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

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== Solution ==
 
== Solution ==
 +
Sorry, I fail to get the diagram up here, someone help me.
 +
 +
Since <math>K</math> is the midpoint of <math>PM, AC</math>.
 +
 +
Thus, <math>AK=CK,PK=MK</math> and the opposite angles are congruent.
 +
 +
Therefore, triangle <math>AMK</math> is congruent to triangle <math>CPK</math>
 +
 +
angle <math>KMA</math> is congruent to <math>KPA</math> because of CPCTC
 +
 +
That shows <math>AM</math> is parallel to <math>CP</math> (also <math>CL</math>)
 +
 +
That makes triangle <math>AMB</math> congruent to <math>LPB</math>
 +
 +
Thus, <math>\frac {AM}{LP}=\frac {AB}{LB}</math>
 +
 +
<cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath>
 +
 +
Now let apply angle bisector thm.
 +
 +
<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>
 +
 +
<cmath>\frac {180}{LP}=frac {5}{2}</cmath>
 +
 +
<cmath>LP=\boxed {072}</cmath>
  
 
== 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 22:31, 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 $CPK$

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 (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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