2006 AMC 12B Problems/Problem 21

Problem

Rectange $ABCD$ 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 $ab\pi$ where $2a$ and $2b$ are the lengths of the axes.)

Solution

Let the rectangle have side lengths $l$ and $w$ . Let the axis of the ellipse on which the foci lie have length $2a$ , and let the other axis have length $2b$ . We have $lw=ab=2006$ From the definition of an ellipse, $l+w=2a\Longrightarrow \frac{l+w}{2}=a$ . Also, the diagonal of the rectangle has length $\sqrt{l^2+w^2}$ . Comparing the lengths of the axes and the distance from the foci to the center, we have $a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b^2=\sqrt{1003}$ Since $ab=2006$ , we now know $a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}$ and because $a=\frac{l+w}{2}$ , or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of $\boxed{8\sqrt{1003}}$ .

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2006 AMC 12B Problems/Problem 21

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