2011 AIME I Problems/Problem 4

Revision as of 03:28, 7 November 2015 by Yangmystery (talk | contribs) (Solution)

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$MN=\boxed{56}$.

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

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