2006 AMC 10A Problem 24

Attempt Problem 24 of the 2006 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10A solutions, or check the answer key.

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24.

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

18\dfrac{1}{8}

16\dfrac{1}{6}

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

Answer: B
Concepts:3D geometrypyramidvolume
Difficulty rating: 1760
Solution:

The six face centers form a regular octahedron, viewed as two congruent square pyramids sharing a base. Adjacent face centers are 22\frac{\sqrt2}{2} apart, so the square base has area (22)2=12.\left(\frac{\sqrt2}{2}\right)^2 = \frac12.

Each pyramid has height 12,\frac12, so its volume is 131212=112.\frac13 \cdot \frac12 \cdot \frac12 = \frac{1}{12}. The octahedron has volume 2112=16.2 \cdot \frac{1}{12} = \frac16.

Thus, the correct answer is B.

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