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

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(Solution)
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<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>
 
<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>
  
<cmath>\frac {AM}{LP}=frac {AL}{LB}+1=frac {5}{2}</cmath>
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<cmath>\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}</cmath>
  
 
<cmath>\frac {180}{LP}=frac {5}{2}</cmath>
 
<cmath>\frac {180}{LP}=frac {5}{2}</cmath>

Revision as of 23:32, 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

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

Since $K$ is the midpoint of $PM, AC$.

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

Therefore, triangle $AMK$ is congruent to triangle $CPK$ because of SAS

angle $KMA$ is congruent to $KPA$ because of CPCTC

That shows $AM$ is parallel to $CP$ (also $CL$)

That makes triangle $AMB$ congruent to $LPB$

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.

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

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

\[\frac {180}{LP}=frac {5}{2}\]

\[LP=\boxed {072}\]

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