On faster convergence of the bisection method for certain 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 ${\Delta _{11}},{\Delta _{12}}$. 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 ${T_j} = \{ {\Delta _{ji}}:1 \leqslant i \leqslant {2^j}\}$. It is known that the mesh of ${T_j}$ tends to zero as $j \to \infty$. It is shown here that if ${\Delta _{01}}$ satisfies any of four certain properties, the rate of convergence of the mesh to zero is much faster than that predicted by the general case.