Difference between revisions of "1992 AHSME Problems/Problem 19"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | <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?
Solution
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 
, we can find the volume of one of the tetrahedra, then multiply it by 8, and subtract from 1 to get 
, which is closest to 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
