Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Triangles and sides.
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
Angles
Triangles
Side Length
Answer: is 72.
AIME I, 2009, Question 5
Geometry Vol I to IV by Hall and Stevens
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.
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.
Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.
Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.