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.)

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

$\dfrac{1003}{4}$

$8\sqrt{1003}$

$6\sqrt{2006}$

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

Solution:

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.

2006 AMC 12B Problem 21

Solution:

Problem 21 in Other Years