2011 AIME I Problems/Problem 4

Revision as of 14:22, 6 April 2011 by Cliu0301 (talk | contribs) (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 $58.5 + 60 - 62.5 = \boxed {56}$.

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS