2009 AIME I Problems/Problem 5: Difference between revisions

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 $CPK$ 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.

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

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

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

$LP=\boxed {072}$

@@ Line 11: / Line 11: @@
 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>
+Therefore, triangle <math>AMK</math> is congruent to triangle <math>CPK</math> because of SAS
 angle <math>KMA</math> is congruent to <math>KPA</math> because of CPCTC

2009 AIME I (Problems • Answer Key • Resources)
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2009 AIME I Problems/Problem 5: Difference between revisions

Revision as of 22:31, 20 March 2009

Problem

Solution

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