Difference between revisions of "2000 AMC 12 Problems/Problem 25"

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== See also ==
 
== See also ==
{{AMC12 box|year=2000|num-b=23|num-a=25}}
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 22:32, 4 January 2008

Problem

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)

$\text {(A)}\ 210 \qquad \text {(B)}\ 560 \qquad \text {(C)}\ 840 \qquad \text {(D)}\ 1260 \qquad \text {(E)}\ 1680$


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Solution

We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube.

Select any vertex and call it $A$; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$, we pick another vertex $B$ adjacent to $A$. There are seven color choices for $B$, but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last question
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All AMC 12 Problems and Solutions