## Sonobe puzzler 2: How many units?

This is the second of three puzzlers for the Sonobe lover.  The first is here; for partial background, see my earlier posts [1, 2, 3, gallery] on the Sonobe unit.

Consider the following origami construction.  How many Sonobe units is it made of?

Keep reading for a few hints.

Hint #1: four of the corners have broken underlying geometry.  (See the links at the top of this post to understand what this means.)  In particular, the dark blue units have been folded perpendicular to their ribs, along their pockets, instead of vice-versa.  Without this twist, the construction would look like a cube.

Hint #2: following Hint #1, let’s talk about cubes.  There are several different ways to make a cube from Sonobe units:

In these images, the leftmost cube is made from six units; its underlying geometry is that of the tetrahedron.  The second and third cubes from the left are made from twelve units each.  In this case, the underlying geometry is actually that of a cube, and we use the flat square faces mentioned in the third image here.  In the second cube, the units are “inside-out,” with the pockets on the interior of the cube and the ribs visible along the edges.  In the third cube, they are “right-side out.”  (If you try to make this yourself, which is not difficult, you’ll notice that this requires you to fold the “wrong way” along the rib.)  In the rightmost cube, made from twenty-four units, the underlying geometry is that of a cuboctahedron.

The flat square faces and the faces of the triangular pyramids constructed on the equilateral triangular faces are coplanar.

The cube that underlies the mystery construction is the same size as the twenty-four unit cube just mentioned, but it’s easy to see that it’s constructed differently.  In fact, it uses more than twenty-four units.

Hint #3: one can understand the geometry of this polyhedron by understanding the A3 lattice, the tetrahedral-octahedral honeycomb, or the hyperplane arrangement consisting of the planes $\pm x \pm y \pm z = 1$ (or the larger arrangement $x \pm y \pm z \in 2\mathbb{Z} + 1$).