Difference between revisions of "2005 AIME II Problems/Problem 14"

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== Problem ==
 
== Problem ==
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In triangle <math> ABC, AB=13, BC=15, </math> and <math>\displaystyle CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math>
 
In triangle <math> ABC, AB=13, BC=15, </math> and <math>\displaystyle CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
 
{{solution}}
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== See also ==
 
== See also ==
  
*[[2005 AIME II Problems/Problem 13| Previous problem]]
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* [[2005 AIME II Problems/Problem 13| Previous problem]]
*[[2005 AIME II Problems/Problem 15| Next problem]]
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* [[2005 AIME II Problems/Problem 15| Next problem]]
 
* [[2005 AIME II Problems]]
 
* [[2005 AIME II Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:34, 7 September 2006

Problem

In triangle $ABC, AB=13, BC=15,$ and $\displaystyle CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

Solution

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See also