Difference between revisions of "2004 AIME II Problems/Problem 13"

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== Problem ==
 
== Problem ==
Let <math> ABCDE </math> be a convex pentagon with <math> AB || CE, BC || AD, AC || DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>\displaystyle DE = 15. </math> Given that the ratio between the area of triangle <math> ABC </math> and the area of triangle <math> EBD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math>
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Let <math> ABCDE </math> be a convex pentagon with <math> AB || CE, BC || AD, AC || DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the ratio between the area of triangle <math> ABC </math> and the area of triangle <math> EBD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math>
  
 
== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
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== See also ==
 
== See also ==
* [[2004 AIME II Problems/Problem 12 | Previous problem]]
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{{AIME box|year=2004|n=II|num-b=12|num-a=14}}
* [[2004 AIME II Problems/Problem 14 | Next problem]]
 
* [[2004 AIME II Problems]]
 

Revision as of 12:25, 19 April 2008

Problem

Let $ABCDE$ be a convex pentagon with $AB || CE, BC || AD, AC || DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

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See also

2004 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions