The South African Mathematical Olympiad Third Round

Note that $\angle AZW=90^{\circ }-\angle AWZ=\angle BWX$ and $\angle AWZ=90^{\circ }-\angle BWX=\angle BXW$. Moreover, $WX=WZ$, so the two triangles $AWZ$ and $BXW$ are congruent. Let $BW=AZ=x$, so that $BX=AW=5-x$. Pythagoras' theorem gives us $$x^2+(5-x{)}^2=13,$$ which simplifies to $x^2-5x+6=0$. The two solutions are $x=2$ and $x=3$, and by symmetry we can assume that $x=2$. So $BX=3$, $XC=7$, $AZ=2$ and $ZD=8$. Now let $CY=y$, so that $YD=5-y$. Applying Pythagoras' theorem again (and making use of the fact that $XY=YZ$), we get $$7^2+y^2=8^2+(5-y{)}^2.$$ This simplifies to $10y=40$, so $y=4$. Now we finally find that $XY=\sqrt{7^2+4^2}=\sqrt{65}$.