Planes of a dodecahedron divide the space $\mathbb R^3$ In how many disjunct parts divide the planes of a regular dodecaheadron the space $\mathbb R^3$?
I encountered this task at university where we discussed it a lot, but nobody found a solution.
Does anybody know the solution? May it work with graph theory?
 A: Here's a graphical count, based on intersecting the configuration with spheres of different sizes -- though in order to make things visible actually I'm normalizing the size of the spheres, such that the images here actually shows a sphere intersected by the planes of a dodecahedron of increasing size.
The 12 planes of the dodecahedron come in 6 parallel pairs, and in the figures, the regions of each sphere are color-coded according to (and labeled with) the number of pairs where the region is between the two parallel planes.


The first image shows a sphere far from the dodecahedron. Each of the 6 pairs of planes become a narrow band around the sphere, and the bands create the graph of an icosidodecahedron. Each face, edge, and vertex of the icosidodecahedron corresponds to an unbounded region of space -- there are $12$ of type 0, $20$ of type 0', $60$ of type 1, and $30$ of type 2.
As we zoom in towards the dodecahedron (or the dodecahedron become larger), the edges of each band move away from each other. The first change that happens is that the the 0' regions disappear and are replaced by $20$ cells of type 3. These new cells are the triangular spires of the great stellated dodecahedron.


Nothing new happens in the first of these pictures, except that the bands keep growing wider.
In the second picture the bands have just met in groups of five, and the last of the 0 and 1 regions disappear. Instead we get cells that are between 4 and 5 pairs of parallel planes.


The new yellow regions of type 4 merge, and we can see there are $30$ such cells, just as there were thirty 2s before, corresponding to the edges of either an icosahedron or a dodecahedron. In fact each of these cells is an irregular tetrahedron that shares one edge with the central dodecahedron, and the opposite edge with a great dodecahedron (whose edge figure looks like an icosahedron).
Removing the yellow tedrahedra from the great dodecahedron leaves a small stellated dodecahedron, whose pentagonal spires are the $12$ magenta 5 cells that came into being together with the 4s.


Finally a single red 6 cell appears: the original dodecahedron in the middle.
So how many regions were that? Summing everything we get
$$ \underbrace{(12+20)+60+30}_{\text{unbounded}}+\underbrace{20+30+12+1}_{\text{bounded}} = 185 $$
A: There is a nice combinatorial formula for the number of regions in a hyperplane arrangement in $\mathbb{R}^d$.
$$r(\mathcal{A}) = (-1)^d\chi_\mathcal{A}(-1)$$
Where $r(\mathcal{A})$ denotes the number of regions and $\chi_\mathcal{A}(t)$ is the characteristic polynomial of the underlying poset of flats.
In this case we get that $\chi_\mathcal{A}(t) = t^3-12t^2+60t-112$, so $$r(\mathcal{A}) = (-1)^3(-1-12-60-112)= 185$$
This setup also lets you read off the number of bounded regions by the formula:
$$b(\mathcal{A}) = \pm\chi_\mathcal{A}(1)$$
So we get 63 bounded regions. 
For a reference on all this I'll recommend Richard Stanley's lecture notes on hyperplane arrangements available here: http://www.cis.upenn.edu/~cis610/sp06stanley.pdf
A: OK, Henning Makholm did the explanation, and did it good, but I'll do my part anyway.
First we extend the faces of our dodecahedron just a bit beyond the edges, until it grows nice little spikes on all 12 faces and thus becomes a small stellated dodecahedron.

Then we grow them a bit further, until each valley between the two neighboring spikes is filled. There were as many valleys as edges in a dodecahedron, so this gives us 30 more regions and creates the thing known as great dodecahedron.

Now we grow them further yet until we see new, sharper spikes where the depressions used to be. These correspond to the vertices of the original guy, so we add 20 of them and behold, here is your great stellated dodecahedron.

All in all, this gives us $1+12+30+20=63$ bounded regions.
So it goes.
A: I found 432 in the book
"Les nombres remarquables" by Francois Le Lionnais (Herman, 1983)
(which is a list of interesting numbers), but there is no reference or proof. Among these 432 parts 192 are bounded.
