2018 AMC 12B Problem 25

Question

Circles

\omega_1,

\omega_2,

and

\omega_3

each have radius

4

and are placed in the plane so that each circle is externally tangent to the other two. Points

P_1,

P_2,

and

P_3

lie on

\omega_1,

\omega_2,

and

\omega_3,

respectively, so that

P_1P_2=P_2P_3=P_3P_1

and line

P_iP_{i+1}

is tangent to

\omega_i

for each

i=1,

2,

3,

where

P_4=P_1.

See the figure below. The area of

\triangle P_1P_2P_3

can be written in the form

\sqrt{a}+\sqrt{b},

where

a

and

b

are positive integers. What is

a+b?

546

548

550

552

554

Answer: D

Concepts:tangent circles special right triangle law of cosines equilateral triangle

Difficulty rating: 2840

Small hint:

Since

P_iP_{i+1}

is tangent to

\omega_i

at

P_i,

the radius

O_iP_i

is perpendicular to

P_iP_{i+1}

Big hint:

Let

K=O_1P_1\cap O_2P_2;

with

d=P_1K,

apply the Law of Cosines in

\triangle O_1KO_2

Solution:

Let

O_i

be the center of

\omega_i,

and let

K

be the intersection of lines

O_1P_1

and

O_2P_2.

Because

\angle P_1P_2P_3=60^\circ,

triangle

P_2KP_1

is a

30

-

60

-

90^\circ

triangle. With

d=P_1K,

we get

P_2K=2d

and

P_1P_2=\sqrt3\,d.

The Law of Cosines in

\triangle O_1KO_2

(with

O_1O_2=8

) gives

\begin{gathered} 8^2=(d+4)^2+(2d-4)^2 \\ {}-2(d+4)(2d-4)\cos60^\circ, \end{gathered}

which simplifies to

3d^2-12d-16=0,

so

d=2+\tfrac23\sqrt{21}.

Then

P_1P_2=\sqrt3\,d=2\sqrt3+2\sqrt7,

and the area is

\begin{gathered} \dfrac{\sqrt3}{4}\left(2\sqrt3+2\sqrt7\right)^2 \\ =10\sqrt3+6\sqrt7 \\ =\sqrt{300}+\sqrt{252}. \end{gathered}

So

a+b=300+252=552.

Thus, the correct answer is D.