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In triangle $ABC$, let $I$ be the incenter of triangle $ABC$. The line through $I$ parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. If $AB = 5$, $AC = 5$, and $BC = 8$, then find the area of triangle $AMN$.

 Mar 22, 2024
 #1
avatar+128406 
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(1/2)BC = 4

Call the altitude  AD =   sqrt [ 5^2 - 4^2]  = 3

tan ( ABC)  = (3/4)

tan (ABC / 2)  =    1 -cos (ABC) / sin (ABC)  =  [ 1  - 4/5 ] / [(3/5 ] = 1/3

 

Equation  of angle bisector  of  ABC  =   y =(1/3)x

 

Intersection of   this line with  angle bisector BAC {line  x = 4 }

 

y= (1/3) (4)  = 4/3

 

Height of triangle  AMN  =  3 - 4/3  =  5/3

 

Area  of  [ABC]  =  (1/2) AD * BC  =   (1/2) (3) (8)  = 12

 

Triangle AMN  similar to triangle ABC

 

Scale factor  of  AMN/ ABC  =  (5/3) / 3 =  5/9

 

[ AMN ]  =  12 *  [ (5/9 ] ^2  = 100 / 27   ≈  3.7

 

cool cool cool

 Mar 23, 2024
edited by CPhill  Mar 23, 2024
edited by CPhill  Mar 23, 2024

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