Difference between revisions of "2006 AMC 12B Problems/Problem 21"
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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.) | ||
== Solution == | == Solution == | ||
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| + | ''This solution needs a picture. Please help add it.'' | ||
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Let the rectangle have side lengths <math>l</math> and <math>w</math>. Let the axis of the ellipse on which the foci lie have length <math>2a</math>, and let the other axis have length <math>2b</math>. We have | Let the rectangle have side lengths <math>l</math> and <math>w</math>. Let the axis of the ellipse on which the foci lie have length <math>2a</math>, and let the other axis have length <math>2b</math>. We have | ||
<cmath>lw=ab=2006</cmath> | <cmath>lw=ab=2006</cmath> | ||
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<cmath>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}</cmath> | <cmath>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}</cmath> | ||
Since <math>ab=2006</math>, we now know <math>a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}</math> and because <math>a=\frac{l+w}{2}</math>, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of <math>\boxed{8\sqrt{1003}}</math>. | Since <math>ab=2006</math>, we now know <math>a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}</math> and because <math>a=\frac{l+w}{2}</math>, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of <math>\boxed{8\sqrt{1003}}</math>. | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2006|ab=B|num-b=20|num-a=22}} | ||
Revision as of 18:18, 22 February 2009
Problem
Rectange 
has area 
. An ellipse with area 
passes through 
and 
and has foci at 
and 
. What is the perimeter of the rectangle? (The area of an ellipse is 
where 
and 
are the lengths of the axes.)
Solution
This solution needs a picture. Please help add it.
Let the rectangle have side lengths 
and 
. Let the axis of the ellipse on which the foci lie have length , and let the other axis have length . We have
![\[lw=ab=2006\]](http://latex.artofproblemsolving.com/6/4/c/64ca5cb3ac715ad0642585f852b24b5a23c6ba4f.png)
From the definition of an ellipse, 
. Also, the diagonal of the rectangle has length 
. Comparing the lengths of the axes and the distance from the foci to the center, we have
![\[a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b^2=\sqrt{1003}\]](http://latex.artofproblemsolving.com/4/b/d/4bdbb683608f55e62374580be8359d2804d4a084.png)
Since 
, we now know 
and because 
, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of 
.
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 | |
