2006 AMC 12B Problem 21

Question

Rectangle

ABCD

has area

2006.

An ellipse with area

2006\pi

passes through

A

and

C

and has foci at

B

and

D.

What is the perimeter of the rectangle? (The area of an ellipse is

\pi ab,

where

2a

and

2b

are the lengths of its axes.)

\dfrac{16\sqrt{2006}}{\pi}

\dfrac{1003}{4}

8\sqrt{1003}

6\sqrt{2006}

\dfrac{32\sqrt{1003}}{\pi}

For the ellipse, the sum of distances from

A

to the foci

B, D

equals a rectangle diagonal plus a side

Let the sides be

x, y;

then

x + y = 2a

and the diagonal

\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}

Let the rectangle’s sides be

x

and

y.

Point

A

is on the ellipse with foci

B

and

D,

so

x + y = AB + AD = 2a.

The distance between the foci is the diagonal, so

\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}.

Then

2xy = (x + y)^2 - (x^2 + y^2)

= 4a^2 - (4a^2 - 4b^2)

= 4b^2,

so

xy = 2b^2.

The area gives

2b^2 = 2006,

hence

b^2 = 1003.

The ellipse area gives

\pi ab = 2006\pi,

so

ab = 2006

and

a = \dfrac{2006}{\sqrt{1003}} = 2\sqrt{1003}.

The perimeter is

2(x + y) = 4a = 8\sqrt{1003}.

Thus, the correct answer is C.

Problem 21 in other years