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 ${CM}$ and ${CN}$ such that they intersects lines ${AB}$ at points $P$ and $Q$ , respectively. Since ${BM}$ is the angle bisector of angle B,and ${CM}$ is perpendicular to ${BM}$ ,so , $BP=BC=120$ , M is the midpoint of ${CP}$ .For the same reason, $AQ=AC=117$ ,N is the midpoint of ${CQ}$ . Hence $MN=\frac{PQ}{2}.But$ PQ=BP+AQ-AB=120+117-125=112,so $MN=\boxed{56}$ .

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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