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Example 5 A public bus company, after two technological innovations, adjusted the seating from New Year's Day 1957, allowing each bus to carry 6 more people, so that the number of people each 5 trips could carry exceeded 270. From New Year's Day 1958, a trailer was added, allowing each bus to carry 98 more people than in 1957, so that the number of people each 3 trips could carry exceeded the number each 8 trips could carry in 1957. How many times more people can each trip carry after New Year's Day 1958 compared to before New Year's Day 1957?
(1964, Chengdu City Senior High School Mathematics Competition)
|
Solution: Let the number of people each car can carry before New Year's Day 1957 be $x$, then on New Year's Day 1957 and 1958, each car can carry $x+6$ and $x+98$ people, respectively. According to the problem, we have $\left\{\begin{array}{l}5(x+6)>270, \\ 3(x+98)>8(x+6) .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}x>48, \\ x<49.2 .\end{array}\right.$
Therefore, $x=49$ (people).
Thus, on New Year's Day 1958, each car can carry $49+98=147$ (people). Hence, on New Year's Day 1958, each trip can carry twice as many people as before New Year's Day 1957.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. A pipe burst occurred in a low-lying area by the riverbank, and river water is continuously gushing out, assuming the water gushing out per minute is constant. If two water pumps are used to pump out the water, it takes 40 minutes to finish; if four water pumps are used, it takes 16 minutes to finish. If the water needs to be pumped out within 10 minutes, then at least $\qquad$ water pumps are required.
|
12.6 .
Let the amount of water that has already gushed out before the pumping starts be $u$ cubic meters, the rate of water gushing out per minute be $b$ cubic meters, and the amount of water each pump can extract per minute be $c$ cubic meters $(c \neq 0)$. From the given conditions, we have
$$
\left\{\begin{array}{l}
a+40 b-2 \times 40 a, \\
a: 16 b=4 \times 16
\end{array}\right\}\left\{\begin{array}{l}
a=\frac{160}{3} c, \\
b=\frac{2}{3}
\end{array}\right.
$$
The number of pumps is
$$
\frac{a+10 b}{10 c}=\frac{\frac{100}{3} c+\frac{20}{3} c}{10 c}=6 .
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, let $x, y$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, and satisfy the condition: when $x>y$, $x-y \geqslant \frac{x}{15}$. Find the maximum number of these natural numbers $n$.
|
$$
\begin{array}{l}
\text { Let's assume } a_{1}=1, d_{4} \geqslant 2 ; \\
a_{5}=a_{4}+d_{4} \geqslant 6, d_{5} \geqslant \frac{6^{2}}{19-6}>2, d_{5} \geqslant 3 ; \\
a_{6} \geqslant 9, d_{6} \geqslant \frac{9^{2}}{19-9}>8, d_{6} \geqslant 9 ; \\
a_{7} \geqslant 18, d_{7} \geqslant \frac{18^{2}}{19-18}=324 ; \\
a_{8} \geqslant 342>19 .
\end{array}
$$
From this, we can see that $n-1 \leqslant 7, n \leqslant 8$.
On the other hand, the 8 natural numbers $1,2,3,4,6,9,18,342$ meet the conditions of the problem.
Therefore, the maximum value of $n$ is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Suppose $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=$ 0 , and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is (1990, Hefei Junior High School Mathematics Competition)
|
Given $a^{2}+2 a-1=0$, we know $a \neq 0$.
$$
\therefore\left(\frac{1}{a}\right)^{2}-2\left(\frac{1}{a}\right)-1=0 \text {. }
$$
From $b^{4}-2 b^{2}-1=0$, we get
$$
\left(b^{2}\right)^{2}-2\left(b^{2}\right)-1=0 \text {. }
$$
From $1-a b^{2} \neq 0$, we know $\frac{1}{a} \neq b^{2}$.
From (1) and (2), we know $\frac{1}{a}$ and $b^{2}$ are the two roots of the equation $x^{2}-2 x-1=0$, so,
$$
\frac{1}{a}+b^{2}=2, \frac{b^{2}}{a}=-1 \text {. }
$$
Therefore, $\frac{a b^{2}+b^{2}+1}{a}=\left(\frac{1}{a}+b^{2}\right)+\frac{b^{2}}{a}$
$$
=2-1=1 \text {. }
$$
Thus, the value of the original expression is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in the figure, circle O is inscribed in $\triangle ABC$ touching sides $AC$, $BC$, and $AB$ at points $D$, $E$, and $F$ respectively, with $\angle C=90^{\circ}$. The area of $\triangle ABC$ is 6. Then $AF \cdot BF=$ $\qquad$
|
4.6 .
Let $A C=b, B C=a, A B=c, \odot O$ have a radius of $r$. Then
$$
\begin{array}{l}
r=\frac{a+b-c}{2} \cdot a^{2}+i^{2}=c^{2} . \\
\therefore A F \cdot B F=A C \cdot B E \\
=(b-r)(a-r) \\
=\frac{b+c-a}{2} \cdot \frac{a+c-b}{2} \\
=\frac{1}{4}\left[c^{2}-(a-b)^{2}\right] \\
=\frac{1}{2} a b=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{5}+1999(x-1)=2519, \\
(y-1)^{5}+1999(y-1)=-2519 .
\end{array}\right.
$$
Then the value of $x+y$ is $\qquad$
|
1.2.
Let $f(t)=t^{5}+1999 t$. Then $f(t)$ is increasing on $R$, and $f(x-1)=-f(y-1)$. Since $f(y-1)=-f(1-y)$, then
$$
\begin{array}{l}
f(x-1)=f(1-y) . \\
\therefore x-1=1-y \Rightarrow x+y=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. 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 $P, Q, R$ on 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
|
5.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
\begin{aligned}
w+z & =2 \bar{z}+(1+\mathrm{i}) z, \text { i.e., } w=2 \bar{z}+\mathrm{i} z . \\
\therefore|w|^{2} & =(2 \bar{z}+\mathrm{i} z)(2 z-\mathrm{i} \bar{z}) \\
& =4+1+2 \mathrm{i}\left(z^{2}-\bar{z}^{2}\right)=5-4 \sin 2 \theta \\
& \leqslant 5+4=9(\text { when } \theta=135 \text {, the equality holds }) .
\end{aligned}
$$
Thus, $|w|_{\text{max}}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) There is a quantity $W$, after "modeling" the relationship is given by
$$
W=\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1-t^{2}}}\right),
$$
where $a, b, c, u, t$ are all positive, $u<1, t<1$, and satisfy $a t+b u=c, a^{2}+2 b c u=b^{2}+c^{2}$. Please design a method to find the minimum value of the quantity $W$.
|
Given $a^{2}=b^{2}+c^{2}-2 b c u$. All $a, b, c, u$ are positive, and $ub^{2}+c^{2}-2 b c=(b-c)^{2}$.
From this, we get $|b-c|<a<b+c$.
Therefore, the positive numbers $a, b, c$ can be the lengths of the three sides of a triangle. Let the vertices opposite these sides be $A, B, C$, respectively. By transforming the constraint, we get
$$
u=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\cos A.
$$
Substituting into the other constraint, we get
$$
a t+b \cos A=c.
$$
Using the projection theorem $c=a \cos B+b \cos A$, we get
$$
t=\cos B.
$$
Since $u$ and $t$ are positive, $\angle A$ and $\angle B$ are acute angles, so,
$$
\begin{array}{l}
\sqrt{1-u^{2}}=\sqrt{1-\cos ^{2} A}=\sin A, \\
\sqrt{1-t^{2}}=\sqrt{1-\cos ^{2} B}=\sin B.
\end{array}
$$
Let the circumradius of $\triangle ABC$ be $R$, then
$$
\begin{array}{l}
a=2 R \sin A, b=2 R \sin B, c=2 R \sin C. \\
\begin{aligned}
\therefore W & =\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1}-\overline{t^{2}}}\right) \\
& =\frac{1}{2 R \sin C}\left(\frac{6 R \sin A}{2}+\frac{2 R \sin B}{\sin B}\right) \\
& =\frac{8 R}{2 R \sin C}=\frac{4}{\sin C} \geqslant 4.
\end{aligned}
\end{array}
$$
The equality holds if and only if $\angle C$ is a right angle, i.e., $a^{2}+b^{2}=c^{2}$. Therefore, the minimum value of $W$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $x_{1}, x_{2}, \cdots, x_{9}$ all be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is $\qquad$
$(1992$, National Junior High School Mathematics League)
|
Solution: From $x_{1}+x_{2}+\cdots+x_{9}=220$,
we know $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>110$,
or $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
From (1), then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant$ $27, x_{8} \geqslant 28, x_{9} \geqslant 29$. We get
$$
\begin{aligned}
\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)+\left(x_{6}+x_{7}+\right. \\
\left.x_{8}+x_{9}\right)>110+110=220 .
\end{aligned}
$$
This contradicts the given condition.
From (2), we can take $x_{1}=20, x_{2}=21, x_{3}=22$, $x_{4}=23, x_{5}=24$, so that $x_{1}+x_{2}+\cdots+x_{5}=110$ reaches its maximum value, and $x_{1}$ also reaches its maximum value. At this point, taking $x_{6}$ $=26, x_{7}=27, x_{8}=28, x_{9}=29$ satisfies all conditions. Therefore, the minimum value of $x_{9}-x_{1}$ is $29-20=9$.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $p^{3}+q^{3}=2$, where $p, q$ are real numbers. Then the maximum value of $p+q$ is $\qquad$
(1987, Jiangsu Province Junior High School Mathematics Competition)
|
Solution: Let $s=p+q$.
From $p^{3}+q^{3}=2$ we get
$(p+q)\left(p^{2}+q^{2}-p q\right)=2$,
$(p+q)^{2}-3 p q=\frac{2}{s}$,
Thus, $p q=\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)$.
From (1) and (2), $p$ and $q$ are the two real roots of the equation
$$
x^{2}-s x+\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)=0
$$
We know $\Delta=s^{2}+\frac{4}{s}\left(s^{2}-\frac{2}{s}\right) \geqslant 0$.
Simplifying, we get $s^{2} \leqslant 8$.
Therefore, $s \leqslant 2$.
Taking $p=q=1$, the maximum value of $p+q$ is 2.
4 Using the definition of absolute value
That is, $|x|= \pm x,|x|^{2}=x^{2}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $5 x, y, z$ are real numbers, and satisfy $x+y+z=0, xyz=2$. Find the minimum value of $|x|+|y|+|z|$.
(1990, Beijing Junior High School Mathematics Competition).
|
Solution: From the problem, we know that among $x, y, z$, there are 2 negative numbers and 1 positive number. Without loss of generality, let $x>0, y<0, z<0$. When $x>0, -y>0, -z>0$, we have
$(-y)(-z) \leqslant \left[\frac{(-y)+(-z)}{2}\right]^{2}=\frac{x^{2}}{4}$,
which means $2 x=\frac{4}{(-y)(-z)} \geqslant \frac{4}{\frac{(-y)+(-z)}{2}}$
$$
=\frac{16}{x^{2}}.
$$
Thus, $x^{3} \geqslant 8$, solving which gives $x \geqslant 2$.
Therefore, $|x|+|y|+|z|=2 x \geqslant 4$,
which means the minimum value of $|x|+|y|+|z|$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd Hope Cup for High School Grade 2)
|
Solution: Completing the square, we get
$$
u=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2 \text {. }
$$
Consider points $P_{1}\left(x, \frac{9}{x}\right)$, $P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate axes as asymptotes. When $y \in \mathbf{R},|y| \leqslant \sqrt{2}$, the trajectory of $P_{2}$ is the lower half of a circle centered at the origin with a radius of $\sqrt{2}$. Geometrically, it is evident that the minimum value of $\left|P_{1} P_{2}\right|$ is $3 \sqrt{2}-\sqrt{2}$, from which we can derive
$$
u_{\text {min }}=(3 \sqrt{2}-\sqrt{2})^{2}-2=6 \text {. }
$$
Note: A slight variation of this example can yield the 1983 Putnam Mathematical Competition problem for college students:
Find the minimum value of $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u}-\frac{9}{v}\right)^{2}$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 If $x, y$ are real numbers, find the minimum value of $S=5 x^{2}-4 x y$ $+y^{2}-10 x+6 y+5$.
|
Solution: From the given, we have
$$
\begin{array}{l}
5 x^{2}-(10+4 y) x+y^{2}+6 y+5-S=0 . \\
\Delta_{x}=-4\left(y^{2}+10 y-5 S\right) \geqslant 0 . \\
\text { Hence, } 5 S \geqslant y^{2}+10 y=(y+5)^{2}-25 \\
\quad \geqslant-25 .
\end{array}
$$
Thus, $5 S \geqslant y^{2}+10 y=(y+5)^{2}-25$
$\therefore S \geqslant-5$, so the minimum value of $S$ is -5.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Let $x, y$ be positive numbers, and $x+y=1$. Find the minimum value of the function $W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$. (Adapted from the 3rd Canadian Mathematical Competition)
|
Solution: Without loss of generality, let $x \geqslant y$, and set $x=\frac{1}{2}+t$, then $y=$
$$
\begin{array}{l}
\frac{1}{2}-t, 0 \leqslant t<\frac{1}{2} . \\
\therefore W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \\
=\frac{3+2 t}{1+2 t} \cdot \frac{3-2 t}{1-2 t}=\frac{9-4 t^{2}}{1-4 t^{2}} \\
\geqslant \frac{9-36 t^{2}}{1-4 t^{2}}=9 .
\end{array}
$$
When $t=0$, i.e., $x=y$, the minimum value sought is 9.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Find the smallest integer \( n (n \geq 4) \), such that from any \( n \) integers, four different numbers \( a, b, c, d \) can be selected, making \( a + b - c - d \) divisible by 20.
|
Solution: First, we consider the different residue classes modulo 20. For a set with $k$ elements, there are $\frac{1}{2} k(k - 1)$ integer pairs. If $\frac{1}{2} k(k - 1) > 20$, i.e., $k \geq 7$, then there exist two pairs $(a, b)$ and $(c, d)$ such that $a + b \equiv c + d \pmod{20}$, and $a, b, c, d$ are all distinct.
In general, consider a set of 9 different elements. If this set has 7 or more elements belonging to different residue classes modulo 20, then from the previous derivation, we can find four distinct numbers $a, b, c, d$ such that $a + b \equiv c + d \pmod{20}$. Assuming that in this set, at most six elements belong to the same residue class modulo 20, we can still find $a, b, c, d$ such that $a + b - c - d$ is divisible by 20.
For a set of 8 elements, such as
$$
\{0, 20, 40, 1, 2, 4, 7, 12\},
$$
we prove that it does not satisfy the required property. The residues of these numbers modulo 20 are $0, 0, 0, 1, 2, 4, 7, 12$. These residues have the property that each number is greater than the sum of any two smaller numbers, and the sum of any two numbers is less than 20. Let $a, b, c, d$ be four distinct numbers from this set, treating 20 and 40 as 0. Assume $a$ is the largest of the four chosen numbers. Then $a + b - c - d < a + b - c < a + b < 20$. Therefore, $a + b - c - d$ is not divisible by 20.
The minimum value of $n$ is 9.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $x$ be a real number, and $f(x)=|x+1|+|x+2|+|x+3|+|x+4|+|x+5|$. Find the minimum value of $f(x)$.
|
Analysis: According to the geometric meaning of absolute value, draw the points $A, B, C, D, E$ corresponding to the real numbers $-1, -2, -3, -4, -5$ on the number line, as shown in Figure 1. Let $x$ correspond to the moving point $P$, then $f(x)=|PA|+|PB|+|PC|+|PD|+|PE| \geqslant |CB|+|CD|+|CA|+|CE|=2+4=6$, that is, when point $P$ coincides with $C$, the sum is minimized to 6.
Therefore, when $x=-3$, the minimum value of the function $f(x)$ is
Figure 1 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given $f(x)=|1-2 x|, x \in[0$, 1]. How many real solutions does the equation $f(f(f(x)))=\frac{x}{2}$ have?
|
Solution: To find the number of real solutions using the graphical method. First, draw the graph of $f(x) = |2x - 1|$, and then double the y-coordinates of all points while keeping the x-coordinates unchanged. Next, shift the obtained graph down by 1 unit (as shown in Figure 2), and then reflect the part of the graph below the x-axis to above the x-axis, to get the graph of $f(f(x)) = |2f(x) - 1|$ (as shown in Figure 3).
Using the same method, we can draw the graph of the function $f(f(f(x))) = |2f(f(x)) - 1|$ (as shown in Figure 4), which intersects the line $y = \frac{x}{2}$ at 8 points in the interval $[0,1]$. Therefore, the original equation has 8 real solutions.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let the function $f_{3}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$
|
Analysis: First, draw the graph of $f_{0}(x)=|x|$ and translate it downward by 1 unit to get the graph of $y=f_{0}(x)-1$, from which we obtain the graph of $f_{1}(x)=\left|f_{0}(x)-1\right|$ (as shown in Figure 7).
Next, translate the graph of $f_{1}(x)$ downward by 2 units to get the graph of $y=f_{1}(x)-2$, and reflect the part of the graph below the $x$-axis to above the $x$-axis, to obtain the graph of $f_{2}(x)=\mid f_{1}(x)-2 |$ (as shown in Figure 8). The area of the closed figure formed with the $x$-axis is
$$
\begin{array}{l}
\quad S=S_{\text {and }}-S \\
=\frac{6+2}{2} \times 2-\frac{1}{2} \times 2 \times 1 \\
=7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7: A town has five primary schools along a circular road, sequentially named First, Second, Third, Fourth, and Fifth Primary Schools. They have 15, 7, 11, 3, and 14 computers, respectively. To make the number of computers equal in each school, some computers need to be transferred to neighboring schools: First to Second, Second to Third, Third to Fourth, Fourth to Fifth, and Fifth to First. If School A gives -3 computers to School B, it means School B gives 3 computers to School A. To minimize the total number of computers moved, how should the transfers be arranged?
|
Analysis: Let $A, B, C,$
$D, E$ represent the five primary schools in a clockwise order, and let them sequentially transfer $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ computers to their neighboring schools, as shown in Figure 9. Then,
\[
\begin{array}{c}
7 + x_{1} - x_{2} = 11 + x_{2} \\
- x_{3} = 3 + x_{3} - x_{4} = 14 + x_{4} - x_{5} = 15 \div x_{5} \\
- x_{1} = \frac{1}{5}(15 + 7 + 11 + 3 + 14). \\
\therefore x_{2} = x_{1} - 3, x_{3} = x_{1} - 2, \\
x_{4} = x_{1} - 9, x_{5} = x_{1} - 5.
\end{array}
\]
Since the total number of computers transferred should be minimized, we need to find the minimum value of the function
\[
\begin{aligned}
y = & \left|x_{1}\right| + \left|x_{2}\right| + \left|x_{3}\right| + \left|x_{4}\right| + \left|x_{5}\right| \\
= & \left|x_{1}\right| + \left|x_{1} - 3\right| + \left|x_{1} - 2\right| + \left|x_{1} - 9\right| \\
& + \left|x_{1} - 5\right|
\end{aligned}
\]
From the conclusion in Example 1, we know that when $x_{1} = 3$, the function $y_{\text {min }} = 12$. At this point, we have $x_{2} = 0, x_{3} = 1, x_{4} = -6, x_{5} = -2$.
Therefore, Primary School 1 transfers 3 computers to Primary School 2, Primary School 2 transfers 1 computer to Primary School 4, Primary School 5 transfers 6 computers to Primary School 4, and Primary School 1 transfers 2 computers to Primary School 5. This way, the total number of computers transferred is minimized and is 12.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 On a circular road, there are four middle schools arranged in sequence: $A_{1}, A_{2}, A_{3}, A_{4}$. 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? Find the minimum total number of TVs transferred.
|
Solution: 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 $r_{1}$ color TVs to $A_{1}$ high school. The same applies below), $A_{2}$ high school transfers $x_{2}$ color TVs to $A_{3}$ high school, $A_{3}$ high school transfers $x_{3}$ color TVs to $A_{4}$ high school, and $A_{4}$ high school transfers $x_{4}$ color TVs to $A_{1}$ high school. Since there are 40 color TVs in total, with an average of 10 per school, therefore,
$$
\begin{array}{c}
15-x_{1}+x_{4}=10,8-x_{2}+x_{1}=10, \\
5-x_{3}+x_{2}=10,12-x_{4}+x_{3}=10 . \\
\therefore x_{4}=x_{1}-5, x_{1}=x_{2}+2, x_{2}=x_{3}+5, \\
x_{3}=x_{4}-2, x_{3}=\left(x_{1}-5\right)-2=x_{1}-7, x_{2}= \\
\left(x_{1}-7\right)+5=x_{1}-2 .
\end{array}
$$
This problem requires finding the minimum value of $y=\left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|+$ $\left|x_{4}\right|=\left|x_{1}\right|+|x_{1}-2|+| x_{1}-7 |+$ $\left|x_{1}-5\right|$. Here, $x_{1}$ is an integer satisfying $-8 \leqslant x_{1} \leqslant 15$.
Let $x_{i}=x$.
Consider the function defined on $-8 \quad 15$
$$
y=|x|+|x-2|+|x-7|+|x-5| \text {. }
$$
$|x|+|x-7|$ represents the sum of the distances from $x$ to 0 and 7. When $0 \leq x \leq 7$, $|x|+|x-7|$ takes the minimum value of 7; similarly, when $2 \leqslant x \leqslant 5$, $|x-2|+|x-5|$ takes the minimum value of 3.
Therefore, when $2 \leqslant x \leqslant 5$, $y$ takes the minimum value of 10. In other words, when $x_{1}=2,3,4,5$, $\left|x_{1}\right|+\left|x_{1}-2\right|+$ $\left|x_{1}-7\right|+\left|x_{1}-5\right|$ takes the minimum value of 10.
Thus, the minimum total number of color TVs to be transferred is 10, and the transfer schemes are as follows:
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 (Fill in the blank Question 1) Given $\frac{1}{4}(b-c)^{2}=$ $(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
|
Solution 1: From the given, we have
$$
\begin{aligned}
0 & =\frac{1}{4}[-(a-b)-(c-a)]^{2}-(a-b)(c-a) \\
& =\frac{1}{4}\left[(a-b)^{2}+(c-a)^{2}-2(a-b)(c-a)\right] \\
& =\frac{1}{4}[(a-b)-(c-a)]^{2} \\
& =\frac{1}{4}(2 a-b-c)^{2} .
\end{aligned}
$$
Thus, $\frac{b+c}{a}=2$.
Solution 2: The given condition indicates that the quadratic equation with roots $\frac{a-b}{2}, \frac{c-a}{2}$ is
$$
\begin{array}{l}
\left(x-\frac{a-b}{2}\right)\left(x-\frac{c-a}{2}\right)=0 \\
\Leftrightarrow x^{2}+\frac{1}{2}(b-c) x+\frac{1}{4}(a-b)(c-a)=0 .
\end{array}
$$
We have $\Delta=\frac{1}{4}(b-c)^{2}-(a-b)(c-a)=0$.
Therefore, the two roots are equal, i.e., $\frac{a-b}{2}=\frac{c-a}{2}$.
Thus, $\frac{b+c}{a}=2$.
Analysis: Let $x=a-b, y=c-a$. The background knowledge of this problem is: in the identity
$$
\frac{1}{4}(x+y)^{2}-x y=\frac{1}{4}(x-y)^{2}
$$
if the left side is 0, then the right side is also 0.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that $[x]$ represents the greatest integer not exceeding $x$. Then the number of solutions to the equation
$$
3^{2 x}-\left[10 \times 3^{x+1}\right]+\sqrt{3^{2 x}-10 \times 3^{x+1}+82}=-80
$$
is
|
5.2.
The original equation can be transformed into
$$
\begin{array}{l}
3^{2 x}-\left[10 \times 3^{x+1}\right]+82 \\
+\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-2=0 . \\
\therefore\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}+2\right) \\
\quad \cdot\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-1\right)=0 . \\
\therefore 3^{2 x}-\left[10 \times 3^{x+1}\right]+81=0,
\end{array}
$$
It is only necessary that $3^{2 x}-10 \times 3^{x+1}+81 \leqslant 0$,
which means $\left(3^{x}-3\right)\left(3^{x}-27\right) \leqslant 0 \Rightarrow 3 \leqslant 3^{x} \leqslant 27$.
$$
\therefore 1 \leqslant x \leqslant 3 \text {. }
$$
When $x=1,3$, the original equation holds. When $x=2$, the original equation has no solution. Therefore, the original equation has two solutions.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{r}
\text { 5. Given a sequence } z_{0}, z_{1}, \cdots, z_{n}, \cdots \text { satisfying } z_{0}=0, z_{1} \\
=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2,
\end{array}
$$
5. Given a sequence of complex numbers $z_{0}, z_{1}, \cdots, z_{n}, \cdots$ satisfying $z_{0}=0, z_{1}$ $=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2$, $\cdots$. The number of $z_{n}$ contained within the circle $\mid z \mid=10$ is
|
5.5.
Since $z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right)=\alpha^{2}\left(z_{n-1}-z_{n-2}\right)$ $=\cdots=\alpha^{n}\left(z_{1}-z_{0}\right)=\alpha^{n}$, therefore,
$$
z_{n}-z_{n-1}=\alpha^{n-1}, \cdots, z_{1}-z_{0}=\alpha^{0}=1 \text {. }
$$
Adding the above $n$ equations, we get
$$
z_{n}=\alpha^{n-1}+\cdots+\alpha+1=\frac{\alpha^{n}-1}{\alpha-1} \text {. }
$$
Since $\alpha=1+\sqrt{3} i=2\left(\cos \frac{\pi}{3}+\sin \frac{\pi}{3}\right)$,
we have, from $\left|z_{n}\right|<10$,
$$
\left|\frac{2^{n}\left(\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}\right)-1}{\sqrt{3} \mathrm{i}}\right|<10,
$$
which means $\left|2^{n} \cos \frac{n \pi}{3}-1+\mathrm{i} 2^{n} \sin \frac{n \pi}{3}\right|<10 \sqrt{3}$.
Thus, $2^{2 n}-2^{n+1} \cos \frac{n \pi}{3}+1<300$.
Noting that $\left(2^{n}-1\right)^{2}=2^{2 n}-2^{n+1}+1 \leqslant 2^{2 n}-2^{n+1}$ $\cdot \cos \frac{n \pi}{3}+1 \leqslant 300$, we can conclude that $n \leqslant 4$, meaning there are 5 points $z_{n}$ inside the circle $|z|<10$, namely $z_{0}, z_{1}, z_{2}, z_{3}$, $z_{4}$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 20 points) For every pair of real numbers $x, y$, the function $f(t)$ satisfies $f(x+y)=f(x)+f(y)+xy+1$. If $f(-2)=-2$, find the number of integer solutions $a$ that satisfy $f(a)=a$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let $x=y=0$, we get $f(0)=-1$; let $x=y=-1$, from $f(-2)=-2$ we get $f(-1)=-2$. Also, let $x=1$, $y=-1$ to get $f(1)=1$. Then let $x=1$, we have
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Therefore, $f(y+1)-f(y)=y+2$, which means when $y$ is a positive integer, $f(y+1)-f(y)>0$.
From $f(1)=1$, we know that for all positive integers $y, f(y)>0$. Thus, when $y \in \mathbf{N}$, $f(y+1)=f(y)+y+2>y+1$, meaning for all positive numbers $t$ greater than 1, $f(t)>t$.
Also from equation $(1), f(-3)=-1, f(-4)=1$.
Next, we prove that when the integer $t \leqslant-4$, $f(t)>0$. Since $t \leqslant$
-4, hence $-(t+2)0 \text {, }
$$
which means $\square$
$$
\begin{array}{l}
f(-5)-f(-4)>0, \\
f(-6)-f(-5)>0, \cdots, \\
f(t+1)-f(t+2)>0, \\
f(t)-f(t+1)>0 .
\end{array}
$$
Adding them up, we get $f(t)-f(-4)>0$,
which means $f(t)>f(-4)=1>0$.
Since $t \leqslant-4$, hence $f(t)>t$.
In summary, the integers that satisfy $f(t)=t$ are only $t=1,-2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) A county is located in a desert area, and people have been engaged in a tenacious struggle against nature for a long time. By the end of 1998, the county's greening rate had reached $30 \%$. Starting from 1999, the following situation will occur every year: $16 \%$ of the original desert area will be greened, while at the same time, due to various reasons, $4 \%$ of the original greened area will be re-sandified.
(1) Let the total area of the county be 1, and the total greened area at the end of 1998 be $a_{1}=\frac{3}{10}$. After $n$ years, the total greened area is $a_{n+1}$. Prove: $a_{n+1} = \frac{4}{5} a_{n} + \frac{4}{25}$.
(2) How many years of effort are needed at least to achieve a county-wide greening rate?
|
(1) Let the current desert area be $b_{1}$, and after $n$ years, the desert area be $b_{n+1}$. Thus, $a_{1}+b_{1}=1, a_{n}+b_{n}=1$.
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} a_{n}$, which is $\frac{96}{100} a_{n}$; the other part is the newly greened area $\frac{16}{100} b_{n}$. Therefore,
$$
\begin{array}{l}
a_{n+1}=\frac{96}{100} a_{n}+\frac{16}{100} b_{n} \\
=\frac{96}{100} a_{n}+\frac{16}{100}\left(1-a_{n}\right)=\frac{4}{5} a_{n}+\frac{4}{25} .
\end{array}
$$
(2) From (1), we get
$$
\begin{array}{l}
a_{n+1}-\frac{4}{5}=\frac{4}{5}\left(a_{n}-\frac{4}{5}\right)=\left(\frac{4}{5}\right)^{2}\left(a_{n-1}-\frac{4}{5}\right) \\
=\cdots=\left(\frac{4}{5}\right)^{n}\left(a_{1}-\frac{4}{5}\right)=-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n} .
\end{array}
$$
Therefore, $a_{n+1}=\frac{4}{5}-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n}$.
Note that $a_{n+1} \geqslant 60 \%$, i.e., $\frac{4}{5}-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n}>\frac{3}{5}$. Simplifying, we get $\left(\frac{4}{5}\right)^{n}\frac{1-2 \lg 2}{1-3 \lg 2}>4$. Therefore, it will take at least 5 years to make the county's greening rate exceed $60 \%$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given three real
numbers $x_{1}, x_{2}, x_{3}$, any one of them plus five times the product of the other two equals 6. The number of such triples $\left(x_{1}, x_{2}\right.$, $x_{3}$ ) is.
|
4.5
From the problem, we have
$$
\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{1} x_{3}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}
$$
(1) - (2) gives $\left(x_{1}-x_{2}\right)\left(5 x_{3}-1\right)=0$.
Thus, $x_{1}=x_{2}$ or $x_{3}=\frac{1}{5}$.
Similarly, $x_{2}=x_{3}$ or $x_{1}=\frac{1}{5}$, $x_{3}=x_{1}$ or $x_{2}=\frac{1}{5}$.
Therefore, $\left(x_{1}, x_{2}, x_{3}\right)$ has 5 sets.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 If $f(x)=\left(2 x^{5}+2 x^{4}-53 x^{3}-\right.$ $57 x+54)^{1998}$, then $f\left(\frac{\sqrt{1}-\frac{1}{2}-i}{2}\right)=$ $\qquad$ .
|
Let $\frac{\sqrt{111}-1}{2}=x$, then we have
$$
\begin{array}{l}
2 x^{2}+2 x-55=0 \\
\because 2 x^{5}+2 x^{4}-53 x^{3}-57 x+54 \\
=x^{3}\left(2 x^{2}+2 x-55\right)+x\left(2 x^{2}+2 x\right. \\
\quad-55)-\left(2 x^{2}+2 x-55\right)-1 \\
\therefore f\left(\frac{\sqrt{111}-1}{2}\right)=(-1)^{1998}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
保留源文本的换行和格式，直接输出翻译结果如下：
```
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
```
|
Solution: Let $f(x)=\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}\right.$
$$
\begin{aligned}
& -3 x-3)^{2001} \\
= & a_{0}+a_{1} x+a_{2} x^{2}+\cdots \\
& +a_{4001} x^{4001}+a_{4002} x^{4002} .
\end{aligned}
$$
Let $x=1$ and $x=-1$, we get
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+\cdots+a_{4001}+a_{4002} \\
=f(1)=0, \\
a_{0}-a_{1}+a_{2}-\cdots-a_{4001}+a_{4002} \\
=f(-1)=2 .
\end{array}
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
2\left(a_{1}+a_{3}+\cdots+a_{4001}\right)=-2 . \\
\therefore a_{1}+a_{3}+\cdots+a_{4001}=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+1994^{2}$.
|
Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of
$$
\underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\cdots+9^{2}}_{\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2}
$$
The unit digits follow a periodic pattern:
$$
1,4,9,6,5,6,9,4,1 \text {, }
$$
The unit digit of the sum of each group is 5, so the unit digit of the sum of 199 groups is also 5. Considering the unit digit sum of the remaining four numbers is 0, the unit digit of the required sum is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Let $1995 x^{3}=1996 y^{3}=1997 z^{3}$,
$$
\begin{array}{l}
x y z>0, \text { and } \sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}} \\
=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} . \\
\quad \text { Then } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=
\end{array}
$$
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
|
Solution: Let $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, obviously $k \neq 0$. Then we have
$$
1995=\frac{k}{x^{3}}, 1996=\frac{k}{y^{3}}, 1997=\frac{k}{z^{3}} .
$$
From the given, we get
$$
\sqrt{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}}>0 .
$$
Thus, $\sqrt[3]{k} \times \sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$
$$
\begin{array}{l}
=\sqrt[3]{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) . \\
\because k \neq 0, \\
\therefore \sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
From the given, we have $x>0, y>0, z>0$, hence $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given real numbers $a, b$ satisfy $3 a^{4}+2 a^{2}-4=0$ and $b^{4}+b^{2}-3=0$. Then $4 a^{-4}+b^{4}=$ ( ).
(A) 7
(B) 8
(C) 9
(D) 10
|
Solution: Transform the equation $3 a^{4}+2 a^{2}-4=0$ into $\frac{4}{a^{4}}-$ $\frac{2}{a^{2}}-3=0$, and it is known that $-\frac{2}{a^{2}}$ and $b^{2}$ are the two distinct real roots of the equation $x^{2}+x-3=0$.
Let $-\frac{2}{a^{2}}=x_{1}, b^{2}=x_{2}$. By Vieta's formulas, we have $x_{1}+x_{2}=-1, x_{1} x_{2}=-3$.
$\therefore 4 a^{-4}+b^{4}=\left(-\frac{2}{a^{2}}\right)+\left(b^{2}\right)^{2}$ $=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=7$.
|
7
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 Given $a+b+c=0, a^{3}+b^{3}+c^{3}$ $=0$. Find the value of $a^{15}+b^{15}+c^{15}$.
|
Solution: From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$ - $\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)$, we have $a b c=0$. Therefore, at least one of $a, b, c$ is 0.
Assume $c=0$, then $a, b$ are opposites,
$$
\therefore a^{15}+b^{15}+c^{15}=0 \text {. }
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example: 15 Given positive integers $x, y, z$ satisfy $x^{3}-y^{3}-z^{3}=3 x y z, x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original text's line breaks and format.
|
$$
\begin{array}{l}
\text { Solution: } 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\right. \\
\left.+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) \\
=\frac{1}{2}(x-y-z)\left[(x+y)^{2}+(y-z)^{2}\right. \\
\left.+(z-x)^{2}\right] \text {. } \\
\end{array}
$$
From the given, we have $x-y-z=0$,
$$
\therefore x=y+z \text {. }
$$
Substituting $x^{2}=2(y+z)$, we get $x=2, y+z=2$. Given $y \geqslant 1, z \geqslant 1$, we have $y=z=1$.
$$
\therefore x y+y z+z x=5 \text {. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let $a, b, c, d$ be four distinct real numbers such that
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=b d$.
Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$.
|
Three, let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we know $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$. Thus, the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$.
Let $x+\frac{1}{x}=e, y+\frac{1}{y}=f$, then $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}=e f$.
$\because$ When $t>0$, $t+\frac{1}{t} \geqslant 2$;
When $t<0, y<0$, then
$$
f \leqslant-2, e=4-f \geqslant 6, e f \leqslant-12 \text{. }
$$
When and only when $y=-1, x=3 \pm 2 \sqrt{2}$, the equality holds. Specifically, when $a=3+2 \sqrt{2}, b=1, c=-1, d=-(3+2 \sqrt{2})$, the equality holds.
$$
\therefore(e f)_{\max }=-12 \text{. }
$$
|
-12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Simplify $\frac{a+1}{a+1-\sqrt{1-a^{2}}}+\frac{a-1}{\sqrt{1-a^{2}}+a-1}$ $(0<|a|<1)$ The result is $\qquad$
|
$\begin{array}{l}\text { Original expression }=\frac{(\sqrt{a+1})^{2}}{\sqrt{a+1}(\sqrt{a+1}-\sqrt{1-a})} \\ \quad+\frac{-(\sqrt{1-a})^{2}}{\sqrt{1-a}(\sqrt{1+a}-\sqrt{1-a})}=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $(x-1)^{2}$ divides the polynomial $x^{4}+a x^{3}-$ $3 x^{2}+b x+3$ with a remainder of $x+1$. Then $a b=$ $\qquad$ .
|
II, 1.0.
From the given, we have
$$
\begin{array}{l}
x^{4}+a x^{3}-3 x^{2}+b x+3 \\
=(x-1)^{2}\left(x^{2}+\alpha x+\beta\right)+x+1 \\
=x^{4}+(\alpha-2) x^{3}+(\beta+1-2 \alpha) x^{2}+(1+\alpha- \\
2 \beta) x+1+\beta .
\end{array}
$$
Then $a=\alpha-2$,
$$
\begin{array}{l}
-3=\beta+1-2 \alpha, \\
b=1+\alpha-2 \beta, \\
3=1+\beta .
\end{array}
$$
Solving, we get $\alpha=3, \beta=2, a=1, b=0$.
$$
\therefore a b=0 \text {. }
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3: A certain area currently has 10,000 hectares of arable land. It is planned that in 10 years, the grain yield per unit area will increase by $22\%$, and the per capita grain possession will increase by $10\%$. If the annual population growth rate is $1\%$, try to find the maximum number of hectares by which the arable land can decrease on average each year.
|
Solution: Let the average annual reduction of arable land be $x$ hectares, and let the current population of the region be $p$ people, with a grain yield of $M$ tons/hectare. Then
$$
\begin{array}{l}
\frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\
\geqslant \frac{M \times 10^{4}}{p} \times(1+10 \%) .
\end{array}
$$
Simplifying, we get
$$
\begin{aligned}
x \leqslant & 10^{3} \times\left[1-\frac{1.1 \times(1+0.01)^{10}}{1.22}\right] \\
= & 10^{3} \times\left[1-\frac{1.1}{1.22}\left(1+\mathrm{C}_{10}^{1} \times 0.01\right.\right. \\
& \left.\left.+\mathrm{C}_{10}^{2} \times 0.01^{2}+\cdots\right)\right] \\
\approx & 10^{3} \times\left[1-\frac{1.1}{1.22} \times 1.1045\right] \approx 4.1 .
\end{aligned}
$$
Therefore, the maximum average annual reduction of arable land in hectares is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $D$ be a point on the side $AB$ of $\triangle ABC$, point $D$ moves along a direction parallel to $BC$ to point $E$ on side $AC$; then from point $E$ along a direction parallel to $AB$ to point $F$ on side $BC$; then from point $F$ along a direction parallel to $CA$ to point $G$ on side $AB$, $\cdots \cdots$ each move along a side parallel to another side counts as one move. Then, at most $n$ moves, point $D$ can return to its original starting point for the first time. What is the value of $n$?
|
Analysis: We can estimate through graphical experiments that point $D$ can return to the original starting point after 6 times.
In fact, as shown in
Figure 1, by the intercept theorem of parallel lines,
we get
$$
\begin{array}{l}
\frac{A D}{B D}=\frac{A E}{E C} . \\
=\frac{B F}{F C}=\frac{B G}{A G} \\
=\frac{C H}{A H}=\frac{C K}{B K}=\frac{A M}{B M} .
\end{array}
$$
By the theorem of the sum of ratios, we get
$$
\frac{A D+B D}{B D}=\frac{A M+B M}{B M},
$$
which means
$$
\frac{A B}{B D}=\frac{A B}{B M} \text { . }
$$
Therefore, point $D$ coincides with point $M$, so $n=6$. Of course, if we take the midpoint of any side as the starting point, it is obvious that it only takes 3 times to return to the original starting point for the first time.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given three real numbers $x_{1}, x_{2}, x_{3}$, any one of these numbers plus five times the product of the other two always equals 6. The number of such triples $\left(x_{1}, x_{2}, x_{3}\right)$ is $\qquad$.
$(1995$, Dongfang Airlines Cup - Shanghai Junior High School Mathematics Competition)
|
Solving the system of equations given by the problem, we have:
$$
\left\{\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{3} x_{1}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.
$$
(1) - (2) gives
$$
\left(x_{1}-x_{2}\right)\left(1-5 x_{3}\right)=0 \text {. }
$$
(2) - (3) gives
$$
\left(x_{2}-x_{3}\right)\left(1-5 x_{1}\right)=0 \text {. }
$$
From (1), (4), and (5), the original system of equations can be transformed into:
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x _ { 1 } = x _ { 2 } , } \\
{ x _ { 2 } = x _ { 3 } , } \\
{ x _ { 3 } + 5 x _ { 1 } x _ { 2 } = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}=x_{2}, \\
5 x_{1}=1, \\
x_{3}+5 x_{1} x_{2}=6 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 5 x _ { 3 } = 1 , } \\
{ x _ { 2 } = x _ { 3 } , } \\
{ x _ { 3 } + 5 x _ { 1 } x _ { 2 } = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
5 x_{3}=1, \\
5 x_{1}=1, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.\right.
\end{array}
$$
Solving these four systems of equations, we get 5 sets of solutions for $\left(x_{1}, x_{2}, x_{3}\right)$: $\square$
$$
\begin{array}{l}
(1,1,1),\left(-\frac{6}{5},-\frac{6}{5},-\frac{6}{5}\right), \\
\left(\frac{1}{5}, \frac{1}{5}, \frac{29}{5}\right),\left(\frac{1}{5}, \frac{29}{5}, \frac{1}{5}\right),\left(\frac{29}{5}, \frac{1}{5}, \frac{1}{5}\right) .
\end{array}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Real numbers $x, y$ satisfy $x \geqslant y \geqslant 1$ and $2 x^{2}-x y$ $-5 x+y+4=0$. Then $x+y=$ $\qquad$
|
3.4 .
From the given equation, we know that $2 x^{2}-5 x+4=y(x-1) \leqslant x(x-1)$, which leads to $x^{2}-4 x+4 \leqslant 0$, or $(x-2)^{2} \leqslant 0$. Therefore, $x=2$.
Substituting $x=2$ into the given equation yields $y=2$, hence $x+y=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $n$ be a natural number, and $a_{n}=\sqrt[3]{n^{2}+2 n+1}$ $+\sqrt[3]{n^{2}-1}+\sqrt[3]{n^{2}-2 n+1}$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{3}}+\frac{1}{a_{5}}$ $+\cdots+\frac{1}{a_{997}}+\frac{1}{a_{999}}$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
|
4. (C).
$$
\begin{array}{l}
a_{n}=(\sqrt[3]{n+1})^{2}+\sqrt[3]{n+1} \cdot \sqrt[3]{n-1}+(\sqrt[3]{n-1})^{2}, \\
\frac{1}{a_{n}}=\frac{1}{2}(\sqrt[3]{n+1}-\sqrt[3]{n-1}) . \\
\therefore \sum_{i=1}^{300} \frac{1}{a_{2 i}-1}=\frac{1}{2}[(\sqrt[3]{2}-0)+(\sqrt[3]{4}-\sqrt[3]{2}) \\
\quad+\cdots+(\sqrt[3]{1000}-\sqrt[3]{998})]=5 .
\end{array}
$$
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\frac{16}{3}$. Then the maximum value of $a$ is $\qquad$ .
|
4. 2 .
Construct the function
$$
y=3 x^{2}-2(b+c+d) x+\left(b^{2}+c^{2}+d^{2}\right) \text {. }
$$
Since $y(x-b)^{2}+(x-c)^{2}+(x-d)^{2} \geqslant 0$, and the graph
is a parabola opening upwards, we have
$$
\Delta=4(b+c+d)^{2}-12\left(b^{2}+c^{2}+d^{2}\right) \leqslant 0,
$$
which simplifies to $(4-a)^{2}-3\left(\frac{16}{3}-a^{2}\right) \leqslant 0$.
Solving this, we get $0 \leqslant a \leqslant 2$. Therefore, the maximum value of $a$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$. Then $m^{5}+n^{5}=$ $\qquad$ .
(From the riverbed Jiangsu Province Junior High School Mathematics Competition)
|
Solution: Construct a quadratic equation $x^{2}-x-1=0$. From the given conditions, we know that $m$ and $n$ are the two roots of the equation $x^{2}-x-1=0$, so $m+n=1, m n=-1$. Then
$$
\begin{array}{l}
m^{2}+n^{2}=(m+n)^{2}-2 m n=3 . \\
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2}=7 \text {. } \\
\therefore m^{5}+n^{5} \\
=(m+n)\left(m^{4}+n^{4}\right)-m n(m+n) \\
\text { - }\left[(m+n)^{2}-3 m n\right] \\
=7-(-1) \times\left[1^{2}-3 \times(-1)\right]=11 \text {. } \\
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a, b, c$ are non-zero real numbers, and $a+b+c$ $=0$. Then the value of $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ is . $\qquad$
|
2. -3
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 30 points) There are several weights of 9 grams and 13 grams each. To weigh an object of 3 grams on a balance, what is the minimum number of such weights needed? Prove your conclusion.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
On the three pans of a balance, assuming that when the balance is level, 9 grams are used, and the number is an integer. Similarly, assuming that 13 grams of weights are used $|y|$ times. Therefore, when the balance is level and measures a 3-gram object, there should be
$$
9 x+13 y=3 \text {. }
$$
The problem becomes finding the minimum value of $|x|+|y|$.
First, we can take any integer solution of the equation, for example, because $y=$ $\frac{3-9 x}{13}$, when $x=9$, we get $y=-6$. Using this solution, we have
$$
\left\{\begin{array}{l}
9 \times 9-13 \times 6=3, \\
9 x+13 y=3 .
\end{array}\right.
$$
Subtracting the two equations, we get $9(x-9)+13(y+6)=0$,
$$
9(x-9)=-13(y+6) \text {. }
$$
Since 9 and 13 are coprime, $x-9$ must be divisible by 13. Therefore, let $x-9=13 k$, where $k$ is an integer. At this time, we have
$9 \times 13 k=-13(y+6),-(y+6)=9 k$.
In summary, we have $\left\{\begin{array}{l}x=9+13 k, \\ y=-6-9 k .\end{array} \quad k=0, \pm 1, \pm 2, \cdots\right.$
(i) When $k=0$, $x=9, y=-6$.
$$
|x|+|y|=15 \text {; }
$$
(ii) When $k \geqslant 1$, $|x| \geqslant 22,|y|>0$, thus
$$
|x|+|y|>22 \text {; }
$$
(iii) When $k \leqslant-1$, if $k=-1$, then $x=-4, y=3$, $|x|+|y|=7$; if $k12,|x|>0$, thus $|x|+|y|>12$.
From the above, we know that at least 7 such weights are needed, with 4 of them being 9 grams and 3 being 13 grams.
(Provided by Li Shuwen, No. 40 Middle School of Benzhuang City, Shandong Province)
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The integer part of $\frac{1}{3-\sqrt{7}}$ is $a$, and the fractional part is $b$. Then $a^{2}+(1+\sqrt{7}) a b=$ $\qquad$ .
|
3.10 . From $\frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{2}$, we know that $2<\frac{3+\sqrt{7}}{2}<3$, thus $a=2, b=\frac{1}{3-\sqrt{7}}-2=\frac{\sqrt{7}-1}{2}$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In flood control and rescue operations, a depression near the river dike has experienced a pipe burst, with $x$ cubic meters of river water already rushing in, and water continues to flow in at a rate of $y$ cubic meters per minute. Now, a water extraction and plugging project needs to be carried out. If 1 water pump is used, it will take 30 minutes to pump out all the water and start construction; if 2 water pumps work simultaneously, it will take 10 minutes to pump out all the water and start construction. Due to the urgent situation, the command post requires that the water be pumped out within 5 minutes to immediately start construction. At least how many water pumps need to work simultaneously? Assume that each water pump has a pumping rate of $z$ cubic meters per minute ($z>0$).
|
5.4
Let at least $n$ water pumps be required. According to the problem, we have
$$
\left\{\begin{array}{l}
x+30 y=30 z, \\
x+10 y=20 z, \\
x+5 y=n \cdot 5 z .
\end{array}\right.
$$
From equations (1) and (2), we can solve for $x=15 z, y=0.5 z$.
Substituting into equation (3), we get
$15 z+0.5 z \leqslant 5 n z$, hence $n \geqslant 3.1$.
Since $n$ is a positive integer, the minimum value is $n=4$.
That is, at least 4 water pumps need to be organized to work simultaneously.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what is the maximum number of pairs of lines that are perpendicular to each other? (Each pair consists of two lines)
保留源文本的换行和格式，翻译结果如下：
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what is the maximum number of pairs of lines that are perpendicular to each other? (Each pair consists of two lines)
|
Four, prove in three steps that there are at most 6 pairs of mutually perpendicular lines.
(1) 6 pairs can be achieved.
When $V A \perp$ plane $A B C$ and $A B \perp A C$, by the property of line perpendicular to a plane, we have $V A \perp A B, V A \perp B C, V A \perp C A$.
By the theorem of three perpendiculars, we also have $V C \perp A B, V B \perp A C$.
This results in 6 pairs of mutually perpendicular lines (refer to Figure 11).
(2) 8 pairs are impossible.
Since a triangle can have at most one right angle and at least two acute angles, 4 triangular faces must have at least 8 acute angles. Given that 6 lines can form $C_{0}^{2}=15$ pairs of lines, and $15-3=7$, we know that the number of mutually perpendicular lines does not exceed 7.
(3) 7 pairs are impossible.
If otherwise, there are 7 pairs of perpendicular lines, and since there are only 3 pairs of skew lines, at least 4 pairs of perpendicular lines must be coplanar, meaning all 4 faces of the tetrahedron must be right-angled triangles (yielding 4 pairs). Furthermore, the 3 pairs of skew lines must also be perpendicular. In this case, the projection of one vertex of the tetrahedron onto the opposite face must be the orthocenter of that triangle, and the orthocenter of a right-angled triangle is the right-angle vertex. Therefore, one lateral edge of the tetrahedron is perpendicular to the base, and the foot of the perpendicular is the vertex of the right-angled triangle (in Figure 11, $V A \perp$ plane $A B C$). At this point, the opposite face of the right-angle vertex of the base triangle is an acute triangle ($\triangle V B C$ is an acute triangle), which contradicts the fact that all 4 faces are right-angled triangles.
In summary, the maximum number of mutually perpendicular lines is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let $m$ be a real number not less than -1, such that the equation $x^{2}+2(m-2) x+m^{2}-3 m+3=0$ has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$. Find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$.
|
$$
\begin{aligned}
\Delta & =4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
& =-4 m+4>0 .
\end{aligned}
$$
Then $m<1$.
Combining the given conditions, we have $-1 \leqslant m<1$.
$$
\begin{array}{l}
\text { (1) } \because x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=4(m-2)^{2}-2\left(m^{2}-3 m+3\right) \\
=2 m^{2}-10 m+10, \\
\therefore 2 m^{2}-10 m+10=6, \\
\end{array}
$$
Thus, $m^{2}-5 m+2=0$. Solving this, we get $m=\frac{5 \pm \sqrt{17}}{2}$.
Since $-1 \leqslant m<1$, we have $m=\frac{5-\sqrt{17}}{2}$.
$$
\begin{array}{l}
\text { (2) } \frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}} \\
=\frac{m\left[x_{1}^{2}\left(1-x_{2}\right)+x_{2}^{2}\left(1-x_{1}\right)\right]}{\left(1-x_{1}\right)\left(1-x_{2}\right)} \\
=\frac{m\left[x_{1}^{2}+x_{2}^{2}-x_{1} x_{2}\left(x_{1}+x_{2}\right)\right]}{x_{1} x_{2}-x_{1}-x_{2}+1} \\
=\frac{m\left[\left(2 m^{2}-10 m+10\right)+\left(m^{2}-3 m+3\right)(2 m-4)\right]}{\left(m^{2}-3 m+3\right)+(2 m-4)+1} \\
=\frac{m\left(2 m^{3}-8 m^{2}+8 m-2\right)}{m^{2}-m} \\
=\frac{2 m(m-1)\left(m^{2}-3 m+1\right)}{m(m-1)} \\
=2\left(m^{2}-3 m+1\right). \\
\end{array}
$$
Let $y=2\left(m^{2}-3 m+1\right)=2\left(m-\frac{3}{2}\right)^{2}-\frac{5}{2}$, $-1 \leqslant m<1$. Since $y$ is decreasing on $-1 \leqslant m<1$, the maximum value of $y$ is 10 when $m=-1$.
Therefore, the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$ is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (16 points) An engineering
team contracted two projects, Project A and Project B, with the workload of Project A being twice that of Project B. In the first half of the month, all workers worked on Project A, and in the second half of the month, the workers were divided into two equal groups, one group continued working on Project A, and the other group moved to Project B. After one month, Project A was completed, and the remaining workload of Project B was just enough for one worker to complete in one month. If each worker has the same work efficiency, how many workers are there in the engineering team?
|
Three, suppose this construction team has $x$ people, and the monthly work volume per person is 1. Then the work volume for site A is $\frac{\bar{x}}{2}+\frac{1}{2}$ $\frac{x}{2}$, and the work volume for site B is $\frac{x}{2} \times \frac{1}{2}+1$. According to the problem, we get $\frac{x}{2}+\frac{x}{4}=2\left(\frac{x}{4}+1\right)$. Solving for $x$ yields $x=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Find the sum of all solutions to the equation $[3 x+1]=2 x-\frac{1}{2}$.
(1987, All-China Junior High School Mathematics Competition)
|
Solution: We can consider $[3 x+1]$ as a whole, for example, an integer $t$, thus transforming the given equation into an inequality related to $t$.
Let $[3 x+1]=t$, then $t$ is an integer, and
$$
0 \leqslant(3 x+1)-t<1 \text {. }
$$
Thus, the original equation becomes
$$
t=2 x-\frac{1}{2} \text {, }
$$
which means $x=\frac{1}{2} t+\frac{1}{4}$.
Substituting into inequality (1), we get
$$
0 \leqslant \frac{3}{2} t+\frac{3}{4}+1-t<1 \text {, }
$$
which simplifies to $0 \leqslant \frac{1}{2} t+\frac{7}{4}<1$.
$$
\therefore-\frac{7}{2} \leqslant t<-\frac{3}{2} \text {. }
$$
Since $t$ is an integer, the only integers satisfying inequality (3) are $t=-2, t=-3$.
Substituting the values of $t$ into equation (2), we get
$$
x_{1}=-\frac{3}{4}, x_{2}=-\frac{5}{4} \text {. }
$$
Therefore, the sum of the roots is
$$
x_{1}+x_{2}=\left(-\frac{3}{4}\right)+\left(-\frac{5}{4}\right)=-2 \text {. }
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find the smallest positive integer $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution $x$.
(1989, Soviet Union Mathematical Olympiad)
|
(Hint: Using the inequality $\frac{10^{n}}{x}-1<\left[\frac{10^{n}}{x}\right] \leqslant \frac{10^{n}}{x}$, solve for $x=5026$, at this point $n=7$.)
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In 1935, the famous mathematicians P. Erdos and G. Szekeres proved a well-known proposition:
For any positive integer $n \geqslant 3$, there exists a number $f(n)$, such that if and only if $m \geqslant f(n)$, any set of $m$ points in the plane, with no three points collinear, contains $n$ points that form a convex $n$-gon. They conjectured that
$$
f(n)=2^{n-2}+1 \text {. }
$$
Below is a proof for the case $n=5$, which includes an important lemma:
|
Lemma 1: Let the vertex set of an arbitrary convex hull $K$ be denoted as $S$. If there are $m$ points inside the convex hull forming a convex $m$-gon, and this convex $m$-gon has a side $A_{2} A_{3}$, with its adjacent sides $A_{1} A_{2}$ and $A_{3} A_{4}$ (where $A_{1}, A_{2}, A_{3}, A_{4}$ are all vertices), and if the region $T$ (the shaded area in Figure 1) is formed by the extensions of $A_{1} A_{2}$ and $A_{4} A_{3}$ and the side $A_{2} A_{3}$, and there are $n$ points ($\epsilon S$) in $T$, then the convex hull formed by these $n$ points and the convex $m$-gon is a convex $(m+n)$-gon, as shown in Figure 1.
Proof: Let the region formed by the convex $m$-gon be denoted as $A$. Clearly, $A \cup T$ is a convex set. If the number of vertices of the convex hull $B$ formed by the $n$ points and the convex $m$-gon is less than $m+n$, then there must be a point inside $B$. Since the vertices of the $m$-gon are all boundary points of the convex set $A \cup T$ (i.e., $A_{2}, A_{3}$), this point must be among the $n$ points. Thus, this point is in $S$. It is also easy to see that $B$ is in $K$, so this point is also inside $K$. Since this point belongs to $S$, it must be a vertex of $K$, which is a contradiction. Therefore, the lemma is proved.
Next, we will prove the proposition.
First, it is easy to see that there are counterexamples with 8 points. For example, in Figure 2, place 4 points on each branch of a hyperbola, such that no three points are collinear, and in this case, there is no convex pentagon.
Now we prove that there must be a convex pentagon among 9 points. Let the 9 points on the plane be $A_{1}, A_{2}, \ldots, A_{9}$. Clearly, we only need to consider the following two cases:
(1) The convex hull is a quadrilateral, say $A_{1} A_{2} A_{3} A_{4}$, and the points $A_{5}, A_{6}, \ldots, A_{9}$ are inside it:
If $A_{5}, A_{6}, \ldots, A_{9}$ form a convex pentagon, the problem is solved.
Otherwise, among these five points, there must be four points such that one of them is inside the triangle formed by the other three. Without loss of generality, assume $A_{5}$ is inside $\triangle A_{6} A_{7} A_{8}$, as shown in Figure 3.
Connect $A_{5} A_{6}, A_{5} A_{7}, A_{5} A_{8}$ and extend them. Let the region enclosed by the rays $A_{5} A_{6}$ and $A_{5} A_{7}$, excluding $\triangle A_{5} A_{6} A_{7}$, be denoted as $\mathrm{I}$, and similarly define regions II and III, as shown in Figure 3. Clearly, $A_{1}, A_{2}, A_{3}, A_{4}$ are all in these three regions.
Since there are only three regions and four points ($A_{1}, A_{2}, A_{3}, A_{4}$), by the pigeonhole principle, at least one region must contain at least two of these points. Without loss of generality, assume $A_{1}, A_{2}$ are in region I. By the lemma, the points $A_{5}, A_{6}, A_{2}, A_{1}, A_{2}$ form a convex pentagon.
(2) The convex hull is a triangle, say $\triangle A_{1} A_{2} A_{3}$, and the other 6 points $A_{4}, A_{5}, \ldots, A_{9}$ are inside it. For $A_{4}, A_{5}, \ldots, A_{9}$, consider the following two cases:
1) If the convex hull is a quadrilateral, say $A_{4} A_{5} A_{6} A_{7}$, consider extending the line segment $A_{8} A_{9}$ at both ends. If it intersects the adjacent sides of the quadrilateral $A_{4} A_{5} A_{6} A_{7}$, say $A_{4} A_{5}$ and $A_{4} A_{7}$, as shown in Figure 4, then $A_{8}, A_{9}, A_{5}, A_{6}, A_{7}$ form a convex pentagon.
If extending the line segment $A_{8} A_{9}$ at both ends intersects the opposite sides of the quadrilateral $A_{4} A_{5} A_{6} A_{7}$, say $A_{5} A_{6}$ and $A_{4} A_{7}$, as shown in Figure 5, the four regions I, II, III, and IV are shaded. It is easy to see that these four regions cover the part outside the quadrilateral $A_{4} A_{5} A_{6} A_{7}$ (of course, if the rays $A_{8} A_{4}$ and $A_{9} A_{5}$ intersect or the rays $A_{8} A_{7}$ and $A_{9} A_{6}$ intersect, the four regions will overlap).
There are three points $A_{1}, A_{2}, A_{3}$ in these four regions. By the lemma, if there is no convex pentagon, then regions II and IV cannot contain any points. Regions I and III can each contain at most one point, so the four regions can contain at most two points, which contradicts the existence of three points $A_{1}, A_{2}, A_{3}$.
2) If the convex hull is a triangle, say $\triangle A_{4} A_{5} A_{6}$, as shown in Figure 6, and without loss of generality, assume extending $A_{7} A_{8}$ intersects $A_{4} A_{5}$ and $A_{4} A_{6}$. Define regions I, II, and III (it is easy to see that regions I and II overlap). Similarly, $A_{1}, A_{2}, A_{3}$ are in these three regions.
If there is no convex pentagon, then by the lemma, region III cannot contain any points, and regions I and II can each contain at most one point, which contradicts the existence of three points $A_{1}, A_{2}, A_{3}$. Therefore, there must be a convex pentagon. In conclusion, $f(5)=9$.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. A magician has one hundred cards, each with a number from 1 to 100. He places these one hundred cards into three boxes, one red, one white, and one blue. Each box must contain at least one card.
A participant selects two of the three boxes and then picks one card from each of the selected boxes, announcing the sum of the numbers on the two cards. Knowing this sum, the magician can identify which box was not selected.
How many ways are there to place the cards so that the trick always works? (Two methods are considered different if at least one card is placed in a different colored box.)
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
Solution: There are 12 different methods. Consider the integers from 1 to 100. For simplicity, define the color of the box in which integer $i$ is placed as the color of that integer. Use $\mathrm{r}$ to represent red, $\mathrm{w}$ to represent white, and $\mathrm{b}$ to represent blue.
(1) There exists some $i$ such that $i, i+1, i-2$ have mutually different colors, for example, $\mathrm{wb}$. Since $i+(i+3)=(i+1)+(i+2)$, the color of $i \div 3$ cannot be the color of $i+1$, which is $\mathrm{w}$, nor can it be the color of $i+2$, which is $\mathrm{w}$. It can only be $\mathrm{r}$. Therefore, as long as three adjacent numbers have mutually different colors, the color of the next number can be determined. Furthermore, this pattern of three numbers' colors must repeat: after rwb, it must be $r$, then $w, b, \cdots$ and so on. Similarly, this process also holds for the opposite direction: before rwb, it must be $b, \cdots$ and so on.
Therefore, it is only necessary to determine the colors of 1, 2, and 3. There are 6 different ways to do this, and all 6 methods can make the magic trick successful because their sums $r+w, w+b, b+r$ give mutually different remainders modulo 3.
(2) There do not exist three consecutive numbers with mutually different colors. Assume 1 is red. Let $i$ be the smallest number that is not red, and assume $i$ is white. Let $k$ be the smallest blue number. By assumption, $i+11$ must be red, and from $t+99=(t-1)+100$, it follows that $t-1$ is blue, which contradicts the assumption that only 100 is blue.
Therefore, the colors of these numbers must be rww…wwb. This method indeed works:
If the sum of the numbers on the two selected cards $\leqslant 100$, the box from which no card was selected must be blue;
If the sum of the numbers is $101$, the box from which no card was selected must be white;
If the sum of the numbers $>101$, the box from which no card was selected must be red.
Finally, there are 6 different ways to arrange the colors in the above manner. Therefore, the answer is 12.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 䒴 $\frac{4 x}{x^{2}-4}=\frac{a}{x+2}+\frac{b}{x-2}$, then the value of $a^{2}+$ $b^{2}$ is $\qquad$
(1996, Hope Cup National Mathematics Invitational Competition)
|
Solution: From the problem, we know that the given equation is not an equation in $x$, but an identity in $x$. From
$$
\frac{a}{x+2}+\frac{b}{x-2}=\frac{(a+b) x-2(a-b)}{x^{2}-4}
$$
we get the identity
$$
\begin{array}{l}
\frac{4 x}{x^{2}-4}=\frac{(a+b) x-2(a-b)}{x^{2}-4}, \\
4 x=(a+b) x-2(a-b) .
\end{array}
$$
According to the conditions for the identity of polynomials, we get the system of equations about $a$ and $b$
$$
\left\{\begin{array}{l}
a+b=4, \\
a-b=0 .
\end{array}\right.
$$
Solving, we get $a=b=2$.
Therefore, $a^{2}+b^{2}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Suppose $1995 x^{3}=1996 y^{3}=1997 z^{3}$, $x y z>0$, and $\sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}}=$ $\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
(1996, National Junior High School Mathematics League)
|
Given that there are three independent equations (equations) and exactly three letters, it should be possible to solve for $x, y, z$, but in practice, it is quite difficult. Inspired by the "chain equality" type problem from the previous question, we can introduce the letter $k$ - try.
Assume $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, then
$$
x=\sqrt[3]{\frac{k}{1995}}, y=\sqrt[3]{\frac{k}{1996}}, z=\sqrt[3]{\frac{k}{1997}} .
$$
From the given conditions, $x, y, z, k$ have the same sign, and since $x y z>0$, we know $x, y, z, k>0$. Substituting into the last equation given, we get
$$
\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}} .
$$
Rearranging, we get
$$
\sqrt[3]{k}\left(\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)=0 \text{. }
$$
Since $k>0$, we know $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is a root of $\sqrt[3]{u}-u=0$.
Therefore, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. (The other two roots $0, -1$ do not meet the conditions)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given that $x, y, z$ are 3 non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $s=2x+y-z$, then what is the sum of the maximum and minimum values of $s$?
(1996, Tianjin Junior High School Mathematics Competition)
|
The given equations and the expression for $s$ can be viewed as a system of equations in $x$, $y$, and $z$:
$$
\left\{\begin{array}{l}
3 x+2 y+z=5, \\
x+y-z=2, \\
2 x+y-z=s .
\end{array}\right.
$$
Solving this system, we get $\left\{\begin{array}{l}x=s-2, \\ y=5-\frac{4 s}{3}, \\ z=-\frac{s}{3}+1 .\end{array}\right.$
Since $x$, $y$, and $z$ are non-negative rational numbers, we have $\left\{\begin{array}{l}s-2 \geqslant 0, \\ 5-\frac{4 s}{3} \geqslant 0, \\ -\frac{s}{3}+1 \geqslant 0 .\end{array} \quad\right.$ Solving these inequalities, we get $\left\{\begin{array}{l}s \geqslant 2, \\ s \leqslant \frac{15}{4}, \\ s \leqslant 3 .\end{array}\right.$
Therefore, the sum of the maximum and minimum values of $s$ is $3+2=$
5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 There are two rivers, A and B, each with a flow rate of $300 \mathrm{~m}^{3} / \mathrm{s}$, which converge at a certain point and continuously mix. Their sediment contents are $2 \mathrm{~kg} / \mathrm{m}^{3}$ and $0.2 \mathrm{~kg} / \mathrm{m}^{3}$, respectively. Assume that from the convergence point, there are several observation points along the shore. During the flow between two adjacent observation points, the mixing effect of the two streams is equivalent to the exchange of $100 \mathrm{~m}^{3}$ of water between the two streams in 1 second, i.e., $100 \mathrm{~m}^{3}$ of water flows from stream A to stream B, mixes, and then $100 \mathrm{~m}^{3}$ of water flows from stream B back to stream A, and mixes together. Question: From which observation point onwards will the difference in sediment content between the two streams be less than $0.01 \mathrm{~kg} / \mathrm{m}^{3}$ (ignoring sedimentation)?
(1999, Comprehensive Test for University Admission of Outstanding Students)
|
Solution: Let the sand content of two water flows be $a \, \text{kg} / \text{m}^{3}$ and $b \, \text{kg} / \text{m}^{3}$, and the water volumes flowing through in a unit time be $p \, \text{m}^{3}$ and $q \, \text{m}^{3}$, respectively. Then the sand content after mixing is
$$
\frac{a p + b q}{p + q} \, \text{kg} / \text{m}^{3}.
$$
At the $n$-th observation point, the sand content of water flow $A$ is $a_{n} \, \text{kg} / \text{m}^{3}$, and the sand content of water flow $B$ is $b_{n} \, \text{kg} / \text{m}^{3}$, where $n = 1, 2, \cdots$. Then
$$
\begin{aligned}
a_{1} & = 2, b_{1} = 0.2, \\
b_{n} & = \frac{1}{400} \left(300 b_{n-1} + 100 a_{n-1}\right) \\
& = \frac{1}{4} \left(3 b_{n-1} + a_{n-1}\right). \\
a_{n} & = \frac{1}{300} \left(200 a_{n-1} + 100 b_{n}\right) \\
& = \frac{1}{4} \left(3 a_{n-1} + b_{n-1}\right).
\end{aligned}
$$
Thus, $a_{n} - b_{n} = \frac{1}{2} \left(a_{n-1} - b_{n-1}\right)$
$$
= \cdots = \frac{1}{2^{n-1}} \left(a_{1} - b_{1}\right) = \frac{1}{2^{n-1}}.
$$
$$
2^{n-1} > 180, \quad n \geqslant 9.
$$
From this, we can see that starting from the 9th observation point, the difference in sand content between the two water flows is less than $0.01 \, \text{kg} / \text{m}^{3}$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 19 drawers arranged in a row, place $n$ identical balls into them such that the number of balls in each drawer does not exceed the number of balls in the drawer to its left. Let the number of such arrangements be denoted by $F_{m, n}$.
(1) Find $F_{1, n}$;
(2) If $F_{m, 0}=1$, prove:
$$
F_{m, n}=\left\{\begin{array}{ll}
F_{n, n}, & (m>n \geqslant 1) \\
F_{m-1, n}+F_{m, n-m} . & (1<m \leqslant n)
\end{array}\right.
$$
(3) Calculate $F_{3,8}$.
(1991, Jiangsu Province Mathematical Summer Camp)
|
Solution: (1) When $m=1$, there is obviously only one way, i.e., $F_{1, n}=1$.
(2) When $m>n \geqslant 1$, for any method that meets the conditions, there will be no balls in all the drawers to the right of the $n$-th drawer, so $F_{m, n}=F_{n, n}$.
When $1$ 0, if we reduce the number of balls in each drawer by 1, we clearly get a method of placing $n-m$ balls into $m$ drawers. Conversely, for any method of placing $n-m$ balls into $m$ drawers, if we increase the number of balls in each drawer by 1, we get a method of placing $n$ balls into $m$ drawers such that $x_{m}>0$. Therefore, $F_{m, n}=F_{m-1, n}+F_{m, n-m}$.
(3) From the result in (2), it is not difficult to deduce that $F_{3,8}=10$.
Example 11 Let $P_{1}$ be a point on the side $AB$ of the equilateral $\triangle ABC$. Draw a perpendicular from $P_{1}$ to the side $BC$, with the foot of the perpendicular being $Q_{1}$. Draw a perpendicular from $Q_{1}$ to the side $CA$, with the foot of the perpendicular being $R_{1}$. Draw a perpendicular from $R_{1}$ to the side $AB$, with the foot of the perpendicular being $P_{2}$, and so on, to get points $Q_{2}, R_{2}, P_{3}, Q_{3}, R_{3}, \cdots$. When $n$ $\rightarrow \infty$, which point does $P_{n}$ approach infinitely?
Solution: As shown in Figure 2, let $B P_{n}=x_{n}, A B=a$, then $B P_{n+1}=x_{n+1}$. Find the relationship between $x_{n}$ and $x_{n+1}$:
$$
\begin{array}{l}
B P_{n}=x_{n} \\
\rightarrow B Q_{n}=\frac{1}{2} x_{n} \\
\rightarrow Q_{n} C=a-\frac{1}{2} x_{n} \\
\rightarrow C R_{n}=\frac{1}{2}\left(a-\frac{1}{2} x_{n}\right) \\
\rightarrow A R_{n}=a-C R_{n}=\frac{1}{2} a+\frac{1}{4} x_{n} \\
\rightarrow A P_{n+1}=\frac{1}{2} A R_{n}=\frac{1}{4} a+\frac{1}{8} x_{n} \\
\rightarrow B P_{n+1}=x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a, \\
\text { hence } x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a .(n \geqslant 1)
\end{array}
$$
Thus, $x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a \cdot(n \geqslant 1)$
It is easy to get $x_{n}=\frac{2}{3} a+\left(x_{1}-\frac{2}{3} a\right)\left(-\frac{1}{8}\right)^{n-1}$
Therefore, $\lim x_{n}=\frac{2}{3} a$, which means that point $P_{n}$ approaches infinitely close to the point that divides $AB$ in the ratio $1:2$.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-$ $a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics Competition)
|
Solution: According to the characteristics of the required expression, the given equation can be regarded as a quadratic equation in $a$ or $b+c$. Therefore, there are two methods of solution.
Solution One: Transform the given equation into a quadratic equation in $a$
$$
a^{2}-(b+c) a+\frac{1}{4}(b+c)^{2}=0 .
$$
Its discriminant: $\Delta=(b+c)^{2}-4 \times \frac{1}{4}(b+c)^{2}=0$. Therefore, equation (1) has two equal roots $a=\frac{b+c}{2}$. Thus, $\frac{b+c}{a}=2$.
Solution Two: Transform the given equation into a quadratic equation in $(b+c)$
$$
(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
$$
Its discriminant $\Delta=16 a^{2}-4 \times 4 a^{2}=0$. Therefore, equation (2) has two equal roots $b+c=2 a$. Thus, $\frac{b+c}{a}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$. Try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$
and write down the derivation process.
(Xu Yichao, provided)
|
II. The answer to this question is 0. Below, we will prove this combinatorial identity by constructing polynomials and using interpolation and difference methods.
Method 1: Construct the polynomial
$$
\begin{aligned}
f(x) & =\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \\
& \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
\end{aligned}
$$
We will appropriately choose the coefficients $a_{i}(0 \leqslant i \leqslant n)$ such that
$$
f(x) \equiv 0 .
$$
Notice that $f(x)$ is a polynomial of degree at most $n$, so $f(x) \equiv 0$ if and only if $f(x)$ has $n+1$ roots $0, -1, -2, \cdots, -n$. Therefore, for $0 \leqslant i \leqslant n$, we should have
$$
\begin{aligned}
0= & f(-i) \\
= & a_{i}(-i)(-i+1) \cdots(-i+i-1)(-i+i+ \\
& 1) \cdots(-i+n)-(-i-m-1) \cdots(-i \\
& -m-n) \cdot \\
= & (-1)^{i} i!(n-i)!a_{i}-(-1)^{n} \\
& \cdot \frac{(m+n+i)!}{m+i)!},
\end{aligned}
$$
which implies $a_{i}:=(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!}$. Thus, we obtain the algebraic identity
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} x(x+1) \cdots \\
\quad \cdot(x+i-1)(x+i+1) \cdots(x+n) \\
=(x-m-1) \cdots(x-m-n) .
\end{array}
$$
In particular, substituting $x=n+k$ in the above equation, and given $1 \leqslant k \leqslant m \leqslant n$, we have $m+1 \leqslant n+k \leqslant m+n$, so the right-hand side of the equation is 0. Therefore,
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{n+i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \\
\cdot k(k+1) \cdots(n+k+i-1)(n+k+i) \\
\cdot(n+k+i+1) \cdots(2 n+k)=0 .
\end{array}
$$
By canceling the factor independent of $i$,
$$
(-1)^{n} k(k+1) \cdots(2 n+k)
$$
we obtain the desired result.
Method 2: Let
$$
g(x)=\frac{(x+m+1)(x+m+2) \cdots(x+m+n)}{x+n+k} .
$$
Given $1 \leqslant k \leqslant m \leqslant n$, we have $m+1 \leqslant n+k \leqslant m+n$. Therefore, $g(x)$ is a polynomial of degree $n-1$.
By the difference formula, the $n$-th difference of $g(x)$ at 0 is
$$
\begin{array}{l}
\Delta^{n} g(0)=\sum_{i=0}^{n}(-1)^{i} \mathrm{C}_{\mathrm{m}}^{i} g(i) \\
=\sum_{i=0}^{n}(-1)^{i} \frac{n!}{i!(n-i)!} \cdot \frac{(m+n+i)!}{(m+i)!} \cdot \frac{1}{n+k+i} .
\end{array}
$$
We know that the $n$-th difference of an $(n-1)$-degree polynomial is always 0, hence
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \cdot \frac{1}{n+k+i} \\
=\frac{\Delta^{n} g(0)}{n!}=0 .
\end{array}
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Given real numbers $x, y, z$ satisfy $x+y=5$, $z^{2}=x y+y-9$. Then, $x+2 y+3 z=$ $\qquad$
(1995, Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
|
Solution: The problem provides two equations with three unknowns. Generally speaking, when the number of unknowns exceeds the number of equations, it is impossible to solve for each unknown individually. However, under the condition that $x$, $y$, and $z$ are all real numbers, a solution may still be possible. In fact, from $x+y=5$, we get $x=5-y$, substituting this into the second equation, we have
$$
z^{2}=y(5-y)+y-9=-(y-3)^{2}.
$$
The left side of the equation is a non-negative real number, so the right side must also be a non-negative real number. However, $-(y-3)^{2}$ is a non-positive real number. Therefore, $y-3=0$. Thus, $y=3$. We can then find
$$
\begin{array}{l}
x=5-y=2, z^{2}=0, z=0 . \\
\text { Hence } x+2 y+3 z=2+2 \times 3+3 \times 0=8 .
\end{array}
$$
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a<b<c<d$. If variables $x, y, z, t$ are a permutation of $a, b, c, d$, then the expression
$$
\begin{array}{l}
n(x, y, z, t)=(x-y)^{2}+(y-z)^{2} \\
\quad+(z-t)^{2}+(t-x)^{2}
\end{array}
$$
can take different values.
|
$=、 1.3$.
If we add two more terms $(x-z)^{2}$ and $(y-t)^{2}$ to $n(x, y, z, t)$, then $n(x, y, z, t)+(x-z)^{2}+(y-t)^{2}$ becomes a fully symmetric expression in terms of $x, y, z, t$. Therefore, the different values of $n(x, y, z, t)$ depend only on the different values of $(x-z)^{2}+(y-t)^{2}=\left(x^{2}+y^{2}+z^{2}+t^{2}\right)-2(x z+y t)$. The first term on the right side, $\left(x^{2}+y^{2}+z^{2}+t^{2}\right)$, is also fully symmetric, so $n$ takes different values only depending on $x z+y t$, which has exactly three different values: $a b+c d, a c+b d, a d+b c$. In fact,
$$
\begin{array}{l}
(a b+c d)-(a c+b d)=a(b-c)+d(c-b) \\
=(b-c)(a-d)>0 .
\end{array}
$$
Thus, $a b+c d>a c+b d$.
Similarly, $a c+b d>a d+b c$.
Therefore, $n(x, y, z, t)$ can take three different values.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the positive direction of the $y$-axis at point $B$. Find the number of isosceles right triangles inscribed in the ellipse with point $B$ as the right-angle vertex.
---
Please note that the translation preserves the original format and line breaks.
|
Let the right-angled isosceles triangle inscribed in the ellipse be $\triangle A B C$, and suppose the equation of $A B$ is
\[
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=b+t \sin \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (1) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+2 a^{2} b t \sin \alpha=0.
\]
$t=0$ corresponds to point $B$, which we discard. Therefore,
\[
|A B|=-t=\frac{2 a^{2} b \sin \alpha}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}.
\]
Suppose the equation of $B C$ is
\[
\left\{\begin{array}{l}
x=t \cos \left(\alpha+90^{\circ}\right)=-t \sin \alpha, \\
y=b+t \sin \left(\alpha+90^{\circ}\right)=b+t \cos \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (3) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha\right) t^{2}+2 a^{2} b t \cos \alpha=0.
\]
$t=0$ corresponds to point $B$, which we discard. Therefore,
\[
|B C|=-t=\frac{2 a^{2} b \cos \alpha}{b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha}.
\]
Since $|B A|=|B C|$, from (2) and (4) we have
\[
\frac{2 a^{2} b \sin \alpha}{b^{2} \cdot \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}=\frac{2 a^{2} b \cos \alpha}{b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha}.
\]
Rearranging the above equation, we get
\[
(\tan \alpha-1)\left[b^{2} \tan ^{2} \alpha+\left(b^{2}-a^{2}\right) \tan \alpha+b^{2}\right]=0.
\]
From $\tan \alpha-1=0$, we have $\alpha=45^{\circ}$, indicating that $\triangle A B C$ is an isosceles right-angled triangle symmetric about the y-axis;
From $b^{2} \tan ^{2} \alpha+\left(b^{2}-a^{2}\right) \tan \alpha+b^{2}=0$, the discriminant $\Delta=\left(b^{2}-a^{2}\right)^{2}-4 b^{4}=\left(a^{2}-3 b^{2}\right)\left(a^{2}+b^{2}\right)$. When $\Delta>0$, i.e., $a>\sqrt{3} b$, there are two such triangles; when $\Delta=0$, i.e., $a^{2}=3 b^{2}$, we have $\tan \alpha=1$.
Therefore, when $a>\sqrt{3} b$, there are two such triangles; when $b<a \leqslant \sqrt{3} b$, there is one such triangle.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 50 points) On a straight ruler of length $36 \mathrm{~cm}$, mark $n$ graduations so that the ruler can measure any integer $\mathrm{cm}$ length in the range $[1,36]$ in one go. Find the minimum value of $n$.
|
Three, if the ruler is marked with 7 (or fewer than 7) graduations, we can prove that it is impossible to measure any integer length in the range [1, 6] cm in a single measurement.
In fact, 7 graduations, including the two end lines of the ruler, total 9 lines, which have 36 different combinations. Therefore, 7 graduations can measure at most 36 different lengths.
If there are two segments of the same length among the 8 segments created by 7 graduations, the number of different lengths that can be measured will be less than 36, clearly failing to meet the measurement requirement.
If none of the 8 segments created by 7 graduations have the same length, note that \(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\). It is easy to see that at this point, there is one segment of each length from \(1 \mathrm{~cm}\) to \(8 \mathrm{~cm}\).
\(1^{\circ}\) If the segment of length \(1 \mathrm{~cm}\) is not at one end of the ruler, then it cannot measure \(35 \mathrm{~cm}\); if the segment of length \(2 \mathrm{~cm}\) is not at one end of the ruler, then it cannot measure \(34 \mathrm{~cm}\).
\(2^{\circ}\) If the segment of length \(1 \mathrm{~cm}\) is at one end of the ruler and the segment of length \(2 \mathrm{~cm}\) is at the other end, then if the segment of length \(3 \mathrm{~cm}\) is not adjacent to the segment of length \(1 \mathrm{~cm}\), it cannot measure \(32 \mathrm{~cm}\); if the segment of length \(3 \mathrm{~cm}\) is adjacent to the segment of length \(1 \mathrm{~cm}\), it cannot measure \(31 \mathrm{~cm}\).
From this, it is evident that to meet the measurement requirement, the number of graduations on the ruler must be at least 8.
8 graduations divide the ruler into 9 segments. When the lengths of these 9 segments are \(1, 2, 3, 7, 7, 7, 4, 4, 1\), it is known that it can achieve the measurement of any integer length in the range \([1, 36]\) cm in a single measurement.
Therefore, the conclusion of this problem is: the minimum value of \(n\) is 8, meaning that at least 8 graduations must be marked on a 36 cm ruler to possibly achieve (the given graduation distribution is one method) the measurement of any integer length in the range \([1, 36]\) cm in a single measurement.
(Contributor: Liu Yue, Tongling No. 3 Middle School, Anhui Province, 244000)
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that the bases of two congruent regular triangular pyramids are glued together, exactly forming a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
(1996, National High School League)
|
Analysis: There are two difficulties in this problem: one is to determine which of $AC$ and $CD$ is 2 in length; the other is to determine which of the line segments $AC$, $CD$, and $AB$ represents the distance between the farthest two vertices.
Solution: As shown in Figure 6, construct $CE \perp AD$, and connect $EF$. It is easy to prove that $EF \perp AD$, so $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let point $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed by plane $ACD$ and plane $BCD$. Given that $\angle CEF = \angle AGB$. Let the side length of the base $\triangle CDF$ be $2a$, and the length of the side edge $AC$ be $b$. In $\triangle ACD$, $CE \cdot b = AG \cdot 2a$, thus
$$
CE = \frac{AG \cdot 2a}{b} = \frac{\sqrt{b^2 - a^2} \cdot 2a}{b}.
$$
In $\triangle ABC$, it is easy to find that
$$
\begin{aligned}
AB & = 2 \sqrt{b^2 - \left(\frac{2}{3} \sqrt{3} a\right)^2} \\
& = 2 \sqrt{b^2 - \frac{4}{3} a^2}.
\end{aligned}
$$
From $\triangle CEF \sim \triangle AGB$, we get
$$
\frac{AB}{CF} = \frac{AG}{CE},
$$
which means $\frac{2 \sqrt{b^2 - \frac{4}{3} a^2}}{2a} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{b^2 - a^2} \cdot 2a}{b}}$.
Thus, $b = \frac{4}{3} a = \frac{2}{3} \times 2a$.
Therefore, $AC < CD$.
Given that $b = 2$, we have $2a = 3$, and $AB = 2$.
Thus, $AC = AB < CD$.
Hence, the distance between the farthest two vertices is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Select $k$ edges and face diagonals from a cube such that any two line segments are skew lines. What is the maximum value of $k$?
|
(提示:考察线段 $A C 、 B C_{1} 、 D_{1} B_{1} 、 A_{1} D$, 它们所在的直线两两都是异面直线. 若存在 5 条或 5 条以上满足条件的线段, 则它们的端点相异, 且不少于 10 个, 这与正方体只有 8 个端点矛盾, 故 $k$ 的最大值是 4.)
(Translation: Consider the line segments $A C, B C_{1}, D_{1} B_{1}, A_{1} D$, the lines on which they lie are pairwise skew lines. If there are 5 or more line segments that satisfy the condition, then their endpoints are distinct and there are at least 10 of them, which contradicts the fact that a cube has only 8 vertices. Therefore, the maximum value of $k$ is 4.)
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.5. Let $M$ be a finite set of numbers. It is known that from any 3 elements of it, two numbers can be found whose sum belongs to $M$. How many elements can $M$ have at most?
|
10.5.7.
An example of a set of numbers consisting of 7 elements is: $\{-3,-2, -1,0,1,2,3\}$.
We will prove that for $m \geqslant 8$, any set of numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ does not have the required property. Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>\cdots>a_{m}$ and $a_{4}>0$ (since multiplying each number by -1 does not change our property). Thus, $a_{1}+a_{2}>a_{1}+a_{3}>a_{1}+a_{4}>a_{1}$, so the sums $a_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}$ do not belong to the set $A$. Furthermore, the sums $a_{2}+a_{3}$ and $a_{2}+a_{4}$ cannot both belong to the set $A$, because $a_{2}+a_{3}>a_{2}, a_{2}+a_{4}>a_{2}$, and $a_{2}+a_{3} \neq a_{2}+a_{4}$. Therefore, for the tuples $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(a_{1}, a_{2}, a_{4}\right)$, at least one of the tuples has the property that the sum of any two of its elements is not an element of $A$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$, and the quadratic equation with $\alpha^{2}, \beta^{2}$ as roots is still $x^{2}-p x+q=0$. Then the number of pairs $(p, q)$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 0
|
5. (B).
From the problem,
$\left\{\begin{array}{l}\alpha+\beta=p, \\ \alpha \beta=q\end{array}\right.$ and $\left\{\begin{array}{l}\alpha^{2}+\beta^{2}=p, \\ \alpha^{2} \beta^{2}=q .\end{array}\right.$
Thus, $q^{2}=q$. Therefore, $q=0$ or $q=1$.
And we have $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$, which means $p^{2}=p+2 q$.
(1) When $q=1$, $p=2$ or $p=-1$;
(2) When $q=0$, $p=0$ or $p=1$.
But $\alpha^{2}+\beta^{2}=p \geqslant 0$, so $p=-1$ (discard).
Therefore, the pairs $(p, q)$ that satisfy the problem can be $(2,1),(0,0)$, $(1,0)$, a total of 3 pairs.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $a, b, c$ are all non-zero, and $a+b+c=$
0. Then
$$
=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)
$$
|
\begin{aligned} \text { II.1. } & -3 . \\ \text { Original expression }= & a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)+b\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \\ & +c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ = & \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c)-3=-3 .\end{aligned}
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2,
Square $A B C D$ has a side length of $1, E$ is a point on the extension of $C B$, connect $E D$ intersecting $A B$ at $P$, and $P E$ $=\sqrt{3}$. Then the value of $B E-P B$ is $\qquad$
|
4.1.
Let $B E=x, P B=y$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=(\sqrt{3})^{2}, \\
\frac{y}{1}=\frac{x}{x+1} .
\end{array}\right.
$$
From (2), we have $x-y=x y$.
From (1) and (3), we have
$$
\begin{array}{l}
(x-y)^{2}+2(x-y)-3=0, \\
(x-y+3)(x-y-1)=0 .
\end{array}
$$
Clearly, $x>y, x-y+3>0$.
Therefore, $x-y-1=0, x-y=1$.
Thus, the value of $B E-P B$ is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{100} a_{n}}$ $=\frac{q}{p}$, where $p, q$ are integers, and $(p, q)=1$. Then $p+q=(\quad)$.
(A) 3
(B) 1023
(C) 2000
(D) 2001
|
-1 (A).
$$
\begin{array}{l}
\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n} \\
=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right) . \\
\text{Since } a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}, \\
\text{therefore } a_{2} a_{3} \cdots a_{1023}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdots \cdots \frac{\lg 1024}{\lg 1023}=\frac{\lg 1024}{\lg 2} \\
=\log _{2} 1024=10 . \\
\text{Hence } \sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\log _{100} 10=\frac{1}{2}, \text{ i.e., } \frac{q}{p}=\frac{1}{2} . \\
\text{Since } (p, q)=1, \\
\text{therefore } p=2, q=1 .
\end{array}
$$
Thus $p+q=3$.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that the polynomial $3 x^{3}+a x^{2}+b x+1$ can be divided by $x^{2}+1$, and the quotient is $3 x+1$. Then, the value of $(-a)^{i}$ is $\qquad$
（Dinghe Hesheng Junior High School Math Competition）
|
Solution: According to the polynomial identity, we have
$$
3 x^{3}+a x^{2}+b x+1=\left(x^{2}+1\right)(3 x+1) \text {. }
$$
Taking $x=1$ gives $a+b+4=8$.
Taking $x=-1$ gives $a-b-2=-4$.
Solving these, we get $a=1, b=3$.
$$
\therefore(-a)^{b}=(-1)^{3}=-1 \text {. }
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Third question: There are $n$ people, and it is known that any two of them make at most one phone call. Any $n-2$ of them have the same total number of phone calls, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
---
The translation maintains the original text's format and line breaks.
|
Solution 1: Clearly, $n \geqslant 5$. Let the $n$ people be $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}$ and $A_{j}$ make a phone call, then connect $A_{i} A_{j}$. Therefore, there must be line segments among these $n$ points. Without loss of generality, assume there is a line segment between $A_{1} A_{2}$:
If there is no line segment between $A_{1} A_{3}$, consider the $n-2$ points $A_{1} A_{4} A_{5} \cdots A_{n}$, $A_{2} A_{4} A_{5} \cdots A_{n}$, and $A_{3} A_{4} A_{5} \cdots A_{n}$. By the problem's condition, the total number of line segments between $A_{1}$, $A_{2}$, and $A_{3}$ and $A_{4}$, $A_{5}$, $\cdots$, $A_{n}$ are all equal, denoted as $m$.
Adding $A_{2}$ to $A_{1} A_{4} A_{5} \cdots A_{n}$, the total number of line segments among these $n-1$ points is $s=3^{k}+m+1$. Removing any one point from these $n-1$ points, the total number of line segments among the remaining $n-2$ points is $3^{k}$, so each point is connected to the other $n-2$ points by $m+1$ line segments. Thus,
$$
s=\frac{1}{2}(n-1)(m+1).
$$
Adding $A_{3}$ to $A_{1} A_{4} A_{5} \cdots A_{n}$, the total number of line segments among these $n-1$ points is $t=3^{k}+m$. Similarly, we get $t=\frac{1}{2}(n-1) m$.
Therefore, from $s=t+1$ we have
$\frac{1}{2}(n-1)(m+1)=\frac{1}{2}(n-1) m+1$, which implies $n=3$, a contradiction. So there must be a line segment between $A_{1} A_{3}$.
Similarly, there must be a line segment between $A_{2} A_{3}$. Therefore, $A_{1}$ and $A_{2}$ are connected to all $A_{i}(i=3,4, \cdots, n)$.
For $A_{i}$ and $A_{j}(i \neq j)$, since $A_{i}$ is connected to $A_{1}$, similarly $A_{i}$ is connected to $A_{j}$. Therefore, $A_{i}$ and $A_{j}$ $(i, j=1,2, \cdots, n)$ are all connected.
Thus, $3^{k}=\frac{1}{2}(n-2)(n-3)$. Hence, $n=5$.
(Provided by Class 1, Senior 3, Xinhua High School, Tianjin, 300201, Tutor: Wang Hao)
Solution 2: Clearly, $n \geqslant 5$. Let the $n$ people be $A_{1}$, $A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $d\left(A_{i}\right), i=1,2, \cdots, n$.
If $d\left(A_{1}\right)=0$, among $A_{3}, A_{4}, \cdots, A_{n}$, there must be one person, say $A_{3}$, who has called at least one of $A_{4}$, $A_{5}, \cdots, A_{n}$. Thus, the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ and $A_{3}, A_{4}, \cdots, A_{n}$ are not equal, a contradiction. Therefore, $d\left(A_{i}\right) \neq 0(i=1,2, \cdots, n)$.
If $d\left(A_{1}\right) \neq d\left(A_{2}\right)$, assume without loss of generality that $d\left(A_{1}\right)<d\left(A_{2}\right)$. Among $A_{3}, A_{4}, \cdots, A_{n}$, if there is one person, say $A_{3}$, who has called both $A_{1}$ and $A_{2}$ or neither, then the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ and $A_{2}, A_{4}, A_{5}, \cdots, A_{n}$ are not equal, a contradiction. If each person in $A_{3}, A_{4}, \cdots, A_{n}$ either calls $A_{1}$ but not $A_{2}$ or calls $A_{2}$ but not $A_{1}$, when all $A_{3}, A_{4}, \cdots, A_{n}$ call $A_{2}$, $A_{1}$ can only call $A_{2}$, so the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ is less than that between $A_{2}, A_{4}, \cdots, A_{n}$. When there is one person, say $A_{3}$, who calls $A_{1}$ but not $A_{2}$, the total number of calls between $A_{1}, A_{4}, \cdots, A_{n}$ is still less than that between $A_{2}, A_{4}, A_{5}, \cdots, A_{n}$, a contradiction. Therefore, $d\left(A_{i}\right)=d\left(A_{j}\right), i \neq j, i, j=1,2, \cdots, n$. Let this common value be $d$.
If $1 \leqslant a \leqslant n-2$, then there exist $A_{i}, A_{j}(i \neq j)$ who have not called each other, and there exist $A_{s}, A_{t}(s \neq t)$ who have called each other. The total number of calls among the $n-2$ people excluding $A_{i}, A_{j}$ is $\frac{1}{2} n d-2 d$, and the total number of calls among the $n-2$ people excluding $A_{s}, A_{t}$ is $\frac{1}{2} n d-2 d+1$, a contradiction. Therefore, $d=n-1$.
Thus, $\frac{1}{2} n(n-1)-2(n-1)+1=3^{k}$, i.e., $(n-2)(n-3)=2 \times 3^{k}$. Hence, $n=5$.
(Provided by Zhu Hengjie, Zibo Teaching Research Office, Shandong Province, 255033)
Solution 3: Clearly, $n \geqslant 5$. Since $n$ people can form $\mathrm{C}_{n}^{n-2}=\mathrm{C}_{n}^{2}$ "groups of $n-2$ people", the total number of calls among $n$ people, counted with multiplicity, is $3^{k} C_{n}^{2}$.
If two people make a call, this call is counted once in $\mathrm{C}_{n-2}^{n-4}=\mathrm{C}_{n-2}^{2}$ "groups of $n-2$ people". Therefore, the multiplicity of each call is $\mathrm{C}_{n-2}^{2}$. Hence, the actual total number of calls among $n$ people is $l=\frac{C_{n}^{2} 3^{k}}{\mathrm{C}_{n-2}^{2}}$, where $l$ is a natural number, i.e.,
$$
\frac{(n-2)(n-3)}{2} l=\frac{n(n-1)}{2} 3^{k}.
$$
Since one of $n$ and $n-2$ or $n-1$ and $n-3$ is a multiple of 4 and the other is an even number but not a multiple of 4, $\frac{1}{2}(n-2)(n-3)$ and $\frac{1}{2} n(n-1)$ are one odd and one even.
When $n_{1}$ is odd, $\frac{n_{1}-1}{2}=1, 3 n_{1}-2=7$, $7 \mid 3^{k}$, a contradiction. When $n_{1}$ is even, $n_{1}-1=1$, $\frac{1}{2}\left(3 n_{1}-2\right)=2,2 \mid 3^{k}$, a contradiction. Hence, $d=1$, and we have
$$
\left.\frac{1}{2}(n-2)(n-3) \right\rvert\, 3^{k}.
$$
If $n$ is even, since $\left(\frac{n-2}{2}, n-3\right)=1$, we have $\frac{n-2}{2}=1, n=4$, a contradiction.
If $n$ is odd, we have $\frac{n-3}{2}=1, n=5$.
When $n=5$, the condition is satisfied.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$. Then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$
(1990, National High School Mathematics Competition)
|
Solution: $\because a, b>0$,
$$
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1 \text {. }
$$
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}}$ $\geqslant 1$.
When $a=b=1$, the above expression $=1$, hence the minimum value is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 When $s$ and $t$ take all real numbers, then the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can achieve is $\qquad$
(1989, National High School Mathematics Competition)
|
Solution: As shown in Figure 1, the distance squared between any point on the line
$$
\left\{\begin{array}{l}
x=s+5, \\
y=s
\end{array}\right.
$$
and any point on the elliptical arc
$$
\left\{\begin{array}{l}
x=3|\cos t|, \\
y=2|\sin t|
\end{array}\right.
$$
is what we are looking for.
Indeed, the shortest distance is the perpendicular distance from point $A$ to the line, i.e., the distance squared from the point $(-3,0)$ to the line. Its value is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The ground floor of a hotel has 5 fewer rooms than the second floor. A tour group has 48 people. If all are arranged to stay on the ground floor, each room can accommodate 4 people, but there are not enough rooms; if each room accommodates 5 people, some rooms are not fully occupied. If all are arranged to stay on the second floor, each room can accommodate 3 people, but there are not enough rooms; if each room accommodates 4 people, some rooms are not fully occupied: The ground floor of this hotel has ( ) rooms.
(A) 9
(B) 10
(C) 11
(D) 12
|
5. (B).
Let the number of guest rooms on the ground floor be $x$, then the number of rooms on the second floor is $x+5$. According to the problem, we have the system of inequalities
$$
\left\{\begin{array}{l}
\frac{48}{5}<x<\frac{48}{4}, \\
\frac{48}{4}<x+5<\frac{48}{3} .
\end{array}\right.
$$
which simplifies to $\left\{\begin{array}{l}9.6<x<12, \\ 7<x<11\end{array} \Rightarrow 9.6<x<11\right.$.
Solving this, we get $x=10$.
|
10
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Cut a wire of length $143 \mathrm{~cm}$ into $n$ small segments $(n \geqslant 3)$, with each segment no less than $1 \mathrm{~cm}$. If no three segments can form a triangle, the maximum value of $n$ is
|
2.1.10.
Each segment should be as small as possible, and the sum of any two segments
should not exceed the third segment. A $143 \mathrm{~cm}$ wire should be divided into 10
such segments:
$1 \mathrm{~cm}, 1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}$,
$13 \mathrm{~cm}, 21 \mathrm{~cm}, 34 \mathrm{~cm}, 55 \mathrm{~cm}$
This is acceptable.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $x_{1999}-x_{601}=$ $\qquad$ .
|
3.0 .
From $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$ we get $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$.
Let $x_{n}=\tan \alpha_{n}$, then
$$
x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) \text {. }
$$
Therefore, $x_{n+6}=x_{n}$,
which means the sequence $\left\{x_{n}\right\}$ is a periodic sequence.
From $1999=333 \times 6+1,601=100 \times 6+1$ we get
$$
x_{1999}-x_{601}=0 \text {. }
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the surface area and volume of a circular cone are divided into upper and lower parts by a plane parallel to the base in the ratio $k$, then the minimum value of $c$ that makes $k c>1$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
4.7.
As shown in Figure 3, let $C O_{1}=r_{1}, A O=r_{2}$, the surface area of the smaller cone is $S_{1}$. The lateral surface area of the frustum is $S_{2}$, then
$$
\frac{S_{2}+S_{\text {base }}}{S_{1}}=\frac{1}{k},
$$
which means
$$
\frac{S_{1}}{S_{\text {lateral }}+S_{\text {base }}}=\frac{k}{k+1} \text {. }
$$
Given $S_{1}=\frac{\pi r_{1}^{2}}{\sin \theta}, S_{\text {lateral }}=\frac{\pi r_{2}^{2}}{\sin \theta}, S_{\text {base }}=\pi r_{2}^{2}$, we have
$$
\frac{k}{k+1}=\frac{r_{1}^{2}}{r_{2}^{2}} \cdot \frac{1}{1+\sin \theta} \text {. }
$$
Also, $\frac{V_{\text {frustum }}}{V_{\text {small cone }}}=\frac{1}{k}$, then
$$
\frac{V_{\text {cone }}}{V_{\text {small cone }}}=\frac{1+k}{k} \text {, }
$$
which means $\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{k}{k+1}$.
Thus,
$$
\begin{array}{l}
\left(\frac{k}{k+1}(1+\sin \theta)\right)^{\frac{1}{2}} \\
=\frac{r_{1}}{r_{2}}=\left(\frac{\dot{k}}{k+\dot{\mathrm{i}}}\right)^{\frac{1}{3}} .
\end{array}
$$
Therefore, $\frac{1}{k}=(1+\sin \theta)^{3}-1\frac{1}{k}=(1+\sin \theta)^{3}-1$.
Hence, the minimum value of $c$ is 7.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.12 friends have a weekly dinner together, each week they are divided into three groups, each group 4 people, and different groups sit at different tables. If it is required that any two of these friends sit at the same table at least once, then at least how many weeks are needed.
|
6.5.
First, for any individual, sitting with 3 different people each week, it would take at least 4 weeks.
Second, there are $C_{12}^{2}=66$ pairs among 12 people. Each table has $\mathrm{C}_{4}^{2}=6$ pairs, so in the first week, $3 \times 6=18$ pairs get to know each other.
Since 4 people sit at 3 tables, after the first week, at least two people at each table must have sat together in the first week. Therefore, the maximum number of new pairs that can form each week is $6-1=5$ pairs per table, totaling 15 pairs.
Since $18+15+15+15=63$, it is impossible to have all 66 pairs sit together in 4 weeks, so it must take 5 weeks.
Using 5 weeks is feasible. For example, 18 pairs in the first week, and 12 pairs each in the remaining 4 weeks, totaling $18+12 \times 4=66$ pairs. Below is a specific seating arrangement:
\begin{tabular}{crrrrrrrrrrrrrr}
Week & & \multicolumn{5}{c}{ Table 1} & & & & Table 2 & & & \multicolumn{5}{c}{ Table 3} & \\
1 & 1 & 2 & 3 & 4 & 5 & 6 & 9 & 10 & 7 & 8 & 11 & 12 \\
2 & 1 & 2 & 5 & 6 & 3 & 4 & 7 & 8 & 9 & 10 & 11 & 12 \\
3 & 1 & 2 & 7 & 8 & 3 & 4 & 9 & 10 & 5 & 6 & 11 & 12 \\
4 & 1 & 2 & 9 & 10 & 3 & 4 & 11 & 12 & 5 & 6 & 7 & 8 \\
5 & 1 & 2 & 11 & 12 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10
\end{tabular}
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Does there exist a prime number that remains prime when 16 and 20 are added to it? If so, can the number of such primes be determined?
|
Solution: Testing with prime numbers $2, 3, 5, 7, 11, 13 \cdots$, we find that 3 meets the requirement. Below, we use the elimination method to prove that apart from 3, no other prime number satisfies the requirement.
Divide the natural numbers into three categories: $3n, 3n+1, 3n+2$ $(n \in \mathbb{N})$,
$\because 2n+1+20 = 2(n+7)$ is a composite number,
$3r+2+16 = 3(n+6)$ is a composite number,
thus numbers of the form $3n+1$ and $3n+2$ do not become prime numbers when 16 and 20 are added.
Among the remaining numbers of the form $3n$, it is clear that except for 3, when $n>1$, they are all composite numbers.
Therefore, the only prime number that meets the condition is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given $a_{1}=1, a_{2}=\frac{5}{2}, a_{n+1}=$ $\frac{a_{n}+b_{n}}{2}, b_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$. Prove:
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .
$$
|
Proof: From the characteristic of the simultaneous recurrence relations, we know that $\left\{a_{n}\right\}$ is an arithmetic mean sequence, and $\left\{b_{n}\right\}$ is a harmonic mean sequence. By considering the inequality of means, we can insert the geometric mean between $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$. Therefore, we construct the auxiliary sequence $c_{n}=a_{n} b_{n}$.
From the simultaneous recurrence relations, we have
$$
\begin{aligned}
c_{n+1} & =a_{n+1} b_{n+1}=\frac{a_{n}+b_{n}}{2} \cdot \frac{2 a_{n} b_{n}}{a_{n}+b_{n}} \\
& =a_{n} b_{n}=c_{n} .
\end{aligned}
$$
$\therefore\left\{c_{n}\right\}$ is a constant sequence, with $c_{n}=c_{1}=a_{1} b_{1}$. From the initial conditions,
$$
\frac{5}{2}=a_{2}=\frac{a_{1}+b_{1}}{2}=\frac{1+b_{1}}{2},
$$
we get $b_{1}=4, a_{1}=1$.
$$
\begin{array}{l}
\therefore c_{n}=4, b_{n}=\frac{c_{n}}{a_{n}}=\frac{4}{a_{n}}, a_{n+1}=\frac{a_{n}^{2}+4}{2 a_{n}} . \\
\therefore \frac{a_{n+1}+2}{a_{n+1}-2}=\left(\frac{a_{n}+2}{a_{n}-2}\right)^{2}=\left(\frac{a_{n-1}+2}{a_{n-1}-2}\right)^{2^{2}} \\
=\cdots=\left(\frac{a_{1}+2}{a_{1}-2}\right)^{2^{n}}=\left(\frac{1+2}{1-2}\right)^{2^{n}}=3^{2^{n}} .
\end{array}
$$
Thus, $a_{n}+2=3^{2^{n-1}}\left(a_{n}-2\right)$.
Solving for $a_{n}$, we get $a_{n}=2 \frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1}$,
$$
\begin{array}{l}
b_{n}=\frac{4}{a_{n}}=2 \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1} . \\
\therefore \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} 2 \frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1}=2, \\
\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} 2 \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1}=2 .
\end{array}
$$
Therefore, $\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}$.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
As shown in Figure $3, D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $F$ is a point on the extension of $BC$. $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the radius of the circumcircle of $\triangle ABD$ is twice the radius of the circumcircle of $\triangle AEC$, and $BC=6, AB=4$, find the length of $BE$.
|
Three, (1) Two circles are externally tangent.
Draw the tangent line $l$ of $\odot A B D$, then $\angle 1=\angle B$.
$$
\begin{array}{l}
\because \angle 3=\angle B+\angle C, \\
\therefore \angle 3=\angle 1+\angle C . \\
\because \angle 1+\angle 2=\angle 3=\angle 1+\angle C, \\
\therefore \angle 2=\angle C .
\end{array}
$$
Draw $A P \perp l$ through $A$, intersecting $\odot A E C$ at point $P$, and connect $P E$.
$\because \angle P=\angle A C E$, then $\angle 2=\angle P$.
$$
\therefore \angle P A E+\angle P=90^{\circ} \text {. }
$$
Thus, $\angle A E P=90^{\circ}$, so $A P$ is the tangent line of $\odot A E C$, i.e., the circle is tangent at point $A$.
(2) Extend $D A$ to intersect $\odot A E C$ at $G$ (assuming $F$ is on $\odot A E C$), and connect $(S F$.
$$
\text { Since } \angle 4=\angle D A E+\angle A E D=\angle 3+\angle A F C \text {, }
$$
we have $\angle 4+\angle 5=180^{\circ}$, then $\angle 4=\angle A G F$.
$\therefore \triangle A D B \backsim \triangle A G F$.
$\therefore A B: A F=2$ (i.e., equal to the ratio of the radii of the two circles).
But $A B=4$,
$\therefore A F=2$. (This can also be done using the sine rule)
$\because B A \cdot B F=B E \cdot B C$,
$\therefore B E=4$.
(Provided by Yu Qiao)
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $m$ and $n$ are integers, the equation
$$
x^{2}+(n-2) \sqrt{n-1} x+m+18=0
$$
has two distinct real roots, and the equation
$$
x^{2}-(n-6) \sqrt{n-1} x+m-37=0
$$
has two equal real roots. Find the minimum value of $n$, and explain the reasoning.
|
1. $\left\{\begin{array}{l}n \geqslant 1, \\ (n-2)^{2}(n-1)-4(m+18)>0, \\ (n-6)^{2}(n-1)-4(m-37)=0 .\end{array}\right.$
(2) - (3), and rearrange to get
$$
(n-4)(n-1)>27.5 \text {, }
$$
then $n \geqslant 8$.
When $n=8$, $m=44$, which satisfies the given conditions. Therefore, the minimum value of $x$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a 、 b \in \mathbf{C}$. Then $a+a^{2}+a^{3}+\cdots+a^{2000}+b+b^{2}$ $+b^{3}+\cdots+b^{2000}=$
|
3. -2 .
From the known $f(-x)=f(x), g(-x)=-g(x)$ we get
$$
\begin{array}{l}
\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b x \\
=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x . \\
2^{-x}+\frac{a+b}{2^{-x}}=-\left(2^{x}+\frac{a+b}{2^{x}}\right) .
\end{array}
$$
Simplifying, from equation (1) we get $a b=1$, and from equation (2) we get $a+b=-1$. Therefore, $a$ and $b$ are a pair of conjugate complex roots of the equation $x^{2}+x+1=0$, which are the imaginary cube roots of 1. Then
$$
\begin{aligned}
a & +a^{2}+a^{3}+\cdots+a^{1999}+a^{2000}+b+b^{2} \\
& +b^{3}+\cdots+b^{1999}+b^{2000} \\
= & a^{1999}+a^{2000}+b^{1999}+b^{2000} . \\
= & a+a^{2}+b+b^{2}=-1-1=-2 .
\end{aligned}
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, if
the three sides of $\triangle A B C$ are
$n+x, n+2 x, n$
$+3 x$, and the height $A D$ from $B C$ is $n$, where $n$
is a positive integer, and $0<x \leqslant$
1. Then the number of triangles that satisfy the above conditions is $\qquad$.
|
6.12 .
Let $a=n+x, b=n+2 x, c=n+3 x, s=\frac{1}{2}(a+b+c)$. By Heron's formula and the general area formula, we have
$$
\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2} n(n+2 x) .
$$
Simplifying, we get $12 x=n$.
And $0<x=\frac{n}{12} \leqslant 1$,
so $0<n \leqslant 12$,
which means $n$ has exactly 12 possibilities (correspondingly giving $x$ values).
Therefore, the number of triangles sought is 12 .
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $x, y \in \mathbf{R}$. If $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$ .
|
2.4.
Given that $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
y=3-2 x \text{. }
$$
From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, it follows that
$$
|x+1|+|x-1|=2 \text{. }
$$
Thus, $y+3, 2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text{. }
$$
Substituting (1) into (2), we get $2 x^{2}-6 x+4=0$.
Solving this, we find $x_{1}=2$ (discard) and $x_{2}=1$. Hence, $y=1$.
Therefore, $(x+1)(y+1)=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R$, where $O$ is the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\qquad$ .
|
5.2.
Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter ).
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta}$.
Then $|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$.
In (1), let $x=0$, we get $t=\frac{a}{\cos \theta}$.
Then $|A R|=\frac{a}{|\cos \theta|}$.
Also, let $O P:\left\{\begin{array}{l}x=t^{\prime} \cos \theta \\ y=t^{\prime} \sin \theta\end{array}\right.$ ( $t^{\prime}$ is a parameter).
Substitute (2) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t^{\prime 2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$.
Then $|O P|^{2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$.
Therefore, $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=\ldots=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the equation about $x$
Figure 3 $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7)\left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1}=\frac{3}{11}$, find the value of $a$.
|
15.1) Let $\frac{x}{x-1}=t$, then $t \neq 1$. The original equation can be transformed into $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes
$-9 t+1=0$ or $-5 t+1=0$,
which means $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Thus, $x=-\frac{1}{8}$ or $x=\frac{1}{4}$.
Therefore, when $a= \pm 1$, the original equation has real roots.
When $a \neq \pm 1$, the original equation has real roots if $\Delta \geqslant 0$.
By $\Delta=[-(2 a+7)]^{2}-4\left(a^{2}-1\right) \geqslant 0$,
we get $a \geqslant-\frac{53}{28}$.
In summary, when $a \geqslant-\frac{53}{28}$, the original equation has real solutions.
(2) From the given, $\frac{x_{1}}{x_{1}-1}$ and $\frac{x_{2}}{x_{2}-1}$ are the two roots of the equation
$$
\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0
$$
Using the relationship between roots and coefficients, we get $\frac{2 a+7}{a^{2}-1}=\frac{3}{11}$. Thus,
$$
3 a^{2}-22 a-80=0,
$$
Solving this, we get $a_{1}=10, a_{2}=-\frac{8}{3}$.
From (1), we know that $a \geqslant-\frac{53}{28}$.
Since $a_{2}=-\frac{8}{3}<-\frac{53}{28}$, $a_{2}$ should be discarded.
Therefore, $a=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (16 points) (1) In a $4 \times 4$ grid paper, some small squares are painted red, and then 2 rows and 2 columns are crossed out. If no matter how they are crossed out, at least one red small square remains uncrossed, how many small squares must be painted red at least? Prove your conclusion.
(2) If the “$4 \times 4$ grid paper” in the previous problem is changed to “$n \times n$ grid paper $(n \geqslant 5)$,” with other conditions unchanged, how many small squares must be painted red at least? Prove your conclusion.
|
Three, (1) At least 7 cells need to be colored.
If the number of colored cells $\leqslant 4$, then by appropriately crossing out 2 rows and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 5, then at least one row has 2 cells colored. By crossing out this row, the remaining colored cells do not exceed 3. Then, by crossing out 1 more row and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 6, then at least one row has 3 cells colored, or at least two rows each have 2 cells colored. Therefore, by crossing out 2 rows, at least 4 colored cells can be crossed out, leaving no more than 2 colored cells. By crossing out 2 more columns, they can all be crossed out.
Color 7 cells as shown in Figure 4, then by crossing out 2 rows, at most 4 colored cells can be crossed out, and the remaining colored cells are in different 3 columns. By crossing out 2 more columns, they cannot all be crossed out.
(2) At least 5 cells need to be colored.
This is because, if the number of colored cells $\leqslant 4$, then by crossing out 2 rows and 2 columns, they can all be crossed out.
Color 5 cells as shown in Figure 5, then by arbitrarily crossing out 2 rows and 2 columns, there will be colored cells that are not crossed out.
|
5
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 There is a type of sports competition with $M$ events, and athletes $A$, $B$, and $C$ participate. In each event, the first, second, and third places receive $p_{1}$, $p_{2}$, and $p_{3}$ points respectively, where $p_{1}$, $p_{2}$, $p_{3} \in \mathbf{Z}^{+}$, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 22 points, $B$ and $C$ both score 9 points. It is also known that $B$ came first in the 100-meter dash. Find the value of $M$.
(18th Canadian High School Mathematics Competition)
|
Solution: Consider the total score of three people, we have
$$
\begin{array}{l}
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40 . \\
\because p_{1}, p_{2}, p_{3} \in \mathbf{Z}^{+}, \text { and } p_{1}>p_{2}>p_{3}, \\
\therefore p_{1}+p_{2}+p_{3} \geqslant 3+2+1=6 .
\end{array}
$$
Thus, $6 M \leqslant 40, M \leqslant 6$.
Since $p_{1}>p_{2}>p_{3}$, and $B$ and $C$ have the same score, then $M \neq 1$. In addition, $M$ must divide 40, so $M$ can be $2, 4, 5$.
(1) If $M=2$, i.e., there is one more competition besides the 100-meter dash. Since $B$ is the first in the 100-meter dash and has a total score of only 9, it must be that $9 \geqslant p_{1}+p_{3}$. Therefore, $p_{1} \leqslant 8$. This way, $A$ cannot score 22 points in two competitions.
(2) If $M=4$, then for $B$'s score, we have $9 \geqslant p_{1} + 3 p_{3}$. So $p_{1} \leqslant 6$.
If $p_{1} \leqslant 5$, then $A$ can score at most 19 points in 4 competitions (since $B$ is the first in the 100-meter dash), but $A$ actually scored 22 points, so $p_{1}=6, p_{3}=1$.
$$
\begin{array}{l}
\because 4\left(p_{1}+p_{2}+p_{3}\right)=40, \\
\therefore p_{2}+p_{3}=4 . \\
\text { Thus, } p_{2}=3 .
\end{array}
$$
But $A$ can win at most 3 first places (since $B$ is already the first in the 100-meter dash) and 1 second place, scoring at most $3 \times 6 + 3 = 21$. This contradicts the given conditions.
(3) If $M=5$, from $5\left(p_{1}+p_{2}+p_{3}\right)=40$, we get $p_{1}+p_{2}+p_{3}=8$.
If $p_{3} \geqslant 2$, then $p_{1}+p_{2}+p_{3} \geqslant 4+3+2=9$, which is a contradiction.
So, $p_{3}=1$.
Also, $p_{1}$ must be at least 5, otherwise $A$ can score at most $4 \times 4 + 3 = 19$ points in 5 competitions, which is a contradiction.
So, $p_{1} \geqslant 5$.
If $p_{1} \geqslant 6$, then $p_{2}+p_{3} \leqslant 2$, which is also impossible. Thus, only $p_{1}=5$, and hence, $p_{2}=2, p_{3}=1$.
In conclusion, $M=5$.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example $1\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}=$ $\qquad$
(1999, National Junior High School Mathematics League Wuhan Selection Competition)
|
Let $3^{999}=a, 5^{099}=b, 7^{999}=c$, then
$$
\begin{array}{l}
\text { Original expression }=\frac{c}{a} \sqrt{\frac{a^{2}+(a b)^{2}}{c^{2}+(b c)^{2}}} \\
=\frac{c}{a} \sqrt{\frac{a^{2}\left(1+b^{2}\right)}{c^{2}\left(1+b^{2}\right)}} . \\
=\frac{c}{a} \cdot \frac{a}{c}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $a^{2}+a+1=0$. Then, $a^{1992}+$ $a^{322}+a^{2}=$ $\qquad$
(Harbin 15th Junior High School Mathematics Competition)
|
Given the condition $a \neq 1$, from $(a-1)(a^2 + a + 1) = 0$, we get $a^3 - 1 = 0$, hence $a^3 = 1$.
Substituting 1 for $a^3$ and 0 for $a^2 + a + 1$, we get
$$
\begin{aligned}
\text { Original expression } & =\left(a^{3}\right)^{664}+\left(a^{3}\right)^{107} a+a^{2} \\
& =1^{664}+1^{107} a+a^{2} \\
& =1+a+a^{2}=0 .
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In a $3 \times 3$ square grid, fill in the numbers as shown in the table below. The operation on the table is as follows: each operation involves adding a number to two adjacent numbers in the grid (adjacent means two small squares that share a common edge).
\begin{tabular}{|l|l|l|}
\hline 0 & 3 & 2 \\
\hline 6 & 7 & 0 \\
\hline 4 & 9 & 5 \\
\hline
\end{tabular}
Can it be done after several operations such that
(1) all the numbers in the grid are 0;
(2) the numbers in the four corners are all 1, and the rest are 0.
|
(Tip: (1) After 5 operations, all cells can be 0.
(2) As shown in the table below, the invariant $S=a-$ $b+c-d+e-f+g-h+k$.
\begin{tabular}{|l|l|l|}
\hline$a$ & $b$ & $c$ \\
\hline$d$ & $e$ & $f$ \\
\hline$g$ & $h$ & $k$ \\
\hline
\end{tabular}
The initial state is
$$
S=0-3+2-6+7-0+4-9+5=0,
$$
The target state is
$$
S=1-0+1-0+0-0+1-0+1=4 \text {. }
$$
Therefore, it is impossible.)
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The solution to the equation $\sqrt{13-\sqrt{13+x}}$ $=x$ is $\qquad$
|
3.3.
Let $\sqrt{13+x}=y$ then
$$
\left\{\begin{array}{l}
\sqrt{13-y}=x, \\
\sqrt{13+x}=y .
\end{array}\right.
$$
(2) $)^{2}-(1)^{2}$ gives $(x+y)(x-y+1)=0$.
From the original equation, we know $x>0$, and from (2), $y>0$, so $x+y \neq 0$. Thus, $x-y+1=0$, which means $y=x+1$. Therefore, it is easy to get $x=3$ (discard $x=-4$).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) A chemical plant, starting from January this year, if it does not improve its production environment and continues to produce as it is, will earn 700,000 yuan per month. At the same time, it will receive penalties from the environmental protection department, with the first month's penalty being 30,000 yuan, and increasing by 20,000 yuan each subsequent month. If the plant invests 5,000,000 yuan to add recycling and purification equipment (assuming no time for equipment renovation), it can not only improve the environment but also significantly reduce raw material costs. According to estimates, the cumulative net income for the first 5 months after production is a quadratic function of production time \( n \) (in months). The cumulative income for the first 1, first 2, and first 3 months can reach 1,010,000 yuan, 2,040,000 yuan, and 3,090,000 yuan, respectively, and will stabilize at the level of the fifth month thereafter. At the same time, the plant will not only avoid penalties but also receive a one-time reward of 1,000,000 yuan from the environmental protection department. How many months will it take for the investment to start paying off, i.e., for the net income after the investment to exceed the net income without the investment?
|
Three, let the cumulative income for $n$ months without modifying the equipment, under the original conditions, be $a(n)$, and let the cumulative income for $n$ months after modifying the equipment be $b(n)$. From the given conditions, we have $a(n)=70 n$, and we can assume
$$
b(n)=a n^{2}+b n+c, \quad n \leqslant 5.
$$
Thus,
$$
\left\{\begin{array}{l}
b(1)=a+b+c=101, \\
b(2)=4 a+2 b+c=204, \\
b(3)=9 a+3 b+c=309.
\end{array}\right.
$$
Solving these, we get $a=1, b=100, c=0$.
$$
\therefore b(n)=\left\{\begin{array}{ll}
n^{2}+100 n, & (n \leqslant 5) \\
b(5)+(n-5 \llbracket b(5)-b(4)], & (n>5)
\end{array}\right.
$$
That is $\square$
$$
\left.\begin{array}{l}
b(n)=\left\{\begin{array}{l}
n^{2}+100 n, \quad(n \leqslant 5) \\
109 n-20. \quad(n>5)
\end{array}\right. \\
\because b(5)-500+100=125a(n)-\left[3 n+\frac{2 n(n-1)}{2}\right](n>5)
\end{array}
\right.
$$
That is $n^{2}+41 n-420>0$.
When $n \geqslant 9$, the above inequality holds.
Therefore, the investment will start to pay off after 9 months.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x, y, z \in \mathbf{R}, x y+y z+z x=-1$. Then the minimum value of $x^{2}+5 y^{2}+8 z^{2}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3.4 .
$$
\begin{array}{l}
\text { Given }(x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0 , \\
x^{2}+5 y^{2}+8 z^{2} \geqslant-4(x y+y z+z x)=4 ,
\end{array}
$$
we have $x=\frac{3}{2} , y=-\frac{1}{2} , z=-\frac{1}{4}$ ,
or $x=-\frac{3}{2}, y=\frac{1}{2}, z=\frac{1}{4}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $S_{n}$ be the sum of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \cdots, \frac{1}{2^{n-1}}\right\}$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=$ $\qquad$ .
|
6.1 .
For any element in set $A$, it appears in a subset containing 3 elements $\mathrm{C}_{n-1}^{2}$ times, then
$$
\begin{array}{l}
S_{n}=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{n-1}}\right) \mathrm{C}_{n-1}^{2} . \\
\text { Therefore, } \lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=\lim _{n \rightarrow \infty} \frac{\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}} \cdot \frac{(n-1)(n-2)}{2}}{n^{2}}=1 .
\end{array}
$$
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=$ $b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, $(n=1,2 \ldots)$. Find the value of $a_{2001}^{2}-2 b_{2001}^{2}$.
|
$$
\begin{array}{l}
a_{0}^{2}-2 b_{0}^{2}=-1, \\
a_{1}^{2}-2 b_{1}^{2}=1, \\
\ldots \ldots
\end{array}
$$
Conjecture $a_{n}^{2}-2 b_{n}^{2}=(-1)^{n+1}(n=0,1 \ldots)$.
We will prove this by mathematical induction.
Assume the proposition holds for $n=k$, that is,
$$
a_{k}^{2}-2 b_{k}^{2}=(-1)^{k+1} \text {. }
$$
Then for $n=k+1$,
$$
\begin{array}{l}
a_{k+1}^{2}-2 b_{k+1}^{2}=\left(a_{k}+2 b_{k}\right)^{2}-2\left(a_{k}+b_{k}\right)^{2} \\
=-\left(a_{k}^{2}-2 b_{k}^{2}\right)=(-1)^{k+2} .
\end{array}
$$
Therefore, by mathematical induction, the conjecture is true.
Thus, $a_{2001}^{2}-2 b_{2001}^{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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