A lower bound on the angles of triangles constructed by bisecting the longest side

Abstract:

Let $\Delta {A^1}{A^2}{A^3}$ be a triangle with vertices at ${A^1},{A^2}$ and ${A^3}$. The process of "bisecting $\Delta {A^1}{A^2}{A^3}$" is defined as follows. We first locate the longest edge, ${A^i}{A^{i + 1}}$ of $\Delta {A^1}{A^2}{A^3}$ where ${A^{i + 3}} = {A^i}$, set $D = ({A^i} + {A^{i + 1}})/2$, and then define two new triangles, $\Delta {A^i}D{A^{i + 2}}$ and $\Delta D{A^{i + 1}}{A^{i + 2}}$. Let ${\Delta _{00}}$ be a given triangle, with smallest interior angle $\alpha > 0$. Bisect ${\Delta _{00}}$ into two new triangles, ${\Delta _{1i}},i = 1,2$. Next, bisect each triangle ${\Delta _{1i}}$, to form four new triangles ${\Delta _{2i}},i = 1,2,3,4$, and so on, to form an infinite sequence T of triangles. It is shown that if $\Delta \in T$, and $\theta$ is any interior angle of $\Delta$, then $\theta \geqslant \alpha /2$.