STATS: 28 people solved it during the exam

Invert around $\gamma$. $\{B, X\}$ and $\{C, Y\}$ get swapped. So, $X, Y, B, C$ are concyclic. Now, because of inversion $AT^2 = AC\cdot AY$ and thus, $\angle ATY = \angle ACT$. Now, angle chasing gives us $AXTC$ is cyclic, similarly $AYTC$ is cyclic. So, $\overline{CX}\cap \overline{BY}$ is the radical center of circles $XYBC$, $AXTC$ and $BASY$.

My solution (Same as official solution): Let $CX \cap \odot (BAS) =S_1,BY \cap \odot (CAT)=T_1, BY \cap CX = K \odot (ABC) \cap \gamma =U,V$. Note that $XS \cdot XS_1 = XA \cdot XB = XU \cdot XV$ $\Rightarrow SUS_1V$ is cyclic $\Rightarrow S_1$ lies on $\gamma$. Similarly, $T_1$ lies on $\gamma$. Thus, $KS \cdot KS_1 =KT \cdot KT_1$ $\Rightarrow K$ lies on the radical axis of $\odot (BAS)$ and $\odot (CAT)$ $\Rightarrow K$ lies on $AP$. $\blacksquare$ REMARK: This problem is true even if $\gamma$ is not centered at $A$

Vrangr wrote: A triangle $ABC$ is given. A circle $\gamma$ centered at $A$ meets segments $AB$ and $AC$. The common chord of $\gamma$ and the circumcircle of $ABC$ meets $AB$ and $AC$ at $X$ and $Y$, respectively. The segments $CX$ and $BY$ meet $\gamma$ at point $S$ and $T$, respectively. The circumcircles of triangles $ACT$ and $BAS$ meet at points $A$ and $P$. Prove that $CX, BY$ and $AP$ concur. Extend lines $\overline{CX}$ and $\overline{BY}$ to meet $\gamma$ again at points $S'$ and $T'$ respectively. Apply radical axes theorem for $\{\odot(ABC), \gamma, \odot(ABS)\}, \{\odot(ABC), \gamma, \odot(ACT)\}, \{\odot(ABS), \odot(ACT), \gamma\}$ to conclude $ASBS', ATCT'$ cyclic and $\overline{SS'}, \overline{TT'}, \overline{PA}$ concur.