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Difference between revisions of "2006 AMC 12A Problems/Problem 20"

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== Solution ==
 
== Solution ==
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Let us count the good paths.  The bug starts at an arbitrary [[vertex]], moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways).  So, [[without loss of generality]], let the [[cube]] have vertex <math>ABCDEFGH</math> such that <math>ABCD</math> and <math>EFGH</math> are two opposite [[face]]s with <math>A</math> above <math>E</math> and <math>B</math> above <math>F</math>.  The bug starts at <math>A</math> and moves first to <math>B</math>, then to <math>C</math>.
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Let us count the good paths.  The bug starts at an arbitrary [[vertex]], moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways).  So, [[without loss of generality]], let the [[cube (geometry) | cube]] have vertex <math>ABCDEFGH</math> such that <math>ABCD</math> and <math>EFGH</math> are two opposite [[face]]s with <math>A</math> above <math>E</math> and <math>B</math> above <math>F</math>.  The bug starts at <math>A</math> and moves first to <math>B</math>, then to <math>C</math>.
  
 
From this point, there are two cases.
 
From this point, there are two cases.

Revision as of 11:21, 30 October 2006

Problem

A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

$\mathrm{(A) \ } \frac{1}{2187}\qquad \mathrm{(B) \ } \frac{1}{729}\qquad \mathrm{(C) \ } \frac{2}{243}\qquad \mathrm{(D) \ } \frac{1}{81}$$\mathrm{(E) \ } \frac{5}{243}$

Solution

Let us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways). So, without loss of generality, let the cube have vertex $ABCDEFGH$ such that $ABCD$ and $EFGH$ are two opposite faces with $A$ above $E$ and $B$ above $F$. The bug starts at $A$ and moves first to $B$, then to $C$.

From this point, there are two cases. Case 1: the bug moves to $D$. From $D$, there is only one good move available, to $H$. From $H$, there are two ways to finish the trip, either by going $H \to G \to F \to E$ or $\displaystyle H \to E \to F \to G$. So there are 2 good paths in this case.

Case 2: the bug moves to $G$. Case 2a: the bug moves $G \to H$. In this case, there are 0 good paths because it will not be possible to visit both $D$ and $E$ without double-visiting some vertex. Case 2b: the bug moves $G \to F$. There is a unique good path in this case, $F \to E \to G \to H$.


Thus, all told we have 3 good paths after the first two moves, for a total of $3\cdot 3 \cdot 2 = 18$ good paths. There were $3^7 = 2187$ possible paths the bug could have taken, so the probability a random path is good is the ratio of good paths to total paths, $\frac{18}{2187} = \frac2{243}$, which is answer choice $\mathrm (C)$.


See also

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