Difference between revisions of "2006 AMC 12B Problems/Problem 21"
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== Problem == | == Problem == | ||
+ | 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.) | ||
+ | |||
+ | <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 == | ||
+ | |||
+ | <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 <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> | ||
+ | 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 | ||
+ | <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> | ||
+ | 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 == | ||
− | + | {{AMC12 box|year=2006|ab=B|num-b=20|num-a=22}} | |
+ | {{MAA Notice}} |
Latest revision as of 13:15, 12 July 2015
Problem
Rectangle 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
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 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.