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

Latest revision as of 21:48, 12 August 2024

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?

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

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.

Case $1$ : 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.

Case $2$ : 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 $\frac{18}{2187} = \boxed{\textbf{(C) }\frac{2}{243}}.$

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 $1$ good path in this case, $F \to E \to H \to D$ .

2006 AMC 12A Problems/Problem 20 (section) this solution is extremely good and very hard to learn if you learn this solution you are officially a math god, 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 of a random path is good is $\frac{18}{2187} = \boxed{\textbf{(C) }\frac2{243}}$ .

Solution 3 (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 $\boxed{\textbf{(C) }\frac2{243}}$ .

Solution 4 (possibly wrong but still)

Pick 1 vertex to start from. Notice that on all the odd moves the bug will move to 1 of 4 other vertices. Notice that on all even moves the bug must go to 1 of 3 other vertices (4 including the starting vertex). Thus, the bug will end on one of the odd vertices. There are 3 choices for the bug's first move from the starting vertex, and 2 choices for the bug's second move (it can't go back to the previous vertices). Then, notice that the bug has 3 choices of an odd vertex to end on not including the vertex the bug went to on move 1. Now pick 1 of the 3 ending vertices to end on. Notice that there are only 6 valid paths from start to finish (1 valid path after moves 1 and 2) to that point because all the moves are forced after the first 2 moves (try it out). Thus, there are 3x2x3 = 18 valid paths the bug can take. There are 3^7 = 2187 total paths, so the answer is 18/2187 = $\boxed{\textbf{(C) }\frac2{243}}$ .

@@ Line 1: / Line 1: @@
+{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}}
 == 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?
-== Solution ==
+<math> \textbf{(A) } \frac{1}{2187}\qquad \textbf{(B) } \frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) }  \frac{5}{243}</math>
+==Solution 1==
+Call this cube <math>ABCDEFGH</math>, with <math>A</math> being the starting point.
+Because in <math>7</math> moves the bug has to visit the other vertices, the bug cannot revisit any vertex.
+Therefore, starting at <math>A</math>, the bug has a <math>\frac{3}{3}</math> chance of finding a good path to the next vertex, and call it <math>B</math>.
+Then, the bug has a <math>\frac{2}{3}</math> chance of reaching a new vertex next. Call this <math>C</math>. <math>A, B,</math> and <math>C</math> are always on the same plane.
+Now, we split cases.
+Case <math>1</math>: The bug goes to the vertex <math>E</math> opposite <math>A</math> on the space diagonal with probability <math>\frac{1}{3}</math>.
+Then, the bug has to visit <math>D</math> on the plane of <math>ABC</math> last, as there is no way in and out from <math>D</math>.
+Therefore, there is only <math>1</math> way out of <math>81</math> to get to <math>D</math> last.
+Therefore, there is a <math>\frac{1}{243} \cdot \frac{6}{9} = \frac{6}{2187}</math> chance of finding a good path in this case.
+Case <math>2</math>: The bug goes to vertex <math>D</math> on plane <math>ABC</math> with a chance of <math>\frac{1}{3}</math>.
+The bug then has only <math>1</math> way to go to a point <math>E</math> on the opposite face, therefore having a <math>\frac{1}{3}</math> probability.
+Then, the bug has a choice of two vertices on the face opposite to <math>ABCD</math>.
+This results in a <math>\frac{2}{3}</math> probability of finding a good path to a point <math>F</math>.
+Then, there is only <math>1</math> way out of <math>9</math> to visit both other vertices on that face in <math>2</math> moves.
+Multiply the probabilities for this case to get <math>\frac{2}{243} \cdot \frac{6}{9} = \frac{12}{2187}</math>.
+Add the probabilities of these two cases together to get <math>\frac{18}{2187} = \boxed{\textbf{(C) }\frac{2}{243}}.</math>
+==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 (<math>3</math> ways), and then to a new neighbor (<math>2</math> more ways).  So, without loss of generality, let the cube have vertices <math>ABCDEFGH</math> such that <math>ABCD</math> and <math>EFGH</math> are two opposite faces 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.
+Case <math>1</math>: the bug moves to <math>D</math>.  From <math>D</math>, there is only one good move available, to <math>H</math>.  From <math>H</math>, there are two ways to finish the trip, either by going <math>H \to G \to F \to E</math> or <math>H \to E \to F \to G</math>.  So there are <math>2</math> good paths in this case.
+Case <math>2</math>: the bug moves to <math>G</math>.
+Case <math>2a</math>: the bug moves <math>G \to H</math>.  In this case, there are <math>0</math> good paths because it will not be possible to visit both <math>D</math> and <math>E</math> without double-visiting some vertex.
+Case <math>2b</math>: the bug moves <math>G \to F</math>.  There is <math>1</math> good path in this case, <math>F \to E \to H \to D</math>.
+AMC 12A Problems/Problem 20 (section) this solution is extremely good and very hard to learn if you learn this solution you are officially a math god, 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 of a random path is good is <math>\frac{18}{2187} = \boxed{\textbf{(C) }\frac2{243}}</math>.
+==Solution 3 (answer choices)==
+As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in <math>3</math> ways. After this, the bug can move to a new neighbor in <math>2</math> ways (it cannot return to the first vertex). The total number of paths (as stated above) is <math>3^7</math> or <math>2187</math>. Therefore, the probability of the bug following a good path is equal to <math>\frac{6x}{2187}</math> for some positive integer <math>x</math>. The only answer choice which can be expressed in this form is <math>\boxed{\textbf{(C) }\frac2{243}}</math>.
+==Solution 4 (possibly wrong but still)==
+Pick 1 vertex to start from. Notice that on all the odd moves the bug will move to 1 of 4 other vertices. Notice that on all even moves the bug must go to 1 of 3 other vertices (4 including the starting vertex). Thus, the bug will end on one of the odd vertices. There are 3 choices for the bug's first move from the starting vertex, and 2 choices for the bug's second move (it can't go back to the previous vertices). Then, notice that the bug has 3 choices of an odd vertex to end on not including the vertex the bug went to on move 1. Now pick 1 of the 3 ending vertices to end on. Notice that there are only 6 valid paths from start to finish (1 valid path after moves 1 and 2) to that point because all the moves are forced after the first 2 moves  (try it out). Thus, there are 3x2x3 = 18 valid paths the bug can take. There are 3^7 = 2187 total paths, so the answer is 18/2187 = <math>\boxed{\textbf{(C) }\frac2{243}}</math>.
 == See also ==
-* [[2006 AMC 12A Problems]]
+{{AMC12 box|year=2006|ab=A|num-b=19|num-a=21}}
+{{AMC10 box|year=2006|ab=A|num-b=24|after=Last Question}}
+{{MAA Notice}}
+[[Category:Introductory Combinatorics Problems]]

2006 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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