Difference between revisions of "2009 AIME I Problems/Problem 5"

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== Solution ==
 
== Solution ==
<geogebra>ccb891f12b378421b058e68a3346c4c3bcd3d124</geogebra>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2009|n=I|num-b=4|num-a=6}}

Revision as of 22:51, 20 March 2009

Problem

Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$, and $\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$. If $AM = 180$, find $LP$.

Solution

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions