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Difference between revisions of "2006 AMC 12B Problems/Problem 21"

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== Problem ==
 
== Problem ==
 
Rectange <math>ABCD</math> has area <math>2006</math>.  An ellipse with area <math>2006\pi</math> passes through <math>A</math> and <math>C</math> and has foci at <math>B</math> and <math>D</math>.  What is the perimeter of the rectangle? (The area of an ellipse is <math>ab\pi</math> where <math>2a</math> and <math>2b</math> are the lengths of the axes.)
 
Rectange <math>ABCD</math> has area <math>2006</math>.  An ellipse with area <math>2006\pi</math> passes through <math>A</math> and <math>C</math> and has foci at <math>B</math> and <math>D</math>.  What is the perimeter of the rectangle? (The area of an ellipse is <math>ab\pi</math> where <math>2a</math> and <math>2b</math> are the lengths of the axes.)
 +
 +
<math>
 +
\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
 +
</math>
 +
 
== Solution ==
 
== Solution ==
  

Revision as of 20:58, 1 January 2012

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.)

$\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi} \qquad \mathrm{(B)}\ \frac {1003}4 \qquad \mathrm{(C)}\ 8\sqrt {1003} \qquad \mathrm{(D)}\ 6\sqrt {2006} \qquad \mathrm{(E)}\ \frac {32\sqrt {1003}}\pi$

Solution

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

See also

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
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