Difference between revisions of "2006 AMC 12A Problems/Problem 20"
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==Solution 3 (answer choices)== | ==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>. | 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 == | == See also == |
Latest revision as of 22:48, 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.
Contents
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 1
Call this cube , with
being the starting point.
Because in moves the bug has to visit the other vertices, the bug cannot revisit any vertex.
Therefore, starting at , the bug has a
chance of finding a good path to the next vertex, and call it
.
Then, the bug has a chance of reaching a new vertex next. Call this
.
and
are always on the same plane.
Now, we split cases.
Case : The bug goes to the vertex
opposite
on the space diagonal with probability
.
Then, the bug has to visit on the plane of
last, as there is no way in and out from
.
Therefore, there is only way out of
to get to
last.
Therefore, there is a chance of finding a good path in this case.
Case : The bug goes to vertex
on plane
with a chance of
.
The bug then has only way to go to a point
on the opposite face, therefore having a
probability.
Then, the bug has a choice of two vertices on the face opposite to .
This results in a probability of finding a good path to a point
.
Then, there is only way out of
to visit both other vertices on that face in
moves.
Multiply the probabilities for this case to get .
Add the probabilities of these two cases together to get
Solution 2
Let us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex ( ways), and then to a new neighbor (
more ways). So, without loss of generality, let the cube have vertices
such that
and
are two opposite faces with
above
and
above
. The bug starts at
and moves first to
, then to
.
From this point, there are two cases.
Case : the bug moves to
. From
, there is only one good move available, to
. From
, there are two ways to finish the trip, either by going
or
. So there are
good paths in this case.
Case : the bug moves to
.
Case : the bug moves
. In this case, there are
good paths because it will not be possible to visit both
and
without double-visiting some vertex.
Case : the bug moves
. There is
good path in this case,
.
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 good paths. There were
possible paths the bug could have taken, so the probability of a random path is good is
.
Solution 3 (answer choices)
As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in ways. After this, the bug can move to a new neighbor in
ways (it cannot return to the first vertex). The total number of paths (as stated above) is
or
. Therefore, the probability of the bug following a good path is equal to
for some positive integer
. The only answer choice which can be expressed in this form is
.
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 = .
See also
2006 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 |
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.