Platonic solids are made of squares, triangles and one of them is made of pentagons.
If the faces of the cube are taken while preserving its vertices, each face can be divided into smaller squares successively: this way the faces remain flat and the shape of the cube is also maintained.
Same happens with the tetrahedron: if their faces are divided into smaller triangles, these faces will remain flat and the shape of the tetrahedron is kept. The same with the other regular bodies with triangular faces. However, that is not the case of the dodecahedron.
If a pentagonal face is taken while keeping the vertices fixed and it is divided it into smaller pentagons, there are two possibilities: either the pentagons are deformed, or the surface rises outward to preserve the regular pentagons (forming a cap).
Fig 1
Fig 2
The subdivision could be carried on: the process is equivalent to replacing each pentagonal face with a smaller cap that shares the same original vertices but with smaller faces.
Possibly, the adjacent faces of a previous step could detach from each other and form holes because what only is fixed in a step are the initial vertices. And maybe there is possibility that the faces could intersect each other.
If this goes on, a fractal would be formed!
In addition, if, among the different cap trajectories, the one that rises vertically on one face was chosen, the face aligned with the vertical at the end would converge at a point. Therefore, in the final limit sphere, the vertices of the icosahedron should also belong to the fractal.
If $N_n$ is the number of faces in step n are: $$N_0=12$$ $$N_n=6N_{n-1}=N_0\cdot6^n$$ If $r_n$ is the radius from the center of the face to a vertex, and $f_n$ is the area of the face then: $$r_n=\alpha r_{n-1}=\alpha^nr_0$$ $$f_n=\beta r_n^2=\alpha^{2n}f_0$$
Where alpha and beta are constants of proportionality. If $A_n$ is the total area in step n: $$A_n=N_nf_n=N_0f_0(6\alpha^2)^n$$
In the cap, the base where it rests forms a pentagon (major), and the upper face is another pentagon (minor). If d is the distance between the center of the larger pentagon and the smaller pentagon of the first cap, the maximum growth $h$ would be: $$h=d+\alpha d+\alpha^2d+\cdots=\frac{1}{1-\alpha}\cdot d$$
Valid because alpha is less than one. With this and with the coordinates of the dodecahedron given on this site: https://en.wikipedia.org/wiki/Regular_dodecahedron
I calculated with an Excel sheet: $$\alpha=0.618$$ $$Final\,outer\,sphere\,radius / Initial\,sphere\, radius =2.384$$ The initial sphere is the one that touches the vertices of the dodecahedron, the final one is the one with the maximum height. $$A_n/A_{n-1}=2.292$$ So the area tends to infinity, but can be enclosed in a finite volume.
Questions:
This figure already exists? Maybe was already described before.
If it is a fractal, what dimension will it have?
Can something similar be done with hexagons?
Comments:
I assumed that the cap was rising, but could also have assumed that it sank, or alternated between one and the other.
I think it could look similar to this other fractal of an stellated dodecahedron:
http://clowder.net/hop/Keplrfrct/Keplrfrct.html