Difference between revisions of "2016 AIME I Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths from that point down to the third plane. Within the third plane, there are 9 paths as well (consider the logic from the second plane) Finally, there is only one way down to the bottom vertex. Therefore there are <math>5 | + | Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths from that point down to the third plane. Within the third plane, there are 9 paths as well (consider the logic from the second plane) Finally, there is only one way down to the bottom vertex. Therefore there are <math>5 \cdot 9 \cdot 2 \cdot 9 \cdot 1=\boxed{810}</math> paths. |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=2|num-a=4}} | {{AIME box|year=2016|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:38, 28 February 2017
Problem 3
A regular icosahedron is a 
-faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
![[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G--H--L--I--J--I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy]](http://latex.artofproblemsolving.com/6/0/1/60163f4cb994a55653ffde09be32f5100795ddd4.png)
Solution
Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths from that point down to the third plane. Within the third plane, there are 9 paths as well (consider the logic from the second plane) Finally, there is only one way down to the bottom vertex. Therefore there are 
paths.
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
