jmc

Assume $O$ to be the center of triangle $ABC$, $OT$ cross $BC$ at $M$, link $XM$, $YM$. Let $P$ be the middle point of $BT$ and $Q$ be the middle point of $CT$, so we have $MT=3\sqrt{15}$. Since $\angle A=\angle CBT=\angle BCT$, we have $\cos A=\frac{11}{16}$. Notice that $\angle XTY=180^{\circ }-A$, so $\cos XYT=-\cos A$, and this gives us $1143-2X{Y}^2=\frac{-11}{8}XT\cdot YT$. Since $TM$ is perpendicular to $BC$, $BXTM$ and $CYTM$ cocycle (respectively), so ${\theta }_1=\angle ABC=\angle MTX$ and ${\theta }_2=\angle ACB=\angle YTM$. So $\angle XPM=2{\theta }_1$, so$$\frac{\frac{XM}{2}}{XP}=\sin {\theta }_1$$, which yields $XM=2XP\sin {\theta }_1=BT(=CT)\sin {\theta }_1=TY.$ So same we have $YM=XT$. Apply Ptolemy theorem in $BXTM$ we have $16TY=11TX+3\sqrt{15}BX$, and use Pythagoras theorem we have $B{X}^2+X{T}^2=16^2$. Same in $YTMC$ and triangle $CYT$ we have $16TX=11TY+3\sqrt{15}CY$ and $C{Y}^2+Y{T}^2=16^2$. Solve this for $XT$ and $TY$ and submit into the equation about $\cos XYT$, we can obtain the result $X{Y}^2=\boxed{717}$.