Question

In triangle $ABC,$ $AB = 125,$ $AC = 117,$ and $BC = 120.$ The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L,$ and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K.$ Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL},$ respectively. Find $MN.$

Answer

Solution:

Extend $\overline{CM}$ and $\overline{CN}$ to meet $\overline{AB}$ at $P$ and $Q,$ respectively. In triangle $BCP,$ the segment $BM$ is both an angle bisector and an altitude, so the triangle is isosceles with $BP = BC = 120,$ and $M$ is the midpoint of $\overline{CP}.$ Similarly, triangle $ACQ$ is isosceles with $AQ = AC = 117,$ and $N$ is the midpoint of $\overline{CQ}.$

Hence $\overline{MN}$ is a midline of triangle $CPQ,$ so $MN = \frac{PQ}{2}.$ Since $PQ = BP + AQ - AB = 120 + 117 - 125 = 112,$ we conclude $MN = 56.$

2011 AIME I Problem 4

Solution:

Problem 4 in Other Years