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

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== Problem ==
 
== Problem ==
 
Eight congruent [[equilateral triangle]]s, 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.)  
 
Eight congruent [[equilateral triangle]]s, 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.)  
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<math>\text {(A)}\ 210 \qquad \text {(B)}\ 560 \qquad \text {(C)}\ 840 \qquad \text {(D)}\ 1260 \qquad \text {(E)}\ 1680</math>
 
  
 
<center><asy>
 
<center><asy>
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</asy></center>
 
</asy></center>
  
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== Solution 1 ==
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<math>\text {(A)}\ 210 \qquad \text {(B)}\ 560 \qquad \text {(C)}\ 840 \qquad \text {(D)}\ 1260 \qquad \text {(E)}\ 1680</math>
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 +
== Solutions ==
 +
=== Solution 1 ===
 
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
 
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
 
<center><asy>
 
<center><asy>
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</asy></center>
 
</asy></center>
  
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== Solution 2 ==
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=== Solution 2 ===
 
We consider the dual of the octahedron, the [[cube (geometry)|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.
 
We consider the dual of the octahedron, the [[cube (geometry)|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.
  
 
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 <math>B</math> adjacent to <math>A</math>. 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 and their locations are now fixed. Thus the total number of ways is <math>\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}</math>.
 
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 <math>B</math> adjacent to <math>A</math>. 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 and their locations are now fixed. 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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== Solution 3 ==
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=== Solution 3 ===
 
There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and <math>8!/24 = 1680 \Rightarrow \mathrm{(E)}</math>
 
There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and <math>8!/24 = 1680 \Rightarrow \mathrm{(E)}</math>
  
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== See also ==
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== See Also ==
 
{{AMC12 box|year=2000|num-b=24|after=Last question}}
 
{{AMC12 box|year=2000|num-b=24|after=Last question}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:55, 19 October 2020

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.)

[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]

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

Solutions

Solution 1

Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.

[asy] size(8cm); 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)); draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);[/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]

Solution 2

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

Solution 3

There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and $8!/24 = 1680 \Rightarrow \mathrm{(E)}$

See Also

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

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