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|---|---|---|---|---|---|---|---|---|
Five, let the sum of the squares of two complex numbers x, $\mathrm{y}$ be 7, and their sum of cubes be 10. The largest real value that $x+y$ can take is what?
---
Translation:
Five, let the sum of the squares of two complex numbers x, $\mathrm{y}$ be 7, and their sum of cubes be 10. The largest real value that $x+y$ can take is what?
|
$$
\begin{array}{l}
\because x^{2}+y^{2}=7, \\
\therefore x^{3}+y^{3}=10=(x+y)\left(x^{2}+y^{2}-x y\right) \\
=(x+y)\left[7-\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2}\right] \\
=(x+y)\left[7-\frac{\left.(x+y)^{2}-7\right] .}{2}\right] .
\end{array}
$$
Let $x+y=t$ , then
$$
\begin{array}{c}
10=\frac{21 t-t^{3}}{2}, \\
t^{3}-21 t+20=0, \\
(t-1)(t-4)(t+5)=0 . \\
\therefore t_{1}=1, t_{2}=4, t_{3}=-5 .
\end{array}
$$
That is, the largest real value that $\mathrm{x}+\mathrm{y}$ can take is 4 .
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Seven, the 25 knights of King Arthur sat at their round table, and three knights (any of whom were chosen with equal probability) were selected to slay the dragon. Let $\mathrm{p}$ be the probability that at least two of the three selected knights are seated next to each other. If $\mathrm{p}$ is written as a reduced fraction, find the sum of the numerator and the denominator.
|
The number of ways to choose 3 people from 25 is
$$
C_{25}^{3}=\frac{25 \cdot 24 \cdot 23}{3 \cdot 2 \cdot 1}=25 \cdot 23 \cdot 4 \text { (ways). }
$$
The number of ways to choose 3 people from 25 such that at least 2 are adjacent can be calculated as follows: treat the adjacent two people as one, then there are 25 ways to choose, and the other person can be chosen in 23 ways. However, this method counts the cases where all three are adjacent twice, and there are 25 ways to choose three people who are all adjacent. Therefore, the number of ways to choose 3 people from 25 such that at least 2 are adjacent is
$$
25 \cdot 23-25=25 \cdot 22 \text { (ways). }
$$
Thus, the required probability is $\frac{25 \cdot 22}{25 \cdot 23 \cdot 4}=\frac{11}{46}$.
Therefore, the sum of the numerator and the denominator is 57.
|
57
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l} \text { Eight, the largest two-digit number factor of the number } \mathrm{n}=\binom{200}{100} \text { is } \\ \text { what? } \quad \text { [Note: }\binom{200}{100} \text { is } \mathrm{C}_{2}^{1} 00 \% \text { ] }\end{array}$
|
Solve $\mathrm{n}=\mathrm{C}_{200}^{+} \times 0=\frac{200 \cdot 199 \cdot 198 \cdots 102 \cdot 101}{100!}$
$$
=2^{50} \cdot \frac{199 \cdot 197 \cdot 195 \cdots 103 \cdot 101}{50!} .
$$
This number is an integer, and the numerator $199 \cdot 197 \cdot 195 \cdots 103 \cdot 101$
is the product of three-digit odd numbers, so the largest two-digit prime factor we are looking for can only be found among the three-digit composite numbers.
Since any composite number less than 200 must have a prime factor less than $\sqrt{200}$, the prime factors of the odd composite numbers between 101 and 199 must include one of $3, 5, 7, 11, 13$. And $\frac{199}{3}$ $=66 \frac{1}{3}$, so the largest two-digit prime factor we are looking for must be less than 66, and the largest prime number less than 66 is 61, and $61 \times 3=183$ is exactly one of the factors in the numerator of $\mathrm{C}_{200}^{100}$, so the largest prime factor we are looking for is 61.
|
61
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Nine, find the minimum value of $f(x)=\frac{9 x^{2} s \sin ^{2} x+4}{x \sin x}(0<x<\pi)$.
|
Let $\mathbf{x} \sin \mathrm{x}=\mathrm{t}$, then $\mathrm{t}$ is a real number.
Thus $\quad \mathrm{f}(\mathrm{x})=\frac{9 \mathrm{t}^{2}+4}{\mathrm{t}} \geq \frac{2 \cdot 3 \cdot 2}{t}=12$.
$\therefore \mathrm{f}(\mathrm{x})$'s minimum value is 12.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Ten, $1447, 1005, 1231$ have many things in common: they are all four-digit numbers, their highest digit is 1, and they each have exactly two identical digits. How many such numbers are there?
|
For the same digit being 1, since the highest digit is 1, the other 1 has three positions to choose from, and the remaining positions have $\mathrm{P}_{8}^{2}$ arrangements, i.e., $3 \cdot P_{0}^{*}=3 \times 72=216$ (ways).
For the same digit not being 1, there are 9 choices for the same digit, 8 choices for the other digit, and the other digit can be placed to the left, right, or in between the two identical digits, giving $3 \cdot 9 \cdot 8=216$ ways,
$\therefore$ there are a total of $216+216=432$ numbers that meet the requirements of the problem.
|
432
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Twelve, as shown in Figure 5, the length of the diameter AB of a circle is a two-digit integer (in decimal). By reversing the two digits, we get the length of the chord CD, which is perpendicular to the diameter AB. The distance from the intersection point H to the center O of the circle is a positive rational number. Find the length of AB.
|
Let $\mathrm{AB}=10 \mathrm{x}+\mathrm{y}$, where $\mathrm{x}, \mathrm{y}=0,1,2, \cdots, 9$, and $x \neq 0$.
Then $C D=10 y+x$.
$$
\begin{aligned}
\mathrm{OH}^{2} & =\left(\frac{10 \mathrm{x}+\mathrm{y}}{2}\right)^{2}-\left(\frac{10 \mathrm{y}+\mathrm{x}}{2}\right)^{2} \\
& =\frac{9}{4} \cdot 11\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) .
\end{aligned}
$$
Since $0 H$ is a rational number, then $11\left(x^{2}-y^{2}\right)$ must be a perfect square. Let
$$
11\left(x^{2}-y^{2}\right)=(11 k)^{2},
$$
then $x^{2}-y^{2}=(x+y)(x-y)=11 k^{2}$.
Obviously $x+y \geqslant x-y$, and $0<x+y \leqslant 18$, $0<x-y<8$,
thus we can only have $x+y=11, x-y=1$.
Solving gives $\mathrm{x}=6, \mathrm{y}=5$,
i.e., $\mathrm{AB}=65$.
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Thirteen, for $\{1,2,, 3 \cdots, n\}$ and each of its non-empty subsets, we define the alternating sum as follows: arrange the numbers in the subset in descending order, then alternately add and subtract the numbers starting from the largest (for example, the alternating sum of $\{1,2,4,6,9\}$ is $9-6+4-2+1=6$, and the alternating sum of $\{5\}$ is 5). For $n=7$, find the total of all these alternating sums.
|
Solution: Obviously, the number of non-empty subsets of $\{1,2,3,4,5,6,7\}$ is $2^{7}-1=127$. Each element appears 64 times in these subsets. According to the problem's requirements, the numbers $1,2,3,4,5,6$ each appear 32 times in odd positions and 32 times in even positions, so the alternating sum of these numbers in the subsets is 0. The number 7 also appears 64 times and always takes a positive value, so the total alternating sum of all subsets is $7 \times 64=448$.
|
448
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Fourteen, in Figure 6, the radii of the two circles are 8 and 6, the distance between the two centers is 12, a line passing through one of the intersection points of the two circles intercepts equal chords $\mathrm{QP}$ and $\mathrm{PR}$ on the two circles, find the square of the length of $QP$.
|
Solve as shown in Figure 7, construct $\mathbf{O}_{1} M \perp Q \mathbf{Q}$ at $M, \mathbf{O}_{2} \mathbf{N} \perp$ $Q R$ at. Given $Q P=$
$\mathrm{PR}$, then $\mathrm{QM}=\mathrm{NR}=\frac{1}{2} \mathrm{Q} P$.
Let $Q \mathrm{M}=\mathrm{x}$, then
$$
\begin{array}{c}
\mathrm{O}_{1} \mathrm{M}=\sqrt{ } \overline{\mathrm{Q}^{2}-\mathrm{O}_{1} \mathrm{M}^{2}} \\
=\sqrt{8^{2}-\overline{x^{2}}},
\end{array}
$$
$$
\mathbf{O}_{2} \mathbf{N}=\sqrt{ } \mathrm{O}_{2}^{-} \mathrm{R}^{2}-\mathrm{NR}^{2}-\therefore 1 / \overline{6^{2}}-x^{2}
$$
Then $\mathrm{O}: \mathrm{X}=\mathrm{M}-\mathrm{O}_{2} \mathrm{~N}, \mathrm{O}_{2} \mathrm{~K}=\mathrm{MN}=2 \mathrm{x}$.
By the Pythagorean theorem, we get
$$
\mathrm{O}_{1} \mathrm{O}_{2}{ }^{2}=\mathrm{O}_{1} \mathrm{~K}^{2}+\mathrm{O}_{2} \mathrm{~K}^{2},
$$
which is $144=\left(\sqrt{ } 8^{2}-x^{2}-\sqrt{ } 6^{2}-x^{2}\right)^{2}+(2 x)^{2}$. Solving this, we get $4 \mathrm{x}^{2}=130$, which means $Q \mathrm{P}^{2}=130$.
|
130
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Fifteen, as shown in Figure 8, if two intersecting chords of a circle, with $\mathrm{B}$ on the smaller arc $\mathrm{AD}$, and the circle's radius is $5, \mathrm{BC} = 6$, $\mathrm{AD}$ is bisected by $\mathrm{BC}$. Also, from $\mathrm{A}$, the chord $\mathrm{AD}$ is the only one that can be bisected by $\mathrm{BC}$. This implies that the sine of the smaller arc $\mathrm{AB}$ is a rational number. If this rational number is expressed in its simplest form as $-\frac{m}{n}$, find $mn$.
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。
|
Solve as shown in Figure 9, let $\mathrm{AD}$ and $\mathrm{BC}$ intersect at $P$, connect $O P$, then $O P \perp A D$. Connect $\mathrm{OA}, \mathrm{O} 3, \mathrm{O} C$, draw $\mathrm{OE} \perp . \mathrm{BC}$ at $E$, and let $P A=P D=y, P E=$ $x$, then $B P=3-x$. Also, from the given conditions, $\mathrm{OE}=4$.
By the intersecting chords theorem
$$
\mathrm{AP} \cdot \mathrm{PD}=\mathrm{BP} \cdot \mathrm{PC} \text {, }
$$
which is
$$
y^{2}=(3-x)(3+x)=9-x^{2} .
$$
Since point $P$ is unique, the circle with $O A$ as the radius is tangent to $B C$, thus the angle between the tangent and the chord $\angle \mathrm{OPE}=\angle:$ :
Because $\cos \mathbf{A}=\frac{\rho \mathbf{A}}{0 \mathbf{A}}=\frac{Z}{5}$,
$$
\cos O P E=\frac{P E}{O P}=\frac{x}{\sqrt{16+x^{2}}},
$$
thus, $\frac{y}{5}=-\frac{x}{\sqrt{16+x^{2}}}$.
Squaring and substituting $y^{2}=9^{2}-x^{2}$, we can solve to get $x=2$.
Therefore, $C P=x+3=5$.
Thus, $\triangle \mathrm{CPO}$ is an isosceles triangle, $\angle S \mathrm{PO}=\angle C O P$.
Since $\angle \mathrm{A}+\angle \mathrm{AOP}=90^{\circ}$, so
$$
\angle \mathrm{COP}+\angle \mathrm{AOP}=90^{\circ} \text {, }
$$
thus $O A \perp O C$.
$$
\text { Therefore, } \begin{array}{l}
\angle \mathrm{AOB}+\angle \mathrm{BO}=90^{\circ}, \\
{ }_{\sin} \mathrm{AOB}=\cos \mathrm{BOC} .
\end{array}
$$
And $\cos B O C=\frac{\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{BC}^{3}}{2 \cdot \mathrm{OB} \cdot \mathbf{O C}}=\frac{7}{25}$.
which is a rational number,
$\therefore \sin A O B=\frac{7}{25}$ is also a rational number,
i.e., $\frac{\mathrm{m}}{\mathrm{n}}-=\frac{7}{25}, \mathrm{mn}=25 \times 7=175$.
|
175
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Let the integer part of $\frac{1}{3-\sqrt{7}}$ be $a$, and the decimal part be $b$, find the value of $a^{2}+(1+\sqrt{7}) a b$.
|
$$
\begin{aligned}
\because \frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{(3-\sqrt{7})(3+\sqrt{7})} \\
\quad=\frac{3+\sqrt{7}}{2},
\end{aligned}
$$
and $2<\sqrt{7}<3,0<\frac{\sqrt{7}-1}{2}<1$,
$$
\begin{array}{l}
\therefore \quad \frac{1}{3-\sqrt{7}}=\frac{3+1+\sqrt{7}-1}{2} \\
=2+\frac{\sqrt{7}-1}{2} .
\end{array}
$$
Therefore, $a=2, b=\frac{\sqrt{7}-1}{2}$,
$$
\begin{array}{l}
\text { Hence } \mathrm{a}^{2}+(1+\sqrt{7}) \mathrm{ab}=2^{2} \\
+\frac{2(1+\sqrt{7})(\sqrt{7}-1)}{2}=4+6=10 .
\end{array}
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Given $1 x=(3+2 \sqrt{2})^{-1}$, $y=(3-2 \sqrt{2})^{-1}$, find the value of $(x+1)^{-1}+(y+1)^{-1}$.
|
$$
\begin{array}{c}
\text { Sol } \because x=(3+2 \sqrt{2})^{-1} \\
=\frac{3-2 \sqrt{2}}{(3+2 \sqrt{2})(3-2 \sqrt{2})}=3-2 \sqrt{2},
\end{array}
$$
$\therefore x, y$ are conjugate radicals, thus we have
$$
\begin{array}{l}
y=x^{-1}=3+2 \sqrt{2} . \\
\therefore(x+1)^{-1}+(y+1)^{-1}=(4-2 \sqrt{2})^{-1} \\
\quad+(4+2 \sqrt{2})^{-1}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
二、 1 . The sum of the first 15 terms of an arithmetic sequence is 225, find the 8th term.
|
二、1, Solve $\because 15 \mathrm{a}_{1}+\frac{15 \times 14}{2} \mathrm{~d}=225$, i.e., $15\left(a_{1}+7 d\right)=225, \quad \therefore a_{3}=a_{1}+7 d=\frac{225}{15}=15$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The sum of the first term and the second term of a geometric sequence is 30, the sum of the third term and the fourth term is 120, find the sum of the fifth term and the sixth term.
|
2. Solution Let the first term be $\mathrm{a}$, and the common ratio be $\mathrm{q}$. Then
$$
\begin{array}{l}
a+a q=30, \\
a q^{2}+a q^{3}=120 .
\end{array}
$$
From (1) and (2), we get $q= \pm 2$. When $q=2$, $a=10$, at this time the sum of the 5th and 6th terms is 480. When $q=-2$, $a=-30$, at this time the sum of the 5th and 6th terms is 480.
|
480
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the last three digits of $769^{304}$.
untranslated text remains in its original form, preserving the line breaks and format.
|
Solution: Since $769^{304}=769^{301} \times 769^{3}$, applying Theorem Six we get
$769^{301}=769^{4 \times 8^{2} \times 3+1}$, whose last three digits are 769.
It is not difficult to calculate that the last three digits of $769^{8}$ are 609.
Therefore, the last three digits determined by $769 \times 609$ are 321,
which means the last three digits of $769^{304}$ are 321.
|
321
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sequence $\left\{\frac{n+1}{n+2}\right\}$, for any $e>0$, take $\mathbf{N}=$ $\qquad$ $\sim$, then when $n>N$, it always holds that $\qquad$ $<\varepsilon$, so the limit of the sequence $\left\{\frac{n+1}{n+2}\right\}$ is 1.
|
1. The integer part of $\left(\frac{1}{\varepsilon}-2\right)$, $\left|\frac{n+1}{n+2}-1\right|$,
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, in an arithmetic sequence, if the 11th term is 20, find the sum of the first 21 terms.
|
Three, Solution From $a_{1}+10 \mathrm{~d}=20$, we have
$$
S_{21}=21 a_{1}+\frac{21 \cdot 20}{2} d=21\left(a_{1}+10 d\right)=420 .
$$
|
420
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 1. Find a prime number that remains prime when 10 or 14 is added to it.
|
Because $2+10=12,2+14=16$, the prime number 2 does not meet the requirement;
Because $3+10=13,3+14=17$, the prime number 3 meets the requirement,
Because $5+10=15,5+14=19$, the prime number 5 does not meet the requirement;
Because $7+10=17,7+14=21$, the prime number 7 does not meet the requirement;
Because $11+10=21,11+14=25$, the prime number 11 does not meet the requirement.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. A newly built city has purchased a large number of public buses to solve the city's public transportation problems. They plan to set up bus stops at 1983 different locations and connect them by opening several bus routes. Their wishes are:
(i) to open as many routes as possible,
(ii) every two routes should have at least one common bus stop;
(iii) each bus stop should be shared by at most two different routes. How many bus routes can they open at most according to these wishes? How many bus stops should each route have at least?
|
Let a total of $k$ bus routes be established, and let the set of 1983 bus stops be denoted as $\mathrm{A}$. The sets of bus stops served by each route are respectively denoted as
$$
A_{1}, \hat{H}_{2}, \cdots, \hat{A}_{K},
$$
It can be known from the problem statement that these subsets have the following properties:
(i) $A_{i_{1}} \cap A_{i_{2}} \neq \phi, i_{1} \neq i_{2}$;
(ii) $A_{i_{1}} \cap A_{i_{2}} \cap A_{i_{3}} = \phi, i_{1}, i_{2}, i_{3}$ are pairwise distinct. Therefore, each subset $A_{i}$ contains at least $k-1$ elements, because by (i) it has at least one common element with each of the other subsets, and by (ii) these elements are pairwise distinct. Thus, the sum of the number of elements in these $k$ subsets is at least $k(k-1)$.
However, in the above calculation, each element is counted twice, so the number of elements in $\bigcup_{i=1}^{k} A_{i}$ is at least $\frac{k(k-1)}{2}$. But we should have $\bigcup_{i=1}^{k} A_{i} = A$, so $\frac{k(k-1)}{2} \leq 1983$.
The largest integer satisfying this inequality is $k=63$, so they can establish at most 63 bus routes, and each route should serve at least 62 bus stops.
|
63
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Let the point $(x, y)$ vary within the set $\left\{\begin{array}{l}y-x+1 \geqslant 0, \\ 2 x-y-3 \geqslant 0\end{array}\right.$, find the minimum value of $x+2 y$.
|
This problem originally belongs to the category of quantitative relationship, but the given set $\left\{\begin{array}{l}y-x+1 \geqslant 0, \\ 2 x-y-3 \geqslant 0\end{array}\right.$ can be represented on a coordinate plane, as shown in Figure 4. If we let $x+2 y=k$, then $y=-\frac{1}{2} x+\frac{k}{2}$, which represents a family of parallel lines with a slope of $-\frac{1}{2}$. Thus, the original problem is transformed into finding when the y-intercept is minimized as 1 varies over the region $G$. Since the range of region $G$ is $x-1 \leqslant y \leqslant 2 x-3, 2 \leqslant x<+\infty$, it is clear that the y-intercept of the family of parallel lines? is minimized at the point $(2,1)$. Therefore, the minimum value of $\mathrm{x}+2 \mathrm{y}$ is
$$
\mathrm{k}=2+2 \times 1=4
$$
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. In isosceles trapezoid $\mathrm{ABCD}$, the lower base $\mathrm{AB}$ is 20, the upper base $\mathrm{CD}$ is 12, and the height is $8 \sqrt{2}$. By folding the trapezoid along the perpendicular bisector MN (Figure 1) of the bases to form a dihedral angle of $120^{\circ}$, find the length of $\mathrm{AC}$ at this time.
|
```
Figure 2 is the folded shape. Figure 3 is the shape after Figure 2 is supported, where $\mathrm{AE}$ $\mathrm{FM}-\mathrm{PQCN}$ is a right quadrilateral prism, with the two bases being parallel parallelograms (the figure is intuitive, and the construction method is not described again, the same below).
\[
\begin{array}{c}
\mathrm{AM}=10, \\
\mathrm{MF}=\mathrm{NC}=6, \\
\angle \mathrm{AMF}=120^{\circ} \\
\mathrm{AF}^{2}-\mathrm{AM}^{2}+\mathrm{MF}^{2}-2 \mathrm{AM} \cdot \mathrm{MF} \cdot \cos 120^{\circ} \\
=14^{2} \cdot \mathrm{CF}=\mathrm{MN}=8 \sqrt{2}, \\
\therefore \quad \mathrm{AC}=\sqrt{\mathrm{AF}^{2}+\mathrm{CF}^{2}}=18 .
\end{array}
\]
Cosine Rule
If the angle between the skew lines $A M$ and $C N$ is $\theta$, and $\mathrm{MN}$ is the distance between the two skew lines, then
\[
\mathrm{AC}=\sqrt{\mathrm{MN}^{2}+\mathrm{AM}^{2}+\mathrm{CN}^{2}-2 \mathrm{AM} \cdot \mathrm{CN} \cos \theta}
\]
This result is consistent with the formula on page 46 of the stereometry textbook for key middle schools with a six-year system.
\[
\begin{array}{l}
M B=2 \sqrt{2}, \\
Q B=4, \\
\text { Pythagorean Theorem } \\
\mathrm{MB} / / \mathrm{PA} \perp \mathrm{QB} \\
Q M^{2}=24 \\
\mathrm{MR}=2, \\
Q R=2 \sqrt{3}, \\
\text { Chord Theorem } \\
\angle \mathrm{QRM}=\theta \\
Q M^{2}=16-8 \sqrt{3} \cos \theta. \\
\mathrm{PA}=4 \sqrt{2}, \\
\angle \mathrm{PLA} \text { is a right angle, } \because \mathrm{PL}=4, \\
\angle \mathrm{PAL}=45^{\circ} \\
\left.\begin{array}{l}
\mathrm{BR}=2, \\
\mathrm{AB}=6
\end{array}\right\} \Rightarrow \mathrm{LR}=4. \\
\therefore \quad \mathrm{PQ}=\sqrt{\mathrm{LR}^{2}+\mathrm{PL}^{2}+\mathrm{QR}^{2}-2 \mathrm{PL} \cdot \mathrm{QR} \cos \theta}=2 \sqrt{15}. \\
\end{array}
\]
```
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Let $f(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x<0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$ Find $\int_{-1}^{1} f(x) d x$.
untranslated text is kept in its original format and alignment.
|
Solve $\because F(x)=\left\{\begin{array}{ll}0, & -1 \leqslant x \leqslant 0, \\ 1, & 0<x \leqslant 1 .\end{array}\right.$
By formula (*), we get
$$
\begin{array}{c}
\int_{-1}^{1} f(x) d x=\int_{-1}^{0} 0 \cdot d x+\int_{0}^{1} 1 \cdot d x \\
=F(0)-F(0+0)+F(0-0)-F(-1)=1 .
\end{array}
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, if $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, find the extremum of $\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}$, and the corresponding values of $\alpha$ and $\beta$.
|
$$
\begin{array}{l}
\geqslant 2 \sqrt{\sin ^{2} \alpha} \cos ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta \\
=\frac{8}{\sin 2 \alpha \sin 2 \beta} \geqslant 8 \text {, it is easy to list the above formula to get the minimum value } \\
\end{array}
$$
The condition for 8 is $\sin 2 \alpha=\sin 2 \beta=1$, at this time $\alpha=\beta=\frac{\pi}{4}$. Note: This answer is incorrect.
Because when $\alpha=\beta=\frac{\pi}{4}$, $\frac{1}{\cos ^{2} \alpha}+$
$$
\overline{\sin } \frac{1}{\sin ^{2} \beta \cos ^{2} \beta}=2+8=10 \neq 8 \text {. As for }
$$
where the mistake lies, it is left for the reader to consider.
Since the variables in the problem are "symmetric," but the conclusion does not conform to the above "rule," it indicates that this "rule" is not reliable.
The correct solution is:
$$
\begin{array}{l}
\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}=\frac{1}{\cos ^{2} \alpha} \\
+\frac{4}{\sin ^{2} \alpha \sin ^{2} 2 \beta} \stackrel{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha} \\
=1+\operatorname{tg}^{2} \alpha+4\left(1+\operatorname{ctg}^{2} \alpha\right)=5+\operatorname{tg}^{2} \alpha \\
+4 \cot \approx^{2} \in \geqslant 5+2 \sqrt{1 g^{2}} \bar{c} \cdot 1 \operatorname{cog}^{2} \alpha=5+2 \times 2 \\
=9 \text {, so } \frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \\
\geqslant 9 \text {. } \\
\end{array}
$$
(1) The equality in (1) holds only when $\sin 2 \beta=1$, i.e., $\beta=\frac{\pi}{4}$; (2) The equality in (2) holds only when $\operatorname{tg}^{2} \alpha=4 \operatorname{ctg}^{2} \alpha$, i.e., $\operatorname{tg}^{2} \alpha=2, \quad \operatorname{tg} \alpha=\sqrt{2}, \quad \alpha=\operatorname{arctg} \sqrt{2}$.
$\therefore$ When $\boldsymbol{\beta}=\frac{\pi}{4}, \alpha=\operatorname{arctg} \sqrt{2}$, the extremum in the original expression is 9, and at this time $\alpha \neq \beta$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given $\mathrm{A}_{0}, \mathrm{~A}_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ are five equally spaced points on the unit circle.
Prove: $\left(\mathrm{A}_{0} \mathrm{~A}\right.$,
$$
\left.\mathrm{A}_{0} \mathrm{~A}_{2}\right)^{2}=5
$$
|
Prove that $A_{0} A_{1}$, $A_{0}$
$A_{2}$ are chords of a circle with a common endpoint at $A_{0}$. Taking $A_{0}$ as the pole and the diameter $A_{0} A$ as the polar axis, we establish a polar coordinate system. According to the problem, $\angle A_{1} A_{0} A_{2} = 36^{\circ}$, $\angle A_{2} A_{0} X = 18^{\circ}$, and the equation of circle $\odot O$ is $P = 2 \cos \theta$. Let $\theta$ be $54^{\circ}$ and $18^{\circ}$, respectively, then $A_{0} A_{1} = 2 \cos 54^{\circ}$, $A_{0} A_{2} = 2 \cos 18^{\circ}$. Therefore, $\left(A_{0} A_{1} \cdot A_{0} A_{2}\right)^{2} = \left(2 \cos 54^{\circ} \cdot 2 \cos 18^{\circ}\right)^{2} = \left(4 \sin 36^{\circ} \cdot \cos 18^{\circ}\right)^{2} = \left(8 \sin 18^{\circ} \cos^{2} 18^{\circ}\right)^{2} = 8^{2}\left(\frac{\sqrt{5}-1}{4}\right)^{2} - \left[1-\left(\frac{\sqrt{5}-1}{4}\right)^{2}\right] = 5$.
|
5
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. A function of the natural number $\mathrm{n}$ satisfies the following conditions:
(1) $f(n)=f(n-1)+a^{n}(n \geqslant 2$, $a$ is a non-zero constant $)$;
(2) $f(1)=18$.
Try to find: (1) the analytical expression of $\mathrm{f}(\mathrm{n})$;
(2) when $a=1$, what is the value of $f(1983)$?
|
From the given recurrence relation, we get
$$
\begin{array}{l}
f(2)=f(1)+a^{2}, \\
f(3)=f(2)+a^{3}, \\
f(4)=f(3)+a^{4}, \\
\cdots \cdots \\
f(n-1)=f(n-2)+a^{n-1}, \\
f(n)=f(n-1)+a^{n} .
\end{array}
$$
By adding both sides of the above equations, we get
$$
\begin{array}{l}
f(n)=f(1)+a^{2}+a^{3}+a^{4}+\cdots+a^{n}, \text { thus } \\
f(n)=\left\{\begin{array}{l}
f(1)+\frac{a^{2}\left(a^{n-1}-1\right)}{a-1},(a \neq 1) \\
f(1)+(n-1) \cdot(a=1)
\end{array}\right.
\end{array}
$$
(2) $\mathrm{f}(1983)=18+(1983-1)=2000$.
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$2, \mathrm{n}$ is the smallest integer with the following property: it is a multiple of 15, and each of its digits is 0 or 8. Find
$$
\frac{\mathrm{n}}{15}
$$
|
Solve $n$ should be a common multiple of 5 and 3. The last digit of a multiple of 5 can only be 0 or 5, and since $n$ consists only of the digits 0 and 8, the last digit of $n$ must be 0. As a multiple of 3, the number of 8s in $n$ must be a multiple of 3. Given the requirement for $n$ to be the smallest, we should have $n=8880$, thus $\frac{n}{15}=592$.
|
592
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in the figure, $P$ is a point inside $\triangle ABC$. Lines are drawn through $P$ parallel to the sides of $\triangle ABC$.
The smaller triangles
$t_{1}, t_{2}$, and $t_{3}$
have areas of 4, 9,
and 49, respectively. Find the area of $\triangle ABC$.
|
Given the figure, let $\mathrm{MP}=\mathrm{p}, \mathrm{PN}=\mathrm{q}, \mathrm{RT}=\mathrm{r}$. Since triangles $t_{1}, t_{2}, t_{3}$ are all similar to $\triangle A B C$, the ratio of their areas is equal to the square of the ratio of corresponding sides.
Let $\mathrm{AB}=c, \triangle \mathrm{ABC}$ have an area of $\mathrm{S}$, then we have:
$$
\sqrt{2}=\frac{q}{c}, \frac{3}{\sqrt{S}}=\frac{p}{c}, \frac{7}{\sqrt{S}}=\frac{r}{c}
$$
Adding the three equations, we get: $\frac{2+3+7}{\sqrt{\bar{S}}}=\frac{\mathrm{p}+\mathrm{q}+\mathrm{r}}{\mathrm{c}}$
$$
\begin{aligned}
=\frac{c}{c} & =1 \\
\therefore \sqrt{S} & =12, S=144
\end{aligned}
$$
|
144
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let S be a table consisting of positive integers - the table can contain the same numbers - and it contains the number 68. The mean (arithmetic average) of the numbers in S is 56. However, if 68 is removed, the mean of the remaining numbers drops to 55. What is the largest number that could appear in S?
|
Let $\mathrm{n}$ denote the number of positive integers in $\mathrm{S}$, then the sum of these $\mathrm{n}$ numbers is $56 n$. After removing 68, the sum of the remaining $n-1$ numbers is $55(n-1)$, thus we get $56 n-55(n-1)$ $=68$, solving for $\mathrm{n}$ gives $\mathrm{n}=13$. After removing 68, the sum of the remaining 12 numbers in $\mathrm{S}$ is $55 \times 12=660$. To make one of these 12 numbers the largest, the other numbers should be as small as possible. For this, $\mathrm{S}$ should have 11 ones, and the largest possible number would be $660-11=649$.
|
649
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $\log _{8} a+\log _{4} b^{2}=5$, and $\log _{8} b$ $+\log _{4} \mathrm{a}^{2}=7$. Find $\mathrm{ab}$.
|
Solving: Add the two equations given in the problem, and by the properties of logarithms, we get:
$$
\log _{8} a b+\log _{4} a^{2} b^{2}=12 .
$$
By changing the base, we get $\frac{\log _{2} a b}{3}+\log _{2} a b=12$,
$$
\begin{array}{l}
\frac{4}{3} \log _{2} \mathrm{ab}=12, \log _{2} \mathrm{ab}=9, \\
\therefore \mathrm{ab}=2^{9}=512 .
\end{array}
$$
|
512
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6 . Three circles, each with a radius of 3. The centers are at $(14,92)$, $(17,76)$, and $(19,84)$. Draw a line through the point $(17,76)$ such that the sum of the areas of the parts of the three circles on one side of the line equals the sum of the areas of the parts of the three circles on the other side of the line. Find the absolute value of the slope of this line.
|
First, note that these three circles are mutually externally tangent equal circles. As shown in the figure, the centers are $\mathrm{O}_{1}(14,92), \mathrm{O}_{2}$ $(19,84), \mathrm{O}_{3}(17, 76)$. Since any line passing through $\mathrm{O}_{3}$ always divides $\odot \mathrm{O}_{3}$ into two equal areas, we only need to consider how to draw a line through a certain point on the plane so that for $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$, the areas on both sides of the line are equal. Clearly, the midpoint $\mathrm{M}(16.5,88)$ of $\mathrm{O}_{1} \mathrm{O}_{2}$ is the center of symmetry for the two equal circles $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$. Any line drawn through this point will meet the above requirement. Therefore, the line passing through $\mathrm{M}(16.5,88)$ and $\mathrm{O}_{3}(17,76)$ is the one we are looking for. Its slope $\mathrm{k}=\frac{88-76}{16.5-17}=-24$. From the figure, we can see that the required line is unique, and the absolute value of its slope is 24.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The function $f$ is defined on the set of integers, satisfying
$$
f(n)=\left\{\begin{array}{ll}
n-3, & \text { when } n \geqslant 1000, \\
f(f(n+5)), & \text { when } n<1000 .
\end{array}\right.
$$
Find $f(84)$.
|
Solve: First, calculate $f(n)$ for $n$ values close to 1000, and observe the pattern when $\mathrm{n}<1000$. Using the recursive definition of $\mathrm{f}$, we can compute:
$$
\begin{array}{l}
\mathrm{f}(999)=\mathrm{f}(\mathrm{f}(1004))=\mathrm{f}(1001)=998, \\
\mathrm{f}(998)=\mathrm{f}(\mathrm{f}(1003))=\mathrm{f}(1000)=997, \\
\mathrm{f}(997)=\mathrm{f}(\mathrm{f}(1002))=\mathrm{f}(999)=998, \\
\mathrm{f}(996)=\mathrm{f}(\mathrm{f}(1001))=\mathrm{f}(998)=997, \\
\mathrm{f}(995)=\mathrm{f}(\mathrm{f}(1000))=\mathrm{f}(997)=998 .
\end{array}
$$
We can conjecture that
$$
f(n)=\left\{\begin{array}{l}
997, \text { when } n \text { is an even number less than } 1000; \quad (2) \\
998, \text { when } n \text { is an odd number less than } 1000.
\end{array}\right.
$$
We prove (2) by backward induction. Since the definition of $f(n)$ involves $f(n+5)$, we first verify that (2) holds for $n=999, 993, 997, 996, 995$. This has already been verified in (1). Assume that (2) holds for $\mathrm{n}<\mathrm{m}<1000$ for some $m$ value, we need to prove that (2) also holds for $f(n)$. Indeed, $\mathrm{n}+5$ is some $\mathrm{m}$ value, so we have $f(n)=f(f(n+5))$
$$
=\left\{\begin{array}{l}
f(997)=998, \text { when } n+5 \text { is even; } \\
f(998)=997, \text { when } n+5 \text { is odd. }
\end{array}\right.
$$
Since $n+5$ is odd when $n$ is even, and $n+5$ is even when $n$ is odd. This proves that (2) holds. In particular, we should have $\mathrm{f}(84)=997$.
|
997
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The equation $\mathrm{z}^{8}+\mathrm{z}^{3}+1=0$ has a complex root, and the argument of this root on the complex plane is between $90^{\circ}$ and $180^{\circ}$. Find the degree of the argument.
|
Let $\omega=z^{3}$, the given equation transforms into $\omega^{2}+\omega+1=0$. Its solutions are $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{-1-\sqrt{3}i}{2}$, with arguments of $120^{\circ}$ and $240^{\circ}$, respectively. Since $z=\sqrt[3]{\omega}$, we can find the six arguments of $z$ to be $\frac{120^{\circ}}{3}, \frac{120^{\circ}+360^{\circ}}{3}, \frac{120^{\circ}+720^{\circ}}{3}$, $\frac{240^{\circ}}{3}, \frac{240^{\circ}+360^{\circ}}{3}, \frac{240^{\circ}+720^{\circ}}{3}$. Clearly, only the second one, which is $160^{\circ}$, lies between $90^{\circ}$ and $180^{\circ}$, and is the required value of $\theta$ in degrees.
|
160
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. In tetrahedron $\mathrm{ABCD}$, the length of edge $\mathrm{AB}$ is $3 \mathrm{~cm}$, the area of face $\mathrm{ABC}$ is $15 \mathrm{~cm}^{2}$, and the area of face $\mathrm{ABD}$ is $12 \mathrm{~cm}^{2}$. The angle between these two faces is $30^{\circ}$. Find the volume of the tetrahedron (in $\mathrm{cm}^{3}$).
|
Let $\mathrm{V}$ be the volume of tetrahedron $\mathrm{ABCD}$, and $\mathrm{h}$ be the height from $\mathrm{D}$ to the base $\mathrm{ABC}$. Then, $\mathrm{V}=\frac{1}{3} \mathrm{hS} \triangle \mathrm{ABC}$. To determine $V$, we only need to determine $h$. Draw $\mathrm{DK} \perp \mathrm{AB}$ at $\mathrm{K}$, and connect $\mathrm{KH}$. By the inverse of the three perpendiculars theorem, $\mathrm{HK} \perp \mathrm{AB}$, thus $\angle \mathrm{DKH}=30^{\circ}$.
From $\mathrm{S} \triangle A B D=\frac{1}{2} \mathrm{DK} \cdot \mathrm{AB}$, we get
$$
\begin{aligned}
\mathrm{DK} & =\frac{2 \mathrm{~S} \triangle A B D}{3}=8 . \\
\therefore \quad \mathrm{h} & =8 \sin 30^{\circ}=4, \\
\mathrm{~V} & =\frac{1}{3} 4 \times 15=20\left(\mathrm{~cm}^{3}\right) .
\end{aligned}
$$
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Xiao Zhi scored above 80 points in an American middle school math test. He told Xiao Jiang his score, and Xiao Jiang could correctly deduce how many questions Xiao Zhi answered. If Xiao Ma scored a bit lower but still above 80 points, Xiao Jiang would not be able to deduce it. How many points did Xiao Ma score?
(This test has 30 multiple-choice questions, and the scoring formula is $\mathrm{S}=30+4 \mathrm{c}-\mathrm{w}$, where $\mathrm{S}$ is the score, $\mathrm{c}$ is the number of correct answers, and $\mathrm{w}$ is the number of wrong answers, with no points for unanswered questions)
|
Solve for the minimum such that $S=30+40-w>80$. The problem requires finding the smallest such $S$, for which the corresponding $c$ is unique. First, note that when $c$ increases by 1, $w$ decreases by 4, and the value of $S$ remains unchanged, but it must satisfy $(c+1)+(w+4) \leqslant 30$, i.e., $c+w \leqslant 25$. When $c+w \leqslant 25$, for each $S$, the value of $c$ cannot be uniquely determined. Therefore, only when $c+w \geqslant 26$ (1) can $c$ be uniquely determined. Secondly, it should be that $w \leqslant 3$ (2), otherwise, if $w>3$, $w$ can be reduced by 4 and $c$ by 1, and the value of $S$ remains unchanged (since from $S>80$ we get $c \geqslant 13$, which is also possible). Thus, to make $S$ as small as possible, we should, within the range allowed by inequalities (1) and (2), minimize $c$ and maximize $w$. This leads to $w=3, c=23$. Therefore, $S=30+4 \times 23-3=119$, meaning the horse scored 119 points.
|
119
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. A gardener randomly arranges three maple trees, four oak trees, and five birch trees in a row. Each different arrangement is equally likely.
Let $\mathrm{m}$ represent the probability that no two birch trees are adjacent (expressed as a simplified fraction). Find $\mathrm{m}+\mathrm{n}$.
|
There are 12! different arrangements of 12 trees. Let $\mathrm{k}$ be the number of arrangements where no two birch trees are adjacent. The required probability is $\frac{\mathrm{k}}{12!}$. Now, we need to find $\mathrm{k}$.
Using $t$ $\mathrm{N}$ to represent non-birch trees (maple or oak), we have (1)N (2)N (3)N (4)N (5)N (6)N (7)N (8)N, with 8 gaps "..." where 5 birch trees can be placed. The arrangement of the 7 non-birch trees has 7! ways, and for each arrangement, there are $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ ways to arrange the birch trees. Therefore, $\mathrm{k}=(7!) 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$.
The required probability $\frac{\mathrm{k}}{12!}=(7!) \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{12!}$
$$
\begin{array}{l}
=\frac{7}{99} \text {, i.e., } \frac{\mathrm{m}}{\mathrm{n}}=\frac{7}{99} \\
\therefore \mathrm{m}+\mathrm{n}=106 .
\end{array}
$$
|
106
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Function f is defined on the real number domain, and satisfies the following conditions: For any real number $\mathrm{x}, \mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})$, and $f(7+x)=f(7-x)$. If $x=0$ is a root of $f(x)=0$, find the minimum number of roots of $f(x)=0$ in the interval -1000 $\leqslant x \leqslant 1000$.
|
Molybdenum First, find the numbers in the domain for which $f$ is periodic from the given equations.
We have
$$
\begin{array}{l}
f(x)=f(2+(x-2))=f(2-(x- \\
2))=f(4-x),(1) \\
f(4-x)=f(7-(x+3))=f(7+ \\
(x+3))=f(x+10) .(2)
\end{array}
$$
From (1) and (2), we get $f(x+10)=f(x)$
Substituting $x+10$ and $x-10$ for $x$ in (3), we get
$$
f(x+10)=f(x+20) \text { and } f(x-10)
$$
$=f(x)$, and continuing this process, we have
$f(x+10n)=f(x) \quad(n= \pm 1$, $\pm 2, \pm 3, \cdots)$.
Since $f(0)=0$, from (4) we get
$f( \pm 10)=f( \pm 20)=\ldots$
$=f( \pm 1000)=0$.
Thus, we have $f(x)=0$ at 201 roots in the interval $[-1000, 1000]$. Are there any other roots? Setting $x=0$ in (1), we get $f(4)=f(0)$ $=0$, and setting $x=4$ in (4), we get
$\mathrm{f}(4 \pm 10)=\mathrm{f}(4 \pm 20)=\cdots$
$=f(4 \pm 1000)=0$. This gives us another 200 roots of $f(x)=0$ in the interval $[-1000,1000]$, different from the previously found roots. These roots are $\mathrm{x}=-996,-986, \cdots$, $-6,4,14, \cdots, 994$. Since the piecewise linear function shown in the figure satisfies the given conditions and there are no other roots, the solution to the problem is 401.
|
401
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Find the value of $10 \operatorname{ctg}(\operatorname{arcctg} 3+\operatorname{arcctg} 7$ $+\operatorname{arcctg} 13+\operatorname{arcctg} 21)$.
|
$\begin{array}{l}\text { Let } f(n)=1+n+n^{2}, \text { it is easy to know that } \\ f(1)=3, f(2)=7, f(3) \\ =13, f(4)=21. \\ \text { It is easy to prove that } \\ \operatorname{arcctg}\left(1+n+n^{2}\right) \\ =\operatorname{arctg}(n+1)-\operatorname{arctg} n. \\ \text { Therefore, } \operatorname{arcctg} 3=\operatorname{arctg} 2-\operatorname{arctg} 1, \\ \operatorname{arcctg} 7=\operatorname{arctg} 3-\operatorname{arctg} 2, \\ \operatorname{arcctg} 13=\operatorname{arctg} 4-\operatorname{arctg} 3, \\ \text { arctg } 21=\operatorname{arctg} 5-\operatorname{arctg} 4. \\ \text { Adding the four equations, and let } \alpha=\operatorname{arcctg} 3+\operatorname{arcctg} 7 \\ +\operatorname{arcctg} 13+\operatorname{arcctg} 21, \text { we get: } \\ \alpha=\operatorname{arctg} 5-\operatorname{arctg} 1. \\ \therefore \operatorname{tg} \alpha=\frac{5-1}{1+5 \cdot 1}=\frac{4}{6}, \\ \operatorname{ctg} \alpha=\frac{6}{4}=\frac{3}{2}. \\ \therefore \quad 10 \operatorname{ctg} \alpha=10 \operatorname{ctg}(\operatorname{arcctg} 3 \\ +\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21) \\ =15. \\\end{array}$
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. The largest even number that cannot be expressed as the sum of two odd composite numbers is what?
Will the translated text be used for further discussion or do you need more information on this topic?
|
First, we prove: If $\mathrm{k} \geqslant 40, \mathrm{k}$ is an even number, then $\mathrm{k}$ can be expressed as the sum of two odd composite numbers. For this, if $\mathrm{n}$ is an odd number greater than 1, then $5 \mathrm{n}$ is an odd composite number with a units digit of 5. To express even numbers greater than 40 as the sum of two odd composite numbers, it is only necessary to find some smaller odd composite numbers with units digits of 5, 7, 9, 1, 3, and add them to odd composite numbers of the form $5 \mathrm{n}$. It is easy to see that 15, 27, 9, 21, 33 are such odd composite numbers.
In the following cases, odd numbers $\mathrm{n}>1$ can always be found such that even numbers not less than 40 can be expressed as the sum of two odd composite numbers.
If the units digit of $\mathrm{k}$ is 0 (i.e., 40, 50,
60, ...), then $k=15+5 n$;
If the units digit of $\mathrm{k}$ is 2 (i.e., 42, 52,
62, ...), then $k=27+5 n_{3}$
If the units digit of $\mathrm{k}$ is 4 (i.e., 44, 54,
64, ...), then $k=9+5 n$;
If the units digit of $\mathrm{k}$ is 6 (i.e., 46, 56, 66,
...), then $\mathrm{k}=21+5 \mathrm{n}_{3}$
If the units digit of $\mathrm{k}$ is 8 (i.e., 48, 58,
68, ...), then $\mathrm{k}=33+5 n$.
Therefore, the largest even number that cannot be written as the sum of two odd composite numbers might be 38. Upon inspection, 38 indeed meets this requirement.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 15. Find the value of } x^{2}+y^{2}+z^{2}+w^{3} \text { . If } \\ \frac{x^{2}}{\varepsilon^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}} \\ +\frac{w^{2}}{2^{2}-7^{2}}=1, \\ \frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}} \\ +\frac{w^{2}}{4^{2}-7^{2}}=1, \\ \frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\ \frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}} \\ +\frac{w^{2}}{8^{2}-7^{2}}=1 .\end{array}$
|
Solving for $\mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{w}$ to satisfy the given system of equations is equivalent to $t=4,16,36,64$ satisfying
$$
\frac{x^{2}}{t-1}+\frac{y^{2}}{t-9}+\frac{z^{2}}{t-25}=\frac{w^{2}}{t-49}=1. \quad (1)
$$
Multiplying both sides of (1) by the denominators, we find that for meaningful $t$ (i.e., $t \neq 1,9,25,49$), (1) is equivalent to the polynomial equation:
$$
\begin{array}{c}
(t-1)(t-9)(t-25)(t-49) \\
-x^{2}(t-9)(t-25)(t-49)-y^{2}(t-1) \\
-(t-25)(t-49)-z^{2}(t-1)(t-9)(t-
\end{array}
$$
49) $-w^{2}(t-1)(t-9)(t-25)=0. \quad (2)$
The left side is a fourth-degree polynomial in $\mathrm{t}$. Since $\mathrm{t}=4,16,36,64$ are four known roots of this equation, and a fourth-degree polynomial can have at most four roots, these are all the roots of equation (2). Therefore, (2) is equivalent to $(t-4)(t-16)(t-36)(t-64)=0. \quad (3)$
In (2) and (3), the coefficient of $t^{4}$ is 1, so the coefficients of the other powers of $t$ should also be equal. In particular, the coefficients of $t^{3}$ in both equations should be equal. By the relationship between roots and coefficients and comparing terms, we get
$$
\begin{array}{l}
1+9+25+49+x^{2}+y^{2}+z^{2}+w^{2} \\
=4+16+36+64.
\end{array}
$$
From this, we find
$$
\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+\mathrm{w}^{2}=36.
$$
(Supplied by Ma Xiwen and Zhou Chunlu, Beijing Mathematical Society)
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Given the quadratic function $\mathrm{y}=3 \mathrm{p} \mathrm{x}^{2}-6 \mathrm{px} +3 p-2$, for what value of $p$ will the distance between the two intersection points of this parabola with the $x$-axis be $\frac{\sqrt{6}}{3}$.
---
The translation maintains the original text's line breaks and formatting.
|
$$
\begin{array}{l}
\text { Solve } \frac{\sqrt{\triangle}}{|3 p|}=\frac{\sqrt{(-6 p)^{2}-4 \cdot 3 p(3 p-2)}}{|3 p|} \\
=\frac{\sqrt{24 p}}{|3 p|}=\frac{\sqrt{6}}{3} . \\
\end{array}
$$
Solving for $\mathrm{p}$, we get $\mathrm{p}=4$ .
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.
untranslated text:
上为自然数, 试求拋物 线族 $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$ 在 $x$ 轴上所截得的线段长度的总和。
translated text:
Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.
|
Solve for the x-coordinates $x_{1}, x_{2}$ of the intersection points of the parabola (family) with the $x$-axis, which are the roots of the equation $\left(n^{2}+n\right) x^{2}-(4 n+2) x+4=0$. Without loss of generality, assume $x_{2}>x_{1}$. Then $x_{2}-x_{1}=\frac{\sqrt{\triangle}}{|a|}=\frac{2}{n}-\frac{2}{n+1}$.
At this point, for $\mathrm{n} \in \mathrm{N}$, the total length of the segments intercepted on the $\mathrm{x}$-axis by the family of parabolas is
$$
\begin{aligned}
s & =\sum_{1}^{\infty}\left(\frac{2}{n}-\frac{2}{n+1}=\lim _{n \rightarrow \infty}\left[\left(\frac{2}{1}\right.\right.\right. \\
\left.-\frac{2}{2}\right) & \left.+\left(\frac{2}{2}-\frac{2}{3}\right)+\cdots+\left(\frac{2}{n}-\frac{2}{n+1}\right)\right] \\
& =\lim _{n \rightarrow \infty}\left(\frac{2}{1}-\frac{2}{n+1}\right)=2 .
\end{aligned}
$$
(Author's affiliation: Yongchun County Finance Bureau, Fujian Province)
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Use the digits $1$, $9$, $8$, and $5$ to fill in the four boxes below so that their product is maximized.
$$
(\square+\square) \times(\square+\square) \text {. }
$$
(Wenzhou Li Fangyue)
|
4. $(8+5) \times(1+9)=130$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
4. $(8+5) \times(1+9)=130$.
|
130
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A person has a probability of 0.6 of hitting the target with one shot. How many times at least should they shoot to make the probability of hitting the target at least once greater than 0.95?
|
(Given $\lg 2=$
0.3010 ) (Answer: $\mathrm{n}=4$ )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Given: $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{d}}{\mathrm{a}}$, find the value of $\frac{a+b+c+d}{b+a+c-d}$.
|
Solution 1: By the ratio theorem, we have:
$$
\begin{array}{c}
\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a} \\
=\frac{a+b+c+d}{a+b+c+d}=1, \text { so } a=b=c=d, \\
\therefore \frac{a+b+c+d}{a+b+c-d}=\frac{4 d}{2 d}=2 .
\end{array}
$$
Solution 2: Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, then
$a=bk, b=ck, c=dk, d=ak$,
$$
\therefore a=ak^{4}, k^{4}=1, k= \pm 1 \text{. }
$$
When $k=1$, we have $a=b=c=d$,
so $\frac{a+b+c+d}{a+b+c-d}=2$,
When $k=-1$, we have $a=-b=c=-d$,
$$
\therefore \frac{a+b+c+d}{a+b+c-d}=0 \text{. }
$$
Thus, we see that Solution 1 is not comprehensive, as it only considers the case $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Father's age is 48 years old, son's age is 20 years old. How many years later will the father's age be 3 times the son's age?
|
Solution: Let $\mathrm{x}$ years later, the father's age is 3 times the son's age.
We have $48+x=3(20+x)$. Solving for $\mathrm{x}$, we get $\mathrm{x}=-6$.
This means that 6 years ago, the father's age was 3 times the son's age. Conventionally, there is no such expression as “-6 years later.” However, according to the meaning of negative numbers, we can interpret “-6 years later” as 6 years ago. Therefore, this conclusion is correct, and the answer should be written as:
Answer: 6 years ago, the father's age was 3 times the son's age.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Select 4 people from 6 boys and 4 girls to participate in an extracurricular interest group. How many ways are there to select them?
(1) At least one boy and one girl participate;
(2) At most 3 boys participate.
|
Analysis: (1) Consider the selection of 4 students from 10 as the whole. The whole can be divided into two parts: one part is that the selected 4 students include at least one boy and one girl; the other part is that all 4 are either boys or girls, and these two parts are mutually exclusive. Therefore, the solution is: $\mathrm{C}_{10}^{4}-\mathrm{C}_{8}^{4}-\mathrm{C}_{4}^{4}=194$ (ways) 3
(2) The opposite of "at most 3 boys" is "all 4 selected are boys." Together, these two scenarios cover the whole. Thus, the solution is: $\mathrm{C}_{10}^{4}-\mathrm{C}_{0}^{4}=195$ (ways).
|
195
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Find the length of the common chord of the circles $x^{2}+y^{2}-10 x-10 y=0$ and $x^{2}+y^{2}+6 x+2 y-40=0$.
|
Subtracting the two equations yields $4 x+3 y-10=0$. (Equation of the common chord)
Completing the square for $x^{2}+y^{2}-10 x-10 y=0$, we get $(x-5)^{2}+(y-5)^{2}=50$.
Thus, the center of the circle is $\mathbf{C}(5,5)$.
The distance from the center of the circle to the common chord is
$$
\mathrm{d}=\frac{|20+15-10|}{5}=5 \text {. }
$$
By the Pythagorean theorem, the length of the chord is
$$
\mathrm{l}=2 \sqrt{50-25}=10 \text {. }
$$
$2^{n}$ cannot be divisible by 32. However, by assumption, $2^{n}(n \geqslant 6)$ can be divided by $2^{6}=64$, leading to a contradiction.
Assume $2^{n}(n \geqslant 6)$ meets the problem's requirements and ends with the digit 2. In this case, the last few digits of $2^{\mathbf{n}}$ could be 02, 22, 012, or it could be of the form
$$
2^{\mathbf{n}}=10 \underbrace{0 \cdots 0112}_{\mathrm{m} \text { digits }} .(\mathrm{m} \geqslant 0)
$$
Since 02 and 22 are not divisible by 4, and 012 is not divisible by 8, in these cases, $2^{n}$ cannot be divisible by 64. Therefore, the first two scenarios are impossible. For the last scenario, it is also impossible. Because $\left.1112=2^{3} \cdot 1306 \mathrm{~m}=0\right)$, $10112=2^{7} \cdot 79(\mathrm{~m}=1)$, it is clear that they are not powers of 2. When $\mathrm{m} \geqslant 2$, $10^{\mathrm{m}+3}$ is divisible by 32, but 112 is not divisible by 32. This also leads to a contradiction. In summary, only $n=5$ meets the requirements.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are 51 cities distributed in a square area with a side length of 1000 kilometers. It is planned to lay down 11000 kilometers of road network in the area. Can all the cities be connected?
|
(Continued from page 49) Therefore, the total length of all these roads does not exceed $1000 \times 6 + 100 \times 50 = 11000$ kilometers. This road network can connect all 51 cities in the region.
5. In the tetrahedron $\mathrm{ABCD}$, take points $\mathrm{A}^{\prime}$, $\mathrm{B}^{\prime}$, $\mathrm{C}^{\prime}$, and $\mathrm{D}^{\prime}$ on the faces $\mathrm{BCD}$, $\mathrm{ACD}$, $\mathrm{ABD}$, and $\mathrm{ABC}$ respectively, such that each pair of these points lies in the same plane with two vertices of the tetrahedron. Prove that the lines $\mathrm{AA}^{\prime}$, $\mathrm{BB}^{\prime}$, $\mathrm{CC}^{\prime}$, and $\mathrm{DD}^{\prime}$ intersect at one point.
Proof: We know that: if three lines intersect each other pairwise and are not coplanar, then they have only one common point. In fact, if these three lines intersect at three different points, then these three points determine a plane, and each line has two points in this plane, so these three lines will be coplanar, which contradicts the given condition of not being coplanar.
Now consider points $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ (Figure 7). They lie in the same plane with two vertices of the tetrahedron, which can only be $\mathrm{A}$ and $\mathrm{B}$. Otherwise, this plane would be one of the faces of the tetrahedron, and thus $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ would be on the same face, which is a contradiction. Suppose $\mathrm{A}$, $\mathrm{B}$, $\mathrm{A}^{\prime}$, and $\mathrm{B}^{\prime}$ lie in the plane $\mathrm{ABM}$, where $\mathrm{M}$ is on the edge $\mathrm{CD}$ (Figure 7). Since $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ lie on $\mathrm{BM}$ and $\mathrm{AM}$ respectively, $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$ intersect. Similarly, we can prove that $\mathrm{AA}^{\prime}$, $\mathrm{BB}^{\prime}$, $\mathrm{CC}^{\prime}$, and $\mathrm{DD}^{\prime}$ intersect pairwise, and any three of them are not coplanar. By the previous proof, these four lines have only one common point.
Solution: It can be done, and we provide a scheme for laying out the road network.
First, lay a road parallel to the boundary of the square and from one side to the other through one of the cities in the region, with a length of 1000 kilometers (Figure 6, cities are represented by small circles). Then, take two points 100 kilometers from each end of this road; between these two points, take a point every 200 kilometers, making a total of three points, plus the two initial points, making five points in total. Lay five roads perpendicular to the first road through these five points, with their ends also reaching the boundary. Now, from each city, lay a road along the shortest route to connect with one of these five roads. The length of each of these additional roads does not exceed 100 kilometers, and there are no more than 50 such roads. Thus,
|
11000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. What is the maximum value of the difference between a three-digit number and the sum of the cubes of its digits? What kind of three-digit number can achieve this maximum value? What is the smallest positive value of this difference?
|
Consider any three-digit number $\overline{\mathrm{bbc}}$, and let
$$
f(\overline{a b c})=\overline{a b c}-\left(a^{3}+b^{3}+c^{3}\right)
$$
Then $f(\overline{a b c})=100 a+10 b+c-\left(a^{3}+b^{3}\right.$
$$
\begin{aligned}
+ & \left.c^{3}\right) \\
= & \left(100 a-a^{3}\right)+\left(10 b-b^{3}\right) \\
& +\left(c-c^{3}\right) \\
= & f_{1}(a)+f_{2}(b)+f_{3}(c) .
\end{aligned}
$$
where $\hat{i}_{1}(a)=100 \mathrm{a}-\mathrm{a}^{3}, \mathrm{f}_{2}(\mathrm{~b})=10 \mathrm{~b}$
$$
-\hat{b}^{3}, f_{3}(c)=c - c^{3} \text{. When and only when } f_{1}(a) \text{, }
$$
$\left.f_{2}(b): f_{3}(c)\right)$ take their maximum values, $f(\overline{(a b c})$ takes the maximum value. Since $f_{1}(1)=99, f_{1}(2)=192$,
$$
\begin{array}{l}
f_{1}(3)=273, f_{1}(4)=336, f_{1}(5)=375, \\
f_{1}(6)=384, f_{1}(7)=357, f_{1}(8)=288, \\
f_{1}(9)=171 .
\end{array}
$$
Therefore, when $a=b$, $f_{1}(a)$ takes the maximum value of 384.
Similarly, it can be found that:
$b=2$ when $f_{2}(b)$ takes the maximum value of 12;
$c=0$ or 1 when $f_{3}(c)$ takes the maximum value of 0.
Thus, when $\overline{\mathrm{abc}}=620$ or 621, $\mathrm{f}(\overline{\mathrm{abc}})$ takes the maximum value of $384+12+0=396$.
Furthermore, $f_{1}(a)=99 a+\left(a-a^{3}\right)=99 a-(a-1) \cdot a \cdot(a+1)$;
$\mathrm{f}_{2}(\mathrm{~b})=9 \mathrm{~b}+\left(\mathrm{b}-\mathrm{b}^{3}\right)=9 \mathrm{~b}-(\mathrm{b}-1) \cdot b \cdot(b+1)$;
$f_{3}(c)=(c-1) \cdot c \cdot(c+1)$.
Since the product of two consecutive integers is divisible by 3, $f_{1}(a)$, $f_{2}(b)$, and $f_{3}(c)$ are all divisible by 3, thus $\mathrm{f}(\overline{\mathrm{abc}})$ is also divisible by 3. Therefore, the smallest positive value of $\overline{\mathrm{abc}}-\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)$ is no less than 3, and in fact, 3 is the minimum value.
It is recommended that readers verify: $\mathrm{f}(437)=\mathrm{f}(474)=f(856)=3$.
|
396
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $a+b+c=0, a b c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
|
\[\begin{array}{l}\text { Solve the original expression }=\frac{a^{2}+b^{2}+c^{2}}{3 a b c} \\ +\frac{2(a b+b c+c a)}{3 a b c}=\frac{(a+b+c)^{2}}{3 a b c}=0 .\end{array}\]
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Calculate $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$
|
Let the original expression $=x$, then $x^{3}=40+6 x$, its positive root $x=4$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Let the original expression $=x$, then $x^{3}=40+6 x$, its positive root $x=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Given $\sum_{j=1}^{n} a_{j} \cos \alpha_{j}=\sum_{j=1}^{n} a_{5} \cdot \cos \left(\alpha_{j}+1\right)=0$, find the value of $\sum_{j=1}^{n} a_{j} \cos \left(\alpha_{j}+\frac{\pi}{10}\right)$.
|
Consider the function $f(x)=\sum_{j=1}^{n} a_{j} \cos \left(\alpha_{j}+x\right)$. It is easy to see that $f(0)=f(1)=0$. Since $\cos \left(a_{j}+x\right) = \cos \alpha_{j} \cos x - \sin \alpha_{j} \sin x$, we have
$$
\begin{array}{l}
f(x)=\cos x \sum_{\mathrm{j}=1}^{n} a_{j} \cos \alpha_{j} + \sin x \sum_{\mathrm{j}=1}^{n} a_{j} \cos \left(\alpha_{j}+\frac{\pi}{2}\right) \\
=\cos x f(0) + \sin x f\left(\frac{\pi}{2}\right) \\
=\sin x f\left(\frac{\pi}{2}\right) .
\end{array}
$$
Let $x=1, f(1)=0, \sin 1 \neq 0$, hence
$$
\begin{array}{l}
f\left(\frac{\pi}{2}\right)=0 . \\
\therefore f(x)=0, \text{ thus } f\left(\frac{\pi}{10}\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a-b=2+\sqrt{3}, b-c=2-\sqrt{3}$, then the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is
|
15.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 (Zhou Jiansan). The line 1 passing through point $A_{2}(6,4)$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ at points $\mathrm{M}, \mathrm{N}$. Find the maximum and minimum values of |AM|・|AN|.
|
Let the inclination angle be $\alpha$, then its parametric equation is
$$
\left\{\begin{array}{l}
x=6+t \sin \alpha, \\
y=4+t \cos \alpha .
\end{array}\right. \text { (t is a parameter) }
$$
Substituting into the ellipse equation, and simplifying, we get
$$
\begin{array}{l}
\left(3 \sin ^{2} \alpha+1\right) t^{2}+4(3 \cos \alpha+8 \sin \alpha)+84 \\
=0 .
\end{array}
$$
Thus,
$|A M| \cdot|A N|=\left|t_{1} t_{2}\right|$
$$
=\frac{84}{1+3 \sin ^{2} \alpha} \text {, }
$$
$\therefore(|\mathrm{AM}| \cdot|\mathrm{AN}|)_{\max }=84$
$(\alpha=0),(|\mathrm{AM}| \cdot|\mathrm{AN}|)_{\mathrm{m} 1 \mathrm{n}}=21$
$$
\left(\alpha=\frac{\pi}{2}\right) \text {. }
$$
Analysis: The point $(6,4)$ is outside the ellipse, so when $l$ intersects the ellipse, $\alpha$ should satisfy $\Delta=[4(3 \cos \alpha$
$$
\begin{array}{l}
+8 \sin \alpha)]^{2}-4\left(3 \sin ^{2} \alpha+1\right) \cdot 84 \geqslant 0(0 \leqslant \alpha \\
<\pi) .
\end{array}
$$
$<\pi$ ).
That is, $\frac{2}{\sqrt{29}} \leqslant \sin \alpha \leqslant \frac{2}{\sqrt{5}}$.
When finding the maximum and minimum values, this point was overlooked.
|
84
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The volume of the cone formed by rotating the right-angled triangle around one of its legs is $800 \pi \mathrm{cm}^{3}$, and the volume of the cone formed by rotating around the other leg is $1920 \pi$ $\mathrm{cm}^{3}$. What is the length of the hypotenuse of this triangle? (in cm)
|
The volume of a cone with a base radius of $\mathrm{r}$ and height $\mathrm{h}$ is $\frac{\pi}{3} \mathrm{hr}$. Let $\mathrm{a}, \mathrm{b}$ represent the two legs of a right triangle, then we have
$$
\frac{\pi}{3} \mathrm{ba}^{2}=800 \pi \text { and } \frac{\pi}{3} \mathrm{ab}^{2}=1920 \pi \text {. }
$$
By comparing the two equations, we get $\frac{\mathrm{a}}{\mathrm{b}}=\frac{5}{12}$, which means $\mathrm{a}=\frac{5}{12} \mathrm{~b}$.
$$
\begin{array}{l}
\text { Therefore, } \mathrm{a}^{3}=\left(\frac{5}{12} \mathrm{~b}\right) \mathrm{a}^{2}=\frac{5}{12}\left(\mathrm{ba}^{2}\right) \\
=\frac{5}{12}(800)(3)=1000 .
\end{array}
$$
Thus, we find: $a=10, b=24$. Using the Pythagorean theorem, the hypotenuse of the triangle is $\sqrt{10^{2}+24^{2}}=26(\mathrm{~cm})$.
|
26
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 3. If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are positive integers, satisfying $\mathrm{c}=$ $(a+b i)^{3}-107 i$, find $c$. (where $i^{2}=-1$)
|
$$
\text { Solve } \begin{array}{l}
c=(a+b i)^{3}-107 i \text { can be written as } \\
c=a\left(a^{2}-3 b^{2}\right)+i\left[b \left(3 a^{2}\right.\right. \\
\left.\left.-b^{2}\right)-107\right] .
\end{array}
$$
Since $c$ is a real number, the imaginary part must be 0, i.e.,
$$
\text { b }\left(3 a^{2}-b^{2}\right)-107 \text {. }
$$
Since $a$ and $b$ are positive integers, and 107 is a prime number, from (2) we can get one of the following two cases: $b=107$ and $3 a^{2}-b^{2}$ $=1$ or $b=1$ and $3 a^{2}-b^{2}=107$. In the first case, $3 a^{2}=107^{2}+1$. This is impossible, because $107^{2}+1$ is not a multiple of three. Therefore, the first case should be excluded. In the second case, we can find $a=6$. From (1), we get $c=a\left(a^{2}-3 b^{2}\right)=6\left(6^{2}-3.1^{2}\right)=198$.
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 4. No figure, divide each side of a single square into $\mathrm{n}$ equal parts, then connect each vertex to the nearest division point of the opposite vertex, thus creating a smaller square inside the square. If the area of the smaller square (shaded in the figure) is exactly $\frac{1}{1985}$, find the value of $\mathrm{n}$.
|
Solve as shown in the figure. $\mathrm{DM}=\frac{\mathrm{n}-1}{\mathrm{n}}-$
$$
\begin{array}{l}
\therefore A M= \\
\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} \\
=\frac{1}{\cos \angle D E H=\cos \angle D A M=\frac{D A}{A M}} \\
\sqrt{1+\left(1-\frac{1}{n}\right)^{2}}
\end{array}
$$
The area of the small square in the middle $\mathrm{S}=\mathrm{GF}^{2}=\mathrm{EH}^{2}=(\mathrm{DE} \cdot \cos \angle \mathrm{DEH})^{2}$
$$
=\frac{1}{1+\left(1-\frac{1}{n}\right)^{2}} \cdot \frac{1}{n^{2}}=\frac{1}{1985},
$$
which gives $2 n^{2}-2 n+1=1985$. Solving this, we get $n_{1}=32$, $\mathrm{n}_{2}=-31$ (discard).
Therefore, $\mathrm{n}=32$.
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Select an integer sequence $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \cdots$, such that:
$$
a_{n}=a_{n-1}-a_{n-2} \text { (where } n \geqslant 3 \text { ) }
$$
If the sum of the first 1492 terms is 1985, and the sum of the first 1935 terms is 1492, what is the sum of the first 2001 terms?
|
Solve We calculate the first eight terms of this sequence, $a_{1}$,
$$
\begin{array}{l}
a_{2}, a_{3}=a_{2}-a_{1}, a_{4}=a_{3}-a_{2}=a_{2}-a_{1} \\
-a_{2}=-a_{1}, a_{5}=a_{4}-a_{3}=-a_{1}-a_{2}+a_{1} \\
=-a_{2}, a_{6}=a_{5}-a_{4}=a_{1}-a_{2}, a_{7}=a_{0}-a_{5} \\
=\left(a_{1}-a_{2}\right)-\left(-a_{2}\right)=a_{1}, a_{8}=a_{7}-a_{0} \\
=a_{1}-\left(a_{1}-a_{2}\right)=a_{2}, \cdots \text { and we find that every six }
\end{array}
$$
terms form a cycle. For any $n$, we have $a_{n+8}=a_{n}$. It is also easy to prove that in this sequence, the sum of every six consecutive terms is 0. Note that 1492 leaves a remainder of 4 when divided by 6, 1985 leaves a remainder of 5 when divided by 6, and 2001 leaves a remainder of 3 when divided by 6, so
$$
\begin{aligned}
S_{1035} & =a_{1081}+a_{1082}+\cdots+a_{1258} \\
=a_{1}+a_{2} & +\cdots+a_{5} \\
& =a_{1}+a_{2}+\left(a_{2} \cdots a_{1}\right)+\left(-a_{1}\right) \\
& +\left(-a_{1}\right)=a_{2}-a_{1}=1492 . \quad(1)
\end{aligned}
$$
$$
\begin{aligned}
S_{1402} & =a_{140}+a_{1400}+a_{1401}+a_{242} \\
& =a_{1}+a_{2}+a_{3}+a_{4} \\
& =a_{1}+a_{2}+\left(a_{2}-a_{1}\right)+\left(-a_{1}\right) \\
& =2 a_{2}-a_{1}=1985 .
\end{aligned}
$$
From (1) and (2), we solve to get $a_{1}=-999, a_{2}=493$.
$$
\begin{aligned}
S_{2001} & =a_{1090}+a_{2000}+a_{2001} \\
& =a_{1}+a_{2}+a_{3} \\
& =a_{1}+a_{2}+\left(a_{2}-a_{1}\right)=2 a_{2} \\
& =986 .
\end{aligned}
$$
|
986
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B6. As shown in the figure, $\triangle \mathrm{ABC}$ is divided into six smaller triangles by three lines passing through its three vertices and an interior point. The areas of four of the smaller triangles are marked in the figure. Find the area of $\triangle \mathrm{ABC}$.
untranslated part:
The areas of four of the smaller triangles are marked in the figure, 求 $\triangle \mathrm{ABC}$ 的面积。
Note: The last sentence is a mix of English and Chinese. Here is the fully translated version:
The areas of four of the smaller triangles are marked in the figure. Find the area of $\triangle \mathrm{ABC}$.
|
If the heights of two triangles are the same, then the ratio of their areas is equal to the ratio of their corresponding base lengths. Therefore, from the figure, we get
$$
\frac{40}{30}=\frac{40+y+84}{30+35+x}, \quad \frac{35}{x}=\frac{35+30+40}{x+84+y},
$$
$\frac{84}{y}=\frac{84+x+35}{y+40+30}$. Solving the first two equations yields $\mathbf{x}=70, \mathrm{y}=56$, which also satisfies the third equation. The area of $\triangle \mathrm{ABC}$ is $30+35+70+84+56+40=315$.
|
315
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 7. Let $a, b, c, d$ be positive integers, satisfying $a^{8}=b^{4}, c^{3}=d^{2}$, and $c-a=19$. Find $d-b$.
The text above is translated into English, preserving the original text's line breaks and format.
|
Since the prime factorization of a positive integer is unique, and because 4 and 5 are coprime, 2 and 3 are coprime, it follows that there exist two positive integers \( m \) and \( n \) such that
$$
a=m^{4}, b=m^{6}, c=n^{2}, d=n^{3}.
$$
Thus, \( 19=c-a=n^{2}-m^{4} \):
$$
=\left(n-m^{2}\right)\left(n+m^{2}\right) \text{. }
$$
Since 19 is a prime number, and \( n-m^{2} < n+m^{2} \), it must be that
$$
n-m^{2}=1, n+m^{2}=19 \text{. }
$$
Therefore, \( m=3, n=10, d=1000, b=243 \).
$$
d-b=1000-243=757
$$
|
757
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 9. In a certain circle, parallel chords of lengths $2, 3, 4$ correspond to central angles $\alpha, \beta, \alpha+\beta$. Here $\alpha+\beta<\pi$. If $\cos \alpha$ (which is a positive rational number) is expressed as a reduced (simplest) fraction, what is the sum of the numerator and the denominator?
|
Solution: Because in the same circle, equal chords subtend equal central angles, the parallelism of these chords is irrelevant. We can select points A, B, C such that AB = 2, BC = 3. Since \( A \widehat{C} = \alpha + \beta \), it follows that \( AC = 4 \). From \( \angle AC3 = \frac{\alpha}{2} \), using the cosine rule, we get
Thus, \( \cos \alpha = 2 \cos^2 \frac{\alpha}{2} - 1 = 2 \cdot \frac{49}{54} - 1 = \frac{17}{32} \). This is an irreducible fraction, so the sum of the numerator and the denominator is \( 17 + 32 = 49 \).
|
49
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
How many of the first 1000 positive integers can be expressed as:
$$
\lfloor 2 x\rfloor+\lfloor 4 x\rfloor+\lfloor 6 x\rfloor+\lfloor 8 x\rfloor \text { of }
$$
form?
where $\mathbf{x}$ is some real number, and $\mathrm{Lz}\rfloor$ denotes the greatest integer not exceeding $z$.
|
$$
\begin{array}{l}
\text { Let } f(x)=\lfloor 2 x\rfloor+\lfloor 4 x\rfloor \\
+\lfloor\lfloor 6 x\rfloor+\lfloor 8 x\rfloor .
\end{array}
$$
It can be seen that if $n$ is an integer, then from (1) we have:
$$
f(x+n)=f(x)+20 n .
$$
If some integer $k$ can be expressed as $f\left(x_{0}\right)$, where $x_{0}$ is some real number, then by (2), for $n=1,2,3, \cdots$, $k+20 n=f\left(x_{0}\right)+20 n=f\left(x_{0}+n\right)$, i.e., $k+20 n$ can also be expressed in the form (i). Therefore, we focus on the values that $f(x)$ can produce when $x$ varies in $(0,1]$. Furthermore, if $x$ increases, $f(x)$ will only change when $2 x, 4 x, 6 x, 8 x$ reach some integer value, and it will always change to a larger integer. In $(0,1]$, this change occurs precisely when $x$ has the form $m / n$, where $1 \leqslant m \leqslant n$, and $n=2,4,6$ or 8. There are 12 such fractions, arranged from smallest to largest:
$$
\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{2}{3},
$$
$\frac{3}{4}, \frac{5}{6}, \frac{7}{8}$ and 1.
Thus, only 12 of the first 20 positive integers can be expressed in the required form, because $1000=50 \cdot 20$, by (2) we know that the first 1000 positive integers are divided into 50 segments, each consisting of 20 consecutive positive integers, of which only 12 can be expressed in the required form. Therefore, the total number of integers that meet the requirement is $50 \times 12=600$.
|
600
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 12. Let the four vertices of a regular tetrahedron be A, B, C, and D, with each edge length being 1 meter. There is a small insect that starts from point A and moves forward according to the following rules: at each vertex, it chooses one of the three edges connected to that vertex with equal probability and crawls all the way to the end of that edge. Suppose the probability that it is exactly at vertex A after crawling 7 meters is \( P = \frac{n}{729} \). Find the value of \( n \).
|
For $\mathrm{n}=0,1,2, \cdots$, let $\mathrm{a}_{\mathrm{n}}$ denote the probability that the bug returns to point $\mathrm{A}$ after walking $\mathrm{n}$ meters. Then we have $\mathrm{a}_{\mathrm{n}+1} = \frac{1}{3} (1 - a_{n})$ (1), because the necessary and sufficient condition for the bug to reach point $\mathrm{A}$ after walking $\mathrm{n}+1$ meters is: (a) after walking $n$ meters, it is not at point $\mathrm{A}$ (probability is $1 - a_{n}$), and (b) it moves towards $\mathrm{A}$ from a point other than $\mathrm{A}$ (probability is $\frac{1}{3}$). Now $\mathrm{a}_{0}=1$ (i.e., the bug starts from point $\mathrm{A}$), and from equation (1), we can find $a_{1}=a, a_{2}=\frac{1}{3}, a_{3}=\frac{2}{9}, a_{4}=\frac{7}{27}, a_{5}=\frac{20}{81}, a_{6}=\frac{6!}{243}$, and BP = $a_{7} = \frac{182}{729}$. Therefore, $\mathrm{n}=182$.
|
182
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 13. The general term of the sequence $101, 104, 109, 116, \cdots$ is $a_{n}=100+n^{2}$. Here $n=1,2,3$, $\cdots$. For each $n$, let $d_{n}$ denote the greatest common divisor of $a_{n}$ and $a_{n+1}$. Find the maximum value of $\mathrm{d}_{\mathrm{n}}$, where $n$ takes all positive integers.
|
We can prove more generally: If $a$ is a positive integer, and $d_{n}$ is the greatest common divisor of $a+n^{2}$ and $a+(n+1)^{2}$, then when $n=2a$, $d_{n}$ reaches its maximum value of $4a+1$. From this, the answer is $4(100)+1 = 401$.
To prove the above proposition, first note that if $d_{n}$ divides both $a+(n+1)^{2}$ and $a+n^{2}$, then it also divides their difference, i.e., $d_{n} \mid (2n+1)$.
Since $2(a+n^{2}) = n(2n+1) + (2a-n)$, from (1) we know $d_{n} \mid (2a-n)$, (2) and from (1) and (2) we have
$$
d_{n} \mid [(2n+1) + 2(2a-n)]
$$
i.e., $d_{n} \mid (4a+1)$.
Therefore, $1 \leqslant d_{n} \leqslant 4a+1$, so $d_{n}$ cannot exceed $4a+1$. Moreover, $d_{n}$ can indeed reach this value, for example, when $n=2a$, we have
$$
\begin{array}{l}
a+n^{2} = a + (2a)^{2} = a(4a+1) \\
a+(n+1)^{2} = a + (2a+1)^{2} \\
= (a+1) \quad (4a+1).
\end{array}
$$
Thus, $d_{n}$ reaches its maximum value of $4a+1$. In particular, when $a=100$, the maximum value of $d_{n}$ is 401.
|
401
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In a certain competition, each player plays exactly one game with every other player. The winner of each game gets 1 point, the loser gets 0 points. If it's a tie, each gets $\frac{1}{2}$ point. After the competition, it is found that each player's score is exactly half from the games played against the ten lowest-scoring players. (The ten lowest-scoring players got half of their points from games played against each other) Find the total number of players in the competition.
|
Let's assume there are $n$ participants in the competition, and we will find two quantities related to $n$: one is the total points of the scores, and the other is the total points of the "losers" (let's call this sum). We need to note that if $k$ participants compete, the total number of matches should be $\frac{k(k-1)}{2}$, and according to the scoring rules, their total score is $\frac{k(k-1)}{2}$ points.
From this, we know that $n$ participants have a total score of $\frac{n(n-1)}{2}$ points. The 10 "losers" have scored a total of $10 \cdot 9 / 2 = 45$ points through their matches with each other. Since this is half of their total score, they have a total of 90 points. The remaining $n-10$ participants have scored a total of $(n-10)(n-11) / 2$ points through their matches with each other. This is also half of their total score, so they have a total of $(n-10)(n-11)$ points. Therefore, we get the equation
$$
\mathbf{n}(\mathbf{n}-1) / 2 = 90 + (\mathbf{n}-10)(\mathbf{n}-11),
$$
which is equivalent to $n^2 - 41n + 400 = 0$. That is,
$$
(n-16) \quad (n-25) = 0.
$$
$n_1 = 25, n_2 = 16$. Upon verification, $n_2 = 16$ should be discarded. Because if there are only 16 participants, there would only be 6 "winners," and their total score would be 30, averaging 5 points per person, which is clearly less than the average score of the "losers," $90 / 10 = 9$. Therefore, $n = 25$, and there were 25 participants in the competition.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 15. Three $12 \mathrm{~cm} \times 12 \mathrm{~cm}$ squares are each divided into two pieces, $A$ and $B$, by a line connecting the midpoints of two adjacent sides, as shown in the first figure. The six pieces are then attached to the outside of a regular hexagon, as shown in the second figure. The pieces are then folded to form a polyhedron. Find the volume of this polyhedron (in $\mathrm{cm}^3$).
|
Fold into a polyhedron as shown in the first figure, where the vertices marked with $P, Q, R, S$ have three right angles. From this, we can recognize that this polyhedron is part of a cube. We can imagine obtaining a cube with an edge length of $12 \mathrm{~cm}$, as shown in the second figure, making a section (through the midpoints of the relevant six edges) which divides the cube into two polyhedra like the one in the first figure. The volume of the cube is $12^{3}=1728\left(\mathrm{~cm}^{3}\right)$. Therefore, the volume of the polyhedron in question is $\frac{1728}{2}=864 \mathrm{~cm}^{3}$.
|
864
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. In space, there are 10 points, 4 of which lie on the same plane, and no other set of 4 points are coplanar; find the number of circular cones (not necessarily right circular cones) with one of the points as the vertex and a circle passing through 3 other points as the base.
|
Analysis: From the problem, we can infer that among the 10 points in space, no 3 points are collinear. Otherwise, each of the remaining 7 points would be coplanar with the 3 collinear points, which contradicts the problem statement. Now, let's denote the plane where 4 points are coplanar as \( a \). We need to consider two cases.
First, without considering the restriction "4 points are coplanar," the 10 points in space, where no 4 points are coplanar, can determine \( C_{10}^{1} C_{0}^{3} \) cones.
( I ) The 4 points on plane \( \alpha \) are concyclic.
In this case, under the restriction, the number of cones that cannot be determined is \( C_{4}^{1} C_{3}^{3} + (C_{4}^{3} - 1) C_{6}^{1} \). According to the complement principle, the number of cones that can be determined is \( C_{10}^{1} C_{0}^{3} - [C_{4}^{1} C_{3}^{3} + (C_{4}^{3} - 1) C_{6}^{1}] \).
(II) The 4 points on plane \( \alpha \) are not concyclic.
In this case, we only need to exclude the cases where the vertex and the 3 points determining the base circle are all on plane \( \alpha \), which is \( C_{4}^{1} C_{3}^{3} \). According to the complement principle, the number of cones that can be determined is \( C_{10}^{1} C_{8} - C_{4}^{1} C_{3}^{3} \).
Solution: Let the plane where 4 points are coplanar in the 10 points in space be \( \alpha \). If no 4 points are coplanar, the 10 points can determine \( C_{10}^{1} C_{9}^{3} \) cones.
( I ) The 4 points on plane \( \alpha \) are concyclic. In this case, under the restriction, the number of cones that cannot be determined is \( C_{4}^{1} C_{3}^{3} + (C_{4}^{3} - 1) C_{8}^{1} \). Therefore, the number of cones that can be determined is
\[
\begin{array}{l}
C_{10}^{1} C_{9}^{3} - [C_{4}^{1} C_{3}^{3} + (C_{4}^{3} - 1) C_{8}^{1}] \\
= 840 - [4 - 18] = 818.
\end{array}
\]
(II) The 4 points on plane \( \alpha \) are not concyclic. In this case, if the vertex and the 3 points determining the base circle are all on \( \alpha \), the number of cones that cannot be determined is \( C_{4}^{1} C_{3}^{3} \). Therefore, the number of cones that can be determined is \( C_{10}^{1} C_{8} - C_{4}^{1} C_{3}^{3} \)
\[
= 840 - 4 = 836.
\]
|
836
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. A parade team whose number of people is a multiple of 5 and no less than 1000, if lined up in rows of 4, is short of 3 people; if lined up in rows of 3, is short of 2 people; if lined up in rows of 2, is short of 1 person. Find the minimum number of people in this parade team.
|
Analysis: Given that the total number of people in the parade is a multiple of 5, let the total number of people be $5n$. "When arranged in rows of 4, there are 3 people short," which can be understood as "1 person more" from the opposite perspective. Similarly, when arranged in rows of 3 or 2, it can also be understood as "1 person more." Clearly, the problem can be transformed into a congruence problem, where $5n$ leaves a remainder of 1 when divided by 4, 3, and 2, i.e., $5n-1$ is a multiple of 12. Given that the total number of people is no less than 1000, the solution to the problem can be found.
Let the total number of people in the parade be $5n (n \in N)$, i.e., $5n-1$ is a common multiple of 4, 3, and 2. Thus, we can let $5n-1=12m (m \in N)$, which gives $n=\frac{12m+1}{5}$.
To minimize the number of people in the parade, the $m$ in equation (A) should be the smallest positive integer and a multiple of 5, which is 2. Therefore, we can let $m=5p+2 (p \in N)$. This gives:
$$
n=\frac{1}{5}[12(5p+2)+1]=12p+5,
$$
i.e., $5n \geqslant 60p+25$.
$$
\therefore 60p+25 \geqslant 1000, \quad p \geqslant 16 \frac{1}{4}.
$$
Taking $p=17$, substituting into equation (B) gives $5n=60 \times 17+25=1045$.
Therefore, the minimum total number of people in the parade is 1045.
|
1045
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Calculate: $\operatorname{tg} 5^{\circ}+\operatorname{ctg} 5^{\circ}-2 \sec 80^{\circ}$. (79 National College Entrance Examination Supplementary Question)
|
$$
\begin{aligned}
\text { Original expression } & =\frac{2}{\sin \left(2 \times 5^{\circ}\right)}-\frac{2}{\cos 80^{\circ}} \\
& =\frac{2}{\sin 10^{\circ}}-\frac{2}{\sin 10^{\circ}}=0 .
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the value of $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}$ $-\cos \frac{4 \pi}{7}+\cos \frac{5 \pi}{7}-\cos \frac{6 \pi}{7}$.
|
Solve: In (1), let $\alpha=0, n=6$, and transform to get
$$
1+\sum_{k=1}^{0} \cos k \beta=\frac{\cos 3 \beta \sin \frac{7}{2} \beta}{\sin \frac{\beta}{2}} \text {. }
$$
Let $\beta=\pi-\frac{\pi}{7}, \cos k \beta=\cos k \left( \pi -\frac{\pi}{7}\right)=(-1)^{k} \cos \frac{k \pi}{7}$. Therefore,
$$
\begin{array}{l}
\sum_{k=1}^{\infty}(-1)^{k-1} \cos k \frac{\pi}{7} \\
=1-\frac{\cos 3\left(\pi-\frac{\pi}{7}\right) \sin \frac{7}{2}\left(\pi-\frac{\pi}{7}\right)}{\sin \frac{1}{2}\left(\pi-\frac{\pi}{7}\right)} \\
=1-0=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In the figure below, $ABCD$ is a trapezoid, $AD // BC$, $\frac{BC}{AD}=5$, and the area of $\triangle OAD$ is $S$.
(1) Express the area of $\triangle OBC$ in terms of $S$;
(2) How many times the area of trapezoid $ABCD$ is the area of $\triangle OAD$?
|
7. (1) $25 S$; (2) 36 times.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Find the extremum of $y=\frac{\sqrt{3} x+1}{\sqrt{x^{2}+1}}+2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Solve using trigonometric substitution. Let $x=\operatorname{tg} \theta$, $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$,
we get $y=\frac{\sqrt{3} \operatorname{tg} \theta+1}{\sec \theta}+2=\sqrt{3} \sin \theta+\cos \theta+2=2 \sin \left(\theta+\frac{\pi}{6}\right)+2$, $-\frac{\pi}{2}+\frac{\pi}{6}<\theta<\frac{\pi}{2}+\frac{\pi}{6}$. When $\theta=\frac{\pi}{3}$, $y$ can achieve a maximum value of 4, since $\theta$ takes values in the open interval $\left(-\frac{\pi}{3}, \frac{2\pi}{3}\right)$, there are no other extremum values.
If the range of $\theta$ is not restricted, maintaining a one-to-one correspondence between $\operatorname{tg} \theta$ and $x$, it might lead to the incorrect conclusion that the minimum value of $y$ is 0 from $y=2 \sin \left(\theta+\frac{\pi}{6}\right)+2$.
In comprehensive trigonometric problems, these constraints often bring some inconvenience, which can increase the difficulty of solving problems. However, these constraints also confine the problem to a specific range, which can sometimes be beneficial to our problem-solving.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { Example 4. Find } S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\text {. } \\ +\frac{1}{\sqrt{1002001}} \text { the integer part of } S. \\\end{array}$
|
Solve the inequality:
$$
\begin{array}{l}
2(\sqrt{k+1}-\sqrt{k})<\frac{1}{\sqrt{k}} \\
<2(\sqrt{k}-\sqrt{k-1}) \\
\end{array}
$$
By setting \( k=1,2, \cdots, 1002001 \), we get:
$$
\begin{array}{l}
2(\sqrt{2}-\sqrt{1})<\frac{1}{\sqrt{1}} \leqslant 1, \\
2(\sqrt{3}-\sqrt{2})<\frac{1}{\sqrt{2}} \\
<2(\sqrt{2}-\sqrt{1}), \\
\cdots \cdots \\
2(\sqrt{1002002}-\sqrt{1002001}) \\
<\frac{1}{\sqrt{1002001}}<2(\sqrt{1002001} \\
-\sqrt{1002000}) .
\end{array}
$$
By summing up, we know \( 2000 < S < 2001 \). Therefore, the integer part of \( S \) is 2000.
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 4. $m, n$ are positive integers, and in the polynomial $f(x)=(1+x)^{m}+(1+x)^{n}$, the coefficient of $x$ is 19.
1) Try to find the minimum value of the coefficient of $x^{2}$ in $f(x)$;
2) For the $m, n$ that make the coefficient of $x^{2}$ in $f(x)$ the smallest, find the term containing $x^{7}$ at this time.
|
Solve for the coefficient of $x$ in $f(x)=(1+x)^{m}+(1+x)^{n}$:
$$
\begin{array}{r}
C_{m}^{1}+C_{n}^{1}=m+n=19 . \\
\therefore \quad m=19-n .(1 \leqslant n \leqslant 18)
\end{array}
$$
1) The coefficient of $x^{2}$ in $f(x)$ is:
$$
\begin{aligned}
C_{m}^{2} & +C_{n}^{2}=\frac{1}{2}[m(m-1)+n(n \\
& -1)] \\
& =\frac{1}{2}[(19-n)(18-n) \\
& +n(n-1)] \\
& =\left(n-\frac{19}{2}\right)^{2}+\frac{323}{4} .
\end{aligned}
$$
$\because 1 \leqslant n \leqslant 18$, and $n \in N$,
$\therefore$ when $n=9$ or 10, the minimum value of $C_{m}^{2}+C_{n}^{2}$ is:
$$
C_{0}^{2}+C_{10}^{2}=81 \text {. }
$$
2) When the coefficient of $x^{2}$ in $f(x)$ is minimized,
$$
\left\{\begin{array} { l }
{ m = 1 0 , } \\
{ n = 9 }
\end{array} \text { or } \left\{\begin{array}{l}
m=9, \\
n=10 .
\end{array}\right.\right.
$$
Thus, the coefficient of $x^{7}$ is $C_{0}^{7}+C_{10}^{7}=C_{0}^{6}+C_{10}^{3}=156$.
|
156
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{aligned}
\text { 23. Let } N=69^{5}+5 \cdot 69^{4}+10 \cdot 69^{3}+10 \cdot 69^{2} \\
+5 \cdot 69+1
\end{aligned}
$$
How many positive integers are divisors of $N$?
(A) 3; (B) 5 ; (C) 69; () 125; (D) 216 .
|
$\frac{23}{\mathrm{E}}$
23. By the binomial theorem, $N=(69+1)^{5}$ $=(2,5,7)^{5}=2^{5}, 5^{5}, 7^{5},$ the number of factors of $N$ is $(5+1)(5+1)(5+1)=6^{3}=216$.
|
216
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the sum of all roots of the following equation:
$$
\sqrt[4]{x}=\frac{12}{7-\sqrt[4]{x}}
$$
|
1. Let $y=\sqrt[4]{x}$, then the original equation can be transformed into $y^{2}-7 y+12=0$, whose roots are 3 and 4, so the solution is $s^{4}+4^{4}$ $=337$.
|
337
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6} \\
\quad-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7}) \\
\cdot(-\sqrt{5}+\sqrt{6}+\sqrt{7}) .
\end{array}
$$
|
$$
\begin{array}{l}
=x^{2}-y^{2} \text {, we have: } \\
(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6} \\
-\sqrt{7})=(\sqrt{5}+\sqrt{6})^{2}-(\sqrt{7})^{2} \\
=11+2 \sqrt{30}-7=4+2 \sqrt{30}, \\
(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6} \\
+\sqrt{7})=(\sqrt{7})^{2}-(\sqrt{5}-\sqrt{6})^{2} \\
=7-(5-2 \sqrt{30})=-4 + 2 \sqrt{30}. \\
\end{array}
$$
$$
\begin{array}{l}
\text { and }(4+2 \sqrt{30})(-4 + 2 \sqrt{30}) \\
=(2 \sqrt{30})^{2}-4^{2}=120-16=104. \\
\end{array}
$$
|
104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 3. If } \operatorname{tg} x+\operatorname{tg} y=25, \text { and } \\ \quad \operatorname{ctg} x+\operatorname{ctg} y=30, \\ \text { find } \quad \operatorname{tg}(x+y) .\end{array}$
|
$\begin{array}{l}\text { 3. From ctg } x+\operatorname{ctg} y=30 \text { we get } \frac{1}{\operatorname{tg} x}+\frac{1}{\operatorname{tg} y} \\ =30 \Rightarrow \operatorname{tg} x+\operatorname{tg} y=30 \operatorname{tg} x \cdot \operatorname{tg} y . \\ \text { That is } \operatorname{tg} x \cdot \operatorname{tg} y=\frac{\operatorname{tg} x+\operatorname{tg} y}{30} \\ =\frac{25}{30}=\frac{5}{6} \text {. } \\ \text { Therefore, }(x+y)=\frac{\operatorname{tg} x+\operatorname{tg} y}{1-\operatorname{tg} x \operatorname{tg} y} \\ =\frac{25}{1-\frac{5}{6}}=15 \text {. } \\\end{array}$
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the following system of equations
$$
\cdot\left\{\begin{array}{c}
2 x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=6, \\
x_{1}+2 x_{2}+x_{3}+x_{4}+x_{5}=12, \\
x_{1}+x_{2}+2 x_{3}+x_{4}+x_{5}=24, \\
x_{1}+x_{2}+x_{3}+2 x_{4}+x_{5}=48, \\
x_{1}+x_{2}+x_{3}+x_{4}+2 x_{5}=96
\end{array}\right.
$$
Find: $3 x_{4}+2 x_{5}$.
|
4. Adding up the 5 equations, and then dividing both sides by 6, we get:
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=31 .
$$
By subtracting (1) from the 4th and 5th equations respectively, we obtain
$$
x_{4}=17, x_{5}=65 \text {, }
$$
Therefore, $3 x_{4}+2 x_{5}=51+130=181$.
|
181
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5 . Find the largest positive integer $n$, such that $n^{3}+100$ can be divided by $n+10$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
5 . Find the largest positive integer $n$, such that $n^{3}+100$ can be divided by $n+10$.
|
5 . By the division algorithm we get $\left.n^{3}+10\right)=(n+10)$
$$
\text { - }\left(n^{2}-10 n+100\right)-900 .
$$
If $n+10$ divides $n^{3}+100$, it must also divide 900. By the maximality of $n$, we have $n+10=903$, so $n=890$.
|
890
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The page numbers of a book are from 1 to $n$. When these page numbers were added up, one page number was mistakenly added one extra time. As a result, the incorrect sum obtained was 1986. What is the page number that was added one extra time?
|
6. Let $k$ be the page number that was added one extra time, then $0<k<n+1$, so $1+2+\cdots+n+k$ is between $1+2+\cdots+n$ and $1+2+\cdots+n+(n+1)$, thus
$$
\frac{n(n+1)}{2}<1988<\frac{(n+1)(n+2)}{2},
$$
which means $n(n+1)<3972<(n+1)(n+2)$. By trial (since $n$ is slightly greater than 60), we get
$$
\begin{array}{l}
n=62, \text{ so } k=1986-\frac{(62)(63)}{2} \\
=1986-1953=33 .
\end{array}
$$
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For each proper divisor of $1,000,000$, take the logarithm to the base 10, add these logarithmic values together, to get the sum $S$, find the integer closest to $S$.
(Proper divisors of a natural number $n$ are the divisors of $n$ that are neither 1 nor $n$.)
|
8. Since $1000000=2^{\circ} \cdot 5^{\circ}$ has $(6+1)$ - $(6+1)=49$ factors (since its factors must be of the form $2^{i} \cdot 5$, where $0 \leqslant i, j \leqslant 6$), excluding 1000, the remaining 48 factors form 24 pairs, each with a product of $1000000=10^{6}$, so the product of all factors of 1000000 is $1000 \cdot\left(10^{8}\right)^{24}=10^{147}$, so the product of all proper factors of 1000000 is $10^{141}$. Since the number $S$ is the logarithm of the product of the proper factors, hence $S=141$. The integer closest to $S$ is of course 141.
|
141
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in the figure, in $\triangle A B C$, $A B$ $=425, B C=450, C A=510, P$ is inside the triangle, $D E 、 F G 、 H I$ all pass through $P$, have the same length $d$, and are parallel to $A B 、 B C 、 C A$ respectively. Find $d$.
|
9. As shown in the figure,
$$
\begin{array}{l}
E H=B C-(B E+H C)=B C-(F P \\
+P G)=450-d .
\end{array}
$$
Similarly, we get $G D=510-d$,
From the similarity of $\triangle D P G$ and $\triangle A B C$, we have
$$
\begin{aligned}
D P= & \frac{A B}{C A} \cdot G D=\frac{425}{510}(510-d) \\
& =425-\frac{5}{6} d
\end{aligned}
$$
Again, from the similarity of $\triangle P E H$ and $\triangle A B C$, we get:
$$
\begin{aligned}
P E & =\frac{A B}{B C} \cdot E H=\frac{425}{450}(450-d) \\
& =425-\frac{17}{18} d .
\end{aligned}
$$
Since $d=D P+P E$, adding (1) and (2) gives,
$$
d=850-\frac{16}{9} d
$$
Thus, $d=306$.
|
306
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In a game, the "magician" asks a person to think of a three-digit number $(a b c)$ (where $a, b, c$ are the digits of the number in base 10), and asks this person to select 5 numbers $(a c b), (b a c), (b c a), (c a b)$, and $(c b a)$, and to find the sum $N$ of these 5 numbers, and tell the sum $N$ to the magician. Then the magician can reveal the number $(a b c)$ that the person thought of.
Now, let $N=3194$, please be the magician and find the number $(a b c)$.
|
10. Add (abc) to $N$, since $a$, $b$, and $c$ each appear exactly twice in every position, we get
$$
N+(a b c)=222(a+b+c) .
$$
Therefore, we should look for a multiple of 222, $222k$, which should be greater than $N$ (since $(abc) \neq 0$) and less than $N+1000$ (since $(abc)$ is a three-digit number), and make equation (1) hold, i.e., the sum of the digits of $222k - N$ equals $k$. If all these conditions are met, then $222k - N$ is the solution to the problem.
$$
\begin{array}{l}
\text { Since } 14 \cdot 222=3194 \\
+ \text { 1000, }
\end{array}
$$
So $k$ can only be chosen from 15, 16, 17, 18. Among these 4 numbers, only 16 satisfies the above conditions, i.e., $16 \cdot 222 - 3194 = 358$. And $3+5+8=16$, so $(abc) = 358$.
|
358
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let the polynomial $1-x+x^{2}-x^{8}+\cdots+x^{16}$ $-x^{17}$ be written as $a_{0}+a_{1} y+a_{2} y^{2}+a_{3} y^{3}+\cdots$ $+a_{10} y^{16}+a_{17} y^{17}$,
where $y=x+1$, and all $a_{\mathrm{i}}$ are constants, find $a_{2}$.
|
$$
\begin{array}{l}
\text { 11. Replace } x \text { with } y-1 \text {, the polynomial becomes } \\
\text { 1- }(y-1)+(y-1)^{2}-(y-1)^{3} \\
+\cdots+(y-1)^{10}-(y-1)^{17}, \\
\quad 1+(1-y)+(1-y)^{2}+(1-y)^{3} \\
+\cdots+(1-y)^{10}+(1-y)^{17} .
\end{array}
$$
Notice that the coefficient of $y^{2}$ in $(1-y)^{\mathrm{k}}$ is $C_{\mathrm{k}}^{2}$ $(2 \leqslant k \leqslant 17)$, so the required number is
$$
\begin{array}{l}
C_{2}^{2}+C_{3}^{2}+C_{4}^{2}+\cdots+C_{17}^{2} \\
=C_{18}^{3}=816 .
\end{array}
$$
|
816
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. The sum of a set of numbers is the sum of all its elements. Let $S$ be a set composed of positive integers not exceeding 15, such that the sums of any two disjoint subsets of $S$ are not equal, and among all sets with the above property, the sum of $S$ is the largest. Find the set $S$ and its sum.
|
12. First, $S$ can contain at most 5 elements, because otherwise, $S$ would have at least 6 elements, and the number of subsets with elements not exceeding 4 would be at least $C_{8}^{1} + C_{8}^{2} + C_{8}^{3} + C_{8}^{4} = 56$. The sum of each such subset is at most 54 (since $15 + 14 + 13 + 12 = 54$). By the pigeonhole principle, at least two subsets would have the same sum. If these two subsets are disjoint, it contradicts the condition that "the sums of any two disjoint subsets of $S$ are not equal"; if they intersect, removing their common elements would still lead to a contradiction. Therefore, $S$ can contain at most 5 elements.
Next, construct a 5-element subset with the largest possible elements. After including $15, 14, 13$, $S$ cannot include 12 (since $15 + 12 = 14 + 13$). $S$ can include 11. If $S$ includes $15, 14, 13, 11$, it cannot include 10 or 9 (since $10 + 15 = 14 + 11$ and $9 + 15 = 13 + 11$). Therefore, the fifth element of $S$ can only be 8. It is not difficult to verify that $S = \{8, 11, 13, 14, 15\}$ satisfies the problem's conditions, so we conjecture that $8 + 11 + 13 + 14 + 15 = 61$ is the solution.
Finally, prove that the maximum value is indeed 61. If a set $S$ satisfying the conditions can be constructed with a sum greater than 61, then $S$ must include $15, 14, 13$ (because even if it does not include the smallest of these, 13, to exceed a sum of 61, it would need to include $10, 11, 12$, which would result in $15 + 11 = 14 + 12$, a contradiction). Thus, $S$ cannot include 12. If $S$ includes 11, it would lead to the set with a sum of 61, a contradiction. If $S$ does not include 11, even if $S$ includes 10 and 9, it can only make the sum reach $15 + 14 + 13 + 10 + 9 = 61$, a contradiction. Therefore, 61 is indeed the maximum value.
Thus, the sum of the set $S$ is 61.
|
61
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $\triangle A B C$ be a right triangle on the $x y$ plane, with $C$ as the right angle, and the chord $A B$ has a length of 60. The medians drawn from $A$ and $B$ lie on the lines $y=x+3$ and $y=2 x+4$, respectively. Find the area of $\triangle A B C$.
|
15. In this problem, $r=\frac{60}{2}=30$.
From figure (1), it is easy to calculate $\operatorname{tg} \theta=\frac{1}{3}, \sin \theta=\frac{1}{\sqrt{10}}$. As shown in figure (2), taking the midpoints of the three sides as $M, N, O$, with $D$ as the incenter, $p, q, s, t$ are defined as shown in the figure, $\angle \theta=\angle A D M$.
$\operatorname{tg} \theta=\frac{1}{3}, \sin \theta=\frac{1}{\sqrt{10}}$ have been calculated as above.
Since $S_{\triangle A B C}=\delta S_{\triangle A D M}$
Therefore, $S_{\triangle A B C}=6 s t \sin \theta=\frac{6}{10} s t$.
Also, since $\angle C=90^{\circ}$,
(2)
By the Pythagorean theorem, we have
$$
\begin{array}{l}
4 p^{2}+4 Q^{2}=A B^{2}=3600, \\
p^{2}+4 q^{2}=9 s^{2}, \\
4 p^{2}+q^{2}=9 t^{2} .
\end{array}
$$
From (1) and (2), we get
$$
\begin{array}{l}
\left(S_{\triangle A B C}\right)^{2}=\frac{18}{5} s^{2} t^{2} \\
\quad=\frac{18}{5} \cdot \frac{p^{2}+4 q^{2}}{9} \cdot \frac{4 q^{2}+q^{2}}{9} \\
\quad=\frac{2}{45}\left(\left(2 p^{2}+2 q^{2}\right)^{2}+9 p^{2} q^{2}\right) \\
\quad=\frac{2}{45}\left(\left(\frac{3600}{2}\right)^{2}+\frac{9}{4}\left(S_{\triangle A B C}\right)^{2}\right) \\
\quad=144000+\frac{1}{10}\left(S_{\triangle A B C}^{2} .\right. \\
\therefore \frac{9}{10}\left(S_{\triangle A B C}=2=144000 .\right.
\end{array}
$$
$\therefore$ The area of $\triangle A B C$ is 400.
(Edited by Tang Shouwen and Zhi Chunlu, Beijing Mathematical Society)
|
400
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Find $\lim _{x \rightarrow 0} \frac{\ln \left(\sin ^{2} x+e^{x}\right)-x}{\ln \left(x^{2}+e^{2 x}\right)-2 x}$.
|
$$
\begin{array}{l}
\text { Solve the original expression }=\lim _{x \rightarrow 0} \ln \left(\sin ^{2} x+e^{x}\right)-\ln e^{x}\left(x^{2}+e^{2 x}\right)-\ln e^{2 x} \\
=\lim _{x \rightarrow 0} \frac{\ln \left(\frac{\sin ^{2} x+e^{x}}{e^{x}}\right)}{\ln \left(\frac{x^{2}+e^{2 x}}{e^{2 x}}\right)} \\
=\lim _{x \rightarrow 0} \frac{\ln \left(1+\frac{\sin ^{2} x}{e^{x}}\right)}{\ln \left(1+\frac{x^{2}}{e^{2 x}}\right)} \\
=\lim _{x \rightarrow 0} \frac{\frac{\sin ^{2} x}{e^{x}}}{\frac{x^{2}}{e^{2 x}}}=\lim _{x \rightarrow 0} e^{x} \frac{x^{2}}{x^{2}}=1 . \\
\end{array}
$$
(Because as $x \rightarrow 0$, $\ln \left(1+\frac{\sin ^{2} x}{e^{x}}\right)$
$$
\sim \frac{\sin ^{2} x}{e^{x}}, \ln \left(1+\frac{x^{2}}{e^{2 x}}\right) \sim \frac{x^{2}}{e^{2 x}})
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Find $\lim _{y \rightarrow 0} \frac{e^{y}+\sin y-1}{\ln (1+y)}$.
|
Since $\lim _{y \rightarrow 0} \frac{e^{y}-1}{y}=1, e^{y}-1 \sim y$
$$
\lim _{\substack{(y \rightarrow 0 \\(y \rightarrow 0 \text { when) }}} \frac{e^{y}-1+\sin y}{2 y}=1, e^{y}-1+\sin y \sim 2 y
$$
Therefore, the original expression $=\lim _{y \rightarrow 0} \frac{2 y}{y}=2$.
It is worth noting that when using equivalent infinitesimal substitution, only local equivalent infinitesimal substitution is performed, without overall verification, which may lead to incorrect results.
For example, when finding $\lim _{x \rightarrow 0} \frac{\operatorname{tg} x-\sin x}{x^{3}}$, only noticing $\sin x \sim x$ and $\operatorname{tg} x \sim x (x \rightarrow 0$ when $)$, and making the following substitution:
Original expression $=\lim _{x \rightarrow 0} \frac{x-x}{x}=0$, this result is incorrect.
In fact, the original expression $=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{\left(\frac{1}{\cos x}-1\right)}{x^{2}}$
$$
=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} \cdot \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}=\frac{1}{2} \text {. }
$$
In finding limits, using equivalent infinitesimals for substitution is often effective.
Let $\lim \alpha(x)=\infty, \lim \beta(x)=\infty$ $\left(x \rightarrow x_{0}\right.$ or $x \rightarrow \infty$ when $)$, if $\lim \frac{\beta(x)}{\alpha(x)}=1$, then $\beta(x)$ and $\alpha(x)$ are said to be equivalent infinitesimals, denoted as $a(x) \sim \beta(x) \quad\left(x \rightarrow x_{0}\right.$ or $x \rightarrow \infty$ when $)$.
Similarly, $\lim \frac{\beta(x)}{\alpha(x)}=\lim \frac{\beta_{1}(x)}{\alpha_{1}(x)}$ $\left(\alpha(x) \sim a_{1}(x), \beta(x) \sim \beta_{1}(x)\right.$, when $x \rightarrow 0$ or $x \rightarrow \infty$).
For example, $\lim _{x \rightarrow+\infty} \frac{e^{x}-e^{-2 x}}{3^{x}+\frac{2}{-x}}$.
Since $e^{x} \sim e^{x}-e^{-2 x}, \quad 3^{x} \sim 3^{x}+2^{-x}$ $(x \rightarrow+\infty$ when $)$, the original expression $=\lim _{x \rightarrow+\infty} \frac{e^{x}}{3^{x}}$ $=\lim _{x \rightarrow+\infty}\left(\frac{e}{3}\right)^{x}=0$.
(Author's unit: Department of Mathematics, Tianjin Normal University)
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Determine the non-negative roots of the equation $\sin x=\frac{1}{100} x$
|
The roots of the equation are the x-coordinates of the intersection points of the graphs of $y=\sin x$ and $y=\frac{1}{100} x$. Therefore, we only need to find the number of intersection points of the two functions in the first quadrant. This can be solved by observing the graphs.
Solution: The curve $y=\sin x$ and the line $y=\frac{1}{100} x$
intersect in the first quadrant within the rectangle bounded by the lines $x=0, x=100$, $y=0, y=1$.
Since $31 \pi < 100 < 32 \pi$, the interval $[0,100]$ can be divided into 32 subintervals: $[0, \pi], [\pi, 2 \pi], \cdots, [30 \pi, 31 \pi], [31 \pi, 100]$. We can construct 32 rectangles using these intervals as their bases.
Furthermore, the intersection points of the line $y=\frac{1}{100} x$ and the curve $y=\sin x$ are only within the 1st, 3rd, ..., 31st rectangles, and each rectangle has only two intersection points. Therefore, the equation has 32 non-negative roots.
From the above, the observations made from the graphs serve as a "guide" for finding the solution. However, only by being imaginative based on these observations can one open the floodgates of thinking.
|
32
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. In an arithmetic sequence with the first term $a_{1}>0$, the sum of the first 9 terms is equal to the sum of the first 17 terms. How many terms have the maximum sum?
|
This is a common problem, usually solved by completing the square or solving a system of inequalities, which can be cumbersome. Let's first analyze the implicit conditions in the problem: (1) It is a decreasing sequence (2) $a_{10}+a_{11}+a_{12}+a_{13}+a_{14}+a_{1 \mathrm{~s}}+a_{16}+a_{17}=0$. By the properties of an arithmetic sequence, we know that $a_{10}+a_{17}=a_{11}+a_{16}=a_{12}+a_{15}=a_{1 \mathrm{~s}}+a_{14}$. Therefore, the equation in (2) can be simplified to $4\left(a_{13}+a_{14}\right)=0$. Observing the structure of this equation, and given that $a_{1}>0$ and $\left\{a_{n}\right\}$ is decreasing, we can deduce that $a_{13}>0$ and $a_{14}<0$. Therefore, the sum of the first 13 terms is the maximum.
In the process of solving this problem, we successively identified and utilized three special conditions that emerged during the solution, simplifying the steps.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. If $A=20^{\circ}, B=25^{\circ}$, find $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.
untranslated part:
- $(1+\operatorname{tg} B)$ 的值.
Note: The untranslated part is kept as is because "tg" is a notation for the tangent function, and the phrase "的值" means "the value of" in this context. However, for a complete translation, it should read:
- the value of $(1+\operatorname{tg} B)$.
Thus, the fully translated text is:
Example 5. If $A=20^{\circ}, B=25^{\circ}$, find the value of $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.
|
$$
\text{Solve } \begin{aligned}
& (1+\tan A) \cdot(1+\tan B) \\
= & \left(1+\tan 20^{\circ}\right) \cdot\left(1+\tan 25^{\circ}\right) \\
= & 1+\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ} .
\end{aligned}
$$
Using the tangent addition formula in reverse $1=\tan 45^{\circ}$
$$
\begin{array}{l}
= \frac{\tan 20^{\circ}+\tan 25^{\circ}}{1-\tan 20^{\circ} \cdot \tan 25^{\circ}} \text{, we get } \\
1-\tan 20^{\circ} \cdot \tan 25^{\circ}=\tan 20^{\circ}+\tan 25^{\circ} .
\end{array}
$$
$$
\begin{array}{l}
\text{Therefore, } \tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ}=1 \text{. } \\
\text{Thus, }(1+\tan A) \cdot(1+\tan B)=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. What is the last digit of $11^{6}+14^{6}+16^{8}$?
|
1. The last digit of the given sum can be determined by adding the last digits of each power: the last digit of any power of a number ending in 1 and 6 remains 1 and 6, the last digit of an even power of a number ending in 4 is 6. Therefore, the last digit of the given sum is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. What is the last digit of $1! +2!+3!+\cdots+8!+9!$?
|
7. The last digit of this sum is 3. Because from 5! onwards, the last digit of each term is 0, and $1!+$ $2!+3!+4!=1+2+6+24=33$
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Which of the following expressions are meaningful? Among the meaningful radicals, which are arithmetic roots? Express the radicals that are not arithmetic roots using arithmetic roots;
(1) $\sqrt[3]{\frac{1}{8}}$,
(2) $\sqrt{-9}$
(3) $\sqrt[3]{-8}$
|
Solution: (1) is an arithmetic root;
(2) is meaningless;
(3) is meaningful, but not an arithmetic root,
$$
\sqrt[3]{-8}=-\sqrt[3]{8}=-2 .
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. The Louka Problem: A shipping company has a ship leaving Harvard for New York and a ship leaving New York for Harvard every noon. The journey takes seven days and seven nights in both directions. How many ships leaving New York will the ship that departs from Harvard at noon today encounter on its journey?
|
Solving the family of time-position broken lines shown in Figure 3, it is easy to see that, under the condition of continuous daily ship departures, starting from the 6th day, each ship departing thereafter will encounter 13 ships coming from the opposite direction.
In the case of continuous daily ship departures, starting from the 6th day, each ship departing thereafter will encounter 13 ships coming from the opposite direction.
|
13
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13. Given that the characteristic of $\lg x^{\frac{3}{2}}$ is the solution to the inequality
$$
a^{2}-8 a+15<0
$$
and the mantissa is the sum of the solutions to the equation
$$
\frac{1}{4}\left|y+\frac{1}{2}\right|+\frac{6}{121} \sqrt{z-1}=0
$$
find the value.
|
Solution: Let $\lg x^{\frac{3}{2}}=p+q$
(where $p$ is the characteristic, and $q$ is the mantissa). Solving the inequality $a^{2}-8 a+15<0$, we get
$$
3<a<5 \text {. }
$$
Since the characteristic of a logarithm of a number must be an integer,
$$
\therefore \quad p=a=4 \text {. }
$$
From equation (2), we get
$$
y=-\frac{1}{2}, \quad \text { and } \quad z=1 .
$$
According to the problem, we have
$$
q=y+z=-\frac{1}{2}+1=\frac{1}{2} .
$$
Substituting $p$ and $q$ into the formula, we get
$$
\begin{aligned}
& \lg x^{\frac{4}{2}}=1+\frac{1}{2}=\frac{9}{2}, \\
\therefore \quad & \lg x=\frac{9}{2} \div \frac{3}{2}=3 .
\end{aligned}
$$
Therefore, $x=1000$.
|
1000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14. Let $x$ be a natural number from 1 to 10, the remainder of $x$ divided by 3 is $f(x)$, $y$ is a one-digit natural number, the remainder of $y$ divided by 4 is $g(y)$, then what is the maximum value of $x+2y$ when $f(x)+2g(y)=0$?
|
Solution: From the problem, we can obtain
$$
\begin{array}{l}
g(y)=\left\{\begin{array}{l}
0, y=4,8 ; \\
1, y=1,5,9 ; \\
2, y=2,6 ; \\
2, y=3,7 .
\end{array}\right. \\
\end{array}
$$
$\therefore f(x), g(y)$ are both non-negative.
To make $f(x)+2 g(y)=0$,
it is necessary that $f(x)=0$ and $g(y)=0$.
At this time, $x=3,6,3$,
$$
y=4,8 .
$$
The maximum value of $x$ is $9$, and the maximum value of $y$ is $8$.
$\therefore$ The maximum value of $x+2 y$ is $9+2 \times 8=25$.
|
25
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $f(x)=\frac{4^{x}}{4^{x}+2}$, then the sum
$$
f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)
$$
$+f\left(\frac{3}{1001}\right)+\cdots+f\left(\frac{1000}{1001}\right)$ is
|
The original sum is $f\left(\frac{1}{1986}\right)+f\left(\frac{2}{1986}\right)$
$+\cdots+f\left(\frac{1985}{1986}\right)$. Considering that the number 1986 is non-essential, and given that using year numbers in problems is common, to avoid unnecessary confusion, this number has been modified. This problem is tricky, and brute force calculation is not feasible. Inspired by Gauss's method of summing, one should naturally first calculate $f\left(\frac{1}{1001}\right)+f\left(\frac{1000}{1001}\right)$. Generally, if $x+y=1$, then calculate $f(x)+f(y)=\frac{4^{x}}{4^{x}+2}$
$+\frac{4^{y}}{4^{y}+2}=1$. The sum has 1000 terms, so the answer should be 500. In fact, we can find the sum in the general case:
$$
\begin{array}{l}
f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right) \\
=\frac{n-1}{2} .
\end{array}
$$
To standardize the answer, we chose a not too small number $n=1001$, hinting that candidates must "extract" the solution.
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Is there a natural number $n$ such that the sum of the digits of $n^{2}$ equals 1983? Equals 1984?
|
Solution: The remainder of $n$ divided by 9 may be $0,1,2,3$, $4,5,6,7,8$, so the remainder of $n^{2}$ divided by 9 may be $0,1,4,7$. Therefore, the remainder of the sum of the digits of $n^{2}$ divided by 9 can only be $0,1,4,7$.
Since 1983 divided by 9 leaves a remainder of 3, the product of the digits of $n^{2}$ cannot be 1983.
The sum of the digits of $a^{2}$ can be 1984. In fact, when
$$
n=10^{22}-3
$$
then,
$$
\begin{aligned}
n^{2} & =10^{44}-6 \times 10^{21}+9 \\
& 219 \text{ nines} \\
& =99 \cdots 940 \cdots 09,
\end{aligned}
$$
the sum of its digits is exactly 1984.
|
1984
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $a$ and $b$ are real numbers, and
$$
\sqrt{2 a-1}+|b+1|=0 \text {. }
$$
Find the value of $a^{-2}+b^{-1987}$.
|
Solution: $\because \sqrt{2 a-1} \geqslant 0,|b+1| \geqslant 0$, according to property 5, we have
$$
\begin{array}{l}
\sqrt{2 a-1}=0 \text { and }|b+1|=0, \\
\therefore a=\frac{1}{2} \text { and } b=-1 \text {. Therefore, } \\
a^{-2}+b^{-1987}=\left(\frac{1}{2}\right)^{-2}+(-1)^{-1987} \\
=4-1=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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