I am studying refinements of triangular and tetrahedral meshes, in particular Longest-Edge (LE) partition methods. In any triangle there is obviously a longest edge and, on this edge there is a point which divide it into two equal parts. The LE-bisection of the triangle is obtained by joining this point with the opposite vertex to the longest-edge. Two new smaller triangles are obtained. The LE-bisection can be applied to tetrahedra too .
The LE-bisection can be applied to any of the two triangles (or tetrahedra) obtained. The iterative application of LE-bisection to every triangle generates a finite number of similar triangles: For example the right isosceles triangle generates only triangles similar to it, or the equilateral triangle generates three diferents triangles. The general statement was settled by Stynes in 1980. There is an example of tetrahedra with similar finitiness property for LE-bisection (Senechal, 1981). But it is the unique example I know.
If the lodgest-edge is divided in three equal parts, the LE-trisection is obtained . I know only a triangle with a finite number of similar triangles generates by sucessive applications of LE-trisection, exactly the triangle obtained from a DIN-A paper by the cut along the diagonal. DIN-A paper is a rectangle with $1:\sqrt{2}$ proportion.
But I don't know a tetrahedra with the finitiness property for LE-trisection. I'm studying this shapes because the other triangles or tetrahedra are not far from them in a very precise sense when a LE-partition is applied.
I ask for tetrahedra which has the finitiness property for LE-trisection especially that you can know. Thanks in advance, Hideyuki.
