symmetric about an axis, with the equation of the axis of symmetry being $x=1$.
---
Note: The translation preserves the original text's line breaks and formatting. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$, then the number of such sequences is
(A) 2
(B) 3
(C) 4
(D) 5 | (C)
3. 【Analysis and Solution】Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. Then, according to the problem, we have
$$
n a+\frac{n(n-1)}{2} d=97^{2} \text {, i.e., }[2 a+(n-1) d] n=2 \times 97^{2}
$$
Since $n$ is a natural number not less than 3, and 97 is a prime number, the ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
13. (2002 National High School Mathematics Competition) Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$, and for any $x \in \mathbf{R}$, we have $f(x+5) \geqslant f(x)+5$ and $f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, find the value of $g(2002)$. | 13. From $g(x)=f(x)+1-x$ we get $f(x)=g(x)+x-1$.
Then $g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5, g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1$.
Thus $g(x+5) \geqslant g(x), g(x+1) \leqslant g(x)$.
Therefore $g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x)$.
So $g(x+1)=g(x... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
88. Divide the numbers $1, 2, 3, 4, 5, 6, 7, 8$ into three groups, and calculate the sum of the numbers in each group. If these three sums are all different, and the largest sum is twice the smallest sum, then the smallest sum is $\qquad$. | Answer: 8 | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18. Given vectors $\boldsymbol{m}=(\sin A, \cos A), \boldsymbol{n}=(\cos B, \sin B), \boldsymbol{m} \cdot \boldsymbol{n}=\sin 2 C$, and $A, B, C$ are the angles opposite to sides $a, b, c$ of $\triangle A B C$.
(1) Find the size of $C$;
(2) If $\sin A, \sin C, \sin B$ form a geometric sequence, and $\overrightarrow{C A... | (1) $\sin A \cos B+\cos A \sin B=\sin (A+B)=\sin C=\sin 2 C \Rightarrow \cos C=\frac{1}{2} \Rightarrow C=60^{\circ}$.
(2) $\sin A \cdot \sin B=\sin ^{2} C, a b \cos C=18 \Rightarrow a b=36$, by the Law of Sines we get $c^{2}=a b=36 \Rightarrow c=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For example, among 211 theater troupes, some troupes perform each day while others watch (those who perform cannot watch). If each troupe has seen performances by 10 other troupes, how many days are needed for the performances at least?
Translating the above text into English, please keep the original text's line br... | Let the total number of performance days be $n$, and the set of days on which the $i$-th troupe does not perform be $A_{i}$ $(i=1,2, \cdots, 11)$. Then $A_{1}, A_{2}, \cdots, A_{11}$ are all subsets of $X=\{1, 2, \cdots, n\}$.
Since each troupe has seen the performances of all other troupes, $A_{i}, A_{j} (1 \leqslant... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Let $x=\cos \frac{2}{5} \pi+i \sin \frac{2}{5} \pi$, then $1+x^{4}+x^{8}+x^{12}+x^{16}=$ | 7. 0
$x$ is a 5th root of unity, so $1+x^{4}+x^{8}+x^{12}+x^{16}=1+x^{4}+x^{3}+x^{2}+x=0$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in the figure, the teacher wrote 9 numbers on the blackboard, and asked to select 3 numbers to form a three-digit number, with any two selected numbers not coming from the same row or the same column. A total of $\qquad$ three-digit numbers that can be divided by 4 can be formed. | $8$ | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. On a rotating round table, there are 8 white teacups and 7 black teacups, and around the table sit 15 dwarfs, each wearing a hat, with 8 white hats and 7 black hats. Each dwarf places a teacup of the same color as their hat in front of them. After the table rotates randomly, what is the maximum number of teacups tha... | 9.7.
First, observe any arrangement of the teacups and write their colors in a row. Below this row of colors, write down all its different cyclic shifts, a total of 14. Then count how many times the colors match at the same position in the initial arrangement and during the shifts. For each black teacup, there are 6 s... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Calculate: $\sum_{k=1}^{2 n-1}(-1)^{k-1} \cdot \frac{n-k}{C_{2 n}^{k}}$. | 11. $\sum_{k=1}^{2 n-1}(-1)^{k-1} \frac{n-k}{C_{2 n}^{k}}=\sum_{k=1}^{n-1}(-1)^{k-1} \frac{n-k}{C_{2 n}^{k}}-\sum_{i=1}^{n-1}(-1)^{i-1} \frac{n-i}{C_{2 n}^{i}}=0$. | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.83 On the ground, there are 10 birds pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are on the circle with the most birds, at minimum? | [Solution 1] Use 10 points to represent 10 birds. If any 4 points among the 10 points are concyclic, then the 10 points are all on the same circle. Below, let points $A, B, C, D$ not be concyclic. At this time, a circle can be drawn through any 3 non-collinear points among the 4 points, and at most 4 different circles ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
46 Sequence $\left\{a_{n}\right\}: a_{1}=3, a_{n}=3^{a_{n-1}}(n \geqslant 2), a_{1990}$ The last digit of is
A. 3
B. 9
C. 7
D. 1 | 46 C. When $n=1$, $a_{1}=4 \times 0+3$; when $n=2$, $a_{2}=4 \times 6+3$. By mathematical induction and the binomial theorem, it can be proven that $a_{n}=4 m+3, m \in \mathbf{N} \cup\{0\}$. Therefore, $a_{1990}=3^{a_{1} 9_{89}}=3^{4 m+3}=\left(3^{4}\right)^{m} - 3^{3}=(81)^{m} \cdot 27$, where $m$ is a non-negative in... | 7 | Number Theory | MCQ | Yes | Yes | olympiads | false |
10. As shown in the figure, in isosceles $\triangle ABC$, $AB=AC$, point $P$ is on the altitude $AD$ of $BC$, and $\frac{AP}{PD}=\frac{1}{3}$. The extension of $BP$ intersects $AC$ at $E$. If $S_{\triangle ABC}=21$, then $S_{\triangle DEC}=$ $\qquad$ . | $9$ | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The last digit of $1+2^{2}+3^{3}+4^{4}+\cdots+1992^{1992}$ is ( ).
A. 8
B. 2
C. 4
D. 0 | 4. D.
For the last digit of $a_{n}\left(n \in \mathbf{N}_{+}\right)$, it can be proven that it cycles with a period of 4 (or 2 or 1), and it only depends on the last digit of $a$. It can be verified that the sum of the last digits of $a^{n}$ for any 10 consecutive $n$ is 7. Therefore, the last digit of the original ex... | 0 | Number Theory | MCQ | Yes | Yes | olympiads | false |
12. $f(x)=a x^{2}+b x+c, a, b, c \in \mathbf{R}$, and when $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then when $|x| \leqslant 1$, the maximum value of $|a x+b|$ is $\qquad$ . | 12. $2 f(0)=c, f(1)=a-b+c, f(-1)=a-b+c \cdot a=\frac{1}{2}(f(1)+f(-1)-2 f(0)), b=\frac{1}{2}$ $(f(1)-f(-1)),|a x+b|=\left|\left(\frac{1}{2} x+\frac{1}{2}\right) f(1)+\left(\frac{1}{2} x-\frac{1}{2}\right) f(-1)-x f(0)\right| \leqslant \frac{1}{2}|x+1|+$ $\frac{1}{2}|x-1|+|x| \leqslant 2$. The function $f(x)=2 x^{2}-1$ ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 105$ The roots of the equation $64 x^{3}-144 x^{2}+92 x-15=0$ form an arithmetic sequence, the difference between the largest and smallest root is
(A) 2 .
(B) 1 .
(C) $\frac{1}{2}$.
(D) $\frac{3}{8}$.
(E) $\frac{1}{4}$.
(18th American High School Mathematics Examination, 1967) | The original equation can be transformed into
$$
x^{3}-\frac{9}{4} x^{2}+\frac{23}{16} x-\frac{15}{64}=0 .
$$
Let the three roots of the equation be $a-d, a, a+d$. By Vieta's formulas,
$$
\left\{\begin{array} { l }
{ ( a - d ) + a + ( a + d ) = \frac { 9 } { 4 } , } \\
{ ( a - d ) a ( a + d ) = \frac { 1 5 } { 6 4 } ... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example $\mathbf{3}$ Given the quadratic equation in $x$, $a(1+\mathrm{i}) x^{2}+\left(1+a^{2} \mathrm{i}\right) x+a^{2}+\mathrm{i}=0$, has real roots, find the value of the real number $a$.
The equation is $a(1+\mathrm{i}) x^{2}+\left(1+a^{2} \mathrm{i}\right) x+a^{2}+\mathrm{i}=0$. | Analysis: Let the real root of the equation be $x_{0}$, substitute it into the original equation, and according to the necessary and sufficient conditions for the equality of complex numbers, transform it into a real number problem to solve.
Solution: Let the real root of the equation be $x_{0}$, then $a(1+\mathrm{i})... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1 Let real numbers $a$, $b$, $c$ satisfy $a^{2}+2 b^{2}+3 c^{2}=\frac{3}{2}$, prove: $3^{a}+9^{a}+27^{c} \geqslant 1$. (Li Shenghong's problem) | By the Cauchy-Schwarz inequality, $(a+2 b+3 c)^{2} \leqslant\left(\sqrt{1}^{2}+\sqrt{2}^{2}+\right.$ $\left.\sqrt{3}^{2}\right)\left[(\sqrt{1} a)^{2}+(\sqrt{2} b)^{2}+(\sqrt{3} c)^{2}\right]=9$, so $a+2 b+3 c \leqslant 3$, hence $3^{a}+9^{b}+27^{c} \geqslant 3 \sqrt[3]{3^{-(a-2 b+3 c}} \geqslant 3 \sqrt[3]{3^{-3}}=1$. | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
(10) (15 points) Given the sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}, n=2$,
$3, \cdots$, where $m$ is a real parameter. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$. | 10 Because $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}, a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}$, $a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, a_{6}=\frac{\frac{a_{0}}{m a_{1}}}{m \frac{1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. There are 8 people, each receiving a unique message at the same time. They inform each other of all the messages they know by making phone calls, with each call taking exactly 3 minutes. To ensure that everyone knows all the messages, at least ( ) minutes are needed.
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。 | 【Answer】9 minutes.
【Analysis】Key point: Optimal strategy problem
To minimize the time, note 2 points: (1) Do not call people who already know the information they have;
(2) Divide into 4 groups and call simultaneously.
Let the 8 people be $\mathrm{ABCDEFGH}$, each holding (1) (2) (3) (4) (5) (6) (7) (8) information (li... | 9 | Number Theory | proof | Yes | Yes | olympiads | false |
Example 7 Given that in any 3 people among $n$ people, there must be 2 who know each other, if there must be 4 people who all know each other, find the minimum value of $n$.
| Solve: Use $n$ points to represent $n$ people. If two people know each other, the line connecting the corresponding points is colored red; if two people do not know each other, the line connecting the corresponding points is colored blue. Thus, the problem is transformed into finding the smallest positive integer $n$ s... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Given the sets $A=\left\{2,4, a^{3}-2 a^{2}-a+7\right\}, B=\left\{-4, a+3, a^{2}-2 a+2, a^{3}+a^{2}+3 a+7\right\}$, and $A \cap B=\{2,5\}$, then the set of possible values of $a$ is $\qquad$ . | 7. $a=-1$ is obtained by solving $a^{3}-2 a^{2}-a+7=5$, which yields $a=-1,1,2$. Calculate the sets $A$ and $B$ respectively and check if they satisfy $A \cap B=\{2,5\}$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. We call integers composed of the digit 8 "lucky numbers", such as $8, 88, 888$, etc. A number obtained by adding 8 "lucky numbers" is called a "happy number", such as $8+8+8+8+8+8+88+88=224$. The number of "happy numbers" less than 1000 is $\qquad$. | $10$ | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Given that $P$ is a point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, $F_{1}$ is its left focus, and $Q$ is on $P F_{1}$ such that
$$
\overrightarrow{O Q}=\frac{1}{2}\left(\overrightarrow{O P}+\overrightarrow{O F_{1}}\right),|\overrightarrow{O Q}|=3 \text {. }
$$
Then the distance from point $P$ to the lef... | 4.5.
From the problem, we know that $Q$ is the midpoint of side $F_{1} P$. Let the right focus of the ellipse be $F_{2}$, and connect $P F_{2}$. Then
$$
\begin{array}{l}
|O Q|=\frac{1}{2}\left|P F_{2}\right|=3 \\
\Rightarrow\left|P F_{2}\right|=6 \Rightarrow\left|P F_{1}\right|=4 .
\end{array}
$$
Let the distance fro... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Xiao Ming and Xiao Qiang are playing a number game. Xiao Ming chose a number $x$ (between 0 and 9), and then said: “I am thinking of a three-digit number (the hundreds place can be 0), where the hundreds digit is $x$, the tens digit is 3, and it is divisible by 11. Can you find the units digit of this three-digit n... | 【Analysis】Let's assume this three-digit number is $\overline{x 3 y}$. If this three-digit number is divisible by 11, then $11 \mid x+y-3$. According to the problem, no matter which of the ten digits from 0 to 9 $y$ is, this three-digit number cannot be divisible by 11. Therefore, we should have $1 \leq x+y-3 \leq 10$, ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. As shown in Figure 12-1, Xiao Ming goes from A to B, each time walking three squares in one direction, then turning 90 degrees and walking one more square, for example, in Figure 12-2, starting from point $\mathrm{C}$, he can reach eight positions. How many times at least does Xiao Ming need to walk to get from poi... | 【Analysis】The answer is as shown in the figure, at least 5 times. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) Two people, A and B, take turns selecting numbers from the integers 1 to 17, with the rule: they cannot select numbers that have already been chosen by either party, they cannot select a number that is twice an already chosen number, and they cannot select a number that is half of an already chosen numb... | 【Answer】Solution:
According to the above analysis, if B chooses 6, then A cannot choose 3 or 12 anymore. The following six pairs can be chosen randomly.
5, 10
7, 14
$1, 2$
9
11
13
15
17
Therefore, the answer to this question should be 6. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. As shown in the figure, in the right trapezoid $A B C D$, $A B=7$. $A D=2, B C=3$, if point $P$ on side $A B$ makes the triangle with vertices $P, A, D$ similar to the triangle with vertices $P, B, C$, then the number of such points $P$ is
(A) 1;
(B) 2;
(C) 3;
(D) 4. | 3. (C)
Solution Let $A P=x$, then $P B=7-x$,
(1) If $\triangle P A D \backsim \triangle P B C$, then $\frac{x}{7-x}=\frac{2}{3}, \quad \therefore x=\frac{14}{5}<7$. This meets the condition.
(2) If $\triangle P A D \backsim \triangle C B P$, then
$$
\frac{x}{3}=\frac{2}{7-x},
$$
$\therefore x_{1}=1, x_{2}=6$ also meet... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example: 510 people go to the bookstore to buy books, it is known that (1) each person bought 3 types of books; (2) any two people have at least one book in common. How many people at most bought the most purchased book? Explain your reasoning.
The above text is translated into English, preserving the original text's ... | Solution: In the right figure, there are 5 triangles formed by the center and two adjacent vertices of the regular pentagon, and there are also 5 triangles formed by 3 non-adjacent vertices of the regular pentagon. It is not difficult to see that any two of these 10 triangles share at least one common vertex. Write out... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10 Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$ and for any $x \in$ $\mathbf{R}$, we have
$$
\begin{array}{l}
f(x+5) \geqslant f(x)+5 \\
f(x+1) \leqslant f(x)+1 .
\end{array}
$$
If $g(x)=f(x)+1-x$, then $g(2002)=$ | Analyze: First determine $f(2002)$ from the given conditions, then we can find $g(2002)$.
Solution: From the given conditions, we have
$$
\begin{aligned}
f(x)+5 & \leqslant f(x+5) \leqslant f(x+4)+1 \\
& \leqslant f(x+3)+2 \leqslant f(x+2)+3 \\
& \leqslant f(x+1)+4 \leqslant f(x)+5,
\end{aligned}
$$
Therefore, all the... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $\triangle A B C$ be an acute-angled triangle with the lengths of the sides opposite to its internal angles being $a$, $b$, and $c$ respectively. If $2 a^{2}=2 b^{2}+c^{2}$, then the minimum value of $\tan A+\tan B+\tan C$ is $\qquad$. | 7.6.
$$
\begin{array}{l}
\text { Given } 2 a^{2}=2 b^{2}+c^{2} \\
\Rightarrow 2 \sin ^{2} A=2 \sin ^{2} B+\sin ^{2} C \\
\Rightarrow \cos 2 B-\cos 2 A=\sin ^{2} C \\
\Rightarrow 2 \sin (A-B) \cdot \sin (A+B)=\sin ^{2} C \\
\Rightarrow 2 \sin (A-B)=\sin C=\sin (A+B) \\
\Rightarrow \sin A \cdot \cos B=3 \cos A \cdot \sin... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given $x y z+y+z=12$, then the maximum value of $\log _{4} x+\log _{2} y+\log _{2} z$ is $\qquad$ | Solve:
$12=x y z+y+z \geqslant 3 \sqrt[3]{x y^{2} z^{2}} \Rightarrow x y^{2} z^{2} \leqslant 64$, equality holds when $x=\frac{1}{4}, y=z=4$. Therefore, $\log _{4} x+\log _{2} y+\log _{2} z=\log _{4}\left(x y^{2} z^{2}\right) \leqslant 3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13-26 Given $|y| \leqslant 1$, and $2 x+y=1$, then the minimum value of $2 x^{2}+16 x+3 y^{2}$ is
(A) $19 / 7$.
(B) 3 .
(C) $27 / 7$.
(D) 13 .
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1997) | [Solution] Given $-1 \leqslant y \leqslant 1$, and $2 x+y=1$, thus $0 \leqslant x \leqslant 1$.
Also, $2 x^{2}+16 x+3 y^{2}=2 x^{2}+16 x+3(1-2 x)^{2}$
$$
=14 x^{2}+4 x+3,
$$
we know that $z=14 x^{2}+4 x+3$ is a parabola opening upwards with the axis of symmetry at $x=-\frac{1}{7}$. In the interval $0 \leqslant x \leqs... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
2・18 Let $x_{1}, x_{2}$ be the two roots of the quadratic equation $x^{2}+x-3=0$. Then the value of $x_{1}^{3}-4 x_{2}^{2}+19$ is
(A) -4.
(B) 8.
(C) 6.
(D) 0.
(China Junior High School Mathematics League, 1996) | [Solution] $\because x_{1}, x_{2}$ are the two roots of the quadratic equation $x^{2}+x-3=0$,
$$
\therefore \quad x_{1}^{2}+x_{1}-3=0, x_{2}^{2}+x_{2}-3=0 \text {, }
$$
i.e., $x_{1}^{2}=3-x_{1}, x_{2}^{2}=3-x_{2}$.
By the relationship between roots and coefficients, we know $x_{1}+x_{2}=-1$, thus we have
$$
\begin{ali... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
D Given that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$, find the sum of the coefficients of this polynomial. (Supplied by Zhu Huawei) | Method One: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1, \alpha \beta=1$, and the sum of the coefficients sought is $S_{k}$. From
$$
\begin{aligned}
& (\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
= & \left(\alpha^{k}+\beta^{k}\right)+\alpha \beta\left(\alpha^{k-2}+\beta^{k-2}\right),
\end{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $y=f(x)$ be a strictly monotonically increasing function, and its inverse function be $y=g(x)$. Let $x_{1}, x_{2}$ be the solutions of the equations $f(x)+x=2$ and $g(x)+x=2$, respectively, then $x_{1}+x_{2}=$ $\qquad$ . | Let $f\left(x_{1}\right)=x_{2}$, then $g\left(x_{2}\right)=x_{1}$. Substituting into the equation, we get $\left\{\begin{array}{l}f\left(x_{1}\right)+x_{1}=x_{1}+x_{2}=2 \text {, } \\ g\left(x_{2}\right)+x_{2}=x_{1}+x_{2}=2 \text {. }\end{array}\right.$
Also, $y=f(x)$ is a strictly increasing function, and $x_{1}, x_{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. For $\triangle A B C$ with side lengths $a, b, c(a \leqslant b \leqslant c)$, the following conditions are satisfied: (1) $a, b, c$ are all integers; (2) $a, b, c$ form a geometric sequence; (3) at least one of $a$ and $c$ equals 100. The number of all possible triples $(a, b, c)$ is $\qquad$ | 8. 10 Detailed Explanation: According to the problem, positive integers $a, b, c$ satisfy $a \leqslant b \leqslant c, a+b>c, b^{2}=a c$, and at least one of $a, c$ is 100. (i) If $a=100$, then $b^{2}=100 c$, so $10 \mid b$. We can take $100+b>c=\frac{b^{2}}{100}, b^{2}-100 b-100^{2}100, b^{2}+100 b-100^{2}>0$, hence $5... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.99 Let each side of a regular octagon be colored either blue or yellow. The following operation is allowed: simultaneously modify the coloring of each side, changing a side to blue if its two adjacent sides are of different colors, and to yellow if they are of the same color. Prove that after several steps of operati... | [Solution] Correspond the blue edges and yellow edges to the numbers -1 and +1, respectively. Thus, the allowed operation in the problem is to change the number on each edge to the product of the numbers on its two adjacent edges. Let the numbers on the 8 edges at the beginning be $a_{1}, a_{2}, \cdots, a_{8}$. After t... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false |
15. In the magical world of Hogwarts, a new operation $\triangle$ is defined, stipulating that $a \triangle b=(a+b) \div b$, so: $3 \frac{4}{5} \triangle \frac{19}{20}=$ $\qquad$ . | $5$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. If positive numbers $x, y$ satisfy $x+3 y=5 x y$, then the minimum value of $3 x+4 y$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve
$x+3 y=5 x y \Rightarrow \frac{3}{x}+\frac{1}{y}=5$, so $3 x+4 y=\frac{1}{5}(3 x+4 y)\left(\frac{3}{x}+\frac{1}{y}\right)=\frac{1}{5}\left(9+4+\frac{3 x}{y}+\frac{12 y}{x}\right) \geqslant 5$, equality holds when $x=2 y \Rightarrow x=1, y=\frac{1}{2}$. Therefore, the minimum value of $3 x+4 y$ is 5. | 5 | Algebra | proof | Yes | Yes | olympiads | false |
Example 9 Consider a square on the complex plane, whose four vertices correspond to the four roots of a certain monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square.
Translate the above text into English, please retain the original text's... | Analysis: After translating the origin of the complex plane to the center of a square, the vertices of the square are uniformly distributed on a circle. Therefore, in the original complex plane, the corresponding complex numbers of these vertices are the solutions to the equation $(x-a)^{4}=b$, where $a, b \in C$. Expa... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 11 Let $p(x)$ be a polynomial of degree $3n$, such that $P(0)=P(3) \cdots=P(3n)=2, P(1)=$ $P(4)=\cdots=P(3n-2)=1, P(2)=P(5)=\cdots=P(3n-1)=0, P(3n+1)=730$. Determine $n$.
(13th US Olympiad Problem) | Notice that as $x$ increases from 0 to $3n$ by 1, $P(x)-1$ cyclically takes the values $1,0,-1$. Mimicking this feature, consider the 3rd root of unity $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$. Note that for $n=0,1,2, \cdots$, the sequence $\left\{\omega^{n}\right\}=\{1, \omega, \omega^{2}, 1, \omega, \omega... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
37. As shown in the figure, a line passing through the origin intersects the graph of the inverse proportion function $y=-\frac{6}{x}$ at points $A, C$. Perpendiculars are drawn from points $A$, $C$ to the $x$-axis, with feet of the perpendiculars at $B, D$, respectively. The area of quadrilateral $A B C D$ is $\qquad$... | answer: 12 | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. From 1 AM to 1 PM on the same day, the minute hand catches up with the hour hand $\qquad$ times. | $11$ | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Example Mountain (2003 China National Training Team) In $\triangle A B C$, $A C>A B, P$ is the intersection of the perpendicular bisector of $B C$ and the internal angle bisector of $\angle A$, construct $P X \perp A B$, intersecting the extension of $A B$ at point $X, P Y \perp A C$ intersecting $A C$ at point $Y, Z$ ... | Consider the circumcircle $\odot O$ of $\triangle ABC$, and let the midpoint of arc $BC$ (not containing $A$) be $D$, and the midpoint of chord $BC$ be $M$.
Since $\overparen{BD} = \overparen{CD}$, we have
$$
BD = DC, \angle BAD = \angle CAD,
$$
and $D$ lies on the angle bisector of $\angle BAC$ and the perpendicular ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Given the set $M=\{1,2, \cdots, 99\}$. Now, nine elements are randomly selected from $M$ to form a subset, and the smallest number in this subset is denoted as $\xi$. Then $\mathrm{E} \xi=$ $\qquad$. | 10. 10 .
For a set $M$, the number of nine-element subsets with the smallest number being a positive integer $k (1 \leqslant k \leqslant 99)$ is $\mathrm{C}_{99-k}^{8}$.
$$
\begin{array}{l}
\text { Hence } P(\xi=k)=\frac{\mathrm{C}_{99-k}^{8}}{\mathrm{C}_{99}^{9}} \\
\Rightarrow \mathrm{E} \xi=\sum_{k=1}^{91} \frac{k ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10th "Hope Cup" Invitational Competition Question) Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}+x+1=0$, then the value of the sum $\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{1}}{x_{2}}\right)^{3}+\cdots+\left(\frac{x_{1}}{x_{2}}\right)^{1998}$ is $\qquad$ $ـ$. | 8. Fill 0. Reason: Since $1998=3 \cdot 666$, the sum can be transformed into the sum of 666 three-term expressions. These three-term expressions are
$$
\begin{array}{l}
\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{1}}{x_{2}}\right)^{3},\left(\frac{x_{1}}{x_{2}}\right)^{4}+\left(\frac{x_{1}}{x... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
One. (20 points) Given the equation in terms of $x$
$$
\frac{x^{2}+(5-2 m) x+m-3}{x-1}=2 x+m
$$
has no positive real roots. Find the range or value of $m$.
| $$
\begin{array}{l}
x^{2}+(3 m-7) x+(3-2 m)=0 . \\
\text { By } \Delta=(3 m-7)^{2}-4(3-2 m) \\
=\left(3 m-\frac{17}{3}\right)^{2}+\frac{44}{9}>0,
\end{array}
$$
we know that equation (1) always has two distinct real roots. Let the two roots be $x_{1}$ and $x_{2}$. Then
$$
x_{1}+x_{2}=7-3 m, x_{1} x_{2}=3-2 m \text {. ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find t... | The function $f(x)$ is symmetric about the line $x=-1$, and $1 \leqslant f(1) \leqslant 1 \Rightarrow f(1)=1$, then $f(x)=a(x+1)^{2} \Rightarrow f(1)=4 a=1 \Rightarrow a=\frac{1}{4} \Rightarrow f(x)=\frac{(x+1)^{2}}{4}$.
By the AM-GM inequality, $f(x)=\frac{(x+1)^{2}}{4}=\frac{x^{2}+1+2 x}{4} \geqslant x$, thus $f(x)=\... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=2, a_{2}=6, \\
a_{n+1}=\frac{a_{n}^{2}-2 a_{n}}{a_{n-1}}(n=2,3, \cdots) .
\end{array}
$$
Then $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}+n}\right\}=$ $\qquad$ | 7.1 .
Notice that, $a_{1}=1 \times 2, a_{2}=2 \times 3$.
Assume $a_{n-1}=(n-1) n, a_{n}=n(n+1)$. Then
$$
\begin{array}{l}
a_{n+1}=\frac{n(n+1)(n(n+1)-2)}{(n-1) n} \\
=\frac{n(n+1)(n+2)(n-1)}{(n-1) n} \\
=(n+1)(n+2) .
\end{array}
$$
Thus, $a_{n}=n^{2}+n$ always holds.
Also, $\sqrt{n^{2}+2 n} \in(n, n+1)$, so
$\left\{\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 10】
Red pens cost 5 yuan each, blue pens cost 7 yuan each, spending 102 yuan in total for 16 pens, bought $\qquad$ blue pens. | 【Analysis and Solution】
Chicken and rabbit in the same cage.
(Method One)
Assume 102 pens are all red pens, then it would cost $5 \times 16=80$ yuan; which is $102-80=22$ yuan less than the actual amount;
One red pen costs $7-5=2$ yuan less than one blue pen;
The number of blue pens bought is $22 \div 2=11$.
(Method Tw... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1.6 ** The terms of an arithmetic sequence are all real numbers, the common difference is 4, and the square of any term plus the sum of the remaining terms does not exceed 100. Find the maximum number of terms in this sequence. | Let the real number sequence $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence satisfying the given conditions, then
i.e., $\square$
$$
\begin{array}{c}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100, \\
a_{1}^{2}+\frac{\left(a_{1}+4\right) \mid\left[a_{1}+(n-1) \cdot 4\right]}{2} \cdot(n-1) \leqslant 100,
\end{array... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
80. In a junior high school, 100 students and teachers from two graduating classes are taking a graduation photo on the steps. The photographer wants to arrange them in a trapezoidal formation with fewer people in the front rows and more in the back rows (number of rows $\geqslant 3$), and requires that the number of p... | answer: 2 | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. There are 9 people lost in the mountains, and the food they have is only enough to last for 5 days. One day later, these 9 people encounter another group of lost people who have no food at all. After calculation, if the two groups share the food, with each person eating the same amount of food per day, the food wil... | 【Analysis】Let's assume each person eats 1 portion of food per day, then the original amount of food is 45 portions. After 1 day, 36 portions remain. If this can last for 3 days, then each day they eat $36 \div 3=12$ portions, meaning there are 12 people at this time, so the number of people from the second team who got... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. If the function $f(x)=\frac{c x}{2 x+3}, x \neq-\frac{3}{2}$, and for all real numbers $x$ except $-\frac{3}{2}$, $f[f(x)]=x$, then $c=$
A. -3
B. $-\frac{3}{2}$
C. $\frac{3}{2}$
D. 3 | 2. A $f(f(x))=\frac{c^{2} x}{2 c x+6 x+9}=x,(2 c+6) x+\left(9-c^{2}\right)=0, c=-3$ | -3 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. (5 points) It is stipulated: the product of $n$ $a$s is denoted as: $\underbrace{\mathrm{a} \times \mathrm{a} \times \cdots \times \mathrm{a} \times \bar{c}}_{\mathrm{n} \text { a's }}=a^{n}$
Observe the following calculations: $8^{1}=8 ; 8^{2}=64 ; 8^{3}=512 ; 8^{4}=4096 ; 8^{5}=32768 ; 8^{6}=262144 ; 8^{7}=2097152... | 【Analysis】Observing “$8^{1}=8 ; 8^{2}=64 ; 8^{3}=512 ; 8^{4}=4096 ; 8^{5}=32768 ; 8^{6}=262144 ; 8^{7}=2097152$; $8^{8}=16777216, \cdots$” we know that, except for the first term, the remainder of the subsequent terms when divided by 10 is $4, 2, 6, 8, 4, 2, 6, 8, \cdots$; that is, they cycle in a group of 4, 2, 6, 8. ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given the function $f(x)=a \ln x+x^{2}$ has an extremum at $x=1$, then the value of the real number $a$ is
A. -2
B. -1
C. 1
D. 2 | Solve:
$f^{\prime}(x)=\frac{a}{x}+2 x \Rightarrow f^{\prime}(1)=a+2=0 \Rightarrow a=-2$, so choose $A$.
| -2 | Calculus | MCQ | Yes | Yes | olympiads | false |
Given $f(x)=a x^{2}+b x+c(a, b, c$ are real numbers) and its absolute value on $[-1,1]$ is $\leqslant 1$, find the maximum value of $|a|+|b|+|c|$. | - $c=f(0)$.
$$
\begin{array}{l}
a+b+c=f(1) \\
a-b+c=f(-1) .
\end{array}
$$
If $a$ and $b$ have the same sign, then
$$
\begin{aligned}
|a|+|b|+|c| & =|a+b|+|c|=|f(1)-c|+|c| \\
& \leqslant|f(1)|+|c|+|c| \\
& \leqslant|f(1)|+2|f(0)| \leqslant 3 .
\end{aligned}
$$
If $a$ and $b$ have opposite signs, then
$$
|a|+|b|+|c|=|... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Using $1$, $2$, and $3$, we can form 6 different three-digit numbers without repeating digits. Arrange these three-digit numbers in ascending order, and find the difference (larger minus smaller) between adjacent numbers. There are $\qquad$ different differences. | $3$ | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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