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 $M$?

 A: For Q1:  Let $M$ be the midpoint of $BC$; then $AB\sin\angle BAM = AC\sin\angle CAM$.  So if we take a triangle with $\angle A=90^\circ$ and, say, $AB=1$ and $AC=10000$, we'll have $\angle BAM$ much larger than $\angle CAM$, and so the median $AM$ won't lie in the middle third of $\angle A$.
For Q2: There is no upper bound.  Take $A=(0,0)$, $B=(0,\frac1n)$, $C=(2n,0)$, where $n$ is a positive integer.  This is a right triangle with area 1, and its centroid has $x$-coordinate $\frac23 n$.  But its Morley triangle lies both in the middle third of $\angle A$ and in the strip $0\le y\le 1$, and the rightmost point satisfying both those conditions is $(\cot\frac\pi6,1)$; so the distance between the centroid and the Morley triangle is at least $\frac23 n - \cot\frac\pi6$, which is unbounded.
(An underlying issue in Q2 is that the space of triangles of unit area is not compact (under Hausdorff distance, say), so there's not usually any reason to expect a functional to be bounded on that set.  You might have better luck considering triangles of unit diameter.)
A:           


@StevenTaschuk's construction, with right triangle
$T$ base $1$ and altitude $2$.
Centroid of $T$ is the purple dot.
A: 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.
