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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.)  
−
 
−
<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>
Line 21: Line 19:
 
draw(F--D--E--cycle,dotted+linewidth(0.7));
 
draw(F--D--E--cycle,dotted+linewidth(0.7));
 
</asy></center>
 
</asy></center>
−
== Solution ==
 
−
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.
 
  
−
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>.
+
<math>\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680</math>
  
−
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.
+
== Solution 1 ==
 +
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
 
<center><asy>
 
<center><asy>
−
unitsize(1.5cm);
+
size(8cm);
 
defaultpen(0.5);
 
defaultpen(0.5);
 
import three;
 
import three;
Line 43: Line 40:
 
draw(F--C--B--cycle);
 
draw(F--C--B--cycle);
 
draw(F--D--E--cycle,dotted+linewidth(0.7));
 
draw(F--D--E--cycle,dotted+linewidth(0.7));
−
fill(A--B--C--cycle,rgb(1,.6,.6));
+
draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);</asy></center>
−
</asy></center>
 
 
There are <math>7!</math> ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is <math>7!/3 = 1680</math>.
 
There are <math>7!</math> ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is <math>7!/3 = 1680</math>.
 
<center><asy>
 
<center><asy>
Line 51: Line 47:
 
import three;
 
import three;
 
import math;
 
import math;
−
currentprojection=orthographic(2,0.2,1);
+
currentprojection=orthographic(2,0,1);
 
triple A=(0,0,1);
 
triple A=(0,0,1);
 
triple B=(sqrt(2)/2,sqrt(2)/2,0);
 
triple B=(sqrt(2)/2,sqrt(2)/2,0);
Line 58: Line 54:
 
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
 
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
 
triple F=(0,0,-1);
 
triple F=(0,0,-1);
 +
triple right=(0,1,0);
 
picture p = new picture, r = new picture, s = new picture;
 
picture p = new picture, r = new picture, s = new picture;
 
draw(p,A--B--E--cycle);
 
draw(p,A--B--E--cycle);
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draw(p,F--C--B--cycle);
 
draw(p,F--C--B--cycle);
 
draw(p,F--D--E--cycle,dotted+linewidth(0.7));
 
draw(p,F--D--E--cycle,dotted+linewidth(0.7));
−
fill(p,A--B--C--cycle,rgb(1,.6,.6));
+
draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
−
fill(p,A--B--E--cycle,rgb(1,1,.6));
+
draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight);
−
add(scale(2.2)*p);
+
add(scale3(2.2)*p);
 
draw(r,A--B--E--cycle);
 
draw(r,A--B--E--cycle);
 
draw(r,A--C--D--cycle);
 
draw(r,A--C--D--cycle);
 
draw(r,F--C--B--cycle);
 
draw(r,F--C--B--cycle);
 
draw(r,F--D--E--cycle,dotted+linewidth(0.7));
 
draw(r,F--D--E--cycle,dotted+linewidth(0.7));
−
fill(r,A--B--C--cycle,rgb(1,.6,.6));
+
draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
−
fill(r,A--C--D--cycle,rgb(1,1,.6));
+
draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight);
−
add(scale(2.2)*shift(2*right)*r);
+
add(scale3(2.2)*shift(2*right)*r);
 
draw(s,A--B--E--cycle);
 
draw(s,A--B--E--cycle);
 
draw(s,A--C--D--cycle);
 
draw(s,A--C--D--cycle);
 
draw(s,F--C--B--cycle);
 
draw(s,F--C--B--cycle);
 
draw(s,F--D--E--cycle,dotted+linewidth(0.7));
 
draw(s,F--D--E--cycle,dotted+linewidth(0.7));
−
fill(s,A--B--C--cycle,rgb(1,.6,.6));
+
draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
−
fill(s,B--C--F--cycle,rgb(1,1,.6));
+
draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight);
−
add(scale(2.2)*shift(4*right)*s);
+
add(scale3(2.2)*shift(4*right)*s);
 
</asy></center>
 
</asy></center>
  
−
== See also ==
+
== 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.
 +
 
 +
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>.
 +
 
 +
== 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 <math>6\cdot 4 = 24</math> ways to orient an octahedron, and <math>8!/24 = 1680 \Rightarrow \mathrm{(E)}</math>
 +
 
 +
== Solution 4 ==
 +
If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0<math>^{\circ}</math> rotation, 120<math>^{\circ}</math> rotation, and 240<math>^{\circ}</math> rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is <math>8 \cdot 3 = 24</math>. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be <math>8!/24 \Rightarrow \mathrm{(E)}</math>
 +
 
 +
~kempwood
 +
 
 +
==Solution 5 (Graph Theory)==
 +
 
 +
[[File:2000AMC12P25.png|500px|center]]
 +
 
 +
This problem can be approached by [https://en.wikipedia.org/wiki/Graph_theory Graph Theory]. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.
 +
 
 +
Now the problem becomes how many distinguishable ways to color the <math>8</math> vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.
 +
 
 +
Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.
 +
 
 +
There are <math>8</math> colors and <math>4</math> outer vertices, therefore there are <math>\binom{8}{4}</math> ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are <math>4</math> colors left, therefore there are <math>4!</math> ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is <math>\binom{8}{4} \cdot 4! = \boxed{\textbf{(E) } 1680 }</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 6==
 +
Let the colors be <math>1</math> to <math>8</math> inclusive, then rotate the octahedron such that color <math>1</math> is on top. You have <math>7</math> choices of what color is on the bottom, WLOG <math>2</math>. Then, there's two rings of each <math>3</math> colors on the top and bottom. For the top ring, you can choose any <math>3</math> out of the <math>6</math> remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have <math>3!</math> ways to put the three remaining colors in three spaces.
 +
In total this is <math>7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{\textbf{(E) } 1680 }</math>
 +
 
 +
-mathfan2020
 +
 
 +
== Video Solution ==
 +
https://youtu.be/8WRpCVwQNBo
 +
 
 +
== 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}}

Latest revision as of 14:31, 23 August 2024

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]

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

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 $6\cdot 4 = 24$ ways to orient an octahedron, and $8!/24 = 1680 \Rightarrow \mathrm{(E)}$

Solution 4

If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0$^{\circ}$ rotation, 120$^{\circ}$ rotation, and 240$^{\circ}$ rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is $8 \cdot 3 = 24$. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be $8!/24 \Rightarrow \mathrm{(E)}$

~kempwood

Solution 5 (Graph Theory)

2000AMC12P25.png

This problem can be approached by Graph Theory. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.

Now the problem becomes how many distinguishable ways to color the $8$ vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.

Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.

There are $8$ colors and $4$ outer vertices, therefore there are $\binom{8}{4}$ ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are $4$ colors left, therefore there are $4!$ ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is $\binom{8}{4} \cdot 4! = \boxed{\textbf{(E) } 1680 }$

~isabelchen

Solution 6

Let the colors be $1$ to $8$ inclusive, then rotate the octahedron such that color $1$ is on top. You have $7$ choices of what color is on the bottom, WLOG $2$. Then, there's two rings of each $3$ colors on the top and bottom. For the top ring, you can choose any $3$ out of the $6$ remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have $3!$ ways to put the three remaining colors in three spaces. In total this is $7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{\textbf{(E) } 1680 }$

-mathfan2020

Video Solution

https://youtu.be/8WRpCVwQNBo

See Also

2000 AMC 12 (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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