Yeah, I realized the torus wasn’t convex after I posted (d’oh)… seems like a infinite tube would count as convex, though. It’s certainly convex at any point a finite distance from the start.
Regardless, I came up with an example (I think). See this picture. Each color represents an equal-length line (green and dotted green are different). Note that you can either fold the B triangles or the C triangles to be next to A, and each results in a different shape. I believe that both shapes are convex, or can be made so with small tweaks (I’m not 100% certain about the angle between B and C).
I think I have worked out a general solution to get six vertices of an octahedron with each face an a, b, c triangle, but I’m not sure if any right triangle assumptions snuck in there. I’ll post it in case anyone is interested in verifying it.
Given a triangle with sides a, b, c name its angles alpha, beta, and gamma in the usual way and let A denote its area.
Let x be the smallest real root of the polynomial z^4 - (b^2 + c^2) z^2 + 4A^2.
Then the following points:
(x, 0, 0)
(-x, 0, 0)
(0, sqrt(b^2 - x^2), 0)
(0, -sqrt(c^2 - x^2), 0)
(0, (ab cos(gamma)-(x^2))/(sqrt(b^2 - x^2)), x)
(0, (ab cos(gamma)-(x^2))/(sqrt(b^2 - x^2)), -x)
Form an octahedron with each face an a, b, c triangle.
Ooh, thanks! That still doesn’t get me everything I want, but it gets me close enough that I should be able to get the rest of the way on my own (cross-products between edges to get normals, and dot-products between normals to get dihedral angles). Or maybe I should just figure out how to use those points directly (I’m sure that’s possible, in Autocad).
How did you find those?
EDIT:
For the general case, you probably need to put in some stipulation about which sides are shorter or longer. Some arrangements won’t produce a valid solid (I think this would mean no real roots of your quartic).
The quartic is guaranteed to have four real roots for all triangles a, b, c. This I can prove.
I don’t know if there are constraints on a, b, c. I sketched a few cases and they all seemed to work, but this is pretty far from a proof.
One thing I should note is that all four edges of length a are adjacent while there are two pairs each of adjacent b’s and c’s. I’m not crazy about the way I worded the previous sentence, but I hope you get the idea.
Well, if nothing else, angle alpha must be acute, since you have four copies of it meeting at each of the “pole” vertices. Trying it with alpha as a right angle would make some faces coplanar and turn it into a tetrahedron, and an obtuse alpha just wouldn’t fit for a convex figure.
OK, I just took a stab at doing those crosses and dots, and there’s clearly something wrong, here. For your x = sqrt((41 - sqrt(1105))/2) , I get 4.93753277 …, compared to your figure of 1.969576061 .
On the other hand, if we can fix that, the relevant products promise to be quite simple, owing to the edge lengths being small integers.
