Difference between revisions of "2000 AMC 12 Problems/Problem 25"
(→Solution: solution) |
m (→Solution: typo fix) |
||
| Line 8: | Line 8: | ||
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. | 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 adjacent <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>. |
== See also == | == See also == | ||
Revision as of 22:32, 4 January 2008
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.)
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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 
; 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 
.
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 | |
