Difference between revisions of "2009 AIME I Problems/Problem 5"
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<cmath>\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}</cmath> | <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> |
<cmath>LP=\boxed {072}</cmath> | <cmath>LP=\boxed {072}</cmath> | ||
Revision as of 22:32, 20 March 2009
Problem
Triangle 
has 
and 
. Points 
and 
are located on 
and 
respectively so that 
, and 
is the angle bisector of angle 
. Let 
be the point of intersection of 
and , and let 
be the point on line 
for which is the midpoint of 
. If 
, find 
.
Solution
Sorry, I fail to get the diagram up here, someone help me.
Since is the midpoint of 
.
Thus, 
and the opposite angles are congruent.
Therefore, triangle 
is congruent to triangle 
because of SAS
angle 
is congruent to 
because of CPCTC
That shows 
is parallel to 
(also 
)
That makes triangle 
congruent to 
Thus, 
![\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]](http://latex.artofproblemsolving.com/e/1/c/e1c643180d89cafc631e9d0edee4f27ece659e43.png)
Now let apply angle bisector thm.
![\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]](http://latex.artofproblemsolving.com/2/2/2/222902e6a7c83208fd7a690061448410a76bd342.png)
![\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]](http://latex.artofproblemsolving.com/9/e/3/9e3fdaedab91b63a87671f6e257cc8f6165c61fe.png)
![\[\frac {180}{LP}=\frac {5}{2}\]](http://latex.artofproblemsolving.com/c/e/6/ce6572ed77cc85768d9fcc93c86781f17423b393.png)
![\[LP=\boxed {072}\]](http://latex.artofproblemsolving.com/2/9/f/29f3056ccdf0b94f38196093cd9a898f77a5beaf.png)
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 | ||
