<s>[INST] Common region between an icosahedron and a dodecahedron This is admittedly one of the hard problems I've come across.  It involves the common region (intersection) between two dual platonic solids: icosahedron, and dodecahedron.
The question is as follows:
A dodecahedron and an icosahedron intersect as shown below. Their edge lengths are such that the intersection (i.e. taking the points that are inside the dodecahedron and inside the icosahedron) results in the polyhedron shown below (second image), where all faces are regular pentagons and hexagons of side length $1$. Find the volume of this polyhedron which is almost like a soccer ball.


 [/INST] As @David K observed, by considering the edge of the icosahedron, it is segmented into 3 equal parts, each being of length $1$.  Hence the edge length of the icosahedron is $3$.
The second concern is the edge length of the dodecahedron.  Consider a face of the dodecahedron.  Suppose the edge length is $a$, and the small pentagon contained within it is of side length $1$.  Therefore the distance $w$ between a vertex of the small pentagon and the edge of the dodecahedron is
$w = (\dfrac{a}{2}) \cot(36^\circ) -(\dfrac{1}{2}) \csc(36^\circ) \hspace{25pt}(1)$
and we have
$ 2 w \sin\left( \frac{1}{2} \cos^{-1}\left(\dfrac{1}{\sqrt{5}} \right) \right) = 1\hspace{25pt}(2)$
Using $(1), (2)$:
$ c_1 a + c_2 = 1 $
where
$ c_1 = 1.17082039 $
$ c_2 = - 1.44721359$
Therefore the edge length of the dodecahedron is
$ a = \dfrac{ 1 - c_2 }{c_1} = 2.09016994$
Finally we note from the figure, that for the intersection solid, there are $12$ unit pentagons and $20 $ unit hexagons.  Their areas are
$A_{Pent} = 5 (\frac{1}{2}) ( (\frac{1}{2} \csc(36^\circ) )^2 \sin(72^\circ) =  1.72047740 $
$A_{Hex} = 6 ( \frac{1}{2}) (1)^2 \sin(60^\circ) ) = 2.59807621$
The distance (pyramid height) between the center of the solid and the hexagons is $3$ time the inradius of the unit icosahedron, and this comes to
$ h_{Hex} = 3 \left( \dfrac{\sqrt{3}}{12}(3+\sqrt{5}) )   \right)  =2.26728394$
and the distance (pyramid height) between the center of the solid and the pentagons is
$a$ times the inradius of the unit dodecahedron,
$h_{Pent} =(2.09016994) \left( \frac{1}{2} \sqrt{ \dfrac{5}{2} + \dfrac{11}{10} \sqrt{5}} \right) = 2.32743843 $
Putting it all together, the volume is
$ \text{V} = \left(\dfrac{1}{3}\right) ( 12 A_{Pent} h_{Pent} + 20 A_{Hex} h_{Hex} ) = \boxed{55.2877} $
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