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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 30, 2024
 #1
avatar+128399 
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The triangle is isosceles

 

Let B = (0,0)   C =(8,0)  A = (3,4)

 

Height of triangle ABC  =  sqrt [5^2 -4^2] =  sqrt [9]  =  3

 

Area of ABC = (1/2) (8)(3) = (1/2) BC * height =   12

 

The incenter =  (4, y)

 

To  find y

 

[ 3* 8 + 0 * 5 + 0 * 5 ]  / [ 5 + 5 + 8 ]  =  24 / 18  =  4/3

 

Height of AMN =  3 - 4/3  = 5/3

 

AMN  similar to ABC

 

Scale factor  of AMN  to ABC =  [5/3]/ 3 =  5/9

 

[AMN ] =  (5/9)^2  [ ABC  ]     =  (25/81) [ 12]  =   100 / 27

 

 

cool cool cool

 Apr 1, 2024

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