2015 AMC 10B Problem 19

The center of the circle through $X,Y,Z,W$ lies on the perpendicular bisectors of $XY$ and $ZW$. These are also the perpendicular bisectors of $AB$ and $AC$, so the same point is the circumcenter of right triangle $ABC$. Therefore the center is the midpoint $O$ of hypotenuse $AB$, so $OA=OB=OC=6$. Let $a=\frac12BC$ and $b=\frac12CA$. Then $a^2+b^2=6^2$. From the square on $AB$, $OX^2=6^2+12^2=180$. From the square on $AC$, the corresponding radius also gives $OW^2=b^2+(a+2b)^2$. Hence $$b^2+(a+2b)^2=180.$$ Subtracting $a^2+b^2=36$ from this equation gives $b(a+b)=36$. But $a^2+b^2=36$ as well, so $ab=a^2$. Since $a>0$, we get $a=b=3\sqrt{2}$. Thus $AC=BC=6\sqrt{2}$, and the perimeter is $12+12\sqrt{2}$. Thus, the correct answer is C.