Difference between revisions of "2006 AMC 10A Problems/Problem 24"
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== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
| + | *not confirmed. My own solution. | ||
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| + | We can first break the octahedron into two tetrahedrons. since we know the cube has sides of length one, we can solve for the side length of the tetrahedron's base: (1/2)(sqrt(2))=(sqrt(2)/2) | ||
| + | We also know that the height of a tetrahedron is (1/2) from the fact that this is a unit cube. Using the area formula of a tetrahdron: A=(1/3)(b)(h) where b is the base side length, we find that the area of one tetrahedron is (1/12). The whole octahedron is twice this area. (1/12)(2)=(1/6) | ||
| + | |||
| + | (B) 1/6 | ||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 20:48, 14 February 2007
Problem
Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
- not confirmed. My own solution.
We can first break the octahedron into two tetrahedrons. since we know the cube has sides of length one, we can solve for the side length of the tetrahedron's base: (1/2)(sqrt(2))=(sqrt(2)/2) We also know that the height of a tetrahedron is (1/2) from the fact that this is a unit cube. Using the area formula of a tetrahdron: A=(1/3)(b)(h) where b is the base side length, we find that the area of one tetrahedron is (1/12). The whole octahedron is twice this area. (1/12)(2)=(1/6)
(B) 1/6
