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

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

Key Concepts


Angles

Triangles

Side Length

Check the Answer


But try the problem first…

Answer: is 72.

Source
Suggested Reading

AIME I, 2009, Question 5

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

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

Second Hint

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

Final Step

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

Subscribe to Cheenta at Youtube