Triangle Center Midpoint Consider the following construction of a triangle center: (The method could also be easily generalized to any shape with finite perimeter)
For each point $X$ on the triangle, find point $X'$ such that $X$ and $X'$ split the triangle into two sections of equal length (each of length equal to the triangle semiperimeter). Then, let $Y$ be the midpoint of $X$ and $X'$. The locus of all such $Y$ is also triangular, so we can repeat the process on the triangle created by the locus of all such $Y$, ad infinitum until the triangles converge to a point.
Through some computer simulation, I was able to determine that this convergence point isn't one of the common triangle centers. Searching through the Encyclopedia of Triangle Centers, I also wasn't able to find this triangle center listed. Does this center have an established name, or is there literature available on the topic?
 A: Not an answer, but information that may help lead to one.

With the help of Mathematica to push some symbols, I've determined that the coordinates of the "Draksis Triangle" $\triangle A^\prime B^\prime C^\prime$ is obtained from those of $\triangle ABC$ thusly:
$$\begin{align}
A^\prime &= \frac{1}{2}\;A + \frac{\cos(\beta/2) \sin(\gamma/2)}{2\cos(\alpha/2)}\;B + \frac{\sin(\beta/2)\cos(\gamma/2)}{2\cos(\alpha/2)}\;C \\[4pt]
B^\prime &= \frac{\sin(\gamma/2)\cos(\alpha/2)}{2\cos(\beta/2)}\;A + \frac{1}{2}\;B + \frac{\cos(\gamma/2) \sin(\alpha/2)}{2\cos(\beta/2)}\;C \\[4pt]
C^\prime &= \frac{\cos(\alpha/2) \sin(\beta/2)}{2\cos(\gamma/2)}\;A + \frac{\sin(\alpha/2)\cos(\beta/2)}{2\cos(\gamma/2)}\;B + \frac{1}{2}\;C
\end{align}$$
where $\alpha := \angle BAC$, $\beta := \angle CBA$, $\gamma := \angle ACB$.
Here are some metrics of the Draksis Triangle: 
$$a^\prime = a\sin(\beta/2) \sin(\gamma/2) \qquad
b^\prime = b\sin(\gamma/2) \sin(\alpha/2) \qquad
c^\prime = c\sin(\alpha/2) \sin(\beta/2)$$
$$\alpha^\prime = \frac{\pi - \alpha}{2} \qquad
\beta^\prime = \frac{\pi - \beta}{2} \qquad
\gamma^\prime = \frac{\pi - \gamma}{2}$$
where $a := |\overline{BC}|$ and $a^\prime := |\overline{B^\prime C^\prime}|$, etc. Note that the angle formula confirms @MvG's assertion that iterated Draksis triangles converge on an equilateral: If $\alpha = \frac13\pi + \theta$, then $\alpha^\prime = \frac13\pi - \frac12\theta$.
Iterating the construction to get at the "Draksis Point" seems daunting, but given the relations above, there may yet be hope.
