Problem 1941

Method 1:
Let $A^*$ be the antipode of $A$ with respect to $\odot(ABC)$, let $M, N$ be the midpoints of $\overline{A^*E}$, $\overline{A^*F}$ respectively, then $\overline{AC} \perp \overline{A^*C}$ and $\overline{AB} \perp \overline{A^*B}$, so $M, N$ lie on the perpendicular bisectors of $\overline{CE}$, $\overline{BF}$ respectively. From $\angle OMP = \angle ONP = 90^\circ$, we know that $M, N, O, P$ are concyclic. Since $MN$ is parallel to $EF$, we obtain
$$90^\circ - \angle TRQ = \angle OPN = \angle OMN = \angle AEF = 90^\circ - \angle TSQ,$$
that is, $Q, R, S, T$ are concyclic.

Method 2:
Let $D$ be the intersection of $BC$ and $EF$, let $M$ be the second intersection point of $\odot(ABC)$ and $\odot(AEF)$, and let $O', O_B, O_C$ be the circumcenters of $\triangle AEF, \triangle ABF, \triangle ACE$ respectively. Then $M$ is the Miquel point of the complete quadrilateral $(BC, CA, AB, EF)$, and it is well known that $M, O, O', O_B, O_C$ lie on a common circle $\Gamma$ (i.e., the Miquel circle). Note that $PO_B, PO_C, OO_B, OO_C$ are the perpendicular bisectors of $\overline{BF}, \overline{CE}, \overline{MB}, \overline{MC}$ respectively, so
$$\angle O_B PO_C = \angle (BF, CE) = \angle BAC = \angle BMC = \angle O_B OO_C,$$
that is, $P$ also lies on $\Gamma$. Since $O'O, O'O_B$ are the perpendicular bisectors of $\overline{MA}, \overline{MF}$ respectively, we have
$$90^\circ - \angle TRQ = \angle OPO_B = \angle OO'O_B = \angle AMF = \angle AEF = 90^\circ - \angle TSQ,$$
that is, $Q, R, S, T$ are concyclic. $\square$

It suffices to show that $QR$ and $ST$ are anti-parallel with respect to $AB$ and $AC$. Let $l_1$ and $l_2$ be two lines through $A$ that are parallel to $QR$ and $ST$, respectively. Then it suffices to show that $l_1$ and $l_2$ are isogonal with respect to $AE$ and $AF$. Note that $l_2$ is perpendicular to $EF$, so it suffices to show that $l_1$ passes through the circumcenter $O'$ of $AEF$. In other words, it suffices to show that $AO'$ is parallel to $OP$. This is true as the lengths of the projections of $AO'$ and $OP$ to $AE$ are both equal to $\frac{1}{2}AE$, and those to $AF$ are both equal to $\frac{1}{2}AF$, showing that $AO'PO$ is a parallelogram.