Difference between revisions of "2015 AMC 10B Problems/Problem 20"
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If we define a "move" as each time Erin crawls along a single edge from 1 vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be 1 unit away from <math>A</math> (since Erin cannot crawl back to <math>A</math>), this vertex must be <math>4</math>. | If we define a "move" as each time Erin crawls along a single edge from 1 vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be 1 unit away from <math>A</math> (since Erin cannot crawl back to <math>A</math>), this vertex must be <math>4</math>. | ||
| − | Therefore, we now just need to count the number of paths from <math>A</math> to <math>4</math>. To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to <math>4</math>, and 2 choices for which vertex Erin was at 2 moves before <math>4</math>. All of Erin's previous moves were forced, so the total number of legal paths from <math>A</math> to <math>4</math> is <math>3 \cdot 2 = \boxed{\textbf{( | + | Therefore, we now just need to count the number of paths from <math>A</math> to <math>4</math>. To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to <math>4</math>, and 2 choices for which vertex Erin was at 2 moves before <math>4</math>. All of Erin's previous moves were forced, so the total number of legal paths from <math>A</math> to <math>4</math> is <math>3 \cdot 2 = \boxed{\textbf{(A)}\; 6}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2015|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:59, 17 August 2017
Problem
Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?
Solution
![[asy]import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); label("2",(0,0,0),S); label("A",(1,0,0),W); label("B",(0,0,1),N); label("1",(1,0,1),NW); label("3",(1,1,0),S); label("C",(0,1,0),E); label("D",(1,1,1),SE); label("4",(0,1,1),NE); [/asy]](http://latex.artofproblemsolving.com/a/5/1/a5174d0579a4c17fc276202d601a81fa38ebc899.png)
We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled 
.
If we define a "move" as each time Erin crawls along a single edge from 1 vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be 1 unit away from (since Erin cannot crawl back to ), this vertex must be 
.
Therefore, we now just need to count the number of paths from to . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to , and 2 choices for which vertex Erin was at 2 moves before . All of Erin's previous moves were forced, so the total number of legal paths from to is 
.
See Also
| 2015 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
