2006 AMC 12B Problem 21

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.