2015 AMC 12B 第 19 题

The circle’s center $O$ is equidistant from $X, Y, Z, W,$ so it is the circumcenter of $\triangle ABC$

Since $\angle C = 90^\circ,$ $O$ is the midpoint of $AB;$ set $a = \tfrac12 BC,$ $b = \tfrac12 CA$ and use $OX = OW$

The center $O$ of the circle lies on the perpendicular bisectors of $XY$ and $ZW,$ which are the same as those of $AB$ and $AC.$ So $O$ is the circumcenter of $\triangle ABC,$ and since $\angle C = 90^\circ,$ $O$ is the midpoint of $AB.$

Let $a = \tfrac12 BC$ and $b = \tfrac12 CA.$ Then $a^2 + b^2 = 6^2,$ and computing $OX^2 = OW^2$ gives $12^2 + 6^2 = b^2 + (a + 2b)^2.$ Because the left side is $5(a^2+b^2),$ subtracting gives $4a(b-a)=0.$ Thus $a=b,$ and $2a^2=36$ gives $a=b=3\sqrt2.$ Therefore $BC=CA=6\sqrt2,$ and the perimeter is $12+12\sqrt2.$

Thus, the correct answer is C.