Let $B_4$ be the second intersection of $\Omega$ with the line parallel to $AC$ trough $B$, and similarly define $C_4$. If we let $B_4'$ be the second intersection of $B_1B_3$ and $\Omega$ we notice that we must have $\measuredangle B_4'B_3B_2=\measuredangle B_4B_3B_2=\frac{\pi}{2}$, meaning that $B_4'$ is the antipode of $B_2$ in $\Omega$, which is well known to be $B_4$. It is also known (see the ISL2011/G4 configuration) that $B_1B_4$ passes through the barycenter $G$ of $ABC$, so it follows that $B_1B_3$ and $C_1C_3$ concur on $G$, meaning that $X=G$, which lies on the $A$-median of $ABC$. Another point which lies on this median $m$ is the point $Z$ such that $ABZC$ is a parallelogram, which is in fact also the intersection of $BB_4$ and $CC_4$ (it just follows from parallel lines). Now, consider the point $Y'=m\cap BC_3$, and consider the map $f$, which is the composition of three projective involutions on $\Omega$, centered at $Z,X,Y'$ respectively. Since these three points lie on the common line $m$, it follows by ping pong lemma that this is also an involution (with center on $m$). Since $C_4=CZ\cap \Omega$, $C_3=C_4X\cap \Omega$ and $B=C_3Y'\cap \Omega$, it follows that $f(C)=B$. Since $f$ is an involution, it also follows that $f(B)=C$. Since $BZ\cap\Omega=B_4$ and $B_4X\cap\Omega=B_3$, it finally follows that $B_3Y'\cap \Omega=C$, or in other words $Y'\in B_3C$. Thus, $Y=BC_3\cap B_3C=Y'\in m$, meaning that $X,Y$ both belong to $m$, which of course being the median bisects $BC$ in the point $M$ [which in fact is the center of the involution $f$].

After getting that $X$ is the centroid, you can finish by Pascal on $BC_3C_4CB_3B_4$.

Let $M$ be the midpoint of $BC$, the reflection of $A$ over $M$ be $A_1$, the second intersection between $A_1B$ and $\Omega$ be $B'$, and $A_1C$ meet $\Omega$ again at $C'$. Because $ABA_1C$ is a parallelogram, we know $\overline{BB'A_1} \parallel AC \perp \overline{BB_1B_2}$, so $B'B_2$ is a diameter of $\Omega$. This gives $$\angle B_2B_3B' = 90^{\circ} = \angle B_2B_3B_1$$which means $B' \in B_1B_3$. Similarly, we deduce $C' \in C_1C_3$. Now, ISL 2011/G4 implies $X$ is the centroid of $ABC$, so $X \in \overline{AMA_1}$. On the other hand, Pascal's on $BC_3C'CB_3B'$ yields $Y \in XA_1$, which clearly suffices. $\blacksquare$ Remarks: This problem is quite underwhelming. Also, I haven't applied Pascal's in a long time.

Here is another proof of $X$ being centroid. Let $H$ and $O$ be the orthocenter and circumcenter of $ABC$, respectively. Define $B_4$ to be midpoint of $B_1B_2$ and $C_4$ similarly. Note that $B_1C_1 \parallel B_4C_4, B_4O \parallel XB_3 \perp B_3B_2, C_4O \parallel XC_3 \perp C_3C_2.$ Since $C_4C_1$ intersects $B_4B_1$ at $H$ and $HC_1 : C_1C_4 = 2:1$ We get that $H, X, O$ collinear with $HX : XO = 2:1$, thus $X$ is the centroid of $ABC$