|
Choose three faces of a cube that meet at a vertex, and draw a diagonal across each face so that the lines all meet at the common vertex. Now draw three lines to connect the other ends of these diagonals. All six lines are the diagonal of a different face of the cube, so they are all equal in length, and therefore form the edges of a regular tetrahedron. Thus the smallest cube that can encompass a tetrahedron with edges of length S is a cube with edges of length S/sqrt(2). For such a tetrahedron, the base has an area
1/2 * B * H = 1/2 * S * S*sqrt(3)/2 = S^2*sqrt(3)/4 To find the height of the tetrahedron, we first find the distance of the centroid of the base from any edge of the base. From symmetry, the centroid falls on the bisector of the three angles, and the distance of the centroid from an edge is S/(2*sqrt(3)) This distance and the height of the tetrahedron form the two legs of a right triangle whose hypoteneus is the height H of a triangular side, already found to be S*sqrt(3)/2. From the Pythagorean theorem, the height x of the tetrahedron is
x = sqrt[(S*sqrt(3)/2)^2 - (S/(2*sqrt(3))^2]
x = S*sqrt(2/3) The volume Vt of the tetrahedron is
Vt = 1/3 * B * H
Vt = 1/3 * S^2*sqrt(3)/4 * S*sqrt(2/3)
Vt = 1/3 * S^3/sqrt(8) The volume Vc of the cube is
Vc = [S/sqrt(2)]^3
Vc = S^3/sqrt(8) Thus the ratio of volumes of a regular tetrahedron and the smallest cube that can encompass it is 1:3. Amazing how the most complex-seeming geometries yield the most uncomplicated answers. It makes me think there is a more elegant solution to this problem.
|
