Question

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 $LP.$

Answer

Answer: 72

Concepts:parallelogram similarity angle bisector theorem

Difficulty rating: 2510

Solution:

Because $AK = CK$ and $K$ is the midpoint of $\overline{PM},$ the diagonals of quadrilateral $APCM$ bisect each other, so $APCM$ is a parallelogram and $AM \parallel CP.$ Since $P$ lies on line $CL$ and $B,$ $P,$ $M$ all lie on line $BK,$ triangles $BLP$ and $BAM$ are similar.

Thus $\frac{LP}{AM} = \frac{BL}{BA}.$ The angle bisector theorem gives $\frac{AL}{LB} = \frac{AC}{BC} = \frac{450}{300} = \frac{3}{2},$ so $\frac{BL}{BA} = \frac{2}{2 + 3} = \frac{2}{5}.$

Therefore $LP = \frac{2}{5} \cdot AM = \frac{2}{5} \cdot 180 = 72.$

2009 AIME I Problem 5

Answer: 72

Difficulty rating: 2510

Solution:

Problem 5 in Other Years