Question

A triangle has corners at `(5 , 5 )`, `(9 ,4 )`, and `(1 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`r = { 8 } / {\sqrt{17} + 4 sqrt{5} + 5 }`

Explanation:

We call the corners vertices.

Let `r` be the radius of the incircle with incenter I. The perpendicular from I to each side is the radius `r`. That forms the altitude of a triangle whose base is a side. The three triangles together make the original trangle, so its area `mathcal{A}` is

`mathcal{A} = 1/2 r(a+b+c)`

We have
`a^2 = (9-5)^2 + (4-5)^2=17`

`b^2 = (9-1)^2 + (8-4)^2=80`

`c^2 = (5-1)^2+(8-5)^2=25`

The area `mathcal{A}` of a triangle with sides `a,b,c` satisfies

`16mathcal{A}^2 = 4a^2 b^2 - (c^2 - a^2 - b^2)^2`

`16 mathcal{A}^2 = 4(17)(80) - (25 - 17 - 80)^2 = 256`

`mathcal{A} = sqrt{256/16} = 4`

`r = {2 mathcal{A} } / (a+b+c)`

`r = { 8 } / {\sqrt{17} + sqrt{80} + sqrt{25} }`