1992 AMC 12 Problem 19

Question

For each vertex of a solid cube, consider the tetrahedron determined by the vertex and the midpoints of the three edges that meet at that vertex. The portion of the cube that remains when these eight tetrahedra are cut away is called a cuboctahedron. The ratio of the volume of the cuboctahedron to the volume of the original cube is closest to which of these?

75\%

78\%

81\%

84\%

87\%

Answer: D

Concepts:solid geometry tetrahedron volume volume ratio

Difficulty rating: 1960

Small hint:

Choose a convenient side length for the cube and compute one removed corner tetrahedron

Big hint:

Each removed tetrahedron has three mutually perpendicular edges equal to half the cube’s side

Solution:

Take the cube’s side length to be

2.

Its volume is

8.

Each corner tetrahedron has three perpendicular edges of length

1,

so its volume is

\frac16.

The eight removed tetrahedra have total volume

\frac43,

leaving

\frac{20}{3}.

The ratio is

\frac{\frac{20}{3}}{8}=\frac56=83\tfrac13\%,

which is closest to

84\%.

Thus the correct answer is D.