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# I need some help! Thank you!!

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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! Nov 21, 2020
edited by Guest  Nov 21, 2020
edited by Guest  Nov 21, 2020

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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......   Nov 21, 2020
edited by CPhill  Nov 21, 2020