problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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13. Given the function $f(x)=a x-\frac{3}{2} x^{2}$ has a maximum value no greater than $\frac{1}{6}$, and when $x \in\left[\frac{1}{4}, \frac{1}{2}\right]$, $f(x) \geqslant \frac{1}{8}$, find the value of $a$. | 13. Since the maximum value of $f(x)=a x-\frac{3}{2} x^{2}$ is no greater than $\frac{1}{6}$, we have $f\left(\frac{a}{3}\right)=\frac{a^{2}}{6} \leqslant \frac{1}{6}$, so $a^{2} \leqslant 1$. Also, when $x \in\left[\frac{1}{4}, \frac{1}{2}\right]$, $f(x) \geqslant \frac{1}{8}$, so $\left\{\begin{array}{l}f\left(\frac{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In an acute triangle $ABC$, $\angle A=30^{\circ}$. A circle is drawn with $BC$ as its diameter, intersecting $AB$ and $AC$ at points $D$ and $E$ respectively. Connecting $D$ and $E$, the triangle $ABC$ is divided into triangle $ADE$ and quadrilateral $BDEC$. Let the areas of these shapes be $S_{1}$ and $S_{2}$ respe... | 4. 3
As shown in the figure, $BC$ is the diameter of the circle,
$$
\begin{array}{l}
\angle AEB = 180^{\circ} - \angle BEC = 90^{\circ}, \\
\therefore \quad \frac{AE}{AB} = \cos A = \cos 30^{\circ} = \frac{\sqrt{3}}{2}. \\
\text{Also, } \triangle ADE \sim \triangle ABC, \\
\therefore \quad \frac{AD}{AC} = \frac{AE}{AB... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given the imaginary number $z$ satisfies $z^{3}+1=0, z \neq-1$. Then $\left(\frac{z}{z-1}\right)^{2018}+\left(\frac{1}{z-1}\right)^{2018}=$ $\qquad$ . | 5. -1 .
$$
\begin{array}{l}
\text { Given } z^{3}+1=0 \\
\Rightarrow(z+1)\left(z^{2}-z+1\right)=0 \\
\Rightarrow z^{2}-z+1=0 . \\
\text { Then }\left(\frac{z}{z-1}\right)^{2018}+\left(\frac{1}{z-1}\right)^{2018} \\
=\frac{z^{2018}+1}{\left(z^{2}\right)^{2018}}=\frac{z^{2}+1}{z^{4}}=\frac{z^{2}+1}{-z}=-1 .
\end{array}
$... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question 53, Let the area of $\triangle A B C$ be $1, \angle A$ be opposite to the side length $a$, try to find the minimum value of $\mathrm{a}^{2}+\frac{1}{\sin \mathrm{A}}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Question 53, Solution: Let the other two sides of $\triangle A B C$ be $b$ and $c$, respectively. According to the given conditions, we have $b c=\frac{2}{\sin A}$. Thus,
$$
\begin{array}{l}
a^{2}+\frac{1}{\sin A}=b^{2}+c^{2}-2 b c \cdot \cos A+\frac{1}{\sin A} \geq 2 b c-2 b c \cdot \cos A+\frac{1}{\sin A} \\
=2 b c(1... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (5 points) This year, Dandan is 4 years old, and Dandan's father is 28 years old. In $a$ years, the father's age will be 3 times Dandan's age. Then the value of $a$ is . $\qquad$ | 【Analysis】According to "Dandan is 4 years old this year, and Dandan's father is 28 years old", we know that the age difference between Dandan's father and Dandan this year is $28-4=24$ years. Since the age difference does not change with time, using the difference multiple formula, we can find Dandan's age when her fat... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. (IMO-31 Shortlist) Find all natural numbers $n$, such that every natural number represented in decimal notation with $n-1$ digits 1 and one digit 7 is a prime number. | 12. A natural number $N$ composed of $n-1$ ones and one 7 can be expressed as
$N=A_{n}+6 \cdot 10^{k}$, where $A_{n}$ is a natural number composed of $n$ ones, and $0 \leqslant k \leqslant n$.
When $3 \mid n$, the sum of the digits of $A_{n}$ is divisible by 3, so
$3 \mid A_{n}$.
Thus, $3 \mid N$.
Since $N>3$, $N$ is n... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$17 \cdot 161$ As shown in the figure, on the sides $BC, CA$, and $AB$ of the equilateral $\triangle ABC$, there are internal division points $D, E, F$, dividing the sides in the ratio $2:(n-2)$ (where $n>4$). The segments $AD, BE, CF$ intersect to form $\triangle PQR$, whose area is $\frac{1}{7}$ of the area of $\tria... | [Solution] As shown in the figure, $B D: D C=C E: E A=A F: F B=2:(n-2)$.
Notice that line $B P E$ intersects the sides or their extensions of $\triangle A D C$ at $P, B, E$. By Menelaus' theorem, we have
$$
\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A}=\frac{A P}{P D} \cdot \frac{2}{n} \cdot \frac{2}{n-2}... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. The complex number $z$ satisfies $|z|(3 z+2 i)=2(i z-6)$, then $|z|$ equals
保留了源文本的换行和格式。 | 5. 2
Analysis: Let $r=|z|$, then the original equation can be transformed into: $12+2 r i=2 i z-3 r z$, so $2 \sqrt{36+r^{2}}=r \sqrt{9 r^{2}+4}$, yielding $r=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (10 points) From a point $P$ inside $\triangle A B C$, perpendiculars are drawn to the sides $B C, C A, A B$, with feet of the perpendiculars being $D, E, F$ respectively. Semi-circles are constructed outwardly on $A F, B F, B D, C D, C E, A E$ as diameters. As shown in the figure, the areas of these six semi-circle... | 3. (10 points) From a point $P$ inside $\triangle ABC$, perpendiculars are drawn to sides $BC$, $CA$, and $AB$, with feet of the perpendiculars being $D$, $E$, and $F$ respectively. Semi-circles are constructed outward with diameters $AF$, $BF$, $BD$, $CD$, $CE$, and $AE$. The areas of these six semi-circles are denote... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Let the line $y=k x+m$ passing through any point $P$ on the ellipse $\frac{x^{2}}{4}+y^{2}=1$ intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ at points $A$ and $B$, and let the ray $P O$ intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ at point $Q$. Then the value of $S_{\triangle A B Q}: S_{\tr... | 4.3.
Let $d_{(X, Y Z)}$ denote the distance from point $X$ to line $Y Z$.
Notice that, $\frac{S_{\triangle A B Q}}{S_{\triangle A B O}}=\frac{d_{(Q, A B)}}{d_{(O, A B)}}=\frac{|P Q|}{|O P|}$.
Let $\frac{|O Q|}{|O P|}=\lambda$. Then $\overrightarrow{O Q}=-\lambda \overrightarrow{O P}$.
Let $P\left(x_{0}, y_{0}\right)$.... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$31 \cdot 50$ satisfies the system of equations $\left\{\begin{array}{l}a b+b c=44 \\ a c+b c=23\end{array}\right.$ the number of positive integer triples $(a, b, c)$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(35th American High School Mathematics Examination, 1984) | [Solution] From the given conditions and the requirement to find the positive integer solutions of the system of equations, from the second equation $a c + b c = 23$, we have $c(a + b) = 23$.
Noting that 23 is a prime number, we can conclude that one of the two factors on the left side must be 1, and the other must be... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is | 5. $\{x \mid x=2\}$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
39. Express 15 as the sum of three different natural numbers (excluding 0), there are $\qquad$ ways to do so. | answer: 12 | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. The number of real solutions to the equation $\sin \left(x-\frac{\pi}{4}\right)=\frac{1}{4} x$ is | 18. From the graphical method, we know the number of intersection points (i.e., the solutions to the equation) is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question 225, Find the smallest positive integer $n(n \geq 4)$, such that from any $n$ different integers, one can always select four distinct numbers $a, b, c, d$, satisfying $20 \mid (a+b-c-d)$. | Question 225, Solution: First, consider the different residue classes modulo 20. Let $\mathrm{M}$ be a set of $\mathrm{k}$ elements that are mutually incongruent modulo 20. Define the set $A=\left\{a_{i}+a_{j} \mid a_{i}, a_{j} \in M, 1 \leq i < j \leq k\right\}$. Since there are more than 20 elements in $A$, by the pi... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) Given
the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$
with the left and right endpoints of the major axis being $A$ and $B$, and the left and right foci being $F_{1}$ and $F_{2}$. If there exists a point $P$ on the ellipse $\Gamma$ such that $P F_{1}$ and $P F_{2}$ trisect $\angle... | 9. From $S_{\triangle P A F_{1}}=S_{\triangle P B F_{2}}$, we get $P A \cdot P F_{1}=P B \cdot P F_{2}$.
Therefore, $P$ must be the endpoint of the minor axis, let's assume point $P(0, b)$. The slopes of lines $P A$ and $P F_{1}$ are $\frac{b}{a}$ and $\frac{b}{c}$, respectively, so $\tan \angle A P F_{1}=\frac{\frac{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A$ and $B$. A point $P$ on the ellipse makes the area of $\triangle P A B$ equal to 3. How many such points $P$ are there?
A. 1
B. 2
C. 3
D. 4 | 2. B Let point $Q(4 \cos t, 3 \sin t)$ be on the "minor arc" opposite to line $A B$. It is known that the area of triangle $A B Q$ is $6 \sin t + 6 \cos t - 6 = 6 \sqrt{2} \sin \left(t + \frac{\pi}{4}\right) - 6 \leqslant 6 \sqrt{2} - 6 < 3$, which means point $P$ has 2 solutions. | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. If the irreducible fraction $\frac{p}{q}\left(p, q \in \mathbf{N}^{*}\right)$ is converted to a decimal as $0.18 \cdots$, then when $q$ is the smallest, $p=(\quad)$
A: 9
B: 7
$\mathrm{C}: 5$
$\mathrm{D}$ :
2 | Answer D.
Analysis From the problem, we get
$$
0.18<\frac{p}{a}<0.19 \Leftrightarrow \frac{100 p}{19}<q<\frac{100 p}{18}
$$
It can be seen that as $p$ increases, the lower bound of $q$ keeps increasing.
When $p=1$, there is no integer $q$ that satisfies the condition; when $p=2$, $q=11$ satisfies the condition, so $q_... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
57. If $a, b, c$ are all multiples of 5, $a<b<c, c=a+10$, then $\frac{(a-b)(a-c)}{b-c}=$ | Answer: -10.
Solution: From the given conditions, we get
$$
b=a+5,
$$
Therefore,
$$
a-b=-5, a-c=-10, \quad b-c=-5 \text {, }
$$
Thus,
$$
\text { the original expression }=\frac{(-5) \times(-10)}{-5}=-10 \text {. }
$$ | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides $BC$, $CA$, $AB$ of the equilateral $\triangle ABC$, there are internal division points $D$, $E$, $F$ dividing the sides in the ratio $3:(n-3)$ (where $n>6$). The segments $AD$, $BE$, $CF$ intersect to form $\triangle PQR$ (where $BE$ intersects $AD$ at $P$ and $FC$ at $Q$). When the area of $\triangle ... | 3. Given $B D: D C=C E: E A=A F: F B=3:(n-3)$, applying Menelaus' theorem to $\triangle A D C$ and the transversal $B P E$, we have $\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A}=\frac{A P}{P D} \cdot \frac{3}{n} \cdot \frac{3}{n-3}=1$, thus $\frac{A P}{P D}=\frac{n(n-3)}{9}, \frac{A P}{A D}=\frac{n(n-3)}... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$14 \cdot 35$ Find the units digit of $\left[\frac{10^{20000}}{10^{100}+3}\right]$.
(47th Putnam Mathematical Competition, 1986) | [Solution] Let $10^{100}=t$, then
$$
\begin{aligned}
A & =\left[\frac{10^{20000}}{10^{100}+3}\right] \\
& =\left[\frac{t^{200}}{t+3}\right] \\
& =\left[\frac{t^{200}-3^{200}}{t+3}+\frac{3^{200}}{t+3}\right],
\end{aligned}
$$
Since $3^{200}=9^{100}<t+3$, we have
$$
0<\frac{3^{200}}{t+3}<1 \text {. }
$$
Also, because $... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. When $n$ is a positive integer, the function $f$ satisfies
$$
f(n+3)=\frac{f(n)-1}{f(n)+1}, f(1) \neq 0 \text {, and } f(1) \neq \pm 1 \text {. }
$$
Then $f(1) f(2023)=$ $\qquad$ | 4. -1 .
Notice that,
$$
\begin{array}{c}
f(n+6)=\frac{f(n+3)-1}{f(n+3)+1} \\
=\frac{\frac{f(n)-1}{f(n)+1}-1}{\frac{f(n)-1}{f(n)+1}+1}=\frac{-1}{f(n)}, \\
f(n+12)=\frac{-1}{f(n+6)}=f(n) .
\end{array}
$$
Thus, $f(n+12k)=f(n)(k \in \mathbf{N})$.
Given $2023=168 \times 12+7$, we have
$$
\begin{array}{l}
f(2023)=f(7)=\frac... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For angles $\alpha$ and $\beta$, if $\sin \alpha \sin \beta+\cos \alpha \cos \beta=0$ then the value of $\sin 2 \alpha+\sin 2 \beta$ is
A. 0
B. 1
C. $\sqrt{2}$
D. Greater than $\sqrt{2}$ | 4. From $\sin \alpha \sin \beta+\cos \alpha \cos \beta=0$ we get $\cos (\alpha-\beta)=0, \sin 2 \alpha+\sin 2 \beta=2 \sin (\alpha+\beta) \cos (\alpha-\beta)=0$ | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 5 Suppose there are two triangles with side lengths $3,4,5$, four triangles with side lengths $4,5, \sqrt{41}$, and six triangles with side lengths $\frac{5}{6} \sqrt{2}, 4,5$. Using these triangles as faces, how many tetrahedra can be formed? | The three given triangles are sequentially numbered as $1, 2, 3$. It is easy to see that triangles 1 and 2 are right triangles, and triangle 3 is an obtuse triangle. Since a tetrahedron has 4 faces, at least $\left[\frac{4-1}{3}\right]+1=2$ faces must be of the same type. Furthermore, it is not difficult to prove that ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (1997 Shanghai High School Mathematics Competition) Let $S=\{1,2,3,4\}$, and the sequence $a_{1}, a_{2}, \cdots, a_{n}$ has the following property: for any non-empty subset $B$ of $S$, there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$. | 1. Since a binary subset containing a fixed element of $S$ has 3 elements, any element of $S$ appears at least twice in the sequence. Therefore, the minimum value of $n$ is calculated to be 8. On the other hand, an 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the minimum value of $n$ is 8. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-233 Let $a, b, c, d$ be odd numbers, $0<a<b<c<d$, and $a d=b c$. Prove that if $a+d=2^{k}, b+c=2^{m}$, where $k, m$ are integers, then $a=1$.
The text is translated while preserving the original line breaks and format. | [Proof] Since $a d=b c$, we have
$$
\begin{aligned}
a[(a+d)-(b+c)] & =a^{2}+a d-a b-a c \\
& =a^{2}+b c-a b-a c \\
& =(a-b)(a-c) \\
& >0,
\end{aligned}
$$
thus $a+d>b+c$,
which implies $2^{k}>2^{m}$,
hence $k>m$.
Also, $b \cdot 2^{m}-a \cdot 2^{k}=b(b+c)-a(a+d)$
$=b^{2}+b c-a^{2}-a d$
$$
=b^{2}-a^{2},
$$
i.e., $2^{m}... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
$31 \cdot 17$ The number of integer pairs $(m, n)$ that satisfy the equation $m+n=mn$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(E) greater than 4.
(28th American High School Mathematics Examination, 1977) | [Solution] If $m+n=mn$, then $m+n-mn=0$, i.e., $\square$ we get
$$
\begin{array}{l}
m+n(1-m)=0, \\
n=\frac{m}{m-1} \quad(m \neq 1) .
\end{array}
$$
When $m=1$, there is no solution; only when $m=0$ or 2, the solution $\left[m, \frac{m}{m-1}\right]$ is an integer pair, i.e., there are two solutions $(0,0)$ and $(2,2)$.... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. If $M=\left\{(x, y)|| \tan \pi y \mid+\sin ^{2} \pi x=0\right\}, N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 2\right\}$, then $|M \cap N|=$ ( )
A. 4
B. 5
C. 8
D. 9 | 2. D From $\tan \pi y=0$ we get $y=k, k \in \mathbf{Z}$; from $\sin \pi x=0$ we get $x=k', k' \in \mathbf{Z}$; also, $x^{2}+y^{2} \leqslant 2$, so $k=0$, $-1,1 ; k'=0,-1,1$. Therefore, $M \cap N$ contains a total of 9 points. | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. In the lottery box of Green Green Park, there are golden, silver, and colorful balls. The lottery participant can randomly draw 3 balls from the lottery box. Xiao Yangyang hopes to get 3 golden balls, Mei Yangyang hopes to get 2 golden balls and 1 silver ball, and Lan Yangyang hopes to get one of each kind of ball. ... | $8$ | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. (20 points) If the equation about $x$
$$
x^{2}+k x-12=0 \text { and } 3 x^{2}-8 x-3 k=0
$$
have a common root, find all possible values of the real number $k$. | Three, 11. Let $x=x_{0}$ be the common root given in the problem.
Then $\left\{\begin{array}{l}x_{0}^{2}+k x_{0}-12=0, \\ 3 x_{0}^{2}-8 x_{0}-3 k=0 .\end{array}\right.$
(1) $\times 3$ - (2) gives $(3 k+8) x_{0}=36-3 k$.
Obviously, $k \neq-\frac{8}{3}$.
Thus, $x_{0}=\frac{36-3 k}{3 k+8}$.
Substituting equation (3) into... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let the function $f(x)=3 \sin x+2 \cos x+1$. If real numbers $a, b, c$ make $a f(x)+b f(x-c)=1$ hold for any real number $x$, then the value of $\frac{b \cos c}{a}$ is
A. $-\frac{1}{2}$
B. $\frac{1}{2}$
C. -1
D. 1 | $$
\begin{array}{l}
\text { According to the problem, } 3 a \sin x+2 a \cos x+a+3 b \sin (x-c)+2 b \cos (x-c)+b \\
=\sqrt{13} a \sin (x+\varphi)+\sqrt{13} b \sin (x-c+\varphi)+a+b \\
=\sqrt{13} a \sin (x+\varphi)+\sqrt{13} b[\sin (x+\varphi) \cos c-\cos (x+\varphi) \sin c]+a+b \\
=\sqrt{13}[(a+b \cos c) \sin (x+\varphi... | -1 | Algebra | MCQ | Yes | Yes | olympiads | false |
13. Person A and Person B use the same amount of money to buy the same type of candy, A buys the ones in tin boxes, and B buys the ones in paper boxes. Both try to buy as much as possible, and as a result, A buys 4 fewer boxes than B and has 6 yuan left, while B uses up all the money they brought. If A uses three times... | 【Answer】 12 yuan, 10 yuan
【Key Points】Divisors and Multiples, Solving Application Problems with Equations
【Analysis】
If A uses the original money to buy iron boxes and has 6 yuan left, then using 3 times the money to buy iron boxes theoretically should leave $6 \times 3=18$ yuan, but still leaves 6 yuan, which means $1... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 If 5 vertices of a regular nonagon are painted red, what is the minimum number of pairs of congruent triangles, all of whose vertices are red points?
(1992 Tianjin Team Test Question) | Solving: There are $C_{5}^{3}=10$ triangles with 5 red points as vertices, which we call red triangles. Let the circumference of the circumcircle of a regular nonagon be 9, and use the length of the arc opposite the chord to represent the chord length. In a regular nonagon, there are only 7 types of non-congruent trian... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Let the set $M=\{1,2, \cdots, 10\}$ have five-element subsets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy the condition: any two elements in $M$ appear in at most two subsets $A_{i}$ and $A_{j}(i \neq j)$. Find the maximum value of $k$. | 4. Solution Let $i \in M(i=1,2, \cdots, 10)$, the number of times $i$ appears in $A_{1}, A_{2}, \cdots, A_{k}$ is denoted as $d(i)$. First, we prove that $d(i) \leqslant 4(i=1,2, \cdots, 10)$.
In fact, for $i \in M$, the pair $(i, j)$ formed with any other 9 elements in $M$ (where $j \neq i$) appears at most twice in ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (5th "Hope Cup" Invitational Competition Question) The radius of the upper base of a frustum is 5 cm, the radius of the lower base is 10 cm, and the slant height $AB$ is 20 cm (where $B$ is on the circumference of the lower base). A rope is pulled from the midpoint $M$ of the slant height $AB$, wraps around the lat... | 10. C. Reason: The lateral development of the frustum is shown in the figure. $AB=20$ cm, $\widetilde{AC}=10\pi$ cm, $\overparen{BD}=20\pi$ cm. Let the center of the sector be $O$, and let $OA=x$ cm, $\angle AOC=\theta$, then $\overparen{AC}=x\theta, \overparen{BD}=(x+20)\theta$. Solving the system of equations $\left\... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Given sets
$$
A=\{1,3,5,7,9\}, B=\{2,4,6,8,10\} \text {. }
$$
If set $C=\{x \mid x=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is $\qquad$ | $-1.9$
Since $a$ is odd and $b$ is even, all elements in set $C$ are odd.
Furthermore, the minimum value of $a+b$ is 3, and the maximum value is 19, and all odd numbers between 3 and 19 can be obtained, thus, set $C$ contains 9 elements. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. In the Cartesian coordinate system $x O y$, given two points $M(-1,2)$ and $N(1,4)$, point $P$ moves on the $x$-axis. When $\angle M P N$ takes its maximum value, the x-coordinate of point $P$ is $\qquad$ | Let $\odot C$ be the circle passing through $M, N$ and tangent to the $x$-axis, so point $P$ is the point of tangency between $\odot C$ and the $x$-axis. Since $k_{M N}=1$, then $M N: y=x+3 \Rightarrow A(-3,0)$. Using the power of a point theorem, we know $|A P|^{2}=|A M| \cdot|A N|=2 \cdot 2 \cdot 4=16 \Rightarrow x_{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. Fill in each blank with a number not equal to 1, so that the equation holds. Then, the number of different ways to fill in is $\qquad$.
$$
[\mathrm{A} \times(\overline{1 \mathrm{~B}}+\mathrm{C})]^{2}=\overline{9 \mathrm{DE} 5}
$$ | 【Answer】 8
Analysis: $[\mathrm{A} \times(\overline{1 \mathrm{~B}}+\mathrm{C})]^{2}=\overline{9 \mathrm{DE} 5}=95^{2}$
$\mathrm{A} \times(\overline{1 \mathrm{~B}}+\mathrm{C})=95=5 \times 19$
When $A=5$, $19=10+9=12+7=13+6=14+5=15+4=16+3=17+2=19+0$ for a total of 8 ways | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. If the 5th term in the expansion of $\left(x \sqrt{x}-\frac{1}{x}\right)^{6}$ is $\frac{15}{2}$, then $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right)=$ | 1. $\because T_{5}=C_{6}^{1}(x \sqrt{x})^{2} \cdot\left(-\frac{1}{x}\right)^{4}=\frac{15}{x}$, from $\frac{15}{x}=\frac{15}{2}$, we get $x^{-1}=\frac{1}{2}$. Therefore, $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right)=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The sum of the digits of the product of 45 and 40 is ( ).
( A ) 9
( B ) 11
(C) 13
( D ) 15 | 【Answer】A
【Analysis】 $45 \times 40=1800,1+8=9$
【Difficulty】F
【Knowledge Point】Multiplication of two-digit numbers | 9 | Number Theory | MCQ | Yes | Yes | olympiads | false |
16 In four-dimensional space, the distance between point $A\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ and point $B\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ is defined as $A B=\sqrt{\sum_{i=1}^{4}\left(a_{i}-b_{i}\right)^{2}}$. Consider the set of points
$$
I=\left\{P\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \mid c_{i}=0 \te... | (16) Construct the following 8 points: $P_{1}(0,0,0,0), P_{2}(0,1,0,0), P_{3}(0,0,0,1), P_{4}(0,0,1,1), P_{5}(1,1,0,0), P_{6}(1,1,1,0), P_{7}(1,1,1,1), P_{8}(1,0,1,1)$. By calculation, it is known that no three of these points can form an equilateral triangle, so $n_{\text {min }} \geqslant 9$.
When $n=9$, let
$$
S_{m}... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12$\cdot$150 Given that $x, y$ are coprime positive integers, and $k$ is a positive integer greater than 1, find all natural numbers $n$ that satisfy $3^{n}=x^{k}+y^{k}$, and provide a proof.
(22nd All-Russian Mathematical Olympiad, 1996) | [Proof] Let $3^{n}=x^{k}+y^{k},(x, y)=1, x, y \in \mathbb{N}$, without loss of generality, assume $x>y, k>1, k \in \mathbb{N}, n \in \mathbb{N}$.
Clearly, neither $x$ nor $y$ can be divisible by 3.
If $k$ is even, then
$$
x^{k} \equiv 1(\bmod 3), y^{k} \equiv 1(\bmod 3),
$$
Thus, $x^{k}+y^{k} \equiv 2(\bmod 3)$,
Clear... | 2 | Number Theory | proof | Yes | Yes | olympiads | false |
17. On the table are 0 - 9, these 10 number cards. Three people, A, B, and C, each take three of them, and sum all the different three-digit numbers that can be formed with the three number cards they have. The results for A, B, and C are $1554, 1688, 4662$, respectively. The remaining number card is (
）(Note: 6 or 9 c... | 【Analysis】If the three cards do not have 0, then the sum of the 6 three-digit numbers formed is $222 \times(a+b+c)$. If the three cards include 0, then only 4 three-digit numbers can be formed, with a sum of $211(a+b) \quad$, and $(222,211)=1$. Therefore, if the sum is a multiple of 222, there must be no 0, and if the ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given a positive integer $m$ such that $m^{4}+16 m+8$ can be expressed as the product of two or more consecutive integers. Find the maximum value of $m$.
(Proposed by Indonesia) | 4. If $m \equiv 0(\bmod 3)$, then
$$
m^{4}+16 m+8 \equiv 2(\bmod 3) \text {; }
$$
If $m \equiv 1(\bmod 3)$, then
$$
m^{4}+16 m+8 \equiv 1(\bmod 3) \text {; }
$$
If $m \equiv 2(\bmod 3)$, then
$$
m^{4}+16 m+8 \equiv 2(\bmod 3) \text {. }
$$
Thus, regardless of the value of $m$, $m^{4}+16 m+8$ is never a multiple of 3... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.69 In a $7 \times 7$ grid, 19 cells are painted red. A row or a column is called red if it contains at least 4 red cells. How many red rows and columns can there be in the grid at most? | [Solution] First, we point out that the number of red rows and columns is no more than 8. Otherwise, if there are at least 9 red rows and columns, then there must be at least 5 red rows or 5 red columns, let's assume the former. Since each red row contains at least 4 red cells, it follows that there are at least 20 red... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. The computer screen displays a series of numbers according to the following rule: if the screen displays $a, b$, then the third display is $ab-1$. Initially, the screen displays $1, 2$. Find the number displayed on the screen at the 2018th display. | 10. From the problem, we know the first ten numbers displayed on the screen are
$$
1,2,1,1,0,-1,-1,0,-1,-1 \text {. }
$$
Obviously, the numbers displayed on the screen at the 8th, 9th, and 10th times are the same as those displayed at the 5th, 6th, and 7th times.
Notice that, $2018=5+671 \times 3$.
Therefore, the numb... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
86. Chen Gang and Zhang Qiang started from their respective homes at the same time and walked towards each other. They first met at a point 5 kilometers away from Chen Gang's home. After meeting, they continued to walk at their original speeds. Chen Gang reached Zhang Qiang's home, and Zhang Qiang reached Chen Gang's h... | Reference answer: 12 | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let positive real numbers $x, y$ satisfy $x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4}$. Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is | 7. 6 .
By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x}, \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$
Adding... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. As shown in the figure, in $\triangle A B C$, point $O$ is the midpoint of $B C$, a line through point $O$ intersects lines $A B$ and $A C$ at two different points $M$ and $N$, respectively. If $\overrightarrow{A B}=m \overrightarrow{A M}, \overrightarrow{A C}=$ $n \overrightarrow{A N}$, then the value of $m+n$ is $... | 2. 2 Detailed Explanation: $\overrightarrow{A O}=\frac{1}{2} \overrightarrow{A B}+\frac{1}{2} \overrightarrow{A C}=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N}$, since $M, O, N$ are collinear, we have $\frac{m}{2}+\frac{n}{2}=1, m+n=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given $x, y > 0$. If
$$
f(x, y)=\left(x^{2}+y^{2}+2\right)\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \text {, }
$$
then the minimum value of $f(x, y)$ is . $\qquad$ | 3.4.
By completing the square, we get
$$
x^{2}+y^{2}+2 \geqslant(x+y)+(x y+1) \text {. }
$$
Then $f(x, y)$
$$
\begin{array}{l}
\geqslant((x+y)+(x y+1))\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \\
\geqslant(1+1)^{2}=4 .
\end{array}
$$
Equality holds if and only if $x=y=1$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example. Let $m, n$ be positive integers, find the minimum value of $\left|12^{m}-5^{n}\right|$. | Solution: First, when $m=n=1$, $\left|12^{m}-5^{n}\right|=7$. The following uses the method of exhaustion to prove that 7 is the minimum value sought.
(1) It is clear that $\left|12^{m}-5^{n}\right|$ is an odd number, and cannot be $2,4,6$.
(2) Noting that $3 \mid 12^{m}, 3 \nmid 5^{n}$ and $5 \nmid 12^{m}$, it is easy... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21. Given trapezoid $A B C D$ with the product of the upper base $A B$ and the height being $\sqrt{2}+1, P$ is a moving point on the lower base $C D$, and line $P A$ intersects the diagonal $B D$ at point $M$. The sum of the areas of $\triangle A M B$ and $\triangle P M D$ is denoted as $S$. Then the minimum value of $... | $1$ | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
* Find all prime numbers $p$ such that $4 p^{2}+1$ and $6 p^{2}+1$ are also prime. | When $p \equiv \pm 1(\bmod 5)$, $5 \mid 4 p^{2}+1$. When $p \equiv \pm 2(\bmod 5)$, $5 \mid 6 p^{2}+1$. Therefore, the only solution to this problem is $p=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.27 In what base, 16324 is the square of 125.
(Kyiv Mathematical Olympiad, 1954) | ［Solution］Let the base of the numeral system be $b$.
By the problem, it is clear that $\quad b>6$.
At this point, we can obtain the equation
$$
b^{4}+6 b^{3}+3 b^{2}+2 b+4=\left(b^{2}+2 b+5\right)^{2},
$$
After rearrangement, we get
$$
\begin{array}{c}
2 b^{3}-11 b^{2}-18 b-21=0, \\
(b-7)\left(2 b^{2}+3 b+3\right)=0 .... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6 Let $\alpha \in\left(0, \frac{\pi}{2}\right)$, then the minimum value of $\frac{\sin ^{3} \alpha}{\cos \alpha}+\frac{\cos ^{3} \alpha}{\sin \alpha}$ is ( ).
(A) $\frac{27}{64}$
(B) $\frac{3 \sqrt{2}}{5}$
(C) 1
(D) $\frac{5 \sqrt{3}}{6}$ | 6
$$
\begin{aligned}
\frac{\sin ^{3} \alpha}{\cos \alpha}+\frac{\cos ^{3} \alpha}{\sin \alpha} & =\frac{\sin ^{4} \alpha+\cos ^{4} \alpha}{\sin \alpha \cos \alpha} \\
& =\frac{1-2(\sin \alpha \cos \alpha)^{2}}{\sin \alpha \cos \alpha} .
\end{aligned}
$$
Let $t=\sin \alpha \cos \alpha=\frac{1}{2} \sin 2 \alpha \in\left... | 1 | Inequalities | MCQ | Yes | Yes | olympiads | false |
2 Let $C$ be a circle with radius $r$, centered at the point $(\sqrt{2}, \sqrt{3})$, where $r$ is a positive real number. When $x, y$ are both rational numbers, the point $(x, y)$ is called a rational point. The maximum number of rational points on circle $C$ is
A. 0
B. 1
C. 2
D. 3 | 2 B Take a rational point $A$ arbitrarily. Draw a circle passing through $A$ with center at $(\sqrt{2}, \sqrt{3})$, then there is at least one rational point on the circle. If there are two rational points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, then the perpendicular bisector of the line segment connect... | 0 | Number Theory | MCQ | Yes | Yes | olympiads | false |
$32 \cdot 65$ The difference in area between two similar triangles is 18 square feet, and the ratio of the area of the larger triangle to the smaller triangle is the square of an integer. The area of the smaller triangle (in square feet) is an integer, and one side is 3 feet, then the length of the corresponding side o... | [Solution] Let the areas of the two triangles be $S$ and $S+18$, and the corresponding side length of the larger triangle be $x$. By the given and the properties of similar figures, we have
$$
\frac{S+18}{S}=\frac{x^{2}}{3^{2}}=\left(\frac{x}{3}\right)^{2},
$$
Solving for $S$, we get
$$
S=\frac{18}{\left[\left(\frac{x... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. If real numbers $x, y$ satisfy $4 x^{2}-4 x y+2 y^{2}=1$, then the sum of the maximum and minimum values of $3 x^{2}+x y+y^{2}$ is $\qquad$ | Given $3 x^{2}+x y+y^{2}=k=k\left(4 x^{2}-4 x y+2 y^{2}\right)$, then $(4 k-3) x^{2}-(4 k+1) x y+(2 k-1) y^{2}=0$. If $k=\frac{3}{4}, y^{2}=8 x y$, at this time $(x, y)=\left( \pm \frac{1}{2}, 0\right)$ or $\left( \pm \frac{1}{10}, \pm \frac{4}{5}\right)$; if $k \neq \frac{3}{4}$, by $y \neq 0 \Rightarrow(4 k-3)\left(\... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (7th Balkan Mathematical Olympiad 1990) For a finite set $A$, there exists a function $f: N-A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j), N=\{1,2, \cdots\}$, find the minimum number of elements in the finite set $A$.
| 11. Starting from 1, the smallest natural number whose difference with it is a prime number is 3; the smallest natural number whose difference with the first two is a prime number is 6; the smallest natural number whose difference with the first three is a prime number is 8. This means that for the set $M=\{1,3,6,8\} \... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. $a_{1}, a_{2}, a_{3}, \cdots, a_{2020}$ are distinct non-zero real numbers, so $\left|\frac{a_{1}}{\left|a_{1}\right|}+\frac{\left|a_{2}\right|}{a_{2}}+\cdots+\frac{a_{2019}}{\left|a_{2019}\right|}+\frac{\left|a_{2020}\right|}{a_{2020}}\right|$ has the minimum value of $\qquad$ | $0$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. There are several socks of black, white, and yellow colors. To ensure you can form two pairs of the same color in the dark, you need to take out at least $\qquad$ socks (for example: one pair black and one pair yellow does not meet the requirement). | 【Analysis】(4-1)×3+1=10 (only)
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。
【Analysis】(4-1)×3+1=10 (only) | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
37. Given that the integer part of the real number $\frac{2+\sqrt{2}}{2-\sqrt{2}}$ is $a$, and the fractional part is $1-b$. Then the value of $\frac{(b-1) \cdot(5-b)}{\sqrt{a^{2}-3^{2}}}$ is | Answer: -1
Solution:
$$
\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2 \sqrt{2}
$$
Since
$$
2 \sqrt{2}=\sqrt{8} \text {, }
$$
Therefore
$$
\begin{array}{c}
2<2 \sqrt{2}<3, \\
5<3+2 \sqrt{2}<6,
\end{array}
$$
We get
$$
a=5,1-b=(3+2 \sqrt{2})-5=2 \sqrt{2}-2, \quad b=3-2 \sqrt{2} \text {, }
$$
Thus
$$
\frac{(b-1) \cdot(5-b)}{\sqr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The terms of the geometric sequence $\left\{a_{n}\right\}$ are all positive, and $a_{1} a_{3}+a_{2} a_{6}+2 a_{3}^{2}=36$, then the value of $a_{2}+a_{4}$ is . $\qquad$ | Answer: 6.
Solution: Since $36=a_{1} a_{3}+a_{2} a_{6}+2 a_{3}^{2}=a_{2}^{2}+a_{4}^{2}+2 a_{2} a_{4}=\left(a_{2}+a_{4}\right)^{2}$, and $a_{2}+a_{4}>0$, thus $a_{2}+a_{4}=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. To promote the development of freshwater fish farming in a certain area, the government controls the price within an appropriate range and provides subsidies for freshwater fish farming. Let the market price of freshwater fish be $x$ yuan/kg, and the government subsidy be 1 yuan/kg. According to market research, wh... | 12. (1) $1000(x+t-8)=500 \sqrt{40-(x-8)^{2}}$, i.e., $5 x^{2}+(8 t-80) x+\left(4 t^{2}-64 t+\right.$ $280 = 0$. From $\Delta \geqslant 0$ we get $t^{2} \leqslant 50$, at this time $x=8-\frac{4}{5} t \pm \frac{2}{5} \sqrt{50-t^{2}}$. From $\Delta \geqslant 0, t \geqslant 0, 8 \leqslant x \leqslant 14$ we get $\left\{\be... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) Calculate: $(98 \times 76-679 \times 8) \div(24 \times 6+25 \times 25 \times 3-3)=$ | 【Analysis】There are parentheses, so calculate what's inside the parentheses first, then calculate what's outside the parentheses, and solve accordingly.
【Solution】Solution: $(98 \times 76-679 \times 8) \div(24 \times 6+25 \times 25 \times 3-3)$
$$
\begin{array}{l}
=(7448-5432) \div(144+1875-3) \\
=2016 \div 2016 \\
=1 ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
65. There is a cup of salt water. If 200 grams of water is added, its concentration becomes half of the original; if 25 grams of salt is added, its concentration becomes twice the original. The original concentration of this cup of salt water is $\qquad$ . | Answer: $10 \%$ | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.3.7 ** 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.
Fin... | Given that $f(x-4)=f(2-x), x \in \mathbf{R}$, we can determine that the axis of symmetry of the quadratic function $f(x)$ is $x = -1$. From (3), we know that $f(x)$ opens upwards, i.e., $a > 0$, so we have $f(x) = a(x + 1)^2 (a > 0)$.
From (1), we get $f(1) \geq 1$, and from (2), we get $f(1) \leq \left(\frac{1+1}{2}\... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7 Find the minimum value of the function $f(x)=\sqrt{2 x^{2}-3 x+4}+\sqrt{x^{2}-2 x}$. | Solve: First, consider the domain of $f(x)$. From $2 x^{2}-3 x+4 \geqslant 0, x^{2}-2 x \geqslant 0$, we have $x \in(-\infty, 0] \cup$ $[2,+\infty)$
It is easy to see that $y=2 x^{2}-3 x+4$ and $y=x^{2}-2 x$ are both decreasing on $(-\infty, 0]$ and increasing on $[2,+\infty)$. Therefore, $f(x)$ is decreasing on $(-\i... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The zoo keeps some sika deer and ostriches. The number of ostrich feet is 20 more than the number of sika deer feet. When each ostrich lifts one foot, the number of ostrich feet lifted is 10 less than the number of sika deer feet. The zoo has $\qquad$ sika deer. | $10$ | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Point $P$ is a moving point on circle $C:(x+2)^{2}+y^{2}=4$, and it is not at the origin. The fixed point $A$ has coordinates $(2,0)$. The perpendicular bisector of line segment $AP$ intersects line $CP$ at point $Q$. Let $M(-1,0), N(1,0)$, then the product of the slopes of lines $MQ$ and $NQ$ is $\qquad$. | Given, $|Q C|-|Q A|=2 \Rightarrow$ the locus of point $Q$ is the right branch of a hyperbola with foci at $A, C$ and a real axis length of 2, the equation of the locus is $x^{2}-\frac{y^{2}}{3}=1(x>1)$.
Therefore, $k_{Q M} \cdot k_{Q N}=\frac{y_{Q}^{2}}{x_{Q}^{2}-1}=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) In the division equation, the dividend is 2016, and the remainder is 7, then the number of divisors that satisfy the equation is ( ).
A. 3
B. 4
C. 5
D. 6 | 【Analysis】Divisor $\times$ Quotient $=2016-7=2009$, then factorize 2009, and according to the remainder being less than the divisor, the number of divisors that satisfy the equation can be determined.
【Solution】Solution: $2016-7=2009$,
$$
2009=7 \times 287=49 \times 41=1 \times 2009
$$
So the divisors that satisfy th... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
6. For the equation about integers $a, b$: $\sqrt{a-1}+\sqrt{b-1}=\sqrt{a b+k}(k \in \mathbb{Z})$, if there is only one ordered real solution, then $k=$ $\qquad$ | Solution: 0.
From the given equation, squaring both sides and rearranging yields $(a-1)(b-1)-2 \sqrt{(a-1)(b-1)}+1+k \neq 0$
$$
\text { we get }(\sqrt{(a-1)(b-1)}-1)^{2}=-k
$$
When $k \geq 1$, there is no solution.
When $k=0$, there is only one solution $(a, b)=(2,2)$.
When $k \leq-1$, $\sqrt{(a-1)(b-1)}=1 \pm \sqrt{-... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. In a $4 \times 4$ grid, there are 16 chess pieces, each with one black side and one white side. The rule is: flipping the 4 pieces in a $2 \times 2$ subgrid once is called “one operation”. Initially, all 16 pieces have their black sides up, as shown in the left figure. The minimum number of operations required to a... | $6$ | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. This year, Xiaobing is 7 years old, and Xiaobing's mother is 35 years old, (）years later, the mother's age will be 3 times Xiaobing's age. | 【Answer】 7 years later.
【Analysis】Exam point: Age problem
Age difference: $35-7=28$ (years) One unit quantity: $28 \div(3-1)=14$ (years) Need to wait: 14-7=7 (years) | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. Wu Yu wrote four letters and prepared four envelopes. He needs to put each letter into the corresponding envelope, with only one letter in each envelope. The number of ways all four letters can be placed incorrectly is $\qquad$ kinds. | 【Analysis】 $P_{4}^{4}-C_{4}^{1} \times P_{3}^{3}+C_{4}^{2} \times P_{2}^{2}-C_{4}^{3} \times P_{1}^{1}+C_{4}^{4} \times P_{0}^{0}=9$ types | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{3}-n$ $\left(n \in \mathbf{Z}_{+}\right)$, and the terms in this sequence whose unit digit is 0, arranged in ascending order, form the sequence $\left\{b_{n}\right\}$. Then the remainder when $b_{2018}$ is divided by 7 is $\qquad$ | 8. 4 .
Since $a_{n}=n^{3}-n=n(n-1)(n+1)$, thus, $a_{n}$ has a units digit of 0 if and only if the units digit of $n$ is $1, 4, 5, 6, 9, 0$. Therefore, in any consecutive 10 terms of the sequence $\left\{a_{n}\right\}$, there are 6 terms whose units digit is 0.
Since $2018=336 \times 6+2, 336 \times 10=3360$, the rema... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$7 \cdot 58$ For the set $S=\left\{\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right) \mid a_{i}=0\right.$ or $1, i=1,2,3$, $4,5\}$, for any two elements $\left(\bar{a}_{1}, \bar{a}_{2}, \bar{a}_{3}, \bar{a}_{4}, \bar{a}_{5}\right)$ and $\left(\bar{b}_{1}, \bar{b}_{2}, \bar{b}_{3}, \bar{b}_{4}, \bar{b}_{5}\right)$, define ... | ［Solution］This subset can contain at most 4 elements.
For convenience, in the set $S$, we call $a_{i}(i=1,2,3,4,5)$ the $i$-th component of the element $A\left(a_{1}\right.$, $\left.a_{2}, a_{3}, a_{4}, a_{5}\right)$. Clearly, any two elements in the set $S$ with a distance greater than 2 can have at most 2 components ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Let $x_{1}, x_{2}, x_{3}$ be non-negative real numbers, satisfying $x_{1}+x_{2}+x_{3}=1$. Find the minimum and maximum values of
$$
\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right)
$$ | 10. By Cauchy's inequality, we have
$$
\begin{array}{l}
\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \\
\geqslant\left(\sqrt{x_{1}} \sqrt{x_{1}}+\sqrt{3 x_{2}} \sqrt{\frac{x_{2}}{3}}+\sqrt{5 x_{3}} \sqrt{\frac{x_{3}}{5}}\right)^{2} \\
=\left(x_{1}+x_{2}+x_{3}\right)^{2}=1,
\end{a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 13 In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B C$ respectively. Find the distance from point $D$ to the plane $B_{1} E F$.
(2002 Hunan Province High School Competition Question) | Solve: Establish a spatial rectangular coordinate system as shown in Figure 20-12, then $D(0,0,0) 、 B_{1}(1,1,1) 、 E\left(1, \frac{1}{2}, 0\right) 、 F\left(\frac{1}{2}, 1,0\right)$.
Thus, $\overrightarrow{B_{1} E}=\left(0,-\frac{1}{2},-1\right), \overrightarrow{B_{1} F}=\left(-\frac{1}{2}, 0,-1\right)$.
Let the normal ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Given any 5 points on a plane, where no 3 points are collinear and no 4 points are concyclic, if a circle passes through 3 of these points and the other two points are one inside and one outside the circle, it is called a "good circle". If the number of good circles is denoted as $n$, find all possible values... | Consider any two points $A, B$ among the 5 points, and draw a line through $A, B$. If the other three points $C, D, E$ are on the same side of line $AB$, then examine $\angle ACB, \angle ADB, \angle AEB$. Without loss of generality, assume $\angle ACB < 180^{\circ}$. Then the circle $ADB$ is the unique good circle; if ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A geometric solid formed by a quadrilateral prism and a quadrilateral pyramid has at least ( ) faces.
A. 6
B. 7
C. 8
D. 9 | 3. $\mathrm{B}$
As shown in Figure 2, there are at least 7 faces. | 7 | Geometry | MCQ | Yes | Yes | olympiads | false |
8. Let non-negative real numbers $a, b, c$ satisfy $a b+b c+c a=a+b+c>0$, then the minimum value of $\sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ is
A. 2
B. 3
C. $\sqrt{3}$
D. $2 \sqrt{2}$ | $$
\begin{array}{l}
(a+b+c)(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) \\
=(a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a}+c \sqrt{a b}+a \sqrt{b c}+b \sqrt{c a} \\
\geqslant(a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a} \\
\geqslant 2(a b+b c+c a)=2(a+b+c) \Rightarrow \sqrt{a b}+\sqrt{b c}+\sqrt{c a} \geqslant 2,
\end{a... | 2 | Inequalities | MCQ | Yes | Yes | olympiads | false |
16. (10 points) A bus has 78 seats, and it starts its journey empty. At the first stop, one passenger boards, at the second stop, two passengers board, at the third stop, three passengers board, and so on. After how many stops will the bus be full of passengers? | 【Solution】Solution: Let the bus be full of passengers after the $n$-th station, we get:
$$
\begin{array}{l}
{[1+1+(n-1) \times 1] \times n \div 2=78} \\
{[2+n-1] \times n \div 2=78,} \\
{[1+n] \times n \div 2=78,} \\
\quad(1+n) \times n=156,
\end{array}
$$
Since $12 \times 13=156$,
i.e., $n=12$.
Answer: After 12 stati... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. (15 points) Master Wang originally planned to start from point $A$ on a circular road with a circumference of 400 meters, and install a street lamp every 50 meters. Each lamp requires digging a hole to bury the lamp post.
（1）According to the original plan, how many holes does Master Wang need to dig?
（2）After diggi... | 【Analysis】(1) Divide 400 by the interval of 50 meters;
(2) Divide 400 by 40 to find the total number of holes to be dug, then find the least common multiple of 50 and 40, which is 200, so the number of holes that do not need to be moved is $400 \div 200=2$, so the number of additional holes needed is: $400 \div 40-2=8$... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5、 "24-point game" is a familiar math game to many people, the game process is as follows: arbitrarily draw 4 cards from 52 playing cards (excluding the jokers), use the numbers on these 4 playing cards (A=1, J=11, Q=12, K=13) to get 24 through addition, subtraction, multiplication, and division. The person who finds t... | $\begin{array}{l}6, 3)(8, 7, 5, 4)(8, 6, 6, 4)(8, 6, 5, 5) \text { total } 9 \text { groups } \\\end{array}$ | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (8 points) The calculation result that equals 9 is ( )
A. $3 \times 3 \div 3+3$
B. $3 \div 3+3 \times 3$
C. $3 \times 3-3+3$
D. $3 \div 3+3 \div 3$ | 【Answer】Solution: $A 、 3 \times 3 \div 3+3$
$$
\begin{array}{l}
=3+3 \\
=6 ;
\end{array}
$$
$$
\begin{array}{l}
B 、 3 \div 3+3 \times 3 \\
=1+9 \\
=10 ;
\end{array}
$$
$$
\begin{array}{l}
\text { C. } 3 \times 3-3+3 \\
=9-3+3 \\
=9 ;
\end{array}
$$
$$
\begin{array}{l}
\text { D. } 3 \div 3+3 \div 3 \\
=1+1 \\
=2 ;
\end... | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 4 Given the sequence $\left\{a_{n}\right\}$ :
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$
Determine the periodicity of the sequence $\left\{a_{n}\right\}$. | Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$
which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
47. Person A and Person B start from the same point A on a circular track, running in opposite directions. Their speeds are 4 meters per second and 6 meters per second, respectively. If they start at the same time and end when they meet again at point A, then they meet a total of $\qquad$ times from the start to the en... | answer: 5 | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7 (1) Find the number of positive integer solutions $(m, n, r)$ to the indeterminate equation $m n+n r+m r=2(m+n+r)$;
(2) For a given integer $k>1$, prove that the indeterminate equation $m n+n r+m r=k(m+n+r)$ has at least $3 k+1$ sets of positive integer solutions $(m, n, r)$. | (1) If $m, n, r \geqslant 2$, then
$$
m n \geqslant 2 m, n r \geqslant 2 n, m r \geqslant 2 r
$$
Thus,
$$
m n+n r+m r \geqslant 2(m+n+r),
$$
Therefore, the above inequalities all hold with equality, so $m=n=r=2$.
If $1 \in\{m, n, r\}$, assume without loss of generality that $m=1$, then $n r+n+r=2(1+n+r)$, hence $(n-1... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) There are 12 students playing a card game, with 4 participants each time, and any 2 students can participate together at most 1 time. How many times can they play $\qquad$ at most. | 【Analysis】First, analyze that the students can be labeled, and then enumerate.
【Solution】According to the problem, we know:
Number the students from 1 to 12.
If the groups are 1-4, 5-8, and 9-12, the next group will not meet the requirements, so we need to form as many groups as possible.
The first group is $1,2,3,4$.... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The sequence $\lg 1000, \lg \left(1000 \cos \frac{\pi}{3}\right), \lg \left(1000 \cos ^{2} \frac{\pi}{3}\right), \cdots, \lg \left(1000 \cos ^{n-1} \frac{\pi}{3}\right)$, $\cdots$, when the sum of the first $n$ terms is maximized, the value of $n$ is ( ).
A. 9
B. 10
C. 11
D. Does not exist | 3. B. $a_{n}=\lg \left(1000 \cos ^{n-1} \frac{\pi}{3}\right)=\lg \frac{1000}{2^{n-1}}$ is decreasing, $a_{10}>0, a_{11}<0$, so $S_{10}$ is the maximum. | 10 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 4 Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms $S_{n}=p^{n}+q(p \neq 0, p \neq 1)$, find the necessary and sufficient condition for the sequence $\left\{a_{n}\right\}$ to be a geometric sequence. | Solve $a_{1}=S_{1}=p+q$, when $n \geqslant 2$, $a_{n}=S_{n}-S_{n-1}=p^{n-1}(p-1)$.
Since $p \neq 0, p \neq 1$, we have $\frac{a_{n+1}}{a_{n}}=\frac{p^{n}(p-1)}{p^{n-1}(p-1)}=p$.
If $\left\{a_{n}\right\}$ is a geometric sequence, then $\frac{a_{2}}{a_{1}}=\frac{a_{n+1}}{a_{n}}=p$, so $\frac{p(p-1)}{p+q}=p$.
Since $p \ne... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Expand the binomial $\left(\frac{\sqrt{x+1}}{2 \sqrt[4]{x}}\right)^{n}$ in descending powers of $x$. If the coefficients of the first two terms form an arithmetic sequence, then the number of terms in the expansion where the power of $x$ is an integer is $\qquad$ . | 3
8.【Analysis and Solution】It is not difficult to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$, respectively. Since these three numbers form an arithmetic sequence, we have $2-\frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get: $n=8$ and $n=1$ (discard).
When $n... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Calculate: $128 \div(64 \div 32) \div(32 \div 16) \div(16 \div 8) \div(8 \div 4)=$ | $8$ | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17.91 Given that the equation $x^{2}-8 x+m+6=0$ has two equal real roots, and its root is the length of side $B C$ of $\triangle A B C$. If the area of $\triangle A B C$ is 6, then the distance from the centroid $G$ to $B C$ is
(A) 2.
(B) 1.
(C) 3.
(D) $1 \frac{1}{2}$.
(China Zhejiang Province Junior High School Mathem... | ［Solution］From the given equation with equal roots, we have
$$
\Delta=(-8)^{2}-4(m+6)=0 \text {, which gives } m=10 \text {. }
$$
Thus, the equation becomes $x^{2}-8 x+16=0$.
Let $A E$ be the altitude of $\triangle A B C$, and $A D$ be the median of $B C$ (as shown in the figure).
Also, $B C=x_{1}=x_{2}=4$, and $\frac... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Given that $a$ is a given real number, the number of subsets of the set $M=\left\{x \mid x^{2}-3 x-a^{2}+2=0, x \in \mathbf{R}\right\}$ is
(A) 1
(B) 2
(C) 4
(D) Uncertain | (C)
I. Multiple Choice Question
1.【Analysis and Solution】 $M$ represents the solution set of the equation $x^{2}-3 x-a^{2}+2=0$ in the real number range. Since $\Delta=1+4 a^{2}>$ 0, $M$ contains 2 elements. Therefore, the set $M$ has $2^{2}=4$ subsets, choose C. | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is | 1
II. Fill-in-the-blank Questions
1. 【Analysis and Solution $\because a b>0$, $\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1$. Therefore, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$. When $a=b=1$, the above expression $=1$, henc... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Prove: If $\alpha, \beta, \gamma$ are acute angles, then
$$
\tan \alpha(\cot \beta+\cot \gamma)+\tan \beta(\cot \gamma+\cot \alpha)+\tan \gamma(\cot \alpha+\cot \beta) \geqslant 6 .
$$ | 2. Left side greater than or equal to
$$
6 \sqrt[6]{\tan ^{2} \alpha \cot \beta \cot \gamma \cdot \tan ^{2} \beta \cot \gamma \cot \alpha \cdot \tan ^{2} \gamma \cot \alpha \cot \beta}=6 .
$$ | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
$8 \cdot 39$ If $|x-8 y|+(4 y-1)^{2}=0$, then the value of $\log _{2} y^{x}$ is
(A) 0.
(B) $\frac{1}{4}$.
(C) $\frac{1}{16}$.
(D) -4.
(E) $\sqrt{2}$.
(China Beijing Junior High School Mathematics Competition, 1982) | [Solution] From the given, we know that $x-8y=0$, and $4y-1=0$ which gives $x=2, y=\frac{1}{4}$. Therefore, $\log _{2} y^{x}=\log _{2}\left(\frac{1}{4}\right)^{2}=-4$.
So the answer is $(D)$. | -4 | Algebra | MCQ | Yes | Yes | olympiads | false |
26-24 If $r$ is a positive number, and the line with equation $x+y=r$ is tangent to the circle with equation $x^{2}+y^{2}=r$, then $r$ equals
(A) $\frac{1}{2}$.
(B) 1 .
(C) 2 .
(D) $\sqrt{2}$.
(E) $2 \sqrt{2}$.
(29th American High School Mathematics Examination, 1978) | [Solution] If the line with the equation $x+y=r$ is tangent to the circle with the equation $x^{2}+y^{2}=r$, it is easy to find that the point of tangency is $\left(\frac{r}{2}, \frac{r}{2}\right)$, and the distance between this point and the origin is $\sqrt{r}$.
$\therefore\left(\frac{r}{2}\right)^{2}+\left(\frac{r}{... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
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