please help on this problem!

Hey there, HT!

So...

Assume O to be the centre of triangle ABC, OT cross BC at M, link XM and 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 $1143-2XY^2=\frac{-11}{8}XT\cdot YT.$ 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 the Pythagoras theorem we have $BX^2+XT^2=16^2$. Same in YTMC and triangle CYT we have $16TX=11TY+3\sqrt{15}CY$ and $CY^2+YT^2=16^2$. Solve this for XT and TY and submit it into the equation about $\cos XYT$, we can obtain the result $XY^2=\boxed{717}$.

Hope this helped! :)

( ﾟдﾟ)つ Bye