Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.9 Junior Find the answer

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 cubeoctahedron. The ratio of the volume of the cubeoctahedron to the volume of the original cube is closest to which of these?
$\text{(A) } 75\%$ $\text{(B) } 78\%$ $\text{(C) } 81\%$ $\text{(D) } 84\%$ $\text{(E) } 87\%$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

$\fbox{D}$ Let the cube have side length 1, and place the cube in the coordinate plane. Then we can pick any vertex, get the coordinates of the midpoints, and hence find the three vectors that define the tetrahedron (the vectors from the chosen vertex to each midpoint). Now using $\frac{1}{6}\|a.(b\times c)\|$ , we can find the volume of one of the tetrahedra, then multiply it by 8, and subtract from 1 to get $\frac{5}{6}$ , which is closest to $84\%$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.