Taiwan IMO Selection Camp, 2021

Let $O$ be the circumcenter of $\triangle ABC$, $J$ the intersection point of $EF$ and $OI$, and $L$ the second intersection point of $AJ$ with $\Omega$ (other than $A$).

Claim. The five points $O, M_a, X, J, L$ are concyclic.

Proof. Since $JO$ is the angle bisector of $\angle XJA$ and $XO = LO$, it follows that $J, L, O, X$ are concyclic. On the other hand, let $M$ be the midpoint of $AX$ on $OI$, and let $Y$ be a point on $\Omega$ such that $XY \parallel BC$. Then since $A, F, I, M, E$ are concyclic and $AFIE$ is a harmonic quadrilateral, we know that
$$-1 = (F, E; A, I) \stackrel{M}{\cong} (F, E; MA \cap EF, J) \stackrel{A}{\cong} (B, C; X, L) \stackrel{Y}{\cong} (B, C; \infty_{BC}, BC \cup YL)$$
Therefore $Y, M_a, L$ are collinear, that is, $OM_a$ is the angle bisector of $\angle KM_aX$, which means $M_a, X, J, O$ are concyclic, and hence the five points $O, M_a, X, J, L$ are concyclic. $\square$

Let $N$ be the midpoint of arc $BC$ (not containing $A$). By the Claim:
$$\angle XJM_a = \angle XOM_a = 2\angle XAN = 2\angle XAI = 2\angle (OI, ED),$$
and since $AJ$ and $XJ$ are symmetric with respect to $OI$, we have that $AJ$ and $M_aJ$ are symmetric with respect to $EF$. Note that the midpoints of segments $AT, M_bM_c, EF$ are collinear (the Newton line of the complete quadrilateral $\{CA, AB, EM_c, FM_b\}$); applying a homothety centered at $A$ with ratio $2$ shows that $TM_a$ passes through the reflection of $A$ with respect to $EF$, hence $M_a, J, T$ are collinear. Then by the Claim again:
$$\angle TM_aX = \angle JM_aX = \angle JLX = \angle ALX = \angle ASX = \angle TSX,$$
that is, $X, M_a, S, T$ are concyclic.