1989 AMC 12 Problem 26

Attempt Problem 26 of the 1989 AMC 12 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 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

26.

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is
312\frac{\sqrt3}{12}
616\frac{\sqrt6}{16}
16\frac16
28\frac{\sqrt2}{8}
14\frac14
Answer: C
Concepts:cube geometrypolyhedronvolume
Small hint:
Scale the cube to side length 22 and place its center at the origin
Big hint:
The octahedron has vertices (±1,0,0),(0,±1,0),(0,0,±1)(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)
Solution:
Take a cube of side 2,2, so its volume is 8.8. The face centers are (±1,0,0),(0,±1,0),(0,0,±1).(\pm1,0,0),(0,\pm1,0),(0,0,\pm1). In each octant, the octahedron cuts out a tetrahedron with three perpendicular unit edges and volume 16.\frac{1}{6}. Thus its volume is 8(16)=43,8(\frac{1}{6})=\frac{4}{3}, and the ratio is 438=16.\frac{\frac{4}{3}}{8}=\frac{1}{6}.
Thus the correct answer is C.
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