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

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== Solution ==
 
== Solution ==
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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>
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<math>\fbox{D}</math> 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 <math>\frac{1}{6}\|a.(b\times c)\|</math>, we can find the volume of one of the tetrahedra, then multiply it by 8, and subtract from 1 to get <math>\frac{5}{6}</math>, which is closest to <math>84\%</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 02:11, 20 February 2018

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

$\fbox{D}$ 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 $84\%$.

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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