Question

In $\triangle ABC,$ $\angle C = 90^\circ$ and $AB = 12.$ Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X,$ $Y,$ $Z,$ and $W$ lie on a circle. What is the perimeter of the triangle?

$12 + 9\sqrt3$

$18 + 6\sqrt3$

$12 + 12\sqrt2$

$30$

$32$

Solution:

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 $12^2 + 6^2 = b^2 + (a + 2b)^2.$ Solving yields $a = b = 3\sqrt2,$ so $BC = CA = 6\sqrt2$ and the perimeter is $12 + 12\sqrt2.$

Thus, the correct answer is C.

2015 AMC 12B Problem 19

Solution:

Problem 19 in Other Years