> Let's say that you have two tetrahedra, call them A > and > B. > > Tetrahedron A has an altitude of 1H. You insert > tetrahedron B into A an rotate it so that all of its > Vertices are pointing in the opposite director of the > vertices of A, and all of its faces are parallel to > the vertices of A. You find that the distance > between one face of A is 4.6 H away from the parallels > face of B. > > What is the altitude of B? Thanks in advance. If you replace tetrahedron by triangle how does this problem change?