2000 AMC 12 Problems/Problem 25

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 $\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2000 AMC 12 Problems/Problem 25

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