Difference between revisions of "2000 AMC 12 Problems/Problem 25"
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<math>\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680</math> | <math>\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680</math> | ||
− | + | == 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> | ||
− | + | == 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>. | ||
− | + | == 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> | ||
− | + | == 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> | 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> | ||
~Hithere22702 | ~Hithere22702 | ||
+ | |||
+ | ==Solution 5 (Graph Theory)== | ||
+ | |||
+ | [[File:2000AMC12P25.png|500px|center]] | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== Video Solution == | == Video Solution == |
Revision as of 22:55, 26 August 2022
Contents
[hide]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.)
Solution 1
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
There are ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is .
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 ; there are color choices for this vertex, but this vertex can be rotated to any of locations. After fixing , we pick another vertex adjacent to . There are seven color choices for , but there are only three locations to which can be rotated to (since there are three edges from ). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is .
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
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 rotation, 120 rotation, and 240 rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is . There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be
~Hithere22702
Solution 5 (Graph Theory)
Video Solution
See Also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
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 |
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