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 positive integers, find .
Solution:
Without loss of generality, let the edges of have length . Note that can be formed by adjoining two square pyramids at their bases. Consider an altitude from a vertex of one of these pyramids to the square base. This altitude is also a leg of a right triangle whose other leg joins the center of the square and a vertex of the square, and whose hypotenuse is an edge of . The length of the altitude is therefore , and so the volume of is . To find the volume of , consider a triangle, one of whose vertices is a vertex of not on the square and whose other two vertices, and , are midpoints of opposite sides of the square. The line segment that joins the centers of the faces containing and , respectively, is a diagonal of a face of . Because these centers are two-thirds of the way from to and from to , respectively, the length of the face diagonal joining them is two-thirds of . But , so the length of each of the edges of is . Hence the volume of is . The requested ratio is thus , so .
The six vertices of are equidistant from its center, and the diagonals that join the three pairs of opposite vertices are mutually perpendicular. Without loss of generality, let the length of each of these three diagonals be . It is possible to place a coordinate system so that the coordinates of the vertices of are , and . Because is composed of two square pyramids, its volume is . The vertices of are the centroids of the faces of , so the coordinates of the vertices of are . Thus the length of each of the edges of is , and the volume of is . The ratio of the volumes is , so .
The problems on this page are the property of the MAA's American Mathematics Competitions