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# Triangles and sides | AIME I, 2009 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Triangles and sides.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Triangles and sides.

## Triangles and sides – AIME I, 2009

Triangle ABC AC=450 BC=300 points K and L are on AC and AB such that AK=CK and CL is angle bisectors of angle C. let P be the point of intersection of Bk and CL and let M be a point on line Bk for which K is the mid point of PM AM=180, find LP

• is 107
• is 72
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Triangles

Side Length

AIME I, 2009, Question 5

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

since K is mid point of PM and AC quadrilateral AMCP is a parallelogram which implies AM parallel LP and triangle AMB is similar to triangle LPB

then $\frac{AM}{LP}=\frac{AB}{LB}=\frac{AL+LB}{LB}=\frac{AL}{LB}+1$

from angle bisector theorem, $\frac{AL}{LB}=\frac{AC}{BC}=\frac{450}{300}=\frac{3}{2}$ then $\frac{AM}{LP}=\frac{AL}{LB}+1=\frac{5}{2}$

$\frac{180}{LP}=\frac{5}{2}$ then LP=72.