In an effort to make the torus polyhedron plots a little easier to interpret, the outer cube might be flattened to a unit square cuboid of height c say [so c = 1 earlier]. It apparently remains true that if either type of corner is "flat" (developable) then both types are. The Helenius relation for an origami torus now generalises to 2*h^2 + q^2 + c^2*p^2 = 2*(1+c^2), where 2*(p+q-1) = p*q, recast to show the symmetry between p,q [and this time, properly checked!] As far as proving these conjectures algebraically goes, notice that the squares of the sines (and cosines) of the face angles are rational in the coordinates of the corners. So there's a good chance that FWH's relation can be fed into something like sin^4(A+B) - 2 sin^2(A+B) (1 + sin^2(A) + sin^2(B) + 2 sin^2(A) sin^2(B)) + (1 + sin^2(A) + sin^2(B))^2 = 0 [unchecked!] to show that the U/V corners have sin(face angle-sum) = 0 [two face angles are already known to be \pi/2]; then using continuity from the (improperly flattened) case q = 1 would clinch it for U/V corners. If it could also be proved that both or neither are developable, then we're home and dry. Otherwise, we need some rational addition formula like the above, but for sums of 4 angles --- maybe RWG might know how to do this? Fred Lunnon