# 2006 AMC 12B Problems/Problem 21

(Redirected from 2006 AMC 12B Problem 21)

## Problem

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 $ab\pi$ where $2a$ and $2b$ are the lengths of the axes.) $\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi} \qquad \mathrm{(B)}\ \frac {1003}4 \qquad \mathrm{(C)}\ 8\sqrt {1003} \qquad \mathrm{(D)}\ 6\sqrt {2006} \qquad \mathrm{(E)}\ \frac {32\sqrt {1003}}\pi$

## Solution $[asy] size(7cm); real l=10, w=7, ang=asin(w/sqrt(l*l+w*w))*180/pi; draw((-l,-w)--(l,-w)--(l,w)--(-l,w)--cycle); draw(rotate(ang)*ellipse((0,0),2*l+2*w,l*w*2/sqrt(l^2+w^2))); draw(rotate(ang)*((0,0)--(2l+2w,0)),red); draw(rotate(ang+90)*((0,0)--(l*w*2/sqrt(l^2+w^2),0)),red); label("A",(-l,w),NW); label("B",(-l,-w),SW); label("C",(l,-w),SE); label("D",(l,w),SE); // Made by chezbgone2 [/asy]$

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 $\boxed{8\sqrt{1003}}$.

## See also

 2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 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

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