On faster convergence of the bisection method for all triangles

Abstract:

Let $\Delta ABC$ be a triangle with vertices A, B, and C. It is "bisected" as follows: choose a/the longest side (say AB) of $\Delta ABC$, let D be the midpoint of AB, then replace $\Delta ABC$ by two triangles $\Delta ADC$ and $\Delta DBC$. Let ${\Delta _{01}}$ be a given triangle. Bisect ${\Delta _{01}}$ into two triangles ${\Delta _{11}}$ and ${\Delta _{12}}$. Next bisect each ${\Delta _{1i}},\;i = 1,2$, forming four new triangles ${\Delta _{2i}},\;i = 1,2,3,4$. Continue thus, forming an infinite sequence ${T_j},\;j = 0,1,2, \ldots$, of sets of triangles, where ${T_j} = \left \{ {{\Delta _{ji}}:1 \leqslant i \leqslant {2^j}} \right \}$. Let ${m_j}$ denote the mesh of ${T_j}$. It is shown that there exists $N = N({\Delta _{01}})$ such that, for $j \geqslant N$, ${m_{2j}} \leqslant {(\sqrt 3 /2)^N}{(1/2)^{j - N}}{m_0}$, thus greatly improving the previous best known bound of ${m_{2j}} \leqslant {(\sqrt 3 /2)^j}{m_0}$. It is also shown that only a finite number of distinct shapes occur among the triangles produced, and that, as the method proceeds, ${\Delta _{01}}$ tends to become covered by triangles which are approximately equilateral in a certain sense.