A triangle, whose sides are 6, 8, and 10, and a circle, whose radius is r, are drawn so that no part of the triangle lies outside the circle. How small can r be?

Question

anonymous · Answer

@AccessDenied

anonymous · Answer

I'll try that again.  Hold on.

anonymous · Answer

This is harder than I thought at first.  I'm sure I could get the answer, but not quickly.

anonymous · Answer

Here's what I tried to do:

|dw:1355968207731:dw|

anonymous · Answer

I was thinking that I could make a 30,60,90 triangle to solve for r, but this picture doesn't make sense because this isn't an equilateral triangle and therefore the triangle angles by the vertices cannot all be 60

anonymous · Answer

Except that you are not going to get pairs of 30 degree angles.  Just got your post and you came up with the same thing.  I have another idea we might be able to work on together.  I'll have to draw it.

anonymous · Answer

|dw:1355968497779:dw|This is crude and unfinished, but I think you are going to be constrained by the length of the longer leg of this 6, 8, 10 right triangle.  Notice how the endpoints of the side of 8 hits the circle, but the side of 6 doesn't make it quite to the circle on one side.  I'm not sure yet, but I think this is what we are going to be faced with.  The side of 10 is not along the diameter.