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Revision as of 19:37, 3 July 2013

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.)

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

[asy] import three; import math; unitsize(1.5cm); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); [/asy]

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

Though the cube may be easier to think about, the octahedron can be directly considered. Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.

[asy] unitsize(1.5cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); fill(A--B--C--cycle,rgb(1,.6,.6)); [/asy]

There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$.

[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); triple right=(0,1,0); picture p = new picture, r = new picture, s = new picture; draw(p,A--B--E--cycle); draw(p,A--C--D--cycle); draw(p,F--C--B--cycle); draw(p,F--D--E--cycle,dotted+linewidth(0.7)); draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*p); draw(r,A--B--E--cycle); draw(r,A--C--D--cycle); draw(r,F--C--B--cycle); draw(r,F--D--E--cycle,dotted+linewidth(0.7)); draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); [/asy]

See also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
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All AMC 12 Problems and Solutions

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