The original problem was to construct a polyhedron, homeomorphic
to a torus, with angular defect zero at every corner (and everywhere else):
properly embedded in 3-space, so that the interiors are homeomorphic.
I found a 2-parameter family of "origami polytore" solutions with
20 faces (4 square + 16 triangular), 12 vertices, 32 edges,
each comprising: a square cuboid of depth c with corners (+/-1,+/-1,+/-c),
deprived of its top and bottom faces, with 8 edges and 8 corners connected
by triangle faces to the skew quadrangle with corners
(+q,0,+h), (0,-q,-h), (-q,0,+h), (0,+q,-h).
Given the depth c, what radii q and heights h are possible?
Theorem: when the polytore is origami,
either
h^2 = -(q^2-2)*( 1/2 + c^2/(q-2)^2 ) --- the "oval" branch,
or
h^2 = -(q^2)*( 1/2 + c^2/((q-2)^2-2) ) --- the "cusp" branch.
There are just two congruency classes of corner: cuboid, and quadrangle.
By the Gauss-Bonnet theorem, the sum of the angular defects vanishes;
so we need ensure only that they vanish at cuboid corners. The latter
having two square faces, it remains to ensure that the other three face
angles A,B,C sum to \pi.
>From the coordinates of the corners, the squares of the edge lengths can
be found by Pythagoras; then in turn by the cosine rule, the squared sines
SA,SB,SC of the face angles are expressible as _rational_ functions of c,q,h.
The squared sine SABC of the sum angle satisfies a quartic
polynomial equation with coefficients polynomial in SA,SB,SC
[of which I am inordinately proud --- see "Horrid trig identity"],
which we need to have root zero, so constant coefficient zero.
The latter turns out to factorise as the square of the product of
the polynomials implicit in the theorem; so we have a necessary
condition for origami-city.
For sufficiency, in practice, all we need to do is check that
the sum is correct at some convenient point, such as q = 1, then
rely on continuity of the root as a function of q. In theory, I'm not
entirely happy about this: just exactly _how_ can we be sure that
some other root of the quartic does not intrude?
That caveat aside, the interesting upshot of this frivolous investigation
was that there is another branch of solutions (cusp) besides those
originally found (oval) by trial-and-lots-of-error. Both branches are plotted
for c = 1 at <hq_plot.jpg>; those solutions giving proper embeddings
(rather than mere immersions) lie between 0 < q < 1.
The new branch is less well house-trained than the old: at q = 2-sqrt2,
h roars off to oo. At one point the branches actually cross: although
somehow I keep feeling this solution should have some interesting
property, in fact it looks just like most of the others <polytore.jpg>.
There is available a Maple worksheet <polytore.mw> on this topic,
incorporating the above still in rotatable form, and an animation
of the polyhedron as q varies.
Further projects suggested by this investigation include the extension
to regular base polygons with any even number edges (rather than just
squares, as above); and an animation of the polyhedron as folding up
from a planar map.
The latter poses a nontrivial robotics question: is it possible to fold the
polyhedron from a sheet on which opposite edges have already been
joined [forcing it to be folded flat, but four layers thick]?
My attempts to find somewhere public to put all this stuff having been
so far cruelly stymied, I must again offer to send the files to anybody
expressing an interest.
Fred Lunnon [18/08/09]
