Having clambered out of one elephant trap, I tumble straight into another. The story so far --- We have a family of "polytores", polyhedra of genus 1 having the symmetry of a square, parameterised by inner quadrangle radius q, inner quadrangle height h, outer cuboid height g (formerly c). Provided 0 < q < 1 and h,g > 0 these solids are properly embedded in Euclidean 3-space; and provided q,h,g satisfy certain other constraints, they are "planar" --- isometrically embedded, with face-angles summing to 2\pi at the corners, and unfoldable into a flat net which tiles the plane. Necessary for planarity is the polynomial equation f_k(a, b, c, ...) = 0 [see "Horrid trig identity"], where A,B,C,... denote face-angles, a = sin^2(A) etc, and k varies depending on the number of faces at the corner under inspection. When f_k = 0, we are assured that, for some choice of the signs, A (+/-) B (+/-) C (+/-) ... = 0 mod \pi . There are two classes of polytore corner, "U/V" with 5 faces and "W" with 6. For either class, we find f_k has two distinct quartic factors, christened "oval" and "cusp" after the shapes of their plane curves when g = 1 is fixed. The first class is characterised by the factor (due originally to Fred Helenius) oval(q,h,g) = 2(q - 2)^2 h^2 + (q^2 - 2)(q^2 - 4q + 2g^2 + 4) ; the second by the factor (whose author now prefers to remain anonymous) cusp(q,h,g) = 2(q^2 - 4q + 2)h^2 + q^2(q^2 - 4q + 2g^2 + 2) . The constraint oval(q,h,g) = 0 appears to be sufficient, as has been verified numerically, and by building models of examples. Just exactly how might these be proved sufficient? For g fixed, oval(q,h) = 0 defines a plane curve. If at some point along this curve --- q = 1 is a promising candidate --- we can verify that all the signs are positive --- more precisely A + B + C + ... = 2\pi --- then by continuity of the sum and of the curve, this will remain true as q varies along the remainder of the curve. I earlier remarked
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? Well, it seemed to work alright, so it must be OK, mustn't it ...
The model built (thanks to Michael Kleber) was of a particular example:
The two branches cross where q = 0.674013, h = 1.285291, g = 1; Somehow it seems that this case should have an interesting property, though it's not clear what. This wasn't very observant: besides the two obvious right-angles at a U/V corner, there is a third belonging to the UWW face; and at a W corner, in the flat net there are two less obvious right-angles between UW and VW edges.
Hmmm --- awful lot of (1/2)\pi angles there --- this "cross-over" case really deserves to be renamed "right-angled" ... Anyway pressing on, since it satisfies both constraints, we have for free a point on the second curve corresponding to a planar polytore; now by continuity, there should be an entire second family of planar polytors satisfying cusp = 0. But 'ang 'abaht --- as q approaches 2-sqrt2 along the cusp curve, h approaches oo. And while the outer U/V cuboid remains fixed, as each W corner disappears into the distance, the sum of the face-angles around its ever-sharpening spike quite plainly approaches zero --- and cannot possibly be planar! So what has gone wrong? Seeing that things seem a bit slow on the list at the moment, I'll leave this for now as an exercise for the student. Fred Lunnon