Difference between revisions of "2006 AMC 12B Problems/Problem 21"
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== 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 | 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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== 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}} | ||
+ | {{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.