Difference between revisions of "2003 AIME II Problems/Problem 11"

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== Problem ==
 
== Problem ==
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Triangle <math>ABC</math> is a right triangle with <math>AC = 7,</math> <math>BC = 24,</math> and right angle at <math>C.</math> Point <math>M</math> is the midpoint of <math>AB,</math> and <math>D</math> is on the same side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math>
  
 
== Solution ==
 
== Solution ==
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== See also ==
 
== See also ==
* [[2003 AIME II Problems/Problem 10| Previous problem]]
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{{AIME box|year=2003|n=II|num-b=10|num-a=12}}
 
 
* [[2003 AIME II Problems/Problem 12| Next problem]]
 
 
 
* [[2003 AIME II Problems]]
 

Revision as of 14:40, 21 November 2007

Problem

Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

Solution

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See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions