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Five. (Full marks: 15 points) From the 91 natural numbers $1,2,3, \cdots, 90,91$, select $k$ numbers such that there must be two natural numbers $p, q$ satisfying $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$. Determine the minimum value of the natural number $k$, and explain your reasoning.
|
Five, divide the 91 natural numbers from 1 to 91 into nine groups, such that the ratio of any two natural numbers in each group is no less than $\frac{2}{3}$ and no more than $\frac{3}{2}$, and the division is as follows:
$$
\begin{array}{l}
A_{1}=\{1\}, A_{2}=\{2,3\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{7,8,9,10\}, \\
A_{5}=\{11,12,13,14,15,16\}, \\
A_{6}=\{17,18,19,20,21,22,23,24,25\}, \\
A_{7}=\{26,27,28,29, \cdots, 39\}, \\
A_{8}=\{40,41,42,43, \cdots, 58,59,60\}, \\
A_{9}=\{61,62,63, \cdots, 89,90,91\} .
\end{array}
$$
If any nine numbers are taken from these 91 numbers, for example, one number from each of $A_{1}, A_{2}, \cdots, A_{9}$, such as $1,3,6,10,16,25,39,60,91$, the ratio of any two of these nine numbers is less than $\frac{2}{3}$ or greater than $\frac{3}{2}$. Therefore, $k > 9$. So $k \leqslant 10$. When $k=10$, if any 10 numbers are taken from the 91 natural numbers from 1 to 91, according to the pigeonhole principle, there must be two numbers in the same $A_{i}$. For example, these two numbers are $p$ and $q$, and $p<q$, then $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$ holds.
Therefore, the minimum value of $k$ is 10.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 If $x, y, z$ are real numbers, and
$$
\begin{aligned}
(y-z)^{2} & +(z-x)^{2}+(x-y)^{2} \\
= & (y+z-2 x)^{2}+(z+x-2 y)^{2} \\
& +(x+y-2 z)^{2},
\end{aligned}
$$
find the value of $M=\frac{(y z+1)(z x+1)(x y+1)}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}$.
|
Solution: The condition can be simplified to
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .
$$
Then $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$,
which implies $x=y=z$.
Therefore, $M=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, in a chess tournament, there are an odd number of participants, and each participant plays one game against every other participant. The scoring system is as follows: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. It is known that two of the participants together scored 8 points, and the average score of the others is an integer. How many participants are there in the tournament?
Will the text above be translated into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Five, suppose there are $(n+2)$ players in total, except for 2 people who get 8 points, $n$ people on average get $k$ points each ($k$ is an integer).
$\because$ Each person plays one match with everyone else, and there are $(n+2)$ people,
$\therefore$ A total of $\frac{(n+1)(n+2)}{2}$ matches are played.
Since each match results in 1 point, we can set up the equation
$$
\frac{1}{2}(n+1)(n+2)=8+n k \text {, }
$$
Rearranging gives $n^{2}+(3-2 k) n-14=0$.
$\because$ The number of people is a positive integer, and $3-2 k$ is an integer,
$\therefore$ The number of people can only be $1, 2, 7, 14$.
Also, $\because$ the number of people is odd, $\therefore n=1$ or 7,
But when $n=1$, $k<0$ which is a contradiction, $\therefore n=7$.
At this time, $n+2=9$, i.e., a total of 9 people participate in the competition.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 As shown, in Rt $\triangle ABC$, the hypotenuse $AB=5, CD \perp AB$. It is known that $BC, AC$ are the two roots of the quadratic equation $x^{2}-(2 m-1) x+4(m-1)=0$. Then the value of $m$ is $\qquad$.
|
Solution: Let $A C=b$, $B C=a$. By Vieta's formulas, we get $a+b=2 m-$
$$
\begin{array}{l}
1, a b=4(m-1) . \\
\begin{aligned}
\therefore A B^{2} & =a^{2}+b^{2}=(a+b)^{2}-2 a b \\
& =(2 m-1)^{2}-2 \times 4(m-1)=5^{2},
\end{aligned}
\end{array}
$$
i.e., $m^{2}-3 m-4=0$.
$$
\therefore m=4 \text { or } m=-1 \text {. }
$$
Since $a$ and $b$ are the lengths of the sides of a triangle, we have $a>0, b>0$. Therefore, $a+b=2 m-1>0$, i.e., $m>\frac{1}{2}$.
$$
\therefore m=4 \text {. }
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in the figure, in the right trapezoid $A B C D$, the base $A B=13$, $C D=8, A D \perp A B$, and $A D=$ 12. Then the distance from $A$ to the side $B C$ is ( ).
(A) 12
(B) 13
(C) 10
(D) $\frac{12 \times 21}{13}$
|
6. (A).
Draw $C C^{\vee} \perp A B$, then
$$
B C=13-8=5 \text {. }
$$
Let the distance from $A$ to $B C$ be $h$,
Connect AC.
$$
\begin{aligned}
& \because S_{\triangle A C D}+S_{\triangle A B C} \\
& =S_{\text {UEABCD }}, \\
\therefore & \frac{1}{2} \times 8 \times 12+\frac{1}{2} \times h \times 13=\frac{1}{2}(8+13) \times 12
\end{aligned}
$$
Therefore, $h=12$.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $|x| \leqslant 2$, the sum of the maximum and minimum values of the function $y=x-|1+x|$ is $\qquad$ .
|
2. -4 .
$$
\begin{array}{l}
\because|x| \leqslant 2, \text { i.e., }-2 \leqslant x \leqslant 2, \\
\therefore y=x-|1+x| \\
\quad=\left\{\begin{array}{ll}
2 x+1, & \text { when }-2 \leqslant x<-1, \\
-1, & \text { when }-1 \leqslant x \leqslant 2.
\end{array}\right.
\end{array}
$$
As shown in the figure.
Indeed, when $x=-2$, $y$
has the minimum value -3; when $-1 \leqslant$ $x \leqslant 2$, $y$ has the maximum value -1.
Therefore, $-3+(-1)$ $=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $M=\cos 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \sin 35^{\circ}, N=$ $\sin 5^{\circ} \cos 15^{\circ} \cos 25^{\circ} \cos 35^{\circ}$. Then $\frac{M}{N}=$ $\qquad$ .
|
$=1.1$.
$$
\begin{aligned}
\frac{M}{N} & =\frac{\frac{1}{2}\left(\sin 20^{\circ}+\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}-\cos 60^{\circ}\right)}{\frac{1}{2}\left(\sin 20^{\circ}-\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}+\cos 60^{\circ}\right)} \\
& =\frac{\left(\sin 20^{\circ}+\sin 10^{\circ}\right)\left(\cos 10^{\circ}-\frac{1}{2}\right)}{\left(\sin 20^{\circ}-\sin 10^{\circ}\right)\left(\cos 10^{\circ}+\frac{1}{2}\right)} \\
= & \frac{\sin 20^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 20^{\circ}+\sin 10^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}}{\sin 20^{\circ} \cos 10^{\circ}+\frac{1}{2} \sin 20^{\circ}-\sin 10^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}} \\
= & \frac{\sin 20^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}}{\sin 20^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}}=1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $a+\lg a=10, b+10^{b}=10$. Then $a+b$
|
4. 10.
Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$
Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10-a}>0 \text{. }
$$
Contradiction.
Therefore, $10-a-b=0$, which means $a+b=10$.
Thought 2: The geometric meaning of this problem is: $a$ is the x-coordinate of the intersection point $A$ of $y=\lg x$ and $y=10-x$, and $b$ is the x-coordinate of the intersection point $B$ of $y=10^{x}$ and $y=10-x$. By the symmetry of the graphs of two inverse functions, $A$ and $B$ are symmetric about the line $y=x$. Solving the system of equations
$$
\left\{\begin{array}{l}
y=x \\
y=10-x
\end{array}\right.
$$
we get $C(5,5)$. Since $C$ is the midpoint of $A$ and $B$, we have $\frac{a+b}{2}=5$, which means $a+b=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $a, b$ be unequal real numbers, and $a^{2}+2 a-5$ $=0, b^{2}+2 b-5=0$. Then $a^{2} b+a b^{2}=$ $\qquad$ .
|
（Answer:10）
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most bought the book that was purchased by the fewest people?
|
Solution: Let the number of people who bought the most popular book be $x$. Among the 10 people, person A bought three books. Since the other 9 people each have at least one book in common with A, and $9 \div 3=3$, it follows that among A's three books, the most popular one must have been bought by at least 4 people, so $x \geqslant 4$.
If $x=4$, then each of A's three books was bought by 4 people. Similarly, each book bought by the other 9 people was also bought by 4 people. Therefore, the total number of books bought by the 10 people should be a multiple of 4, i.e., $4 \mid 30$, which is a contradiction. Thus, $x \geqslant 5$.
When $x=5$, let $A_{i}$ represent different books. The scenario where the most popular book is bought by exactly 5 people is as follows:
$$
\begin{array}{l}
\left\{A_{1} A_{2} A_{3}\right\} 、\left\{A_{1} A_{2} A_{6}\right\} 、\left\{A_{2} A_{3} A_{4}\right\} 、 \\
\left\{A_{1} A_{4} A_{6}\right\} 、\left\{A_{1} A_{4} A_{5}\right\} 、\left\{A_{2} A_{4} A_{5}\right\} 、 \\
\left\{A_{1} A_{3} A_{5}\right\} 、\left\{A_{2} A_{5} A_{6}\right\} 、\left\{A_{3} A_{5} A_{6}\right\} 、 \\
\left\{A_{3} A_{4} A_{6}\right\} .
\end{array}
$$
In conclusion, the most popular book was bought by at least 5 people.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that $x$, $y$, $z$ are positive integers, and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(y+z)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Explanation: This is a typical example of constructing a geometric figure to solve a problem. As shown in Figure 1, construct $\triangle ABC$, where the lengths of the three sides are
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then its area is
$$
\begin{aligned}
\triangle & =\sqrt{p(p-a)(p-b)(p-c)} \\
& =\sqrt{(x+y+z) x y z}=1 .
\end{aligned}
$$
On the other hand,
$$
(x+y)(y+z)=a b=\frac{2 \triangle}{\sin C} \geqslant 2 .
$$
Therefore, when and only when $\angle C=90^{\circ}(\sin C=1)$, $a b$ takes the minimum value of 2. At this time, by the Pythagorean theorem, we have
$$
(x+y)^{2}+(y+z)^{2}=(x+z)^{2} \text {, }
$$
which means $y(x+y+z)=x z$ when $(x+y)(y+z)$ takes the minimum value. For example, when $x=z=1, y=\sqrt{2}-1$, $(x+y)(y+z)=2$.
This construction is indeed quite ingenious and should be recognized for its training value, but for solving this problem, a direct algebraic method is much simpler.
The following two problems from the former Soviet Union's math competitions are also classic examples of constructing geometric figures to solve problems, and they can both find simpler algebraic solutions.
1. Let $a, b, c$ be positive numbers, and
$$
\left\{\begin{array}{l}
a^{2}+a b+\frac{b^{2}}{3}=25, \\
\frac{b^{2}}{3}+c^{2}=9, \\
a^{2}+a c+c^{2}=16 .
\end{array}\right.
$$
Find the value of $a b+2 b c+3 a c$.
2. Positive numbers $a, b, c, A, B, C$ satisfy the conditions $a+A=$ $b+B=c+C=k$. Prove that $a B+b C+c A<k^{2}$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ .
|
Solution: Since $m \neq n$, by the definition of roots, $m, n$ are two distinct roots of the equation $x^{2}-x-1=0$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=1, m n=-1 . \\
\because m^{2}+n^{2}=(m+n)^{2}-2 m n \\
\quad=1^{2}-2 \times(-1)=3, \\
\quad m^{3}+n^{3}=(m+n)^{3}-3 m n(m+n) \\
\quad=1^{3}-3 \times(-1) \times 1=4, \\
\therefore m^{5}+n^{5} \\
=-\left(m^{3}+n^{3}\right)\left(m^{2}+n^{2}\right)-(m n)^{2}(m+n) \\
\quad=4 \times 3-(-1)^{2} \times 1=11 .
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}$, $a_{2}, \cdots, a_{n}$ have the following property, for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
|
Three, the minimum value of $n$ is 8.
First, prove that each number in $S$ appears at least 2 times in the sequence $a_{1}, a_{2}, \cdots, a_{n}$. This is because, if a number in $S$ appears only once in this sequence, since there are 3 two-element subsets containing this number, but in the sequence, the adjacent pairs containing this number can have at most two different combinations, it is impossible for all 3 two-element subsets containing this number to appear as adjacent pairs in the sequence.
Therefore, $n \geqslant 8$.
On the other hand, the 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the required minimum value is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in the figure, the side lengths of $\triangle A B C$ are $A B=14, B C$ $=16, A C=26, P$ is a point on the angle bisector $A D$ of $\angle A$, and $B P \perp A D, M$ is the midpoint of $B C$. Find the value of $P M$ $\qquad$
|
3. 6 .
From the figure, take $B^{\prime}$ on $A C$ such that $A B^{\prime}=A B=14$, then $B^{\prime} C=12$. Since $\triangle A B B^{\prime}$ is an isosceles triangle, we know that the intersection point of $B B^{\prime}$ and $A D$ is $P$ (concurrency of five lines), so $P$ is the midpoint of $B B^{\prime}$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1 \\
(y-1)^{3}+1997(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ $\qquad$ (Proposed by the Problem Group)
|
$=、 1.2$.
The original system of equations is transformed into
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1, \\
(1-y)^{3}+1997(1-y)=-1 .
\end{array}\right.
$$
Since $f(t)=t^{3}+1997 t$ is monotonically increasing on $(-\infty,+\infty)$, and $f(x-1)=f(1-y)$, it follows that $x-1=1-y$, i.e., $x+y=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, a line $l$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly 3 lines $l$, then $\lambda=$
(Proposed by the Problem Committee)
|
2. 4 .
First, note the following conclusion: For a chord passing through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ and intersecting the right branch at two points, the chord attains its minimum length $\frac{2 b^{2}}{a}=4$ if and only if the chord is perpendicular to the $x$-axis. (In fact, the polar equation of this hyperbola is $\rho=\frac{2}{1-\sqrt{3} \cos \theta}$. Suppose $A B$ is a chord passing through the right focus $F$ and intersecting only the right branch, with $A\left(\rho_{1}, \theta\right), B\left(\rho_{1}, \pi+\theta\right)$ $\left(\rho_{1}>0, \rho_{2}>0\right)$, then $|A B|=\rho_{1}+\rho_{2}=\frac{2}{1-\sqrt{3} \cos \theta}+$ $\frac{2}{1+\sqrt{3} \cos \theta}=\frac{4}{1-\sqrt{3} \cos ^{2} \theta} \geqslant 4$, with equality when $\theta=\frac{\pi}{2}$.)
Second, when there are exactly three lines satisfying the given conditions, there are only two possibilities:
(1) Only one line intersects both the left and right branches of the hyperbola, and two lines intersect only the right branch. In this case, the line intersecting both branches must be the $x$-axis, and the distance between its two intersection points is $2 a=2$. However, the length of the two chords intersecting only the right branch, $\lambda>4$, does not satisfy the given conditions;
(2) Two lines intersect both the left and right branches of the hyperbola, and only one line intersects only the right branch, and this chord must be perpendicular to the $x$-axis (otherwise, by symmetry, there would be two chords intersecting only the right branch). In this case, $|A B|=\lambda=4$, and the length of the chords intersecting both branches can also satisfy this condition, so $\lambda=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $x=\frac{-\sqrt[1]{17}+\sqrt{\sqrt{17}+4 \sqrt{15}}}{2 \sqrt{3}}$. Then, $3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5=$
|
2. 0 .
From the given, we have
$$
2 \sqrt{3} x+\sqrt[4]{17}=\sqrt{\sqrt{17}+4 \sqrt{15}} \text {. }
$$
Squaring and rearranging, we get
$$
\begin{array}{l}
\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}=0 . \\
\text { Also, } 3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5 \\
=\left(3 x^{4}-2 \sqrt{3} \cdot \sqrt{5} x^{2}-5\right)-1-\sqrt{17} x^{2} \\
=\left(\sqrt{3} x^{2}-\sqrt{5}\right)^{2}-(\sqrt[4]{17} x)^{2} \\
=\left(\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}\right)\left(\sqrt{3} x^{2}-\right. \\
\sqrt[4]{17} x-\sqrt{5}) \\
\stackrel{1}{=} 0 .
\end{array}
$$
Explanation: Generally, if $x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$, then
$$
\begin{array}{l}
a^{2} x^{4}+\left(2 a c-b^{2}\right) x^{2}+c^{2} \\
=\left(a x^{2}+b x+c\right)\left(a x^{2}-b x+c\right) \\
=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the integer $n$ is not a multiple of 5. Then the remainder when $n^{4}+4$ is divided by 5 is $\qquad$ .
|
4. 0 .
$$
\text { Given } \begin{aligned}
& n^{4}+4=\left(n^{4}-1\right)+5 \\
= & \left(n^{2}+1\right)\left(n^{2}-1\right)+5 \\
= & \left(n^{2}-4\right)\left(n^{2}-1\right)+5\left(n^{2}-1\right)+5 \\
= & (n+2)(n-2)(n+1)(n-1) \\
& +5\left(n^{2}-1\right)+5,
\end{aligned}
$$
and $n$ is not a multiple of 5, so among $n+2, n-2, n+1, n-1$, there must be one that is a multiple of 5, thus $n^{4}+4$ is also a multiple of 5. Therefore, $n^{4}+4 \equiv 0(\bmod 5)$
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For a cube wooden block $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with an edge length of 1, take points $P, Q, R$ on the three edges passing through vertex $A_{1}$, such that $A_{1} P=A_{1} Q=A_{1} R$. After cutting off the tetrahedron $A_{1}-P Q R$, use the section $\triangle P Q R$ as the base to drill a triangular prism-shaped hole in the cube, making the sides of the prism parallel to the body diagonal $A_{1} C$. When the hole is drilled through, vertex $C$ is cut off, and the exit is a spatial polygon. How many sides does this spatial polygon have?
(A) 3
(B) 6
(C) 8
(D) 9
|
6. (B).
Through point $R$, a line parallel to $A_{1} C$ intersects $A C$ at a point $R^{\prime}$. Through point $Q$, a line parallel to $A_{1} C$ intersects $B_{1} C$ at a point $Q^{\prime}$. A plane through $R Q$ and parallel to $A_{1} C$ intersects the side face of the triangular prism at a point $C_{0}$ on the edge $C_{1} C$. Thus, the plane through $R Q$ and parallel to $A_{1} C$ cuts out two segments $R^{\prime} C_{0}$ and $Q^{\prime} C_{0}$ near point $C$. Similarly, planes through $P Q$ and $R P$ and parallel to $A^{\prime} C$ each cut out two segments near point $C$.
Therefore, the spatial polygon cut out at the exit has a total of 6 sides.
|
6
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $n=\underbrace{111 \cdots 11}_{1999 \uparrow 1}, f(n)=90 n^{2000}+20 n+$ 1997. Then the remainder when $f(n)$ is divided by 3 is
|
Ni.1.1.
A natural number $a$ has the same remainder when divided by 3 as its digit sum $S(a)$ when divided by 3. Therefore, $n=\underbrace{11 \cdots 111}_{1999 \uparrow 1}$ has the same remainder when divided by 3 as 1999, which is 1.
$90 n^{2000}$ has a remainder of 0 when divided by 3, and $20 n$ has a remainder of 2 when divided by 3.
Thus, the remainder of $f(n)=90 n^{2000}+20 n+1997$ when divided by 3 is the same as the remainder of $2+1997=1999$ when divided by 3, which is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $a$, $b$, and $c$ are the lengths of the three sides of a right triangle, and for a natural number $n$ greater than 2, the following holds:
$$
\left(a^{n}+b^{n}+c^{n}\right)^{2}=2\left(a^{2 n}+b^{2 n}+c^{2 n}\right) .
$$
Then $n=$
|
6. 4 .
Let $x=a^{\frac{n}{2}}, y=b^{\frac{n}{2}}, z=c^{\frac{\pi}{2}}$, then
$$
\begin{aligned}
0= & 2\left(a^{2 n}+b^{2 n}+c^{2 n}\right)-\left(a^{n}+b^{n}+c^{n}\right)^{2} \\
= & 2\left(x^{4}+y^{4}+z^{4}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
= & x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 x^{2} z^{2}-2 y^{2} z^{2} \\
= & -(x+y+z)(x+y-z)(y+z-x)(z+x \\
& -y) .
\end{aligned}
$$
Assume $c$ is the hypotenuse, then $z>x, z>y$. It follows that
$$
x+y+z>0, y+z-x>0, z+x-y>0 \text {. }
$$
$\therefore(*)$ is equivalent to $z=x+y$,
which means $\left(\frac{a}{c}\right)^{\frac{n}{2}}+\left(\frac{b}{c}\right)^{\frac{n}{2}}=1$.
On the other hand, $a^{2}+b^{2}=c^{2}$ holds, or $\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}=$ 1.
Since $0<\frac{a}{c}<1,0<\frac{b}{c}<1, y=\left(\frac{a}{c}\right)^{x}+\left(\frac{b}{c}\right)^{x}$ is a monotonically decreasing function, and it takes $y=1$ at only one point $x$, therefore, $\frac{n}{2}=2, n=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (12 points) On the hypotenuse $AB$ of the right triangle $\triangle ABC$, color the points $P$ that satisfy $PC^{2} = PA \cdot PB$ in red. How many red points are there at least, and at most, on the hypotenuse?
|
Six, as shown, let $B P=x$, then $A P=c-x$. Draw the altitude $C H$ on the hypotenuse, then $C H=\frac{a b}{c}, B H$ $=\frac{a^{2}}{b}$. Therefore,
$$
\left.A P=\frac{a^{2}}{c}-x \right| \, \text { ( } P \text { can be between } A \text { and } Y \text {, also please consider } H \text { and }
$$
$B$ points.
In the right triangle $\triangle C$, we have
$$
\begin{aligned}
P C^{2} & =C H^{2}+H P^{2} \\
& =\frac{a^{2} b^{2}}{c^{2}}=\left(\frac{a^{2}}{c}-x\right)^{2} \\
& =x^{2}-\frac{2 a^{2}}{c} x+a^{2} .
\end{aligned}
$$
But from the given, we also have
$$
P C^{2}=P A \cdot P B=x(c-x) .
$$
From (1) and (2), we get
$$
x^{2}-\frac{2 a^{2}}{c} x+a^{2}=x(c-x) \text {, }
$$
which simplifies to $2 x^{2}-\frac{2 a^{2}+c^{2}}{c} x+a^{2}=0$.
Factoring gives $(2 x-c)\left(x-\frac{a^{2}}{c}\right)=0$.
Thus, $x_{1}=\frac{c}{2}$ (the midpoint of the hypotenuse),
$x_{2}=\frac{a^{2}}{c}$ (the foot of the altitude on the hypotenuse).
Since a quadratic equation has at most two real roots, the hypotenuse has at most two red points; when the right triangle $\triangle A B C$ is isosceles, the midpoint and the foot of the altitude coincide, so the hypotenuse has at least one red point.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the Cartesian coordinate system, points of the form $\left(m, n^{2}\right)$ are painted red (where $m, n$ are integers), referred to as red points, and their surrounding points are not colored. Then, the parabola $y=x^{2}-$ $196 x+9612$ has $\qquad$ red points.
|
2. 2 .
Let $\left(m, n^{2}\right)$ be on the parabola $y=x^{2}-196 x+9612$, then $n^{2}=m^{2}-196 m+9612$.
Completing the square and factoring, we get
$$
\begin{array}{l}
(n+m-98)(n-m+98) \\
=8=2 \times 4=(-2) \times(-4) .
\end{array}
$$
$\because m, n$ are integers,
$\therefore n+m-98$ and $n-m+98$ are of the same parity.
When $n+m-98$ takes $-4,-2,2,4$ in turn, $n-m+98$ takes $-2,-4,4,2$ in turn. Therefore, $\left(m, n^{2}\right)=(97,9)$ or $(99,9)$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given, as shown in the figure, a semicircle $O$ with a diameter of $20 \mathrm{~cm}$ has two points $P$ and $Q$, $P C \perp A B$ at $C, Q D$ $\perp A B$ at $D, Q E \perp$ $O P$ at $E, A C=4 \mathrm{~cm}$. Then $D E=$
|
$3.8 \mathrm{~cm}$.
Take the midpoint $M$ of $O P$ and the midpoint $N$ of $O Q$, and connect $C M, D N$, and $E N$. Then
$$
\begin{array}{c}
M C=P E=\frac{1}{2} O P \\
=\frac{1}{2} O Q=E N=D N, \\
\angle P M C=\angle M C O+ \\
\angle M O C=2 \angle M O C .
\end{array}
$$
Since $O, D, Q, E$ are concyclic, with $N$ as the center,
$$
\begin{array}{l}
\therefore \angle D N E=2 \angle D Q E=2 \angle M O C=\angle P M C . \\
\therefore \triangle P C M \cong \triangle D E N .
\end{array}
$$
Thus $D E=P C=\sqrt{O P^{2}-O C^{2}}$
$$
=\sqrt{10^{2}-(10-4)^{2}}=8 .
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers 1, 2, $\cdots, n$. Let $f(n)$ be the number of such permutations that satisfy: (1) $a_{1}=1$; (2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1$, $2, \cdots, n-1$. Determine whether $f(1996)$ is divisible by 3.
|
Solution: It is easy to find that $f(1)=f(2)=1, f(3)=2$.
When $n \geqslant 4$, we have $a_{1}=1, a_{2}=2$ or 3.
(a) When $a_{2}=2$, we proceed as follows: delete $a_{1}$, $b_{1}=a_{2}-1, b_{2}=a_{3}-1, \cdots, b_{n-1}=a_{n}-1$, to get a permutation $b_{1}, \cdots, b_{n-1}$ that meets the conditions, and the number of such permutations is $f(n-1)$.
(b) When $a_{2}=3$, for $a_{3}$ we have two cases:
( $\left.1^{\circ}\right) a_{3}=2$, then $a_{4}=4$. By the same process as (a), the number of permutations that meet the conditions is $f(n-3)$;
$\left(2^{\circ}\right) a_{3} \neq 2$, then 2 must be after 4. This leads to the conclusion that the odd numbers are arranged in reverse order, i.e., $1357 \cdots 8642$, which is one permutation.
Therefore, $f(n)=f(n-1)+f(n-3)+1$.
For $n \geqslant 1$, let $r(n)$ be the remainder when $f(n)$ is divided by 3, then we have
$$
\begin{array}{l}
r(1)=r(2)=1, r(3)=2, \\
r(n)=r(n-1)+r(n-3)+1 .
\end{array}
$$
Thus, $r(n+8)=r(n+7)+r(n+5)+1$
$$
\begin{aligned}
= & {[r(n+6)+r(n+4)+1] } \\
& +[r(n+4)+r(n+2)+1]+1 \\
= & r(n+6)+2 r(n+4)+r(n+2)+3 \\
= & {[r(n+5)+r(n+3)+1] } \\
& +2 r(n+4)+r(n+2)+3 \\
= & \{[(r(n+4)+r(n+2)+1] \\
& +[r(n+2)+r(n)+1]+1\} \\
& +2 r(n+4)+r(n+2)+3 \\
= & 3 r(n+4)+3 r(n+2)+6+r(n),
\end{aligned}
$$
which means $r(n+8)$ has the same remainder as $r(n)$ when divided by 3, i.e., $r(n+8)=r(n)$.
Therefore, $r(1996)=r(4+249 \times 8)=r(4)$.
And $r(3)+r(1)+1=4$,
so $r(4)=1$,
which means the remainder when $f(1996)$ is divided by 3 is 1. Therefore, $f(1996)$ is not divisible by 3.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the largest integer $n$, such that all non-zero solutions of the equation $(z+1)^{n}=z^{n}+1$ lie on the unit circle.
|
$$
\begin{array}{l}
\text { Solution: Using the binomial theorem, the equation can be transformed into } \\
z\left(C_{n}^{1} z^{n-2}+C_{n}^{2} z^{n-3}+C_{n}^{3} z^{n-4}+\cdots+C_{n}^{n-2} z+\right. \\
\left.C_{n}^{n-1}\right)=0(n>3) .
\end{array}
$$
Let the non-zero solutions of the equation be $z_{i}(i=1,2, \cdots, n-2)$. By Vieta's formulas, we have
$$
\begin{aligned}
S_{1} & =\sum_{i=1}^{n-2} z_{i}=-\frac{C_{n}^{2}}{C_{n}^{1}}=-\frac{n-1}{2}, \\
S_{2} & =\sum_{1 \leqslant i4$ satisfies the given conditions, then $z_{i} \cdot \bar{z}_{i}=\left|z_{i}\right|^{2}$ $=1, x_{i}=z_{i}+\bar{z}_{i}$ is a real number $(i=1,2, \cdots, n-2)$. Since the coefficients of the equation are all real, the roots appear in conjugate pairs, so the non-zero solutions can also be represented as $\bar{z} i=1,2, \cdots, n-2)$. Therefore,
$$
\begin{array}{l}
t_{1}=\sum_{i=1}^{n-2} x_{i}=\sum_{i=1}^{n-2}\left(z_{i}+\bar{z}_{i}\right)=2 S_{1}=1-n, \\
t_{2}=\sum_{1: i<j \leqslant n-i} x_{i} x_{i}:=\frac{1}{2}\left(t_{1}^{2}-\sum_{i=1}^{n-2} x_{i}^{2}\right), \\
\sum_{i=2}^{n-2} x_{i}^{2}=\sum_{i=1}^{n-2}\left(z_{i}^{2}+\bar{z}_{i}^{2}+2 z_{i} \cdot \bar{z}_{i}\right) \\
=\sum_{i=1}^{n-2}\left(z_{i}^{2}+\bar{z}_{i}^{2}\right)+2 \sum_{i=1}^{n-2} z_{i} \cdot \bar{z}_{i} \\
=2 \sum_{i=1}^{n-2} z_{i}^{2}+2 \sum_{i=1}^{n-2} 1 \\
=2\left(S_{1}^{2}-2 S_{2}\right)+2(n-2) .
\end{array}
$$
Therefore, $t_{2}=\frac{1}{2}\left[\left(2 S_{1}\right)^{2}-2\left(S_{1}^{2}-2 S_{2}\right)\right.$
$$
\begin{aligned}
& -2(n-2)] \\
= & \frac{1}{12}\left(7 n^{2}-30 n+35\right) .
\end{aligned}
$$
By Vieta's formulas, the real numbers $x_{i}(i=1,2, \cdots, n-2)$ are the $n-2$ roots of the real-coefficient equation $x^{n-2}-t_{1} x^{n-3}+t_{2} x^{n-4}+\cdots+$ $t_{n-3} x+t_{n-2}=0$. Using Theorem (1), we have
$$
\Delta_{1}=(n-3)\left(-t_{1}\right)^{2}-2(n-2) t_{2} \geqslant 0,
$$
which is $(n-3)(n-1)^{2}-2(n-2) \cdot$
$$
\frac{7 n^{2}-30 n+35}{12} \geqslant 0 \text {. }
$$
Simplifying, we get
$$
(n-4)\left[(n-5)^{2}-12\right] \leqslant 0 .
$$
Solving, we get $n \leqslant 5+\sqrt{12}<9$, i.e., $n \leqslant 8$.
When $n=8$, the equation becomes
$$
\begin{array}{l}
\left(8 z^{6}+28 z^{5}+56 z^{4}+70 z^{3}+56 z^{2}\right. \\
+28 z+8) z=0 .
\end{array}
$$
Its non-zero solutions are the 6 roots of the following equation (*):
$$
\begin{array}{l}
4 z^{6}+14 z^{5}+28 z^{4}+35 z^{3}+28 z^{2}+14 z+4 \\
=0 . \\
\text { The equation }(*) \text { can be transformed into } \\
4\left(z^{3}+z^{-3}\right)+14\left(z^{2}+z^{-2}\right)+28\left(z+z^{-1}\right) \\
+35=0, \\
4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right) \\
+7=0 .
\end{array}
$$
The equation (*) can be transformed into
$4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right)$
$\Delta_{2}=(3-1) \cdot 16^{2}-2 \cdot 3 \cdot 14 \cdot 7=$ $-76<0$, by Theorem (2), the roots of the equation $4 x^{3}+14 x^{2}+$ $16 x+7=0$ are not all real, i.e., there exists a root $z_{i}$ of equation (*) such that $z_{i}+z_{i}^{-1}$ is not a real number. Thus, $z_{i}+z_{i}^{-1} \neq z_{i}+\bar{z}_{i}$, i.e., $\left|z_{i}\right| \neq 1$. Therefore, $n=8$ does not satisfy the given conditions.
When $n=7$, the equation becomes
$$
\begin{aligned}
0 & =(z+1)^{7}-\left(z^{7}+1\right) \\
& =(z+1) \sum_{i=0}^{6} C_{6}^{i} z^{i}-(z+1) \sum_{i=0}^{6}(-1)^{i} z^{6} \\
& =(z+1) \sum_{i=0}^{6}\left[C_{6}^{i}-(-1)^{i}\right] z^{i} \\
& =7(z+1) \approx\left(z^{2}+z+1\right)^{2} .
\end{aligned}
$$
Its non-zero solutions are -1 and $\cos 120^{\circ} \pm i \sin 120^{\circ}$, all on the unit circle.
In summary, the largest integer $n$ that satisfies the given conditions is
$$
\boxed{7}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $S=1^{2}-2^{2}+3^{2}-1^{2}+\cdots+99^{2}-$ $100^{2}+101^{2}$. Then the remainder when $S$ is divided by 103 is $\qquad$
|
2. 1 .
$$
\begin{aligned}
S= & 1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(99^{2}-98^{2}\right) \\
& +\left(101^{2}-100^{2}\right) \\
= & 1+2+3+\cdots+100+101 \\
= & \frac{101 \times 102}{2}=5151=103 \times 50+1 .
\end{aligned}
$$
Therefore, the required remainder is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Locations A and B are respectively upstream and downstream on a lake. Each day, there is a boat that departs on time from both locations and travels at a constant speed towards each other, usually meeting at 11:00 AM on the way. ... Due to a delay, the boat from location A left 40 minutes late, and as a result, the two boats met at 11:15 AM on the way. It is known that the boat departing from location A travels at a speed of 44 kilometers per hour in still water, and the boat departing from location B travels at a speed of $v$ kilometers per hour in still water. Then $v=$ $\qquad$ kilometers per hour.
|
3. 6 .
As shown in the figure, the two ships usually meet at point $A$. On the day when ship B is late, they meet at point $B$. According to the problem, ship A takes $15^{\prime}$ to navigate segment $AB$, and ship C takes
$$
\begin{array}{l}
(40-15)^{\prime}=25^{\prime} . \\
\therefore \frac{15}{60}(44+v)=\frac{25}{60}\left(v^{2}-v\right),
\end{array}
$$
which simplifies to $5 v^{2}-8 v-132=0$.
$$
(v-6)(5 v+22)=0 \text {. }
$$
Solving this, we get $v=6$ (the negative root is discarded).
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $a$ is an integer, the two real roots of the equation $x^{2}+(2 a-1) x+$ $a^{2}=0$ are $x_{1}$ and $x_{2}$. Then $\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=$
|
4. 1.
From the problem, the discriminant $\Delta=-4 a+1 \geqslant 0$, so $a \leqslant 0$.
$$
\begin{array}{l}
\left(\sqrt{x_{1}}-\sqrt{x_{2}}\right)^{2}=\left(x_{1}+x_{2}\right)-2 \sqrt{x_{1} x_{2}} \\
=1-2 a+2 a=1 \\
\therefore\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the equation $\left(a^{2}-1\right) x^{2}-2(5 a+1) x+24=0$ has two distinct negative integer roots. Then the integer value of $a$ is $\qquad$ .
（The 1st Zu Chongzhi Cup Junior High School Mathematics Competition）
|
(Solution: $a=-2$ )
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $k$ is an integer, and the equation $\left(k^{2}-1\right) x^{2}-$ $3(3 k-1) x+18=0$ has two distinct positive integer roots. Then $k=$ $\qquad$
（4th Hope Forest Junior High School Mathematics Competition）
|
(Solution: $k=2$ )
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $a$ is an integer, the equation $x^{2}+(2 a+1) x+a^{2}=0$ has integer roots $x_{1} 、 x_{2}, x_{1}>x_{2}$. Try to find $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$.
(1991, Nanchang City Junior High School Mathematics Competition)
|
(Given $a>0$, we know $0>x_{1}>x_{2}$, the result is -1)
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 For any $a>1, b>1$,
we have $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$.
(1992, 26th Commonwealth Mathematics Olympiad (10th grade))
|
To prove for $a:=1+t_{1}, b:=1+t_{2}$, where $t_{1}, t_{2}>0$. The inequality to be proven becomes
$$
\begin{array}{l}
\frac{\left(1+t_{1}\right)^{2}}{t_{2}}+\frac{\left(1+t_{2}\right)^{2}}{t_{1}} \geqslant 8 . \\
\text { The left side of the above equation } \geqslant \frac{2\left(1+t_{1}\right)\left(1+t_{2}\right)}{\sqrt{t_{1} t_{2}}} \\
=\frac{2+2 t_{1} t_{2}+2\left(t_{1}+t_{2}\right)}{\sqrt{t_{1} t_{2}}} \\
\geqslant \frac{2}{\sqrt{t_{1} t_{2}}}+2 \sqrt{t_{1} t_{2}}+4 \geqslant 4+4=8 .
\end{array}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
5. Product $\prod_{k=1}^{7}\left(1+2 \cos \frac{2 k \pi}{7}\right)=$ $\qquad$
|
5.3 .
$$
\begin{array}{l}
\text { Let } \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} \text {. Then } \omega^{7}=1 . \\
\omega^{k}=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}, \\
\omega^{-k}=\cos \frac{2 k \pi}{7}-i \sin \frac{2 k \pi}{7}, \\
\therefore \omega^{k}+\omega^{-k}=2 \cos \frac{2 k \pi}{7} . \\
\text { The original expression }=\prod_{k=1}^{7}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{7} \omega^{-k}\left(1+\omega^{k}+\omega^{2 k}\right) \\
=\omega^{-1 \cdots 2 \cdots \cdots-7} \cdot \prod_{k=1}^{6}\left(1+\omega^{k}+\omega^{2 k}\right) \cdot\left(1+\omega^{7}\right. \\
\left.\quad+\omega^{14}\right) \\
=\left(\omega^{7}\right)^{-4} \cdot 3 \prod_{k=1}^{6} \frac{1-\omega^{3 k}}{1-\omega^{k}} \\
=3 \cdot \frac{1-\omega^{3}}{1-\omega} \cdot \frac{1-\omega^{6}}{1-\omega^{2}} \cdot \frac{1-\omega^{9}}{1-\omega^{3}} \cdot \frac{1-\omega^{12}}{1-\omega^{4}} \cdot \frac{1-\omega^{15}}{1-\omega^{5}} \\
\cdot \frac{1-\omega^{18}}{1-\omega^{6}}=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, on the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$ there are 16 points, sequentially $P_{1}, P_{2}, \cdots, P_{16}, F$ is the left focus, and the angles between each adjacent pair of points and $F$ are equal $\left(\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{16} F P_{1}\right)$. Let the distance from $P_{i}$ to the left directrix be $d_{i}(i=1,2, \cdots, 16)$. Find $\sum_{i=1}^{16} \frac{1}{d_{i}}$.
|
$$
\begin{array}{l}
a=5, b \\
=4, c=3 . \text { Let } \angle X F P_{1} \\
=\alpha \cdot, \angle P_{1} F P_{2}= \\
\angle P_{2} F P_{3}=\cdots= \\
\angle P_{16} F P_{1}=\frac{\pi}{8} . \\
F M=d:-P_{i} F \cos \left[(i-1) \frac{\pi}{8}+\alpha\right] \\
\quad=\frac{a^{2}}{c}-c=\frac{16}{3} .
\end{array}
$$
By the definition of an ellipse, $P_{i} F=e d_{i}=\frac{c}{a} d_{i}=\frac{3}{5} d_{i}$, substituting into (1), we have
$$
\begin{array}{l}
d_{i}-\frac{3}{5} d_{i} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right]=\frac{16}{3} . \\
\text { Thus } \frac{1}{d_{i}}=\frac{3}{16} \cdot \frac{1}{5}\left[5-3 \cos \left(\frac{(i-1) \pi}{8}+\alpha\right)\right] \\
\text { ( } i=1,2, \cdots, 16 \text { ). } \\
\therefore \sum_{i=1}^{16} \frac{1}{d}=\frac{3}{80}\left[\sum_{i=1}^{16} 5-3 \sum_{i=1}^{16} \cos \left(\frac{(i-1) \pi}{8}+\alpha\right)\right] \\
=3-\frac{9}{80} \sum_{i=1}^{16} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right] \text {. } \\
\because \sum_{i=1}^{16} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right] \\
=\sum_{i=1}^{i 6} \frac{2 \sin \frac{\pi}{16} \cos \left[(i-1)-\frac{\pi}{3}+\alpha\right]}{2 \sin \frac{\pi}{16}} \\
=\frac{1}{2 \sin \frac{\pi}{16}} \sum_{i=1}^{16}\left(\sin \left[\frac{\pi}{16}+(i-1) \frac{\pi}{8}+\alpha\right]\right. \\
\left.-\sin \left[(i-1) \frac{\pi}{8}+\alpha-\frac{\pi}{16}\right]\right) \\
=\frac{1}{2 \sin \frac{\pi}{16}}\left[\sin \left(\alpha+\frac{16 \pi}{8}-\frac{\pi}{16}\right)-\sin \left(x-\frac{\pi}{16}\right)\right] \\
=0 \text {, } \\
\therefore \sum_{i=1}^{16} \frac{1}{d_{i}}=3 \text {. } \\
\end{array}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
63. Let $a_{1}=1997^{1997^{1997}}{ }^{.197}$ (1997 sevens in total), the sum of the digits in the decimal representation of $a_{1}$ is $a_{2}$, the sum of the digits of $a_{2}$ is $a_{3}$, and so on. Find $a_{2000}$.
|
Solution: Let $x_{n}=10000^{10000^{10000}}$ (with $n$ 10000s), $n=1,2, \cdots$, then for $n \geqslant 2$, we have
$$
\begin{array}{l}
x_{n}=10000^{x_{n-1}}=10^{4 x_{n-1}} . \\
\therefore a_{1}<x_{1997}=10^{4 x_{1996}}, \\
a_{2} \leqslant 9 \times 4 x_{1996}<100 x_{1996} \\
=100 \cdot 10^{4 x 1995}=10^{4 x_{1995}+2}, \\
a_{3} \leqslant 9\left(4 x_{1995}+2\right)<9 \times 6 x_{1995}<100 x_{1995} \\
=10^{4 x_{1994}+2},
\end{array}
$$
Thus, for any $k$, when $1 \leqslant k \leqslant 1996$, we have
$$
a_{k}<10^{4 x} 1997-k{ }^{+2} \text {. }
$$
When $k=1996$, i.e.,
$$
\begin{array}{l}
a_{1996}<10^{4 x_{1}+2}, \\
\therefore a_{1997}<9 \times\left(4 x_{1}+2\right)=9 \times 40002<10^{6}, \\
a_{1998}<9 \times 6=54, \\
a_{1999} \leqslant 13, \\
a_{2000} \leqslant 9 .
\end{array}
$$
Since $a_{2000} \geqslant 1$,
$$
\therefore 1 \leqslant a_{20000} \leqslant 9 \text {. }
$$
From the generation of $a_{2}, a_{3}, \cdots, a_{2000}$, we know that
$$
a_{1} \equiv a_{2} \equiv a_{3} \equiv \cdots \equiv a_{2000}(\bmod 9) \text {. }
$$
It is easy to see that $1997=9 m-1(m=222)$, and the exponent of $a_{1}$ is odd, denoted as $r$.
$$
\begin{aligned}
\therefore a_{1} & =1997^{r}=(9 m-1)^{r} \\
& =9 s+(-1)^{r}=9 s-1, \\
a_{2000} & =a_{1}=-1(\bmod 9) .
\end{aligned}
$$
Therefore, $a_{23000}=8$.
(Qin Zongci, Jiangsu Zhenjiang Teachers College, 212003)
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $m=\sqrt{5}+1$. Then the integer part of $m+\frac{1}{m}$ is $\qquad$ .
|
\begin{array}{l}\text { 1. } m=\sqrt{5}+1, \frac{1}{m}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}, \\ \therefore m+\frac{1}{m}=\frac{5}{4} \sqrt{5}+\frac{3}{4},\left[m+\frac{1}{m}\right]=3 .\end{array}
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x^{2}-x-1=0$. Then, the value of the algebraic expression $x^{3}$ $-2 x+1$ is $\qquad$ .
|
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
The translation is as follows:
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $m$ and $n$ are rational numbers, and the equation $x^{2}+$ $m x+n=0$ has a root $\sqrt{5}-2$. Then the value of $m+n$ is $\qquad$ .
|
4. Since $m, n$ are rational, the other root is $-\sqrt{5}-2$, thus by Vieta's formulas,
$$
\begin{array}{l}
w:=4, n=-1 . \\
\therefore m+n=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The number of integer pairs $(m, n)$ that satisfy $1998^{2}+m^{2}=1997^{2}+n^{2}(0<m$ $<n<1998)$ is $\qquad$.
|
6. $n^{2}-m^{2}=3995=5 \times 17 \times 47,(n-m)(n+m)=5 \times 17 \times 47$, obviously any integer factorization of 3995 can yield $(m, n)$, given the condition $(0<m<n<1998)$, thus there are 3 integer pairs $(m, n)$ that satisfy the condition.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $a, b$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-$ $a-2 b$ is $\qquad$.
|
11 .
$$
\begin{array}{l}
a^{2}+a b+b^{2}-a-2 b \\
=a^{2}+(b-1) a+b^{2}-2 b \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4} b^{2}-\frac{3}{2} b-\frac{1}{4} \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 .
\end{array}
$$
When $a+\frac{b-1}{2}=0, b-1=0$,
i.e., $a=0, b=1$, the equality in the above inequality holds, so the minimum value sought is -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Every book has an international book number:
ABCDEFGHIJ
where $A B C D E F G H I$ are composed of nine digits, and $J$ is the check digit.
$$
\text { Let } \begin{aligned}
S= & 10 A+9 B+8 C+7 D+6 E+5 F \\
& +4 G+3 H+2 I,
\end{aligned}
$$
$r$ is the remainder when $S$ is divided by 11. If $r$ is not 0 or 1, then $J=11-r$ (if $r=0$, then $J=0$; if $r=1$, then $J$ is represented by $x$). If a book's number is $962 y 707015$, then $y=$ $\qquad$
|
$$
\text { 15. } \begin{aligned}
S=9 \times 10 & +6 \times 9+2 \times 8+y \times 7+7 \times 6 \\
& +0 \times 5+7 \times 4+0 \times 3+1 \times 2
\end{aligned}
$$
$\therefore S$ the remainder when divided by 11 is equal to the remainder when $7 y+1$ is divided by 11.
From the check digit, we know that the remainder when $S$ is divided by 11 is 11 $-5=6$, so the remainder when $7 y$ is divided by 11 is $6-1=5$. Therefore, $y$ $=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the line $y=-2 x+3$ intersects the parabola $y=$ $x^{2}$ at points $A$ and $B$, and $O$ is the origin. Then, the area of $\triangle O A B$ is $\qquad$ .
|
7.6 .
As shown in the figure, the line $y=$
$-2 x+3$ intersects the parabola $y$ $=x^{2}$ at points $A(1,1), B(-3,9)$.
Construct $A A_{1}, B B_{1}$ perpendicular to the $x$-axis, with feet of the perpendiculars at
$$
\begin{array}{l}
A_{1} 、 B_{1} . \\
\therefore S_{\triangle O A B}=S_{\text {trapezoid } A A_{1} B_{1} B}-S_{\triangle A A_{1} O}-S_{\triangle B B_{1} O} O \\
=\frac{1}{2} \times(1+9) \times(1+3)-\frac{1}{2} \times 1 \times 1 \\
\quad-\frac{1}{2} \times 9 \times 3=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $n$ be a positive integer. $0<x \leqslant 1$. In $\triangle A B C$, if $A B=n+x, B C=n+2 x, C A=n+3 x, B C$ has a height $A D=n$. Then, the number of such triangles is ( ).
(A) 10
(B) 11
(C) 12
(D) infinitely many
|
6. (C).
Let $\triangle A B C$ be the condition; then $c=n+1, a=n+2 x$, $b=n+3 x, p=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 n+6 x)$. By Heron's formula, we get
$$
\begin{array}{l}
\sqrt{\frac{3 n+6 x}{2} \cdot \frac{n+4 x}{2} \cdot \frac{n+2 x}{2} \cdot \frac{n}{2}}=\frac{n(n+2 x)}{2} . \\
\therefore \frac{3 n(n+2 x)^{2}(n+4 x)}{16}=\frac{n^{2}(n+2 x)^{2}}{4},
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
\frac{3(n+4 x)}{4}=n . \\
\therefore n=12 x . \\
\because 0<x \leqslant 1, n \text { is a positive integer. } \\
\therefore(x, n)=\left(\frac{1}{12}, 1\right),\left(\frac{2}{12}, 2\right), \cdots,\left(\frac{12}{12}, 12\right) .
\end{array}
$$
Therefore, there are 12 such triangles.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
(1995, National High School Mathematics Competition)
|
Analysis: The difficulty of this problem lies in the uncertainty of $[\lg x]$. Since $[\lg x] \leqslant \lg x$, the original problem is first transformed into finding the values of $x$ that satisfy the inequality $\lg ^{2} x-\lg x-2 \leqslant 0$.
Solving this, we get $-1 \leqslant \lg x \leqslant 2$.
Therefore, the number of roots of the equation can be found by solving for $[\lg x]$ taking the values -1, 0, 1, 2, respectively, yielding the three roots as $x_{1}=\frac{1}{10}, x_{2}=10^{\sqrt{3}}, x_{3}=100$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. For all real numbers $x, y$, if the function $f$ satisfies:
$$
f(x y)=f(x) \cdot f(y)
$$
and $f(0) \neq 0$, then $f(1998)=$ $\qquad$ .
|
$\approx 、 1.1$.
Let $f(x y)=f(x) \cdot f(y)$, set $y=0$, we have $f(x \cdot 0)=$ $f(x) \cdot f(0)$, which means $f(0)=f(x) \cdot f(0)$, and since $f(0) \neq 0$, it follows that $f(x)=1$, thus $f(1998)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Quadrilateral $ABCD$ is inscribed in a circle, $BC=CD=4$, $AC$ and $BD$ intersect at $E$, $AE=6$, and the lengths of $BE$ and $DE$ are both integers. Then the length of $BD$ is $\qquad$
|
4.7.
From $\overparen{B C}=\overparen{C D}, \angle B D C=\angle C A D, \angle A C D=\angle A C D$, we know $\triangle D C E \backsim \triangle A C D$, so $\frac{C E}{C D}=\frac{C D}{A C}$, which means $C E \cdot A C=C D^{2}$. Therefore, $C E(C E+6)=16$, solving this gives $C E=2$. Also, from $B E \cdot D E=A E \cdot C E=12$, and $B E+C E>B C$, then $B E>$ 2. Similarly, $D E>2$. Since $B E$ and $D E$ are integers, then $B E=3, D E=4$ (or $B E=4, D E=3$). Hence, $B D=B E+D E=7$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (Full marks 25 points) As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $M$, and the angle between their two external common tangents is $60^{\circ}$. The line connecting the centers intersects $\odot O_{1}$ and $\odot O_{2}$ at $A$ and $B$ (different from $M$), respectively. A line through $B$ intersects $\odot O_{1}$ at points $C$ and $D$. Find the value of $\operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D$.
|
From the given conditions, we have $O_{1} E \perp P E, O_{1} F \perp P F, O_{1} E= O_{1} F$, thus $O_{1}$ lies on the bisector of $\angle E P F$. Similarly, $O_{2}$ lies on the bisector of $\angle E P F$. Therefore, $P A$ is the bisector of $\angle E P F$. Since $O_{2} Q / / P E$, we have $\angle Q O_{2} O_{1}=\angle E P O_{1}=30^{\circ}$, hence $O_{1} Q=\frac{1}{2} O_{1} O_{2}$.
Let the radii of the larger and smaller circles be $R$ and $r$. Then $O_{1} Q=R-r, O_{1} O_{2}=R+r$, so $R-r=\frac{1}{2}(R+r)$, thus $R=3 r$.
$\because \angle A C M=\angle A D M=90^{\circ}$,
$\therefore \operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D=\frac{A C}{C M} \cdot \frac{A D}{D M}$.
From $\triangle B A C \sim \triangle B D M, \triangle B A D \sim \triangle B C M$ we get
$$
\frac{A C}{D M}=\frac{A B}{D B}, \frac{A D}{C M}=\frac{D B}{M B} \text {. }
$$
Therefore, $\operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D=\frac{A B}{D B} \cdot \frac{D B}{M B}=\frac{8 r}{2 r}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $f(x)=x^{10}+2 x^{9}-2 x^{8}-2 x^{7}+x^{6}$ $+3 x^{2}+6 x+1$, then $f(\sqrt{2}-1)=$
|
$=1.4$.
Let $x=\sqrt{2}-1$, then $x+1=\sqrt{2} \Rightarrow (x+1)^{2}=2 \Rightarrow x^{2}+2x-1=0$. That is, $x=\sqrt{2}-1$ is a root of $x^{2}+2x-1=0$. But $f(x)=(x^{8}-x^{6}+3)(x^{2}+2x-1)+4$, so $f(\sqrt{2}-1)=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If
$$
\dot{z}=\frac{(1+i)^{2000}(6+2 i)-(1-i)^{1998}(3-i)}{(1+i)^{1996}(23-7 i)+(1-i)^{1994}(10+2 i)} \text {, }
$$
then $|z|=$ . $\qquad$
|
3.1.
$$
\begin{array}{l}
\text { When }(1+i)^{2}=2 i, (1-i)^{2}=-2 i \text { and } X \in \mathbb{Z}, \\
i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i, i^{4 k}=1. \\
\text { Therefore, }(1+i)^{2000}=(2 i)^{1000}=2^{1000}, \\
(1-i)^{1998}=(-2 i)^{999}=2^{999} \cdot i, \\
(1+i)^{1996}=(2 i)^{998}=-2^{998}, \\
(1-i)^{1994}=(-2 i)^{997}=-2^{997} i. \\
\text { Then } z=\frac{2^{1000}(6+2 i)-2^{999} i(3-i)}{-2^{998}(23-7 i)-2^{997} i(10+2 i)} \\
=\frac{8(6+2 i)-4 i(3-i)}{-2(23-7 i)-i(10+2 i)} \\
=\frac{48+16 i-12 i-4}{-46+14 i-10 i+2} \\
=\frac{44+4 i}{-44+4 i}=-\frac{11+i}{11-i}. \\
\text { Hence }|z|=\frac{|11+i|}{|11-i|}=1. \\
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-\right.$ $3)^{2001}$ is expanded and like terms are combined. The sum of the coefficients of the odd powers of $x$ in the resulting expression is $\qquad$.
|
4. -1 .
$$
\text { Let } \begin{aligned}
f(x)= & \left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-3\right)^{2001} \\
= & A_{0}+A_{1} x+A_{2} x^{2}+\cdots+A_{4001} x^{4001} \\
& +A_{4002} x^{4002} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Then } A_{0}+A_{1}+A_{2}+\cdots+A_{4001}+A_{4002} \\
\quad=f(1)=0, \\
A_{0}-A_{1}+A_{2}-\cdots-A_{4001}+A_{4002} \\
=f(-1)=2 .
\end{array}
$$
Subtracting the two equations gives $2\left(A_{1}+A_{3}+\cdots+A_{4001}\right)=-2$.
$$
\text { Therefore, } A_{1}+A_{3}+\cdots+A_{4001}=-1 \text {. }
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
保留源文本的换行和格式，所以翻译结果如下：
Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
|
Let the arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$, with common difference $d=4$. Then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{2 a_{1}+4 n}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{array}
$$
It holds for at least one value of $a_{1}$, so the discriminant is non-negative. That is,
$$
\begin{array}{l}
(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0 \\
\Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0 \\
\Leftrightarrow-\frac{\sqrt{2} 816}{7} \leqslant n \leqslant \frac{3+\sqrt{2816}}{7} .
\end{array}
$$
But $n \in \mathbb{N}$, it is easy to see that $n \leqslant 8$.
In fact, the arithmetic sequence with a common difference of 4:
$-4,0,4,8,12,16,20,24$ has 8 terms, and
$$
(-4)^{2}+0+4+8+12+16+20+24=100 \text {. }
$$
Therefore, the maximum value of $n$ is 8. That is, such an arithmetic sequence can have at most 8 terms.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initial 65. Given a real-coefficient polynomial function $y=a x^{2}+b x+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$.
|
Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |(A+B) + (A-B)| = |2A| \leqslant 4$.
Now, let's solve the original problem.
When $x=0$, $|y| \leqslant 1$,
i.e., $|c| \leqslant 1$.
For any $|x| \leqslant 1$, we have
\[
|a x^2 + b x| = |y - c| \leqslant |y| + |c| \leqslant 2.
\]
Taking $x=1$, we get $|a + b| \leqslant 2$,
taking $x=-1$, we get $|a - b| \leqslant 2$.
According to the auxiliary proposition, we have
\[
\begin{aligned}
|a| + |b| &= \frac{1}{2}(|2a| + |2b|) \\
&= \frac{1}{2}(|(a+b) + (a-b)| + |(a+b) - (a-b)|) \\
&\leqslant \frac{1}{2} \times 4 = 2.
\end{aligned}
\]
From (1) and (2), we know $|a| + |b| + |c| \leqslant 3$.
Next, we show that $|a| + |b| + |c|$ can achieve the maximum value of 3.
Take $a=2, b=0, c=-1$. In this case, $y = 2x^2 - 1$. It is easy to verify that when $|x| \leqslant 1$, $|y| \leqslant 1$.
Therefore, the maximum value of $|a| + |b| + |c|$ is 3.
(Chen Kuanhong, Nengshi High School, Yueyang County, Hunan Province, 414113)
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 If any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>$ 0 , to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National High School Mathematics Competition)
|
Analysis: The inequality can be transformed into
$$
\begin{array}{l}
k \leqslant-\frac{\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993}{\log _{\frac{x_{0}}{x_{3}}} 1993} \\
=f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \text {. } \\
\end{array}
$$
Thus,
$$
\{k\}_{\text {max }}=\left\{f\left(x_{0}, x_{1}, x_{2}, x_{3}\right)\right\}_{\text {min }} .
$$
By the harmonic mean inequality, we get
$$
\begin{array}{l}
f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \geqslant \\
\frac{3}{\left(\log _{1993} \frac{x_{0}}{x_{1}}+\log _{1993} \frac{x_{1}}{x_{2}}+\log _{1993} \frac{x_{2}}{x_{3}}\right) \log _{\frac{x_{0}}{x_{3}}} 1993} \\
\end{array}
$$
$$
=3 \text {. }
$$
Therefore, $k_{\max }=3$.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=$ 0. Then $\cos (x+2 y)=$ $\qquad$
(1994, National High School Mathematics Competition)
|
Analysis: Since $2 a=x^{3}+\sin x=(-2 y)^{3}+$ $\sin (-2 y)$, if we let $f(t)=t^{3}+\sin t$, then we have $f(x)$ $=f(-2 y)$.
And $f(t)$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so by monotonicity we have
$$
x=-2 y \text {, hence } x+2 y=0 \text {. }
$$
Thus $\cos (x+2 y)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
Let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
|
$$
\text { Three, } \because 1^{3}=1,2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125 \text {. }
$$
$\therefore(x, y, z)$ has the following 10 possibilities:
(1) $(1,8,7) ;(2)(1,8,4) ;(3)(1,8,5)$;
(4) $(1,7,4) ;(5)(1,7,5) ;(6)(1,4,5)$;
(7) $(8,7,4) ;(8)(8,7,5) ;(9)(8,4,5)$;
$(10)(7,4,5)$.
Then the values of $(x-y)(y-z)(z-x)$ are
$$
-42,-84,-84,-54,-48,12,-12,-6 \text {, }
$$
12,6 .
Therefore, the minimum value of $(x-y)(y-z)(z-x)$ is -84.
At this time, $(x, y, z)=(1,8,4)$ or $(1,8,5)$.
The corresponding $a b c=1 \cdot 2 \cdot 4=8$ or $a b c=1 \cdot 2 \cdot 5=10$. Hence the maximum value of $a b c$ is 10.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, on a circular road, there are 4 middle schools arranged clockwise: $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferred to minimize the total number of TVs transferred? And what is the minimum total number of TVs transferred?
|
Let $A_{1}$ High School transfer $x_{1}$ color TVs to $A_{2}$ High School (if $x_{1}$ is negative, it means $A_{2}$ High School transfers $|x_{1}|$ color TVs to $A_{1}$ High School. The same applies below), $A_{2}$ High School transfer $x_{2}$ color TVs to $A_{3}$ High School, $A_{3}$ High School transfer $x_{3}$ color TVs to $A_{4}$ High School, and $A_{4}$ High School transfer $x_{4}$ color TVs to $A_{1}$ High School.
Since there are 40 color TVs in total, with an average of 10 per school, therefore,
$$
\begin{array}{l}
15-x_{1}+x_{4}=10,8-x_{2}+x_{1}=10, \\
5-x_{3}+x_{2}=10,12-x_{4}+x_{3}=10 . \\
\text { Then }=x_{1}-5, x_{1}=x_{2}+2, x_{2}=x_{3}+5, x_{3}= \\
x_{3}-2, x_{3}=\left(x_{1}-5\right)-2=x_{1}-7, x_{2}=\left(x_{1}-7\right)+5 \\
=x_{:}-2 .
\end{array}
$$
Then $x_{4}=x_{1}-5, x_{1}:=x_{2}+2, \tilde{x}_{2}=x_{3}+5, x_{3}=$ $=x:-\hat{2}$.
The problem requires $y=|x_{1}|+|x_{2}|+|x_{3}|+|x_{4}|$
$$
=|x_{1}|+|x_{1}-2|+|x_{1}-7|+|x_{1}
$$
$-5|$ to be minimized,
where $x_{1}$ is an integer satisfying $-8 \leqslant x_{1} \leqslant 15$.
Let $x_{1}=x$, and consider the function defined on $-8 \leqslant x \leqslant 15$
$$
y=|x|+|x-2|+|x-7|+|x-5| \text {, }
$$
$|x|+|x-7|$ represents the sum of the distances from $x$ to 0 and 7. When $0 \leqslant x \leqslant 7$, $|x|+|x-7|$ takes the minimum value 7; similarly, when $2 \leqslant x \leqslant 5$, $|x-2|+|x-5|$ takes the minimum value 3.
Therefore, when $2 \leqslant x \leqslant 5$, $y$ takes the minimum value 10. In other words, when $x_{1}=2,3,4,5$, $|x_{1}|+|x_{1}-2|+|x_{1}-7|+|x_{1}-5|$ takes the minimum value 10.
Thus, the minimum total number of color TVs transferred is 10, and there are four allocation schemes:
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: Xiao Zhang is riding a bicycle on a road next to a double-track railway. He notices that every 12 minutes, a train catches up with him from behind, and every 4 minutes, a train comes towards him from the opposite direction. If the intervals between each train are constant, the speeds are the same, and both the train and bicycle speeds are uniform, find the interval in minutes at which trains depart from the stations in front of and behind Xiao Zhang.
(1990, Xianyang City Junior High School Mathematics Competition Selection)
Analysis: Let the interval be $x$ minutes at which trains depart from the stations in front of and behind Xiao Zhang. Let Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$. Let $AB$ be the distance between two adjacent trains in the same direction, then $|AB| = x v_{2}$.
(1) First, consider the trains coming towards Xiao Zhang from the opposite direction. Assume Xiao Zhang meets the first train at point $A$, then when he meets the next train, both the train and Xiao Zhang have traveled the distance $|AB|$, so we have $4(v_{1} + v_{2}) = |AB|$. As shown in Figure 1.
(2) Similarly,
consider the trains
catching up with
Xiao Zhang. At point
$B$, he meets the
first train, and the next train is at point $A$ at that moment. To catch up with Xiao Zhang, the next train travels an additional distance of $|AB|$, so we have $12(v_{2} - v_{1}) = |AB|$. As shown in Figure 2.
|
Solution: Let the trains depart from the station ahead and behind Xiao Zhang every $x$ minutes, Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$, then
$$
\left\{\begin{array}{l}
4\left(v_{1}+v_{2}\right)=x v_{2}, \\
12\left(v_{2}-v_{1}\right)=x v_{2} .
\end{array}\right.
$$
Solving, we get $x=6$ (minutes).
Answer: Omitted.
(2) Time equivalence relationship
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 A person walks from place A to place B, and there are regular buses running between A and B, with equal intervals for departures from both places. He notices that a bus going to A passes by every 6 minutes, and a bus going to B passes by every 12 minutes. How often do the buses depart from their respective starting stations? (Assume that each bus travels at the same speed)
|
Analysis: Let the distance between two consecutive buses in the same direction be $s$, and the time be $t$ minutes, then the speed of the bus is $\frac{s}{t}$.
(1) A person's speed relative to the bus going to location A is $\frac{s}{6}$, so the person's speed is $\frac{s}{6}-\frac{s}{t}$.
(2) A person's speed relative to the bus going to location B is $\frac{s}{12}$, so the person's speed is $\frac{s}{t}-\frac{s}{12}$.
$$
\therefore \frac{s}{6}-\frac{s}{t}=\frac{s}{t}-\frac{s}{12} \text {. }
$$
Solving for $t$ gives $t=8$ (minutes).
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (Full marks 20 points) Given that $x$, $y$, $z$ are positive integers, and satisfy $x^{3}-y^{3}-z^{3}=3 x y z$, $x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original format and line breaks.
|
$$
\begin{aligned}
- & \because x^{3}-y^{3}-z^{3}-3 x y z \\
= & x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
= & (x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z+z^{2}\right) \\
& -3 y z(x-y-z) \\
= & (x-y-z)\left(x^{2}+y^{2}+z^{2}-x y-y z+x z\right),
\end{aligned}
$$
X. $x^{3}-y^{3}-z^{3}=3 x y z$,
$$
\begin{array}{l}
\therefore(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
\quad=0 .
\end{array}
$$
Since $x, y, z$ are all natural numbers,
$$
\begin{array}{l}
\text { and } x^{2}+y^{2}+z^{2}+x y-y z+x z \\
=\frac{1}{2}\left[(x+y)^{2}+(y-z)^{2}+(z+x)^{2} \neq 0,\right. \\
\therefore x-y-z=0,
\end{array}
$$
i.e., $x=y+z$.
Substituting (1) into $x^{2}=2(y+z)$, we get $x^{2}=2 x$.
$$
\therefore x=2, y+z=2 \text {. }
$$
Given $y \geqslant 1$ and $z \geqslant 1$, we know $y=z=1$.
$$
\therefore x y+y z+z x=5 \text {. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 25 points) On the blackboard, all natural numbers from 1 to 1997 are written. Students $A$ and $B$ take turns to perform the following operations: Student $A$ subtracts the same natural number from each number on the blackboard (the number subtracted can be different in different operations); Student $B$ erases two numbers from the blackboard and writes down their sum. Student $A$ goes first, and the operations continue until only one number remains on the blackboard. If this number is non-negative, find this number.
|
Three, because after each operation by student $A$ and student $B$, the number of numbers written on the blackboard decreases by 1. Since students $A$ and $B$ take turns operating, when $B$ completes the last operation, only one number remains on the blackboard, and both have performed 1996 operations. Let $d_{k} (k=1,2, \cdots, 1996)$ be the natural number removed by student $A$ during the $k$-th operation. Since in the $k$-th operation, the sum of the numbers on the blackboard decreases by $(1998-k) d_{k}$, and student $B$'s operation does not change this sum, after 1996 alternating operations, the number written on the blackboard is
\[
\begin{aligned}
x= & (1+2+\cdots+1997)-1997 d_{1}-1996 d_{2}-\cdots \\
& -2 d_{1996} \\
= & 1997\left(1-d_{1}\right)+1996\left(1-d_{2}\right)+\cdots+(1998 \\
& -k)\left(1-d_{k}\right)+\cdots+2\left(1-d_{1996}\right)+1 .
\end{aligned}
\]
For all $k=1,2, \cdots, 1996$, the number $1-d_{k}$ is non-positive.
If for some $k$, $d_{k} \geqslant 2$, then
\[
(1998-k)\left(d_{k}-1\right) \geqslant 2 \text{. }
\]
Thus, $x \leqslant(1998-k)\left(1-d_{k}\right)+1 \leqslant-1$.
This contradicts the conditions given in the problem. Therefore, for all $k=1,2, \cdots, 1996$, $d_{k}=1$. This means $x=1$.
(243000, Anhui Province, Maguan Second Middle School, Cheng Lihu)
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. From the center of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two perpendicular chords $A C$ and $B D$ are drawn. Connecting $A, B, C, D$ in sequence forms a quadrilateral. Then, the maximum value of the area $S$ of quadrilateral $A B C D$ is
|
3. 12.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Given $O A \perp O B$, we have
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=0 . \\
\begin{aligned}
\therefore S & =4 S_{\triangle A O B}=2|O A| \cdot|O B| \\
& =2 \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)} \\
& =2 \sqrt{\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}+\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2}} \\
& =2 \sqrt{\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2}} \\
& =12 \sqrt{\left(\frac{x_{1}}{3} \cdot \frac{y_{2}}{2}-\frac{x_{2}}{3} \cdot \frac{y_{1}}{2}\right)^{2}} \\
& =12 \sqrt{\left(\frac{x_{1}^{2}}{9}+\frac{y_{1}^{2}}{4}\right)\left(\frac{x_{2}^{2}}{9}+\frac{y_{2}^{2}}{4}\right)-\left(\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}\right)^{2}} \\
& =12 \sqrt{1-\left(\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}\right)^{2}} \leqslant 12 .
\end{aligned}
\end{array}
$$
Equality holds if and only if $\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}=0$. Given $x_{1} x_{2}+$ $y_{1} y_{2}=0$, it follows that $x_{1} x_{2}=0, y_{1} y_{2}=0$. In this case, points $A$ and $B$ are the endpoints of the major and minor axes of the ellipse, respectively.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given a pyramid $S-ABCD$ inscribed in a sphere with the base being a rectangle $ABCD$, and $SA=4, SB=8, SD=7$, $\angle SAC=\angle SBC=\angle SDC$. Then the length of $BD$ is
|
5. $B D=9$.
As shown in the figure, $\because \angle S A C=\angle S B C=\angle S D C$,
$$
\therefore \frac{S C}{\sin \angle S A C}=\frac{S C}{\sin \angle S B C}=\frac{S C}{\sin \angle S D C} \text {. }
$$
Therefore, the circumcircles of $\triangle S A C$,
$\triangle S B C, \triangle S D C$ have equal diameters, meaning the circles formed by the intersection of the sphere with the planes $S A C$, plane $S B C$,
plane $S D C$ are equal.
The plane passing through the midpoint $O$ of $S C$ and perpendicular to $S C$ intersects $S A C, S B C, S D C$ at the midpoints of the arcs $A^{\prime} 、 B^{\prime} 、 D^{\prime}$, and $O A^{\prime}=O B^{\prime}=O D^{\prime}$, thus, $O$ is the circumcenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Hence, $S C$ is the diameter of the sphere, $O$ is the center of the sphere, and $\angle \mathrm{SAC}=\angle \mathrm{SBC}=\angle \mathrm{SDC}=90^{\circ}$.
Let $B D=A C=x, A B=y, B C=z$. Then
$$
S C^{2}=x^{2}+4^{2}=y^{2}+7^{2}=z^{2}+8^{2} \text {, }
$$
which simplifies to $2\left(x^{2}+16\right)=\left(y^{2}+49\right)+\left(z^{2}+64\right)$
$$
=x^{2}+49+64 \text {. }
$$
Solving this, we get $x=9$, i.e., $B D=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 20 points) The complex numbers $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}$ satisfy
$$
\left\{\begin{array}{l}
\left|z_{1}\right| \leqslant 1, \\
\left|z_{2}\right| \leqslant 1, \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{5}-\left(z_{3}+z_{4}\right)\right| \leqslant\left|z_{3}-z_{4}\right| .
\end{array}\right.
$$
Find the maximum value of $\left|z_{5}\right|$.
|
$$
\begin{array}{l}
\text { 3, } \because\left|z_{1}-z_{2}\right| \geqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \\
\quad \geqslant|2| z_{3}|-| z_{1}+z_{2}||, \\
\therefore\left|z_{1}+z_{2}\right|-\left|z_{1}-z_{2}\right| \leqslant 2\left|z_{3}\right| \\
\quad \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right| . \\
\therefore\left|z_{3}\right| \leqslant \frac{\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right|}{2} \\
\quad \leqslant \sqrt{\frac{\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}}{2}} \\
=\sqrt{\frac{2\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}}{2}} \leqslant \sqrt{2} .
\end{array}
$$
Similarly, $\left|z_{4}\right| \leqslant \sqrt{2}$.
$$
\therefore\left|z_{5}\right| \leqslant \sqrt{\left|z_{3}\right|^{2}+\left|z_{4}\right|^{2}} \leqslant 2 \text {. }
$$
The equality in the above formula can sometimes be achieved. For example, when $z_{1}=1$, $z_{2}=$, $z_{3}=1+i$; and when $z_{1}=-1, z_{2}=i$, $z_{4}=-1+i$; when $z_{3}=1+i, z_{4}=-1+i$, $z_{5}=2i$, at this time $\left|z_{5}\right|=2$.
Therefore, $\left|z_{5}\right|_{\max }=2$.
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|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all be natural numbers, and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$. Try to find the maximum value of $x_{5}$.
|
Solution: Without loss of generality, let $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
$$
\because x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} \text {. }
$$
$x_{4} x_{5}$,
$$
\begin{aligned}
\therefore 1= & \frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
& +\frac{1}{x_{1} x_{2} x_{4} x_{5}}+\frac{1}{x_{1} x_{2} x_{3} x_{5}} \\
& +\frac{1}{x_{1} x_{2} x_{3} x_{4}} \\
\leqslant & \frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{5}}+\frac{1}{x_{4}} \\
& =\frac{3+x_{4}+x_{5}}{x_{4} x_{5}} .
\end{aligned}
$$
Thus, $x_{4} x_{5} \leqslant 3+x_{4}+x_{5}$.
Therefore, $\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4$.
If $x_{4}=1$, then $x_{1}=x_{2}=x_{3}=x_{4}=1$. From the given, we have $4+x_{5}=x_{5}$, which is a contradiction. Therefore, $x_{4} \geqslant 2$. Then $x_{5}-1 \leqslant\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4, x_{5} \leqslant 5$.
When $x_{5}=5$, there exist $x_{1}=x_{2}=x_{3}=1$, $x_{4}=2$ that satisfy the equation.
Therefore, the maximum value of $x_{5}$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If $x y=1$, then the minimum value of the algebraic expression $\frac{1}{4}+\frac{1}{4 y^{4}}$ is $\qquad$
|
Solution: From $a+b \geqslant 2 \sqrt{a b}$ we know $\frac{1}{x^{4}}+\frac{1}{4 y^{4}} \geqslant 2 \sqrt{\frac{1}{x^{4} \cdot 4 y^{4}}}=1$. Therefore, the minimum value of $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example: $3 A, B, C, D, E$ five people participate in an exam, with 7 questions, all of which are true or false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 1 records the answers of $A, B, C, D, E$ five people. It is known that $A, B, C, D$ each scored 2 points, how many points should $E$ get? What is the answer to each question?
保留源文本的换行和格式，直接输出翻译结果如下：
Example: $3 A, B, C, D, E$ five people participate in an exam, with 7 questions, all of which are true or false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 1 records the answers of $A, B, C, D, E$ five people. It is known that $A, B, C, D$ each scored 2 points, how many points should $E$ get? What is the answer to each question?
|
Let: Assign $k=1,2, \cdots, 7$. When the conclusion of the $k$-th question is correct, i.e., $x_{k}:=1$, if it is judged as correct (i.e., marked with the symbol “$\checkmark$”), then $x_{k}$ points are scored; if it is judged as incorrect (i.e., marked with the symbol “$X$”), then $-x_{k}$ points are scored. When the conclusion of the $k$-th question is incorrect, i.e., $x_{k}=-1$, if it is judged as correct, then $x_{k}$ points are scored; if it is judged as incorrect, then $-x_{k}$ points are scored. Since $A, B, C, D$ each scored 2 points, we can obtain the system of equations:
$$
\left\{\begin{array}{l}
x_{1}+0 \cdot x_{2}-x_{3}+x_{4}-x_{5}+x_{6}+x_{7}=2, \\
x_{1}-x_{2}+x_{3}+x_{4}-x_{5}-x_{6}+0 \cdot x_{7}=2, \\
0 \cdot x_{1}+x_{2}-x_{3}-x_{4}+x_{5}-x_{6}+x_{7}=2, \\
-x_{1}-x_{2}-x_{3}+x_{4}+x_{5}+0 \cdot x_{6}-x_{7}=2
\end{array}\right.
$$
Adding these four equations, we get
$$
x_{1}-x_{2}-2 x_{3}+2 x_{4}+0 \cdot x_{5}-x_{6}-x_{7}=8.
$$
Noting that $x_{i}= \pm 1(i=1,2, \cdots, 7)$, the left side of the above equation is $\leqslant 8$, and the right side is $=8$, hence $x_{1}=1, x_{2}=-1$, $x_{3}=-1, x_{4}=1, x_{6}=-1, x_{7}=1$. Substituting these results into the first equation of the system, we get $x_{5}=1$. Therefore, the 1st, 4th, 5th, and 7th questions are correct, and the 2nd, 3rd, and 6th questions are incorrect. Thus, according to the problem, $E$ scored 4 points.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given that $x, y, z$ are all positive numbers, and $x y z \cdot (x+y+z)=1$. Then, the minimum value of $(x+y)(y+z)$ is . $\qquad$
|
Analysis: To find the minimum value of the original expression, that is, to find the minimum value of $x y+y^{2}+$ $x z+y z$, we can achieve this by appropriately combining terms so that the sum becomes the sum of two terms, and the product of these two terms should be a constant.
$$
\text { Solution: } \begin{array}{l}
\because(x+y)(y+z)=\left(x y+y^{2}+y z\right) \\
+x z \\
= y(x+y+z)+x z, \\
\text { and } x z>0, y(x+y+z)>0, \\
\therefore y(x+y+z)+x z \\
\geqslant 2 \sqrt{y(x+y+z) \cdot x z}=2, \\
\therefore(x+y)(y+z) \text { has a minimum value of } 2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ in the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by segments $PQ, PR$ is $S$. Then the maximum distance from point $S$ to the origin is $\qquad$.
|
2.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
w+z=2 \bar{z}+(1+i) z \text{, i.e., } w=2 \bar{z}+i z \text{. }
$$
Therefore, $|w|^{2}=(2 \bar{z}+i z)(2 z-i \bar{z})$
$$
\begin{array}{l}
=4+1+2 i\left(z^{2}-\bar{z}^{2}\right) \\
=5-4 \sin 2 A<5 \div 4=9 .
\end{array}
$$
(When $\theta=135^{\circ}$, equality holds.)
Hence, $|w|_{\operatorname{max}}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In an arithmetic sequence with real number terms, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most terms.
|
4.8 .
Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then
$$
\begin{aligned}
& a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2} \cdot(n-1) \\
& \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{aligned}
$$
There exists at least one real number $a_{1}$ satisfying inequality (1) if and only if $\Delta \dot{=}^{\prime}(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0$. Since $\Delta \geqslant 0 \Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0$
$$
\Leftrightarrow n_{1} \leqslant n \leqslant n_{2} \text {, }
$$
where $n_{1}=\frac{3-\sqrt{2816}}{7}<0,8<n_{2}=\frac{3+\sqrt{2816}}{7}<9$.
Therefore, the maximum natural number $n$ satisfying inequality (2) is 8, meaning the sequence can have at most 8 terms.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$
(1993, National Junior High School Competition)
|
Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$. Then,
$$
\left(3-\frac{1}{2} y\right) x^{2}+(6-y) x+(5-y)=0 \text {. }
$$
Since $x$ is a real number, $\Delta \geqslant 0$, so,
$$
y^{2}-10 y+24 \leqslant 0 \text {. }
$$
Thus, $4 \leqslant y \leqslant 6$.
When $y=4$, $x=1$.
Therefore, when $x=1$, the minimum value of the original expression is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Four cities each send 3 political advisors to participate in $k$ group inspection activities (each advisor can participate in several groups), with the rules: (1) advisors from the same city are not in the same group; (2) any two advisors from different cities exactly participate in one activity together. Then the minimum value of $k$ is $\qquad$ .
|
2.9 Group.
First, consider the CPPCC members from two cities, Jia and Yi. Let the members from Jia city be $A_{1}, A_{2}, A_{3}$, and the members from Yi city be $B_{1}, B_{2}, B_{3}$. They can form 9 pairs: $A_{1} B_{1}, A_{1} B_{2}, A_{1} B_{3}, A_{2} B_{1}, A_{2} B_{2}, A_{2} B_{3}$, $A_{3} B_{1}, A_{3} B_{2}, A_{3} B_{3}$. According to the problem, the number of groups $k$ is no less than 9. To minimize the number of groups, the number of people in each group should be as large as possible (up to 4 people), and the following nine groups can be arranged to meet the requirements:
$$
\begin{array}{ll}
\left(A_{1}, B_{1}, C_{1}, D_{1}\right) & \left(A_{1}, B_{2}, C_{2}, D_{2}\right) \\
\left(A_{1}, B_{3}, C_{3}, D_{3}\right) & \left(A_{2}, B_{1}, C_{2}, D_{3}\right) \\
\left(A_{2}, B_{2}, C_{3}, D_{1}\right) & \left(A_{2}, B_{3}, C_{1}, D_{2}\right) \\
\left(A_{3}, B_{1}, C_{3}, D_{2}\right) & \left(A_{3}, B_{2}, C_{1}, D_{3}\right) \\
\left(A_{3}, B_{3}, C_{2}, D_{1}\right) &
\end{array}
$$
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In $\triangle A B C$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a^{2}+b^{2}-7 c^{2}=0$, then $\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=$
|
6.3 .
$$
\begin{array}{l}
\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\cos C \sin A \sin B}{\sin (A+B) \sin C} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}} \\
=\frac{7 c^{2}-c^{2}}{2 c^{2}}=3 .
\end{array}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{c}
\text { II. (50 points) }(1) \text { For } 0 \leqslant x \leqslant 1, \text { find the range of the function } \\
h(x)=(\sqrt{1+x}+\sqrt{1-x}+2) . \\
\left(\sqrt{1-x^{2}}+1\right)
\end{array}
$$
(2) Prove: For $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequality
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{a}}{\beta}
$$
holds for the smallest positive number $\alpha=2$. And find the smallest positive number $\beta$ at this time.
|
$$
\begin{array}{l}
=(1) 00$, the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{a}}{\beta}(x \in[0,1])$ does not hold. Conversely, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{0}}{\beta}$, which means $x^{2 \cdots} \geqslant \frac{h(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get
$$
0 \geqslant \frac{h(0)}{2 \beta} \text {. }
$$
Estimate $h(0)=8$.
This is impossible. This indicates that $\alpha=2$ is the smallest positive number that satisfies the condition.
Find the smallest $\beta>0$ such that the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{2}}{\beta}(x \in[0,1])$, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{2}}{\beta}$
holds, which is equivalent to finding
$$
\beta=\max _{0 \leqslant x \leqslant 1} \frac{1}{2} h(x) \text {. }
$$
$\because$ For any real numbers $u, v$, we have
$$
\sqrt{u}+\sqrt{v} \leqslant \sqrt{2} \bar{u}+v) \text {. }
$$
Let $u=1+x, v=1-x$, we get
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2 .
$$
Thus, when $x \in[0,1]$,
$$
\begin{aligned}
\frac{1}{2} h(x) & =\frac{1}{2}(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right) \\
& \leqslant 2\left(\sqrt{1-x^{2}}+1\right) \leqslant 4 .
\end{aligned}
$$
And $\frac{1}{2} h(0)=4$. Therefore, the maximum value of the function $\frac{1}{2} h(x)$ on $[0,1]$ is 4, i.e., the smallest positive number $\beta$ that satisfies condition (2) is 4.
$$
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For a constant $p \in N$, if the indeterminate equation $x^{2}+y^{2}=$ $p(x y-1)$ has positive integer solutions, prove that $p=5$ must hold.
|
Proof: Let $x_{0}, y_{0}$ be positive integer solutions of the equation. If $x_{0}=y_{0}$, substituting gives $p=(p-2) x_{0}^{2}$.
$$
\therefore x_{0}^{2}=\frac{p}{p-2}=1+\frac{2}{p-2} \in \mathbb{N} \text { : }
$$
Then $x_{0}^{2}=2$ or 3, which contradicts $x_{0} \in \mathbb{N}$.
Assume $x_{0}>y_{0} \geqslant 2$. Consider the original equation as a quadratic equation in $x$: $x^{2} - p y_{0} x + y_{0}^{2} + p = 0$. Its roots $x_{1}, x_{0}$ satisfy $x_{1} \geqslant x_{0} > y_{0} \geqslant 2$, and
$$
\begin{array}{l}
x_{1} + x_{0} = p y_{0}, \\
x_{1} x_{0} = y_{0}^{2} + p .
\end{array}
$$
It is easy to prove: $x_{1} x_{0} > 2 x_{1} \geqslant x_{1} + x_{0}$.
$$
\therefore y_{0}^{2} + p > p y_{0} \text {. }
$$
Thus, $py_{0} + 1$, which is a contradiction.
\end{array}
\end{array}
$$
In summary, it must be that $x_{0} > y_{0} = 1$. Substituting into the equation gives
$$
\begin{array}{l}
x_{0}^{2} - p x_{0} + p + 1 = 0 . \\
\therefore p = \frac{x_{0}^{2} + 1}{x_{0} - 1} = x_{0} + 1 + \frac{2}{x_{0} - 1} .
\end{array}
$$
When $x_{0} = 2$ or 3, $p = 5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 15 Find the minimum value of $\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Solution: Construct right triangles $\triangle P A C$ and $\triangle P B I$ as shown in Figure 1, such that
$$
\begin{array}{l}
A C=1, B D=2, P C \\
=x, C I=4, \text { and }
\end{array}
$$
$P C$ and $P D$ lie on line $l$. Then the problem of finding the minimum value is converted to "finding a point $I$ on line $l$ such that $P A+P B$ is minimized."
Take the symmetric point $A^{\prime}$ of $A$ with respect to $l$. Clearly,
$$
\text { the original expression }=P A+P B \geqslant A^{\prime} B=5 \text {. }
$$
Therefore, the required minimum value is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$
|
Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l}
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$
i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3 b-2$
$$
\begin{array}{l}
=(a+3)(b-4)+10 \\
=\sqrt{13} \times(-\sqrt{13})+10=-3 .
\end{array}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Simplify $\frac{2 b-a-c}{(a-b)(b-c)}$
$$
+\frac{2 c-a-b}{(b-c)(c-a)}+\frac{2 a-b-c}{(c-a)(a-b)} .
$$
|
$\begin{array}{l}\text { Solution: Original expression }=\frac{(b-c)-(a-b)}{(a-b)(b-c)} \\ \quad+\frac{(c-a)-(b-c)}{(b-c)(c-a)}+\frac{(a-b)-(c-a)}{(c-a)(a-b)} \\ =\frac{1}{a-b}-\frac{1}{b-c}+\frac{1}{b-c}-\frac{1}{c-a} \\ \quad+\frac{1}{c-a}-\frac{1}{a-b}=0 .\end{array}$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 If $x=\sqrt{19-8 \sqrt{3}}$. Then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$
Preserve the original text's line breaks and format, output the translation result directly.
|
Given: From the above, $x=\sqrt{(4-\sqrt{3})^{2}}=4-\sqrt{3}$, thus $x-4=-\sqrt{3}$.
Squaring and simplifying the above equation yields
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore, the denominator $=\left(x^{2}-8 x+13\right)+2=2$.
By long division, we get
the numerator $=\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10$.
Substituting (1) into the above equation gives
the numerator $=10$.
Thus, the original fraction $=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For a $4 n+2$-sided polygon $A_{1} A_{2} \cdots A_{4 n-2}$ (where $n$ is a natural number), each interior angle is an integer multiple of $30^{\circ}$, and $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$, then the possible values of $n$ are $\qquad$
|
4. 1 .
From the problem, we get
$$
\begin{array}{l}
90^{\circ} \times 3+(4 n+2-3) \times 30^{\circ} \mathrm{k} \\
=(4 n+2-2) \times 180^{\circ},
\end{array}
$$
where $k \geqslant 4 n-1$.
Simplifying, we get $(4 n-1) k=24 n-9$,
which means $24 n-9 \geqslant(4 n-1)^{2}$.
Thus, $8 n^{2}-16 n+5 \leqslant 0$.
Clearly, $0<n<2$, so the integer $n$ can only be:
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The solution set of the inequality $\frac{1}{x-1}+\frac{2}{x-2} \geqslant \frac{3}{2}$, is the union of some non-overlapping intervals with a total length of $\qquad$.
|
5.2.
For $g(x)=(x-1)(x-2)>($ or $\text { (or } \leqslant \text { ) } 0 \text {. }$
The graph of $y=f(x)$ is a parabola opening upwards, and the graph of $y=$ $g(x)$ is a parabola opening upwards.
For $f(x):$ when $x=1$, $y0$.
Therefore, the graph intersects the
$x$-axis at two points
$$
\begin{array}{l}
\left(x_{1}, 0\right),\left(x_{2}, 0\right) \text { and } \\
12,
\end{array}
$$
as shown in Figure 6.
The solution to the inequality is
the set of $x$ values for which both functions are either both positive or both negative,
thus the solution set is $\left(1, x_{1}\right] ;\left(2, x_{2}\right]$.
The total length $L$ of the solution set is
$$
L=\left(x_{1}-1\right)+\left(x_{2}-2\right)=\left(x_{1}+x_{2}\right)-(1+2) \text {. }
$$
Given $x_{1}+x_{2}=1+2+\frac{2}{3}(1+2)$,
$$
\therefore L=\frac{2}{3}(1+2)=2 \text {. }
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, for what real number $x$ does $y=x^{2}-x+1+$ $\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}}$ have a minimum value? What is the minimum value?
|
As shown in Figure 5, let \( P\left(x, x^{2}\right) \) be a point on the parabola \( y=x^{2} \). The distance from \( P \) to the line \( y=x-1 \) is
\[
\begin{aligned}
|P Q| & =\frac{\left|x^{2}-x+1\right|}{\sqrt{(-1)^{2}+1^{2}}} \\
& =\frac{\sqrt{2}}{2} \left| x^{2}-x+1 \right| \\
& =\frac{\sqrt{2}}{2}\left(x^{2}-x+1\right).
\end{aligned}
\]
The distance from \( P \) to \( M(-3,5) \) is
\[
|P M|=\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}}.
\]
When \( |P Q|+|P M| \) is minimized, points \( M, P, Q \) are collinear,
and this line is the line through \( M \) and perpendicular to the line \( y=x-1 \), which is \( y=-x+2 \).
The minimum value of \( |P Q|+|P M| \) is the distance from point \( M \) to the line \( y=x-1 \): \(\frac{|-3-5-1|}{\sqrt{1^{2}+(-1)^{2}}}=\frac{9}{2} \sqrt{2}\).
Since
\[
\begin{aligned}
y & =x^{2}-x+1+\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}} \\
& =\sqrt{2}\left[\frac{\sqrt{2}}{2}\left(x^{2}-x+1\right)+\right. \\
& \left.\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}}\right] \\
& = \sqrt{2}(|P Q|+|P M|).
\end{aligned}
\]
And since the minimum value of \( |P Q|+|P M| \) is \(\frac{9}{2} \sqrt{2}\),
the minimum value of \( y \) is 9.
Since the line \( y=-x+2 \) intersects the parabola \( y=x^{2} \) at points \( P_{1}(-2,4) \) and \( P_{2}(1,1) \),
when \( x=-2 \) or 1, \( y \) has a minimum value of 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most could have bought the book that was purchased by the fewest people?
(1993, China Mathematical Olympiad)
|
Solution: Let's assume that person A bought three books, and since A has at least one book in common with each of the other 9 people, among A's three books, the book purchased by the most people is bought by no less than 4 people.
If the book purchased by the most people is bought by 4 people, then all three books bought by A are each purchased by 4 people, and each book bought by the other 9 people is also purchased by 4 people. Therefore, the total number of books bought by the 10 people is a multiple of 4. That is, $4 \mid 30$, which is a contradiction. Thus, the book purchased by the most people is bought by at least 5 people.
Consider the following purchasing method:
$$
\begin{array}{l}
\left\{B_{1}, B_{2}, B_{3}\right\},\left\{B_{1}, B_{2}, B_{4}\right\},\left\{B_{2}, B_{3}, B_{5}\right\}, \\
\left\{B_{1}, B_{3}, B_{6}\right\},\left\{B_{1}, B_{4}, B_{5}\right\},\left\{B_{2}, B_{4}, B_{6}\right\}, \\
\left\{B_{3}, B_{4}, B_{5}\right\},\left\{B_{1}, B_{5}, B_{6}\right\},\left\{B_{2}, B_{5}, B_{6}\right\}, \\
\left\{B_{3}, B_{4}, B_{6}\right\},
\end{array}
$$
It can be seen that the book purchased by the most people is bought by at least 5 people.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $S=\{1,2,3,4\}$. An $n$-term sequence: $q_{1}, q_{2}, \cdots, q_{n}$ has the following property: For any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted by $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shanghai High School Mathematics Competition)
|
Solution: First, each number in $S$ appears at least twice in the sequence $q_{1}, q_{2}$, $\cdots, q_{n}$, otherwise, since there are 3 binary subsets containing a certain number, but in the sequence, the number of adjacent pairs containing this number is at most 2, therefore, $n \geqslant 8$.
Moreover, the 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition exactly, hence the minimum value of $n$ is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For $n \in \mathbf{N}$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=17$. If there exists a unique $n$ such that $S_{n}$ is also an integer, find the value of $n$.
|
Solution: Based on the structure of $\sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we construct the complex number $z_{k}=(2 k-1)+\sigma_{k} i, k=1,2, \cdots, n$.
Then $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-i)+a_{k} i\right|$.
Thus $\sum_{k=1}^{n} \vee \sqrt{(2 k-1)^{2}+a_{k}^{2}}$
$=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right|$
$=\sum_{k=1}^{n}\left|z_{k}\right| \geqslant\left|\sum_{k=1}^{n} z_{k}\right|$
$=\mid[1+3+\cdots+(2 n-1)]$
$+\left(a_{1}+a_{2}+\cdots+a_{n}\right) i \mid$
$=\left|n^{2}+17 i\right|=\sqrt{n^{4}+17^{2}}$.
Given that $S_{n}$ is the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we have
$$
S_{n}=\sqrt{n^{4}+17^{2}} \text {. }
$$
Since $S_{n}$ is an integer, let $S_{n}=m$. Then we have
$$
\begin{array}{l}
n^{4}+17^{2}=m^{2}, \\
\left(m-n^{2}\right)\left(m+n^{2}\right)=289 . \\
\therefore m-n^{2}=1, m+n^{2}=289 .
\end{array}
$$
Solving these equations, we get $n=12$.
2 Using $|z|^{2}=z \bar{z}$, convert the modulus operation to complex number operations.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 The elements of set $A$ are all integers, the smallest of which is 1, and the largest is 100. Except for 1, each element is equal to the sum of two numbers (which can be the same) in set $A$. Find the minimum number of elements in set $A$.
|
Solution: Construct a set with as many elements as possible to satisfy the conditions, such as $\{1,2,3,5,10,20,25,50, 9\}$.
Extend $\left\{1,2, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 100\right\}$ to also satisfy the conditions, then $x_{1} \leqslant 4, x_{2} \leqslant 8, x_{3} \leqslant 16, x_{4} \leqslant 32, x_{5}$ $\leqslant 64$.
$$
\text { but } \begin{array}{l}
x_{4}+x_{5} \leqslant 96<100, \therefore x_{5}=50, \\
x_{3}+x_{4} \leqslant 48<50, \therefore x_{4}=25, \\
x_{2}+x_{3} \leqslant 24<25, \therefore 25=2 x_{3} .
\end{array}
$$
Contradiction.
Therefore, the minimum number of elements in $A$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 For a finite set $A$, there exists a function $f: N \rightarrow$ $A$, with the following property: if $i, j \in N$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least?
|
Solution: Since the absolute value of the difference between any two numbers among $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, and $f(8)$ are four distinct elements in $A$, thus $|A| \geqslant 4$.
On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: N \rightarrow A$ is defined as: if $x \in N, x=4 k+r$, then $f(x)=r$, where $k \in N \cup\{0\}, r=0,1,2,3$. For any $x, y \in N$, if $|x-y|$ is a prime number, assume $f(x) = f(y)$, then $x \equiv y(\bmod 4)$, thus $4 \mid x-y\rfloor$, which contradicts the fact that $|x-y|$ is a prime number.
Therefore, $A$ contains at least 4 distinct elements.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is $\qquad$ .
|
Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, then we have
$$
\left\{\begin{array}{l}
5^{y}=3^{x}+4^{x}, \\
4^{y}=5^{x}-3^{x}
\end{array}\right.
$$
(1) + (2) gives $5^{y}+4^{y}=5^{x}+4^{x}$.
However, the function $f(x)=5^{x}+4^{x}$ is an increasing function, so from (3) we have
$$
f(y)=f(x) \Leftrightarrow y=x .
$$
Substituting into (1) or (2) gives
$$
5^{x}=3^{x}+4^{x} \text {, }
$$
which simplifies to $\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}$.
Since the function $g(x)=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$ is a decreasing function, we have
$$
g(x)=g(2) \Leftrightarrow x=2 \text {. }
$$
Thus, from the above, we get $x=2$.
Substituting back into the original equation for verification, the solution set of the equation is $\{2\}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the pure imaginary numbers $x_{1}, x_{2}, \cdots, x_{1999}$ have a modulus of 1. Then the remainder when $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1998} x_{1999}+x_{1999} x_{1}$ is divided by 4 is $\qquad$
|
4.1.
Obviously, if $x_{k}$ takes $i$ or $-i$, we have
$$
\begin{array}{l}
\left(i+x_{k}\right)\left(i-x_{k+1}\right) \\
=\left\{\begin{array}{l}
0, \text{ when } x_{k}=-i \text{ or } x_{k+1}=i; \\
-4, \text{ when } x_{k}=i \text{ and } x_{k+1}=-i
\end{array}\right.
\end{array}
$$
Both are multiples of 4, so the sum is also a multiple of 4. Therefore,
$$
\begin{aligned}
4 m= & \left(i+x_{1}\right)\left(i-x_{2}\right)+\left(i+x_{2}\right)\left(i-x_{3}\right) \\
& +\cdots+\left(i+x_{1999}\right)\left(i-x_{1}\right) \\
= & -1999+\left(x_{1}+x_{2}+\cdots+x_{1999}\right) i \\
& -\left(x_{1}+x_{2}+\cdots+x_{1999}\right) i \\
& -\left(x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1999} x_{1}\right) .
\end{aligned}
$$
Thus,
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1999} x_{1} \\
=-4 m-1999=4(-m-500)+1 .
\end{array}
$$
It is evident that the above expression, when divided by 4, leaves a remainder of 1.
Note: As a fill-in-the-blank question, we can take $x_{k}=(-1)^{k-1} i$, then
$$
\text { original expression }=1+1+\cdots+1-1=1997=4 \times 499+1 \text {. }
$$
Or take $x_{1}=x_{2}=\cdots=x_{1999}=i$, then
$$
\text { original expression }=-1999=4 \times(-500)+1 \text {. }
$$
From this, we conjecture that the original expression, when divided by 4, leaves a remainder of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Using $1,2, \cdots, n$ to form an $n$-digit number without repeating digits, where 2 cannot be adjacent to 1 or 3, a total of 2400 different $n$-digit numbers are obtained. Then $n=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
6.7 .
Obviously, $n \geqslant 4$, otherwise 2 must be adjacent to 1 or 3.
When 2 is adjacent to 1, there are $2 \cdot(n-1)$! arrangements, and when 2 is adjacent to 3, there are also $2 \cdot(n-1)$! arrangements. When 2 is adjacent to both 1 and 3, there are $2 \cdot(n-2)$! arrangements. Therefore, the number of arrangements that meet the conditions is
$$
\begin{array}{l}
n!-2 \cdot(n-1)!-2 \cdot(n-1)!+2 \cdot(n-2)! \\
=(n-2)!(n-2)(n-3) \\
=2400<7!.
\end{array}
$$
Moreover, $(n-2)$! $<7$!.
Thus, $n-2<7$. Therefore, $4 \leqslant n<9$.
Substituting $n=4,5,6,7,8$ into (1) in sequence, only when $n=7$
$$
5!\times 5 \times 4=2400
$$
satisfies the condition, hence $n=7$.
|
7
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (Full marks 20 points) As shown in Figure 8, point $O$ represents the sun, $\triangle A B C$ represents a triangular sunshade, and points $A$ and $B$ are two fixed points on the ground in the north-south direction. The sunlight $O C D$ from the due west direction casts the shadow of the sunshade onto the ground, forming the shadow $\triangle A B D$. It is known that the sunlight $O C D$ makes an acute angle $\theta$ with the ground.
(1) At what angle should the sunshade make with the ground to maximize the area of the shadow $\triangle A B D$?
(2) When $A C=3, B C=4, A B=5, \theta=30^{\circ}$, find the maximum area of the shadow $\triangle A B D$.
|
(1) As shown in Figure 17, draw the perpendicular line $O H$ from $O$ to the ground, connect $H D$ to intersect $A B$ at $E$, and connect $C E$. Then $H D$ is the projection of the oblique line $O D$ on the ground, and $\angle C D E = \theta$. Given that $A B$ is in the north-south direction and $C D$ is in the east-west direction, we have
$A B \perp C D$. (Implicit condition)
According to the inverse of the Three Perpendiculars Theorem, $D E \perp A B$. By the property theorem of line perpendicular to a plane, $A B \perp O H$. This implies that $A B$ is perpendicular to the intersecting lines $O H$ and $O D$, so $A B$ is perpendicular to the plane $C D E$, and thus,
$A B \perp C E$.
Hence, $C E$ is the altitude of $\triangle A B C$ on side $A B$, and $D E$ is the altitude of $\triangle A B D$ on side $A B$. The dihedral angle between the sunshade and the ground is $\angle C E D$.
In $\triangle A B D$, to maximize the area, with $A B$ and $\theta$ fixed, $D E$ should be maximized.
In $\triangle C E D$, we have
$$
\begin{aligned}
D E & =\frac{C E}{\sin \theta} \cdot \sin \angle C E D=\frac{2 S_{\triangle A B C}}{A B \sin \theta} \cdot \sin \angle C E D \\
& \leqslant \frac{2 S_{\triangle A B C}}{A B \sin \theta} \text { (a constant). }
\end{aligned}
$$
Therefore, when $\angle D C E = 90^{\circ}$ (the sunshade is perpendicular to the sunlight), $D E$ is maximized. Thus, when $\angle C E D = 90^{\circ} - \theta$, the area of the shadow $\triangle A B D$ is maximized.
(2) When $A C = 3$, $B C = 4$, $A B = 5$, and $\theta = 30^{\circ}$, $\triangle A B C$ is a right triangle, and $S_{\text {SMIX: }} = \frac{1}{2} A C \cdot B C = 6$.
Substituting (1) into the area formula, we get the maximum area as
$$
S_{\triangle A W B} = \frac{1}{2} A B \cdot D E = \frac{S_{\triangle A B C}}{\sin 30^{\circ}} = 12.
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$
|
$$
\begin{array}{l}
\because(b-c)^{2}=4(a-b)(c-a), \\
b^{2}-2 b c+c^{2}=4 a c-4 b c+4 a b-4 a^{2}, \\
\therefore(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
\end{array}
$$
Therefore, $[2 a-(b+c)]^{2}=0$, which means $2 a=b+c$.
$$
\therefore \frac{b+c}{a}=2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$ and $b$ are integers, and satisfy
$$
\left(\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}-\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\right)\left(\frac{1}{a}-\frac{1}{b}\right) \frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{2}{3} \text {. }
$$
Then $a+b=$ $\qquad$ .
|
3.3.
$$
\begin{array}{l}
\text { Left side }=\frac{a b}{a-b}=\frac{2}{3}, \\
\therefore(3 b-2)(3 a-2)=4 .
\end{array}
$$
Given that $a \neq b$ and they are integers, hence $3 b-2, 3 a-2$ can only take the values 1, 4 or -1, -4.
(1) Suppose $3 b-2=1, 3 a-2=4$.
Solving gives $b=1, a=2$. Therefore, $a+b=3$.
(2) Suppose $3 b-2=-1, 3 a-2=-4$.
This results in $a, b$ being fractions, which we discard.
Thus, $a+b=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the 5th term of the expansion of $\left(x \sqrt{x}-\frac{1}{x}\right)^{6}$ is $\frac{15}{2}$, then $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-a}\right)=$
|
$\begin{array}{l}\text { II, 1.1. } \\ \because T_{5}=C_{6}^{4}(x \sqrt{x})^{2} \cdot\left(-\frac{1}{x}\right)^{4}=\frac{15}{x}, \\ \text { from } \frac{15}{x}=\frac{15}{2} \text {, we get } x^{-1}=\frac{1}{2} . \\ \therefore \lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right) \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and
$$
\begin{array}{l}
f(x-y)=f(x) g(y)-g(x) f(y), f(-2) \\
=f(1) \neq 0, \text { then } g(1)+g(-1)=\ldots
\end{array}
$$
(Answer with a number).
|
4. -1 .
$$
\begin{array}{l}
\because f(x-y)=f(x) g(y)-g(x) f(y), \\
\therefore f(y-x)=f(y) g(x)-g(y) f(x) \\
\quad=-[f(x) g(y)-g(x) f(y)]
\end{array}
$$
There is $f(x-y)=-f(y-x)$
$$
=-f[-(x-y)] \text {. }
$$
Then $f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.
$$
\text { Hence } \begin{aligned}
f(1) & =f(-2)=f(-1-1) \\
& =f(-1) g(1)-g(-1) f(1) \\
& =-f(1)[g(1)+g(-1)] . \\
\therefore g(1) & +g(-1)=-1 .
\end{aligned}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: In a certain year, the total coal production of a coal mine, apart from a certain amount of coal used for civilian, export, and other non-industrial purposes each year, the rest is reserved for industrial use. According to the standard of industrial coal consumption of a certain industrial city in that year, it can supply 3 such industrial cities for 6 years, or 4 such industrial cities for 5 years (of course, the fixed amount of coal for non-industrial use is deducted each year). How many years can it supply if it only supplies the industrial coal for this one city?
(1978, Chongqing Mathematical Competition)
|
Solution: Let the total coal production of the mine for the year be $x$, the annual non-industrial coal quota be $y$, and the industrial coal consumption of each industrial city for the year be $z$. Let $p$ be the number of years the coal can supply only one city. According to the problem, we have the system of equations
$$
\left\{\begin{array}{l}
x=6 \times(3 z)+6 y, \\
x=5 \times(4 z)+5 y, \\
p=\frac{x-p y}{2},
\end{array}\right.
$$
which is $\left\{\begin{array}{l}x=18 z+6 y, \\ x=20 z+5 y, \\ p=\frac{x-p y}{2} .\end{array}\right.$
From (1) and (2), we get $y=2 z$.
Then, by eliminating $x$, $y$, and $z$ using equations (1), (3), and (4), we get
$p=30-2 p$. Solving this, we get
$p=10$.
Therefore, the coal can supply only one city for 10 years.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 12 points) At the foot of the mountain is a pond, the scene: a steady flow (i.e., the same amount of water flows into the pond from the river per unit time) continuously flows into the pond. The pond contains a certain depth of water. If one Type A water pump is used, it will take exactly 1 hour to pump out all the water in the pond; if two Type A water pumps are used, it will take exactly 20 minutes to pump out all the water in the pond. If three Type A water pumps are used simultaneously, how long will it take to pump out all the water in the pond exactly?
|
Three, let the amount of water flowing into the pond from the spring every minute be $x$ m$^{3}$, each water pump extracts $y$ m$^{3}$ of water per minute, and the pond originally contains $z$ m$^{3}$ of water. It takes $t$ minutes for three water pumps to drain the pond. According to the problem, we have
$$
\left\{\begin{array}{l}
60 y-60 x=z, \\
20 y+20 y-20 x=z, \\
3 t y-t x=z .
\end{array}\right.
$$
Solving this, we get $t=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given two points $A(0,1), B(6,9)$. If there is an integer point $C$ (Note: A point with both coordinates as integers is called an integer point), such that the area of $\triangle A B C$ is minimized. Then the minimum value of the area of $\triangle A B C$ is $\qquad$
|
5.1.
Since the slope of line $AB$ is $k=\frac{3}{4}$, the line passing through point $C$ and parallel to $AB$ can be expressed as $3 y-4 x=m$. Let the coordinates of point $C$ be $\left(x_{0}, y_{0}\right)$, where $x_{0}, y_{0}$ are integers. Therefore, we have
$$
3 y_{0}-4 x_{0}=m \text {. }
$$
Thus, $m$ is an integer.
The equation of line $AB$ is
$$
3 y-4 x-3=0 \text {. }
$$
The distance from point $C$ to line $AB$ is
$$
d=\frac{\left|3 y_{0}-4 x_{0}-3\right|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{1}{5}|m-3| \text {. }
$$
When $d$ is at its minimum, the area of $\triangle ABC$ is minimized.
At this time, $m=2$ or 4.
Thus, $d_{\text {min }}=\frac{1}{5}$.
$$
S_{\triangle A B C}=\frac{1}{2} \cdot A B \cdot d_{\min }=\frac{1}{2} \times 10 \times \frac{1}{5}=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) Given $0<\alpha_{i}<\frac{\pi}{4}(i=1$, $2,3,4)$, and $\sum_{i=1}^{4} \sin ^{2} \alpha_{i}=1$. Prove: $\sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 \alpha_{i}} \geqslant$ 2.
|
Let $a_{i}=\sin ^{2} \alpha_{i}$, then $0<a_{i}<\frac{1}{2}$,
$$
\begin{array}{l}
\cos 2 \alpha_{i}=1-2 \sin ^{2} \alpha_{i}=1-2 a_{i} . \\
\frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}}=\frac{a_{i}}{1-2 a_{i}}=2\left(\frac{a_{i}^{2}}{1-2 a_{i}}+\frac{a_{i}}{2}\right) \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\right)^{2}+2 a_{i}-\frac{1-2 a_{i}}{2}+a_{i} \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\right)^{2}+4 a_{i}-\frac{1}{2}
\end{array}
$$
$\geqslant 4 a_{i}-\frac{1}{2}$. (Equality holds if and only if $a_{i}=\frac{1}{4}$).
$$
\begin{array}{l}
\text { Hence } \sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}} \geqslant \sum_{i=1}^{4}\left(4 a_{i}-\frac{1}{2}\right) \\
=4 \sum_{i=1}^{4} a_{i}-2=2 \text {. } \\
\end{array}
$$
|
2
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $x=\frac{1}{\sqrt{3}+\sqrt{2}}, y=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then, $x^{2}+y^{2}$ is
untranslated part: 轩隹
Note: The term "轩隹" does not have a clear meaning in this context and has been left untranslated. If you can provide more context or clarify the term, I can attempt to translate it accurately.
|
$=, 7.10$.
Let $x=\sqrt{3}-\sqrt{2}, y=\sqrt{3}+\sqrt{2}$, so,
$$
x^{2}+y^{2}=(\sqrt{3}-\sqrt{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}=10 .
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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