竹枝词
刘禹锡 【朝代】唐
山桃红花满上头，蜀江春水拍山流。
花红易衰似郎意，水流无限似侬愁。
竹枝词二首
其一
杨柳青青江水平，闻郎江上唱歌声。
东边日出西边雨，道是无晴却有晴。
https://www.bilibili.com/bangumi/play/ep148473
chapter{AMC 10}
egin{ltbox}egin{example}
(2000 AMC 10 21) If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?
[ extrm{I. All alligators are creepy crawlers.}][ extrm{II. Some ferocious creatures are creepy crawlers.}][ extrm{III. Some alligators are not creepy crawlers.}]
$mathrm{(A)} ext{I only} qquadmathrm{(B)} ext{II only} qquadmathrm{(C)} ext{III only} qquadmathrm{(D)} ext{II and III only} qquadmathrm{(E)} ext{None must be true}$
end{example}end{ltbox}
egin{solution}
We interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.
To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as $A$, $C$, and $F$.
We got the following information:
If $x$ is an $A$, then $x$ is an $F$.
There is some $x$ that is a $C$ and at the same time an $A$.
We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are $A$s, but only Johnny is a $C$" meets both conditions, but the first statement is false.
We CAN conclude that the second statement is true. We know that there is some $x$ that is a $C$ and at the same time an $A$. Pick one such $x$ and call it Bobby. Additionally, we know that if $x$ is an $A$, then $x$ is an $F$. Bobby is an $A$, therefore Bobby is an $F$. And this is enough to prove the second statement -- Bobby is an $F$ that is also a $C$.
We CAN NOT conclude that the third statement is true. For example, consider the situation when $A$, $C$ and $F$ are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.
Therefore the answer is $oxed{ ext{(B) \, II only}}$.
end{solution}
egin{ltbox}egin{example}
(2000 AMC 10 22) One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$ ext {(A)} 3 qquad ext {(B)} 4 qquad ext {(C)} 5 qquad ext {(D)} 6 qquad ext {(E)} 7$
end{example}end{ltbox}
egin{solution}
Solution 1
Let $c$ be the total amount of coffee, $m$ of milk, and $p$ the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so[left(frac{c}{6} + frac{m}{4} ight)p = c + m]Regrouping, we get $2c(6-p)=3m(p-4)$. Since both $c,m$ are positive, it follows that $6-p$ and $p-4$ are also positive, which is only possible when $p = 5 mathrm{(C)}$.
Solution 2 (less rigorous)
One could notice that (since there are only two components to the mixture) Angela must have more than her "fair share" of milk and less then her "fair share" of coffee in order to ensure that everyone has $8$ ounces. The "fair share" is $1/p.$ So,
[frac{1}{6} < frac{1}{p}<frac{1}{4}]
Which requires that $p$ be $p = 5 mathrm{(C)},$ since $p$ is a whole number.
Solution 3
Again, let $c,$ $m,$ and $p$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is $8p,$ and also $c+m.$ Thus, $c+m = 8p,$ so $m = 8p-c$ and $c = 8p-m.$
We also know that the amount Angela drank, which is $frac{c}{6} + frac{m}{4},$ is equal to $8$ ounces, thus $frac{c}{6} + frac{m}{4} = 8.$ Rearranging gives[24p - c = 96.]Now notice that $c > 0$ (by the problem statement). In addition, $m > 0,$ so $c = 8p-m < 8p.$ Therefore, $0 < c < 8p,$ and so $24p > 24p-c > 16p.$ We know that $24p-c = 96,$ so[24p > 96 > 16p.]From the leftmost inequality, we get $p > 4,$ and from the rightmost inequality, we get $p < 6.$ The only possible value of $p$ is $p = 5 mathrm{(C)}$.
Solution 4
Let $c,$ $m,$ and $p$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. $c$ and $m$ obviously can't be $0$. We know $frac{c}{6} + frac{m}{4} = 8$ or $2c + 3m = 96$ and $c + m = 8p$ or $2c + 2m = 16p$. Then,[(2c + 2m) + m = 16p + m = 96]Because $16p$ and $96$ are both divisible by $16$, $m$ must also be divisible by $16$. Let $m = 16k$. Now,[3(16k) + 2c = 48k + 2c = 96]$k$ can't be $0$, otherwise $m$ is $0$, and $k$ can't be $2$, otherwise $c$ is $0$. Therefore $k$ must be $1$, $m = 16$ and $c = 24$. $c + m = 24 + 16 = 40 = 8p$. Therefore, $p = 5 mathrm{(C)}$.
Solution 5
Let $m$, $c$ be the total amounts of milk and coffee, respectively. In order to know the number of people, we first need to find the total amount of mixture $t = m + c$. We are given that[frac{m}{4} + frac{c}{6} = 8]Multiplying the equation by $4$ yields[m + frac{2}{3}c = (m + c) - frac{1}{3}c = t - frac{1}{3}c = 32]Since $frac{1}{3}c > 0$, we have $t > 32$. Now multiplying the equation by $6$ yields[frac{3}{2}m + c = (m+c) + frac{1}{3}m = t + frac{1}{3}m = 48]Since $frac{1}{3}m > 0$, we have $t < 48$. Thus, $32 < t < 48$.
Since $t$ is a multiple of $8$, the only possible value for $t$ in that range is $40$. Therefore, there are $frac{40}{8} = 5$ people in Angela's family. $mathrm{(C)}$.
end{solution}
egin{ltbox}egin{example}
(2000 AMC 10 23) When the mean, median, and mode of the list
[10,2,5,2,4,2,x]
are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $x$?
$ ext {(A)} 3 qquad ext {(B)} 6 qquad ext {(C)} 9 qquad ext {(D)} 17 qquad ext {(E)} 20$
end{example}end{ltbox}
egin{solution}
The mean is $frac{10+2+5+2+4+2+x}{7} = frac{25+x}{7}$.
Arranged in increasing order, the list is $2,2,2,4,5,10$, so the median is either $2,4$ or $x$ depending upon the value of $x$.
The mode is $2$, since it appears three times.
We apply casework upon the median:
If the median is $2$ ($x le 2$), then the arithmetic progression must be constant.
If the median is $4$ ($x ge 4$), because the mode is $2$, the mean can either be $0,3,6$ to form an arithmetic progression. Solving for $x$ yields $-25,-4,17$ respectively, of which only $17$ works because it is larger than $4$.
If the median is $x$ ($2 le x le 4$), we must have the arithmetic progression $2, x, frac{25+x}{7}$.
Thus, we find that $2x=2+frac{25+x}{7}$ so $x=3$.
The answer is $3 + 17 = 20 mathrm{(E)}$.
end{solution}
egin{ltbox}egin{example}
(2000 AMC 10 24) Let $f$ be a function for which $f(dfrac{x}{3}) = x^2 + x + 1$. Find the sum of all values of $z$ for which $f(3z) = 7$.
$ ext {(A)} -1/3 qquad ext {(B)} -1/9 qquad ext {(C)} 0 qquad ext {(D)} 5/9 qquad ext {(E)} 5/3$
end{example}end{ltbox}
egin{solution}
Solution 1
Let $y = frac{x}{3}$; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$. Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$, and $z = -frac{1}{3}, frac{2}{9}$. These sum up to $oxed{ extbf{(B) }-frac19}$.
Solution 2
Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-frac{b}{a}=-frac{9}{81}=oxed{ extbf{(B) }-frac19}$.
Solution 3
Set $f(frac{x}{3}) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $frac{x}{3} = 3z$ gives $x = 9z$), thus the sum of the roots in the equation $f(3z)=7$ is $-frac{1}{9}$ or $oxed{ extbf{(B) }-frac19}$.
Solution 4
Since we have $f(x/3)$, $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$. We set this equal to 7:
$81z^2 + 9z +1 = 7 Longrightarrow 81z^2 + 9z - 6 = 0$. For any quadratic $ax^2 + bx +c = 0$, the sum of the roots is $-frac{b}{a}$. Thus, the sum of the roots of this equation is $-frac{9}{81} = oxed{ extbf{(B) }-frac19}$.
end{solution}
egin{ltbox}egin{example}
(2000 AMC 10 25)
In year $N$, the $300^{ ext{th}}$ day of the year is a Tuesday. In year $N+1$, the $200^{ ext{th}}$ day is also a Tuesday. On what day of the week did the $100$th day of year $N-1$ occur?
$ ext {(A)} ext{Thursday} qquad ext {(B)} ext{Friday}qquad ext {(C)} ext{Saturday}qquad ext {(D)} ext{Sunday}qquad ext {(E)} ext{Monday}$
end{example}end{ltbox}
egin{solution}
Solution 1
There are either[65 + 200 = 265]or[66 + 200 = 266]days between the first two dates depending upon whether or not year $N$ is a leap year. Since $7$ divides into $266$ but not $265$, for both days to be a Tuesday, year $N$ must be a leap year.
Hence, year $N-1$ is not a leap year, and so since there are[265 + 300 = 565]days between the date in years $N, ext{ }N-1$, this leaves a remainder of $5$ upon division by $7$. Since we are subtracting days, we count 5 days before Tuesday, which gives us $oxed{mathbf{(A)} ext{Thursday}.}$
Solution 2
The $300-29cdot 7=97^{ ext{th}}$ day of year $N$ and the $200-15cdot 7=95^{ ext{th}}$ day of year $N+1$ are Tuesdays. If there were the same number of days in year $N$ and year $N-1,$ the $99^{ ext{th}}$ day of year $N-1$ will be a Tuesday. But year $N$ is a leap year because $97-95 cong 366 pmod{7},$ so the $98^{ ext{th}}$ day of year $N-1$ is a Tuesday. It follows that the $100^{ ext{th}}$ day of year $N-1$ is $oxed{mathbf{(A)} ext{Thursday}.}$
end{solution}