Problem: Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$.
Let $\ell$ be a variable line that is tangent to $\omega$ and
meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively.
A point $R$ is chosen such that $PR=PA$ and $QR=QB$.
Find all possible locations of the point $R$, over all choices of $\ell$.

Solution: Let $r$ be the inradius.
Let $T$ be the tangency point of $\ol{PQ}$
on arc $\widehat{DE}$ of the incircle, which we consider varying.
We define $R_1$ and $R_2$ to be the two intersections
of the circle centered at $P$ with radius $PA$,
and the circle centered at $Q$ with radius $QB$.
We choose $R_1$ to lie on the opposite side of $C$ as line $PQ$.

\begin{center}
\begin{asy}
size(11cm);
pair A = dir(150);
pair B = dir(30);
pair C = dir(270);
pair D = midpoint(B--C);
pair E = midpoint(A--C);
pair I = incenter(A, B, C);
draw(incircle(A, B, C), blue);
pair R_1 = dir(70);
pair P = extension(midpoint(A--R_1), I, B, C);
pair Q = extension(midpoint(B--R_1), I, A, C);
pair T = extension(R_1, I, P, Q);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, blue);
draw(P--Q, blue);

pair Ap = 4*D;
pair Bp = 4*E;
pair R_2 = 2*T-R_1;

draw(R_2--P--R_1, gray);
draw(P--A, gray);
draw(R_2--R_1, heavycyan);
draw(A--Ap, deepgreen);
draw(B--Bp, deepgreen);

draw(arc(I,B,A), red+1);
draw(arc(I,Bp,Ap), red+1);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(10));
dot("$E$", E, dir(170));
dot("$I$", I, dir(100));
dot("$R_1$", R_1, dir(R_1));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(290));
dot("$A'$", Ap, dir(Ap));
dot("$B'$", Bp, dir(Bp));
dot("$R_2$", R_2, dir(R_2));

/* TSQ Source:

!size(11cm);
A = dir 150
B = dir 30
C = dir 270
D = midpoint B--C R10
E = midpoint A--C R170
I = incenter A B C R100
incircle A B C blue
R_1 = dir 70
P = extension midpoint A--R_1 I B C
Q = extension midpoint B--R_1 I A C
T = extension R_1 I P Q R290
A--B--C--cycle 0.1 yellow / blue
P--Q blue

A' = 4*D
B' = 4*E
R_2 = 2*T-R_1

R_2--P--R_1 gray
P--A gray
R_2--R_1 heavycyan
A--Ap deepgreen
B--Bp deepgreen

!draw(arc(I,B,A), red+1);
!draw(arc(I,Bp,Ap), red+1);

*/
\end{asy}
\end{center}

\begin{claim*}
  The point $R_1$ is the unique point on ray $TI$ with $R_1I = 2r$.
\end{claim*}
\begin{proof}
  Define $S$ to be the point on ray $TI$ with $SI = 2r$.
  Note that there is a homothety at $I$ which maps $\triangle DTE$
  to $\triangle ASB$, for some point $S$.

  Note that since $TASD$ is an isosceles trapezoid,
  it follows $PA = PS$.
  Similarly, $QB = QS$.
  So it follows that $S = R_1$.
\end{proof}

Since $T$ can be any point on the open arc $\widehat{DE}$,
it follows that the locus of $R_1$
is exactly the open $120\dg$ arc of $\widehat{AB}$
of the circle centered at $I$ with radius $2r$
(i.e.\ the circumcircle of $ABC$).

It remains to characterize $R_2$.
Since $TI = r$, $IR_1 = 2r$, it follows $TR_2 = 3r$ and $IR_2 = 4r$.
Define $A'$ on ray $DI$ such that $A'I = 4r$,
and $B'$ on ray $IE$ such that $B'I = 4r$.
Then it follows, again by homothety,
that the locus of $R_2$ is the $120\dg$ arc $\widehat{A'B'}$
of the circle centered at $I$ with radius $4r$.

In conclusion, the locus of $R$ is the two open $120\dg$ arcs we identified.