How big can a triangle be, whose sides are the perpendiculars to the sides of a triangle from the vertices of its Morley triangle? Given any triangle $\varDelta$, the perpendiculars from the vertices of its (primary) Morley triangle to their respective (nearest) side of $\varDelta$ intersect in a triangle $\varDelta'$, which is similar to $\varDelta$ but on a smaller scale—say with scale factor $s$. (If $\varDelta$ is isosceles, then $\varDelta'$ degenerates to a point, and $s=0$.)
What is the maximum value of $s$, and for what shape of $\varDelta$ is this value attained?
(This question was posted previously on Mathematics Stack Exchange, without response, but is probably more suitable for this site.)

We will perform computation using
normed barycentric coordinates in the given triangle $\Delta ABC$ with side lengths $a,b,c$ and angles
$\hat A=3\alpha$,
$\hat B=3\beta$,
$\hat C=3\gamma$.
We denote by $R$ the circumradius of the circle $\odot(ABC)$, and the area $[ABC]$ by $S$.
Trilinear coordinates of the points $A'$, $B'$, $C'$ (vertices of the first Morley triangle of
$\Delta ABC$) can be extracted from the rows of the matrix:
$$
\begin{bmatrix}
  1 & 2\cos \gamma & 2\cos\beta\\
  2\cos \gamma & 1 & 2\cos\alpha\\
  2\cos \beta & 2\cos\alpha & 1\\
\end{bmatrix}
\ .
$$
(See for instance https://mathworld.wolfram.com/FirstMorleyTriangle.html .)
The (not-normalized) barycentric coordinates can then be extracted from the rows of the matrix:
$$
\tag{$*$}
\begin{bmatrix}
  a             & 2b\cos \gamma & 2c\cos\beta\\
  2a\cos \gamma & b             & 2c\cos\alpha\\
  2a\cos \beta  & 2c\cos\alpha  & c\\
\end{bmatrix}
\ .
$$
Let now $P_A$, $P_B$, $P_C$ be three points with normalized barycentric coordinates
$$
\begin{aligned}
P_A &= (x_A, y_A, z_A)\ ,\\
P_B &= (x_B, y_B, z_B)\ ,\\
P_C &= (x_C, y_C, z_C)\ .
\end{aligned}
$$
Define $Q_A$ to be such that $Q_AP_B\perp AC$ and $Q_AP_C\perp AB$.
Construct in a similar manner $Q_B,Q_C$.
Computer algebra shows that
$$
\begin{aligned}
s^2 :=\ &\frac{[Q_AQ_BQ_C]}{[ABC]}
=\frac{P_BP_C}{a^2}
=\frac{P_CP_A}{b^2}
=\frac{P_AP_B}{c^2}
\qquad\text{ is given by the relation}
\\
16S^2\cdot s^2 =\  
 &\ 
\begin{pmatrix}
 + x_A(b^2-c^2) + a^2(y_A-z_A) \\
 + y_B(c^2-a^2) + b^2(z_B-x_B) \\
 + z_C(a^2-b^2) + c^2(x_C-y_C)
\end{pmatrix}
^2 \ .
\end{aligned}
$$
(Keeping $A$ and exchanging $B\leftrightarrow C$ moves $s$ to $-s$.)
So we have to maximize the expression
$$
\begin{aligned}
s 
&= \frac 1{4S}
\begin{pmatrix}
 + x_A(b^2-c^2) + a^2(y_A-z_A) \\
 + y_B(c^2-a^2) + b^2(z_B-x_B) \\
 + z_C(a^2-b^2) + c^2(x_C-y_C)
\end{pmatrix}
\\
&=\frac 1{4S}\sum x_A(b^2-c^2) + a^2(y_A-z_A)
\\
&=\frac 1{4S}\sum (1-y_A-z_A)(b^2-c^2) + a^2(y_A-z_A)
\\
&=\frac 1{4S}\sum y_A(a^2-b^2+c^2) -z_A(a^2+b^2-c^2)
\\
&=\frac 1{4S}\sum y_A\cdot 2ac\sin B -z_A\cdot 2ab\sin C
\\
&=\frac 1{4RS}\sum y_A\cdot abc -z_A\cdot abc
\\
&=\sum(y_A - z_A)\ .
\\[3mm]
&\qquad\text{Now we plug in the values for $y_A$, $z_A$ from $(*)$:}
\\[3mm]
y_A &= 
\frac 
{2\cdot 2R\sin B\cos\gamma}
{2R(\sin A +2\cdot \sin B\cos \gamma +2\cdot \sin C\cos \beta)}\ ,
\\
z_A &= 
\frac 
{2\cdot 2R\sin C\cos\beta}
{2R(\sin A +2\cdot \sin B\cos \gamma +2\cdot \sin C\cos \beta)}\ ,
\\[3mm]
&\qquad\text{getting:}
\\[3mm]
s 
&=\sum
\frac
{2\cdot \sin B\cos \gamma - 2\sin C\cos\beta}
{\sin A +2\cdot \sin B\cos \gamma +2\cdot \sin C\cos \beta}
% \\
% &=-3 + \sum
% \frac
% {\sin A + 2\cdot \sin B\cos \gamma}
% {\sin A + 2\cdot \sin B\cos \gamma + 2\cdot \sin C\cos \beta}
\ .
\end{aligned}
$$
This is not an easy task now. (Either for the last expression of $s$,
or for the formula in between with a cyclic sum $s=\sum(y_A-z_A)$, if there is some
better geometric interpretation.)
So we are starting to solve a new problem in the problem.

Since time is an issue for me, instead of starting to compute using Lagrange multipliers,
in order to proceed in some few lines
here is a picture of the function to be maximized:

The plot uses $x,y\in[0,1]$ to parametrize the angles $\alpha,\beta,\gamma$ as follows:
$\displaystyle \alpha = \frac\pi3\cdot x$,
$\displaystyle \beta = \frac\pi3\cdot y(1-x)$, and
$\displaystyle \gamma = \frac\pi3\cdot (1-y)(1-x)$.
It turns out that the maximum is in fact a supremum, taken for the case when
$\displaystyle \alpha\nearrow\frac \pi 3$,
and then $\displaystyle \beta,\gamma\searrow 0$, so that
$\displaystyle \alpha+\beta+\gamma=\frac \pi 3$.
In terms of the angles of $\Delta ABC$ we have $\hat A\nearrow\pi$, $\hat B,\hat C\searrow 0$. (But $\hat B$, $\hat C$ may need to be still correlated.)
To compute this supremum, i need a notation that i can better type.
So i will switch from $\alpha,\beta,\gamma$ to $x,y,z$.
Then $\hat B=3y$, $\hat C=3z$, and corresponding sine values
will be approximated by hand waving with $3y+O(y^3)$ and $3z+O(z^3)$,
so that for $\hat A=\pi-(3x+3y)$ we have also a sine value in the shape $3y+3z+O(\dots)$.
Computations will omit below terms in total monomial degree bigger / equal two.
So
$$
\begin{aligned}
s &=
\frac
{2\sin B\cos\frac C3 - 2\sin C\cos \frac B3}
{\sin A + 2\sin B\cos\frac C3 + 2\sin C\cos \frac B3}
\\
&\qquad
+\frac
{2\sin C\cos\frac A3 - 2\sin A\cos \frac C3}
{\sin B + 2\sin C\cos\frac A3 + 2\sin A\cos \frac C3}
\\
&\qquad\qquad
+\frac
{2\sin A\cos\frac B3 - 2\sin B\cos \frac A3}
{\sin C + 2\sin A\cos\frac B3 + 2\sin B\cos \frac A3}
\\
&\sim
\frac{2(3y-3z)}{3y+3z + 2(3y+3z)}\\
&\qquad 
+\frac {2\left(3z\cdot\frac 12 -(3y+3z)\right)}{3y + 2\left(3z\cdot\frac 12  +(3y+3z)\right)}\\
&\qquad\qquad 
+\frac {2\left((3y+3z) - 3y\cdot\frac 12\right)}{3z + 2\left((3x+3y) + 3y\cdot\frac 12\right)}
\\
&=
\frac {2y-2z}{3y+3z}
-
\frac {2y+z}{3y+3z}
+
\frac {y+2z}{3y+3z}
\\
&=\frac{y-z}{3y+3z}\ .
\end{aligned}
$$
So we expect $\displaystyle\color{blue}{\frac 13}$ as a supremum value.
