On Faster Convergence of the Bisection Method

Let AABC be a triangle with vertices A, B, and C. It is "bisected" as follows: choose a/the longest side (say AB) of AABC, let D be the midpoint of AB, then replace AABC by two triangles aA.DC and ADBC. Let Ag| be a given triangle. Bisect Aqj into two triangles A¡¡ and Aj2. Next bisect each Aj(-, /= 1, 2, forming four new triangles A2j-, i = 1,2, 3,4. Continue thus, forming an infinite sequence T-, / = 0, 1,2,..., of sets of triangles, where T¡ = {a.(-: 1 < i < 21}. Let m.denote the mesh of Ti. It is shown that there exists N = JV(Aqj) such that, for / > N, m2¡ < (^3/2)^(1/2y íMq, thus greatly improving the previous best known bound of m2< (\/3/2)'mQ. It is also shown that only a finite number of distinct shapes occur among the triangles produced, and that, as the method proceeds, Anj tends to become covered by triangles which are approximately equilateral in a certain sense.