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

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== Problem ==
 
== Problem ==
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Rectangle <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.)
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<math>
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\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi}
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\qquad
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\mathrm{(B)}\ \frac {1003}4
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\qquad
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\mathrm{(C)}\ 8\sqrt {1003}
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\qquad
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\mathrm{(D)}\ 6\sqrt {2006}
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\qquad
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\mathrm{(E)}\ \frac {32\sqrt {1003}}\pi
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</math>
  
 
== Solution ==
 
== Solution ==
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 +
<asy>
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size(7cm);
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 +
real l=10,
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w=7,
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ang=asin(w/sqrt(l*l+w*w))*180/pi;
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 +
draw((-l,-w)--(l,-w)--(l,w)--(-l,w)--cycle);
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draw(rotate(ang)*ellipse((0,0),2*l+2*w,l*w*2/sqrt(l^2+w^2)));
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draw(rotate(ang)*((0,0)--(2l+2w,0)),red);
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draw(rotate(ang+90)*((0,0)--(l*w*2/sqrt(l^2+w^2),0)),red);
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label("$A$",(-l,w),NW);
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label("$B$",(-l,-w),SW);
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label("$C$",(l,-w),SE);
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label("$D$",(l,w),SE);
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// Made by chezbgone2
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</asy>
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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
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<cmath>lw=ab=2006</cmath>
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From the definition of an ellipse, <math>l+w=2a\Longrightarrow \frac{l+w}{2}=a</math>.  Also, the diagonal of the rectangle has length <math>\sqrt{l^2+w^2}</math>.  Comparing the lengths of the axes and the distance from the foci to the center, we have
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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=\sqrt{1003}</cmath>
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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>.
  
 
== See also ==
 
== See also ==
* [[2006 AMC 12B Problems]]
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{{AMC12 box|year=2006|ab=B|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 13:15, 12 July 2015

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 (ProblemsAnswer KeyResources)
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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