Does the centroid of a triangle ever fall outside of its Morley's triangle? Let $T$ be a triangle, and $M$ its (first) Morley triangle:

          


          

(Image from Bruce Shawyer web page.)



Q1. Does the centroid $c$ of $T$ ever fall outside of $M$?

Let $m$ be the centroid of $M$.

Q2. Among all triangles $T$ of unit area, which achieve the maximum
  separation $\| c-m \|$ between the centroid of $T$ and the centroid of $M$?


An approach to Q2 ...

Divide each angle into sixths, with $A = 6\alpha$, $B = 6 \beta$, $C = 6 \gamma$, so that $\alpha + \beta + \gamma = 30^\circ$. Position $\triangle ABC$ on the coordinate plane with $A$ at the origin and the positive $x$-axis bisecting $\angle A$, as shown. Then, with $r$ the circumradius of $\triangle ABC$, we have
$$\begin{align}
A &= (0,0) \\
B &= c\,(\cos 3\alpha,-\sin 3\alpha) = 2 r \sin 6\gamma\,(\cos 3\alpha,-\sin 3\alpha) \\ 
C &= b\,( \cos 3 \alpha,\phantom{-}\sin 3\alpha ) = 2 r \sin 6\beta\,(\cos 3\alpha,\phantom{-}\sin 3\alpha)
\end{align}$$
The centroid, $K$, is given by

$$\begin{align}
K :&= \frac{1}{3}(A+B+C) \\[4pt]
&= \frac{2r}{3}\left(\; \cos 3\alpha \left( \sin 6\beta + \sin 6\gamma \right), \sin 3\alpha \left( \sin 6\beta - \sin 6\gamma \right) \;\right) \\[4pt]
&= \frac{4r}{3} \left(\; 
  \cos 3\alpha\,\sin 3 (\beta + \gamma)\,\cos 3 (\beta - \gamma)\;, 
  \;\sin 3\alpha\,\sin 3 (\beta - \gamma)\,\cos 3 (\beta + \gamma) 
\;\right) \\[4pt]
&= \frac{4r}{3} \left(\; 
  \sin^2 3(\beta+\gamma)\,\cos 3 (\beta - \gamma)\;, 
  \;\cos^2 3(\beta+\gamma)\,\sin 3 (\beta - \gamma) 
\;\right)\end{align}$$

Converting from the expressions for $BP$ and $BR$ in Cut-the-Knot's Proof 1 of Morley's Miracle, we have, for circumradius $r$,
$$E = |AE|\,(\cos\alpha,\phantom{-}\sin\alpha) = 8 r\,\sin 2\beta\,\sin 2 \gamma\,\sin(60^\circ + 2\beta)\;(\cos\alpha,\phantom{-}\sin\alpha)$$
$$F = |AF|\,(\cos\alpha,-\sin\alpha) = 8 r\,\sin 2\beta\,\sin 2 \gamma\,\sin(60^\circ + 2\gamma)\;(\cos\alpha,-\sin\alpha)$$
and we can deduce
$$D = 4 r \sin 2\beta \, \sin 2\gamma\;(\; 
 \cos(\beta-\gamma) ( 4 \cos^2(\beta+\gamma) - 1 )\;, 
\;\sin(\beta-\gamma) ( 4 \sin^2(\beta+\gamma) - 1 )
\;)$$
so that the Morley center, $M$, is (after some simplification effort)

$$\begin{align}
M :&= \frac{1}{3}(D+E+F) \\[6pt]
&= \frac{16r \sqrt{3}}{3} \; \sin 2 \beta \sin 2 \gamma \;\left(   
  \cos\alpha\,\cos(\beta-\gamma)\,\cos(\beta+\gamma), 
 -\sin\alpha\,\sin(\beta-\gamma)\,\sin(\beta+\gamma)
\right)
\end{align}$$

From here, one can arrive at an expression for $|K-M|$, and set about maximizing the distance subject to the unit-area condition: 

$$1 = |\triangle ABC| = \frac{1}{2} b c \sin A = 2 r^2 \sin 6\alpha\,\sin 6 \beta\,\sin 6 \gamma$$

Unfortunately, the trigonometric manipulations get pretty hairy at this point, and I haven't been able to derive anything useful from them yet. Perhaps this information will give someone else a nice head start on a full solution.
