+817 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 $1.$
+129771 
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
