2009 AIME I Problems/Problem 5

Revision as of 20:52, 20 March 2009 by Moplam (talk | contribs) (Solution)

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

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See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions