2005 AIME II Problems/Problem 10
Problem
Given that is a regular octahedron, that is the cube whose vertices are the centers of the faces of and that the ratio of the volume of to that of is where and are relatively prime integers, find
Solutions
Solution 1
Let the side of the octahedron be of length . Let the vertices of the octahedron be so that and are opposite each other and . The height of the square pyramid is and so it has volume and the whole octahedron has volume .
Let be the midpoint of , be the midpoint of , be the centroid of and be the centroid of . Then and the symmetry ratio is (because the medians of a triangle are trisected by the centroid), so . is also a diagonal of the cube, so the cube has side-length and volume . The ratio of the volumes is then and so the answer is .
Solution 2
Let the octahedron have vertices . Then the vertices of the cube lie at the centroids of the faces, which have coordinates . The cube has volume 8. The region of the octahedron lying in each octant is a tetrahedron with three edges mutually perpendicular and of length 3. Thus the octahedron has volume , so the ratio is and so the answer is .
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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