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?

$12+9\sqrt{3}$

$18+6\sqrt{3}$

$12+12\sqrt{2}$

$30$

$32$

Solution:

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 $b^2+(a+2b)^2=180.$ Together with $a^2+b^2=36$ , this gives $a=b=3\sqrt{2}$ .

Thus $AC=BC=6\sqrt{2}$ , and the perimeter is $12+12\sqrt{2}$ .

Thus, the correct answer is C.

2015 AMC 10B Problem 19

Solution:

Problem 19 in Other Years