2006 AMC 10A Problems/Problem 24

Problem

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

$\mathrm{(A) \ } \frac{1}{8}\qquad\mathrm{(B) \ } \frac{1}{6}\qquad\mathrm{(C) \ } \frac{1}{4}\qquad\mathrm{(D) \ } \frac{1}{3}\qquad\mathrm{(E) \ } \frac{1}{2}\qquad$

Solution

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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

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2006 AMC 10A Problems/Problem 24

Problem

Solution

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