3D Dodecahedron model: Construction question. The image below is a 2D construction that, when cut-out and folded appropriately (hopefully it is intuitively clear how to cut and fold), forms a 3D dodecahedron. It works great: I've successfully used it to build a very nice dodecahedron. The problem I'm trying to solve is, given a rectangular piece of paper (mine happens to be 32" X 40", but any rectangle reasonable for the construction could be substituted), what is the largest such pattern that can be drawn?
As one aid to this, I am basing my construction on Steiner's Porism (using concentric circles with the radii in a given ratio to fill in the annular region with a chain of sequentially tangent circles, each of which is also tangent to the inner and outer circle). This specific part of the construction is significant only in that it makes the problem of constructing the largest possible dodecahedron cut-out within the given rectangular region somewhat simpler (to my mind at least), because one can now draw two circles with the same diameter to fill the page.
The problem is that the relationship between these two circles is not something simple like tangency. They overlap (as the diagram shows), but by how much?
In the diagram, one can imagine the line 'm' as dividing the sheet of paper in half by cutting the longer dimension. I'm taking it as given by the phrase 'reasonable rectangle for this construction' that one of the circles can be centered along the similarly defined dividing line for the shorter dimension. 
My problem, really, is to place the points A, B, and C, where the midpoint of A and B is on the dividing line for the shorter dimension, and making the circles as large as possible while still fitting on the page.
Purely empirically, I measured $\angle BAC = 8$ and the angle between dividing line m and and edge of the right hand "outer" pentagon, as 16 degrees (as labeled on the diagram).

 A: First off: I'm not perfectly clear about how you do the construction, so I might be missing something. For example, in your picture you have some pentagon edges horizontal, but is that a requirement? You could fit things more easily without that. It's not clear to my why the line $AB$ is of any importance. As far as I can tell, that line seems to be defined by the point $A$ and either the vertical direction or one of the pentagon corners. But the angle will be $8°$ in neither of these cases.
If you want to make things as big as possible, I'd rotate the figure, or equivalently rotate the sheet of paper beneath it:

Now you'd maximize the resulting dodecahedron by making the paper merely touch the outermost pentagons. That's the orange rectangle in my picture. But perhaps you need all of the circles on the sheet for the construction to work. In that case, I'd use the pink rectangle, which is the smallest rectangle containing both circles.
To compute its size, I originally used that measured angle of $125°$ between the centers of the circles and one point of intersection in, and based my computation on that. The measurement was done in Cinderella, and a nice round angle for such a regular setup seemed plausible enough. In the meantime I found out that the angle is not exactly $125°$, but instead it's
$$\arccos\frac{3\sqrt5-9}4\approx124.956°$$
The distance between the two circle centers is
$$d=\sqrt{\frac{13-3\sqrt5}2}r\approx1.774r$$
Then the long side of the paper would have to be at least $2r+d\approx3.774r$ while the short side is at least $2r$. Now for $40$ inch as the longer edge you get
$$r=\frac{40''}{2+\sqrt{\frac{13-3\sqrt5}2}}\approx10.60''$$
so you know the radii of the circles and the distance between their centers.
If you instead go for the orange containing rectangle, its edges have lengths
$$\frac{3\sqrt5}2r\approx3.354r
\qquad\text{and}\qquad
\frac{\sqrt{25-2\sqrt5}}2r\approx2.265r$$
Remember that in all of these formulas, $r$ is the radius of one of the two circles you drew. Two other useful quantities are the edge length of one small face of the dodecahedron, or the edge length of the pentagon formed by six such faces. These lengths, expressed as a multiple of $r$, are
$$\sqrt{\frac{25-11\sqrt5}2}r\approx0.449r
\qquad\text{resp.}\qquad
\sqrt{\frac{5-\sqrt5}2}r\approx1.176r$$
How did I find those numbers? Well, I did some computation in Sage, representing points as complex numbers. I started from fifth roots of unity and went outward from there using reflection in the midpoint of an edge as the essential operation. I used exract algebraic numbers, then computed radical expressions from their minimal polynomials.
# To find all vertices
z5 = QQbar.zeta(5)
pts1 = [z5^k for k in range(5)]
pts2 = [pts1[a]+pts1[(a+1)%5]-pts1[(a+1+b)%5] for a in range(5) for b in range(4)]
pts3 = pts2 + [pts2[1]+pts2[2]-z for z in [pts2[0]] + pts2[3:]]

# To see them, either as points or with their indices
points([(z.real(), z.imag()) for z in map(CDF,pts3)], aspect_ratio=1)
sum(text(k, (z.real(), z.imag())) for k, z in enumerate(map(CDF,pts3))).show(aspect_ratio=1)


# To compute the ratio d/r
c2 = pts3[1]+pts3[2]   # center of second circle
r = pts3[2].abs()      # radius of the circles
c2.abs()/r             # 1.773667960400232?
(c2.abs()/r).minpoly() # x^4 - 13*x^2 + 31

Other numbers were computed in a similar fashion.
The angle was computed from $d/r$ using the cosine law.
