2009 AMC 12A Problem 19

Question

Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon (

7

sides). The areas of the two regions were

A

and

B,

respectively. Each polygon had a side length of

2.

Which of the following is true?

A = \dfrac{25}{49}B

A = \dfrac{5}{7}B

A = B

A = \dfrac{7}{5}B

A = \dfrac{49}{25}B

Answer: C

Concepts:regular polygon Pythagorean Theorem circle area

Difficulty rating: 1910

Small hint:

For a regular polygon with side

2,

let

r

and

R

be the inradius and circumradius, and look at the center, a side’s midpoint, and its endpoint

Big hint:

That right triangle has legs

r

and

1

and hypotenuse

R,

so

R^2 - r^2 = 1

regardless of the number of sides

Solution:

For a regular polygon with side length

2,

let

O

be the center,

M

the midpoint of a side, and

N

an endpoint of that side. Then

\triangle OMN

has a right angle at

M,

with

MN = 1,

OM = r

(inradius), and

ON = R

(circumradius).

So

R^2 - r^2 = 1,

and the area between the circles is

\pi(R^2 - r^2) = \pi

for any number of sides. Hence

A = B.

Thus, the correct answer is C.