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

Revision as of 20:25, 1 June 2021

The following problem is from both the 2006 AMC 12A #20 and 2006 AMC 10A #25, so both problems redirect to this page.

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?

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

Solutions

Solution 1

Call this cube $ABCDEFGH$ , with $A$ being the starting point.

Because in $7$ moves the bug has to visit the other vertices, the bug cannot revisit any vertex.

Therefore, starting at A, the bug has a $\frac{3}{3}$ chance of finding a good path to the next vertex, and call it $B$ .

Then, the bug has a $\frac{2}{3}$ chance of reaching a new vertex next. Call this $C$ . $A, B,$ and $C$ are always on the same plane.

Now, we split cases.

In the first case, the bug goes to the vertex $E$ opposite $A$ on the space diagonal with probability $\frac{1}{3}$ .

Then, the bug has to visit $D$ on the plane of $ABC$ last, as there is no way in and out from $D$ .

Therefore, there is only $1$ way out of $81$ to get to $D$ last.

Therefore, there is a $\frac{1}{243} \cdot \frac{6}{9} = \frac{6}{2187}$ chance of finding a good path in this case.

In the second case, the bug goes to vertex $D$ on plane $ABC$ with a chance of $\frac{1}{3}$ .

The bug then has only $1$ way to go to a point $E$ on the opposite face, therefore having a $\frac{1}{3}$ probability.

Then, the bug has a choice of two vertices on the face opposite to $ABCD$ .

This results in a $\frac{2}{3}$ probability of finding a good path to a point $F$ .

Then, there is only $1$ way out of $9$ to visit both other vertices on that face in $2$ moves.

Multiply the probabilities for this case to get $\frac{2}{243} \cdot \frac{6}{9} = \frac{12}{2187}$ .

Add the probabilities of these two cases together to get $\frac{18}{2187} = \frac{2}{243}\Rightarrow \boxed{\mathrm (C)}$

Solution 2

$[asy] import three; unitsize(1cm); size(50); currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); [/asy]$

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 vertices $ABCDEFGH$ 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 $H \to E \to F \to G$ . 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, $F \to E \to H \to D$ .

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, $\frac{18}{2187} = \frac2{243}\Rightarrow \boxed{\mathrm (C)}$ .

Solution 3 (using the answer choices)

As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in 3 ways. After this, the bug can move to a new neighbor in 2 ways (it cannot return to the first vertex). The total number of paths (as stated above) is $3^7$ or $2187$ . Therefore, the probability of the bug following a good path is equal to $\frac{6x}{2187}$ for some positive integer $x$ . The only answer choice which can be expressed in this form is $\frac2{243}\Rightarrow\boxed{\mathrm (C)}$ .

@@ Line 65: / Line 65: @@
 Case 2b: the bug moves <math>G \to F</math>.  There is a unique good path in this case, <math>F \to E \to H \to D</math>.
-Thus, all told we have 3 good paths after the first two move, for a total of <math>3\cdot 3 \cdot 2 = 18</math> good paths.  There were <math>3^7 = 2187</math> possible paths the bug could have taken, so the [[probability]] a random path is good is the [[ratio]] of good paths to total paths, <math>\frac{18}{2187} = \frac2{243}\Rightarrow \boxed{\mathrm (C)}</math>.
+Thus, all told we have 3 good paths after the first two moves, for a total of <math>3\cdot 3 \cdot 2 = 18</math> good paths.  There were <math>3^7 = 2187</math> possible paths the bug could have taken, so the [[probability]] a random path is good is the [[ratio]] of good paths to total paths, <math>\frac{18}{2187} = \frac2{243}\Rightarrow \boxed{\mathrm (C)}</math>.
 ===Solution 3 (using the answer choices)===

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Question
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