Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12A Problems/Problem 20"
Line 40: Line 40:
 
{{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}}
  
−
[Category: INTRODUCTORY SET THEORY PROBLEMS]]
+
[Category: Introductory Set Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:11, 21 April 2020

Problem 20

Let $S$ be the set $\{1,2,3,...,19\}$. For $a,b \in S$, define $a \succ b$ to mean that either $0 < a - b \le 9$ or $b - a > 9$. How many ordered triples $(x,y,z)$ of elements of $S$ have the property that $x \succ y$, $y \succ z$, and $z \succ x$?

$\textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988$

Solution

Imagine 19 numbers are just 19 persons sitting evenly around a circle $C$; each of them is facing to the center.

One may check that $x \succ y$ if and only if $y$ is one of the 9 persons on the left of $x$, and $y \succ x$ if and only if $y$ is one of the 9 persons on the right of $x$. Therefore, "$x \succ y$ and $y \succ z$ and $z \succ x$" implies that $x, y, z$ cuts the circumference of $C$ into three arcs, each of which has no more than $10$ numbers sitting on it (inclusive).

We count the complement: where the cut generated by $(x,y,z)$ has ONE arc that has more than $10$ persons sitting on. Note that there can only be one such arc because there are only $19$ persons in total.

Suppose the number of persons on the longest arc is $k>10$. Then two places of $x,y,z$ are just chosen from the two end-points of the arc, and there are $19-k$ possible places for the third person. Once the three places of $x,y,z$ are chosen, there are three possible ways to put $x,y,z$ into them clockwise. Also, note that for any $k>10$, there are $19$ ways to choose an arc of length $k$. Therefore the total number of ways (of the complement) is

\[\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36\]

So the answer is

\[3\cdot \binom{19}{3} - 3\cdot 19\cdot 36 = 3\cdot 19 \cdot (51 - 36) = 855\]

NOTE: this multiple choice problem can be done even faster -- after we realized the fact that each choice of the three places of $x,y,z$ corresponds to $3$ possible ways to put them in, and that each arc of length $k>10$ has $19$ equitable positions, it is evident that the answer should be divisible by $3\cdot 19$, which can only be $855$ from the five choices.

Solution 2

First, we can find out that the only $x, y, z$ that satisfy the conditions in the problem are $(1 \leq z-y \leq 9 , 1 \leq y-x \leq 9 , z-x \geq 10)$ , $(1 \leq y-x \leq 9 , 1 \leq x-z \leq 9 , y-z \geq 10)$ , and $(1 \leq x-z \leq 9 , 1 \leq z-y \leq 9 , x-y \geq 10)$.

Consider the 1st set of conditions for $x, y, z$. We get that there are

\[45\cdot 10 - \sum_{k=1}^{9} \binom{k+1}{2} = 285\]

cases for the first set of conditions.

Since the 2nd and 3rd set of conditions are simply rotations of the 1st set, the total number of cases is

\[3\cdot 285 = \fbox{(B)855}\]

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

[Category: Introductory Set Theory Problems]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_20&oldid=121333"