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

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== Solution ==
 
== Solution ==
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We consider the dual of the octahedron, the [[cube]]; a cube can be inscribed in an octahedron with each of its [[vertex|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.
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Select any vertex and call it <math>A</math>; there are <math>8</math> color choices for this vertex, but this vertex can be rotated to any of <math>8</math> locations. After fixing <math>A</math>, we pick another vertex adjacent <math>B</math> that one. There are seven color choices for <math>B</math>, but there are only three locations to which <math>B</math> can be rotated to (since there are three edges from <math>A</math>). The remaining six vertices can be colored in any way. Thus the total number of ways is <math>\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}</math>.
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=23|num-a=25}}
 
{{AMC12 box|year=2000|num-b=23|num-a=25}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 23:30, 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 adjacent $B$ that one. 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. 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 23
Followed by
Problem 25
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