2009 AIME I Problems/Problem 5

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

Sorry, I fail to get the diagram up here, someone help me.

Since $K$ is the midpoint of $\overline{PM}, \overline{AC}$ .

Thus, $AK=CK,PK=MK$ and the opposite angles are congruent.

Therefore, $\bigtriangleup{AMK}$ is congruent to $\bigtriangleupCPK$ (Error compiling LaTeX. Unknown error_msg) because of SAS

$\angle{KMA}$ is congruent to $\angle{KPA}$ because of CPCTC

That shows $\overline{AM}$ is parallel to $\overline{CP}$ (also $CL$ )

That makes $\bigtriangleup{AMB}$ congruent to $\bigtriangleup{LPB}$

Thus, $\frac {AM}{LP}=\frac {AB}{LB}$

$\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1$

Now let apply angle bisector thm.

$\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}$

$\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}$

$\frac {180}{LP}=\frac {5}{2}$

$LP=\boxed {072}$

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2009 AIME I Problems/Problem 5

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