problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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(4) In $\triangle A B C$, $D$ is the midpoint of side $B C$. If $\overrightarrow{A D} \cdot \overrightarrow{A C}=0$, then the value of $\tan A+2 \tan C$ is ( ).
(A) $-\frac{\sqrt{3}}{3}$
(B) 0
(C) $\frac{\sqrt{3}}{3}$
(D) 1 | (4) Take the midpoint $E$ of $AB$, then $DE \parallel AC, \angle ADE=90^{\circ}, \angle DAE=A-90^{\circ}$, let $AC=b, AD=m$, then
$$
\tan C=\frac{m}{b}, \tan \left(A-90^{\circ}\right)=\frac{b}{2 m},
$$
which means
$$
-\cot A=\frac{b}{2 m},
$$
so
$$
\tan A=-\frac{2 m}{b},
$$
therefore
$$
\tan A+2 \tan C=0 .
$$
Hence... | 0 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Let $i$ be the imaginary unit, $a$ and $b$ be positive integers, and $|(a+i)(2+i)|=\left|\frac{b-i}{2-i}\right|$, then $a+b=$ | Answer 8.
Analysis From the problem, we get $(2 a-1)^{2}+(a+2)^{2}=\left(\frac{2 b+1}{5}\right)^{2}+\left(\frac{b-2}{5}\right)^{2} \Rightarrow(b+5 a)(b-5 a)=24$.
Since $b+5 a$ and $b-5 a$ are integers with the same parity, we have $\left\{\begin{array}{l}b+5 a=12 \\ b-5 a=2\end{array}\right.$ or $\left\{\begin{array}{l... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. For the geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, and common ratio $q=-\frac{1}{2}$, let $f(n)$ denote the product of its first $n$ terms. What is $n$ when $f(n)$ reaches its maximum value? | 12. The general term formula of the geometric sequence $\left\{a_{n}\right\}$ is $a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1}$, and the product of the first $n$ terms is $f(n)=1536^{n} \times\left(-\frac{1}{2}\right)^{\frac{n(n-1)}{2}}$.
First, consider the absolute value of $f(n)$, denoted as $b_{n}=|f(n)|=1536^{... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12・104 Find all positive integer solutions \(x, y, z\) to the equation \(3^{x}+4^{y}=5^{z}\).
(32nd International Mathematical Olympiad Preliminary Question, 1991) | [Solution] Clearly, $x=y=z=2$ is a positive integer solution to the equation.
Next, we prove that the equation
$$
3^{x}+4^{y}=5^{z}
$$
has no other positive integer solutions besides $x=y=z=2$.
Suppose positive integers $x, y, z$ satisfy equation (1), then
$$
5^{z} \equiv 1 \pmod{3}.
$$
Since $5^{z}=(3+2)^{z}$, then
... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given $x \geqslant \frac{5}{2}$, then $f(x)=\frac{x^{2}-4 x+5}{2 x-4}$ has
A. Maximum value of $\frac{5}{4}$
B. Minimum value of $\frac{5}{4}$
C. Maximum value of 1
D. Minimum value of 1 | 6. $\mathrm{D}$ Hint: The original expression can be transformed into $\frac{1}{2}\left[(x-2)+\frac{1}{x-2}\right] \geqslant 1$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. (3 points) During the summer vacation, Xiaoming went to the park and met four classmates: Jia, Yi, Bing, and Ding. Xiaoming shook hands with all four classmates. Jia shook hands with 3 people, Yi shook hands with 2 people, and Bing shook hands with 1 person. How many people did Ding shake hands with? $\qquad$ | 【Answer】Solution: If everyone shakes hands with each other, then each person needs to shake hands 4 times,
Xiaoming shook hands with four classmates, including Ding and Bing;
Bing shook hands with 1 person, he only shook hands with Xiaoming, not with Jia;
Jia shook hands with 3 people, only one person didn't shake hand... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
16. Let $n$ be a given positive integer, $S_{n} \subseteq\left\{\alpha \mid \alpha=\left(p_{1}, p_{2}, \cdots, p_{n}\right), p_{k} \in\{0,1\}, k=1,2, \cdots, n\right\}$. For any elements $\beta=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ and $\gamma=\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ in the set $S_{n}$, the follow... | (1) Clearly, each element in $S_{5}$ contains an odd number of 1s.
Since $\left\{\alpha \mid \alpha=\left(p_{1}, p_{2}, \cdots, p_{5}\right), p_{k} \in\{0,1\}, k=1,2, \cdots, 5\right\}$ contains $C_{5}^{1}+C_{5}^{3}+C_{5}^{5} = 16$ elements with an odd number of 1s, we can divide them into the following 5 groups:
Grou... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (1996 28th Canadian Mathematical Olympiad) If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-x-1=0$, find the value of $\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$. | 1. From the given, we have
$$
x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma) \text {, }
$$
Therefore,
$$
\left\{\begin{array}{l}
\alpha+\beta+\gamma=0, \\
\alpha \beta+\beta \gamma+\gamma \alpha=-1, \\
\alpha \beta \gamma=1 .
\end{array}\right.
$$
Thus, we have
$$
\begin{aligned}
& \frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. For $n \in \mathbf{N}^{*}$, let $S_{n}$ be the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, where $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}$, and $\sum_{n=1}^{n} a_{k}=17$. If there exists $n$ such that $S_{n}$ is also an integer, find all values of $n$. | 6. From $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=1(2 k-1)+a_{k} i \mid, k=1,2, \cdots, n$, we get $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}=\sum_{k=1}^{n} \mid(2 k$ $-1)+a_{k} i\left|\geqslant 1 \sum_{k=1}^{n}(2 k-1)+\left(\sum_{k=1}^{n} a_{k}\right) i\right|=\left|n^{2}+17 i\right|=\sqrt{n^{4}+17^{2}}$, clearly the equality c... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Given real numbers $x_{1}, x_{2}, x_{3}$ satisfy $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{1} x_{2}+x_{2} x_{3}=2$. Then the maximum value of $\left|x_{2}\right|$ is $\qquad$. | 11. 2 .
From the condition, we have
$$
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+x_{3}^{2}=4 \text {. }
$$
Notice that,
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2} \geqslant \frac{x_{2}^{2}}{2}, \\
x_{3}^{2}+\left(x_{2}+x_{3}\right)^{2} \geqslant \frac{x_{2}^{2}}{2} .
\end{arr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. As shown in the figure, in trapezoid $A B C D$, $A B / / C D, A C \perp B D, A D=3 \sqrt{2}, B C=3 \sqrt{3}, A B: C D$ $=1: 2$, then $C D=$ $\qquad$ | $6$ | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Let set $A=\{2,0,1,3\}$,
$$
B=\left\{x \mid -x \in A, 2-x^{2} \notin A\right\} \text {. }
$$
Then the sum of all elements in set $B$ is ( ).
(A) -4
(B) -5
(C) -6
(D) -7 | 一,1. B.
It is easy to know that $B=\{-2,-3\}$.
Then, the sum of all elements in set $B$ is -5. | -5 | Algebra | MCQ | Yes | Yes | olympiads | false |
* In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is equal to | Let the inradius of $\triangle ABC$ be $r$. According to the Law of Sines, we have $b \sin A = a \sin B$. Also, $a \cos A = b \cos B$, so $\sin 2A = \sin 2B$. Since $A \neq B$, then $A + B = 90^{\circ}$. Therefore, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}$, and $a^2 + b^2 = c^2$.
Given that $c = 10$... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. As shown in Figure $14-12, \angle M O N=20^{\circ}, A$ is a point on $O M$, $O A=4 \sqrt{3}, D$ is a point on $O N$, $O D=8 \sqrt{3}, C$ is any point on $A M$, $B$ is any point on $O D$, then the length of the broken line $A B C D$ is $A B + B C + C D \geqslant 12$. | 1. Taking $O M$ as the axis of symmetry, construct the symmetric point $D^{\prime}$ of point $D$, and taking $O N$ as the axis of symmetry, construct the symmetric point $A^{\prime}$ of point $A$, then $\angle A^{\prime} O D^{\prime}=60^{\circ}, A B+B C+C D=A^{\prime} B+B C+C D^{\prime}$. Therefore,
$$
\begin{array}{l}... | 12 | Geometry | proof | Yes | Yes | olympiads | false |
Example 7 An arithmetic sequence $\left\{a_{n}\right\}, a_{1}>0$, the sum of the first $n$ terms is $S_{n}$, and $S_{9}>0, S_{10}<0$, for what value of $n$ is $S_{n}$ maximized? | Solution: Since $S_{n}=\frac{9\left(a_{1}+a_{9}\right)}{2}=9 a_{5}>0$,
$$
S_{10}=\frac{10\left(a_{1}+a_{10}\right)}{2}=5\left(a_{5}+a_{6}\right)a_{2}>\cdots>a_{5}>0>a_{6}>a_{7}>\cdots$,
thus, when $n=5$, $S_{n}$ is maximized.
The correctness of the result can be verified not only by directly checking the calculation pr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (2002 Hunan Province Competition Question) The number of sets $X$ that satisfy the condition $\{1,2,3\} \subseteq X \subseteq\{1,2,3,4,5,6\}$ is $\qquad$ | 4. Fill 8. Reason: $X$ must include the 3 elements $1, 2, 3$, while the numbers 4, 5, 6 can either belong to $X$ or not belong to $X$, with each number having 2 possibilities. Therefore, the total number of different $X$ is $2^{3}=8$. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$6 \cdot 82$ Find the smallest real number $A$, such that for every quadratic polynomial $f(x)$ satisfying the condition
$$
|f(x)| \leqslant 1 \quad(0 \leqslant x \leqslant 1)
$$
the inequality $f^{\prime}(0) \leqslant A$ holds. | [Solution] Let the quadratic trinomial be $f(x) = ax^2 + bx + c$. When $0 \leq x \leq 1$, it satisfies the inequality
$$
|f(x)| \leq 1.
$$
In particular, the following inequalities should hold:
$$
|f(0)| \leq 1, \left|f\left(\frac{1}{2}\right)\right| \leq 1, |f(1)| \leq 1.
$$
Since
$$
\begin{array}{c}
f(0) = c, f\lef... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 16$ The number of digits of the number $N=2^{12} \cdot 5^{8}$ is
(A) 9.
(B) 10.
(C) 11.
(D) 12.
(E) 20.
(22nd American High School Mathematics Examination, 1971) | ［Solution］ Reorganize the factors
$$
N=2^{4}\left(2^{8} \cdot 5^{8}\right)=16 \cdot 10^{8}=1600000000 \text {, }
$$
It is a ten-digit number.
Therefore, the answer is (B). | 10 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. Given two lines $I_{1}: 3 x+4 y-25=0, I_{2}: 117 x-44 y-175=0$, point $A$ has projections $B, C$ on lines $I_{1}, l_{2}$, respectively. (1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$; if circle $T:\left(x-\frac{39}{5}\right)^{2}+\left(y-\frac{27}{5}\right)^{2}=r^{2}(... | 11. Analysis: (1) Let $A(p, q)$, then $A B=\frac{|3(p-3)+4(q-4)|}{5}, A C=\frac{|117(p-3)-44(q-4)|}{125}$. Let the inclination angles of the lines $\boldsymbol{I}_{1}, \boldsymbol{l}_{2}$ be $\alpha, \beta$, respectively, then the sine of the angle between the lines $\boldsymbol{I}_{1}, \boldsymbol{l}_{2}$ is $\frac{24... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Lei Lei bought some goats and sheep. If she bought 2 more goats, the average price per sheep would increase by 60 yuan. If she bought 2 fewer goats, the average price per sheep would decrease by 90 yuan. Lei Lei bought $\qquad$ sheep in total. | 【Answer】Solution: Assume Leilei bought $x$ sheep, the original average price was $a$ yuan. Buying 2 goats would increase the average price of each sheep by 60 yuan, and the total price would increase by $60 x+2 (a+60)$ yuan;
If 2 fewer goats were bought, then the average price of each sheep would decrease by 90 yuan, ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Ophelia is paid $\$ 51$ for the first week of her part-time summer job. For each week after the first, she is paid $\$ 100$. How many weeks in total does she have to work for her average weekly pay to be $\$ 93$ ? | 3. Solution 1
Suppose that Ophelia works for a total of $n$ weeks.
Then she earns $\$ 51$ for 1 week and $\$ 100$ a week for $(n-1)$ weeks.
Since her average weekly pay is $\$ 93$, then $\frac{51+(n-1) \times 100}{n}=93$.
Therefore,
$$
\begin{aligned}
51+100(n-1) & =93 n \\
51+100 n-100 & =93 n \\
100 n-49 & =93 n \\
1... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Let $\alpha, \beta$ be a pair of conjugate complex numbers. If $|\alpha-\beta|=2 \sqrt{3}$, and $\frac{\alpha}{\beta^{2}}$ is a real number, then $|\alpha|=$ | 2
1.【Analysis and Solution】Let $\alpha=a+b i, \beta=a-b i, a, b \in \mathbf{R}$,
then from $|\alpha-\beta|=2 \sqrt{3}$,
we get $|b|=\sqrt{3}$,
also from $\frac{\alpha}{\beta^{2}}=\frac{\alpha^{3}}{(\alpha \beta)^{2}}$ being a real number and $\alpha \beta$ being a real number, we know that $\alpha^{3}$ is a real number... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6 Proof: The only positive integer solution to the indeterminate equation $8^{x}+15^{y}=17^{z}$ is $x=y=z=2$. | First, we use congruence to prove that $y$ and $z$ are both even numbers.
Taking equation (1) modulo 4, we get $(-1)^{y} \equiv 1(\bmod 4)$,
thus $y$ is even. Taking equation (1) modulo 16, we get $8^{x}+(-1)^{y} \equiv 1(\bmod 16)$, which simplifies to $8^{x} \equiv 0(\bmod 16)$, hence $x \geqslant 2$.
Note that $17^... | 2 | Number Theory | proof | Yes | Yes | olympiads | false |
2 Find the smallest positive real number $k$ such that for any 4 distinct real numbers $a, b, c, d$ not less than $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation
$$
\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)=0
$$
has 4 distinct real roots. (Supplied by Feng Zhigang) | The required minimum positive real number $k=4$.
On one hand, if $k4(d-a)>0, c^{2}-4 b>4(c-b)>0$, so (1) and (2) each have two distinct real roots.
On the other hand, if (1) and (2) have a common real root $\beta$, then $\left\{\begin{array}{l}\beta^{2}+d \beta+a=0, \\ \beta^{2}+c \beta+b=0,\end{array}\right.$ Subtrac... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
95. A rope is folded in half, and then folded in half again. At this point, each fold is 1 meter long, the rope is $\qquad$ meters long.
Translating the question while retaining the original format and line breaks, the answer is as follows:
95. A rope is folded in half, and then folded in half again. At this point... | Answer: 4 | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In the Cartesian coordinate system, a line $l$ passing through the origin $O$ intersects the curve $y=\mathrm{e}^{x-1}$ at two distinct points $A$ and $B$. Parallel lines to the $y$-axis are drawn through $A$ and $B$, intersecting the curve $y=\ln x$ at points $C$ and $D$. The slope of the line $CD$ is $\qquad$. | 3. 1 .
Given the line $l: y=k x(k>0)$, and it intersects the curve $y=\mathrm{e}^{x-1}$ at different points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
$$
\begin{array}{l}
\text { Hence } k x_{1}=\mathrm{e}^{x_{1}-1}, k x_{2}=\mathrm{e}^{x_{2}-1} \\
\Rightarrow x_{1}=\frac{1}{k} \mathrm{e}^{x_{1}-1}, x_{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6 \cdot 91$ Consider the function $f(x)$ on $[0,1]$ satisfying the conditions:
(1) $f(x) \geqslant 0$, for any $x \in[0,1]$;
(2) $f(1)=1$;
(3) $f(x)+f(y) \leqslant f(x+y), x, y, x+y \in[0,1]$
Try to find the smallest constant $c$, such that for any function $f(x)$ satisfying (1) - (3), we have $f(x) \leqslant c x, x ... | [Solution] The minimum value of $c$ is 2.
In fact, consider the function
$$
f(x)=\left\{\begin{array}{l}
0, \quad 0 \leqslant x \leqslant \frac{1}{2} ; \\
1, \quad \frac{1}{2}<x \leqslant 1.
\end{array}\right.
$$
If $x > \frac{1}{2}$, then $f(x) \leqslant 1 < 2x$.
If $0 < x \leqslant \frac{1}{2}$, then we can choose $... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Given that the bases of two congruent regular tetrahedra are glued together, exactly forming a hexahedron with all dihedral angles equal, and the shortest edge of this hexahedron is 2, then the distance between the farthest two vertices is $\qquad$ . | 4. As shown in the figure, construct $C E \perp A D$, connect $E F$, it is easy to prove $E F \perp A D$, then $\angle C E F$ is the plane angle of the dihedral angle formed by plane $A D F$ and plane $A C D$. Let $G$ be the midpoint of $C D$, similarly, $\angle A G B$ is the plane angle of the dihedral angle formed by... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. Use the digits $1,2,3,4,5,6,7,9$ to form 4 two-digit prime numbers (each digit must be used exactly once). How many different combinations of these 4 prime numbers are possible? | 【Answer】 4
【Solution】 It is easy to know that $2, 4, 5, 6$ can only be used as the tens digit. Let these four two-digit prime numbers be $\overline{2 a}, \overline{4 b}, \overline{5 c}, \overline{6 d}$. The remaining four digits are $1, 3, 7, 9$. A simple analysis shows that $a \neq 1,7, b \neq 9, c \neq 1,7, d \neq 3,... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. In the sequence $\left\{a_{n}\right\}$, $a_{7}=\frac{16}{3}$, and when $n \geqslant 2$, $a_{n}=\frac{3 a_{n-1}+4}{7-a_{n-1}}$.
(1) Does there exist a natural number $m$, such that when $n>m$, $a_{n}2$. If it exists, find the value of $m$; if not, explain the reason.
(2) When $n \geqslant 10$, prove that $\frac{a_{n... | 14. (1) Given $a_{7}=\frac{16}{3}$, and the recursive formula, we can get $a_{8}=12, a_{9}=-8$,
$$
a_{n-2}=\frac{3 a_{n-1}+4}{7-a_{n-1}}-2=\frac{3\left(a_{n-1}-2\right)}{7-a_{n-1}} \text {. }
$$
Therefore, if $a_{n-1}2$.
In summary, the $m$ that satisfies the condition exists, and $m=8$.
（2） $a_{10}=-\frac{4}{3}$, it ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (10 points) Two people are playing a table tennis match, a best-of-three games, where the first to 11 points and the opponent has fewer than 10 points wins the game, or after a 10-all tie, the first to gain a 2-point lead wins. The total score of both players is 31 points, and one player won the first game and the ... | 【Analysis】For the sake of explanation, let's set the winner as Party A and the loser as Party B. Party A wins the first round. If Party A wins the second round again, then Party A will definitely have a higher total score, which does not meet the problem's requirement, so Party A loses the second round and wins the thi... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant 180^{\circ}\right)$, and the complex numbers $z, (1+i) z, 2 z$ correspond to points $P, Q, R$ on the complex plane, respectively. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by segments $... | 3
2.【Analysis and Solution】Let the complex number $\omega$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have $\omega+z=2 \bar{z}+(1+i) z$, which simplifies to $\omega=2 \bar{z}+i z$.
Thus, $|\omega|^{2}=(2 \bar{z}+i z)(2 \bar{z}-i z)$
$$
\begin{array}{l}
=4+1+2 i\left(z^{2}-{ }_{-2}^{z}\right) \\
=5-... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. As shown in Figure 1, a carpet has three different colors, and the areas of the three differently colored regions form an arithmetic sequence. The width of the smallest rectangle in the middle is 1 foot, and the surrounding width of the other two shaded areas is also 1 foot. Then the length of the smallest rectangl... | 10. B.
Let the length of the central small rectangle be $x$ feet.
Then the area of the small rectangle is $x \cdot 1$, the area of the middle region is $(x+2) \times 3 - x = 2x + 6$, and the area of the outermost region is $(x+4) \times 5 - (x+2) \times 3 = 2x + 14$. Since the areas of the three parts form an arithmet... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
11. (20 points) Let complex numbers $z_{1}$ and $z_{2}$ satisfy $\operatorname{Re} z_{1}>0$, $\operatorname{Re} z_{2}>0$, and $\operatorname{Re} z_{1}^{2}=\operatorname{Re} z_{2}^{2}=2$. Find
(1) the minimum value of $\operatorname{Re} z_{1} z_{2}$;
(2) the minimum value of $\left|z_{1}+2\right|+\left|\overline{z_{2}}+... | 11. (1) For $k=1,2$, let
$$
z_{k}=x_{k}+y_{k} \mathrm{i}\left(x_{k} \backslash y_{k} \in \mathbf{R}\right) \text {. }
$$
From the conditions, we know
$$
x_{k}=\operatorname{Re} z_{k}>0, x_{k}^{2}-y_{k}^{2}=\operatorname{Re} z_{k}^{2}=2 \text {. }
$$
Then $\operatorname{Re} z_{1} z_{2}=\operatorname{Re}\left(x_{1}+y_{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. If $x+\frac{1}{x}=\sqrt{5}$, then the value of $x^{11}-7 x^{7}+x^{3}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2
(E) $\sqrt{5}$ | 15. B.
$$
\begin{array}{l}
\text { Given } x+\frac{1}{x}=\sqrt{5} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=5 \\
\Rightarrow x^{2}+\frac{1}{x^{2}}=3 \Rightarrow x^{4}+\frac{1}{x^{4}}+2=9 \\
\Rightarrow x^{4}+\frac{1}{x^{4}}=7 \text {. } \\
\text { Therefore } x^{11}-7 x^{7}+x^{3}=x^{7}\left(x^{4}+\frac{1}{x^{4}}-7\right)=0 .... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. (3 points) On the clock tower of a railway station, there is an electronic clock. On the boundary of the circular clock face, there is a small colored light at each minute mark. At 9:35:20 PM, there are $\qquad$ small colored lights within the acute angle formed by the minute and hour hands. | 【Analysis】First, find the angle between the hour hand and the minute hand at 9:35:20 PM; then, according to the clock face being divided into 60 small segments, with each large segment corresponding to an angle of $30^{\circ}$ and each small segment corresponding to an angle of $6^{\circ}$, we can find the minute marks... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Let the dihedral angle between a lateral face and the base of a regular quadrilateral pyramid $P-ABCD$ be $\alpha$, and the dihedral angle between two adjacent lateral faces be $\beta$. Then $2 \cos \beta + \cos 2 \alpha =$ $\qquad$ . | 7. -1 As shown in the figure, draw $PO \perp$ plane $ABCD$, then $O$ is the center of the square $ABCD$. Let the side length of the base $ABCD$ be $a$, and the side length of the lateral edge be $b$.
Since the projection of the lateral face $\triangle PCD$ on the base $ABCD$ is $\triangle OCD$,
by the projection area t... | -1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18. In $\triangle A B C$, the three sides $a$, $b$, $c$ satisfy $2 b=a+c$. Find the value of $5 \cos A-4 \cos A \cos C+5 \cos C$. | $\begin{array}{l}\text { 18. } 2 b=a+c \Leftrightarrow 2 \sin B=\sin A+\sin C \Leftrightarrow 2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2} \Leftrightarrow \tan \frac{A}{2} \tan \frac{C}{2}=\frac{1}{2} \Leftrightarrow \sqrt{\frac{1-\cos A}{1+\cos A}} \\ \cdot \sqrt{\frac{1-\cos C}{1+\cos C}}=\frac{1}{3} \Leftrightarrow 5 \co... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4・ 135 Find all real or complex roots of the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3 .
\end{array}\right.
$$ | [Solution] Let $x, y, z$ be the roots of the cubic equation $r^{3}-a r^{2}+b r-c=0$. Then, by Vieta's formulas, we have: $a=3, b=3$, and it follows that $x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}=0$ (where $n$ is a natural number), which also holds for $y$ and $z$. That is,
$$
\begin{array}{l}
x^{n+3}-a x^{n+2}+b x^{n+1}-c x... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Given the set $A=\{-100,-50,-1,1,2,4,8,16,32,2003\} . A$'s all non-empty subsets are denoted as $A_{1}, i=1,2, \cdots, 1023$. The product of all elements in each $A$, is denoted as $a, i=1,2, \cdots$, 1023. Then $a_{1}+a_{2}+\cdots+a_{1023}$ is
(A) 0
(B) 1
(C) -1
(D) 1023 | 1. (C).
Let the elements of $A$ be $x_{1}, x_{2}, \cdots, x_{10}$, then $a_{1}+a_{2}+\cdots+a_{1023}=\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1-x_{10}\right)$ -1. Note that one of $x_{1}, x_{2}, \cdots, x_{10}$ is -1, hence $a_{1}+a_{2}+\cdots+a_{1023}=-1$. | -1 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$
then $k=$ . $\qquad$ | 4. 4 .
Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\ri... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given the parabola $C: y^{2}=2 p x(p>0)$ and points $A(a, 0), A^{\prime}(-a, 0)(t>0)$. A line passing through point $A^{\prime}$ intersects the parabola at points $P, Q$. Then the sum of the slopes of lines $A P, A Q$ is $k_{A P}+k_{A Q}=$ $\qquad$ | Solution: 0.
Let points $P\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), Q\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right)$.
From the collinearity of $A^{\prime}, P, Q$ we know
$$
\frac{y_{1}}{\frac{y_{1}^{2}}{2 p}+a}=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{2 p}-\frac{y_{1}^{2}}{2 p}}=\frac{2 p}{y_{1}+y_{2}} .
$$
Simplifying, we get... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
As shown in the figure, fold a square piece of paper twice, then cut along the dotted line. Which of the following options is the unfolded shape?
Translate the above text into English, keep the line breaks and format of the source text, and output the translation result directly. | Restore the problem. Therefore, the answer is (4).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (6 points) Xiao Qing is making a dish: "Fragrant Chives Scrambled Eggs", which requires 7 steps, with the following times:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Washing and chopping scallions & Beating eggs & Mixing egg liquid and chives & Washing the pan & Heating the pan & Heating the oil & Cooking the dish \\
\... | 【Answer】Solution: According to the analysis of the question, the overall process can be as follows:
Wash the pan (half a minute) $\rightarrow$ Chop the vegetables (1 minute) $\rightarrow$ Beat the eggs (half a minute) $\rightarrow$ Heat the pan and oil (while stirring the egg and vegetable mixture)
(1 minute) $\rightar... | 5 | Other | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 61$ The algebraic expression $\frac{1}{\log _{2} 100!}+\frac{1}{\log _{3} 100!}+\frac{1}{\log _{4} 100!}+\cdots+\frac{1}{\log _{100} 100!}$ has the value
(A) 0.01 .
(B) 0.1 .
(C) 1 .
(D) 2 .
(E) 0 .
(49th American High School Mathematics Examination, 1998) | [Solution] By the properties of logarithms, we have
$$
\begin{aligned}
\text { Original expression } & =\log _{100!} 2+\log _{100!} 3+\cdots+\log _{100!} 100 \\
& =\log _{100!} 100! \\
& =1
\end{aligned}
$$
Therefore, the correct choice is (C). | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Let $x, y, z$ be non-negative real numbers, and satisfy the equation
$$
4^{\sqrt{5 x+9 y+4 z}}-68 \times 2^{\sqrt{5 x+9 y+4 x}}+256=0 \text {, }
$$
Then the product of the maximum and minimum values of $x+y+z$ is $\qquad$ . | 4. Let $x=2^{\sqrt{5 x+9 y+4 z}}$, then
$$
x^{2}-68 x+256=0 \text {. }
$$
Solving the equation, we get $x=4,64$.
When $5 x+9 y+4 z=4$, from the inequality
$$
4(x+y+z) \leqslant 5 x+9 y+4 z \leqslant 9(x+y+z),
$$
we know
$$
4(x+y+z) \leqslant 4, \quad 9(x+y+z) \geqslant 4 .
$$
Thus, at this time, the minimum value of... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. Combine a regular hexagon with a side length of 1 and an equilateral triangle with a side length of 1, requiring no overlap, and the sides to be completely coincident, then the number of sides of the new shape is $\qquad$ . | answer: 5 | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 66$ Given that the sum of the squares of the two real roots of the equation $2 x^{2}+k x-2 k+1=0$ is $\frac{29}{4}$, then the value of $k$ is
(A) 3.
(B) -11.
(C) 3 or -11.
(D) 11.
(E) None of the above conclusions is correct.
(China Tianjin Junior High School Mathematics Competition; China Junior High School M... | ［Solution］Let the two real roots of the equation be $x_{1}, x_{2}$, then by the problem we have
$$
\left\{\begin{array}{l}
\Delta=k^{2}-4 \cdot 2(-2 k+1)>0, \\
x_{1}^{2}+x_{2}^{2}=\frac{29}{4}, \\
x_{1}+x_{2}=-\frac{k}{2}, \\
x_{1} x_{2}=\frac{1-2 k}{2} .
\end{array}\right.
$$
From (1), we get $k-8+6 \sqrt{2}$.
From (... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. Find the smallest positive integer $n(n \geqslant 4)$, such that from any $n$ different integers, four different numbers $a$, $b$, $c$, $d$ can always be selected, satisfying $a+b-c-d$ is divisible by 20. | 5. Since the remainder of any integer divided by 20 can only be $0,1,2, \cdots, 19$, for a set containing $k$ integers from $\{0,1,2, \cdots, 19 \mid$, there can be $C_{k}^{2}=\frac{1}{2} k(k-1)$ different integer pairs. If $-\frac{1}{2} k(k-1)>20$, i.e., $k \geqslant 7$, then there exist two different pairs $(a, b)$ a... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
21. The number of positive integer solutions to the equation $\sqrt{x}+\sqrt{y}=\sqrt{1000}$ is $\qquad$ groups. | $9$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Let the focus of the parabola $y^{2}=4 x$ be $F$, and draw any line through point $F$ intersecting the parabola at points $A$ and $B$. Does there exist a line $l$ such that for any point $M$ on line $l$, the slopes of lines $M A$, $M F$, and $M B$ always form an arithmetic sequence? If it exists, find the equation ... | 10. Solution: $F(1,0)$, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, the equation of the line passing through $F$ is $x=m y+1$ combined with $y^{2}=4 x$ gives: $y^{2}=4 m y+4, y_{1}+y_{2}=4 m, \quad y_{1} \cdot y_{2}=-4$,
Assume there exists a point $M(a, b)$ on the line $l$ such that $2 k_{M F}=k_{M A... | -1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then the value of $\frac{1}{\sin \alpha}+\cos ^{4} \alpha$ is | Answer: 2. Solution: From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$. Repeatedly using this conclusion and noting that $\cos ^{2} \alpha+\sin ^{2} \alpha=1$,
$$
\begin{array}{l}
\text { we get } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) Fill the numbers $1 \sim 25$ into the $5 \times 5$ table shown below. In each row, select the largest number, and in each column, select the smallest number. This way, we have made 10 selections in total. Among these 10 selected numbers, there are at least $\qquad$ distinct numbers. | 【Analysis】First, according to the problem, determine that there must be a number which is the largest number in its row and the smallest number in its column; then apply the method of assumption, to conclude that: there do not exist two numbers which are both the largest in their row and the smallest in their column, l... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given the sets $A=\{x, x y, x+y\}, B=\{0,|x|, y\}$ and $A=B$, then $x^{2018}+y^{2018}=$ | Answer 2.
Analysis From the fact that $B$ has three elements, we know that $x \neq 0$ and $y \neq 0$, so in $A$, $x+y=0$, which means $x=-y$. Also, $\{x, x y\}=\{|x|, y\}$: If $\left\{\begin{array}{l}|x|=x \\ x y=y\end{array}\right.$, then $\left\{\begin{array}{l}x=1 \\ y=-1\end{array}\right.$, in this case $A=\{1,-1,0... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 40 Find the maximum positive integer $n$, such that $I=\{1,2, \cdots, n\}$ can be partitioned into two disjoint subsets $A$ and $B$ satisfying that no three numbers in the same subset form an arithmetic progression. | When $n=8$, let $A=\{1,4,5,8\}, B=\{2,3,6,7\}$, then any three numbers in the same subset of $A$ or $B$ do not form an arithmetic sequence.
When $n \geqslant 9$, if $I$ can be decomposed into two disjoint subsets $A$ and $B$ such that any 3 numbers in the same subset do not form an arithmetic sequence, then there are ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 8 Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$, try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$
and write down the reasoning process.
(2000 China National Training Team Selection Exam Problem) | Solution 1: Construct a polynomial
$$
\begin{aligned}
f(x)= & \sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \cdots(x+n)-(x-m-1)(x-m-2) . \\
& (x-m-3) \cdots(x-m-n),
\end{aligned}
$$
We can appropriately choose the coefficients \(a_{i}(0 \leqslant i \leqslant n)\) such that \(f(x) \equiv 0\).
Since \(f(x)\) is a pol... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Expand the binomial $\left(\sqrt{x}+\frac{1}{2 \sqrt{x}}\right)^{n}$ in descending powers of $x$. If the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ . | 8. 3
It's not difficult to find that the coefficients of the first 3 terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these 3 numbers form an arithmetic sequence, we have $2 \cdot \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get: $n=8$ and $n=1$ (discard).
When $\pi=8$, $T_{r+1}=C_{8}^{r}\left(\frac{1}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. 5 red card papers can be exchanged for 2 gold card papers, 1 gold card paper can be exchanged for 1 red card paper and 1 silver card paper. Person A has 3 red card papers and 3 gold card papers, and can get ( ) silver card papers. | 【Answer】 7 sheets.
【Analysis】Key point: Equivalent substitution 2 Gold $=5$ Red; 1 Gold $=1$ Red +1 Silver, then 2 Gold $=2$ Red +2 Silver; so 2 Silver $=3$ Red. Party A has 3 Gold, 3 Gold $=3$ Red +3 Silver; originally also has 3 Red, in total 6 Red $=4$ Silver. Therefore, Party A can exchange for a total of $3+4=7$ (... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The last two digits of the integer $\left[\frac{10^{93}}{10^{31}+3}\right]$ are $\qquad$ (write the tens digit first, then the units digit; where $[x]$ denotes the greatest integer not exceeding $x$). | 4.
$$
\begin{aligned}
{\left[\frac{10^{93}}{10^{31}+3}\right] } & =\left[\frac{\left(10^{31}\right)^{3}+3^{3}-3^{3}}{10^{31}+3}\right] \\
& =\left(10^{31}\right)^{2}-3 \times 10^{31}+3^{2}-1 \\
& =\left(10^{31}\right)\left(10^{31}-3\right)+8
\end{aligned}
$$
So its last two digits are 08. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. If the complex number $z$ satisfies $|z-1|+|z-3-2 \mathrm{i}|=2 \sqrt{2}$, then the minimum value of $|z|$ is | 【Analysis】Answer: 1.
Let $z=x+y \mathrm{i}$, then the geometric meaning of $|z-1|+|z-3-2 \mathrm{i}|=2 \sqrt{2}$ is that the sum of the distances from point $P(x, y)$ to points $A(1,0), B(3,2)$ is $2 \sqrt{2}$. Since $|A B|=2 \sqrt{2}$, point $P$ lies on the line segment $A B$, thus: $|O P| \geq 1$. That is, when $z=1$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$24 \cdot 14$ A frustum of a cone with height 8 contains a sphere $O_{1}$ with radius 2, whose center $O_{1}$ is on the axis of the frustum. The sphere $O_{2}$ is tangent to the upper base and the lateral surface of the frustum. The frustum can also contain another sphere $O_{2}$ with radius 3, which is tangent to the ... | [Solution] Make an axial section through $\mathrm{O}_{2}$ of the frustum, as shown in the figure below. Then make a section through $\mathrm{O}_{2}$ perpendicular to the axis of the frustum, intersecting the axis at circle $O$. From the figure below, it is easy to find that $O O_{2}=4$.
This problem is equivalent to: ... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
2. Xiao Ming has several 1 yuan, 2 yuan, and 5 yuan RMB notes. He wants to use no more than 10 RMB notes to buy a kite priced at 18 yuan, and he must use at least two different denominations of RMB. How many different ways can he pay?
(A) 3
(B) 9
(C) 11
(D) 8 | 【Answer】 $\mathrm{C}$
【Question Type】Odd: List Enumeration
【Analysis】
\begin{tabular}{|c|c|c|c|}
\hline 5 Yuan & 2 Yuan & 1 Yuan & Total Number of Notes \\
\hline 3 & 0 & 3 & 6 \\
\hline 3 & 1 & 1 & 5 \\
\hline 2 & 4 & 0 & 6 \\
\hline 2 & 3 & 2 & 7 \\
\hline 2 & 2 & 4 & 8 \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|... | 11 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4. Given that $n$ is even. If the sum of the first $n$ positive integers starting from 1 has a units digit of 8, then the units digit of the sum of the next $n$ positive integers is ( ).
(A) 6
(B) 4
(C) 2
(D) 0 | 4. C.
Let $n=2k$, and let $S$ be the sum of the first $n$ positive integers, and $D$ be the sum of the next $n$ positive integers (from $2k+1$ to $4k$).
$$
\begin{array}{l}
\text { Then } S=k(2k+1), \\
D=\frac{2k(2k+1+4k)}{2}=(6k+1)k .
\end{array}
$$
The last digits of $S$ and $D$ depend only on the last digit of $k$... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
12. Given the function $f(x)=x^{2}-2 \ln x$, if there exist $x_{1}, x_{2}, \cdots, x_{n} \in\left[\frac{1}{\mathrm{e}}, \mathrm{e}\right]$, such that $f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n-1}\right)=f\left(x_{n}\right)$ holds, find the maximum value of $n$. | Solve
$f^{\prime}(x)=2 x-\frac{2}{x}=\frac{2(x+1)(x-1)}{x}$, then $f(x)$ is monotonically decreasing on $\left(\frac{1}{\mathrm{e}}, 1\right)$ and monotonically increasing on $(1, \mathrm{e})$, so $f(x)_{\min }=f(1)=1, f(x)_{\max }=\max \left\{f\left(\frac{1}{\mathrm{e}}\right), f(\mathrm{e})\right\}=\mathrm{e}^{2}-2$.... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
28・15 A student recorded the exact percent frequency distribution of a set of measurements as shown in the table below:
But he forgot to mark the total number of measurements $N$, then the smallest possible value of $N$ is
(A) 5 .
(B) 8 .
(C) 16 .
(D) 25 .
(E) 50 .
(38th American High School Mathematics Examination, 19... | [Solution] Given $12.5=\frac{1}{8}, 50=\frac{4}{8}, 25=\frac{2}{8}, \quad$ and $\frac{1}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}=\frac{8}{8}=100 \%$, then the smallest possible value of $N$ is 8. Therefore, the answer is $(B)$. | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
15. Given the sequence $\left\{a_{n}\right\}, a_{n}+a_{n+1}=n \cdot(-1)^{\frac{n(n+1)}{2}}$, the sum of the first $n$ terms is $S_{n}$, and $m+S_{2015}=-1007, a_{1} m>0$. Then the minimum value of $\frac{1}{a_{1}}+\frac{4}{m}$ is . $\qquad$ | $$
a_{2}+a_{3}=-2, a_{4}+a_{5}=4, a_{6}+a_{7}=-6, a_{8}+a_{9}=8, \cdots, a_{2014}+a_{2015}=-2014 \text {, }
$$
Thus $S_{2015}=a_{1}-1008 \Rightarrow a_{1}+m=1$. Therefore $\frac{1}{a_{1}}+\frac{4}{m} \geqslant \frac{9}{a_{1}+m}=9$, equality holds when $a_{1}=\frac{1}{3}, m=\frac{2}{3}$, so the minimum value of $\frac{... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Now there are 3 1-gram weights, 3 2-gram weights, and 1 5-gram weight. For a 9-gram object, how many different ways can it be weighed (weights are only allowed on one side of the balance)?
A. 6
B. 7
C. 8
D. 9 | 1. C Hint: Consider the generating function $f(t)=\left(1+t+t^{2}+t^{3}\right)\left(1+t^{2}+\left(t^{2}\right)^{2}+\left(t^{2}\right)^{3}\right)\left(1+t^{5}\right)=\cdots+8 t^{9}+\cdots$, the coefficient of $t^{9}$ is 8. | 8 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
2. (10 points) Let $a \triangle b$ and $a \nabla b$ represent taking the minimum and maximum of two numbers $a$ and $b$, respectively. For example, $3 \triangle 4=3, 3 \nabla 4=4$. For different numbers $x$, the number of possible values for $5 \nabla[4 \nabla(x \triangle 4)]$ is $\qquad$. | 【Analysis】Discuss in two cases: $x>4$ and $x \leqslant 4$, and solve accordingly.
【Solution】Solution: Discuss in cases:
(1) When $x \leqslant 4$, $x \triangle 4=x, 4 \nabla x=4,5 \nabla 4=5$;
(2) When $x>4$, $x \triangle 4=4,4 \nabla 4=4,5 \nabla 4=5$.
Therefore, the value of $5 \nabla[4 \nabla(x \triangle 4)]$ has 1 ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(5) Let $n \in \mathbf{N}^{*}$, and $n^{4}+2 n^{3}+5 n^{2}+12 n+5$ is a perfect square, then $n=$ | (5) 1 or 2 Hint: Since
$$
\left(n^{2}+n+2\right)^{2}<n^{4}+2 n^{3}+5 n^{2}+12 n+5<\left(n^{2}+n+4\right)^{2},
$$
so $n^{4}+2 n^{3}+5 n^{2}+12 n+5=\left(n^{2}+n+3\right)^{2}$, solving it yields $n=1,2$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $\left\{\begin{array}{l}x^{3}+\sin x-2 a=0 \\ 4 y^{3}+\sin y \cos y+a=0\end{array}\right.$, then $\cos (x+2 y)=$ | 1
2.【Analysis and Solution】From the given, we have $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$. Let $f(t)=t^{3}+\sin t$, then $f(x)=(-2 y)$. Since the function $f(t)=t^{3}+\sin t$ is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\therefore x=-2 y, x+2 y=0$, hence $\cos (x+2 y)=1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
115 Let $[x]$ denote the greatest integer not exceeding $x$, referred to as the integer part of $x$. Let $\{x\}=x-[x]$ denote the fractional part of $x$. If the integer part of $x$ is the mean proportional between $x$ and $\{x\}$, then the difference between $x$ and the reciprocal of $x$ is . $\qquad$ | 115 1. From the problem, we have
$$
x(x-[x])=[x]^{2}
$$
Rearranging gives
$$
x^{2}-[x] \cdot x-[x]^{2}=0 .
$$
Solving for $x$, we get
$$
x=\frac{1+\sqrt{5}}{2}[x],(x>0) .
$$
Since
$$
[x] \leqslant x<1+[x],
$$
it follows that
$$
[x] \leqslant \frac{1+\sqrt{5}}{2}[x]<1+[x] .
$$
Solving this, we get
$$
0 \leqslant[x]... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
23. Use 4 red beads and 3 blue beads to make a necklace. How many different necklaces can be made? | 23. $D=\{1,2,3,4,5,6,7\} \quad R=\{r, b\}, C$ has $2^{\top}$ coloring methods,
$$
\begin{array}{l}
G=\left\{e, \rho_{7}, \rho_{7}^{2}, \rho_{7}^{3}, \rho_{7}^{4}, \rho_{7}^{5}, \rho_{7}^{6}, \tau_{1}, \tau_{2}, \tau_{3}, \tau_{4}, \tau_{5}, \tau_{6}, \tau_{7}\right\}, \\
P_{G}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. 2 teachers are taking 30 students across a river by boat, but there is only one small boat that can carry 8 people (no boatman). Each time they cross the river, 2 teachers are needed to act as boatmen. To get everyone across, the boat must make at least $\qquad$ trips. (A round trip counts as 2 trips) | $9$ | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
One, (20 points) Given that $m$ and $n$ are integers, point $P\left(m, n^{2}\right)$
lies on the graph of the function
$$
f(x)=\left(x^{2}-4\right)\left(x^{2}+12 x+32\right)+4
$$
Find the number of points $P$ that satisfy the condition. | Notice,
$$
\begin{aligned}
n^{2} & =\left(m^{2}-4\right)\left(m^{2}+12 m+32\right)+4 \\
& =(m+2)(m-2)(m+4)(m+8)+4 \\
& =\left(m^{2}+6 m+8\right)\left(m^{2}+6 m-16\right)+4 .
\end{aligned}
$$
Let $t=m^{2}+6 m-4$.
Then $n^{2}=(t+12)(t-12)+4$, which means
$$
(t+n)(t-n)=140 \text {. }
$$
Assume $n$ is a non-negative inte... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given $f(x)=a \sin x+\sqrt[2019]{x}+1(a \in \mathbf{R})$, $f\left(\lg \log _{2} 10\right)=3$. Then $f(\lg \lg 2)=$ | ,$- 1 .-1$.
Let $g(x)=f(x)-1=a \sin x+\sqrt[2019]{x}$. Since $g(x)$ is an odd function,
$$
\begin{array}{l}
\Rightarrow f(-x)-1=-(f(x)-1) \\
\Rightarrow f(-x)=-f(x)+2 .
\end{array}
$$
Also, $\lg \lg 2=-\lg \log _{2} 10$, so $f(\lg \lg 2)=-f\left(\lg \log _{2} 10\right)+2=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
59. From the integers 1 to 10, at least $\qquad$ numbers must be taken to ensure that there are two numbers among the taken numbers whose sum is 10. | answer: 7 | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
143 If the circle $x^{2}+y^{2}=R^{2}(R>0)$ and the curve $||x|-| y||=1$ have all their common points as the vertices of a regular polygon, then $R$ equals
A. 1
B. 1 or $\sqrt{2}$
C. 1 or $\sqrt{2+\sqrt{2}}$
D. 1 or $2+\sqrt{2}$ | 143 C. First, draw the graph of $||x|-|y||=1$ in the first quadrant: when $x \geqslant 0, y \geqslant 0$, the graph consists of the rays $x-y=1$ and $y-x=1$. By symmetry about the $x$-axis and $y$-axis, the complete graph can be derived (as shown in Figure 4).
Thus, the regular polygon in question is either a square or... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. Given the function $f(x)=\frac{\left(2^{x}+1\right)^{2}}{2^{x} x}+1$ on the interval $[-2018,0) \cup(0,2018]$, the maximum value is $M$ and the minimum value is $N$. Then $M+N=(\quad)$.
(A) 3
(B) 2
(C) 1
(D) 0 | 5. B.
It is known that the graph of $f(x)$ is symmetric about the point $(0,1)$. Therefore, $M+N=f(x)_{\max }+f(x)_{\min }=2$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
(2) Given the function $f(x)=\sqrt{3} \sin 2 x+2 \cos ^{2} x+a$, if the minimum value of $f(x)$ on the interval $\left[0, \frac{\pi}{2}\right]$ is -1, then the value of $a$ is $\qquad$. | (2) -1 Hint: Since
$$
f(x)=\sqrt{3} \sin 2 x+1+\cos 2 x+a=2 \sin \left(2 x+\frac{\pi}{6}\right)+a+1,
$$
and when $x \in\left[0, \frac{\pi}{2}\right]$, $\frac{\pi}{6} \leqslant 2 x+\frac{\pi}{6} \leqslant \frac{7 \pi}{6}$.
Therefore, when $2 x+\frac{\pi}{6}=\frac{7 \pi}{6}$, i.e., $x=\frac{\pi}{2}$, $f(x)$ reaches its ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Find the maximum value $m_{n}$ of the function
$$
\begin{array}{l}
f_{n}\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
= \frac{x_{1}}{\left(1+x_{1}+\cdots+x_{n}\right)^{2}}+ \\
\frac{x_{2}}{\left(1+x_{2}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}
\end{array}
$$
express $m_{n}$ in te... | 【Analysis】Noticing the linear form of the denominator, it is advisable to first make a substitution for the denominator, and then use the method of undetermined coefficients to eliminate variables to find the extremum.
Let $a_{i}=\frac{1}{1+x_{i}+\cdots+x_{n}}, a_{n+1}=1$. Then
$x_{i}=\frac{1}{a_{i}}-\frac{1}{a_{i+1}}$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Find the maximum value of the positive real number $k$ such that for any positive real numbers $a, b$, we have
$$
\sqrt{a^{2}+k b^{2}}+\sqrt{b^{2}+k a^{2}} \geq a+b+(k-1) \sqrt{a b}
$$ | 10. Analysis: $k=3$. When $a=b$, $2 \sqrt{k+1} a \geq(k+1) a$ holds, i.e., $k \leq 3$;
Next, we prove that for any positive real numbers $a, b$, the inequality $\sqrt{a^{2}+3 b^{2}}+\sqrt{b^{2}+3 a^{2}} \geq a+b+2 \sqrt{a b}$ holds.
Introduce the parameter $p$. By the Cauchy-Schwarz inequality, we have:
$\sqrt{\left(a... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
19. (6 points) A 30 cm long iron wire can form $\qquad$ different rectangles (with side lengths in whole centimeters). | 【Answer】Solution: The perimeter of a rectangle $=($ length + width $) \times 2$,
The sum of length and width is: $30 \div 2=15$ (cm),
Since $15=1+14=2+13=3+12=4+11=5+10=6+9=7+8$, we can form 7 different rectangles.
Answer: We can form 7 different rectangles.
Therefore, the answer is: 7. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. The parabolas $y=x^{2}-b x$, $y=-x^{2}+b x$ and the line $y=k x$ can divide the plane into at most
$\qquad$ parts. | 12. 9 .
Two parabolas are symmetric with respect to the $x$-axis, and the origin is their common point. Also, the line $y=k x$ intersects them at most at 3 points, so it can be known that the plane is divided into at most 9 parts. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (20 points) Find the largest real number $k$ such that for any triangle with sides $a$, $b$, and $c$, the following inequality holds:
$$
\frac{b c}{b+c-a}+\frac{a c}{a+c-b}+\frac{a b}{a+b-c} \geqslant k(a+b+c) .
$$ | 11. For an equilateral triangle, i.e., $a=b=c$, we have
$$
3 a \geqslant k \cdot 3 a \Rightarrow k \leqslant 1 \text {. }
$$
For any triangle, let
$$
x=\frac{b+c-a}{2}, y=\frac{c+a-b}{2}, z=\frac{a+b-c}{2} \text {. }
$$
$$
\begin{array}{l}
\text { Then } \frac{b c}{b+c-a}+\frac{c a}{c+a-b}+\frac{a b}{a+b-c} \\
=\frac{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. On the plane, there is a set of points $M$ and seven different circles $C_{1}, C_{2}, \cdots, C_{7}$, where circle $C_{7}$ passes through exactly 7 points in $M$, circle $C_{6}$ passes through exactly 6 points in $M$, $\cdots$, circle $C_{1}$ passes through exactly 1 point in $M$. The minimum number of points in $M$... | The intersection points of $C_{7}$ and $C_{6}$ can be at most 2, the intersection points of $C_{5}$ with $C_{7}$ and $C_{6}$ can be at most 4, so $C_{7}, C_{6}, C_{5}$ must contain at least $7+6+5-2-4=12$ points, meaning the number of points in $M$ is at least 12. Therefore, the answer is $B$. | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
14. In the United States, June 1 is written as 6/1, while in the United Kingdom, June 1 is written as 1/6. Throughout the year, there are $\qquad$ days when the date is written the same way in both countries. | answer: 12 | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Given that the three sides of triangle $A B C$ are all positive integers, if $\angle A=2 \angle B$, and $C A=9$, then the minimum possible value of $B C$ is $\qquad$ . | By the Law of Sines, $\frac{a}{\sin 2 B}=\frac{9}{\sin B}=\frac{c}{\sin 3 B} \Rightarrow \frac{a}{2 \cos B}=\frac{c}{3-4 \sin ^{2} B}=9$ $\Rightarrow\left\{\begin{array}{l}a=18 \cos B, \\ c=36 \cos ^{2} B-9\end{array} \Rightarrow \cos B>\frac{1}{2}\right.$ and the denominator of $\cos B$ is one of $2,3,6$, so $\cos B=\... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$13.1 .22 \star$ Three vertices of a rhombus are located successively on the sides $A B$, $B C$, and $A D$ of a square with side length 1. Find the area of the figure filled by the fourth vertex of the rhombus. | Let $K$, $L$, $N$ be the vertices of the rhombus on the sides $AB$, $BC$, $AD$ of the square (as shown in Figure a). Note that the length of $KB$ is equal to the distance from point $M$ to the line $AD$. Therefore, if point $K$ is fixed, the possible positions of point $M$ fill a line segment $M_1 M_2$ parallel to the ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Jeanne is standing in a single line with several other people. There are four people in front of her and seven people behind her. How many people are in the line? | 1. The line consists of Jeanne herself, the four people in front of Jeanne, and the seven people behind her. Therefore, there are $4+1+7=12$ people in the line.
ANSWER: 12 | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$17 \cdot 158$ An equilateral triangle $ABC$ has an area of 36, and points $P, Q, R$ are on $BC$, $AB$, $CA$ respectively, such that $BP=\frac{1}{3}BC$, $AQ=QB$, and $PR \perp AC$. Then the area of $\triangle PQR$ is
(A) 10.
(B) 12.
(C) 14.
(D) 15.
(China Junior High School Mathematics Correspondence Competition, 1988... | [Solution] Let the side length of equilateral $\triangle A B C$ be $6a$,
then we have $A Q=Q B=3a, B P=2a, P C=4a, C R=2a, R A=4a$.
$$
\begin{array}{ll}
\because & S_{\triangle A B C}=\frac{1}{2}(6a)^{2} \sin 60^{\circ}=36, \\
\therefore & \frac{1}{2} a^{2} \sin 60^{\circ}=1 .
\end{array}
$$
Thus,
$$
\begin{aligned}
&... | 10 | Geometry | MCQ | Yes | Yes | olympiads | false |
10. (10 points) There is a cup filled with a 15% saltwater solution. There are large, medium, and small iron balls, with volume ratios of 10: 5: 3. First, the small ball is submerged in the saltwater cup, causing 10% of the saltwater to overflow. The small ball is then removed. Next, the medium ball is submerged in the... | 【Analysis】The overflow water volume is actually the volume of the large ball, which is $10 \% \times \frac{10}{3}=\frac{1}{3}$ of the entire cup of brine. Therefore, after filling with water, the concentration becomes $\frac{15 \times\left(1-\frac{1}{3}\right)}{100}$. Solve accordingly.
【Solution】Solution: $10 \% \time... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $\sqrt[3]{26+15 \sqrt{3}}+\sqrt[3]{26-15 \sqrt{3}}=$ | $4$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Two small rulers form a set of rulers, and the small rulers can slide along the large ruler. Each unit on the large ruler is marked with a natural number. The first small ruler divides 11 units on the large ruler into 10, and the second small ruler divides 9 units on the large ruler into 10. The starting... | 【Analysis】According to the problem, the unit ratio of the first small ruler to the large ruler is 11:10. The unit 3 of the first small ruler is equivalent to 3.3 units of the large ruler (calculated based on the ratio). 3.3 of the large ruler and 18.7 can be added to get an integer, so the 0 of the small ruler correspo... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 84$ For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by
$$
q_{i}=\left(q_{i-1}-1\right)^{3}+3 \quad(i=1,2, \cdots, n)
$$
If each $q_{i}(i=1,2, \cdots, n)$ is a power of a prime, find the largest possible value of $n$. | [Solution] Since
$$
m^{3}-m=m(m-1)(m+1) \equiv 0(\bmod 3),
$$
then
$$
q_{i}=\left(q_{i-1}-1\right)^{3}+3 \equiv\left(q_{i-1}-1\right)^{3} \equiv q_{i-1}-1(\bmod 3) .
$$
Therefore, among $q_{1}, q_{2}, q_{3}$, there must be one divisible by 3, and it should be a power of 3. However, in $31((q-$ $\left.1)^{3}+3\right)$... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) In the figure on the right, $ABCD$ is a rectangle, $EF$ is parallel to $BC$, the area of quadrilateral $AECF$ is 17.5, the area of triangle $AFD$ is 20, the area of triangle $BCE$ is 15, and the area of triangle $CDF$ is 12.5. What is the area of triangle $ABE$? | 【Analysis】Since: the area of $\triangle A F D$ plus the area of $\triangle B C E$ equals half the area of the rectangle, from this we can find the area of the rectangle, and then find the area of triangle $A B E$.
【Solution】Solution: The area of $\triangle A F D$ + the area of $\triangle B C E$ = half the area of the ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.59 For each pair of real numbers $x, y$, the function $f$ satisfies the functional equation
$$
f(x)+f(y)=f(x+y)-x y-1 .
$$
If $f(1)=1$, then the number of integers $n$ (where $n \neq 1$) that satisfy $f(n)=n$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) infinitely many.
(30th American High School Mathematics Examination, 1979... | [Solution] Substituting $x=1$ into the functional equation, we get
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Since $f(1)=1$, substituting $y=2,3,4, \cdots$ consecutively, we can see that for $y$ being a positive integer, $f(y)>0$. Therefore, for $y$ being a positive integer,
$$
f(y+1)>y+2>y+1 ;
$$
Thus, for integers $n>1$, $f... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$ . | $=, 7.1$.
Connect $O A, D E$.
From $\angle B O C=2 \angle B A C=120^{\circ}$
$$
\begin{array}{l}
\Rightarrow \angle C O E=60^{\circ}=\angle D A E \\
\Rightarrow A, D, O, E \text{ are concyclic} \\
\Rightarrow \angle D E B=\angle D A O=\angle D B E \\
\Rightarrow D B=D E .
\end{array}
$$
Similarly, $C E=D E$.
Thus, $\f... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Given $f(x)=\sum_{k=1}^{2017} \frac{\cos k x}{\cos ^{k} x}$. Then $f\left(\frac{\pi}{2018}\right)=$ | 8. -1 .
Notice that,
$$
\begin{array}{l}
f(x)+\frac{\sin x}{\sin x} \\
=\frac{\sin 2 x}{\sin x \cdot \cos x}+\sum_{k=2}^{2017} \frac{\cos k x}{\cos ^{k} x} \\
=\frac{\sin 3 x}{\sin x \cdot \cos ^{2} x}+\sum_{k=3}^{2017} \frac{\cos k x}{\cos ^{k} x} \\
=\frac{\sin 2018 x}{\sin x \cdot \cos ^{2017} x} .
\end{array}
$$
... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Three. (50 points)
There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made between any $\frac{n}{2}$ people is the same, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
| Obviously, $n \geqslant 5$. Let the $n$ people be $A_{1}, A_{2}, \cdots, A_{n}$.
Suppose the number of calls made by $A_{t}$ is $m_{i}$, and the number of calls between $A_{t}$ and $A_{j}$ is $\lambda_{i, j}, 1 \leqslant i, j \leqslant n$. Then
$$
m_{2}+m_{j}-\lambda_{r, j}=\frac{1}{2} \sum_{s=1}^{\pi} m_{r}-3^{k}=c .
... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
27. 9 racing cars have different speeds, and they need to compete to determine who is faster, but there are no timing tools, so they can only race on the track to see who comes first, and each time a maximum of 3 cars can race. Therefore, the minimum number of races needed to guarantee selecting the 2 fastest cars is $... | $5$ | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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