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

Revision as of 17:45, 1 July 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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\angle{KPA}$ 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.

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

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

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

$LP=\boxed {072}$

@@ Line 4: / Line 4: @@
 == Solution ==
-Sorry, I failed to get the diagram up here, someone help me.
+{{image}}
 Since <math>K</math> is the midpoint of <math>\overline{PM}, \overline{AC}</math>.
@@ Line 34: / Line 34: @@
 == See also ==
 {{AIME box|year=2009|n=I|num-b=4|num-a=6}}
+[[Category:Intermediate Geometry Problems]]

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

Page

Toolbox

Search

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

Revision as of 17:45, 1 July 2009

Problem

Solution

See also