Difference between revisions of "2015 AMC 10B Problems/Problem 20"
Vishnuemani (talk | contribs) (→Solution) |
(→Solution) |
||
Line 3: | Line 3: | ||
<math>\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}</math> | <math>\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}</math> | ||
− | ==Solution== | + | ==Solution 1== |
<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)); | <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("2",(0,0,0),S); | ||
Line 20: | Line 20: | ||
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>. | 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>. | ||
+ | |||
+ | ==Solution 2 (3D Geo)== | ||
+ | Lets say that this cube is an unit cube and the given corner is <math>(0,0,0)</math>. Because Erin can not return back to its starting point, he can not be on <math>(0,0,1)</math>, <math>(0,1,0)</math>, or <math>(1,0,0)</math>. He can not be on <math>(1,1,0)</math> , <math>(1,0,1)</math>, or <math>(0,1,1,)</math> because after <math>7</math> moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at <math>(1,1,1)</math>. Now, we draw a net and see that there are <math>3</math> choices for the first move, <math>2</math> for the second, and the rest are forced. Thus the answer is <math>3*2 = \boxed{\textbf{(A)}\; 6}</math>. | ||
+ | |||
+ | -Lcz | ||
==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}} |
Latest revision as of 09:45, 13 February 2019
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 1
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 .
Solution 2 (3D Geo)
Lets say that this cube is an unit cube and the given corner is . Because Erin can not return back to its starting point, he can not be on , , or . He can not be on , , or because after moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at . Now, we draw a net and see that there are choices for the first move, for the second, and the rest are forced. Thus the answer is .
-Lcz
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.