Difference between revisions of "2013 AMC 12A Problems/Problem 20"

Revision as of 17:48, 22 February 2013

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 $x \succ y$ iff $y$ is one of the 9 persons on the left of $x$ , and $y \succ x$ iff $y$ 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 $x,y,z$ is 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.

@@ Line 1: / Line 1: @@
+== Problem 20 ==
+Let <math>S</math> be the set <math>\{1,2,3,...,19\}</math>. For <math>a,b \in S</math>, define <math>a \succ b</math> to mean that either <math>0 < a - b \le 9</math> or <math>b - a > 9</math>. How many ordered triples <math>(x,y,z)</math> of elements of <math>S</math> have the property that <math>x \succ y</math>, <math>y \succ z</math>, and <math>z \succ x</math>?
+<math> \textbf{(A)} \ 810 \qquad  \textbf{(B)} \ 855 \qquad  \textbf{(C)} \ 900 \qquad  \textbf{(D)} \ 950 \qquad  \textbf{(E)} \ 988 </math>
+==Solution==
 Imagine 19 numbers are just 19 persons sitting evenly around a circle <math>C</math>; each of them is facing to the center.
@@ Line 14: / Line 22: @@
 NOTE: this multiple choice problem can be done even faster -- after we realized the fact that each choice of the three places of <math>x,y,z</math> corresponds to <math>3</math> possible ways to put them in, and that each arc of length <math>k>10</math> has <math>19</math> equitable positions, it is evident that the answer should be divisible by <math>3\cdot 19</math>,  which can only be <math>855</math> from the five choices.
+== See also ==
+{{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}}

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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Difference between revisions of "2013 AMC 12A Problems/Problem 20"

Revision as of 17:48, 22 February 2013

Problem 20

Solution

See also