2010 AMC 10B Problem 20

Question

Two circles lie outside regular hexagon

ABCDEF.

The first is tangent to

\overline{AB},

and the second is tangent to

\overline{DE}.

Both are tangent to lines

BC

and

FA.

What is the ratio of the area of the second circle to that of the first circle?

18

27

36

81

108

Answer: D

Concepts:regular polygon tangent line special right triangle area ratio

Difficulty rating: 1960

Small hint:

Compare the two circle radii

Big hint:

The larger radius comes from a

30

-

60

-

90

triangle

Solution:

Consider the following diagram:

Assume the regular hexagon has side length

1.

The smaller circle is inscribed in an equilateral triangle of side length

1.

The inradius of this equilateral triangle is

\dfrac{\sqrt3}{6}.

The area of the circle is then

\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}.

Let

O

be the center of the larger circle. Drop the perpendicular from

O

to

\overline{GH}

at

J.

Draw

\overline{OG}.

We have that

\triangle OJG

is right. Since

\angle HGI = 60^{\circ},

we also have that

\triangle OJG

is a

30-60-90

triangle.

Let

OJ = r.

Then

OG = 2r.

We also have that

OG

is the sum of the height of the hexagon, equilateral triangle, and radius of the circle.

Then

OG = \dfrac{\sqrt3}{2} + \sqrt3 + r.

Substituting in

OG,

we get

2r = \dfrac{\sqrt3}{2} + \sqrt3 + r.

Simplifying gives us

r = \dfrac{3\sqrt3}{2}.

The area of the larger circle is then

\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi.

The desired ratio is then

\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81.

Thus, D is the correct answer.