# Difference between revisions of "2011 AIME I Problems/Problem 4"

## Problem 4

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 $MN$.

## Solution

Extend ${MN}$ such that it intersects lines ${AC}$ and ${BC}$ at points $O$ and $Q$, respectively. Notice that ${OQ}$ is a midline; it then follows that ${OC} = 58.5$ and ${QC} = 60$. Now notice that $\triangle MQC$ and $\triangle NOC$ are both isosceles. Thus, $ON = OC = 58.5$ and $MQ = QC = 60$. Since $OQ$ is a midline, $OQ = 62.5$. We want to find $MN$, which is just $ON + MQ - OQ$.

Substituting, the answer is $58.5 + 60 - 62.5 = \boxed {56}$.