This is a dodecahedron $D$.
Nested inside the dodecahedron $D$ we have a cube $C$. There are
twelve faces of $D$ and twelve edges of $C$, with one edge of $C$
cutting across each face of $D$. We can think of $D$ as being
formed by adding a tent to each face of $C$. The tents must have
exactly the right dimensions to make everything fit together
neatly. It works out that the eight vertices of $C$ are at the
points $(\pm 1,\pm 1,\pm 1)$ and the remaining twelve vertices of
$D$ are at points $(0,\pm\tau,\pm1/\tau)$ or
$(\pm\tau,\pm 1/\tau,0)$ or $(\pm 1/\tau,0,\pm\tau)$, where
$\tau=(\sqrt{5}+1)/2$.
Nested inside the cube $C$ we have a tetrahedron $T$. There are
six faces of $C$ and six edges of $T$, with one edge of $T$
cutting across each face of $C$. The vertices of $C$ have the
form $(\pm 1,\pm 1,\pm 1)$. The ones with an even number of
$(-1)$'s are the vertices of $T$.
Nested inside the tetrahedron $T$ we have an octahedron $O$.
There are six edges of $T$, and the midpoints of these edges form
the six vertices of $O$. The coordinates of these vertices are
$(\pm 1,0,0)$ and $(0,\pm 1,0)$ and $(0,0,\pm 1)$.
Nested inside the octahedron $O$ we have an icosahedron $I$.
There are twelve edges of $O$, and each of these edges contains one
of the twelve vertices of $I$. The coordinates of these vertices
are $(0,\pm 1/\tau^2,\pm 1/\tau)$ or $(\pm 1/\tau^2,\pm
1/\tau,0)$ or $(\pm 1/\tau,0,\pm 1/\tau^2)$, where
$\tau=(\sqrt{5}+1)/2$.