Construction of a Simson Line Passing Through a Given Point, and Other Related Questions Update.  Part (a) is somewhat solved (i.e., if I blindly believe the paper I have not yet fully digested).  Part (c) is completely solved by timon92.  Part (b) is competely solved.  There are also two bonus questions (Bonus Question 2 has also been answered by timon92).  The only question left is just Bonus Question 1.  Therefore, this question is marked as answered.

Let $\Gamma$ be the circumcircle of a given triangle $ABC$, whereas $H$ is its orthocenter.  For a point $X$ on $\Gamma$, define $s(X)$ to be the Simson line of $X$ with respect to the triangle $ABC$.

(a) For a point $Y$ on the plane, how many points $X$ on $\Gamma$ are there such that $Y$ lies on $s(X)$?
(b) For a (generic) point $Y$ on the plane, is it possible to construct a point $X$ on $\Gamma$ using a straightedge and a compass so that $Y$ lies on $s(X)$?  
(c) Let $D$, $E$, and $F$ be three points on $\Gamma$.  Suppose that $H'$ is the orthocenter of the triangle $DEF$, and $M$ the midpoint of the line segment $HH'$.  The straight lines $s(E)$ and $s(F)$ intersect at $D'$, the straight lines $s(F)$ and $s(D)$ intersects at $E'$, and the straight lines $s(D)$ and $s(E)$ intersect at $F'$.  Prove that $M$ is the circumcenter of the triangle $D'E'F'$.


For Part (a), for each point $Y$ on the plane, let $n(Y)$ denote the number of points $X$ on $\Gamma$ such that $s(X)$ contains $Y$.  Then, I believe that the region consisting of points $Y$ with $n(Y)=3$ is the interior of a Steiner deltoid $\mathcal{T}$ as shown below (see also here and there).  It appears also that the boundary of $\mathcal{T}$ minus the vertices is the set of points $Y$ with $n(Y)=2$.  The three sides of $\mathcal{T}$ seem to be tangent to the sides of the triangle $ABC$.  Outside $\mathcal{T}$ as well as the vertices of $\mathcal{T}$ is the set of points $Y$ with $n(Y)=1$.  (I found the proof of this Steiner deltoid here.  I will have to digest it, but I think Part (a) is more or less settled.  According to the Wolfram link, the nine-point circle of the triangle $ABC$ is the incircle of $\mathcal{T}$, and the circle centered at the nine-point center of the triangle $ABC$ with the radius three times the nine-point radius is the circumcircle of $\mathcal{T}$.)

Bonus Question 1.  Is it possible to construct (with a straightedge and a compass) the vertices of the Steiner deltoid $\mathcal{T}$?  Does $\mathcal{T}$ touch the sides of the triangle $ABC$?  If so, can we construct the points at which $\mathcal{T}$ touch the sides of the triangle $ABC$, using a straightedge and a compass?

It is confirmed by the Wikipedia page that $\mathcal{T}$ indeed touches the sides (or their extensions) of the triangle $ABC$.  (I have not found a proof, so if you have a reference, please let me know.)  The construction of the point of tangency between $\mathcal{T}$ and the straight line $BC$ can be contructed as follows.  Let $A_1$ be the point diametrically opposite to $A$ with respect to $\Gamma$.  Then, the orthogonal projection of $A_1$ onto the line $BC$ gives the point $A_2$ which coincides with the point where $\mathcal{T}$ touches the line $BC$.  The points $B_2$ and $C_2$ where $\mathcal{T}$ touches $CA$ and $AB$ can be constructed similarly.
As a side note, it appears that the vertices of $\mathcal{T}$ are connected to the first Morley triangle of the triangle $ABC$.  Since the construction of the first Morley triangle involves trisecting an angle, it appears that the vertices of $\mathcal{T}$ may not be constructible.  However, the exact reasoning is unclear to me at this moment.

Due to the possibility of having three solutions, I think that finding the point $X\in\Gamma$ such that $Y\in s(X)$ must involve solving a cubic polynomial.  Therefore, I guess that the answer to Part (b) is no.  (That is why I added the tag abstract-algebra to the question.)  In other words, if $m$ denotes the slope of $s(X)$, then $m$ should be a root of a cubic polynomial $f(t)\in\mathbb{R}[t]$.  I think that the interior of $\mathcal{T}$ corresponds to this cubic polynomial having three distinct real roots.  The sides (but not the vertices) of $\mathcal{T}$ are the locations where two real roots coincide and the other is different.  The vertices are where the three real roots coincide.  The exterior of $\mathcal{T}$ is the set where two roots are complex, and one root is real.
Indeed, let $a$, $b$, and $c$ be the complex coordinates of $A$, $B$, and $C$, and suppose that they lie on the unit circle $\big\{z\in\mathbb{C}\,\big|\,|z|=1\big\}$.    from Proposition 4 of this article, we see that the complex coordinates $x$ of $X$ must satisfy
$$x^3+(a+b+c-2y)\,x^2-(bc+ca+ab-2abc\bar{y})\,x
-abc=0\,,$$
where $y$ is the complex coordinate of $Y$.  This is a cubic polynomial which is irreducible over $\mathbb{C}(a,b,c,y,\bar{y})$ (treating $a$, $b$, $c$, $y$, and $\bar{y}$ as indeterminates).  Consequently, it is not possible to construct a point $X$ on $\Gamma$ such that $Y\in s(X)$ using a straightedge and a compass.
For Part (c), I have not made much progress.  However, below is a drawing of the figure for this part.  It feels like this part must be known.  Maybe somebody can give me a reference.

For those who are wondering how Parts (a) and (b) are related to Part (c).  My explanation is as follows.  I was trying to find a condition where three Simson lines concur.  Then, it occurred to me that I would have to find a way to construct a Simson line that passes through a given point.  After playing with the figure, I discovered that the circumcenter of the triangle made from three Simson lines is the midpoint of the two orthocenters.

Bonus Question 2.  What is a necessary and sufficient conditions for points $D$, $E$, and $F$ on $\Gamma$ so that $s(D)$, $s(E)$, and $s(F)$ concur?  

 A: Sketch of proof for c) since I'm typing on my phone (sorry about that). If something needs an additional explanation feel free to ask.
We recall some well-known facts:


*

*Simson line of $X$ bisects $HX$.

*If $XX'$ is a chord of the circumcircle of $ABC$ perpendicular to $BC$ then $AX'$ is parallel to the Simson line of $X$.


Using these we reduce the problem to the following (note that here $D', E', F'$ will be different points than in OP's question!): Let $DD', EE', FF'$ be parallel chords of a circle $\omega$ and let $A\in \omega$. Draw parallels to $AD', AE', AF'$ through $D,E,F$, respectively. These parallels determine a triangle $D''E''F''$. Show that its circumcenter coincides with the orthocenter of $DEF$.
To do this, an easy angle chasing shows that $D''E''F''$ is similar to $DEF$ and moreover the circles $(D''EF), (DE''F), (DEF'')$ all pass through the orthocenter $H'$ of $DEF$. We conclude that $\angle F''D''H'=\angle EFH' = \frac\pi2 -\angle DEF = \frac\pi2 - \angle D''E''F''$ so the line $D''H'$ is isogonal to the $D''$-altitude of $D''E''F''$. Similar results hold for other vertices, hence $H'$ is the circumcenter of $D''E''F''$.
Bonus question 2: Similar ideas as above lead to the following conclusion: Simson lines of $D, E, F$ are concurrent if and only if $s(D)\perp EF$. Hence, changing $D, E, F$ cyclically, this is also equivalent to $s(E) \perp FD$ as well as to $s(F) \perp DE$. 
Interpreting $A, B, C, D, E, F$ as points on the unit circle in $\mathbb C$, this turns out to be equivalent to $ABC=DEF$. In particular we get the following cute fact: Simson lines of $D, E, F$ with respect to $ABC$ are concurrent if and only if Simson lines of $A, B, C$ with respect to $DEF$ are concurrent.
