2015 AMC 12B Problem 19

Problem 19 of 25HarderGeometry

\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?

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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.

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.