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

We can break the octahedron into two tetrahedrons. The cube has sides of length one, so we can solve for the area of the tetrahedron's base: $(\frac{1}{2}\sqrt{2})^2 = \frac{1}{2}$ . We also know that the height of a tetrahedron, which would be half the height of the cube, is $\frac{1}{2}$ . Using the volume formula of a tetrahedron: $A=\frac{1}{3}Bh$ where B is the base side length, we find that the volume of one of the tetrahedrons is $\frac{1}{3}(\frac{1}{2})(\frac{1}{2}) = \frac{1}{12}$ . The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \frac{1}{6} \Rightarrow \mathrm{B}$ .

2006 AMC 10A (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
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2006 AMC 10A Problems/Problem 24

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