Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 16$, $BC = 22$, and $TX^2 + TY^2 + XY^2 = 1143$. Find $XY^2$.
Let $O$ be the circumcenter of $\triangle ABC$; say $OT$ intersects $BC$ at $M$; draw segments $XM$, and $YM$.  We have $MT=3\sqrt{15}$. 

[Fanyuchen.png](https://artofproblemsolving.com/wiki/index.php/File:Fanyuchen.png)
Since $\angle A=\angle CBT=\angle BCT$, we have $\cos A=\tfrac{11}{16}$. Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$, so $\cos XTY=-\cos A$, and the cosine law in $\triangle TXY$ gives \[1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.\]
Since $\triangle BMT \cong \triangle CMT$, we have $TM\perp BC$, and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$) be the midpoint of $BT$ (resp. $CT$). So $P$ (resp. $Q$) is the center of $(BXTM)$ (resp. $CYTM$). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$. So $\angle XPM=2\theta$, so\[\frac{\frac{XM}{2}}{XP}=\sin \theta,\]which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$. Similarly we have $YM=XT$. 
Ptolemy's theorem in $BXTM$ gives \[16TY=11TX+3\sqrt{15}BX,\] while Pythagoras' theorem gives $BX^2+XT^2=16^2$. Similarly, Ptolemy's theorem in $YTMC$ gives\[16TX=11TY+3\sqrt{15}CY\] while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$. Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to obtain the result $XY^2=\boxed{717}$.
(Notice that $MXTY$ is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)
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