A6 and O6 are, loosely put, cubes stretched (A6) and squashed (O6)
along a long diagonal.  All are made from golden rhombi; a golden
rhombus is a planar quadrilateral with four equal sides and whose
diagonals are in golden ratio to one another.  The two acute angles are
arctan(2), about 63.43494882292200998108455678°; the other two are 180°
minus that.

Assemble three rhombi at their acute corners to get a pointy
three-diamond; two of those go together to form an A6.  Similarly,
assemble three rhombi at their obtuse corners to get a nearly-flat
three-diamond; two of those also go together, forming an O6.

Two of the obtuse face-face angles of an O6, plus one acute face-face
angle from an A6, will add up to 360°.  If you take an O6, then take a
second O6 translated so the translated verion has one face co-located
with the orignal's parallel face, then rotate one of them by 180°
around the centre normal to the face where they meet, the touching
faces will once again match, but the resulting shape will be a sort of
open-clam shape.  The gaps can be then filled by two A6s, yielding a
dodecahedron which is quasi-regular: all edges are the same length and
all faces are congruent, but those faces are rhombi, not the pentagons
of a regular dodecahedron.

BILINSKI	B12 = 2 A6 + 2 O6
FEDEROV		F20 = 1 B12 + 3 A6 + 3 O6
KEPLER		K30 = 1 F20 + 1 B12 + 5 A6 + 5 O6

The BILINSKI shape is the dodecahedron described above; the name is
probably because it is the Bilinski dodecahedron.

It is possible to attach three more A6 and three more O6 to a Bilinski
dodecahedron to form the "FEDEROV" shape, presumably named after Evgraf
Federov, who classified convex polyhedra with congruent rhombic faces.

I take the third line to indicate there is a 30-faced shape, formed
from the other two plus five more each of A6 and O6.  I do not know why
he called it "Kepler", nor have I investigated how to form it.