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

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Therefore, <math>\bigtriangleup{AMK}</math> is congruent to <math>\bigtriangleup{CPK}</math> because of SAS
 
Therefore, <math>\bigtriangleup{AMK}</math> is congruent to <math>\bigtriangleup{CPK}</math> because of SAS
  
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<math>\angle{KMA}</math> is congruent to <math>\angle{KPA}</math> because of CPCTC
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<math>\angle{KMA}</math> is congruent to <math>\angle{KPC}</math> because of CPCTC
  
 
That shows <math>\overline{AM}</math> is parallel to <math>\overline{CP}</math> (also <math>CL</math>)
 
That shows <math>\overline{AM}</math> is parallel to <math>\overline{CP}</math> (also <math>CL</math>)

Revision as of 16:31, 27 February 2010

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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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 $\bigtriangleup{CPK}$ because of SAS

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

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

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

Thus,

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

Now lets apply the angle bisector theorem.

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

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

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

\[LP=\boxed {072}\]

See also

2009 AIME I (Problems • Answer Key • Resources)
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
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