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Let $I$ be the incenter of triangle $ABC$.  If $AB = BC = 4$ and $\angle B = 60^\circ$, then find the length $BI$.

Express your answer in the form $a + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers, and $c$ is not divisible by any perfect square other than $1.$

 Aug 21, 2023
 #1
avatar+128732 
+1

 

 

 

 

Since AB = BC and angle B = 60....then angle BAC = angle BCA  =  [180 - 60] / 2 =  60

 

Then the triangle is equilateral

 

The height of the  triangle = 2sqrt (3)

 

The area = (1/2)* 4^2 * sqrt (3)  / 2 =    4sqrt 3

The semi-perimeter =   4*3 / 2 = 6

 

The inradius =  area / semi-perimeter  =  [4sqrt 3 ] /  6 =    2sqrt (3) / 3 =  (2/3)sqrt 3 = IE

 

I =  (2, (2/3)sqrt 3 )

 

B =  (2, 2sqrt 3)

 

BI =  2sqrt (3)  - (2/3)sqrt (3) =   (4/3)sqrt 3 - (2/3)sqrt 3  =  (2/3)sqrt 3

 

cool cool cool

 Aug 22, 2023

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