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In $\triangle ABC,$ $AB=6 $ , $BC = 4$ and $ AC=8.$ A segment parallel to $ \overline{BC}$ and tangent to the incircle of $\triangle ABC$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N$.  Find the length $MN$.

 

 Dec 27, 2020
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Area of triangle ABC 

 

Find the semi-perimeter, S   = ( 8 + 6+ 4)  /2   =  9

 

Area  = sqrt  [  9 (9-8) (9-6) (9-4)  ] =    sqrt  [ 9 * 3 * 5  ] =  3sqrt (15)

 

The circle is an inscribed circle in a triangle....its radius, R, can be found as

 

Area / S  =  R

 

3sqrt (15)  / 9   = R   =   sqrt (15) /  3

 

And  the height of  triangle ABC  can be found as

 

3sqrt (15)  = (1/2)  BC * height

 

6sqrt (15)  =  4 * height

 

(3/2) sqrt (15)  = height  of ABC

 

Height of triangle AMN  =   (3/2)sqrt (15)  - 2R  =  (3/2)sqrt (15) - (2/3)sqrt (15)   = (5/6)sqrt (15)

 

And triangle AMN is similar to triangle ABC....so....

 

MN/ height of AMN  = BC /height of ABC

 

MN/ (5/6)sqrt (3)   = 4 / [(3/2)sqrt (15)]

 

MN  / (5/6)  = 4 /(3/2)

 

MN  = 4 (5/6) / (3/2)

 

MN  = (4)(2/3)(5/6)  = 40/18  =   20/9

 

 

cool cool cool

 Dec 27, 2020

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