Circles on a Line

Consider three identical equilateral triangles,
such that one triangle shares a side
with each of the other two. A circle is drawn inside each triangle, where
all three circles are identical and as large as possible, while keeping
the centers of the three circles collinear (all on the same line.)
If the height of the triangles is 1, what are the radii of the circles?
How far along the side of one triangle is the point of tangency with its
internal circle?

Consider three identical squares, where one shares a side with each of
the other two (in a rightangle configuration.)
As above, identical circles are drawn inside each square such that their
centers are collinear and the circles are as large as possible.
If the side of the squares is 1, what are the radii of the circles?
How far along the side of one square is the point of tangency with its
internal circle?

In a 4x4 grid of squares, identical circles are placed in the outer 12
squares, such that they are as large as possible and
their centers are all on the same large circle (with center at the middle
of the grid.)
If the side of the squares is 1, what are the radii of the circles?
How far along the side of one square is the point of tangency with its
internal circle?

Consider three identical hexagons, where each shares a side with each of
the other two.
As above, identical circles are drawn inside each hexagon such that their
centers are collinear and the circles are as large as possible.
If the height of the hexagons is 1 (between opposite sides),
what are the radii of the circles?
How far along the side of one hexagon is the point of tangency with its
internal circle?

Consider three identical circles, where each is tangent to each of
the other two.
As above, identical circles are drawn inside each circle such that their
centers are collinear and the internal circles are as large as possible.
If the radii of a large circle is 1,
what are the radii of the smaller circles?
Source: Original. The last was suggested by Phillippe Fondanaiche.
Solution
Mail to Ken