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

This solution needs a picture. Please help add it.

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 $a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}$ 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2006 AMC 12B Problems/Problem 21

Problem

Solution

See also