Given isosceles triangles $\triangle ABC$ and $\triangle DBF$ (all spherical chords), identify the chord $\overline {DF}$ so that $|AD| = |DF| = |FC|$ tl;dr: As shown in the image below, find the chord $\overline {DF}$ so that $|\overline {AD}| = |\overline {DF}|$, and have the answer be in the form of the ratio between $|\overline {AC}|$ and $|\overline {DF}|$. Ideally, I'm actually looking for a slightly different ratio (which is explained below), but this is the basic problem I believe that needs to solved. Only the isosceles triangle $\triangle ABC$ may be used as input  (the image is slightly misleading, as they all should be straight lines and $\overline DF$ should be slightly higher than $\triangle ABC$ and not intersecting it).

The actual problem I'm trying to solve is to subdivide a dodecahedron into smaller faces (known as a chamfered dodecahedron), and approximate a sphere using that polyhedron. This is done in 3 steps:

*

*Scale down each face from its center by some scale factor (what I'm trying to solve for) so that the hexagons constructed in the next step are planar (they will not be equiangular, however)

*Construct hexagons in between the newly subdivided faces (goldberg polyhedrons will always have 12 pentagons, no matter how subdivided they are)

*Scale the new face from the origin so that each vertex has the same magnitude of the original face (this is done by normalizing each vertex, then scaling by a specified radius - this will make the hexagons non-planar)

My problem is that I'm stuck on how to identify how to scale factor in the first step, as there doesn't seem to be a linear relationship between the scale factor needed and the number of subdivisions that are performed. This is shown in the two images below, where I'm adding the new faces over the original dodecahedron (radius of 1 in my case). I manually calculated approximately $0.40706$ (or $40.706\%$) for the first subdivision, which allowed the construction of the [green] hexagon to have satisfyingly equivalent sides. However, when I tried scaling the new pentagons again with the same scale factor (shown with the yellow pentagons), the hexagons in-between were not equilateral...

I suspect that the scale factor will change the more I subdivide these faces - indicating that there must be a non-linear relationship between side length and the length of the connecting chord. To attempt to solve my problem, I reconstructed the issue to what I hope is a solvable math problem, shown in the images below (although lines $\overline {AB}$ and $\overline {BC}$ should be chords, not arcs - thanks to @Blue in the comments). However, I frankly have no idea where to start... I feel like solving this problem is beyond my knowledge. I then tried condensing this to a simpler case with an isosceles triangle in 2D space, but I encountered a similar block as well. I know that I need "scale" as the output (specifically when scale was used in the first step), and my known values are the pentagon/hexagon side lengths, its position in 3D space, and the magnitude of each of the points. I would be very grateful for your guidance on this issue!

Edit:
This is the result I'm trying to achieve, using the following manually approximated scale factors (note that the hexagons are not planar):
Subdivision | Pentagons | Hexagons 
------------|-----------|---------
     0      |     -     |    -   
     1      |  0.40706  |    -   
     2      |  0.44950  |  0.4874 
     3      |    ...    |   ... 




 A: This isn't a complete solution but so far I have the essential conventions in place.
To make things more specific, I'll consider the sphere to have radius 1 with $B$ at the north pole, $O$ as the origin, and the points $A,D$ lying in the positive $xz$-half plane. Additionally, I'll denote
\begin{align}
\theta&=\angle ABC=\angle DBF,\\
\phi_1&=\angle BOD =\angle BOD,\\
\phi_2&=\angle AOB=\angle BOF. \\
\end{align}
(An aside: Note that these denote the arc-lengths of the various segments of the spherical figure, which is always larger than the Euclidean distance between the two vertices. This isn't crucial to the problem posed: If two pairs of vertices have the same arc-length then they have the same Euclidean distance, so we might as well consider arc-length. But it will matter if we consider ratios, since the relationship between these distance measures isn't linear.)
As such, the Cartesian coordinates for points $A,B,C,D,F$ are given as
\begin{align}
A&=(\sin\phi_2,0,\cos\phi_2),\\
B&=(0,0,1),\\
C&=(\sin\phi_2\cos\theta,\sin\phi_2\sin\theta,\cos\phi_2),\\
D&=(\sin\phi_1,0,\cos\phi_1)\\
F&=(\sin\phi_1\cos\theta,\sin\phi_1\sin\theta,\cos\phi_1),\\
\end{align}
The condition of interest is then $\angle DOF=\phi_2-\phi_1$, which is imposed by via the dot product of the position vectors for $D,F$:
$$\cos(\phi_2-\phi_1)=\cos \angle DOF=\vec{D}\cdot \vec{F}=\sin^2\phi_1\cos\theta+\cos^2\phi_1.$$
In principle, we can use $\phi_2,\theta$ to determine the angle $\phi_1\in [0,\pi]$.
Similarly, the angle $\angle AOC$ is determined via the vectors for $A,C$:
$$\cos\angle AOC = \vec{A}\cdot \vec{C}=\sin^2\phi_2\cos\theta+\cos^2\phi_2$$
which is determined by $\phi_2,\theta$ directly. So in that sense there is a functional relationship between $\phi_1$, $\phi_2-\phi_1$ and $\angle AOC$, but at present they seem...unpleasant. I'll see if I can find any clever way to reduce this. Alternatively things may simplify if we restrict to a certain case of interest. For instance, when $\theta=2\pi/5$ (as for the icosahedron) one has $\cos\theta=(2\phi)^{-1}$ where $\phi$ is the golden ratio. So it's plausible that a specific case may be more tractable than the generic.
