Difference between revisions of "1992 AHSME Problems/Problem 19"

Revision as of 00:09, 29 August 2016

Problem

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?

$\text{(A) } 75\%\quad \text{(B) } 78\%\quad \text{(C) } 81\%\quad \text{(D) } 84\%\quad \text{(E) } 87\%$

Solution

$\fbox{D}$

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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 == Solution ==
-Call the edge of the cube <math>x.</math> Each side of the eight tetrahedra will be <math>frac{x}{2}</math> and will have a right isosceles triangle as the base.  The volume of all eight tetrahedra then will be <math>frac{8}{3} \cdot frac{x^{2}}{8} \cdot frac{x}{2}.</math> This simplifies to <math>frac{x^{3}}{6}</math>, and the volume of the cubeoctahedron is <math>x^{3} - frac{x^{3}}{6}</math> = <math>frac{5x^{3}}{6}.</math> since frac{5}{6} is approximately 84%, the volume of the cubeoctahedron is approximately 84% of the regular cube. <math>\fbox{D}</math>
+<math>\fbox{D}</math>
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Difference between revisions of "1992 AHSME Problems/Problem 19"

Revision as of 00:09, 29 August 2016

Problem

Solution

See also