2015 AMC 12B Problem 19

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

Answer: C

Concepts:circumcircle, circumcenter, and circumradius right triangle coordinate geometry

Difficulty rating: 2040

Small hint:

The circle’s center

O

is equidistant from

X, Y, Z, W,

so it is the circumcenter of

\triangle ABC

Big hint:

Since

\angle C = 90^\circ,

O

is the midpoint of

AB;

set

a = \tfrac12 BC,

b = \tfrac12 CA

and use

OX = OW

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.

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.