Hi! I'm currently stuck on solving this problem:
"The triangle ABC is an isosceles triangle where AB = 4sqrt(2) and angle B is a right angle. If I is the incenter of triangle ABC, then what is 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."
For this problem, I drew triangle ABC with angle B as the right angle, the incircle, and the angle bisectors, and dropped down lines from the incenter to the points of tangency the circle had with the triangle. I knew those three dropped-down lines all had a length of the inradius and that they were perpendicular to a side of the triangle ABC. Thus, I realized that BI was a diagonal of a square with side lengths of the inradius, which meant that BI was equal to sqrt(2r^2) = r*sqrt(2), where r is the inradius. However, I didn't know what to do next or how to find r. Can someone give me a hint? I don't want the answer; I just want to know how to solve the problem. Thank you so much! 
+128401 We have an isosceles right triangle..... legs = 4sqrt (2) hypotenuse = 8
Perimeter = 8 + 8sqrt (2) = 8 ( 1 + sqrt (2) )
Let B = (0,0)
Let A = (0, 4sqrt(2) )
C = (4sqrt(2) ,0)
Formula to find the incenter =
x = ( side opposite A * x coordinate of A + side opp B * x coordinate of B + side opp C * x coordinate C) / perimeter
So we have
x = ( 4sqrt (2) * 0 + 8 * 0 + 4sqrt (2)*4sqrt (2) )/ [ 8 (1 + sqrt (2))] =
(32) / [ 8 (1 + sqrt (2)] =
4/ ( 1 + sqrt (2) ) = ( multiply by 1 -sqrt (2) on top/bottom and simplify ]
4 ( 1 - sqrt (2) ) / (-1) =
4 (sqrt (2) - 1 ) = 4sqrt (2) - 4 = sqrt (32) - 4
y = ( side opposite A *y coordinate of A + side opp B * y coordinate of B + side opp C * y coordinate of C) /perimeter
If you do this correctly you will find that x = y = sqrt (32) - 4
BI = sqrt ( [ sqrt (32) - 4 - 0 ]^2 + [ sqrt (32) - 4 - 0]^2 ) =
sqrt [ 32 - 8sqrt (32) + 16 + 32 - 8sqrt (2) + 16 ] =
sqrt [ 96 - 16 sqrt (32) ] =
sqrt [ 16 (6 - sqrt (32) ] =
4 sqrt [ 6 - 4sqrt (2 ] ≈ 2.343 units
See the following image :
i is the intersection of the angle bisectors......
