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 $58.5 + 60 - 62.5 = \boxed {56}$ .

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2011 AIME I Problems/Problem 4

Problem 4

Solution

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